1

Lecture #01

Finite Difference-based Numerical Methods in Chemical Engineering

Recommended Books:

1) Numerical Methods for Engineers (SKG) 2) Numerical Methods for Engineering Application (Ferziger) 3) Linear Algebra (Strang)

Objective: Learn a few basic numerical techniques (Finite Difference) to solve simple ODEs/PDEs/algebraic equations derived from a variety of heat and mass transfer related conservation equations.

• Analytical methods yield exact solutions, often in infinite series; use mathematical functions. • Numerical methods yield approximate solutions; use numbers, computations; results are dependent

on the method accuracies and limited by computers machine errors. Course focuses on

• Algebra (Matrix Operation) • Methods (Finite Difference)

Requires knowledge of • Computer Programming(because large # of iterations)

(there are engineering softwares, e.g., Matlab, NAG, IMSL, Polymath) • No numerical analysis in this course!

Examples: (Applications in chemical engineering)

1. Distillation: 𝑉1 𝐿𝑜 1

2 𝑖 − 1

𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑖 𝑖 𝑖 𝑖 + 1 𝑁 Trays 𝑉𝑁+1 𝐿𝑁 Assume: Constant equilibrium flowrates of vapor & liquid. Binary components: A, B

(more volatile)

Species balance on 𝒊𝒕𝒉 plate:

2

𝐿𝑖 = 𝑥𝑖 𝐿 𝑉𝑖 = 𝑦𝑖 𝑉 𝑦𝑖 = 𝑘𝑖𝑥𝑖 (𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚)

Therefore, 𝑉𝑖 = �𝑦𝑖𝑥𝑖� �𝑉

𝐿� 𝐿𝑖

or 𝑉𝑖 = 𝐴𝑖′𝐿𝑖

And, 𝑉𝑖 + 𝐿𝑖 = 𝑉𝑖+1 + 𝐿𝑖−1 (𝐼𝑛 = 𝑜𝑢𝑡)

𝑉𝑖 ↓ 𝑙𝑖−1

𝑉𝑖+1 ↓ 𝑙𝑖 ⋯ 𝑖𝑡ℎ 𝑝𝑙𝑎𝑡𝑒

𝑉𝑖 + 𝑉𝑖 𝐴𝑖′⁄ − 𝑉𝑖−1𝐴𝑖−1′ − 𝑉𝑖+1 = 0

Re-arrange 𝐴𝑖−1𝑉𝑖−1 + (1 − 𝐴𝑖)𝑉𝑖 − 𝑉𝑖+1 = 0

𝑖 = 1,𝑁

Thus, there are ‘N’ equations and (N+2) variables (𝑉𝑜 ⋯⋯𝑉𝑁+1)

However, 𝑉𝑜 𝑎𝑛𝑑 𝐿𝑜 are known (inlet or boundary conditions)

Re- arrange for all trays 𝑖 = 1,𝑁

(1 − 𝐴1)𝑉1 − 𝑉2 = −𝐴𝑜𝑉𝑜(= 𝐿𝑜) (𝑘𝑛𝑜𝑤𝑛)

𝐴1𝑉1 + ( 1 − 𝐴2)𝑉2 − 𝑉3 = 0 (𝑖)

⋮

𝐴𝑁−1𝑉𝑁−1 + ( 1 − 𝐴𝑁+1)𝑉𝑁 = 𝑉𝑁+1 (𝑘𝑛𝑜𝑤𝑛) (𝑁)

We have a set of algebraic equations (linear/non-linear) represented as

𝐴𝑋� = 𝑏�

coefficient matrix column vectors or (𝑁 × 1) 𝑚𝑎𝑡𝑟𝑖𝑥

rows column

𝐴 =

⎣⎢⎢⎡

( 1 − 𝐴1) −1 0 0 0𝐴1 ( 1 − 𝐴2) −1 0 0 00 𝐴2 (1 − 𝐴3) −1 0 00 0 0 0 𝐴𝑁−1 (1 − 𝐴𝑁+1)⎦

⎥⎥⎤

𝑁×𝑁

rows columns

3

𝑋� = �

𝑉1𝑉2⋮𝑉𝑁

� , 𝑏� = �

𝐿0⋮⋮

𝑉𝑁+1

� (Known)

column vectors

Alternatively*,

�

( 1 − 𝐴1) −1 0 0 0𝐴1 ( 1 − 𝐴2) −1 0 0 0⋮ ⋮ ⋮ ⋮ ⋮ ⋮0 0 0 0 𝐴𝑁−1 (1 − 𝐴𝑁+1)

�

𝑁×𝑁

�

𝑉1𝑉2⋮𝑉𝑁

�

𝑁×1

= �

𝐿𝑜

⋮𝑉𝑁+1

�

𝑁×1

∗ represented as matrix multiplication

Note:

o Most of the finite difference methods lead to the discretization of ODEs/PDEs and a set of algebraic equations.

o You have to solve a set of algebraic equations using Gauss Elimination, G- Jordan etc.

Example 2: Particle settling

(𝑑𝑝,𝜌𝑝 , 𝑣𝑝,𝐴𝑝,𝑚𝑝)

𝑔

𝐻 𝑙𝑖𝑞𝑢𝑖𝑑/𝑔𝑎𝑠

𝑡 =?

(𝜌𝑓 , 𝑣𝑓)

Force balance on the particle:

𝑚𝑝𝑑𝑣𝑑𝑡

= 𝑚𝑝𝑔 − 𝑣𝑝𝜌𝑓𝑔 − 𝐶𝐷 �12𝜌𝑓𝑣2�𝐴𝑝

𝑤𝑒𝑖𝑔ℎ𝑡 𝑏𝑢𝑜𝑦𝑎𝑛𝑐𝑦 𝑑𝑟𝑎𝑔

𝑑𝑣𝑑𝑡

= 𝑔 − 𝜌𝑓𝜌𝑝𝑔 − 𝐶𝐷 �1

2𝜌𝑓𝑣2� �𝜋

𝑑𝑝2

4�

4

𝑑𝑣𝑑𝑡

= 𝐴 − 𝐵𝐶𝐷(𝑣)𝑣2 𝐶𝐷 24 𝑅𝑒 ,⁄ 𝑝

= 𝐴 − 𝐵Φ(𝑣) 0.44

𝑡 = 0, 𝑣 = 0 𝑅𝑒 ,𝑝

Therefore, you have to solve an initial value problem (ODE)

Solution: linear or non-linear

𝑣𝑡 homogeneous or non-homogeneous 𝑣 or a set of ODEs.

𝑜 𝑡 Methods: RK-4, Euler Forward, etc.

Example 3: Heating or cooling of a moving particle (𝛁𝑻(𝒓) = 𝟎)

𝑡 = 0 𝑣𝑝 = 0

𝑇 𝑇𝑓 𝑇 = 𝑇𝑜 > 𝑇𝑓(𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡)

𝑟𝑝 ℎ𝑓 (𝑓𝑙𝑢𝑖𝑑 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒)

𝑇(𝑡) =? (𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑣𝑒 ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡)

Assume/ Apply lump body approach (𝑩𝒊 → 𝟎)

1. Energy balance over entire particle ℎ𝑑𝑝𝑘𝑝

→ 0 , 𝑘𝑝 >> ℎ𝑑𝑝

𝑚𝐶𝑝𝑑𝑇𝑑𝑡

= −ℎ𝐴𝑠�𝑇 − 𝑇𝑓�

(𝐴𝑠 = 𝜋𝑑𝑝2) (𝑁𝑜 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑖𝑛 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒)

ℎ = ℎ(𝑁𝑢) ⇒ ℎ𝑑𝑝𝑘𝑓

= 𝑓(𝑅𝑒,𝑝,𝑃𝑟)

𝑜𝑟 ℎ = ℎ �𝑣𝑝 �; 𝑅𝑒 = 𝑣1𝑑𝑝𝜌𝑓𝜇𝑓

,𝑃𝑟 = �𝐶𝑝𝜇𝑘�𝑓

2 Time-dependence of velocity (see previous example) and energy balance equation can be written as

5

�𝑑𝑣𝑝𝑑𝑡 = 𝐴 − 𝐵Φ(𝑣𝑝)𝑑𝑇𝑑𝑡 = 𝐶 𝐷�𝑣𝑝� − 𝐸

� 𝑡 = 0 𝑣𝑝 = 0, 𝑇 = 𝑇𝑜Type equation here.

In this example, two equations are coupled and have to be solved simultaneously:

�𝑑𝑦1𝑑𝑡

= 𝑓(𝑦1)

𝑑𝑦2𝑑𝑡

= Φ(𝑦1,𝑦2) � 𝑇𝑤𝑜 𝑂𝐷𝐸𝑠

Example 4: Heating or cooling of a stationary solid (𝛁𝑻 ≠ 𝟎) in a moving fluid 𝑡 = 0 𝑇 = 𝑇𝑜 > 𝑇𝑓(𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡) 𝑇𝑜 𝑓𝑙𝑢𝑖𝑑 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑟 𝑇(𝑡, 𝑟) =? 𝑟𝑝 𝑇𝑓 ℎ𝑓(𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑣𝑒 ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡) Assume ∇𝑇(𝑟) ≠ 0 𝑖. 𝑒. , 𝑡ℎ𝑒𝑟𝑒 𝑖𝑠 𝑎 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑖𝑛𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 Energy balance within the particle:

𝜌𝐶𝑝𝜕𝑇𝜕𝑡

= 𝑘𝑟2

𝜕𝜕𝑟

(𝑟2 𝜕𝑇𝜕𝑟

)

𝑡 = 0 𝑇 = 𝑇𝑜 𝑟𝑝 ≥ 𝑟 ≥ 0 (𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛)

= 0+ 𝑟 = 0 𝜕𝑇𝜕𝑟

= 0 (𝐵𝑜𝑢𝑛𝑑𝑎𝑟𝑦 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠)

𝑟 = 𝑟𝑝, − 𝑘 𝜕𝑇𝜕𝑟

= ℎ(𝑇 − 𝑇𝑓)

Now, we have to solve a PDE equation (unsteady – state 1D-space problem)

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Example 5: Mass transfer in a tubular liquid flow (steady- state) 𝐶𝐴 = 𝐶𝑖𝑛 𝑅 𝑟 (𝑤𝑎𝑡𝑒𝑟 𝑓𝑙𝑜𝑤𝑠) 𝑉𝑧(𝑟) 𝑡 = 0 z tube length = L

For 𝑅𝑒 < 2100

𝑣𝑧(𝑟) = 2𝑣𝑜(1 − 𝑟2

𝑅2)

𝑐(𝑡, 𝑧, 𝑟) =?

Species (A) balance in the CV (2𝜋𝑟 ∆𝑧 ∆𝑟)

CV

𝜕𝐶𝐴𝜕𝑡

+ 𝑉𝑧.∇𝐶𝐴 = 𝐷∇2𝐶𝐴 + 𝑟(= 0)

(no reaction)

𝜕𝐶𝐴𝜕𝑡

+ 2𝑣𝑜 �1 −𝑟2

𝑅2�𝜕𝐶𝐴𝜕𝑧

+ 𝑉𝑟𝜕𝐶𝐴𝜕𝑟

= 𝐷�𝜕𝐶𝐴𝜕𝑧2

+1𝑟𝜕𝜕𝑟�𝑟𝜕𝐶𝐴𝜕𝑟

��

𝑡 = 0 , 𝐶𝐴 = 0 𝑒𝑣𝑒𝑟𝑦𝑤ℎ𝑒𝑟𝑒

0+ 𝑧 = 0 𝐶𝐴 = 𝐶𝑖𝑛

𝑧 = 𝐿 𝜕𝐶𝐴𝜕𝑋

= 0 (𝑙𝑜𝑛𝑔 𝑡𝑢𝑏𝑒 𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑖𝑜𝑛)

𝑟 = 0 , 𝜕𝐶𝐴𝜕𝑟

= 0 (𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐)

= 𝑅 − 𝜕𝐶𝐴𝜕𝑟

= 0 (𝑛𝑜𝑛 − 𝑟𝑒𝑎𝑐𝑡𝑖𝑣𝑒 𝑤𝑎𝑙𝑙)

In this example, we have a transient/unsteady-state 2D (space) problem or a PDE to solve.

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Lecture #02 Linear Algebra Equations

�𝑓1(𝑋1,𝑋2,⋯⋯𝑋𝑛 = 0𝑓2(𝑋1,𝑋2,⋯⋯𝑋𝑛 = 0⋮

𝑓𝑛(𝑋1,𝑋2,⋯⋯𝑋𝑛 = 0

� 𝑠𝑦𝑠𝑡𝑒𝑚 𝑜𝑓 𝑛 𝑙𝑖𝑛𝑒𝑎𝑟 𝑎𝑙𝑔𝑒𝑏𝑟𝑎𝑖𝑐 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠

or

�

𝑎11𝑋1 + 𝑎12𝑋2 + ⋯⋯+ 𝑎1𝑛𝑋𝑛 = 𝑏1

𝑟𝑜𝑤 𝑐𝑜𝑙𝑢𝑚𝑛 𝑎21𝑋1 + 𝑎22𝑋2 + ⋯⋯+ 𝑎2𝑛𝑋𝑛 = 𝑏2⋮

𝑎𝑛1𝑋1 + 𝑎𝑛2𝑋2 + ⋯⋯+ 𝑎𝑛𝑛𝑋𝑛 = 𝑏𝑛 ⎭⎪⎬

⎪⎫

𝑛 # 𝑜𝑓 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑡𝑜 𝑠𝑜𝑙𝑣𝑒 𝑛 # 𝑜𝑓𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠

or 𝐴𝑋� = 𝑏� or 𝐴𝑋 = 𝑏

𝑤ℎ𝑒𝑟𝑒 𝐴 = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑚𝑎𝑡𝑟𝑖𝑥

= �

𝑎11 𝑎12 ⋯ 𝑎1𝑛𝑎21 𝑎22 ⋯ 𝑎2𝑛⋮ ⋮ ⋮ ⋮𝑎𝑛1 𝑎𝑛2 ⋯ 𝑎𝑛𝑛

�

𝒏×𝒏

(𝑎𝑖𝑗 𝑖𝑠 𝑎 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡, 𝑖, 𝑗 = 1 𝑡𝑜 𝑛)

row column

𝑋� = 𝑐𝑜𝑙𝑢𝑚𝑛 𝑣𝑒𝑐𝑡𝑜𝑟 = �

𝑋1𝑋2⋮𝑋𝑛

� , 𝑏� = 𝑐𝑜𝑙𝑢𝑚𝑛 𝑣𝑒𝑐𝑡𝑜𝑟 = �

𝑏1𝑏2⋮𝑏𝑛

�

or,

𝑋 = 𝑚𝑎𝑡𝑟𝑖𝑥 = �

𝑋1𝑋2⋮𝑋𝑛

�

𝑛×1

𝑏 = 𝑚𝑎𝑡𝑟𝑖𝑥 = �

𝑏1𝑏2⋮𝑏𝑛

�

𝑛×1

2

⎩⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎧

𝑁𝑜𝑡𝑒: 𝐼𝑛 𝑔𝑒𝑛𝑒𝑟𝑎𝑙,𝑎 𝑚𝑎𝑡𝑟𝑖𝑥 ℎ𝑎𝑠 𝑛 × 𝑚 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠

𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑎𝑖𝑗 𝑖 = 1 𝑡𝑜 𝑛 𝑗 = 1 𝑡𝑜 𝑚

𝐼𝑓 𝑛 = 𝑚, 𝑡ℎ𝑒𝑟𝑒 𝑚𝑎𝑦 𝑏𝑒 𝑢𝑛𝑖𝑞𝑢𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝑟𝑎𝑛𝑘 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑡𝑟𝑖𝑥 = 𝑚𝐼𝑛 𝑔𝑒𝑛𝑒𝑟𝑎𝑙, 𝑟𝑎𝑛𝑘 >, =, < 𝑚 (3 𝑐𝑎𝑠𝑒𝑠)(𝐹𝑜𝑟 𝑚𝑜𝑟𝑒, 𝑟𝑒𝑎𝑑 𝑎 𝑏𝑜𝑜𝑘 𝑜𝑛 𝑚𝑎𝑡𝑟𝑖𝑥)

𝐼𝑛 𝑡ℎ𝑖𝑠 𝑐𝑜𝑢𝑟𝑠𝑒,𝑤𝑒 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡𝑒𝑑 𝑖𝑛 𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑤𝑖𝑡ℎ 𝑎𝑠 𝑚𝑎𝑛𝑦 𝑣𝑎𝑟𝑎𝑖𝑏𝑙𝑒𝑠:

�3𝑥 + 2𝑦 = 5𝑥 − 𝑦 = 6 � 𝑜𝑟 �

3𝑥 + 2𝑦 + 𝑧 = 5𝑥 + 𝑦 − 𝑧 = 2

2𝑥 + 2𝑦 − 2𝑧 = 4� 𝑒𝑡𝑐.

�

Some properties of a matrix and operation

(1) Symmetric matrix, 𝑎𝑖𝑗 = 𝑎𝑗𝑖

Square matrix, n = m

Diagonal matrix, �⋱ 00 ⋱� (𝑜𝑛𝑙𝑦 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠)

Banded matrix (tridiagonal)

⎣⎢⎢⎢⎡𝑎11 𝑎12 0 0 ⋯ 0𝑎21 𝑎22 𝑎23 0 ⋯ 00 𝑎31 𝑎32 𝑎33 ⋯ 0⋮ ⋮ ⋮ ⋮ ⋮ ⋮0 0 0 0 𝑎𝑛𝑛−1 𝑎𝑛𝑛⎦

⎥⎥⎥⎤

𝑛×𝑛 𝑚𝑎𝑡𝑟𝑖𝑥

(First and last rows have only two elements and all other rows have 3 elements, one on the diagonal and the other two on each side of the diagonal element) Upper/Lower triangular matrix: (All elements below and above the diagonal are zero, respectively)

UTM:

⎣⎢⎢⎢⎡𝑎11 𝑎12 𝑎13 ⋯ 𝑎1𝑛

𝑎22 𝑎23 ⋯ 𝑎2𝑛⋱ ⋱ ⋮

𝟎 ⋱ 𝑎𝑛−1𝑛𝑎𝑛𝑛 ⎦

⎥⎥⎥⎤

𝒏×𝒏

LTM:

⎣⎢⎢⎢⎡𝑎11𝑎21 𝑎22𝑎31 𝑎32 ⋱ 𝟎⋮ ⋮ ⋱ ⋱𝑎𝑛1 𝑎𝑛2 ⋯ 𝑎𝑛𝑛−1 𝑎𝑛𝑛⎦

⎥⎥⎥⎤

𝒏×𝒏

(2) Operation: Summation: [𝐴]𝑚×𝑛 = [𝐵]𝑚×𝑛 + [𝐶]𝑚×𝑛

3

Multiplication (rules): [𝐴]𝑚×𝑛 × [𝐵]𝑛×𝑙 = [𝐶]𝑚×𝑙 same Therefore, multiplication is not permissible for [𝐵] × [𝐴]

�

𝑛 × 𝑙 𝑚 × 𝑛

𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑙 ≠ 𝑚

�

Also, these multiplications are permissible

�5 13 20 3

�3×2

× �−3 01 0�2×2

; �5 13 20 3

�3×2

× �−4 −3 22 7 −3�2×3

; [1 3 2]1×3 × �05−8

�3×1

Rule: 𝑪𝒊𝒋 = ∑ 𝒂𝒊𝒌 × 𝒃𝒌𝒋𝒌𝒍=𝟏

examaple: �3 18 60 4

�3×2

× �5 67 2�2×2

= �(3 × 5 + 1 × 7) 3 × 6 + 1 × 2(8 × 5 + 6 × 7) (8 × 6 + 6 × 2)0 × 5 + 4 × 7) (0 × 6 + 4 × 2

�

3×2

Inverse of a matrix:

[𝐴]−1 ⇒ [𝐴]𝑚×𝑛 × [𝐴∗]𝑛×𝑚−1 = [𝐼] ≡ �

1 0 00 1 00 0 1

�

Identity matrix For 𝟐 × 𝟐 matrix

[𝐴]−1 = 1|𝐴| �

𝑎22 𝑎12−𝑎21 𝑎11� 𝑤ℎ𝑒𝑟𝑒 𝐴 = �

𝑎11 𝑎12𝑎21 𝑎22�

Note: |𝐴| = (𝑎11 × 𝑎22 − 𝑎12 × 𝑎21) ≠ 0 else matrix A is considered to be singular. For bigger size matrices, there are special methods to determine[𝐴]−1. Transpose of a matrix [𝑨]𝑻:

[𝐴] = �𝑎11 𝑎12 ⋯ 𝑎1𝑛𝑎𝑚1 𝑎𝑚2 ⋯ 𝑎𝑚𝑛� , [𝐴]𝑇 = �

𝑎11 𝑎21 ⋯ 𝑎𝑚1𝑎1𝑛 ⋯ ⋯ 𝑎𝑚𝑛�𝑛×𝑚

(rows have been changed with columns)

4

example: [𝐴] = �

𝑎1𝑎2⋮𝑎𝑛

�

𝑛×1

⇒ [𝐴]𝑇 = [𝑎1 𝑎2 ⋯ 𝑎𝑛]1×𝑛

Let us get back to solving a set of equations

� [𝐴]{𝑋} = {𝑏}𝑜𝑟 [𝐴][𝑋] = [𝑏] 𝑜𝑟 𝐴�⃗� = 𝑏�⃗

� 𝑏𝑇 = [𝑏1, 𝑏2,⋯⋯𝑏𝑛], 𝑋𝑇 = [𝑋1,𝑋2,⋯⋯𝑋𝑛]

𝑿 = 𝑨−𝟏𝒃

⇒ �𝐹𝑖𝑛𝑑𝑖𝑛𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑡𝑜 𝑡ℎ𝑒 𝑠𝑒𝑡 𝑜𝑓 𝑙𝑖𝑛𝑒𝑎𝑟 𝑎𝑙𝑔𝑒𝑏𝑟𝑎𝑖𝑐 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑖𝑠

𝑛𝑜𝑡ℎ𝑖𝑛𝑔 𝑏𝑢𝑡 𝑓𝑖𝑛𝑑𝑖𝑛𝑔 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑚𝑎𝑡𝑟𝑖𝑥, 𝑓𝑜𝑙𝑙𝑜𝑤𝑒𝑑 𝑏𝑦𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ 𝑏.

�

Generally, Cremer’s rule works nicely for small matrix:

𝑿𝒋 =�𝑨𝒋�|𝑨|

𝐴𝑗 is determined by replacing 𝑗𝑡ℎ column of A by b.

Therefore, if 𝐴 = �3 5 72 0 8−1 2 5

� ; 𝑏 = �732�

𝑋2 = �3 7 72 3 8−1 2 5

� |𝐴|� ; |𝐴| ≠ 0

𝑋3 = �3 5 72 0 3−1 2 2

� |𝐴|� , 𝑒𝑡𝑐.

Let us briefly discuss unique solution, singularity, and rank(r) of a (𝑚 × 𝑛) matrix. (Read the textbook for details)

Consider the following set of equations:

(1) 2𝑥1 + 3𝑥2 = 11 4𝑥1 + 6𝑥2 = 22 Note: Two equations are not independent.

and |𝐴| = �2 34 6�2×2

= 0

5

Therefore, there is no unique solution, coefficient matrix is singular and 𝑟 of such matrix = 1 < 2 𝑜𝑟 𝑚(# 𝑜𝑓 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠) < 𝑛(# 𝑜𝑓 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠)

(2) 2𝑥1 + 3𝑥2 = 11 𝑥1 + 𝑥2 = 4 Note: Two equations are independent.

|𝐴| = �2 31 1�2×2

≠ 0; 𝑎𝑛𝑑 𝑚 = 𝑛, 𝑟 = 2(𝑛)

There is a unique solution.

Geometrically, two equations represent two straight lines intersecting at a unique point.

Similarly, 3 independent equations for 𝑋2

3 variables will represent 3 planes on a ⇒

geometrical space; two planes intersecting

will yield a straight line, and two straight lines 𝑋1

intersecting will yield a point (𝑋1,𝑋2,𝑋3), or the unique solution to the equations.

(3) 2𝑥1 + 3𝑥2 − 𝑥3 = 11 𝑥1 + 𝑥2 + 𝑥3 = 4 𝑥1 + 2𝑥2 + 2𝑥3 = 7 Note: There are three linearly independent equations representing three non-parallel equations. In this case m = 𝑛, 𝑟 = 3(𝑛).

1

Lecture #03

Gauss Elimination

It is the most common method to solve a set of linear algebraic equations. The method reduces the original matrix to a upper triangular matrix which can also be used to determine the determinant of the matrix. Interestingly, one can also inspect the intermediate steps/solutions to determine singularity, rank, and number of independent equations.

General steps: 𝐴�̅� = 𝑏� 𝑤ℎ𝑒𝑟𝑒 𝐴 𝑖𝑠 𝑎 3 × 3 𝑚𝑎𝑡𝑟𝑖𝑥

𝑎11𝑥1 + 𝑎12𝑥2 + 𝑎13𝑥3 = 𝑏1 ⋯⋯⋯𝑅1

𝑎21𝑥1 + 𝑎22𝑥2 + 𝑎23𝑥3 = 𝑏2 ⋯⋯⋯𝑅2

𝑎31𝑥1 + 𝑎32𝑥2 + 𝑎33𝑥3 = 𝑏3 ⋯⋯⋯𝑅3

or �𝑎11 𝑎12 𝑎13𝑎21 𝑎22 𝑎23𝑎31 𝑎32 𝑎33

� �𝑥1𝑥2𝑥3� = �

𝑏1𝑏2𝑏3�

𝑅1, first row is called pivot row and first non-zero element (𝑎11) is called pivotal element. Subsequent operations are performed around pivot row and pivotal element.

Step 1: �𝑎11 𝑎12 𝑎130 𝑎22′ 𝑎23′0 𝑎32′ 𝑎33′

� �𝑥1𝑥2𝑥3� = �

𝑏1𝑏2

′

𝑏3′�

where �

𝑎22′ = 𝑎22 − �𝑎21𝑎11�× 𝑎12

𝑎23′ = 𝑎23 − �𝑎21𝑎11�× 𝑎13

𝑏2′ = 𝑏2 − �𝑎21

𝑎11�× 𝑏1 ⎭

⎪⎬

⎪⎫

(2)

𝐵𝑟𝑖𝑒𝑓𝑙𝑦, �irst equation is divided by 𝑎11,𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑒𝑑 𝑏𝑦 𝑎21 𝑎𝑛𝑑 𝑠𝑢𝑏𝑠𝑡𝑟𝑎𝑐𝑡𝑒𝑑𝑓𝑟𝑜𝑚 𝑠𝑒𝑐𝑜𝑛𝑑 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑡𝑜 𝑦𝑖𝑒𝑙𝑑 𝑡ℎ𝑒

𝑚𝑜𝑑𝑖𝑓𝑖𝑒𝑑 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛(2)

𝑎11 ≠ 0

Similarly,

�

𝑎32′ = 𝑎32 − �𝑎31𝑎11�× 𝑎12

𝑎33′ = 𝑎33 − �𝑎31𝑎11�× 𝑎13

𝑏3′ = 𝑏3 − �𝑎31

𝑎11�× 𝑏1 ⎭

⎪⎬

⎪⎫

(3)

Brie�ly, �irst equation has divided by 𝑎11,𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑒𝑑 𝑏𝑦 𝑎31 𝑎𝑛𝑑 𝑠𝑢𝑏𝑠𝑡𝑟𝑎𝑐𝑡𝑒𝑑

3𝑟𝑑 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑡𝑜 𝑦𝑖𝑒𝑙𝑑 𝑡ℎ𝑒 𝑚𝑜𝑑𝑖𝑓𝑖𝑒𝑑 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛(3)

or, in the simplest form 𝑅2′ = 𝑅2 − 𝑅1 �𝑎21𝑎11�

2

𝑅3′ = 𝑅3 − 𝑅1 �𝑎31𝑎11�

Note: Multiplication or division of an equation, or subtraction or addition of one independent equation from/to another independent equation does not yield a new independent equation. Therefore, at the end of step 1 we still have three linearly independent equations, although modified, to solve.

Geometrically, two planes (non-parallel) will intersect to yield a straight line. Therefore, step1 has yielded two straight lines by the intersections of plane/(eq1) with plane2 (eq2) and plane3 (eq3).

Step2: Follow the similar procedure. Now 𝑅2′(2nd row) becomes pivotal equation and 𝑎22′ becomes pivotal element.

�𝑎11 𝑎12 𝑎130 𝑎22′ 𝑎23′0 0 𝑎33′′

� �𝑥1𝑥2𝑥3� = �

𝑏1𝑏2

′

𝑏3′′�

where,

𝑎33′′ = 𝑎33′ − �𝑎32′

𝑎22′� × 𝑎23′

𝑏3′′ = 𝑏3

′ − �𝑎32′

𝑎22′� × 𝑏2

′

or in the simplest form, 𝑅3′′ = 𝑅3′ − 𝑅2′ �𝑎32′

𝑎22′�

Geometrically, two straight lines (𝑅2′ 𝑎𝑛𝑑 𝑅3′) intersect at a point.

Note: At the end of step (2) an upper triangular matrix is obtained. The three modified algebraic equations are as follows:

�𝑎11𝑥1 + 𝑎12𝑥2 + 𝑎13𝑥3 = 𝑏1 𝑎22′𝑥2 + 𝑎23′𝑥3 = 𝑏2

′

𝑎33′′𝑥3 = 𝑏33′′�

This is a new set of three linearly independent algebraic equations to solve! Here ‘new’ means ‘modified’.

Reverse (back) substitution (to determine𝒙�):

𝑥3 = 𝑏3′′ 𝑎3′′⁄ (from last row)

3

𝑥2 = 𝑏2′−𝑎23′𝑥3𝑎22′

(from 2nd last row)

𝑥1 = 𝑏1−𝑎12𝑥2−𝑎13𝑥3𝑎11

(from 1st row)

In general, for ith variable of ‘n’ equations

𝑥𝑖 =𝑏𝑖 − ∑ 𝑎𝑖𝑗

(𝑖−1)𝑥𝑗𝑛 𝑗=𝑖+1

𝑎𝑖𝑖(𝑖−1)

A large number of equations (𝑛 > 3) requires a programming code to solve.

Evaluation of determinant:

Consider a simple 2 × 2 matrix

𝐴 = �𝑎11 𝑎12𝑎21 𝑎22�2×2

⇒ 𝑑𝑒𝑡𝐴 = (𝑎11 × 𝑎22 − 𝑎21 × 𝑎12)

Apply GE method

𝐴′ = �𝑎11 𝑎120 𝑎22′

�2×2

= (𝑎22 − �𝑎21𝑎11� × 𝑎12)

determinant of the modified matrix 𝐴′ = (𝑎11 × 𝑎22 − 𝑎21 × 𝑎12) = 𝑠𝑎𝑚𝑒 𝑎𝑠 det (𝐴)

“Value of determinant is not changed by the forward elimination step of GE”. This must be true because forward step only modifies the equations.

Take 3 × 3 matrix:

�𝑎11 𝑎12 𝑎13𝑎21 𝑎22 𝑎23𝑎31 𝑎32 𝑎33

�𝐺𝐸�� �

𝑎11 𝑎12 𝑎130 𝑎22′ 𝑎23′0 0 𝑎33′′

�

[𝐴] [𝐴′]

𝑑𝑒𝑡[𝐴] = 𝑑𝑒𝑡[𝐴′] = 𝑎11𝑎22′𝑎33′′ (can be checked)

Therefore,

𝑑𝑒𝑡[𝐴] ⇒ 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝑖𝑡𝑠 𝑢𝑝𝑝𝑒𝑟 𝑡𝑟𝑖𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑚𝑎𝑡𝑟𝑖𝑥

Note that swapping of rows of a matrix does not alter values of determinant. However, sign changes: (−1)𝑘, where k is the number of times rows are swapped.

4

Re-visit

𝑑𝑒𝑡[𝐴]3×3 = 𝑎11𝑎22′𝑎33′′ (𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝑈𝑇𝑀 𝑜𝑓 𝐴)

Note: 1. For a matrix to be non-singular 𝑑𝑒𝑡[𝐴] ≠ 0

𝑜𝑟 𝑎11𝑎22′𝑎33′′ ≠ 0

Alternatively, the corresponding three equations are linearly independent; rank of

[𝐴] = #𝑜𝑓 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 = #𝑜𝑓 𝑒𝑞𝑛𝑠 = 3.

Geometrically, three planes intersect at a unique point in space, or none of them is ∥ to other.

Alternatively,

If a33′′ = 0, det[A] = 0 ⇒ the matrix is singular; rank of the matrix = 2(< 3); only first two equations are linearly independent; problem is under-determined and one more equation is required to solve(𝑋1,𝑋2,𝑋3). Algebraically, 3rd plane is ∥ to one of the other two planes.

Pivoting and ill-conditioning:

𝐴𝑋� = 𝑏� ⇒ 𝑋� = 𝐴−1𝑏� ; 𝐴−1 = 𝐴𝑑𝑗|𝐴||𝐴|

Therefore, for 𝑋� to be solved or A to be invertible, |𝐴| ≠ 0. Yet if |𝐴| → 0(𝑜𝑟 𝑖𝑡 𝑖𝑠 𝑎 𝑠𝑚𝑎𝑙𝑙 𝑛𝑢𝑚𝑏𝑒𝑟), a small change in |𝐴| will lead to a large charge in 𝐴−1.

[Note: 𝑌 = 1𝑋� , 𝑑𝑌

𝑑𝑋= − 1

𝑋2

In other words, a small change in X will lead to large change in Y]

In such case, the problem is said to be ill-conditioned. To invert such a matrix, large machine precision may be required, and one has to be careful with round-off errors.

Recall, determinant of a matrix is the product of the diagonal elements of UTM. Therefore, pivotal elements should not only be zero, but also large, or the matrix should be “diagonally strong”.

Example: 0.0001𝑥1 + 𝑥2 = 1 (𝐴)

𝑥1 + 𝑥2 = 2

5

Working on 4th decimal, apply GE with 0.0001 as the pivotal element to obtain 𝑋𝑇 = [0.0000 1.0000] which is wrong. Now, swap the equations as

𝑥1 + 𝑥2 = 2

0.0001𝑥1 + 𝑥2 = 1

Apply GE to obtain 𝑥𝑇 = [1.0000 1.0000] which is 4th decimal accurate.

This is called pivoting. Swap the row, so that the pivotal element is relatively larger. Also, scaling (or multiplying equation (A) by 10000) can be done before applying GE.

Example: Solve by GE:

�1 −1 22 −2 31 1 1

� �𝑥1𝑥2𝑥3� = �

−8−20−2

�

Step 1: �1 −1 20 0 −10 2 −1

� �𝑥1𝑥2𝑥3� = �

−8−4+6

�

Row pivoting is required (or swap 2nd equation with 3rd so that pivotal elemental is non – zero).

�1 −1 20 2 −10 0 −1

� �𝑥1𝑥2𝑥3� = �

−8+6−4

�

We have now UTM. Reverse substitution will yield

𝑥𝑇 = [−11 5 4]

(a) 𝑑𝑒𝑡[𝐴] = −1 × 2 × −1 = 2 ≠ 0 (matrix is singular, rank = 3, all 3 equations are linearly independent ). Also, note that we swapped one time. So, there is the multiplication with -1.

Column pivoting (generally it should be avoided )

�1 2 −10 −1 00 −1 2

� �𝑥1𝑥2𝑥3� = �

−8−4+6

�

Column 2 is replaced with column 3 to make the pivotal element ≠ 0. However, note that the variables 𝑥2 𝑎𝑛𝑑 𝑥3 are also swapped in the column vector �̅�, else you would be solving a different set of equations!

6

�𝑥1 + 2𝑥2 − 𝑥3 = −8

−𝑥2 = −4−𝑥2 + 2𝑥3 = +6

�(left) original equation/(right) swapped equation: �𝑥1 + 2𝑥3 − 𝑥2 = −8

−𝑥3 = −42𝑥2 − 𝑥3 = +6

�

Proceed as follows:

�1 −2 −10 −1 00 0 2

� �𝑥1𝑥3𝑥2� = �

−8−410�

Reverse substitution will yield 𝑥𝑇 = [−11 , 5 , 4]

1

Lecture #04

Gauss-Jordan

1. Procedure is similar to that of GE

2. Normalizer the pivotal element

3. Eliminate the unknown from all rows, below and above the pivot row

4. No back substitution is required as required in GE.

𝐴𝑋� = 𝑏�

�𝑎11 𝑎12 𝑎13𝑎21 𝑎22 𝑎23𝑎31 𝑎32 𝑎33

�3×3

�𝑥1𝑥2𝑥3� = �

𝑏1𝑏2𝑏3�

Step 1: 𝑎1𝑗′ = 𝑎1𝑗𝑎11

𝑗 = 1,3 (𝑛𝑜𝑟𝑚𝑎𝑙𝑖𝑧𝑒 𝑎11(𝑝𝑖𝑣𝑜𝑡𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡) to 1.0)

𝑏1′ = 𝑏1 𝑎11⁄

�

𝑎2𝑗′ = 𝑎2𝑗 − 𝑎𝑖𝑗′ × 𝑎21𝑎3𝑗′ = 𝑎3𝑗 − 𝑎𝑖𝑗′ × 𝑎31𝑏2

′ = 𝑏2 − 𝑏1′ × 𝑎21

𝑏3′ = 𝑏3 − 𝑏1

′ × 𝑎31 ⎭⎪⎬

⎪⎫

𝑠𝑎𝑚𝑒 𝑎𝑠 𝑡ℎ𝑒 𝐺𝐸 𝑚𝑒𝑡ℎ𝑜𝑑;𝑚𝑎𝑘𝑖𝑛𝑔𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑐𝑜𝑙𝑢𝑚𝑛 𝑏𝑒𝑙𝑜𝑤 𝑡ℎ𝑒 𝑝𝑖𝑣𝑜𝑡𝑎𝑙 𝑟𝑜𝑤 𝑡𝑜 𝑏𝑒 0.

�1 𝑎12′ 𝑎13′0 𝑎22′ 𝑎23′0 𝑎32′ 𝑎33′

�

3×3

�𝑋� � = �𝑏1

′

𝑏2′

𝑏3′�

Step2: 𝑎2𝑗′′ = 𝑎2𝑗′

𝑎22′ 𝑗 = 2,3 (𝑛𝑜𝑟𝑚𝑎𝑙𝑖𝑧𝑒 𝑝𝑖𝑣𝑜𝑡𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑜𝑓 2𝑛𝑑 𝑟𝑜𝑤 to 1.0)

𝑏2′′ = 𝑏2

′ 𝑎22′⁄

�𝑎1𝑗′′ = 𝑎1𝑗′ − 𝑎2𝑗′ × 𝑎12′

𝑏1′′ = 𝑏1

′ − 𝑏2′ × 𝑎12′

� 𝑠𝑢𝑏𝑠𝑡𝑟𝑎𝑐𝑡 𝑡ℎ𝑒 𝑟𝑜𝑤 𝑎𝑏𝑜𝑣𝑒 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑝𝑖𝑣𝑜𝑡𝑎𝑙 𝑟𝑜𝑤.𝑵𝒐𝒕𝒆 𝒕𝒉𝒊𝒔 𝒊𝒔 𝒕𝒉𝒆 𝒆𝒙𝒕𝒓𝒂 𝒔𝒕𝒆𝒑 𝒊𝒏 𝑮𝑬.

�𝑎3𝑗′′ = 𝑎3𝑗′ − 𝑎2𝑗′ × 𝑎32′

𝑏3′′ = 𝑏3

′ − 𝑏2′ × 𝑎32′

� 𝑠𝑢𝑏𝑠𝑡𝑟𝑎𝑐𝑡 𝑡ℎ𝑒 𝑟𝑜𝑤 𝑏𝑒𝑙𝑜𝑤 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑝𝑖𝑣𝑜𝑡𝑎𝑙 𝑟𝑜𝑤.𝑇ℎ𝑖𝑠 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑠𝑡𝑒𝑝 𝑎𝑠 𝑖𝑛 𝐺𝐸.

2

�1 0 𝑎13′′0 1 𝑎23′′0 0 𝑎33′′

� 𝑋� = �𝑏1

′′

𝑏2′′

𝑏3′′�

Step 3: 𝑏3′′′ = 𝑏3

′′ 𝑎33′′⁄ (normalize pivotal element to 1)

�𝑏2′′′ = 𝑏2

′′ − 𝑏3′′′𝑎23′′

𝑏1′′′ = 𝑏1

′′ − 𝑏3′′′𝑎13′′

� 𝑠𝑢𝑏𝑠𝑡𝑟𝑎𝑐𝑡 𝑏𝑜𝑡ℎ 𝑟𝑜𝑤𝑠 1 𝑎𝑛𝑑 2 𝑎𝑏𝑜𝑣𝑒 𝑓𝑟𝑜𝑚 𝑝𝑖𝑣𝑜𝑡𝑎𝑙 𝑟𝑜𝑤.𝑇ℎ𝑒𝑟𝑒 𝑖𝑠 𝑛𝑜 𝑟𝑜𝑤 𝑏𝑒𝑙𝑜𝑤 𝑡ℎ𝑒 𝑝𝑖𝑣𝑜𝑡𝑎𝑙 𝑟𝑜𝑤.

�1 0 00 1 00 0 1

� �𝑥1𝑥2𝑥3� = �

𝑏1′′′

𝑏2′′′

𝑏3′′′�

Therefore, 𝑋𝑇 = 𝑏(′′′)𝑇: 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛.

No back substitution is required for a unit matrix!

Make a note that G-J method can also be used to determine inverse of the matrix, if working on the augmented matrix. Check text books for more details.

example: �1 −1 21 1 12 −2 3

� �𝑋� � = �−8−2−20

�

⇒ �1 −1 20 2 −10 0 −1

� �𝑥1𝑥2𝑥3� = �

−8 6−4

� (the pivotal element already has ‘1’ ⇒ no need to normalize!)

⇒ �1 −1 20 1 −0.50 0 −1

� �𝑥1𝑥2𝑥3� = �

−83−4

� (divide pivotal row by 2 to make the pivotal element to be1)

⇒ �1 0 1.50 1 −0.50 0 −1

� �𝑥1𝑥2𝑥3� = �

−5 3−4

� (subtract the rows above and below from the pivotal row)

⇒ �1 0 1.50 1 −0.50 0 1

� �𝑥1𝑥2𝑥3� = �

−534� (make the pivotal element to 1)

⇒ �1 0 00 1 00 0 1

� �𝑥1𝑥2𝑥3� = �

−1154� (subtract the rows 1 & 2 from the pivotal row)

𝑋𝑇 = [−11 5 4]𝑇

You should ensure that the programming code for G-J is written by modifying the code for GE, instead of writing fresh, by changing the indices for i and j.

3

LU Decomposition: Two similar/identical methods

Dolittle Crout’s

𝐴𝑋� = 𝑏�

(This method is considered to be the best if only right hand side 𝑏� changes from problem/case to problem/case)

𝐴 = 𝐿𝑈 ⇒𝑑𝑒𝑐𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛𝑠 𝑖𝑛𝑡𝑜 𝐿 𝑎𝑛𝑑 𝑈

⎩⎪⎪⎨

⎪⎪⎧�

1 0 0𝑙21 1 0𝑙31 𝑙32 1

� �𝑈11 𝑈12 𝑈13

0 𝑈22 𝑈230 0 𝑈33

�

𝐷𝑜𝑙𝑖𝑡𝑡𝑙𝑒

�𝑙11 0 0𝑙21 𝑙22 0𝑙31 𝑙32 𝑙33

� �1 𝑈12 𝑈130 1 𝑈230 0 1

�

𝐶𝑟𝑜𝑢𝑡

�

In either case, steps are as follows:

1. 𝐴 = 𝐿𝑈 (𝑑𝑒𝑐𝑜𝑚𝑝𝑜𝑠𝑒)

2. �𝐿 𝑈 𝑋� = 𝑏�

�̅� � ⇒ 𝐿�̅� = 𝑏�, 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 �̅� 𝑓𝑖𝑟𝑠𝑡

3. 𝑈𝑋� = �̅� ⇒ 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑋�

example: �1 −1 21 1 12 −2 3

� �𝑋� � = �−8−2−20

�

𝐴 = �1 0 0𝑙21 1 0𝑙31 𝑙32 1

�3×3

�𝑈11 𝑈12 𝑈13

0 𝑈22 𝑈230 0 𝑈33

�3×3

=

⎣⎢⎢⎢⎢⎢⎢⎢⎡𝑈11 𝑈12 𝑈13

𝑙21𝑈11 �𝑙21𝑈12

+𝑈22

� �𝑙21𝑈13

+𝑈23

�

𝑙31𝑈11 �𝑙31𝑈12

+𝑙32𝑈22

�

⎝

⎜⎛𝑙31𝑈13

+𝑙32𝑈23

+𝑈33 ⎠

⎟⎞

⎦⎥⎥⎥⎥⎥⎥⎥⎤

3×3

Match,

𝑈11 = 1, 𝑈12 = −1, 𝑈13 = 2

4

𝑙21𝑈11 = 1, 𝑙21𝑈12 + 𝑈12 = 1, 𝑙21𝑈13 + 𝑈23 = 1

𝑙31𝑈11 = 2, 𝑙31𝑈12 + 𝑙32𝑈23 = −2, 𝑙31𝑈13 + 𝑙32𝑈23 + 𝑈33 = 3

Therefore, there are 9 unknown & 9 equations to solve and the problem is well defined.

How do we obtain LU?

Well, use GE to reduce A to an upper triangular matrix U.

However, in addition, store multiplication coefficients of GE as the coefficients 𝑙21, 𝑙31,𝑎𝑛𝑑 𝑙32 of lower triangular matrix

as: 𝑙𝑖𝑘 = 𝑎𝑖𝑘𝑎𝑘𝑘

; � 𝑘 = 1 𝑡𝑜 𝑛 − 1𝑖 = 𝑘 + 1 𝑡𝑜 𝑛�

It is clear that the programming code used for GE to reduce the matrix A to UTM (U) is the same as that for determining U of LU decomposition method. In addition, the coefficients of L are also determined automatically by defining an extra coefficient 𝑙𝑖𝑘(see above) in the same programming code.

Example:

𝐴 = �1 −1 21 1 12 −2 3

�

Step 1:

= �1 0 0𝑙21 1 0𝑙31 𝑙32 1

� �𝑈11 𝑈12 𝑈13

0 𝑈22 𝑈230 0 𝑈33

�

L U

= �1 0 0

𝑙21 = 1 1 0𝑙31 = 2 𝑙32 = 0 1

� �1 −1 20 2 −10 0 −1

�

⎩⎪⎨

⎪⎧

𝑁𝑜𝑡𝑒 𝑙21 𝑖𝑠 𝑡ℎ𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑡𝑜 𝑚𝑎𝑘𝑒 𝑡ℎ𝑒 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑈12 = 0. 𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑖𝑡𝑦,

𝑙31 𝑎𝑛𝑑 𝑙32 𝑎𝑟𝑒 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒𝑑

�

L U

Therefore, �1 −1 21 1 12 −2 3

� = �1 0 01 1 02 0 1

� �1 −1 20 2 −10 0 −1

�

A L U

5

Step 2: 𝐿𝑑 = 𝑏�⃗

�1 0 01 1 02 0 1

� �𝑑1𝑑2𝑑3� = �

−8−2−20

�

𝑑1 = −8 ; 𝑑2 = −2 + 8 = 6; 𝑑3 = −20 + 16 = −4

𝑑𝑇 = [−8 6 − 4]𝑇

(Note: you have done forward substitution or solved a LTM, which is just opposite to solving an UTM or reverse/backward substitution step of GE)

Step 3: 𝑈𝑋� = �̅�

�1 −1 20 2 −10 0 −1

� �𝑋1𝑋2𝑋3� = �

−8 6−4

�

(Note this is the reverse/backward substitution of GE)

𝑋3 = −4, 𝑋2 = (6 + 4) 2⁄ = 5, 𝑋1 = −8 + 5 − 8 = −11

𝑋𝑇 = [−8 5 4]𝑇 Ans.

You should ensure that you are not writing a fresh code for the LU decomposition method. Rather, you simply modify the code you wrote earlier for the 1st part (forward substitution) of GE, or the code to reduce A to UTM. Then, write a code for inverting LTM and use the previously written code for the back-substitution or the reverse substitution of GE. In other words, the code of LU decomposition has three sub-parts.

1

Lecture #05 Matrix Inverse 𝑨 𝑨−𝟏 = 𝑰(𝒅𝒆𝒇𝒊𝒏𝒊𝒕𝒊𝒐𝒏 𝒇𝒐𝒓 𝑨−𝟏)

�1 0 00 1 00 0 0

�

If 𝐴 = �𝑎11 𝑎12 𝑎13𝑎21 𝑎22 𝑎23𝑎31 𝑎32 𝑎33

�, then

�𝑎11 𝑎12 𝑎13𝑎21 𝑎22 𝑎23𝑎31 𝑎32 𝑎33

� �𝑏11 𝑏12 𝑏13𝑏21 𝑏22 𝑏23𝑏31 𝑏32 𝑏33

� = �1 0 00 1 00 0 1

�

𝐴 𝐴−1 𝐼

Making use of the rule of matrix multiplication (row×column), it is clear that

�𝑎11 𝑎12 𝑎13𝑎21 𝑎22 𝑎23𝑎31 𝑎32 𝑎33

� �𝑏11𝑏21𝑏31

� = �100� ⇒ 𝐴𝑏1� = �

100�

and � � �𝑏12𝑏22𝑏32

� = �010� ⇒ 𝐴𝑏2����⃗ = [0 1 0]𝑇

� � �𝑏13𝑏23𝑏33

� = �001� ⇒ 𝐴𝑏3��� = [0 0 1]𝑇

Therefore, 𝑏𝑖𝑗 can be calculated from as many equations.

However, LU decomposition can be used to determine 𝐴−1as well:

𝐴𝑏1� = 𝐿𝑈 𝑏1� = [1 0 0]𝑇

𝑑1���

or 𝐿 𝑑1 = [1 0 0]𝑇 ⇒ 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑑1���

So, 𝑈𝑏1� = 𝑑1��� ⇒ 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑏1�

Similarly,

𝐴𝑏2��� = 𝐿𝑈 𝑏2��� = [0 1 0]𝑇

2

or 𝐿 𝑑2��� = [0 1 0]𝑇 ⇒ 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑑2���

So, 𝑈𝑏2��� = 𝑑2��� ⇒ 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑏2���, 𝑠𝑜 𝑓𝑜𝑟𝑡ℎ,

Example: [𝐴] = �25 5 164 8 1

144 1 1� = 𝐿𝑈

�25 5 164 8 1

144 1 1�

Apply GE (Forward step)

�25 5 10 −4.8 0.560 −16.8 −4.36

� ⇒

𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑢𝑠𝑒𝑑 𝑡𝑜 𝑚𝑎𝑘𝑒1𝑠𝑡 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑜𝑤𝑠 𝑏𝑜𝑡𝑡𝑜𝑚𝑜𝑓 𝑡ℎ𝑒 𝑝𝑖𝑣𝑜𝑡𝑎𝑙 𝑟𝑜𝑤 𝑡𝑜 𝑏𝑒 𝑧𝑒𝑟𝑜

= 6425� 𝑎𝑛𝑑 144

25 𝑜𝑟 2.56 𝑎𝑛𝑑 5.76

�25 5 10 −4.8 −1.560 0 0.7

� ⇒

𝑇ℎ𝑖𝑠 𝑖𝑠 𝑎𝑛 𝑈𝑇𝑀 .𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑢𝑠𝑒𝑑 𝑡𝑜 𝑚𝑎𝑘𝑒 𝑡ℎ𝑒

1𝑠𝑡 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑎𝑠𝑡 𝑟𝑜𝑤𝑡𝑜 𝑏𝑒 𝑧𝑒𝑟𝑜

= 16.8 4.8 = 3.5⁄

Therefore,

𝐿 = �1 0 0

2.56 1 05.76 3.5 1

�

or,

�25 5 164 8 1

144 1 1� = �

1 0 02.56 1 05.76 3.5 1

� �25 5 10 −4.8 −1.560 0 0.7

�

A L U

Now, determine 𝐴−1

Step 1: 𝐿𝑈 𝑏1� = [1 0 0]𝑇

𝑑1���

3

�1 0 0

2.56 1 05.76 3.5 1

� �𝑑1𝑑2𝑑3� = �

100�

L 𝒅𝟏����

𝑑1 = 1, 𝑑2 = 0 − 2.56 = −2.56, 𝑑3 = 0 − 5.76 + 3.50 × 2.56 = 3.2

2. �25 5 10 −4.8 −1.560 0 0.7

� �𝑏11𝑏21𝑏31

� = �𝑑1𝑑2𝑑3�

U 𝒃𝟏���

𝑏31 = 3.20.7� = 4.571, 𝑏21 = (−2.56 − 4.571 × (−1.56))/−4.8 = −0.9524

𝑏11 = (1 − 4.571 − 5 × (−0.9524))/25.0 = 0.04762

or, 𝑏1𝑇 = [0.04762 − 0.9524 4.571]𝑇

3. �1 0 0

2.56 1 05.76 3.5 1

� �𝑑1𝑑2𝑑3� = �

010� ⇒ 𝑑1 = 0, 𝑑2 = 1, 𝑑3 = −3.5

L 𝒅𝟏����

�25 5 10 −4.8 −1.560 0 0.7

� �𝑏12𝑏22𝑏32

� = �01

−3.5�

U 𝒃𝟐���

𝑏�2𝑇 = [−0.0833 1.417 − 0.5]𝑇

Similarly,

𝑏�3𝑇 = [0.03571 − 0.4643 1.429]𝑇

𝐴−1 = �0.04762 −0.0833 0.03571−0.9524 1.417 −0.4643

4.571 −0.5 1.429�

⇒ Methods so far to solve 𝑨𝑿� = 𝒃� and programming codes:

(a) GE: To solve a large number of algebraic equations, a programming code is required. As earlier shown, GE contains two distinct steps: (1) forward elimination to convert 𝑨� to an upper triangular matrix (UTM) (2) Backward/reverse substitution to solve UTM. Therefore, while writing the code;

4

it is recommended that one clearly distinguishes the two steps as “subroutines”. You will later see that the other methods also often require one or both of the steps/codes.

(b) GJ: The programming code for GJ is similar to that of the forward elimination of GE, with some modification. The second part (code) of GE, ie. reverse substitution is not required. The modification or extra step is simple. First, pivotal elements should be normalized to 1; in addition to subtracting the modified pivotal row from the rows below the ‘𝑖𝑡ℎ’ row of GE, the pivotal row is also subtracted from the rows above. Therefore, one extra line is included in the code for GJ.

(c) The code to determine the determinant of the matrix A is the same as that for converting A to UTM, ie. the first or forward elimination step of GE. Further, add a line to determine the ∏ 𝑎𝑖𝑖𝑛

𝑖=1 product of the diagonal elements of the UTM ie. one can put the flag to check if any of the 𝑎𝑖𝑖 is zero, the matrix is singular and cannot be inverted. One can also write a simple code to check the number # of zeros on the diagonal elements and therefore, determine the rank of the matrix as (𝑛 − #).

(d) LU method: The programming code is similar to that of the forward step of GE, to determine UTM. An extra line in the programming loop is required to store the coefficients 𝑙𝑖𝑘 = �𝑎𝑖𝑘

𝑎𝑘𝑘�, used to make the elements of the first column of the

UTM to be zero, as the elements of the empty columns of the LTM.

𝐴 = �𝑙11𝑙21 𝑙22

� × � �

LTM UTM

(e) Inverse of the matrix: First, the programming code of LU is to be used. The forward substitution on LTM will yield/give �̅� (𝑐𝑜𝑙𝑢𝑚𝑛 𝑚𝑎𝑡𝑟𝑖𝑥) and the backward substitution on UTM will yield 𝑏�(𝑐𝑜𝑙𝑢𝑚𝑛 𝑚𝑎𝑡𝑟𝑖𝑥)(already available). Therefore, one new code/subroutine is required for the forward substitution on LTM, which is not very different from that for the reverse substitution on UTM:

• 𝐴 = 𝐿𝑈 (𝑐𝑜𝑑𝑒 # 𝑑) • 𝐹𝑜𝑟𝑤𝑎𝑟𝑑 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 → �̅� (𝑠𝑖𝑚𝑖𝑙𝑎𝑟 𝑡𝑜 𝑐𝑜𝑑𝑒 #𝑎) • 𝐵𝑎𝑐𝑘𝑤𝑎𝑟𝑑 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛 → 𝑏� ( 𝑐𝑜𝑑𝑒 #𝑎)

1

Lecture #06

Thomas Algorithm (Tridiagonal system/matrix)

𝐴𝑋� = 𝑏�

⎣⎢⎢⎢⎢⎡𝑎11 𝑎12 0 ⋯ 0 0 0 0𝑎21 𝑎22 𝑎23 ⋯ 0 0 0 00 𝑎32 𝑎33 𝑎34 0 ⋯ 0 00 0 𝑎43 𝑎44 𝑎45 ⋯ 0 0

0 0 0 0 0 0 𝑎𝑛𝑛−1 𝑎𝑛𝑛⎦⎥⎥⎥⎥⎤

⎩⎪⎨

⎪⎧𝑋1𝑋2𝑋3⋮𝑋𝑛⎭⎪⎬

⎪⎫

=

⎩⎪⎨

⎪⎧𝑏1𝑏2𝑏3⋮𝑏𝑛⎭⎪⎬

⎪⎫

Tridiagonal matrix has 3 non-zero elements in all of its rows, except in the 1st and last row, with one element each on the left and right of the diagonal element. The 1st and last rows have one element on the right and left of the diagonal elements, respectively. It is a banded matrix around its diagonal:

⎣⎢⎢⎢⎡

⎦⎥⎥⎥⎤

⎩⎪⎨

⎪⎧𝑋1𝑋2𝑋3⋮𝑋𝑛⎭⎪⎬

⎪⎫

=

⎩⎪⎨

⎪⎧𝑏1𝑏2𝑏3⋮𝑏𝑛⎭⎪⎬

⎪⎫

Therefore, a tridiagonal matrix can be represented using single subscripted indices for its elements:

⎣⎢⎢⎢⎢⎡𝑏1 𝑐1 0 0 ⋯ 0𝑎2 𝑏2 𝑐2 0 ⋯ 00 𝑎3 𝑏3 𝑐3 ⋯ 0

𝑎4 𝑏4 𝑐4𝑧𝑒𝑟𝑜

𝑎𝑛 𝑏𝑛⎦⎥⎥⎥⎥⎤

Zero

Note that the element in the 𝑖𝑡ℎ row is represented as(𝑎𝑖, 𝑏𝑖, 𝑐𝑖). All bs are on the diagonal.

Therefore, 𝐴𝑋� = �̅�, where 𝐴 is a tridiagonal matrix and it represents the following set of linear algebraic equations:

2

�

𝑏1𝑥1 + 𝑐1𝑥2 = 𝑑1𝑎2𝑥1 + 𝑏2𝑥2 + 𝑐2𝑥3 = 𝑑2𝑎3𝑥2 + 𝑏3𝑥3 + 𝑐3𝑥4 = 𝑑3𝑎4𝑥3 + 𝑏4𝑥4 + 𝑐4𝑥5 = 𝑑4

𝑎𝑛𝑥𝑛−1 + 𝑏𝑛𝑥𝑛 = 𝑑𝑛⎭⎪⎬

⎪⎫

⇒ You must have noted that for the 𝑖𝑡ℎ row, 𝑏𝑖 is the diagonal element multiplied with the variable 𝑥𝑖 in the some row, whereas 𝑎𝑖 is multiplied with 𝑥𝑖−1(the variable above 𝑖𝑡ℎ row) and 𝑐𝑖 is multiplied with 𝑥𝑖+1(the variable below 𝑖𝑡ℎrow). Naturally, the first and last rows have only two variables.

⇒(Tridiagonal system is quite common when using Finite Difference 2nd order method to solve boundary value problems or partial differential 𝑒𝑞𝑛𝑠.)

Thomas algorithm to solve such system:

Step 1:

𝒙𝟏 is eliminated : �𝑏2 −𝑐1𝑏1𝑎2� 𝑥2 + 𝑐2𝑥3 = (𝑑2 −

𝑑1𝑏1𝑎2)

(by dividing 1st row with 𝑏1, multiplying with 𝑐1 and subtracting from row 2)

𝒙𝟐 is eliminated : �𝑏3 −𝑐2𝑏2𝑎3� 𝑥3 + 𝑐3𝑥4 = (𝑑3 −

𝑑2𝑏2𝑎3)

⋮

⋮

𝒙𝒌−𝟏 is eliminated : �𝑏𝑘 −𝑐𝑘−1𝑏𝑘−1

𝑎𝑘� 𝑥𝑘 + 𝑐𝑘𝑥𝑘+1 = (𝑑𝑘 −𝑑𝑘−1𝑏𝑘−1

𝑎𝑘)

It is clear that;

� 𝐴𝑙𝑙 𝑎𝑠 𝑎𝑟𝑒 𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑒𝑑𝐴𝑙𝑙 𝑏𝑠 𝑎𝑟𝑒 𝑚𝑜𝑑𝑖𝑓𝑖𝑒𝑑𝐴𝑙𝑙 𝑐𝑠 𝑎𝑟𝑒 𝑢𝑛𝑎𝑙𝑡𝑒𝑟𝑒𝑑𝐴𝑙𝑙 𝑑𝑠 𝑎𝑟𝑒 𝑚𝑜𝑑𝑖𝑓𝑖𝑒𝑑

�

Last row: 𝑏𝑛 = 𝑏𝑛 −𝑐𝑛−1𝑏𝑛−1

𝑎𝑛; 𝑑𝑛 = 𝑑𝑛 −𝑑𝑛−1𝑏𝑛−1

𝑎𝑛

3

(It is important to note that 𝑘𝑡ℎ row uses the latest modified values of 𝑏𝑘 and 𝑑𝑘 from the previous (k-1) step. Therefore, there is no need to store the previous values of (a, b, d), while writing the code)

Back-substitution: 𝑋𝑛 = 𝑑𝑛𝑏𝑛

(𝑛𝑡ℎ 𝑟𝑜𝑤 ℎ𝑎𝑠 𝑜𝑛𝑙𝑦 𝑏𝑛 𝑎𝑛𝑑 𝑑𝑛)

𝑋𝑛−1 = 𝑑𝑛−1−𝑐𝑛−1𝑥𝑛𝑏𝑛−1

(𝑛 − 1 𝑟𝑜𝑤 ℎ𝑎𝑠 𝑏𝑛−1, 𝑐𝑛−1𝑎𝑛𝑑 𝑑𝑛−1)

⇒ A pseudo programming code can be written as follows:

Tridiagonal (N, a, b, c, d, X)

𝑑𝑜 𝑖 = 1,𝑁

𝑎(𝑖) =

𝑏(𝑖) =

𝑐(𝑖) =

𝑑(𝑖) =

end do

𝑑𝑜 𝑖 = 2,𝑁

�𝑏(𝑖) = 𝑏(𝑖) − 𝑎(𝑖)

𝑐(𝑖 − 1)𝑏(𝑖 − 1)

𝑑(𝑖) = 𝑑(𝑖) − 𝑎(𝑖)𝑑(𝑖 − 1)𝑏(𝑖 − 1)⎭

⎪⎬

⎪⎫

𝑓𝑜𝑟𝑤𝑎𝑟𝑑 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛

end do

𝑋𝑛 = 𝑑(𝑛)𝑏(𝑛)

𝑑𝑜 (𝑖) = 𝑁 − 1, 1,−1

�𝑋𝑖 =𝑑(𝑖) − 𝑐(𝑖)𝑋𝑖+1

𝑏(𝑖)� 𝑟𝑒𝑣𝑒𝑟𝑠𝑒 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛

end do

4

Indirect Methods

GE, GJ, LU decomposition are the direct methods to solve 𝐴𝑋� = 𝑏�. Simple iterations can also be done to solve a set of algebraic equations, Jacobi and Gauss-Seidel being the two commonly used indirect methods.

Ex. �2 1 01 2 10 1 1

� �𝑋1𝑋2𝑋3� = �

124�

Make a guess 𝑋1(1),𝑋2

(1),𝑋3(1)

Jacobi Gauss-Seidel

𝑋1(2) = 1−𝑋2

(1)

2 = 1−𝑋2

(1)

2

𝑋2(2) = 2−𝑋1

(1)−𝑋3(1)

2 = 2−𝑋1

(2)−𝑋3(1)

2

𝑋3(2) = 4−𝑋2

(1)

1 = 4−𝑋2

(2)

1

Takes all 𝑋𝑠(𝑘−1) of the previous iteration Uses the most latest iterated values of 𝑋𝑠.

In general: 𝐴𝑋� = 𝑏�

Jacobi: 𝑋𝑖(𝑘+1) =

𝑏𝑖−∑ 𝑎𝑖𝑗𝑋𝑗(𝑘)𝑛

𝑗=1

𝑎𝑖𝑖(≠0) (𝑗 ≠ 𝑖)

diagonal elements

G-S: 𝑋𝑖(𝑘+1) =

⎩⎪⎪⎨

⎪⎪⎧𝑏𝑖 − ∑ 𝑎𝑖𝑗𝑋𝑗

(𝑘+1) − ∑ 𝑎𝑖𝑗𝑋𝑗(𝑘)𝑛

𝑗=𝑖+1𝑖−1𝑗=1

𝑢𝑝𝑑𝑎𝑡𝑒𝑑 𝑜𝑙𝑑 𝑙𝑎𝑡𝑒𝑠𝑡 𝑣𝑎𝑙𝑢𝑒𝑠𝑣𝑎𝑙𝑢𝑒𝑠

𝑎𝑖𝑖(≠ 0)

�

5

At any 𝑖𝑡ℎ row:

1⋯⋯𝑖 − 1 𝑖 𝑖 + 1⋯⋯ 𝑁

𝑢𝑝𝑑𝑎𝑡𝑒𝑑𝑣𝑎𝑙𝑢𝑒𝑠(𝑘 + 1)

𝑜𝑙𝑑

𝑣𝑎𝑙𝑢𝑒𝑠(𝑘)

Schematically,

elements multiply with the variables 1 to 𝑥𝑖−1

⎣⎢⎢⎢⎢⎢⎡

1 ⋯ 𝑖 − 1 𝑖 𝑖 + 1 ⋯ 𝑛

⎦⎥⎥⎥⎥⎥⎤

⎩⎪⎪⎨

⎪⎪⎧𝑋1𝑋2⋮𝑋𝑖⋮⋮𝑋𝑛⎭⎪⎪⎬

⎪⎪⎫

elements multiply with the

𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝑥𝑖+1 𝑡𝑜 𝑥𝑛.

G-S (modified)

𝑋𝑖 = 𝜆𝑋𝑖(𝑛𝑒𝑤) + (1 − 𝜆)𝑋𝑖(𝑜𝑙𝑑)

𝜆 = 1 (𝑢𝑛𝑚𝑜𝑑𝑖𝑓𝑖𝑒𝑑)

0 < 𝜆 < 1 ⇒ 𝑢𝑛𝑑𝑒𝑟 − 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟

2 > 𝜆 < 2 ⇒ 𝑜𝑣𝑒𝑟 − 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟

(′𝜆′𝑖𝑠 𝑜𝑓𝑡𝑒𝑛 𝑎𝑠𝑠𝑢𝑚𝑒𝑑 𝑑𝑒𝑝𝑒𝑛𝑑𝑖𝑛𝑔 𝑢𝑝𝑜𝑛 𝑔𝑢𝑒𝑠𝑒𝑠, 𝑒𝑡𝑐)

6

Look differently! : LDU method

⎣⎢⎢⎢⎡𝑎11 𝑎12 ⋯ 𝑎1𝑛⋮⋮⋮𝑎𝑛1 ⋯ ⋯ 𝑎𝑛𝑛⎦

⎥⎥⎥⎤

𝑛×𝑛

=

⎣⎢⎢⎢⎡ 0𝑎21 0

00

𝑎𝑛1 ⋯ 𝑎𝑛−1 ⎦⎥⎥⎥⎤

+

⎣⎢⎢⎢⎡𝑎11 ⋱

⋱

𝑎𝑛𝑛⎦⎥⎥⎥⎤

+

⎣⎢⎢⎢⎡

0 ⋯ ⋯ 𝑎1𝑛0

0𝑎𝑛−1𝑛

0 ⎦⎥⎥⎥⎤

strictly lower𝑡𝑟𝑖𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑚𝑎𝑡𝑟𝑖𝑥

(𝐿)

𝐷(𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙

𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑛𝑙𝑦)

𝑠𝑡𝑟𝑖𝑐𝑡𝑙𝑦 𝑢𝑝𝑝𝑒𝑟𝑡𝑟𝑖𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑚𝑎𝑡𝑟𝑖𝑥

(𝑈)

𝐴 = 𝐿 + 𝐷 + 𝑈

𝐴𝑋� = 𝑏�

(𝐿 + 𝐷 + 𝑈)𝑋� = 𝑏�⃗ ⇒ 𝐷𝑋� = 𝑏� − 𝐿𝑋� − 𝑈𝑋�

𝑜𝑟 𝑋� =𝑏� − 𝐿𝑋� − 𝑈𝑋�

𝐷

Jacobi:

𝑋�(𝑘+1) =𝑏� − (𝐿 + 𝑈)𝑋�(𝑘)

𝐷

𝑜𝑟 𝑋𝑖(𝑘+1) =

𝑏𝑖 − ∑ 𝑋𝑗(𝑘)𝑎𝑖𝑗𝑛

𝑗=1

𝑎𝑖𝑖 ; 𝑗 ≠ 𝑖 (𝑗 = 𝑖 𝑤𝑖𝑙𝑙 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡)

𝐆 − 𝐒 ∶ (𝐿 + 𝐷)𝑋� = 𝑏� − 𝑈𝑋�

(𝐿 + 𝐷)𝑋�(𝑘+1) = 𝑏� − 𝑈𝑋�(𝑘)

𝐷𝑋�(𝑘+1) = 𝑏� − 𝑈𝑋�(𝑘) − 𝐿𝑋�(𝑘+1)

𝑋𝑖(𝑘+1) =

𝑏𝑖−∑ 𝑋𝑖(𝑘+1)𝑎𝑖𝑗−∑ 𝑋𝑖

𝑘𝑎𝑖𝑗𝑛𝑗=𝑖+1

𝑖−1𝑗=1

𝑎𝑖𝑖

(𝑠𝑎𝑚𝑒 𝑎𝑠 𝑏𝑒𝑓𝑜𝑟𝑒 ) ( 𝑎𝑖𝑖 ≠ 0 )

Both methods will lead to the same solutions, with different # of iterations.

1

Lecture #07 Homogeneous linear algebraic equations

-Represents a special class of problems, also known as Eigenvalue or Characteristic type of problem

- Mathematically represented as 𝐴�̅� = 0 or 𝑏� = 0 (𝑛𝑢𝑙𝑙 𝑣𝑒𝑐𝑡𝑜𝑟𝑠)

𝑜𝑟 �

𝑎11𝑥1 + 𝑎12𝑥2 + ⋯⋯𝑎1𝑛𝑥𝑛 = 0⋮ ⋮ 𝑎𝑛1𝑥1 + 𝑎𝑛2𝑥2 + ⋯⋯𝑎𝑛𝑛𝑥𝑛 = 0

�

Naturally, a trivial solution is �̅� = 0 . In the simplest geometrical term, two straight lines or three planes intersect at the origin:

� 𝑥1 − 𝑥2 = 0 𝑥2 − 0.3𝑥2 = 0� ⇒ 𝑥2

𝑥1

⇒ More interesting is in seeking a non-trivial solution when 𝑑𝑒𝑡(𝐴) = 0 𝑜𝑟 𝑟 < 𝑚. Such simultaneous set of homogeneous linear equations when 𝑑𝑒𝑡(𝐴) = 0 is common in several engineering and mechanics applications, and also in initial and boundary value problems. Such situation is better known as eigenvalue problem and is represented by eigenvalues and the corresponding eigenvectors.

Compare this situation to the earlier discussed set of non-homogeneous linear algebraic equation 𝐴𝑋� = 𝑏 � , where we sought a unique solution, when 𝑑𝑒𝑡|𝐴| ≠ 0. Comparatively,

Non-homogeneous Homogeneous equations

𝐴𝑋� = 𝑏� 𝐴𝑋� = 0

det (𝐴) ≠ 0 det(𝐴) = 0

𝑟 = 𝑚 𝑟 < 𝑚

Unique solution Non-trivial solution

𝐸𝑖𝑔𝑒𝑛𝑣𝑎𝑙𝑢𝑒 𝑝𝑟𝑜𝑏𝑙𝑒𝑚

2

Let us look at

𝐴𝑋� = 𝜆𝑋� 1 where A is the coefficient matrix and ′𝜆′ is a non-zero number.

matrix scalar quantity

or (𝐴 − 𝜆𝐼)𝑋� = 0 2 where 𝐼 is an identity matrix

𝐼 = �1 0 00 1 00 0 1

�

Equation 1 or 2 represents eigenvalue problem and such equations commonly occur in several

dynamic studies of distillation, adsorption and CSTR, viz. in solving 𝑑𝑦�𝑑𝑡

= 0. Equation (2) can

also be written as

⎩⎪⎨

⎪⎧

(𝑎11 − 𝜆)𝑋1 + 𝑎12𝑋2 + ⋯⋯𝑎1𝑛𝑋𝑛 = 0𝑎21𝑋1 + (𝑎22 − 𝜆)𝑋2 + ⋯⋯𝑎2𝑛𝑋𝑛 = 0⋮ ⋮

𝑎𝑛1𝑋1 + 𝑎𝑛2𝑋2 + ⋯⋯ (𝑎𝑛𝑛 − 𝜆)𝑋𝑛 = 0⎭⎪⎬

⎪⎫

or

�

𝑎11 𝑎12 ⋯ 𝑎1𝑛⋮⋮𝑎𝑛1 𝑎𝑛2 ⋯ 𝑎𝑛𝑛

� �

𝑋1𝑋2⋮𝑋𝑛

� = 𝜆 �

𝑋1𝑋2⋮𝑋𝑛

�

𝐴(𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑚𝑎𝑡𝑟𝑖𝑥) 𝑐𝑜𝑙𝑢𝑚𝑛𝑣𝑒𝑐𝑡𝑜𝑟 𝑎 𝑠𝑐𝑎𝑙𝑎𝑟𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑐𝑜𝑙𝑢𝑚𝑛𝑣𝑒𝑐𝑡𝑜𝑟

Re-visit 𝐴𝑋� = 𝜆𝑋� - eigenvalue problem

Most vectors 𝑋� will not satisfy such an equation. A common vector (𝑖𝑋1 + 𝑗𝑋2) will change direction and magnitude to (𝑖𝑏1 + 𝑗𝑏2) on the transformation by A. Only certain special vectors 𝑋�, called eigenvectors, corresponding to only special numbers 𝜆(𝑒𝑖𝑡ℎ𝑒𝑟 + 𝑣𝑒 𝑜𝑟 − 𝑣𝑒), called eigenvalue will satisfy the above equation. In such case, the eigenvector 𝑋� does not change its direction or does not rotate when transformed by the coefficient matrix A, but is only scaled by the eigenvalue 𝜆. See the geometrical representation below:

3

𝑋2

𝑋2

𝑋�

𝑏� 𝜆�⃗�

𝑋�

𝑋1 𝑋1

𝐴𝑋� = 𝑏� 𝐴𝑋� = 𝜆𝑋�

(𝑛𝑜𝑡 𝑎𝑛 𝑒𝑖𝑔𝑒𝑛𝑣𝑎𝑙𝑢𝑒 𝑝𝑟𝑜𝑏𝑙𝑒𝑚)

(𝑒𝑖𝑔𝑒𝑛𝑣𝑎𝑙𝑢𝑒 𝑝𝑟𝑜𝑏𝑙𝑒𝑚)

(𝑋� 𝑖𝑠 𝑡ℎ𝑒 𝑒𝑖𝑔𝑒𝑛𝑣𝑒𝑡𝑜𝑟 𝑜𝑓 𝐴; 𝑖𝑠 𝑡ℎ𝑒 𝑒𝑖𝑔𝑒𝑛𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝐴)

𝜆

It turns out that (𝑛 × 𝑛) matrix 𝐴 will give ‘𝑛’ 𝜆𝑠(𝑒𝑖𝑔𝑒𝑛𝑣𝑎𝑙𝑢𝑒𝑠) and each eigenvalue will give ‘𝑛’ eigenvectors that will also be linearly independent. As an example

�𝑎11 𝑎12𝑎21 𝑎22�2×2

𝑤𝑖𝑙𝑙 ℎ𝑎𝑣𝑒 2𝜆𝑠 (𝜆1&𝜆2) ⇒ 𝜆1 →{𝑋}1

𝜆2 → {𝑋}2

Both vectors {𝑋}1, & {𝑋}2 will satisfy 𝐴𝑋� = 𝜆𝑋�. They will also be linearly independent so that they can be used as a basis for the space description. They may be orthogonal or non-orthogonal.

𝑋2

𝜆1

𝜆2

𝑋1

In the figure above, dotted and solid lines represent two vectors scaled by the respective eigenvalues1 and 2, respectively.

4

Similarly, (3 × 3) A matrix will give 3𝜆𝑠(𝜆1,𝜆2, 𝜆3). Each 𝜆 will give three independent vectors {𝑋}1, {𝑋}2, {𝑋}3 satisfying 𝐴𝑋� = 𝜆𝑋�.

or (𝐴 − 𝜆𝐼)𝑋� = 0

Linear algebra tells us that for 𝑋� to have a non-trivial solution 𝑑𝑒𝑡|(𝐴 − 𝜆𝐼)| = 0.

Example: 𝐴 = �2 11 2�

𝐴𝑋� = 𝜆𝑋�(𝐴 − 𝜆𝐼)𝑋� = 0

𝑑𝑒𝑡 �2 − 𝜆 11 2 − 𝜆� = 0 ⇒ (2 − 𝜆)2 − 1 = 0

𝜆1 = 1, 3

𝜆1 = 1: �2 11 2� �

𝑋1𝑋2� = 1 �𝑋1𝑋2

�

Both equations yield 𝑋1 = −𝑋2 𝑜𝑟 𝑋1 + 𝑋2 = 0 (𝑟 = 1)

So, 𝑋� (𝜆1 = 1): (𝑖 − 𝑗) ≡ � 1−1�

𝜆2 = 3: �2 11 2� �

𝑋1𝑋2� = 3 �𝑋1𝑋2

�

Both equations yield 𝑋1 = 𝑋2 𝑜𝑟 𝑋1−𝑋2 = 0 (𝑟 = 1)

𝑋� (𝜆2 = 3): (𝑖 + 𝑗) ≡ �11�

Note two vectors (𝑖 − 𝑗) 𝑎𝑛𝑑 (𝑖 + 𝑗) are linearly independent and can be used as a basis for defining 2D space. They need not be orthogonal! They can be used as a linear combination of

any other vector 𝐵�⃗ in 2D space:

𝑋2

𝐵�⃗ = 𝑐1(𝑖 + 𝑗) + 𝑐2(𝑖 − 𝑗)

𝑖 + 𝑗

𝑋1

(𝑖 − 𝑗)

5

System of ODEs

It may represent dynamics, i.e. how a system behaves due to a perturbation around steady state value. Such situation is common in kinetics and heat and mass transfer related problems of chemical engineering.

�

𝑑𝑋1𝑑𝑡

= 𝑓1(𝑋1 … …𝑋𝑛)

𝑑𝑋2𝑑𝑡

= 𝑓2(𝑋1 … …𝑋𝑛)⋮ 𝑑𝑋𝑛𝑑𝑡

= 𝑓𝑛(𝑋1 … …𝑋𝑛)⎭⎪⎬

⎪⎫

⇒ dynamics

𝑋� 𝑋3

𝑑𝑋�

𝑑𝑡= 𝑓(̅𝑋�) [𝑋�]𝑇 ≡ [𝑋1, … … .𝑋𝑛] 𝑋2 𝑋1

under SS, 𝑓(̅𝑋�) = 0 ⇒ 𝑋� = 𝑋�𝑠 0 𝑡

𝑜𝑟

�𝑓1(𝑋1 … …𝑋𝑛) = 0𝑓2(𝑋1 … …𝑋𝑛) = 0⋮ 𝑓𝑛(𝑋1 … …𝑋𝑛) = 0

� 𝑇ℎ𝑒𝑠𝑒 𝑠𝑒𝑡𝑠 𝑜𝑓 𝑎𝑙𝑔𝑒𝑏𝑟𝑎𝑖𝑐 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠𝑚𝑎𝑦 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑟 𝑠𝑝𝑒𝑐𝑖𝑒𝑠

𝑏𝑎𝑙𝑎𝑛𝑐𝑒 𝑢𝑛𝑑𝑒𝑟 𝑆𝑆.

𝑋�(𝑡) = ?

define, 𝜉̅(𝑡) = 𝑋�(𝑡) − 𝑋�𝑠 (𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑑𝑒𝑝𝑎𝑟𝑡𝑢𝑟𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠)⁄

𝐸𝑙𝑒𝑚𝑒𝑛𝑡 𝑤𝑖𝑠𝑒 𝜉1(𝑡) = 𝑋1(𝑡) − 𝑋1𝑠

𝑜𝑟 𝜉𝑖(𝑡) = 𝑋𝑖(𝑡) − 𝑋𝑖𝑠 (𝑖 = 1 … … …𝑛)

𝑜𝑟, 𝑑𝜉�

𝑑𝑡= 𝑑𝑋�(𝑡)

𝑑𝑡= 𝑓(̅𝑋�)

𝑜𝑟, 𝑑𝜉𝑖𝑑𝑡

= 𝑓𝑖(𝑋1 … …𝑋𝑛) (𝑖 = 1 … …𝑛)

= 𝑓𝑖(𝑋1𝑠 + 𝜉1,𝑋2𝑠 + 𝜉2, … …𝑋𝑛𝑠 + 𝜉𝑛)

= 𝑓𝑖(𝑋1𝑠,𝑋2𝑠,𝑋3𝑠, … …𝑋𝑛𝑠) + � 𝜕𝑓𝑖

𝜕𝑋1�𝑋1𝑠

𝜉1 + � 𝜕𝑓𝑖𝜕𝑋2

�𝑋2𝑠

𝜉2 + ⋯⋯ �𝜕𝑓𝑖𝜕𝑋𝑛

�𝑋𝑛𝑠

𝜉𝑛

(𝑆𝑆)0

6

𝜉1 = (𝑋1 − 𝑋1𝑠), 𝜉2 = (𝑋2 − 𝑋2𝑠) … … . , 𝜉𝑛 = (𝑋𝑛 − 𝑋𝑛𝑠)

(Taylor series expansion for multivariables: retain 1st order terms only, i.e. small departure from SS)

⎩⎪⎪⎪⎨

⎪⎪⎪⎧𝑑𝜉1𝑑𝑡....

𝑑𝜉𝑛𝑑𝑡

⎭⎪⎪⎪⎬

⎪⎪⎪⎫

=

⎣⎢⎢⎢⎢⎢⎢⎡𝜕𝑓1𝜕𝑋1

𝜕𝑓1𝜕𝑋2

⋯𝜕𝑓1𝜕𝑋𝑛

𝜕𝑓2𝜕𝑋1

𝜕𝑓2𝜕𝑋2

⋯𝜕𝑓2𝜕𝑋𝑛

⋮𝜕𝑓𝑛𝜕𝑋1

𝜕𝑓𝑛𝜕𝑋2

⋯𝜕𝑓𝑛𝜕𝑋𝑛⎦

⎥⎥⎥⎥⎥⎥⎤

⎩⎪⎪⎨

⎪⎪⎧𝜉1𝜉2⋮ .

. . . 𝜉𝑛⎭

⎪⎪⎬

⎪⎪⎫

(𝑟𝑜𝑤 × 𝑐𝑜𝑙𝑢𝑚𝑛 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑟𝑢𝑙𝑒)

𝑁𝑜𝑡𝑒: 𝐹𝑜𝑟 𝑡𝑤𝑜 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝑋1 𝑎𝑛𝑑 𝑋2

�

𝑑𝜉1𝑑𝑡

= � 𝜕𝑓1𝜕𝑋1

�𝑋1𝑠

(𝑋1 − 𝑋1𝑠) + � 𝜕𝑓1𝜕𝑋2

�𝑋2𝑠

(𝑋2 − 𝑋2𝑠)

𝑑𝜉2𝑑𝑡

= � 𝜕𝑓2𝜕𝑋1

�𝑋1𝑠

(𝑋1 − 𝑋1𝑠) + � 𝜕𝑓2𝜕𝑋2

�𝑋2𝑠

(𝑋2 − 𝑋2𝑠)⎭⎪⎬

⎪⎫

1𝑠𝑡 𝑜𝑟𝑑𝑒𝑟𝑡𝑒𝑟𝑚 𝑜𝑛𝑙𝑦

�

𝑑𝜉1𝑑𝑡𝑑𝜉2𝑑𝑡

� =

⎣⎢⎢⎡𝜕𝑓1𝜕𝑋1

𝜕𝑓1𝜕𝑋2

𝜕𝑓2𝜕𝑋1

𝜕𝑓2𝜕𝑋2⎦

⎥⎥⎤�𝑋1 − 𝑋1𝑠 = 𝜉1𝑋2 − 𝑋2𝑠 = 𝜉2

�

or,

�

𝑑𝜉�

𝑑𝑡= 𝐴𝜉̅ 𝑜𝑟 𝜉̅′ = 𝐴𝜉̅

𝑤ℎ𝑒𝑟𝑒 𝐴 =

⎣⎢⎢⎡𝜕𝑓1𝜕𝑋1

⋯ 𝜕𝑓1𝜕𝑋𝑛

⋮𝜕𝑓𝑛𝜕𝑋1

⋯ 𝜕𝑓𝑛𝜕𝑋𝑛⎦

⎥⎥⎤

; 𝐴 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑡ℎ𝑒 𝐽𝑎𝑐𝑜𝑏𝑖𝑎𝑛 𝑚𝑎𝑡𝑟𝑖𝑥 𝑜𝑓 𝐴.

⎭⎪⎬

⎪⎫

Thus, we started from 𝑑𝑋�𝑑𝑡

= 𝑓̅(𝑋�)

And derived:

𝑑𝜉�

𝑑𝑡= 𝐴𝜉̅

where 𝜉̅ = (𝑋� − 𝑋�𝑠) and A is the Jacobian matrix of 𝑓 ̅as

⎣⎢⎢⎡𝜕𝑓1𝜕𝑋1

⋯ 𝜕𝑓1𝜕𝑋𝑛

⋮𝜕𝑓𝑛𝜕𝑋1

⋯ 𝜕𝑓𝑛𝜕𝑋𝑛⎦

⎥⎥⎤

1

Lecture #08 (continue..)

Seeking solution to 𝑑𝜉�

𝑑𝑡= 𝐴𝜉̅ 𝑜𝑟 𝜉′ = 𝐴𝜉̅

Let us assume, 𝜉𝑖 = 𝑧𝑖𝑒𝜆𝑡, 𝑖 = 1 … …𝑛 (𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛)

(this is the most plausible solution to a simple ODE)

or, 𝜉̅ = 𝑧̅𝑒𝜆𝑡 (𝜆 = 𝑠𝑐𝑎𝑙𝑎𝑟, 𝑧̅ = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑣𝑒𝑐𝑡𝑜𝑟)

Substitute,

𝑧𝑖𝜆𝑒𝜆𝑡 = 𝑎𝑖1𝑧1𝑒𝜆𝑡 + 𝑎𝑖2𝑧2𝑒𝜆𝑡 + ⋯⋯𝑎𝑖𝑛𝑧𝑛𝑒𝜆𝑡 (𝑖 = 1⋯⋯⋯𝑛)

or,

�𝑎11𝑧1 + 𝑎12𝑧2 + ⋯⋯⋯𝑎1𝑛𝑧𝑛 = 𝜆𝑧1⋮ 𝑎𝑛1𝑧1 + 𝑎𝑛2𝑧2 + ⋯⋯⋯𝑎𝑛𝑛𝑧𝑛 = 𝜆𝑧𝑛

� 𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠.

(∑ 𝑎𝑖𝑗𝑧𝑗 = 𝜆𝑧𝑖; 𝑖 = 1, ⋯⋯𝑛)𝑛𝑗=1

or, 𝐴𝑧̅ = 𝜆𝑧̅ 𝑧̅𝑇 = [𝑧1 ⋯⋯𝑧𝑛]

or, (𝐴 − 𝜆𝐼)𝑧̅ = 0 𝑤ℎ𝑒𝑟𝑒 𝐴 = �𝑎11 ⋯ 𝑎1𝑛⋮𝑎𝑛1 ⋯ 𝑎𝑛𝑛

�

Thus, we have got a characteristic equation which describes the characteristics of the system dynamics AND this equation represents an eigenvalue problem. In summary, we started from

𝑎𝑛𝑑 𝑔𝑜𝑡 �

𝑑𝑋�𝑑𝑡

= 𝑓(̅𝑋�)

𝑑𝜉̅𝑑𝑡

= 𝐴𝜉̅

𝑁𝑜𝑤, 𝑤𝑒 ℎ𝑎𝑣𝑒 (𝐴 − 𝜆𝐼)𝑧̅ = 0⎭⎪⎬

⎪⎫

𝑤ℎ𝑒𝑟𝑒 𝐴 = 𝐽𝑎𝑐𝑜𝑏𝑖𝑎𝑛 𝑚𝑎𝑡𝑟𝑖𝑥 𝑜𝑓 𝑓 ̅𝑎𝑛𝑑 𝜉̅ = 𝑧̅𝑒𝜆𝑡

Therefore, for a non-trivial solution

det|𝐴 − 𝜆𝐼| = 0 (𝑟 < 𝑚)

2

or 𝑑𝑒𝑡 �

𝑎11 − 𝜆 𝑎12 ⋯ 𝑎1𝑛⋮⋮𝑎𝑛1 ⋯ ⋯ 𝑎𝑛𝑛−𝜆

� = 0

Solve for 𝑛 𝜆𝑠′ (eigenvalues) from the 𝑛𝑡ℎ degree polynomial on ′𝜆′ after you have expanded the (𝑛 × 𝑛) |𝐴 − 𝜆𝐼|matrix to determine its determinant. Recall, every 𝜆𝑖(𝑖 = 1⋯⋯𝑛) will give one 𝑧̅(𝑖) eigenvector so that 𝐴𝑧̅(𝑖) = 𝜆𝑖𝑧̅(𝑖) . In other words,

�𝜆1 → 𝑧̅(1)𝑒𝜆1𝑡

𝜆2 → 𝑧̅(2)𝑒𝜆2𝑡⋮ 𝜆𝑛 → 𝑧̅(𝑛)𝑒𝜆𝑛𝑡 ⎭

⎬

⎫ 𝐴𝑛𝑑 𝑎𝑙𝑙 𝑡ℎ𝑒𝑠𝑒 𝑒𝑖𝑔𝑒𝑛𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑎𝑟𝑒 𝑙𝑖𝑛𝑒𝑎𝑟𝑙𝑦

𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡

Therefore, we have a general solution:

𝜉̅ = 𝑐1𝑧̅(1)𝑒𝜆1𝑡 + 𝑐2𝑧̅(2)𝑒𝜆2𝑡 ⋯+ 𝑐𝑛𝑧̅(𝑛)𝑒𝜆𝑛𝑡

where 𝑐1, 𝑐2 ⋯𝑐𝑛 are arbitrary constants to be determined from initial conditions.

�

𝜉1𝜉2⋮𝜉𝑛

� = 𝑐1𝑒𝜆1𝑡 �

𝑧1𝑧2⋮𝑧𝑛

�

(1)

+ 𝑐2𝑒𝜆2𝑡 �

𝑧1𝑧2⋮𝑧𝑛

�

(2)

+ ⋯𝑐𝑛𝑒𝜆𝑛𝑡 �

𝑧1𝑧2⋮𝑧𝑛

�

(𝑛)

(𝑁𝑜𝑡𝑒 𝑋� = 𝑋�𝑠 + 𝜉̅)

Numerical values of ′𝜆′ decides how solution will behave i.e. error 𝜉 or perturbation or deviation from SS will grow or decay. If the real part of 𝜆 𝑖𝑠 ′ − 𝑣𝑒′, 𝑦 → 𝑦𝑠 or solution will decay; if ′ + 𝑣𝑒′ solution will grow ⇒ (unstable)

Example: Study the dynamics

� 𝑋1′ = −3𝑋1 + 2𝑋2

𝑋2′ = 𝑋1 − 2𝑋2�

𝐍𝐨𝐭𝐞: If there is a non − homogeneous part inthe equations, ignore it because only homogenous parts

contribute to the growth or decay of the solution.

𝑜𝑟 𝑑𝑋�

𝑑𝑡= 𝐴𝑋�, 𝐴 = �−3 2

1 −2�

Determine 𝜆𝑠: det �−3 − 𝜆 2 1 −2 − 𝜆� ⇒ 𝜆2 + 5𝜆 + 4 = 0 (𝐶ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐𝑠 𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙)

𝜆1 = −1, 𝜆2 = −4

3

Determine corresponding eigenvector:

𝜆1 = −1: �−3 2 1 −2� �

𝑋1𝑋2� = −1 �𝑋1𝑋2

�

� or − 3X1 + 2X2 = −X1and X1 − 2X2 = −X2

� ⇒ 𝑋1 = 𝑋2 ⇒ 𝑋�(1) = �11�(𝑟 = 1 < 𝑚)

𝜆2 = −4 𝑋1 = −2𝑋2 ⇒ 𝑋�(2) = �−2 1 �

Check 𝑋�1(1) 𝑎𝑛𝑑 𝑋�(2) are linearly independent.

⇒ Note: both 𝜆𝑠 are −𝑣𝑒. Therefore, the solutions will decay to the SS values or perturbation will die/decay, or solutions will not grow, or the systems will be stable.

General solution:

𝑋� = 𝐶1𝑋�(1)𝑒𝜆1𝑡 + 𝐶2𝑋�(2)𝑒𝜆2𝑡

𝑜𝑟 �𝑋1𝑋2� = 𝐶1𝑒−𝑡 �11� + 𝐶2𝑒−4𝑡 �−2

1 �

�𝑜𝑟 𝑋1 = 𝐶1𝑒−𝑡 − 2𝐶2𝑒−4𝑡

𝑋2 = 𝐶1𝑒−𝑡 + 𝐶2𝑒−4𝑡�

(Note: For repeat or complex 𝜆, check the books for the methods to determine 𝑋�)

Ex. [𝐴] = � 0 2 3−10 −1 2−2 4 7

� determine all λs and X�s.

From 𝑑𝑒𝑡[𝐴 − 𝜆𝐼] = 0 ⇒ 𝜆1 = 1, 𝜆2 = 2, 𝜆3 = 3

(A 3 × 3 matrix will give 3λs )

𝜆1 = 1 [𝐴 − 𝜆𝐼]𝑋� = 0 ⇒ � −1 2 3−10 −2 2−2 4 6

� �𝑋1𝑋2𝑋3� = 0

Apply Gauss Elimination:

� −1 2 3 0 −22 −28 0 0 0

� �𝑋1𝑋2𝑋3� = 0

You will get only 2 independent 𝑒𝑞𝑛𝑠: (𝑟𝑎𝑛𝑘 𝑟 = 2 < 𝑚, 𝑑𝑒𝑡( ) = 0)

4

� 𝑋1 = 2𝑋2 + 3𝑋3−22𝑋2 − 28𝑋3 = 0� ⇒

choose 𝑋3 = 1 𝑋1 = 5 11⁄

𝑋2 = −14 11⁄

Therefore,

𝑋�(1) = �5 11⁄

− 14 11⁄1

�

Similarly for

𝜆2 = 2: �𝑋�(2)�𝑇

= [1 2 − 1 1⁄ ]

𝜆3 = 3: �𝑋�(3)�𝑇

= [1 2 −3 4⁄ 1⁄ ]

Note: These are all linearly independent (𝑋�0(1) ≠ 𝐶1𝑋�(2) + 𝐶2𝑋�(3)). They may not be

orthogonal.

Spring-problem (2nd order ODEs can also describe/represent an eigenvalue problem)

𝑋1 𝑋2

𝑚1 𝑚2

𝐾𝑋1 𝐾(𝑋1 − 𝑋2)

Force balance or mass 𝑚1 𝑎𝑛𝑑 𝑚2, when the two blocks are displaced by 𝑋1 & 𝑋2 in the right direction from its original position:

5

�𝑚1

𝑑2𝑋1𝑑𝑡2

= −𝐾𝑋1 + 𝐾(𝑋2 − 𝑋1)

𝑚2 𝑑2𝑋2𝑑𝑡2

= −𝐾(𝑋2 − 𝑋1) − 𝐾𝑋2⎭⎬

⎫ 𝑋� = �𝑋1𝑋2

�

𝑋𝑖 = 𝐴𝑖𝑠𝑖𝑛(𝜔𝑡) (𝑎𝑠𝑠𝑢𝑚𝑒 𝑎 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛)

( 𝑜𝑟 𝑋� = �̅� 𝑠𝑖𝑛(𝜔𝑡))

Therefore,

𝑋𝑖′′ = −𝐴𝑖𝜔2𝑠𝑖𝑛𝜔𝑡

Substitute,

��2𝐾𝑚1

− 𝜔2�𝐴1 −𝐾𝑚1𝐴2 = 0

− 𝐾𝑚2𝐴1 + �2𝐾

𝑚2− 𝜔2�𝐴2 = 0

�

𝑜𝑟

���

2𝐾𝑚1

− 𝜔2� −𝐾𝑚1

−𝐾𝑚2

�2𝐾𝑚2

− 𝜔2��� �𝐴1𝐴2� = 0

Therefore, the characteristic equation describes an eigenvalue problem (or spring dynamics is an eigenvalue problem). It has assumed the form of

(𝐴 − 𝜔2𝐼)𝑋� = 0

where 𝜔2 = 𝜆, 𝐴 = �

2𝐾𝑚1

− 𝐾𝑚1

− 𝐾𝑚2

2𝐾𝑚2

� and 𝑋� = �𝑋1𝑋2�

𝑒𝑖𝑔𝑒𝑛 𝑣𝑎𝑙𝑢𝑒 𝑒𝑖𝑔𝑒𝑛𝑣𝑒𝑐𝑡𝑜𝑟

Example (From Kreysig's Math book)

𝑚1 = 𝑚2 = 40 𝑘𝑔 ; K = 200𝑁 𝑚⁄

For a non-trivial solution 𝑑𝑒𝑡(𝐴 − 𝜔2𝐼)𝑋� = 0

𝑒𝑞𝑛𝑠: (10 −𝜔2)𝐴1 − 5𝐴2 = 0

−5𝐴1 + (10 − 𝜔2)𝐴2 = 0

6

𝑑𝑒𝑡( ) : (10 − 𝜔2)2 − 25 = 0

𝜆1 = 15, 𝜆2 = 5 𝑜𝑟 𝜔1 = �𝜆1, 𝜔2 = �𝜆2

𝜆1 = 15 𝐴1 = −𝐴2 𝑜𝑟 𝐴(1) = � 1−1�

𝜆2 = 5 𝐴1 = 𝐴2 𝑜𝑟 𝐴(2) = �11�

General solution:

𝑋� = 𝐶1�̅�(1)𝑠𝑖𝑛𝜔1𝑡 + 𝐶2�̅�(2)𝑠𝑖𝑛𝜔2𝑡

𝑜𝑟 �𝑋1𝑋2� = 𝐶1 �

1−1� 𝑠𝑖𝑛𝜔1𝑡 + 𝐶2 �

11� 𝑠𝑖𝑛𝜔2𝑡

�𝑜𝑟 𝑋1 = 𝐶1𝑠𝑖𝑛𝜔1𝑡 + 𝐶2𝑠𝑖𝑛𝜔2𝑡𝑎𝑛𝑑 𝑋2 = −𝐶1𝑠𝑖𝑛𝜔1𝑡 + 𝐶2𝑠𝑖𝑛𝜔2𝑡

�

Graphical representation of block oscillations:

𝜆1=15𝜔1=√15

𝐴1 𝐴2 𝜆2=5𝜔2=√5

𝐴1 𝐴2

1

Lecture #09 One more example of eigenvalue or characteristic type of problem:

Let us revisit Jacobi iteration to invert a matrix.

𝐴𝑋� = 𝑏�

⇒ (𝐿 + 𝐷 + 𝑈)𝑋� = 𝑏�

⇒ 𝐷𝑋� = 𝑏� − (𝐿 + 𝑈)𝑋�

𝑜𝑟 𝑋� = 𝐷−1𝑏� − 𝐷−1(𝐿 + 𝑈)𝑋�

or 𝑋� = 𝑆𝑋� + 𝐶̅

𝑜𝑟 𝑋�(𝑘+1) = 𝑆𝑋�(𝑘) + 𝐶̅ ∶ 𝐽𝑎𝑐𝑜𝑏𝑖 𝐼𝑡𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟𝑚𝑢𝑙𝑎

S is called stationary matrix because very often, matrix 𝐴 𝑜𝑟 (𝐿,𝐷,𝑈) do not change and are fixed. It is 𝑏� (right hand side) that varies from one problem to the other, as a forcing function.

Define, error vector �̅�(𝑘) = 𝑋�(𝑘) − 𝑋�

Approximate

𝑣𝑎𝑙𝑢𝑒 𝑎𝑡 𝑖𝑡𝑒𝑟𝑎𝑡𝑖𝑜𝑛# 𝑘

𝑇𝑟𝑢𝑒 𝑣𝑎𝑙𝑢𝑒

Similarly,

�̅�(𝑘+1) = 𝑋�(𝑘+1) − 𝑋�

= 𝑆�𝑋�(𝑘) − 𝑋�� = 𝑆�̅�(𝑘)

𝑜𝑟 �̅�(𝑘+1) = 𝑆�̅�(𝑘)

𝑜𝑟 �̅�(𝑘) = 𝑆𝑘 �̅�(0) (𝑦𝑜𝑢 𝑠ℎ𝑜𝑢𝑙𝑑 𝑟𝑒𝑎𝑙𝑖𝑧𝑒 𝑡ℎ𝑎𝑡 𝑖𝑡

𝑖𝑠 𝑎𝑛 𝑒𝑖𝑔𝑒𝑛𝑣𝑎𝑙𝑢𝑒 𝑝𝑟𝑜𝑏𝑙𝑒𝑚 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛⁄𝐴𝑋� = 𝜆𝑋� 𝑤𝑖𝑡ℎ 𝜆 = 1)

For 𝑘 → ∞ 𝑆𝑘 → 0 𝑓𝑜𝑟 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑛𝑐𝑒 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 �̅�(𝑘) → 0 𝑎𝑠 𝑒(0) ≠ 0

Therefore, criteria for the convergence: 𝑆𝑘 → 0 𝑎𝑠 𝑘 → ∞

Let �𝜈𝑗�𝑗=1,𝑛 be the eigenvectors corresponding to the 𝜆𝑗 eigenvalue of the stationary matrix

2

𝑆(𝑛 × 𝑛). Recall eigenvectors are linearly independent. So, they can form the basis of n dimensional space. Alternatively, any vector can be represented as the linear combination of the eigenvectors.

Therefore,

𝑒(0) = ∑ 𝑐𝑗𝜈𝑗 𝑛𝑗=1

where Sνj = λjνjor Sν1 = λ1ν1, etc.

𝑒(1) = 𝑆𝑒(0) = ∑ 𝑐𝑗𝜈𝑗𝜆𝑗 𝑛𝑗=1

𝑒(2) = 𝑆2𝑒(0) = ∑ 𝑐𝑗𝜈𝑗𝜆𝑗2 𝑛

𝑗=1 (𝑢𝑠𝑒 𝑆𝜈𝑗 = 𝜆𝑗𝜈𝑗)

⋮ 𝑒(𝐾) = ∑ 𝑐𝑗𝜈𝑗𝜆𝑗

𝑘 𝑛𝑗=1

For convergence, 𝑘 → ∞ 𝑒(𝑘) → 0 𝑜𝑟 �𝜆𝑗� < 1 (𝑗 = 1⋯𝑛)

For a 𝑛 × 𝑛 S matrix, there will be 𝜆1 ⋯𝜆𝑛 eigenvalues.

If 𝜆𝑚𝑎𝑥 < 1 𝑡ℎ𝑒𝑛 𝜆𝑗 < 1

Therefore, for Jacobi Iteration to converge: |𝜆𝑚𝑎𝑥| < 1

(𝑀𝑎𝑥 (𝜆𝑗) ≡ 𝑠𝑝𝑒𝑐𝑡𝑟𝑎𝑙 𝑟𝑎𝑑𝑖𝑢𝑠 𝜌(𝑆) < 1)

There is the Gershgorin Theorem (Check the book by Strang) which states that in such case

∑�𝑎𝑖𝑗� < |𝑎𝑖𝑖|

𝑛𝑜𝑛 − 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠

𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠

Such matrix is said to be diagonally strong. It is a desirable characteristic of a matrix to be inverted. Compare two extreme situations:

and

non-zero

Note that zero or no iterations are required to solve an identity matrix! Also, a diagonally strong matrix will require relatively fewer number of iterations. On the other hand, the above right hand matrix is a singular matrix and cannot be inverted. Similarly, a matrix with smaller

3

value-numbers on its diagonal relative to the other elements in the same row will require a relatively larger number of iterations to converge.

Power Method

The previous examples have shown that the ‘dynamics’ of a 1st order system can be inspected or characterized by determining 𝜆𝑚𝑎𝑥 𝑎𝑛𝑑 𝜆𝑚𝑖𝑛, instead of all 𝜆𝑠. Therefore, a 𝑛 × 𝑛

matrix will have 𝑛𝜆𝑠, and the overall response of the system will be the least sensitive to 𝜆𝑚𝑖𝑛and most sensitive to 𝜆𝑚𝑎𝑥. The response corresponding to the other 𝜆𝑠 will be intermediate. Power method is commonly used to determine 𝜆𝑚𝑎𝑥 and also, 𝜆𝑚𝑖𝑛.

If [𝐴]𝑛×𝑛 matrix has 𝑛 𝜆𝑠(𝑒𝑖𝑔𝑒𝑛𝑣𝑎𝑙𝑢𝑒𝑠), then one can write |𝜆1| > |𝜆2| > ⋯⋯ |𝜆𝑛|

𝜆𝑚𝑎𝑥 𝜆𝑚𝑖𝑛

And AX� = λmaxX� and AX� = λminX�

Method: Make a guess of 𝑋𝑇 = [1 0 1] 𝑜𝑟 [1 1 0] (say, for a 3 × 3 matrix)

Then 𝐴𝑋�(0) ⇒ 𝑋�(1)

Use 𝑋�(1) = 𝑋�(1)

�𝑋�(1)�= 𝐴𝑋(0)

�𝐴𝑋(0)�

�magnitude (you are scaling or

normalizing the vector,i. e. a�⃗ |a|⁄ or a�⃗ �a12 + a22 + a32⁄

�

𝑋�(𝑘) = 𝐴𝑋�(𝑘−1)

�𝐴𝑋�(𝑘−1)� , 𝑘 = 1, 2,3 (# 𝑜𝑓 𝑖𝑡𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝑠)

Can be shown that when 𝑘 → ∞ �𝐴𝑋�(𝑘−1)� → |𝜆𝑚𝑎𝑥|

In other words, after sufficient # of iterations, one obtains maximum eigenvalue and the vector 𝑋�(𝑘−1) when transformed by ‘A’ matrix does not rotate and is just scaled by 𝜆𝑚𝑎𝑥, i.e.

𝐴𝑋�(𝑘−1) = 𝜆𝑚𝑎𝑥𝑋�(𝑘)

In other wors, X�(k−1) and X�(k) have the same directions.

𝜆𝑚𝑎𝑥 𝑋�(𝑘+1)

𝑋�(𝑘)

4

Example: Determine 𝜆𝑚𝑎𝑥 of 𝐴 = � 0 2 3−10 −1 2−2 4 7

�

𝜈(1) = �100� ≡ [1 0 0]𝑇 (1𝑠𝑡 𝑔𝑢𝑒𝑠𝑠)

𝜈(2) =�

0 2 3−10 −1 2−2 4 7

��100�

�𝐴𝜈(1)�=

�0

−10−2

�

√100+4= �

0.0000−0.9806−0.1961

�

|𝜆| = √104 = 10.198

𝜈(3) =[A]�

0.0000−0.9806−0.1961

�

�Aν(2)�=

�−0.43170.0916−0.8965

�

√.43172+.09162+.89652= � �

𝜆2 = 5.9063

After 20 iterations,

𝑋(20) = �−0.3714 0.556980.7428

� with λ(20) = 3.0014

You will see a gradual convergence from 10.198 to 3.0014.

The basics remains the same:

𝐴𝑋(19) = 𝜆𝑚𝑎𝑥𝑋(20) and 𝑋(19) 𝑎𝑛𝑑 𝑋(20) will have approximately the same direction.

It can be shown that 𝝀𝒎𝒊𝒏𝐨𝐟 𝐀 = 𝟏𝝀𝒎𝒂𝒙� 𝒘𝒉𝒆𝒓𝒆 𝝀𝒎𝒂𝒙 is the highest eigenvalue of the

inverse of matrix 𝑨 𝒐𝒓 𝑨−𝟏:

AX� = λmaxX�

or X� = A−1(λmaxX�) = λmax(A−1X�)

or A−1X� = (1λmax� )X�

or BX� ≡ λ′maxX�

where B is A−1 and λmin of A is (1λ′max� ) of B where λ′max is the highest λ of B.

5

The λmax to λmin ratio is known as stiffness ratio and indicates the stiffness (maximum rate of change in the functional value corresponding to λmax relative to minimum rate corresponding to λmin) of the system. A ratio > 10 indicates the system to be stiff requiring a fine step size for calculations (to be discussed later)

Notes (From Strang's book):

(1) If X is an eigenvector of A corresponding to 𝜆𝑚𝑎𝑥 and A is invertible, then X is also an eigenvector of 𝐴−1 corresponding to the inverse of its 𝜆𝑚𝑎𝑥.

(2) Eigenvectors corresponding to different 𝜆𝑠 are linearly independent.

(3) A matrix and its transpose have the same eigenvalues.

(4) A matrix is singular if and only if it has zero 𝜆 . A non-singular matrix has all non zero 𝜆𝑠.

(5) ∑𝜆𝑖 = 𝑡𝑟𝑎𝑐𝑒 𝑜𝑓 𝐴 = ∑ 𝑎𝑖𝑖𝑛𝑖=1

(6) ∏𝜆𝑖 = det (𝐴)

(7) The 𝜆𝑠 of an upper or lower triangular matrix are the elements on its main diagonal.

[It also follows that det A = det (UTM of A) = ∏(𝑎𝑖𝑖)𝑈𝑇𝑀]

Quiz 1

1

Lecture #10 Nonlinear Algebraic Equations

Solving

�𝐹1(𝑋1⋯⋯𝑋𝑛) = 0𝐹2(𝑋1⋯⋯𝑋𝑛) = 0⋮ 𝐹𝑛(𝑋1⋯⋯𝑋𝑛) = 0

� 𝑜𝑟 𝐹�(𝑋�) = 0

𝑜𝑟 𝐹�(𝑋�) = �𝐹1(𝑋�)⋮ 𝐹𝑛(𝑋�)

� = 0

where 𝐹𝑖(𝑋�) takes the form such as (𝑋1 − sin𝑋1) = 0

�𝑜𝑟 𝑋1 − sin𝑋2 = 0

and 𝑋1 = 𝑒𝑋2 �

Solving a set of nonlinear algebraic equations is naturally difficult.

Let us begin with solving one nonlinear equation:

𝑓(𝛼) = 0 (𝑅𝑜𝑜𝑡 𝑓𝑖𝑛𝑑𝑖𝑛𝑔) 𝛼 =?

Closed Methods

1. Bracketing method (for a monotonically increasing or decreasing function)

𝑓(𝑋)

𝑓(𝛼) = 0

𝛼

𝑋𝑙1 𝑋𝑙2

𝑋𝑚2 𝑋𝑚1 𝑋𝑈 𝑋

Step 1: Corner the root. Find 𝑋𝑙 𝑎𝑛𝑑 𝑋𝑈 such that

𝑓(𝑋𝑙)𝑓(𝑋𝑈) < 0 ∶ All it means is that one root is cornered or bracketed. Some

2

functions may have multiple roots (be careful and corner all three roots separately by plotting the function qualitatively but accurately).

𝑁𝑜𝑡𝑒: 𝑇ℎ𝑖𝑠 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑡ℎ𝑟𝑒𝑒 𝑟𝑜𝑜𝑡𝑠

𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑋𝑙 𝑎𝑛𝑑 𝑋𝑈.

𝑋𝑈

𝑋𝑙

Step 2: 𝑋𝑚(1) = 𝑋𝑈+𝑋𝑙2

= 𝑋𝑜𝑙𝑑

Check f(Xm) = 0. If true, you have the root or Xm = α

If not, check ⇒ if f(Xm)f(Xl) < 0. If true,

XU = Xm (replace XU with Xm)

else Xl = Xm (replace Xl with Xm)

(This way, you are cornering the root or coming closer to the root)

(Note: The other check f(X)f(XU) < 0 will also work. If – ve Xl = Xm else

XU = Xm)

Step 3: Xm(2) = XU+Xl2

= Xnew (XU & Xl are new values)

�you can begin error

checks after 2nditerations only

� Check |ξa| = �Xm

new−Xmold

Xmnew � × 100 <∈ (speci�ied by user)

else go to step 2 (keep iterating)

till �X(i+1)−X(i)

X(i+1) � × 100 <∈

Example: Van der Waals gas law/equation:

�P + a V2� (V − b) = constant

3

It is clear that this equation is polynomial in V with 𝑛 = 3.

Let us assume that the other variables and constants are known and the simplified equation takes the following form:

f(V) = 0

V3 − 0.165V − 3.993 × 10−4 = 0

Calculate V or find the roots

Step 0: Bracket the root: V{0, 0.11} f(V) Xm(1)

Xl = 0, XU = 0.11 by plotting

0.11

1. Check f(Xl)f(XU) < 0 V

So that the root is bracketed (cornered)

2. Xm(1) = 0+0.112

= 0.055 (3 digits accurate after decimal)

f(0.055)f(0) = 3.993 × 10−4 × 6.655 × 10−5 > 0

(you would have checked whether f(0.055) x f(0.11) < 0 or not as well!)

Xl = 0.055, XU = 0.11 (remains the same)

3. Xm(2) = Xl+Xu2

= 0.055+0.0112

= 0.0825 = Xnew

f(Xm(2)) = −1.622 × 10−4

Check f(Xl)f(Xm) < 0

XU = 0.0825, Xl = 0.055

4: Check ∈a= �0.0825−0.0550.0825

� × 100 = 33.33% > 0.2% (∈s)

Therefore, Xm = 0.055+0.08252

= 0.00875

Do 10 iterations to obtain Xm = 0.0625 with ∈a= 0.1721% < 0.2%

4

Ans. V (root) = 0.0625.

You should also prepare a table while solving:

�Xl XU Xm ∈ f(Xm)f(Xl)

> < 0

or f(Xm)f(XU) > 0<

�your

decision

1. Yes 2. Xm No

⋮

10.

2. False – Position method (Regular-Falsi)

(This method is similar to the previous one, except draw a straight line connecting XU and Xm to determine the intersection on X axis as the approximate value for the root)

Step1: Bracket root

f(X) f(XU)f(Xl) < 0

2. Xr(1) = XU −f(XU)

f(Xl) − f(XU)× (Xl − XU)

X � f(XU)XU − Xr

= 0 − f(Xl)Xr − Xl

�

α(root)

Xl Xr(1)

= XUf(Xl) − Xlf(XU)f(Xl) − f(XU)

= Xold

3. Check: f(Xr)f(Xl) < 0 or f(Xr)f(XU) < 0 XU = Xr or Xl = Xr

Xr(2) =XUf(Xl) − Xlf(XU)

f(Xl) − f(XU) = Xnew

Check ∈a<∈s else go to 2; iterate till convergence. Note that it is difficult to choose between two methods. In general, this method is faster. There are always exceptions. Try f(X) = X10 − 1!

5

Open Methods

1. Fixed Point Iteration.

Step1. f(X) = 0 ⇒ Modify transform it to⁄

f(X) = X − g(X) = 0

eg. f(X) = X − cos X

or X − eX

(Try for x2 − 1 as X − g(X) = 0 or X = g(X)

Sometimes it is not straight forward & there may be more than one combinations.

The working formula is

Y = X = g(X) (2 equations and find its interaction)

Y = X f(X) = 0 split into

Y Y = X and Y = g(X)

Y = g(X)

Therefore, a sequence is generated.

X3X2 X1 X0 Xn+1 = g(Xn)

Sometimes, one has to start from left to the root; sometimes there is a divergence or

spiral convergence, depending on type of functions.

Y or

𝑔(𝑋)

𝑋0 𝑋1𝑋2 fast convergence slow convergence divergence

(spiral) (because of a wrong starting value)

Criterion for convergence: |𝐠′(𝐗)| < 𝐘′(𝐗) = 𝟏 at 𝐗 = 𝐗𝐧

6

Note: Closed methods often generate at least one root. Open methods may sometimes result in

divergence depending upon type of function and starting guess Xo. However, if they converge,

they will do more quickly than the closed methods.

Step 2: Start with Xo

X1 = g(Xo)

Step 3: Check ∈a= �X1−X0X1

� <∈s

If not, Xold = Xr(iterated value)

Xr = g(Xold)

Check ∈a= �Xr −XoldXr

� <∈s

Else keep on iterating, keeping in mind the sequence Xn+1 = g(Xn).

2. Newton-Raphson Method

(Most popular method to solve a non-linear algebraic equation)

f(X) = 0 Tangent Step1: Choose

Tangent Xr = Xo (guess)

2. f ′(Xr) = tangent at Xr = f(Xr)X0 − X1

X2 X0 X1 = Xr −f(Xr)f′(Xr)

(Note: f ′(Xr) ≠ 0)

3. If ∈a= �X1−XrX1

� × 100 <∈s stop

α = X1 (root)

else Xr = X1 (repeat the steps)

Sequence: Xi+1 = Xi −f(Xi)f′(Xi)

; f ′(Xi) ≠ 0

and ∈i+1= �Xi+1−XiXi+1

� × 100

7

Notes: ⇒ • f ′(X) ≠ 0 is the major drawback of the method. • Simple to code and convergence is fast (see later) • There are cases of divergence and ‘missing’ roots, depending on the function and

initial guess.

divergence

Y missing root

X0

X1 X3 X2X0 X X1

root

In all such cases, the best strategy is to plot and locate all roots graphically (qualitatively but

accurately), and also, start close to the root.

The NR method can be derived from the Taylor series expansion showing the order of error

in the method.

f(Xi+1) = f(Xi) + f ′(Xi)(Xi+1 − Xi) + f′′(Xi)2!

(Xi+1 − Xi)2 + ⋯⋯

= f(Xi) + f ′(Xi)(Xi+1 − Xi) + 0(h2)

(error)

If f(Xi+1) = 0, root is located

Xi+1 = Xi −f(Xi)f′(Xi)

Example (NR) : 𝑓(𝑋) = 𝑋3 − 0.165𝑋2 + 3.993 × 10−4

𝑓′(𝑋) = 3𝑋2 − 0.33𝑋

Step 1. 𝑋0 = 0.05: 𝑃𝑙𝑜𝑡 𝑓(𝑋)𝑎𝑛𝑑 𝑐ℎ𝑜𝑜𝑠𝑒 𝑋0 𝑐𝑙𝑜𝑠𝑒 𝑡𝑜 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑓(𝑋) 𝑤𝑖𝑡ℎ 𝑋 – axis

2. 𝑓′(𝑋0) = 3𝑋02 − 0.33𝑋0

𝑋1 = 𝑋0 −𝑓(𝑋0)𝑓′(𝑋0) = 0.06242

8

3. ∈ = �0.06242−0.05000.06242

�× 100 = 19.89%

4. 𝑋2 = 𝑋1 −𝑓(𝑋1)𝑓′(𝑋1) = 0.06238

∈ = 0.06411 ⇒ 𝛼 = 0.06238 (𝑟𝑜𝑜𝑡)

(accurate to the 1st significant digit after decimal)

1

Lecture #11 Newton-Raphson (Multi-variables)

�F1(X1, X2 ⋯Xn) = 0F2(X1, X2 ⋯Xn) = 0

⋮Fn(X1, X2 ⋯Xn) = 0

� ⇒F�⃗ (X�) = 0

or Fi(X�) = 0, i = 1⋯n or Fi(Xi) = 0, i = 1⋯n

Taylor-expansion of multi-variables:

𝐹1�𝑋�(𝑘+1)� ≈ 𝐹1�𝑋�(𝑘)� +𝜕𝐹1𝜕𝑋1

�(𝑋1(𝑘+1) − 𝑋1

(𝑘)) +𝑋�(𝑘) ≠ 𝑋1

𝜕𝐹1𝜕𝑋2

� (𝑋2(𝑘+1) − 𝑋2

(𝑘)) +𝑋�(𝑘) ≠ 𝑋2

⋯⋯𝜕𝐹1𝜕𝑋𝑛

(𝑋𝑛(𝑘+1) − 𝑋𝑛

(𝑘)) 𝑋�(𝑘) ≠ 𝑋𝑛

(1𝑠𝑡𝑜𝑟𝑑𝑒𝑟 𝑎𝑐𝑐𝑢𝑟𝑎𝑡𝑒)

𝐹2�𝑋�(𝑘+1)� ≈ 𝐹2�𝑋�(𝑘)� +𝜕𝐹2𝜕𝑋1

�(𝑋1(𝑘+1) − 𝑋1

(𝑘)) +𝑋�(𝑘) ≠ 𝑋1

𝜕𝐹2𝜕𝑋2

� (𝑋2(𝑘+1) − 𝑋2

(𝑘)) +𝑋�(𝑘) ≠ 𝑋2

⋯⋯𝜕𝐹2𝜕𝑋𝑛

(𝑋𝑛(𝑘+1) − 𝑋𝑛

(𝑘)) 𝑋�(𝑘) ≠ 𝑋𝑛

.

.

.

𝐹𝑛�𝑋�(𝑘+1)� ≈ 𝐹𝑛�𝑋�(𝑘)� +𝜕𝐹𝑛𝜕𝑋1

�(𝑋1(𝑘+1) − 𝑋1

(𝑘)) +𝑋�(𝑘) ≠ 𝑋1

𝜕𝐹𝑛𝜕𝑋2

� (𝑋2(𝑘+1) − 𝑋2

(𝑘)) +𝑋�(𝑘) ≠ 𝑋2

⋯⋯𝜕𝐹𝑛𝜕𝑋𝑛

(𝑋𝑛(𝑘+1) − 𝑋𝑛

(𝑘)) 𝑋�(𝑘) ≠ 𝑋𝑛

If 𝑋�(𝑘+1) = 𝛼 (𝑟𝑜𝑜𝑡 𝑎𝑡 (𝑘 + 1)𝑡ℎ 𝑖𝑡𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠) ⇒ 𝐹𝑖�𝑋�(𝑘+1)� = 0,

(𝑖 = 1⋯𝑛)

2

⎩⎪⎨

⎪⎧0

0⋮⋮0⎭⎪⎬

⎪⎫

=

⎩⎪⎨

⎪⎧𝐹1𝐹2⋮⋮𝐹𝑛⎭⎪⎬

⎪⎫

(𝑘)

+

⎣⎢⎢⎢⎢⎢⎡𝜕𝐹1𝜕𝑋1

𝜕𝐹1𝜕𝑋2

⋯ 𝜕𝐹1𝜕𝑋𝑛

𝜕𝐹2𝜕𝑋1

⋯ ⋯ ⋯

⋮⋮

𝜕𝐹𝑛𝜕𝑋1

𝜕𝐹𝑛𝜕𝑋2

⋯ 𝜕𝐹𝑛𝜕𝑋𝑛⎦

⎥⎥⎥⎥⎥⎤

(𝑘)

⎩⎪⎨

⎪⎧∆𝑋1∆𝑋2⋮⋮

∆𝑋𝑛⎭⎪⎬

⎪⎫

(∆𝑋1 = 𝑋1(𝑘+1) − 𝑋1(𝑘), 𝑒𝑡𝑐)

𝑜𝑟

⎣⎢⎢⎡𝜕𝐹1𝜕𝑋1

⋯ 𝜕𝐹1𝜕𝑋𝑛

⋮𝜕𝐹𝑛𝜕𝑋1

⋯ 𝜕𝐹𝑛𝜕𝑋𝑛⎦

⎥⎥⎤

(𝑘)

�∆𝑋1⋮

∆𝑋𝑛� = −�

𝐹1⋮𝐹𝑛�

(𝑘)

𝑜𝑟 𝐽𝐽���∆𝑋�(𝑘) = −𝐹 � (𝑘)

𝑜𝑟 ∆𝑋�(𝑘) = − 𝐽𝐽���−1𝐹 � (𝑘)

𝑜𝑟 𝑋�(𝑘 + 1) = 𝑋�(𝑘) − 𝐽𝐽���−1𝐹 � (𝑘) ⇒ Multi-variable NR method (similar in form to single variable NR method)

Example: 𝐹1(𝑌�) = 4 − 8𝑌1 + 4𝑌2 − 2𝑌13 = 0

𝐹2(𝑌�) = 1 − 4𝑌1 + 3𝑌2 + 𝑌22 = 0

Starting guess values

𝑌� (1) = [𝑌1 𝑌2]𝑇(1) = [0.5 0.5]𝑇

Apply NR multivariable method formula:

𝑋�(𝑘+1) = 𝑋�(𝑘) − 𝐽𝐽�−1𝐹 � (𝑘)

𝑜𝑟 �𝑌1𝑌2�

(𝑘+1)= �𝑌1𝑌2

�(𝑘)

−

⎣⎢⎢⎡𝜕𝐹1𝜕𝑌1

𝜕𝐹1𝜕𝑌2

𝜕𝐹2𝜕𝑌1

𝜕𝐹2𝜕𝑌2⎦

⎥⎥⎤−1

�𝐹1𝐹2�

(𝑘)

3

�𝑌1𝑌2�1

= �0.50.5� − �(−8 − 6𝑌12) 4

−4 (3 + 2𝑌2)�−1

�𝐹1𝐹2�0.5,0.5

= �0.50.5� − �−9.5 4

−4 4�−1�1.750.75�

= �0.50.5� + 1

22� 4 4 −4 −9.5� �

1.750.75�

(Note: For 𝟐 × 𝟐 matrix simple Cramer’s rule was applied to determine the inverse of the matrix, else apply 𝑮 − 𝑱 𝒐𝒓 𝑳𝑼 decomposition can be used for a large size matrix)

𝑌1(1) = 0.5 − 122

(4 × 1.75 + 4 × 0.75) = 0.31818

𝑌2(1) = 0.5 −1

22(−4 × 1.75 − 9.5 × 0.75) = 0.50568

- Continue iteration till there is a convergence. One can prepare a table like this:

# 𝑌1 𝑌2 𝐹1 𝐹2 𝜕𝐹1𝜕𝑌1

𝜕𝐹1𝜕𝑌2

𝜕𝐹2𝜕𝑌1

𝜕𝐹2𝜕𝑌2

𝑌1 𝑌2 ∈𝑎

1

2

Secant Method

(Another common open method; it does not require derivative to compute. Requires two starting adjacent guess values instead)

Step1:

Straight line: Choose 𝑋𝑖−1 𝑎𝑛𝑑 𝑋𝑖 adjacent guess values near the root ′𝛼′.

2. Draw straight line connecting 𝑓(𝑋𝑖−1) 𝑎𝑛𝑑 𝑓(𝑋𝑖). Find

𝑓(𝑥) intersection on X axis → 𝑋𝑖+1

′𝛼′ 𝑓′(𝑋𝑖+1) = 𝑓(𝑋𝑖)𝑋𝑖−𝑋𝑖+1

= 𝑓(𝑋𝑖−1)𝑋𝑖−1−𝑋𝑖+1

𝑋𝑖+1 𝑋 𝑋𝑖+1 = 𝑋𝑖 − 𝑓(𝑋𝑖) × (𝑋𝑖−𝑋𝑖−1)𝑓(𝑋𝑖)−𝑓(𝑋𝑖−1)

𝑜𝑟 𝑋𝑖+1 = 𝑋𝑖 −𝑓(𝑋𝑖)

𝑓(𝑋𝑖)−𝑓(𝑋𝑖−1)𝑋𝑖−𝑋𝑖−1

: sequence

4

Check if 𝑓(𝑋𝑖+1) → (𝑦𝑜𝑢 ℎ𝑎𝑣𝑒 ℎ𝑖𝑡 𝑡ℎ𝑒 𝑟𝑜𝑜𝑡), else repeat with 𝑋𝑖 𝑎𝑛𝑑 𝑋𝑖+1 following the above sequence.

• The method converges faster than Bisection method, but slower that NR. It is preferred over NR to avoid 𝑓′ = 0.

• If 𝑋𝑖 → 𝑋𝑖−1, note 𝑋𝑖+1 = 𝑋𝑖 −𝑓(𝑋𝑖)𝑓′(𝑋𝑖)

(same as NR!)

At this stage, let us look at Taylor’s series. Any function f(X) can be expanded as:

𝑓(𝑋) = 𝑓(𝑋0) + 𝑓′(𝑋0)(𝑋 − 𝑋0) + 𝑓′′(𝑋0)2!

(𝑋 − 𝑋0)2 + ⋯𝑓𝑛(𝑋0)(𝑋−𝑋0)𝑛

𝑛!+ ⋯ 𝑡𝑜 ∞

exact

or, 𝑓(𝑋𝑖+1) = 𝑓(𝑋𝑖) + 𝑓′(𝑋𝑖)(𝑋 − 𝑋𝑖) + 𝑓′′(𝑋𝑖)2!

(𝑋𝑖+1 − 𝑋𝑖)2 + ⋯𝑓𝑛(𝑋𝑖)(𝑋𝑖+1−𝑋𝑖)𝑛

𝑛!+ ⋯ 𝑡𝑜 ∞

exact

≈ 𝑓(𝑋𝑖) + 𝑓′(𝑋𝑖)(𝑋𝑖+1−𝑋𝑖)1!

+ 𝑓′′(𝑋𝑖)(𝑋𝑖+1−𝑋𝑖)2

2!+ ⋯ 0(ℎ3)

approximate

or truncation error is 0(ℎ3) (accurate to 2nd order)

≈ 𝑓(𝑋𝑖) + 𝑓′(𝑋𝑖)(𝑋𝑖+1−𝑋𝑖)1!

+ 0(ℎ2)

approximate

(accurate to 1st order or truncation error is 0(ℎ2))

(error is one order greater than the accuracy of the method)

= 𝑓(𝑋𝑖) + 𝑓′(𝑋𝑖)(𝑋𝑖+1 − 𝑋𝑖) + ⋯𝑓𝑛(𝑋𝑖)(𝑋𝑖+1−𝑋𝑛)𝑛

𝑛!+ 𝑅𝑛+1

exact

where 𝑅𝑛+1 = 𝑓𝑛+1(𝜉)𝑛+1!

(𝑋𝑖+1 − 𝑋𝑖) (𝑐𝑎𝑛 𝑏𝑒 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒𝑑)

Here, 𝜉 is ‘some’ number between 𝑋𝑖+1𝑎𝑛𝑑 𝑋𝑖.

Therefore,

𝑓(𝑋𝑖+1) = 𝑓(𝑋𝑖) + 𝑓′(𝑋𝑖)(𝑋𝑖+1−𝑋𝑖)1!

+ 𝑓2(𝜉)(𝑋𝑖+1−𝑋𝑖)2!

exact

5

= 𝑓(𝑋𝑖) + 𝑓′(𝜉)(𝑋𝑖+1−𝑋𝑖)1!

(1st Mean Value Theorem)

exact

In other words, 𝑓′(𝜉) = 𝑓(𝑋𝑖+1)−𝑓(𝑋𝑖)(𝑋𝑖+1−𝑋𝑖)

All it means there is ′𝜉′ somewhere between 𝑋𝑖𝑎𝑛𝑑 𝑋𝑖+1, where the tangent equals the divided difference between two end points 𝑋𝑖 − 𝑋𝑖+1.

Note: Taylor series is used to determine the error or accuracy of a numerical method, or to determine the convergence of the method.

Convergent Analysis

Fixed Point Iteration method

𝑓(𝑋) = 𝑋 − 𝑔(𝑋)

𝑜𝑟 𝑋(𝑛+1) = 𝑔�𝑋(𝑛)� (𝑖𝑡𝑒𝑟𝑎𝑡𝑖𝑜𝑛)

If 𝛼 is the root then 𝛼 − 𝑔(𝛼) = 0

Error at (𝑛 + 1)𝑡ℎ iteration =|True Value − Approximate Value|

𝑒𝑛+1 = �𝛼 − 𝑋(𝑛+1)�

= �𝑔(𝛼) − 𝑔(𝑋(𝑛))�

= �𝑔′(𝑋�)(𝛼 − 𝑋(𝑛)� ∶ Recall Mean Value Theorem. X� is between

α and X(n)

= �𝑔′(𝑋�)𝑒𝑛�

𝑒𝑛+1

𝑒𝑛= �𝑔′(𝑋�)� < 1 for error to decrease as 𝑛 → ∞ where 𝑋� is {𝛼,𝑋𝑛}

NR Convergent

𝑋𝑘+1 = 𝑋𝑘 − 𝑓𝑘𝑓𝑘′ ; 𝑓𝑘′ ≠ 0

𝑒(𝑘+1) = |𝛼 − 𝑋𝑘+1| = �𝛼 − 𝑋𝑘 + 𝑓𝑘𝑓𝑘′�

6

𝑓(𝑋𝑘+1) = 𝑓(𝑋𝑘) + 𝑓𝑘′(𝑋𝑘+1 − 𝑋𝑘) + 𝑓′′(𝑋�) (𝑋𝑘+1−𝑋𝑘)2

2!

exact (𝑋�: {𝑋𝑘,𝑋𝑘+1})

If 𝑋𝑘+1 is the root (𝛼)

0 = 𝑓(𝑋𝑘) + 𝑓𝑘′(𝛼 − 𝑋𝑘) +

(𝛼 − 𝑋𝑘)2

2!𝑓′′(𝑋�)

∴ 𝑒(𝑘+1) = �𝑓′′(𝑋�)2𝑓′𝑘

(𝛼 − 𝑋𝑘)2�

= �𝑓′′(𝑋�)2𝑓′𝑘

(𝑒𝑘)2�

𝑒(𝑘+1)

(𝑒𝑘)2= �𝑓

′′(𝑋�)2𝑓′(𝛼)

� ; 𝑓′(𝛼) ≠ 0

If 𝑘 → ∞ 𝑒(𝑘+1) decreases by the square of previous step error or. In other words, the method has quadratic convergence (fast convergence), although the method is only 1st order accurate. This is the special feature of NR method.

1

Lecture #12 Function Approximation: Interpolation

Fit a line passing through the data points to

predict the functional value at an intermediate interpolated value

data. The line must pass through the data. The method is used

to interpolate, for example, certain properties such as enthalpy and

vapour pressure, at an intermediate point where the functional 𝑓𝑖

values or properties are not repeated or available.

Therefore, the data provided for the interpolation must

be reliable or accurate. 𝑋𝑖

- Mathematically, interpolating function must be smooth (differential); continuous and pass through all data points.

- Compare it to ‘regression’ (best fit to the data) when the line need not pass through each data. Regression is used for prediction when data are experimentally measured or approximately calculated. Therefore, the regressed data may have errors.

Yi experimental data

𝑓𝑖

𝑋𝑖

(No regression topic to be covered in this course; no extrapolation as well!)

Interpolating function for ‘n’ data points

𝑃𝑛−1(𝑋) = 𝑎𝑛−1𝑋𝑛−1 + 𝑎𝑛−2𝑋𝑛−2 + ⋯𝑎0: a polynomial of degree (𝑛 − 1).

If there are ‘n’ data points, a unique polynomial of degree (n-1) passes through the points:

Ex. 1. 𝑦 = 𝑚𝑥 + 𝑏 (1st order polynomial) passing through two data points. Two unknown & two equations to solve m and b:

2. Three data points P2

2

y = ax2 + bx + c (2nd order polynomial, a quadratic equation)

(3 equations to solve three unknowns 𝑎, 𝑏 & 𝑐)

3. Thus, ‘n’ data points will have a 𝑢𝑛𝑖𝑞𝑢𝑒 polynomial of degree (n − 1) passing through the points.

𝑦 = 𝑎𝑛−1𝑥𝑛−1 + 𝑎𝑛−2𝑥𝑛−2 + ⋯⋯𝑎0

Therefore,

�𝑦1 = 𝑎𝑛−1𝑥1𝑛−1 + 𝑎𝑛−2𝑥1𝑛−2 + ⋯⋯𝑎0𝑦2 = 𝑎𝑛−1𝑥2𝑛−1 + 𝑎𝑛−2𝑥2𝑛−2 + ⋯⋯𝑎0⋮ 𝑦𝑛 = 𝑎𝑛−1𝑥𝑛𝑛−1 + 𝑎𝑛−2𝑥𝑛𝑛−2 + ⋯⋯𝑎0⎭

⎬

⎫ nequations

or

�

𝑦1𝑦2⋮𝑦𝑛

� = ��

1 𝑥1 𝑥12 ⋯ 𝑥1𝑛−1

1 𝑥2 𝑥22 ⋯ 𝑥2𝑛−1

⋮1 𝑥𝑛 𝑥𝑛2 𝑥𝑛𝑛−1

�� �

𝑎0⋮⋮

𝑎𝑛−1

�

Although the problem is well defined, it is tedious to solve (𝑎0⋯⋯𝑎𝑛−1) from a large sized 𝑛 × 𝑛 matrix.

𝑦� = 𝐴𝑛×𝑛𝑎�

The common methods to fit 𝑃𝑛−1 polynomial passing through ‘n’ data points, or 𝑃𝑛 through

(n + 1) data points:

(1) Newton’s divided-difference scheme Consider 2 data points Let a line pass through (𝑋0,𝑌0) & (𝑋1,𝑌1)

𝑌1 − 𝑌0𝑋1 − 𝑋0

=𝑦1 − 𝑦0𝑥 − 𝑥0

or 𝑌 = 𝑌0 + 𝑦1−𝑦0𝑥1−𝑥0

(𝑋 − 𝑋0)

𝑜𝑟 𝑌 = 𝑎0 + 𝑎1(𝑋 − 𝑋0)

⎩⎪⎨

⎪⎧ where a0 = Y0

a1 = y1−y0x1−x0

= Y(X1, X0) de�ined as

Newton First DividedDifference(N1DD)

�

Similarly, define

Y1

Y Y0

X0 X1 X

3

𝑌[𝑋2,𝑋1,𝑋0] ≡ Newton 2nd divided difference

=𝑌[𝑋2,𝑋1]− 𝑌[𝑋1,𝑋0]

𝑋2 − 𝑋0 (𝑁2𝐷𝐷)

𝑌[𝑋𝑛,𝑋𝑛−1,⋯𝑋0] =𝑌[𝑋𝑛,𝑋𝑛−1,⋯𝑋1] − 𝑌[𝑋𝑛−1,𝑋𝑛−2,⋯𝑋0]

𝑋𝑛 − 𝑋0 (𝑁𝑛𝐷𝐷)

Take 3 data points (𝑋0,𝑋1,𝑋2):

𝑌(𝑋) ≡ 𝑃2(𝑥) = 𝑎0 + 𝑎1(𝑥 − 𝑥0) + 𝑎2(𝑥 − 𝑥0)(𝑥 − 𝑥1)

By fitting 𝑃2(𝑥) to three data points, the three coefficients

can be determined as

𝑎0 = 𝑌0, 𝑎1 = 𝑌1−𝑌0𝑋1−𝑋0

≡ 𝑌[𝑋1,𝑋0]

and 𝑎2 = 𝑌[𝑋2,𝑋1]−𝑌[𝑋1,𝑋0]𝑋2−𝑋0

≡ 𝑌[𝑋2,𝑋1,𝑋0]

Similarly for (n+1) data

𝑌(𝑥) 𝑜𝑟 𝑃𝑛(𝑥)

= 𝑎0 + 𝑎1(𝑥 − 𝑥0) + 𝑎2(𝑥 − 𝑥0)(𝑥 − 𝑥1) + ⋯𝑎𝑛(𝑥 − 𝑥0)(𝑥 − 𝑥1)⋯ (𝑥 − 𝑥𝑛−1)

where 𝑎0 = 𝑌0

𝑎1 = 𝑌[𝑋1,𝑋0] = 𝑌1−𝑌0𝑋1−𝑋0

(1st divided difference)

𝑎𝑛 = 𝑌[𝑋𝑛,𝑋𝑛−1⋯𝑋0] � nthdivided difference �

(Substitute x = x0, x1, x2 … xn in Pn(x) to explore the sequence of polynomial)

In the tabular form,

𝑖 𝑥𝑖 𝑦𝑖 1𝑠𝑡 2𝑛𝑑 ⋯⋯ 𝑛𝑡ℎ

⎩⎪⎨

⎪⎧

0 𝑥0 𝑦0 1 𝑥1 𝑦1 2 𝑥2 𝑦2 ⋮

𝑛 − 1 𝑥𝑛−1 𝑦𝑛−1 𝑛 𝑥𝑛 𝑦𝑛

�

𝑎𝑛 = (𝑌[𝑋𝑛,⋯𝑋1] −𝑌[𝑋𝑛−1,⋯𝑋0])

𝑋𝑛 − 𝑋0

X0 X1 X2

Xi

Y0

Y2

Yi Y1

4

2. Lagrangian Interpolation

𝑦(𝑥) = ∑ 𝑙𝑘(𝑥) 𝑦𝑘𝑛𝑘=0 (𝑛 + 1 𝑑𝑎𝑡𝑎)

Lagrangian coefficient of the kth term

• Function must be a linear combination of 𝑦𝑘. • 𝑙𝑘(𝑥) must not depend on 𝑦𝑘 • 𝑙𝑘(𝑥) is defined as 𝑙𝑗(𝑥𝑖) = 𝛿𝑖𝑗 (Kronecker delta) = 1 when i = j and 0 when i ≠ j Propose, lj(𝑥) = 𝐶𝑗(𝑥 − 𝑥0)(𝑥 − 𝑥1)⋯�𝑥 − 𝑥𝑗−1��𝑥 − 𝑥𝑗��𝑥 − 𝑥𝑗+1�⋯ (𝑥 − 𝑥𝑛), and

= 0 (𝑥 ≠ 𝑥𝑗) = 1 (𝑥 = 𝑥𝑗)

Therefore,𝐶𝑗 = �𝑥𝑗 − 𝑥0�−1�𝑥𝑗 − 𝑥1�

−1 ⋯�𝑥𝑗 − 𝑥𝑗−1�−1�𝑥𝑗 − 𝑥𝑗+1�

−1 ⋯ �𝑥𝑗 − 𝑥𝑛�−1

and, 𝑙𝑗(𝑥) =(𝑥 − 𝑥0)(𝑥 − 𝑥1)⋯�𝑥 − 𝑥𝑗−1��𝑥 − 𝑥𝑗+1�⋯ (𝑥 − 𝑥𝑛)

�𝑥𝑗 − 𝑥0��𝑥𝑗 − 𝑥1�⋯�𝑥𝑗 − 𝑥𝑗−1��𝑥𝑗 − 𝑥𝑗+1�⋯ �𝑥𝑗 − 𝑥𝑛�

The interpolating function is as follows:

𝑌𝑛(𝑥) = �𝑙𝑘(𝑥)𝑌𝑘 ≡ 𝑃𝑛

𝑛

𝑘=0

where

𝑙𝑗(𝑥) = �(𝑥 − 𝑥𝑖)�𝑥𝑗 − 𝑥𝑖�

𝑛

𝑖=0(𝑖≠𝑗)

Suppose 2 data points: (𝑋0,𝑌0) and (𝑋1,𝑌1) Y1(x) = l0(x)Y0 + l1(x)Y1 l0 = x−x1

x0−x1, l1 = x−x0

x1−x0

or Y1(x) = � x−x1x0−x1

�Y0 + � x−x0x1−x0

�Y1

(Since 𝑌1(𝑥) is unique, it cannot be different from Newton-Divided difference formula; only forms are different) Suppose there are 3 data points: (X0, Y0), (X1, Y1), and (X2, Y2) 𝑌2(𝑥) = 𝑙0(𝑥)𝑌0 + 𝑙1(𝑥)𝑌1 + 𝑙2(𝑥)𝑌2

𝑙0(𝑥) =(𝑥 − 𝑥1)(𝑥 − 𝑥2)

(𝑥0 − 𝑥1)(𝑥0 − 𝑥2) ; 𝑙1(𝑥) =(𝑥 − 𝑥0)(𝑥 − 𝑥2)

(𝑥1 − 𝑥0)(𝑥1 − 𝑥2) ,

(X,Y) (X1,Y1)

(X0,Y0)

5

𝑌(𝑥) =(𝑥 − 𝑥1)(𝑥 − 𝑥2)

(𝑥0 − 𝑥1)(𝑥0 − 𝑥2)𝑌0 +(𝑥 − 𝑥0)(𝑥 − 𝑥2)

(𝑥1 − 𝑥0)(𝑥1 − 𝑥2)𝑌1 +(𝑥 − 𝑥0)(𝑥 − 𝑥1)

(𝑥2 − 𝑥0)(𝑥2 − 𝑥1)𝑌2

≡ 𝑠𝑎𝑚𝑒 𝑎𝑠 𝑁𝐷𝐷𝐸 𝑜𝑟 𝑎𝑋2 + 𝑏𝑋2 + 𝑐 Ex. 𝑋1 𝑋2 𝑋 𝑋3

Depth (m) -1 0 0.5 1.0

𝑌(𝑡0𝑐) 10.1 11.3 ? 11.9

𝑌2(𝑥) = 𝐶0 + 𝐶1𝑥 + 𝐶2𝑥2 (unique)

3 data points and 3 eqns to solve Cs:

�1 −1 11 0 01 1 1

� �𝐶0𝐶1𝐶2� = �

10.111.311.9

�

Solve by any previously learnt method to obtain

𝐶0 = 11.3, 𝐶1 = 0.9, 𝑎𝑛𝑑 𝐶2 = −0.3 ⇒ 𝑌(0.5) = 11.675

(1) 𝑁𝐷𝐷 1𝑠𝑡𝐷𝐷:𝑌[𝑋0,𝑋1] 2𝑛𝑑𝐷𝐷:𝑌[𝑋0,𝑋1,𝑋2]

𝑋0 − 1 10.1 11.3−10.10+1

= 1.2 1.2−0.6−1−1

= −0.3

𝑋1 0 11.3 11.9−11.31−0

= 0.6

𝑋2 1 11.9

Therefore, Y2(x) = a0 + a1(x − x0) + a2(x − x0)(x − x1)

a0 = Y0 = 10.1, a1 = FDD = 1.2, a2 = SDD = −0.3

= 10.1 + 1.2(x + 1) − 0.3(x + 1)x

- (Same as before, check) Y(0.5) = 11.675

(2) Lagrange Y2(x) = ∑ lkYk2k=0 ; lj(x) = ∏ �x−xi

xj−xi�2

i=0i≠j

l0(x) = (x−x1)(x−x2)(x0−x1)(x0−x2) ; l1(x) = (x−x0)(x−x2)

(x1−x0)(x1−x2) , etc.

Y2(x) =(x − 0)(x − 1)−1 × −2

× 10.1 +(x + 1)(x − 1)

1 × −1× 11.3 +

(x + 1)x2 × 1

× 11.9

- (Same as before, check) Y(0.5) = 11.675

6

Note: ⇒ If ‘n’ is large, (> 4), the polynomial shows oscillation. In principle, one uses ‘piece-wise’ interpolation for large # of data points.

⇒ Less smooth if data points are less or sparse.

⇒ ‘Problem’ at the end points.

⇒ NDD uses the previous calculations, if an extra point is to be included in finding a new polynomial, unlike each coefficient has to be recalculated in ‘LI’. However, it is easy to program ‘LI’.

Error analysis: For (n+1) data points, 𝑃𝑛(𝑜𝑟𝑑𝑒𝑟 𝑛) is the unique polynomial to pass through (n+1) data points. This polynomial can be used to predict 𝑃𝑛(𝑋) at a new data points ‘X’. How much error can be expected in this prediction/interpolation? From engineering point of view, how much confidence an user has in the interpolated functional value? Well, in such case one should use an additional data (𝜉,𝑌(𝜉)) over the range (𝑋0 ⋯𝑋𝑛) and check the convergence in the re-interpolated value at 𝑃𝑛+1(𝑋). Note, now you have (n+2) data points including 𝜉.

Theoretically, it can be shown using Taylor’s series that the error in the interpolation using 𝑃𝑛(𝑋) for (n+1) data is

𝑦𝑛+1(𝜉)𝑛+1!

(𝑥 − 𝑥0)(𝑥 − 𝑥1)⋯⋯ (𝑥 − 𝑥𝑛)

where (𝑥0 < 𝜉 < 𝑥𝑛)

additional data pt.

You should have also noted that

𝑦[𝑥0,𝑥1] = 𝑦(𝑥1)−𝑦(𝑥0)𝑥1−𝑥0

= 𝑦′ (accurate to the 1st order)

𝑦[𝑥0,𝑥1,𝑥2] = ⋯⋯ = 𝑦′′ (accurate to the 2nd order)

and 𝑦[𝜉, 𝑥0, 𝑥1,⋯𝑥𝑛] = 𝑌[𝜉,⋯𝑥𝑛]−𝑌[𝑥0,⋯𝑥𝑛]𝜉−𝑥𝑛

= 𝑦𝑛+1(𝜉)

Therefore, the error calculated is nothing but 𝑅𝑛+1(remainder) of the Taylor’s series. For more on this, read the book by Ferziger.

1

Lecture #13 Similar to Newton’s Divided Difference and Lagrangian interpolation schemes, there is a Newton’s interpolation formula that makes use of ′Δ′ operator (Forward Difference Operator)

Δ𝑦(𝑥) = 𝑦(𝑥 + Δ𝑥) − y(𝑥) ∶ 𝐹𝑜𝑟𝑤𝑎𝑟𝑑 − 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑝𝑒𝑟𝑎𝑡𝑜𝑟

�𝐸 𝑦(𝑥) = 𝑦(𝑥 + Δ𝑥) ∶ 𝑆ℎ𝑖𝑓𝑡 𝑂𝑝𝑒𝑟𝑎𝑡𝑜𝑟𝐸2𝑦(𝑥) = 𝐸[𝑦(𝑥 + Δ𝑥)] = 𝑦(𝑥 + 2Δ𝑥)

𝐸𝑛𝑦(𝑥) = 𝑦(𝑥 + 𝑛Δ𝑥)�

Therefore, Δ𝑦(𝑥) = 𝐸𝑦(𝑥) − 𝑦(𝑥) = (𝐸 − 1)𝑦(𝑥)

𝑜𝑟 𝐸 = 1 + Δ ∶ 𝐸 & Δ operators

𝑜𝑟 𝐸𝛼 = (1 + Δ)𝛼 = 1 + 𝛼Δ +𝛼(𝛼 − 1)

2! Δ2 + ⋯

𝛼(𝛼 − 1)⋯ (𝛼 − 𝑛 + 1)𝑛!

Δn+. ..

𝑜𝑟 𝐸𝛼𝑦(𝑥0) = 𝑦(𝑥0 + 𝛼Δ𝑥) = (1 + Δ)𝛼𝑦(𝑥0)

= 𝑦0 + 𝛼Δ𝑦0 + 𝛼(𝛼−1)2!

Δ2𝑦0 + ⋯𝛼(𝛼−1)⋯(𝛼−𝑛+1)𝑛!

Δn𝑦0 +

If 𝛼 = 𝑥−𝑥0Δ𝑥

𝑦(𝑥) = 𝑦(𝑥0 + 𝛼Δ𝑥) = 𝑦0 + 𝛼(Δ𝑦0) + 𝛼(𝛼−1)2!

Δ2𝑦0 + ⋯𝛼(𝛼−1)⋯(𝛼−𝑛+1)𝑛!

Δn𝑦0 + 𝑅

𝑅 = 𝛼(𝛼−1)(𝛼−𝑛)𝑛!

Δn+1𝑦(𝜉) 𝑤ℎ𝑒𝑟𝑒 𝑋 < 𝜉 < 𝑋𝑛

- Newton Forward-Difference Interpolation formula In general

� Δ𝑦𝑖 = y(𝑥𝑖 + Δ𝑥) − 𝑦(𝑥𝑖) = 𝑦𝑖+1 − 𝑦𝑖

Δ2𝑦𝑖 = Δ(Δ𝑦𝑖) = 𝑦𝑖+2 − 2𝑦𝑖+1 + 𝑦𝑖�

Similar to ′Δ′, ∇ has been defined as backward difference operator:

∇𝑦𝑖 = 𝑦𝑖 − 𝑦𝑖−1 𝑜𝑟 ∇𝑦(𝑥) = 𝑦(𝑥𝑛 + 𝛼Δ𝑥), where α = x−xnΔx

And, therefore, there is another Newton’s interpolation formula!

Newton’s interpolation formula is also often used for error analysis.

2

Splines

We have earlier noted that a high order polynomial (𝑛 ≥ 4) usually shows oscillation. Therefore, a piece-wise polynomial (𝑛 ≤ 3) is usually preferred for most engineering applications.

→ ‘Spline’ is a piece-wise cubic polynomial (𝑦 = 𝑎𝑥3 + 𝑏𝑥2 + 𝑐𝑥 + 𝑑) fitting 2 data points in segments or pieces. Is it unique? No. Because there are two extra degrees of freedom. In other words, two extra constraints or conditions may be imposed.

→ Let us define the problem:

𝑆𝑖 = 𝑎𝑖𝑥3 + 𝑏𝑖𝑥2 + 𝑐𝑖𝑥 + 𝑑𝑖 is the spline fitting the data 𝑥𝑖 𝑎𝑛𝑑 𝑥𝑖+1 . Thus, it is a higher order polynomial for two data points.

(1) Each spline (𝑆𝑖) is a piece-wise cubic. (2) Curve must pass through the points (𝑥𝑖 & 𝑥𝑖+1) (3) 𝑓𝑖′𝑎𝑛𝑑 𝑓𝑖′′ are continuous at all ‘knots’, i.e, curve is smooth and curvature is the same

at all nodes.

or �𝑆𝑖−1(𝑥𝑖) = 𝑆𝑖(𝑥𝑖)

𝑆′𝑖−1(𝑥𝑖) = 𝑆𝑖′(𝑥𝑖) 𝑆𝑖−1′′(𝑥𝑖) = 𝑆𝑖′′(𝑥𝑖)

� ≡ �𝑆+ = 𝑆−𝑆′+ = 𝑆′−𝑆′′+ = 𝑆′′−

�

Therefore, for n data points there are (n-1) spline segments or 4(n-1) unknown quantities (coefficients) to be determined:

Equations: 𝑦𝑖 = 𝑆𝑖(𝑥𝑖) ∶ 𝑛

𝑆+ = 𝑆− ∶ (𝑛 − 2) ∶ only interior pts.

𝑆′+ = 𝑆′− ∶ (𝑛 − 2)

𝑆′′+ = 𝑆′′− ∶ (𝑛 − 2)

4𝑛 − 6

You have two BCs specified at the end data points or knots.

Therefore, there are as many (4n-4) equations as # of unknowns.

The problem is now well defined:

Si-1 Si

Si+1

xi-1 xi xi+1

3

𝑆𝑖(𝑥) = ∑ 𝑎𝑖𝑥23𝑖=0 = ?

Note that 𝑆𝑖"(𝑥) is a piece-wise linear and continuous (i. e. 𝑆𝑖"(𝑥) = 6𝑎𝑖𝑥 + 2𝑏𝑖)

𝑆𝑖"(𝑥)

𝑥𝑖 𝑥𝑖+1

𝑆𝑖(𝑥) ≡ 𝑐𝑢𝑏𝑖𝑐

Therefore, 𝑆𝑖”(𝑥) = ∑ 𝑙𝑘(𝑥)𝑦𝑘1𝑘=0 (𝐿𝑎𝑔𝑟𝑎𝑛𝑔𝑖𝑎𝑛 𝑖𝑛𝑡𝑒𝑟𝑝𝑜𝑙𝑎𝑡𝑖𝑜𝑛)

= 𝑥−𝑥𝑖+1𝑥𝑖 −𝑥𝑖+1

𝑆”(𝑥𝑖) + 𝑥−𝑥𝑖𝑥𝑖+1 −𝑥𝑖

𝑆”(𝑥𝑖+1)

On integrating twice, 𝑆𝑖(𝑥) = �𝑥3

6− 𝑥𝑖+1

𝑥2

2 � 𝑆”(𝑥𝑖)

Δ𝑖+ �𝑥

3

6− 𝑥𝑖

𝑥2

2 � 𝑆”(𝑥𝑖+1)

Δ𝑖+1+ 𝑐1𝑥 + 𝑐2

Apply the conditions

𝑆𝑖(𝑥𝑖) = 𝑦𝑖 and Si(𝑥𝑖+1) = 𝑦𝑖+1

→ 𝑆𝑖(𝑥) =𝑆′′(𝑥𝑖)(𝑥𝑖+1 − 𝑥)3

6Δ𝑖+𝑆′′(𝑥𝑖+1)(𝑥 − 𝑥𝑖)3

6Δ𝑖+ �

𝑦𝑖Δ𝑖−Δ𝑖6𝑆𝑖(𝑥𝑖)� (𝑥𝑖+1 − 𝑥)

+ �𝑦𝑖+1Δ𝑖

−Δ𝑖6𝑆𝑖(𝑥𝑖+1)� (𝑥 − 𝑥𝑖)

Apply another conditions

𝑆𝑖−1′(𝑥𝑖) = 𝑆𝑖′(𝑥𝑖) on the following

𝑆𝑖′(𝑥) = −𝑆′′(𝑥𝑖)(𝑥𝑖+1 − 𝑥)2

2Δ𝑖+𝑆′′(𝑥𝑖+1)(𝑥 − 𝑥𝑖)2

2Δ𝑖− �

𝑦𝑖Δ𝑖−Δ𝑖6𝑆𝑖′′(𝑥𝑖)�

+ ��𝑦𝑖+1Δ𝑖

−Δ𝑖6𝑆𝑖′′(𝑥𝑖+1)��

On re-arrangement it can be shown

Δ𝑖−16

𝑆′′(𝑥𝑖−1) +Δ𝑖−1 + Δ𝑖

3𝑆′′(𝑥𝑖) +

Δ𝑖6𝑆′′(𝑥𝑖+1)

=𝑦𝑖+1 − 𝑦𝑖

Δ𝑖−𝑦𝑖 − 𝑦𝑖−1Δ𝑖−1

, 𝑖

= 2,𝑛 − 1 (𝑖𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝑛𝑜𝑑𝑒𝑠)

(linear)

4

1 2 𝑥𝑖−1 𝑥𝑖 𝑥𝑖+1 𝑛 − 1 𝑛

If 𝑆′′(𝑥𝑖) 𝑎𝑛𝑑 𝑆′′(𝑥𝑛) are pre-known as boundary conditions, we have

𝐴𝑋� = 𝑏� 𝑜𝑟 𝐴𝑆′′(𝑥) = 𝑏� where A is the tridiagonal matrix

For a ‘natural’ spline, 𝑆′′(𝑥1) = 𝑆′′(𝑥𝑛) = 0 OR 𝑆′′(1) = 𝑆′′(𝑛) = 0

If Δ𝑖 = Δ𝑖−1 (𝑒𝑞𝑢𝑎𝑙𝑙𝑦 𝑠𝑝𝑎𝑐𝑒𝑑 𝑑𝑎𝑡𝑎)

** 16

[𝑆′′(𝑥𝑖−1) + 4𝑆′′(𝑥𝑖) + 𝑆′′(𝑥𝑖+1)] = �𝑦𝑖+1−2𝑦𝑖+𝑦𝑖−1Δ2

�

Call subroutine Tridiag�𝑛 − 2, 16� , 2

3� , 16

, 𝑏��

# 𝑜𝑓 𝑒𝑞𝑛𝑠 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑠

𝑜𝑓 �̅� 𝑅𝐻𝑆 𝑜𝑟 𝑏������ 𝑣𝑒𝑐𝑡𝑜𝑟

𝑆𝑖′′(𝑥𝑖), 𝑖 = 2, 𝑛 − 1 are now known on solving the set of eqns.

Thus, 𝑆𝑖(𝑥) is determined from the boxed equation on the preceding page 3.

Example: -2.5 0 2.5

0 1.67 0

Fit a cubic natural spline.

Use ** above

16𝑆′′(𝑥1) + 2

3𝑆′′(𝑥2) + 1

6𝑆′′(𝑥3) = 0−2×1.67+0

2.52

(You have only one linear algebraic eqn)

𝑆′′(𝑥2) = −0.8016 𝑖 = 2 (middle node only)

�𝑆1(𝑥) = −0.05344(𝑥 + 2.5)3 + 1.002(𝑥 + 2.5)𝑆2(𝑥) = 1.667 − 0.4𝑥2 + 0.0533𝑥3)

� 𝐴𝑛𝑠.

5

Numerical Differentiation

P2 approximated

polynomialf(x) ≈ Pn(x)

Given a continuous 𝑓(𝑥), numerical differentiation can be performed to determine gradient 𝑓′(𝑋𝑖) 𝑎𝑡 𝑋𝑖. Fit a polynomial (𝑢𝑠𝑢𝑎𝑙𝑙𝑦 𝑃1−3) and then determine 𝑃1−3′ to estimate 𝑦𝑖′, depending on desired order of accuracy or error. Taylor’s series can be conveniently used to derive 𝑓′ 𝑜𝑟 𝑓′′ because the series terms contain 𝑓′ 𝑜𝑟 𝑓′′ .

Taylor’s series: 𝑓(𝑋 + ℎ) = 𝑓(𝑋) + ℎ𝑓′(𝑋) + ℎ2 2⁄ 𝑓′′(𝑋) + ⋯ (1)

𝑓(𝑋 − ℎ) = 𝑓(𝑋) − ℎ𝑓′(𝑋) + ℎ2 2⁄ 𝑓′′(𝑋) −⋯ (2)

2 pts

⎩⎪⎨

⎪⎧𝑓′(𝑥𝑖) = 𝑓(𝑥𝑖+1)−𝑓(𝑥𝑖)

ℎ+ 0(ℎ2) ∶ 𝐵𝑎𝑐𝑘𝑤𝑎𝑟𝑑 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑆𝑐ℎ𝑒𝑚𝑒 (𝐵𝐷𝑆)

𝑓′(𝑥𝑖) = 𝑓(𝑥𝑖)−𝑓(𝑥𝑖−1)ℎ

+ 0(ℎ2) ∶ 𝐹𝑜𝑟𝑤𝑎𝑟𝑑 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑆𝑐ℎ𝑒𝑚𝑒 (𝐹𝐷𝑆)

𝑓′(𝑥𝑖) = 𝑓(𝑥𝑖+1)−𝑓(𝑥𝑖−1)2ℎ

+ 0(ℎ2) ∶ 𝐶𝑒𝑛𝑡𝑟𝑎𝑙 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑆𝑐ℎ𝑒𝑚𝑒 (𝐶𝐷𝑆)

�

(1) Richardson’s extrapolation technique can be used to take 3 data pts.

𝑓(𝑥𝑖 + 2ℎ) = 𝑓(𝑥) + 2ℎ𝑓′(𝑥) + 4 (ℎ2 2)⁄ 𝑓′′(𝑥) + ⋯⋯ (3)

𝑓(𝑥𝑖 − 2ℎ) = 𝑓(𝑥) − 2ℎ𝑓′(𝑥) + 4 (ℎ2 2)⁄ 𝑓′′(𝑥) + ⋯⋯ (4)

Use 1 an 3 : Multiply eq(1) by 4 and subtract from 3 to eliminate 𝑓′′(𝑥𝑖)

3𝑝𝑜𝑖𝑛𝑡𝑠

⎩⎪⎨

⎪⎧ 𝑓′(𝑥𝑖) =

4𝑓(𝑥𝑖+1) − 3𝑓(𝑥𝑖) − 𝑓(𝑥𝑖+2)2ℎ

+ 0(ℎ3) ∶ 𝐹𝐷𝑆

𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑙𝑦 2 𝑎𝑛𝑑 4 𝑤𝑖𝑙𝑙 𝑔𝑖𝑣𝑒

𝑓′(𝑥𝑖) =−4𝑓(𝑥𝑖−1) + 3𝑓(𝑥𝑖) + 𝑓(𝑥𝑖−2)

2ℎ + 0(ℎ3) ∶ 𝐵𝐷𝑆

�

Higher order 𝑓′(𝑥𝑖) can also be obtained as CDS:

f(x)

y(xi)

xi-1

or

xi xi+1

xi-2 xi-1 xi xi+1 xi+2

(1)

6

4𝑑𝑎𝑡𝑎 𝑝𝑜𝑖𝑛𝑡𝑠 �𝑓

′(𝑥𝑖) =−4𝑓(𝑥𝑖+2) − 8𝑓(𝑥𝑖−1) + 8𝑓(𝑥𝑖+1) + 4𝑓(𝑥𝑖+2)

12ℎ + 0(ℎ4) �

( See the book by Ferziger)

Similarly, use eqns 1 & 2 to obtain

𝑓′′(𝑥𝑖) = 𝑓(𝑥𝑖+1)−2𝑓(𝑥𝑖)+𝑓(𝑥𝑖−1)ℎ2

+ 0(ℎ2)

= 𝑓′′(𝑥𝑖−1) = 𝑓′′(𝑥𝑖+1):

𝑊ℎ𝑦? (𝑌𝑜𝑢 ℎ𝑎𝑣𝑒 2𝑛𝑑 𝑜𝑟𝑑𝑒𝑟 𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 3 𝑑𝑎𝑡𝑎 𝑝𝑜𝑖𝑛𝑡𝑠)

Apply Richardson’s extrapolation technique to obtain even higher order 𝑓′′

𝑓′′(𝑥𝑖) =−𝑓(𝑥𝑖−2) + 16𝑓(𝑥𝑖−1) + 30𝑓(𝑥𝑖) + 16𝑓(𝑥𝑖+1) − 𝑓(𝑥𝑖+2)

12ℎ2 + 0(ℎ4)

(Five data points)

Note the notations: 𝑓(𝑥𝑖) = 𝑦𝑖(𝑥𝑖) and f ′and f ′′ ≈ y′and y′′; 𝑓(𝑥) ≃ 𝑃𝑛(𝑥)

1

Lecture #14-15

Numerical Integration

Similar to numerical differentiation, integration of 𝑓(𝑥), i.e., ∫𝑓(𝑥)𝑑𝑥 may be performed by approximating 𝑓(𝑥) ≈ 𝑃𝑛(𝑥), usually 𝑛 ≤ 3. Therefore, first calculate 𝑦𝑖(𝑥𝑖) = 𝑓(𝑥), then fit a polynomial through the ′𝑛′ data points and determine∫𝑃𝑛𝑑𝑥 𝑜𝑟 area under the approximated curve. Considering usual oscillations set in the fitted polynomial for 𝑛 ≥ 4, piece-wise integration is performed Detailed method and analysis are as follows.

𝑃𝑛(𝑥) is the interpolating 𝑓𝑛 between (𝑥0,⋯𝑥𝑛)

It can be shown/proposed that

∫ 𝑓(𝑥)𝑑𝑥 = ∑ 𝑤𝑘𝑓(𝑥𝑘); 𝑓(𝑥𝑘) = 𝑦(𝑥𝑘)𝑛𝑘=0

𝑏𝑎

where 𝑛 is the number of segments between ‘a’ and ‘b’

and ‘𝑤𝑖’ is the respective weight at 𝑥𝑖 . It is also known as Quadrature formula. Let us look at the proposed formula differently. Apply Lagrange interpolation between (𝑛 + 1) data points or 𝑛 segments:

𝑃𝑛(𝑥) = ∑ 𝑙𝑘(𝑥)𝑦𝑘 (𝑦𝑘 ≡ 𝑦(𝑥𝑘) ≡ 𝑦𝑘(𝑥𝑘))𝑛𝑘=0

∫ 𝑓(𝑥)𝑑𝑥 = ∫ ∑ 𝑙𝑘(𝑥)𝑦𝑘𝑑𝑥 = ∑ ∫ 𝑙𝑘(𝑥)𝑦𝑘𝑑𝑥𝑏𝑎

𝑛𝑘=0

𝑛𝑘=0

𝑏𝑎

𝑏𝑎

= (𝑏 − 𝑎)∑ 𝐶𝑘𝑦𝑘; 𝐶𝑘 = 1(𝑏−𝑎)∫ 𝑙𝑘(𝑥)𝑑𝑥𝑏

𝑎𝑛𝑘=0

Cotes number

- It is the same as the Quadrature formula ; n ≡ parameter(degree of polynomial)

(It can be shown ∑ 𝐶𝑘 = 1 (𝑝𝑢𝑡 𝑓(𝑥) = 1))𝑛𝑘=0

Therefore, we can have a table as follows:

𝑎 𝑏 𝑥𝑖

𝑓(𝑥)

𝑃𝑛(𝑥)

2

Newton-Cotes Coefficients (ends pts included)

𝑛 𝑁 𝑁𝐶0𝑛 𝑁𝐶1𝑛 𝑁𝐶2𝑛 ⋯⋯ 𝑁𝐶0𝑁 𝐸𝑟𝑟𝑜𝑟

12345

268

90288

1117

19

143

3275

13

1250

13250

7

75

19

. 0183Δ3𝑓′′. 0035Δ5𝑓′′′′

. 00016Δ5𝑓′′′′⋯⋯

These are called closed N-C formula because they use functions at the end points also (𝑎, 𝑏). Note, 𝑛 or the number of segments is usually 1 or 2 or 3, or the degree of polynomial is 1 or 2 or 3, respectively. Readers may have recognized that n = 1, corresponds to Trapezoidal rule

with ‘1 2� ’ coefficients of 𝑦𝑘; 𝑛 = 2 corresponds to Simpson’s ‘1 3� 𝑟𝑑’ rule with

(16� , 2

3� , 16� ) coefficients, etc. For more details, you can refer the book by Ferziger or SKG.

The error in the numerical integration can be calculated from the remainder term of the polynomial:

⇒ 𝐼 = �(𝑒𝑟𝑟𝑜𝑟)𝑑𝑥 ⇒ 𝑤ℎ𝑒𝑟𝑒 ′𝑒𝑟𝑟𝑜𝑟′ = 𝑦𝑛+1(𝜉)𝑛 + 1!

(𝑥 − 𝑥0)(𝑥 − 𝑥1)⋯ (𝑥 − 𝑥𝑛)

𝑓𝑜𝑟 (𝑛 + 1)𝑑𝑎𝑡𝑎𝑝𝑜𝑖𝑛𝑡𝑠

⇒ SKG’s book uses Newton’s Interpolation formula to determine the various integration formulae:

𝑦 (𝑥) = 𝑦(𝑥0 + 𝛼Δ𝑥) = 𝑦0 + 𝛼(Δ𝑦0) + 𝛼(𝛼−1)�Δ2𝑦0�2!

+ ⋯⋯𝛼(𝛼−1)(𝛼−2)𝛼⋯(𝛼−𝑛)(Δ𝑛+1𝑦0)𝑛+1!

𝑅𝑛

To obtain the different methods for integration, viz. Trapezoidal’s rule, Simpson’s 1 3�𝑟𝑑

, etc, two approaches must give the identical results because polynomial of the highest degree that can fit to (𝑛 + 1) data points is unique.

Take 𝑛 = 1

∫ 𝑓(𝑥)𝑑𝑥 = (𝑏 − 𝑎)∑ 𝐶𝑘𝑛𝑓(𝑥𝑘)𝑛𝑘=0

𝑏𝑎

= (𝑏 − 𝑎) �12𝑦𝑎 + 1

2𝑦𝑏�

= (𝑏−𝑎)2

(𝑦𝑎 + 𝑦𝑏)

𝑓(𝑥) 𝑃1(𝑥)

𝑓(𝑏)

𝑓(𝑎)

𝑎 𝑥 𝑏

𝑓(𝑥)

3

�𝑜𝑟 = (𝑏−𝑎)2

(𝑓(𝑎) + 𝑓(𝑏))�

(Trapezoidal rule)

𝑛 = 2

� 𝑓(𝑥)𝑑𝑥 =(𝑏 − 𝑎)

6�𝑓(𝑎) + 4𝑓 �

𝑎 + 𝑏2

� + 𝑓(𝑏)�𝑏

𝑎

(𝑁𝑜𝑡𝑒: 𝑃2 ≡ (𝑎 + 𝑏𝑥 + 𝑎𝑥2) ≡ 𝑝𝑎𝑟𝑎𝑏𝑜𝑙𝑎)

= Δ𝑋3�𝑓(𝑎) + 4𝑓 �𝑎+𝑏

2� + 𝑓(𝑏)�

𝑛 = 3

� 𝑓(𝑥)𝑑𝑥 =(𝑏 − 𝑎)

8[𝑓(𝑎) + 3𝑓(𝑐) + 3𝑓(𝑑) + 𝑓(𝑏)]

𝑏

𝑎

= 3Δ𝑋8

[ − 𝑑𝑖𝑡𝑡𝑜 − ], Δ𝑥 = (𝑏−𝑎)3

- Simpson’s 3 8�𝑡ℎ

formula.

In actual practice, applying a single Cotes formula over the entire range is rarely done. Considering large error for large ′Δ𝑋′, interval is broken into sub-intervals & then applying piece-wise quadrature formula and summing it over.

∫ 𝑓(𝑥)𝑑𝑥 = ℎ

2[𝑓(𝑋𝑖−1) + 𝑓(𝑋𝑖)]𝑋𝑖

𝑋𝑖−1

ℎ = (𝑏−𝑎)𝑛

(𝑛 + 1 data points or 𝑛 segments)

Therefore, ∫ 𝑓(𝑥)𝑑𝑥 = ∑ ℎ2

[𝑓(𝑥𝑖−1) + 𝑓(𝑥𝑖)]𝑛𝑖=1

𝑏𝑎

𝑓(𝑥)

𝑆𝑖𝑚𝑝𝑠𝑜𝑛′𝑠 13�𝑟𝑑

𝑓𝑜𝑟𝑚𝑢𝑙𝑎(ℎ𝑖𝑔ℎ𝑒𝑟 𝑜𝑟𝑑𝑒𝑟 𝑎𝑐𝑐𝑢𝑟𝑎𝑡𝑒)

𝑎 𝑏 (𝑎 + 𝑏)

2�

𝑃2(𝑥)

ΔX ΔX

𝑎 𝑏 𝑋𝑖−1 𝑋𝑖 𝑋𝑛 𝑋0

4

= �

ℎ2�𝑓(𝑥0) + 2�𝑓(𝑥𝑘) + 𝑓(𝑥𝑛)

𝑛−1

𝑘=1

�

(𝑏 − 𝑎)𝑛

��𝑓(𝑥𝑘) −12

[𝑓(𝑎) + 𝑓(𝑏)]𝑛

𝑘=0

�⎭⎪⎬

⎪⎫

∗

* Trapezoidal rule

𝑺𝒊𝒎𝒑𝒔𝒐𝒏’𝒔 𝟏 𝟑�𝒓𝒅

𝒓𝒖𝒍𝒆: (intervals must be even or data pts must be odd),

𝑛 𝑁 𝑁𝐶0𝑛 𝑁𝐶1𝑛 𝑁𝐶2𝑛

2 6 1 4 1

∫ 𝑓(𝑥)𝑑𝑥 = �𝑏−𝑎6� (𝑓(𝑎) + 4𝑓(𝑐) + 𝑓(𝑏))𝑏

𝑎

ℎ = (𝑏−𝑎)𝑛

(𝑛 𝑠𝑒𝑔𝑚𝑒𝑛𝑡𝑠 𝑜𝑟

𝑛 + 1 𝑑𝑎𝑡𝑎 𝑝𝑜𝑖𝑛𝑡𝑠)

∫ 𝑓(𝑥)𝑑𝑥 = ℎ3

(𝑓(𝑋0) + ∑⋯+𝑏𝑎 ∑… + 𝑓(𝑋𝑛))

= (𝑏−𝑎)6

(𝑓(𝑋0) + 4∑ +2𝑛−1𝑘=1,3,5 ∑ +𝑓(𝑋𝑛))𝑛−2

𝑘=2,4,6

(Note all even nodes ‘𝑋𝑘’ are summed/counted twice in calculating total area)

What do you do if 𝑛 ≡ odd or data are even? Apply the method to one data less so that n = even. Add the area corresponding to the left-over segment, calculated using Trapezoidal rule!

One open NC method is popular: Mid-point rule, the method is frequently applied when the 𝑓𝑛 is discontinuous at end points: a or b or both.

∫ 𝑓(𝑥)𝑑𝑥 = (𝑏 − 𝑎) �𝑓 �𝑎+𝑏2��𝑏

𝑎

ex. ∫ 11−𝑥2

𝑑𝑥10 (Note 𝑓(1) → ∞)

𝑎 𝑏 (𝑎+𝑏)2

𝑎 𝑏 𝑐

ℎ = (𝑏 − 𝑎)2�

𝑋𝑖 𝑋𝑖 𝑋𝑖+1 ℎ ℎ

𝑋𝑖−1 𝑋𝑖+1 𝑋𝑖 𝑎 𝑏 (𝑋𝑛) (𝑋0)

5

Example:

𝑣 = 2�̅�(1 − 𝑟2

𝑅2)

𝜃 = 𝑇−𝑇𝑤𝑇𝑤

= −4𝑧 − 𝑟2 + 𝑟4

4+ 7

24

(𝑧, 𝑟 are dimensionless quantities: 𝑟 𝑅� 𝑎𝑛𝑑 𝑧 𝐿� )

Consider the classical heat transfer/transport problem of SS heat transfer at constant wall temperature in the fully developed region of a tubular laminar parabolic flow, often discussed in TP lectures. The dimensionless SS temperature 𝜃(𝑧, 𝑟) in the fluid is analytically derived. See the solution above. Determine the ‘mix-cup’ temperature at z = 0.5.

Soln : In a flow-system, mix-cup temperature is the radially emerged temperature, taking into consideration the total heat (cal) carried away by the flowing fluid in unit time across the cross-section of the tube.

𝑇� =

⎝

⎜⎜⎜⎛∫ 𝜌𝐶𝑝𝑣(𝑟′)𝑇(2𝜋𝑟′)𝑑𝑟′𝑅0

𝑘𝑔 𝑚3⁄ 𝑐𝑎𝑙 𝑘𝑔⁄ − 𝐾 𝐾∫ 𝜌𝐶𝑝𝑣(𝑟′)2𝜋𝑟′𝑑𝑟′𝑅0

𝑐𝑎𝑙 𝑠 − 𝐾⁄ ⎠

⎟⎟⎟⎞

(𝑁𝑜𝑡𝑒:𝑑𝑞 = 𝑣2𝜋𝑟𝑑𝑟𝑚3

𝑠)

𝑄 = � 𝑣2𝜋𝑟𝑑𝑟𝑅

0at a ref. temperature

In the dimensionless quantities,

�̅�(𝑧) =∫ 𝜃(𝑟, 𝑧)(1 − 𝑟2)𝑟𝑑𝑟10

∫ (1 − 𝑟2)𝑟𝑑𝑟10

Simpson’s 1 3� rule can be applied by taking odd number (5) of data points over 𝑟 ≡ (0, 1).

Therefore # of segment = 4, step size, ℎ = 1−04

= 0.25, Δ𝑟 ≡ ℎ = 0.25; 𝑟𝑖 = 𝑖Δ𝑟

Numerical integration is performed for both numerator and denominator terms.

𝐼 = ∫ 𝑓(𝑟)𝑑𝑟10 ∫ Φ(𝑟)𝑑𝑟1

0�

Δr T(r) 𝑣(r)

Tw

z

fully developed region (2) for both �low and temperature

r

R

(𝑛) 4 3 2 1 𝑘 = 0

6

where 𝑓(𝑟) = 𝜃(𝑟, 𝑧)(1− 𝑟2)𝑟 & Φ(𝑟) = (1 − 𝑟2)𝑟

For both functions,

𝐼 = ℎ3� �𝑓(𝑋0) + 4∑ +2∑ + 𝑓(𝑋𝑛)𝑛−2

𝑘=2,4𝑛−1𝑘=1,3,5 �

= ℎ3� �𝑓(𝑋0) + 4∑ +2∑ +𝑓(𝑋4)2,41,3 �

Best way is to prepare a table for calculations:

𝑘 𝑟𝑘 𝜃𝑘 (1 − 𝑟𝑘2)𝑟𝑘Φ(𝑟) 𝜃𝑘(1 − 𝑟𝑘2)𝑟𝑘

𝑓(𝑟)

01234

00.250.500.751.00

−1.7083−1.7698−1.9427−2.1917−2.4583

0.0

0.0

0.0

0.0

𝐹𝑖𝑙𝑙 − 𝑢𝑝 𝑡ℎ𝑒𝑡𝑎𝑏𝑙𝑒

Show,

�̅�𝑚𝑖𝑥−𝑐𝑢𝑝(0.5) ≡ −1.9976 = 𝑇−𝑇𝑤𝑇𝑤

(wall is colder than bulk fluid)

What does decide # of data points ‘n’ or the step size ‘h’ ? It is the level of desired accuracy. Repeat the entire calculations for n = 9 and check your requirement of an improved �̅�𝑚𝑖𝑥−𝑐𝑢𝑝 from the previous value -1.9976. Sometimes you have to decrease ‘n’ or increase ‘h’ if you have exceeded accuracy limit!

Error Analysis

Let us take the simplest Trapezoidal rule applied to one segment over (a,b).

Start from Newton’s Difference Interpolation formula:

𝑓(𝑥) = 𝑓(𝑥0) + 𝛼Δ𝑓(𝑥0) +𝛼(𝛼 − 1)

2!Δ2𝑓(𝑥0) + ⋯𝑅𝑛

𝑇𝑤

𝑇𝑓𝑙𝑢𝑖𝑑

𝑇𝑤

7

𝑅𝑛 = 𝛼(𝛼−1)⋯(𝛼−𝑛)𝑛+1!

Δ𝑛+1𝑓(𝜉) 𝑤ℎ𝑒𝑟𝑒 𝑥0 < 𝜉 < 𝑥

𝛼 = 𝑥−𝑥0ℎ

𝑜𝑟 𝑑𝑥 = ℎ 𝑑𝛼 ; 𝑥0 = 𝑎,

If there is only one segment, ℎ = (𝑏 − 𝑎)

𝐼 = ∫ 𝑓(𝑥)𝑑𝑥 = ℎ ∫ �𝑓(𝑥0) + 𝛼Δ𝑓(𝑥0) + 𝛼(𝛼−1)2!

Δ2𝑓(𝜉)�10

𝑏𝑎 𝑑𝑥

𝑃1 = (𝑎𝑥 + 𝑏)

= ℎ �𝑓(𝑎) +12�𝑓(𝑏) − 𝑓(𝑎)� + (

𝛼3

6−𝛼2

4)� ℎ2𝑓′′(𝜉)

1

0�

= ℎ �𝑓(𝑎)+𝑓(𝑏)2

� − 112ℎ3𝑓′′(𝜉) (𝑓′′(𝜉) = Δ2𝑓(𝜉)

2ℎ)

Therefore, error on one segment is 1 2� ℎ3𝑓′′(𝜉) 𝑜𝑟 0(ℎ3)

If there are n segments between (a & b) error should also be summed up to

112ℎ3 ∑ 𝑓′′(𝜉) = 1

12ℎ2𝑓′′�𝜉̅�(𝑏 − 𝑎) 𝑜𝑟 0(ℎ2) 𝑛

𝑁𝑜𝑡𝑒: 𝑓′′�𝜉̅� = ∑𝑓′′(𝜉)𝑛

= ℎ∑𝑓′′(𝜉)(𝑏−𝑎)

.

Similarly, for the error estimation of Simpson’s 1 3� rule start with 𝑃2 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐, 𝑜𝑟

𝑅𝑛 = 𝛼(𝛼−1)(𝛼−2)Δ3𝑓(𝜉)3!

over one segment or 3 data pts.

to show that error ≡ 0(ℎ5)𝑓′′′′(𝜉) and if summed over ‘n’ segments error = 0(ℎ4)𝑓′′′′�𝜉̅�.

Simpson’s 3 8� 𝑟𝑢𝑙𝑒: error(one segment) ≡ 0(h5)error( segments) ≡ 0(h4)

A common question is asked: Show that Simpson’s 1 3� 𝑟𝑢𝑙𝑒 gives the error corresponding to a cubic interpolating function, although it is based on a parabola or a quadratic polynomial. See Chhapra and Cannale's book for the details.

Mid-Term

1

Lecture #15-16 1st order ODE

A set of ODEs takes the following form:

�

𝑑𝑦1𝑑𝑡

= 𝑓1(𝑦1,𝑦2 ⋯𝑦𝑛)⋮

𝑑𝑦𝑛𝑑𝑡

= 𝑓2(𝑦1,𝑦2 ⋯𝑦𝑛)� 𝑜𝑟 𝑑𝑦�

𝑑𝑡= 𝑓(̅𝑦�)

Requires initial conditions to solve 𝑦� .

𝑡 = 0, 𝑦1 = 𝑦10 ⋯𝑦𝑛 = 𝑦𝑛0, 𝑒𝑡𝑐 𝑜𝑟 𝑦� = 𝑦�0

Note that ODEs can also be1st order on x (space) & 𝑑𝑦�

𝑑𝑋= 𝑓(̅𝑦�) with 𝑥 = 0, 𝑦� = 𝑦�0.

⇒ We explore numerical solution to ODEs starting with Euler’s forward and backward methods. These methods are basics, and also set the tone for a stability analysis. Let us begin with one ODE.

Euler’s Method

𝑑𝑦𝑑𝑥

= 𝑓(𝑥,𝑦) ; 𝑦(𝑥0) = 𝑦0

𝑜𝑟 , 𝑦𝑛+1−𝑦𝑛ℎ

= 𝑦(𝑥𝑛,𝑦𝑛) + 0(ℎ2)

(This is the general ‘discretization’ of a derivative at the ‘n’ th grid. Recall the lecture on numerical differentiation)

or 𝑦𝑛+1 = 𝑦𝑛 + ℎ𝑓(𝑥𝑛,𝑦𝑛) + 0(ℎ2)

𝑦𝑛′

Note: The RHS terms are known to be 1st order accurate or have 2nd order error.

From the previous calculations, start from 𝑦(𝑥 = 0 ) = 𝑦0 → 𝑦1 → 𝑦2 ⋯𝑒𝑡𝑐. Therefore, you march forward and the method is called Euler’s forward explicit method.

𝑦0 𝑦1 𝑦𝑛 𝑦𝑛+1

𝑥0 𝑥1 𝑥𝑛 𝑥𝑛+1

2

Graphically,

�Theoretical y(x)

⎭⎪⎪⎬

⎪⎪⎫

It is all aboutcalculating true derivatives

(gradients) at xn asyou march forward:

y0′ = f(x0, y0)y1′ = f(x1, y1)⋮ yn′ = f(xn, yn)

It is clear that smaller ‘h’(step-size) is, less is the error or accuracy is higher: 0(ℎ2). There is a price for long CPU time. Let us look at the stability aspect of the method.

𝑑𝑦𝑑𝑥

= 𝑓(𝑥,𝑦) = 𝑓(𝑥0 + Δ𝑥,𝑦0 + Δ𝑦)

= 𝑓(𝑥0,𝑦0) + �(𝑦 − 𝑦0)𝑑𝑓𝑑𝑦�𝑥0,𝑦0

+ �(𝑥 − 𝑥0)𝑑𝑓𝑑𝑥�𝑥0,𝑦0

+ 0(Δ𝑥2,Δ𝑦2)

Multi-variable Taylor series can be applied to estimate gradient 𝑦′(𝑥0,𝑦0):

= 𝛼𝑦 + 𝛽𝑥 + 𝜈

gives exponential

solution term const. for linear solution

𝑑𝑦𝑑𝑥

= 𝛼𝑦 ∶ Model equation for testing stability.(Note: Stability is governed by thehomogeneous part of the solution)

Let us look at differently: Introduce perturbation or find true value 𝑦 .

Therefore, 𝑑𝑦𝑑𝑥

= 𝛼𝑦� + 𝛽𝑥 + 𝜈

𝑜𝑟, 𝑑𝜉𝑑𝑥

= 𝛼(𝑦 − 𝑦) = 𝛼𝜉 (𝜉 = 𝑦 − 𝑦)

We have the same inference: error satisfies homogeneous part of the solution. The equation describes how ‘𝜉’ can grow because of a small perturbation.

At 𝑥 = 𝑥0, 𝑦 = 𝑦0 ⇒ exact solution 𝑦 = 𝑦0𝑒𝛼𝑥 𝐼𝑓 𝛼 > 0, fngrows,α < 0, fn is bounded.

Now Apply Euler’s method:

Numerical calculations

𝑦1 𝑦2 𝑌

𝑦0 𝑥0 𝑥1 𝑥2 𝑋

3

𝑦𝑛+1 − 𝑦𝑛 = ℎ𝑓(𝑥𝑛,𝑦𝑛) = ℎ𝑦𝑛′ = ℎ(𝛼 𝑦𝑛)

𝑦𝑛+1 = 𝑦𝑛(1 + 𝛼ℎ)

𝑦1 = 𝑦0(1 + 𝛼ℎ), 𝑦2 = 𝑦1(1 + 𝛼ℎ)2

𝑦𝑛 = 𝑦0(1 + 𝛼ℎ)𝑛 ⇒ Numerical solution

� If α > 0 function growsIf α < 0 function may be bounded�

Let us write, 𝛼 = 𝛼𝑟 + 𝑖𝛼𝐼

𝑦𝑛 = 𝑦0(1 + 𝛼𝑟ℎ + 𝑖𝛼𝐼ℎ)𝑛

For function to be bounded

|(1 + 𝛼𝑟ℎ) + 𝑖𝛼𝐼ℎ| < 1

(1 + 𝛼𝑟ℎ)2 + 𝛼𝑖2ℎ2 < 1

0 > 𝛼𝑟ℎ > −2 if αi = 0 for stability

Note: 𝛼 = −𝑣𝑒, the exact solution is always bounded! Therefore, Euler’s forward /explicit method is conditionally stable.

example:

𝑑𝑦𝑑𝑥

= −𝑦 (𝑥 = 0,𝑦 = 1) ⇒ 𝑦 = 𝑒−𝑥

Numerical soln :

𝛼 = −1; 𝛼𝑟 = −1, 𝛼𝐼 = 0

0 > 𝛼𝑟ℎ > −2 for stability

or ℎ < +2

⇒ 𝑦𝑛+1 − 𝑦𝑛 = ℎ𝑓(𝑥𝑛,𝑦𝑛) = −ℎ𝑦𝑛 ⇒ 𝑦𝑛 = (1 − ℎ)𝑛

Plot to see that oscillations start creeping in h > 2

𝑋 6 4 2 0 10−7

𝑌

𝑙𝑜𝑔 1.0 exact(bounded)

ℎ = 0.001 ℎ = 0.1 ℎ = 0.8

error increases ⇒

2.0

1.0

0 −2 −1 𝑋

ℎ = 1

ℎ > 5

(unbounded)

exact solution: 𝑌

𝑋

1.0 ′bounded solution′

𝑅(𝛼𝑟ℎ)

Stable region(depends on ℎ and αr)

𝐼(𝛼𝑖ℎ)

−2 −1 0

4

Backward or Implicit Euler:

𝑑𝑦𝑑𝑥

= 𝑓(𝑥,𝑦), 𝑦(𝑥0) = 𝑦0

𝑦𝑛+1−𝑦𝑛

ℎ= 𝑓(𝑥𝑛+1,𝑦𝑛+1)

(Note this method may require iteration), maybe NR method; you can not march!

𝑦𝑛+1 − 𝑦𝑛 = ℎ𝑓(𝑥𝑛+1,𝑦𝑛+1) ⇒ Test it on the model eqn

dydx

= αy for the stability analysis

𝑦𝑛+1 = 𝑦𝑛 + ℎ(𝛼𝑦𝑛+1)

𝑦𝑛+1 = 𝑦𝑛 (1 − 𝛼ℎ)⁄

= 𝑦0(1 − 𝛼𝑟ℎ − 𝑖𝛼𝑖ℎ)−𝑛

For 𝑓𝑛 to be bounded |(1 − 𝛼𝑟ℎ − 𝑖𝛼𝑖ℎ)| > 1

𝑜𝑟 (1 − 𝛼𝑟ℎ)2 + 𝛼𝑖2ℎ2 > 1

𝑜𝑟 |(1 − 𝛼𝑟ℎ)| > 1

Clearly, 0 > 𝛼𝑟ℎ > 2

Draw the region:

⇒ �

For all (αrh) in the left half of the plane αr < 0 𝑎𝑛𝑑

the method is unconditionallystable

�

For all ℎ outside the circle

the method is stable (or αrh > 2)

It is clear the stability region for the implicit Euler is much more larger than that of the explicit Euler.

Notes: Error decreases with decreasing step sizes (h): 0(ℎ𝑛), however, at the expense of CPU time. This is not surprising. The ‘issue’ is with increasing step sizes. There is a tendency to speed up computation by taking large step-size. An explicit method like Euler Forward may not permit or allow you to do so, even if the level of desired accuracy may be acceptable. Instability may set in at large ‘h’. On the other hand, Euler Backward or an implicit method permits you to use relatively layer ‘h’ without instability, yielding a bounded solution.

(𝛼𝑖ℎ)

(𝛼𝑟ℎ) 0 1 2

5

⇒ ‘Error’ must not be mixed up with the ‘instability’. A method may be relatively higher order accurate, but may show instability at small step-size. In general, implicit methods are more stable than explicit methods, but require iterations to compute 𝑦𝑛+1. Examples:

𝑑𝑦𝑑𝑥

= 𝑓(𝑥,𝑦) = 𝛼𝑦2

Explicit: 𝑦𝑛+1 − 𝑦𝑛 = ℎ𝛼𝑛𝑦𝑛2

Implicit: 𝑦𝑛+1 − 𝑦𝑛 = ℎ 𝑥𝑛+1𝑦𝑛+12 (requires iterations or NR method? )

⇒ " A method is stable if it produces a bounded solution when it is supposed to and is unstable

if it does not." Therefore, while solving 𝑑𝑦𝑑𝑥

= −𝑦, if the numerical method produces a stable

solution, it is stable, else unstable. On the other hand, there is no instability ‘issue’ with 𝑑𝑦𝑑𝑥

= +𝑦 because the exact solution itself is unbounded. Therefore, both Euler

Forward(explicit) or Backward(implicit) will produce an unbounded solutions!

Get back to 𝑑𝑦𝑑𝑥

= −𝑦(𝛼 = −1) and 𝑦𝑛 = (1 + ℎ)−𝑛

Error increases with increasing ‘h’ but the

solution remains stable. Compare to Forward Euler

when 𝛼 < 2 for a stable solution.

⇒ Trapezoidal rule �0(ℎ2)�:

𝑦𝑛+1 − 𝑦𝑛 = ℎ2

[𝑓(𝑥𝑛,𝑦𝑛) + 𝑓(𝑥𝑛+1,𝑦𝑛+1)]

(Implicit method)

Can also be tested on the model equation

𝑑𝑦𝑑𝑥

= 𝛼𝑦

𝑜𝑟 𝑦𝑛+1 − 𝑦𝑛 =𝛼ℎ2

[𝑦𝑛 + 𝑦𝑛+1]

𝑦𝑛+1 = 𝑦𝑛(1+𝛼ℎ 2)⁄(1−𝛼ℎ 2)⁄

𝑜𝑟 𝑦𝑛 = 𝑦0(1 + 𝛼ℎ 2)⁄(1 − 𝛼ℎ 2)⁄

𝑛

error

exact

h=1

0.2 0.1 0

y 1.0

X

6

|𝑧| < 1

For stability �(1+𝛼ℎ 2)⁄(1−𝛼ℎ 2)⁄ � < 1 𝑜𝑟

Each method has a stability region for 𝛼𝑟ℎ

selecting step-size ‘h’ for calculation without instability.

See the textbooks for details. Stability region

.

1

Lecture #17

Single Step Methods

Forward or Backward Euler is the simplest single step method to solve ODEs. The methods are 1st order accurate. A general higher order method can be derived considering that it is all about choosing a suitable gradient at (𝑥𝑛,𝑦𝑛) 𝑜𝑟 (𝑡𝑛,𝑦𝑛) to calculate 𝑦𝑛+1(𝑥𝑛+1):

𝑑𝑦𝑑𝑡

= 𝑓(𝑡, 𝑦); 𝑦(0) = 𝑦0

𝑦𝑛+1 = 𝑦𝑛 + ℎ𝜙(𝑡𝑛,𝑦𝑛,ℎ) (ℎ = 𝑠𝑡𝑒𝑝 𝑠𝑖𝑧𝑒 = Δ𝑡)

It is all about finding: 𝜙(𝑡𝑛,𝑦𝑛,ℎ) = �𝑑𝑦𝑑𝑡�𝑡𝑛,𝑦𝑛

Propose as

𝑦𝑛+1 = 𝑦𝑛 + ℎ[𝑎𝑘1 + 𝑏𝑘2] 1 ⇒ 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑚𝑒𝑡ℎ𝑜𝑑.

Here,𝑘1 and 𝑘2 are the slopes with weights a & b respectively ( Forward Euler may be considered to assume 𝑎 = 1, 𝑏 = 0 and 𝑘1 = 𝑦𝑡𝑛,𝑦𝑛

′ , whereas Backward Euler assumes 𝑎 =0, 𝑏 = 1,𝑘2 = 𝑦𝑡𝑛+1,𝑦𝑛+1

′ )

Here, � 𝑘1 = 𝑓(𝑡𝑛,𝑦𝑛) 𝑘2 = 𝑓(𝑡𝑛 + 𝑝ℎ, 𝑦𝑛 + 𝑞(ℎ𝑘1) �

Thus, the model has4 parameters; a, b, p, q.

⎩⎪⎨

⎪⎧

Thus k2(slope)is calculatedat (tn + ph), fraction less than

(tn + h) or tn+1, as x − coordinate,and yn + q(hk1), fraction less

than (yn + hk1) or yn+1, as y −coordinate.

�

𝑘2 = 𝑓(𝑡𝑛,𝑦𝑛) + 𝑓𝑡(𝑡𝑛,𝑦𝑛).𝑝ℎ + 𝑓𝑌(𝑡𝑛,𝑦𝑛)𝑞ℎ𝑘1 + 0(ℎ2)

-Taylor multi-variable series applied on 𝑘2(𝑡, 𝑦) with Δ𝑡 = 𝑝ℎ, Δ𝑦 = 𝑞ℎ𝑘

Substituting in (1)

𝑦𝑛+1 = 𝑦𝑛 + ℎ[𝑎𝑓(𝑡𝑛,𝑦𝑛) + 𝑏𝑓(𝑡𝑛,𝑦𝑛)] + ℎ2�𝑏𝑝𝑓𝑡(𝑡𝑛,𝑦𝑛) + 𝑏𝑞𝑓𝑓𝑦(𝑡𝑛,𝑦𝑛)� + 0(ℎ3) − (2)

(𝑠𝑙𝑜𝑝𝑒)

𝑦𝑛+1

𝑦𝑛 + 𝑞(ℎ𝑘1) 𝑦𝑛

𝑡𝑛 𝑡𝑛+1 (𝑡𝑛 + 𝑝ℎ)

𝑞(ℎ𝑘1)

ℎ𝑘1 𝑘2

𝑘1 ℎ

𝑦

2

Also, restart from 𝑑𝑦𝑑𝑡

= 𝑓(𝑡,𝑦(𝑡)) as a total derivative on t.

or 𝑦𝑛+1 = 𝑦𝑛 + ℎ𝑓 + ℎ2

2!𝑓′ + 0(ℎ3): (Single variable(t) Taylor − Series)

where 𝑓′ = 𝑑𝑓𝑑𝑡

(𝑡𝑛,𝑦𝑛) = �𝜕𝑓𝜕𝑡

+ 𝜕𝑓𝜕𝑦∙ 𝜕𝑦𝜕𝑡�𝑡𝑛,𝑦𝑛

= �𝑓𝑡 + 𝑓𝑦𝑓�

On substitution,

𝑦𝑛+1 = 𝑦𝑛 + ℎ𝑓 + ℎ2

2𝑓𝑡(𝑡𝑛,𝑦𝑛) + ℎ2

2𝑓𝑦𝑓 + 0(ℎ3) (3)

Thus, model equation (2) can be equated with the fundamental equation (3) derived for 𝑦𝑛+1;

�

𝑎 + 𝑏 = 1𝑏𝑝 = 1

2�

𝑏𝑞 = 12��

Case 1: 𝑎 = 12� , 𝑏 = 1

2� , 𝑝 = 𝑞 = 1

𝑦𝑛+1 = 𝑦𝑛 + ℎ2

[𝑓(𝑡𝑛,𝑦𝑛) + 𝑓(𝑡𝑛 + ℎ, 𝑦𝑛 + ℎ𝑦𝑛′)] + 0(ℎ3)

Write differently,

𝑦𝑛+1∗ = 𝑦𝑛 + ℎ𝑓(𝑥𝑛,𝑦𝑛) predictor full step

�𝑦𝑛+1 = 𝑦𝑛 + ℎ

2[𝑓(𝑡𝑛,𝑦𝑛) + 𝑓(𝑡𝑛 + ℎ, 𝑦𝑛+1∗ )]

� corrector full step

Look at graphically,

Therefore, first predict 𝑦𝑛+1∗ and then correct as 𝑦𝑛+1 by taking the average of the slopes, 𝑘1 and 𝑘2 ∶ 𝑦𝑛+1 = 𝑦𝑛 + ℎ𝑘�

𝑦𝑛

𝑦𝑛+1

𝑦𝑛+1

𝑡𝑛 𝑡𝑛+1 𝑡

𝑛 𝑘1

𝑘2

(𝑘1 + 𝑘2)2� = 𝑘

𝑦

𝑘1 𝑘2

𝑘1 𝑘2

3

- This method is called Heun’s improved method or modified Euler method or Runga-Kutta (RK-2) method.

Case 2: 𝑎 = 0 , 𝑏 = 1, 𝑝 = 𝑞 = 12�

𝑦𝑛+1 = 𝑦𝑛 + ℎ �𝑓 �𝑡𝑛 + ℎ2

,𝑦𝑛 + ℎ2𝑦𝑛′��

Write differently,

𝑦𝑛+1 2�∗ ≡ 𝑦𝑛 + 1

2� ℎ𝑦𝑛′ (Euler predict 12� step)

𝑦𝑛+1 ≡ 𝑦𝑛 + ℎ �𝑓 �𝑡𝑛 + ℎ2� ,𝑦𝑛+1 2�

∗ �� : Mid − point corrector full step

Graphically,

�

⎭⎪⎬

⎪⎫

(1) Start with slope 1 Reach half step.

(2) Calculate slope 2 (3)Re − start with the slope

2 to reach full step.

- This method is called midpoint method.

3. RK-4

𝑦𝑛+1 == 𝑦𝑛 + ℎ6

[𝑘1 + 2𝑘2 + 2𝑘3 + 𝑘4] + 0(ℎ5)

avg of four slopes with

more weights at mid points.

𝑘1 = 𝑓(𝑥𝑛, 𝑦𝑛)

𝑘2 = 𝑓�𝑥𝑛 + ℎ2� ,𝑦𝑛 + 𝑘1 ℎ 2� � 𝑜𝑟 𝑓(𝑥𝑛+1 2� ,𝑦𝑛+1 2�

∗ )

𝑦𝑛+1 2�∗ ∶ 𝐸𝑢𝑙𝑒𝑟 𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑜𝑟 1

2� 𝑠𝑡𝑒𝑝

𝑘3 = 𝑓�𝑥𝑛 + ℎ2� ,𝑦𝑛 + 𝑘2 ℎ 2� � 𝑜𝑟 𝑓(𝑥𝑛+1 2� ,𝑦𝑛+1 2�

∗∗ )

𝑦𝑛

𝑌𝑛+1 2�∗

𝑌𝑛+1∗

𝑡𝑛 𝑡𝑛+1 2� 𝑡𝑛+1 (𝑡𝑛 + ℎ)

𝑡

1

2

𝑦

4

𝑦𝑛+1 2�∗∗ ∶ 𝐵𝑎𝑐𝑘𝑤𝑎𝑟𝑑 𝐸𝑢𝑙𝑒𝑟 1

2� 𝑠𝑡𝑒𝑝 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛

𝑘4 = 𝑓(𝑥𝑛 + ℎ,𝑦𝑛 + 𝑘3ℎ): 𝑀𝑖𝑑 𝑝𝑜𝑖𝑛𝑡 𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑜𝑟 𝑓𝑢𝑙𝑙 𝑠𝑡𝑒𝑝

𝑦𝑛+1∗∗∗

Write differently,

𝑦𝑛+1 2�∗ = 𝑦𝑛 + ℎ

2� 𝑓(𝑥𝑛,𝑦𝑛) ∶ 𝐸𝑃 12 � 𝑠𝑡𝑒𝑝

𝑦𝑛+1 2�∗∗ = 𝑦𝑛 + ℎ

2� 𝑓 �𝑥𝑛+1 2 � ,𝑦𝑛+1 2 �∗ � ∶ 𝐵𝐸 1

2 � 𝑠𝑡𝑒𝑝 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑜𝑟

𝑦𝑛+1∗∗∗ = 𝑦𝑛 + ℎ𝑓 �𝑥𝑛+1 2 � ,𝑦𝑛+1 2 �∗∗ � :𝑀𝑖𝑑 𝑝𝑜𝑖𝑛𝑡 𝑝𝑟𝑒𝑑𝑖𝑐𝑡𝑜𝑟 𝑓𝑢𝑙𝑙 𝑠𝑡𝑒𝑝.

𝑦𝑛+1 = 𝑦𝑛 +ℎ6�𝑓(𝑥𝑛,𝑦𝑛) + 2𝑓(𝑥𝑛+1 2 � ,𝑦𝑛+1 2 �

∗ ) + 2𝑓 �𝑥𝑛+1 2 � ,𝑦𝑛+1 2 �∗∗ �+ 𝑓(𝑥𝑛+1,𝑦𝑛+1∗∗∗ )�

∶ 𝑆𝑖𝑚𝑝𝑠𝑜𝑛 𝑓𝑢𝑙𝑙 𝑠𝑡𝑒𝑝 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑜𝑟.

Graphically,

𝑘� = (𝑘1 + 2𝑘2 + 2𝑘3 + 𝑘4) 6⁄

Notes: (1) RK-4 is a 4th order accurate method!

(2) The method is explicit (does not require iterations to calculate 𝑦𝑛+1). You can march ahead from 𝑥𝑛 𝑡𝑜 𝑥𝑛+1. Yet, considering that it makes use of the value 𝑦𝑛+1 at 𝑥𝑛+1 ahead of 𝑥𝑛 via a predictor step, the method possesses some ‘characteristics’ of an implicit method. Therefore, the method also produces a stable solution at a relatively larger step size.

(3) Coding (program) is simple.

𝑘4

𝑘3

𝑘2

𝑘3

𝑘2 𝑘1

𝑥𝑛 𝑥𝑛+1 2� 𝑥𝑛+1 ℎ

2�

ℎ2�

𝑋

𝑌𝑛+1∗∗∗

𝑌𝑛+1 2�∗

𝑌𝑛+1 2�∗∗

𝑦𝑛

𝑌

𝑦𝑛+1

𝑘�

5

(4) Even if two or more number of simultaneous ODEs are coupled, the method is explicitly applied w/o requiring iterations,

(5) Considering that the method is explicit, non-linearity is not an issue.

See example below:

Solve: 𝑑𝑦1𝑑𝑡

= −100𝑦1 with choose 𝑦1(0) = 2 ℎ = Δ𝑡 = 0.02

𝑑𝑦2𝑑𝑡

= 2𝑦1 − 𝑦2 𝑦2(0) = 1

Write alternatively, solve 𝑑𝑦𝑑𝑡

= 𝑓(𝑥,𝑦) ; 𝑦(𝑡, 0) = 𝑦0

or �𝑦1′

𝑦2′� = �−100 0

2 −1� � 𝑦1 𝑦2� ; �

𝑦1 𝑦2�0

= �21�

or

�𝑦1′

𝑦2′� = �−100𝑦1

2𝑦1 − 𝑦2�

(1) 𝑘�1 = 𝑓(̅𝑡𝑛,𝑦𝑛)

𝑘1 = �𝑘1′

𝑘1′′� = �−100 × 2

2 × 2 − 1� = �−2003 �

(2) 𝑘2 = �𝑘2′

𝑘2′′� ⇒ 𝑘�2 = 𝑓(̅𝑡𝑛+1 2 � ,𝑦𝑛 + 1

2𝑘�1ℎ)

= 𝑓 �0.01, 2 +

12

(−200) × .02

0.01, 1 +12

(3). 02� = 𝑓 �0.01, 0

0.01, 1.03� = � 0−1.03�

(3) 𝑘3 = �𝑘3′

𝑘3′′� ⇒ 𝑘�3 = 𝑓(𝑡𝑛+1 2 � ,𝑦𝑛 + 1

2𝑘�2ℎ)

= 𝑓 �0.01, 2 − 0 ×

. 022

0.01, 1 − 1.03 ×0.02

2

� = 𝑓 �0.01, 20.01, 0.9897� = 𝑓 � −100 × 2

2 × 2 − 0.9897�

= � −2003.0103�

(4) 𝑘4 = �𝑘4′

𝑘4′′� ⇒ 𝑘�4 = 𝑓(𝑡𝑛+1,𝑦𝑛 + ℎ𝑘�3)

6

= 𝑓 �0.02, 2 − 0.02 × 2000.02, 1 + 0.02 × 3.0103� = 𝑓 �0.02, −2

0.02, 1.060206�

= 𝑓 � −100 × −22 × −2 − 1.060206� = � 200

−5.060206�

�𝑦1𝑦2� = �2

1� + 0.02

6� −200 + 2 × 0 − 2 × 200 + 2003 − 2 × 1.03 + 2 × 3.0103 − 5.060206�

= �0.666671.00633� 𝐴𝑛𝑠.

⇒How to choose h? Choose a reasonable ‘h’ depending on the physics of the problem. Solve and

re-do the calculation for ℎ 2 � and check the convergence. Sometimes, one may have to increase the step-size if the accuracy can be relaxed.

⇒ There are R-K-Mersen or R-K-Fehlberg automatic step-size control method where ‘h’ is

adjusted(𝑒𝑖𝑡ℎ𝑒𝑟 ℎ → 2ℎ 𝑜𝑟 ℎ → ℎ2 � ) depending upon the error in the RK-4 solution relative

to that from error in a higher order method proposed by Mersen or Fehlberg.

1

Lecture #18

Example 1 (RK-4)

Consider a special (diameter = 0.1cm) pellet made of steel k = 40 𝑊 𝑚−𝐾, 𝜌 = 8000 𝑘𝑔 𝑚3, 𝐶𝑝 = 400 𝐽 𝑘𝑔 − 𝑘).⁄ ⁄⁄ The initial temperature of the pellet is 300 K. It is immersed in a large oil tank at 400 K. The convective heat transfer coefficient, h at the sphere surface is 3000 𝑤 𝑚2 − 𝑘⁄ . Assume that there is no radial temperature gradient inside the pellet. Use R-K-4 numerical technique to solve the governing first order energy balance equation for predicting temperature of the sphere at 𝑡 = 10 min after it was immersed in the oil tank. Use Δ𝑡 = 10 𝑠. Repeat calculations for Δ𝑡 = 5 𝑠. Draw graphical depiction of the slopes comprising all stages of the RK method. Do calculations upto 5 digits after decimal. Calculate the rate of temperature decrease at 𝑡 = 0, 0.5 and 1.0 min.

The energy balance ( transient/unsteady equation)

𝜌𝐶𝑝𝑑𝑇𝑑𝑡

= +ℎ𝐴(𝑇 − 𝑇0)

(𝑘𝑔 𝑚3⁄ 𝐽 𝑘𝑔 − 𝑘⁄ . 𝑘𝑠) = 𝐽 𝑆 − 𝑚2 − 𝑘⁄ (400 − 𝑇) � 4𝜋𝑅2

43� 𝜋𝑅3

�𝑚2

𝑚3

8000 × 400 × 𝑑𝑇𝑑𝑡

= +3000 × 30.05

(400 − 𝑇)

𝑑𝑇𝑑𝑡

= +0.05625(400 − 𝑇) = 𝑓(𝑡,𝑇)

Δ𝑇 = ℎ = 10 or 5 𝑠

𝑘1 = 𝑓(0, 300) = +0.05625(400 − 300) = +5.6250

𝑘2 = 𝑓(5, 300 + 5.625 × 5) = +0.05625(400 − 328.125) = 4.0429

𝑘3 = 𝑓(5, 300 + 4.0429 × 5) = 0.05625(400 − 320.2145) = 4.4879

↓ 𝐴𝑠

𝑉𝑠

328.125

320.2145

𝑡 = 0 𝑇 = 300 𝐾 𝑇(𝑡)

𝑅

𝑡

ℎ

𝑇𝑜 = 400 𝐾

𝑅 = 0.05𝑐𝑚

2

𝑘4 = 𝑓(10, 300 + 4.4879 × 10) = 0.05625(400 − 344.879) = 3.1005

𝑘 = 16� (5.6206 + 2 × 4.0429 + 2 × 4.4879 + 3.1005) = 4.29795

𝑇 = 300 + 4.29795 × 10 = 𝟑𝟒𝟐.𝟗𝟕𝟗𝟓 𝑲 𝑨𝒏𝒔.

March for the next time steps 𝑡 = 20, 30, 40, 50, 60 𝑠, each time using the calculated values from the previous time steps(viz. 342.9795, etc). Repeat the entire calculations for Δ𝑡 = 5𝑠.

⇒ In a 20-25 min quiz or examination, such problem can be solved for one time step using calculator. One requires to write a code to solve such problem for a long time or several steps calculations.

⇒ Rate at 𝑡 = 0, 0.5 and 1.0 min

�𝑑𝑇𝑑𝑡�0

= 0.05625(400 − 𝑇) = 0.05625 × 100 = 5.625 𝑘 𝑠⁄

Plot the graph as done in the previous lecture and approximately draw the steps for

𝑘1,𝑘2,𝑘3,𝑘4,𝑘, and functional values (𝑇𝑛,𝑇𝑛+1 2�∗ ,𝑇𝑛+1 2�

∗∗ ,𝑇𝑛+1∗∗∗ ,𝑇𝑛+1) on y-axis.

⇒ From chemical engineering point of view, gradient in the steel was neglected considering a large thermal conductivity of the material, 𝑘 ≫ ℎ′𝑅′ or small Biot #. Therefore, the lump body approach was used to determine the sphere temperature, and the resulting energy balance equation is derived to be a 1st order ODE, else.

Ex2. Consider a steady-state 1D reactive flow of a gaseous species A in a long quartz tubular reactor (1D = D , length L)(Re>5000) radiated by UV light. The average velocity in

the tube is 𝑉. The species 𝐴 is converted into B by the 1st order homogeneous reaction (𝑟 = 𝑘𝐶𝐴), as it flows in the tube. The inlet concentration of A is 𝐶𝐴𝑜 . Determine the axial

concentration profiles of A in the reactor. The reaction is exothermic�−Δ𝐻; 𝑐𝑎𝑙 𝑚𝑜𝑙� �.

Inlet temperature is 𝑇𝑜

A species balance can be derived as follows, neglecting radial concentration profiles (Pe, r>>1)

344.8790

𝑇𝑜

𝐶𝐴𝑜

𝑉� 𝐶𝐴(𝑥) =? (𝑟 = 𝑘𝐶𝐴)

𝑥 = 𝐿

3

𝜕𝐶𝐴𝜕𝑡

+ 𝑉�𝑑𝐶𝐴𝑑𝑥

= 𝐷𝑥𝑑2𝐶𝐴𝑑𝑥2

− 𝑘𝐶𝐴

𝑜𝑟, 𝑉�𝑑𝐶𝐴𝑑𝑥

= 𝐷𝑥𝑑2𝐶𝐴𝑑𝑥2

− 𝑘𝐶𝐴 − (1)

Note the reaction is exothermic and the rate constant is temperature dependent. Write down the energy-balance eqn.

𝜌𝐶𝑝 �𝜕𝑇𝜕𝑡

+ 𝑉� 𝜕𝑇𝜕𝑥� = 𝑘 𝜕2𝑇

𝜕𝑥2+ 𝑘𝐶𝐴 (−Δ𝐻) (𝐶𝑎𝑙 𝑠 − 𝑚2� )

𝑉 𝑑𝑇𝑑𝑥

= 𝛼 𝑑2𝑇𝑑𝑥2

+ 𝑘𝜌𝐶𝑝

𝐶𝐴 (+Δ𝐻) − (2)

Further approximation can be made if mass/ thermal diffusion terms are reflected(𝑉∇𝐶𝐴 ≫ 𝐷∇2𝐶𝐴 and 𝑉∇𝑇 ≫ 𝛼∇2𝑇), considering a large axial packet # for mass as well as thermal transport. The equations (1-2) are modified as

𝑉 𝑑𝐶𝐴𝑑𝑥

= −𝑘𝐶𝐴 = −𝑘0 exp�−𝐸 𝑅𝑇� �𝐶𝐴

𝑉 𝑑𝑇𝑑𝑥

= 𝑘0𝑒𝑥𝑝�−𝐸 𝑅𝑇� �𝜌𝐶𝑝

(−Δ𝐻)𝐶𝐴

or 𝑑𝑦1𝑑𝑥

= 𝑓1(𝑦1,𝑦2) = −𝑘0𝑒𝑥𝑝�−𝐸 𝑅𝑦2� �𝑦1

𝑉= 𝐴𝑒𝑥𝑝 �− 𝐵

𝑦2� 𝑦1

𝑑𝑦2𝑑𝑥

= 𝑓2(𝑦1,𝑦2) =𝑘0𝑒𝑥𝑝�−𝐸 𝑅𝑦2� �(−Δ𝐻)𝑦1

𝜌𝐶𝑝= 𝐶𝑒𝑥𝑝 �− 𝐵

𝑦2� 𝑦1

The two 1st order equations are coupled and right hand side terms have non-linear dependence on 𝑦1 and 𝑦2. RK-4 can be applied without any problem, with the conditions

𝑋 = 0 𝑦1 = 𝑦10 𝑎𝑛𝑑 𝑦2 = 𝑦20. You can choose a step-size of Δ𝑥 = 𝐿10� .

Note: -If the axial diffusion terms are not neglected, the governing equations will assume a form of 2nd order ODE, which will be discussed in the next lecture as Boundary Value Problems.

- There are higher-order RK methods and even implicit RK method discussed in literature to solve ODE or initial value problem.

- There are two commonly used multi-step methods for solving ODEs (Note:RK-4 is a single step multi-stage method): Adams-Bashforth and Adams-Moulten. These

(𝑆𝑆)

4

methods use a polynomial on the derivatives of a function, passing through the single or multi-points. Adams-Bashforth (Explicit) 𝑑𝑦𝑑𝑡

= 𝑓(𝑦, 𝑡) ≡ 𝑓(𝑦(𝑡), 𝑡)

𝑡 = 0, 𝑦(0) = 𝑦0 Solve for y(t) = ?

�𝑡0𝑦0𝑦0′�

𝑡1𝑦1𝑦1′

𝑡𝑛𝑦𝑛𝑦𝑛′

𝑡𝑛+1𝑦𝑛+1𝑦𝑛+1′

: solve : gradient

We have,

𝑦𝑛+1 = 𝑦𝑛 + ∫ 𝑦′𝑑𝑡𝑡𝑛+1𝑡𝑛

extraplated value between tn and tn+1

Define 𝛼 = 𝑡−𝑡𝑛ℎ

:⇒ 𝑦𝑛+1 = 𝑦𝑛 + ℎ ∫ 𝑦′(𝛼)𝑑𝛼 (0 ≤ 𝛼 ≤ 1) − (1)10

Fit a jth order polynomial of 𝑦𝑛′ through 𝑋𝑛 and 𝑋𝑛−𝑗 (𝑗 + 1 pts preceding 𝑋𝑛)𝑗 using backward

difference formula (∇𝑦𝑖 = 𝑦𝑖 − 𝑦𝑖−1).

𝑦′(𝛼) = �1 + 𝛼∇ +𝛼(𝛼 + 1)

2!∇2 +⋯

𝛼(𝛼 + 1)⋯ (𝛼 + 𝑗 − 1)𝑗!

∇𝑗� 𝑦𝑛′ + 𝑅(𝜉)

(Thus, 𝑦′(𝛼) is the polynomial between �𝑋𝑛,𝑋𝑛−1,𝑋𝑛−2 ⋯𝑋𝑛−𝑗�, substituting in (1)

𝑦𝑛+1 = 𝑦𝑛 + ℎ �12∇ +

512

∇2 +38∇3 + ⋯𝐶𝑗∇j� 𝑦𝑛′ + ℎ� 𝑅(𝜉)𝑑𝑥

1

0

𝑗 = 0 ∶ you have fitted a constant polynomial of the derivative (𝑦′)

𝑦𝑛+1 = 𝑦𝑛 + ℎ𝑦𝑛′ + ℎ2

2𝑦′′�𝜉� (𝑦𝑛 < 𝜉 < 𝑦𝑛+1)

Recognize it as Euler 1st order formula/explicit

𝑗 = 1, you have fit a linear line(polynomial) between 𝑋𝑛𝑎𝑛𝑑 𝑋𝑛−1

𝑦𝑛+1 = 𝑦𝑛 + ℎ �𝑦𝑛′ +12∇𝑦𝑛′ �+

512

h3𝑓′′(𝜉)

𝑛 𝑛 + 1

𝑦𝑛′ 𝑦𝑛+1′

𝑛 + 1 𝑛 − 1 𝑛

𝑦𝑛−1′ 𝑦𝑛′

𝑦𝑛+1′

5

= 𝑦𝑛 + ℎ2

[3𝑓(𝑦𝑛) − 𝑓(𝑦𝑛−1)] − 2𝑛𝑑 𝑜𝑟𝑑𝑒𝑟 (𝐴𝑑𝑎𝑚 − 𝐵𝑎𝑠ℎ𝑓𝑜𝑟𝑡ℎ)

𝑗 = 2, you have a quadratic polynomial between (Xn, Xn−1, Xn−2)

Therefore, by taking different numbers of points preceding 𝑡𝑛 𝑜𝑟 𝑋𝑛 one can get a set of working formulae having difference 0(ℎ𝑛)

𝑦𝑛+1 = 𝑦𝑛 + ℎ∑ 𝛽𝑘𝑚𝑓𝑛+1−𝑚𝑘𝑚=1

𝛽1𝑚2𝛽2𝑚

12𝛽3𝑚

𝑚 = 113

23

2

−1−16

3

5

4

⋯

1𝑠𝑡 𝑜𝑟𝑑𝑒𝑟2𝑛𝑑 𝑜𝑟𝑑𝑒𝑟3𝑟𝑑 𝑜𝑟𝑑𝑒𝑟

Graphically,

⇒

⇒ From the above description, it is clear that the method has a ‘starting’ problem because only one condition is known: 𝑦(0) = 𝑦0. In practice one starts with Euler’s forward/explicit approach to calculate 𝑦1, then use 2nd order method to predict 𝑦2, and then use (𝑦0,𝑦1,𝑦2) to calculate 𝑦3. Use of the preceding ‘3’ data-points gives 3rd order accurate method.

⇒ Similar to the Adams-Bashforth (explicit), there is Adams-Moulton(Implicit) method based on forward difference interpolation formula. For details, refer the textbooks.

𝑦𝑛−2′ 𝑦𝑛−1′

𝑦𝑛′

𝑦𝑛+1′

𝑛 − 2 𝑛 − 1 𝑛 𝑛 + 1

𝑗 = 1(𝑙𝑖𝑛𝑒𝑎𝑟)

𝑗 = 0(𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡)

𝑦𝑛′

𝑡𝑛−1 𝑡𝑛+1 𝑡𝑛

𝑦𝑛∗

𝑡𝑛−1 𝑡𝑛 𝑡𝑛−1

𝑗 = 1

(𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑖𝑐)𝑗 = 1

𝑗 = 0

(𝑙𝑖𝑛𝑒𝑎𝑟)

1

Lecture #19

Instability and stiffness for a system of ODEs

In the previous lectures, we have shown that a system of ODEs (IV𝑃𝑠) 𝑦𝑖′ = 𝑓𝑖(𝑦1⋯𝑦𝑚) 𝑖 =1,2,⋯𝑚 with 𝑦𝑖(0) = 𝐶0 can be written as the coupled ODEs: 𝑦𝑖

′ = 𝐴𝑦 where A is the Jacobian matrix.

It is also clear that for the 𝑚 × 𝑚 matrix A, 𝐴𝜙𝑘 = 𝜆𝑘𝜙𝑘, where 𝜙𝑘 is the eigenvector corresponding to 𝜆𝑘 eigenvalue. Now make use of linear transformation. One can construct a

matrix S whose column is the eigenvectors of A: �𝜙1⋮𝜙𝑘� having the property 𝑆−1𝐴𝑆 = Λ, where

Λ is the diagonal matrix of the elements which are 𝜆𝑘 of A, as �𝜆1

⋱𝜆𝑘�. Then one can

write,

𝑧̅′(de�intion) = 𝑆−1𝑦�′ = 𝑆−1𝐴𝑦� = (𝑆−1𝐴𝑆)𝑆−1𝑦� = Λz�

or 𝑧𝑘′ = 𝜆k𝑧𝑘, 𝑘 = 1,2,⋯𝑛 − cannonical form.

where 𝑧𝑘′ are uncoupled ODEs. Therefore, all (stability) characteristics of single ODE will also be shown by the system of ODEs.

We have also seen earlier shown that a general solution to the system of ODEs can be written as 𝐶1𝑒𝜆1𝑡{ } + 𝐶2𝑒𝜆2𝑡{ } +⋯ ,where the terms in the parenthesis are the eigenvectors corresponding to the eignevalues. In such case growth or decay of the function depends on 𝜆𝑚𝑖𝑛 − 𝜆𝑚𝑎𝑥. A large numerical value of 𝜆𝑚𝑎𝑥 will cause the function to grow or decay at a faster rate than that corresponding to

𝑦 𝜆𝑚𝑖𝑛

𝜆𝑚𝑎𝑥

𝑡

2

𝜆𝑚𝑖𝑛. In other words, small change in ‘t’ will cause a large change/variation in 𝑦(𝜆𝑚𝑎𝑥). Solving such type of system of ODEs is numerically not trivial because a fine grid or step-size Δ𝑡 𝑜𝑟 ℎ is required to accurately predict change in 𝑦(𝜆𝑚𝑎𝑥), whereas a coarse grid can be used to predict 𝑦(𝜆𝑚𝑖𝑛).

- Such equations are called stiff ODEs. In general,

Stiff ratio(𝑆𝑅) = 𝜆𝑚𝑎𝑥𝜆𝑚𝑖𝑛

> 10 represents a system

of stiff ODEs. - Examples:

(1) A common example is the bounding layer problem in hydrodynamics or heat or mass transport. Functional value changes sharply within the boundary layer, from surface to the region far from the surface, in the potential region.

(2) A combination of series and parallel chemical reactions with low and high rate constants:

ex: 𝑑𝑦𝑑𝑡

= �−100 02 −1� 𝑦 ; 𝑦(0) = [2 1]𝑇

on coefficient matrix ‘A’: (−100− 𝜆)(−1− 𝜆) = 0 ⇒ 𝜆1 = −100,𝜆2 = −1

𝜆1 = −100 ∶ �−100 02 −1� �

𝑌1𝑌2� = �𝑌1𝑌2

� (𝑁𝑜𝑡𝑒: 𝑆𝑅 = 100)

2𝑋1 = −99𝑋2 ⇒ 𝑌(1) = �1

−299� �

𝜆2 = −1 ∶ �−100 02 −1� �

𝑌1𝑌2� = �𝑌𝑌2

�

−100𝑌1 = −𝑌1 and 2𝑌1 − 𝑌2 = −𝑌2 ⇒ 𝑌(2) = �01�

GS: 𝑌 = 𝐶1𝑒−100𝑡 �1

−299� �+ 𝐶2𝑒−𝑡 �

01�

or �𝑌1 = 𝐶1𝑒−100𝑡

𝑌2 = −299𝐶1𝑒−100𝑡 + 𝐶2𝑒−𝑡

�

Apply: 𝑌 = � 2𝑒−100𝑡10399𝑒−𝑡 − 4

99𝑒−100𝑡� 𝐴𝑛𝑎𝑙𝑦𝑡𝑖𝑐𝑎𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛.

𝑦

Δ𝑡2

t Δ𝑡1

−1

100

3

To solve numerically, a fine grid size Δ𝑡 will be initially

required to accurately predict a fast decay in the functional

value of 𝑦1, which can be relaxed, or step-size can be

increased during the later calculations.

In other words, time change in 𝑦2 is gradual (slow). Therefore, one would have used a coarse size for 𝑦2, but for a rapid change in 𝑦1 during initial time of calculations requiring fine grid sizes.

⇒ SKG’s book has given one good example of stiff equations

𝑑𝑦1𝑑𝑡

= 77.27(𝑦2 − 𝑦1𝑦2 + 𝑦1 − 8.375 × 10−6𝑦12)

𝑑𝑦2𝑑𝑡

= −𝑦2−𝑦1𝑦2+𝑦377.23

𝑦(0) = [4 1.1 4]𝑇

𝑑𝑦3𝑑𝑡

= 0.161(𝑦1 − 𝑦3)

In this case, stiffness changes with time, requiring adaptive mesh sizes: coarse when 𝑓𝑛 values slowly change and fine when they change rapidly.

If you use one fixed fine step-size throughout the calculations, it is a wastage of CPU time.

Gear’s Technique: Used to solve stiff ODEs, based on multiple steps and predictor-corrector approach:

Predictor(𝑒𝑥𝑝𝑙𝑖𝑐𝑖𝑡): 𝑦𝑛+1 = 𝛼1𝑦𝑛 + 𝛼2𝑦𝑛−1 + ⋯𝛼𝑘𝑦𝑛−(𝑘−1) + ℎ𝛽0𝑦𝑛

′

Corrector(𝑖𝑚𝑝𝑙𝑖𝑐𝑖𝑡): 𝑦𝑛+1 = 𝛼1𝑦𝑛 + 𝛼2𝑦𝑛−1 + ⋯𝛼𝑘𝑦𝑛−(𝑘−1) + ℎ𝛽0𝑦𝑛+1′

You should note that this method also has a "starting" problem, similar to the previously discussed multi-step methods.

𝑦

2.0

1.0 𝑦2

𝑦1 𝑡 1 2

𝑃𝑙𝑜𝑡:

𝑓𝑖𝑛𝑒 𝑚𝑒𝑠ℎ 𝑠𝑡𝑒𝑝 − 𝑠𝑖𝑧𝑒 𝑖𝑠 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑⁄

𝑓𝑖𝑛𝑒 𝑚𝑒𝑠ℎ 𝑠𝑡𝑒𝑝 − 𝑠𝑖𝑧𝑒 𝑖𝑠 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑⁄ 𝑡

10−2

106

𝑦

4

Predictor Table

𝑘 𝛼1 𝛼2 𝛼3 ⋯ 𝛼6 𝛽0

123⋮⋮6

10

−32�

01

62�

00

−12�

⋯⋯⋯

000

12

62�

Corrector Table

𝑘 𝛼1 𝛼2 𝛼3 ⋯ 𝛼6 𝛽0

123⋮⋮6

14

3�18

11�

0−1

3�−9

11�

00

211�

⋯⋯⋯

000

12

3�6

11�

Quiz II

1

Lecture #20 BVP (Boundary Value Problems)/2nd order ODEs

In this lecture, we will learn how to apply finite difference (direct) methods to solve BVPs. Such problems are common in 1 D steady-state heat and mass transport in solid or fluid. The general form of the BVP assumes a 2nd order ODE:

𝑑2𝑦𝑑𝑥2

= 𝑓(𝑥,𝑦,𝑦′) with two boundary conditions

required at two ends of the domain �𝑥 = 0 and 𝑥 = 𝐿 𝑜𝑟 (0,1)�.

A general form of BVP can also be written as:

𝐴(𝑥) 𝑑2𝑦

𝑑𝑥2+ 𝐵(𝑥) 𝑑𝑦

𝑑𝑥+ 𝐶(𝑥)𝑦 + 𝐷(𝑥) + 𝐸 = 0; (𝐴(𝑥) ≠ 0)

The required boundary conditions at two ends of the domain can assume any form: Drichlet or Danckwarts or Neumann. That is, either functional value or gradient or mixed condition can be specified:

𝑦 = 𝑐, or 𝑦′ = c, or 𝑎𝑦′ + 𝑏𝑦 = 𝑐.

Let us solve a general BVP equation:

𝐴(𝑥) 𝑑2𝑦

𝑑𝑥2+ 𝐵(𝑥) 𝑑𝑦

𝑑𝑥+ 𝐶(𝑥)𝑦 + 𝐷(𝑥) + 𝐸 = 0;

Step1: divide the domain into N equal steps/grids or grids. Therefore, there are (N+1) nodes.

Step 2: discretize the equation or each terms of the equation at ‘ith’ grid. In this course, we will discretize the terms using 2nd order method (recall numerical differentiation).

𝐴(𝑥𝑖)𝑦𝑖−1 − 2𝑦𝑖 + 𝑦𝑖+1

ℎ2+ 𝐵(𝑥𝑖)

𝑦𝑖+1 − 𝑦𝑖−12ℎ

+ 𝐶(𝑥𝑖)𝑦𝑖 + 𝐷(𝑥𝑖) + 𝐸 = 0

Step3: Arrange the terms:

�𝐴(𝑥𝑖)ℎ2

− 𝐵(𝑥𝑖)2ℎ

� 𝑦𝑖−1 − �2𝐴(𝑥𝑖)ℎ2

− 𝐶(𝑥𝑖)� 𝑦𝑖 + �𝐴(𝑥𝑖)ℎ2

+ 𝐵(𝑥𝑖)2ℎ

� = −𝐷(𝑥𝑖) − 𝐸

or 𝑎𝑖𝑦𝑖−1 + 𝑏𝑖𝑦𝑖 + 𝑐𝑖𝑦𝑖+1 = 𝑑𝑖 ← discretized form of equation

0(ℎ2) 0(ℎ2)

0 𝑖 − 1 𝑁 − 1 𝑁 1 2 𝑖 𝑖 + 1

2

In principle, this equation should be valid for 𝑖 = 0⋯𝑁. However, there may be problem (!) at the boundary nodes 𝑖 = 0 and 𝑁, because 𝑦−1 and 𝑦𝑁+1 may not be known or defined or calculated. You should carefully inspect the given boundary conditions and decide to apply the discretized equation over (𝑖 = 0⋯𝑁) 𝑜𝑟 (𝑖 = 1⋯𝑁)𝑜𝑟 (𝑖 = 1⋯𝑁 − 1)𝑜𝑟 (𝑖 = 0⋯𝑁 − 1) so that the number of equations is the same as that many unknowns (𝑦𝑖). For the moment, let us take the simplest boundary conditions, 𝑦(0) = 𝑦0 and 𝑦(𝑁) = 𝑦𝑁.

Apply the discretized equation over 𝑖 = 1⋯𝑁 − 1 as

�

𝑖 = 1: 𝑎1𝑦0 + 𝑏1𝑦1 + 𝑐1𝑦2 = 𝑑1 𝑎2𝑦1 + 𝑏2𝑦2 + 𝑐2𝑦3 = 𝑑2

⋮ 𝑎𝑁−2𝑦𝑁−3 + 𝑏𝑁−2𝑦𝑁−2 + 𝑐𝑁−2𝑦𝑁−1 = 𝑑𝑁−2𝑁 − 1: 𝑎𝑁−1𝑦𝑁−2 + 𝑏𝑁−1𝑦𝑁−1 + 𝑐𝑁−1𝑦𝑁 = 𝑑𝑁−1 ⎭

⎪⎬

⎪⎫

Step 4: Apply the BCs and bring the terms containing y0 and yN to RHS:

�

𝑏1𝑦1 + 𝑐1𝑦2 = (𝑑1 − 𝑎1𝑦0) = 𝑑1′ 𝑎2𝑦1 + 𝑏2𝑦2 + 𝑐2𝑦3 = 𝑑2

⋮ 𝑎𝑁−2𝑦𝑁−3 + 𝑏𝑁−2𝑦𝑁−2 + 𝑐𝑁−2𝑌𝑁−1 = 𝑑𝑁−2

𝑎𝑁−1𝑦𝑁−2 + 𝑏𝑁−1𝑦𝑁−1 = (𝑑𝑁−1 − 𝑐𝑁−1𝑌𝑁) = 𝑑𝑁−1′ ⎭⎪⎬

⎪⎫

There are exactly (N-1) unknown and (N-1) equations. Notably, set of equations can be written as

𝐴𝑦 = 𝑏 where 𝐴 =

⎣⎢⎢⎡𝑏1 𝑐1 0𝑎2 𝑏2 𝑐2 0 ⋯0 0 𝑎𝑁−2 𝑏𝑁−2 𝑐𝑁−2

0 𝑎𝑁−1 𝑏𝑁−1⎦⎥⎥⎤

is the tridiagonal matrix. Step 5: Call Thomas Algorithm (learnt earlier) to invert such matrix: Tridiagonal (𝑁 − 1,𝑎, 𝑏, 𝑐,𝑑,𝑌)

You should be able to clearly recognize that while preparing to call tridiagonal matrix, the following coefficients were changed:

𝑑′1 = 𝑑1 − 𝑎1𝑌0 ∶ 1st row (𝑖 = 1) The other coefficients remained the same: (i = 2, N-2)

𝑎𝑖 =𝐴(𝑥𝑖)ℎ2

−𝐵(𝑥𝑖)

2ℎ; 𝑏𝑖 = 𝐶(𝑥𝑖) −

2𝐴(𝑥𝑖)ℎ2

# 𝑜𝑓 𝑒𝑞𝑛𝑠 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑠𝑜𝑓 𝑚𝑎𝑡𝑟𝑖𝑥 𝐴

𝑜𝑢𝑡𝑝𝑢𝑡 (𝑖 = 1,⋯𝑁 − 1)

3

𝑐𝑖 =𝐴(𝑥𝑖)ℎ2

+𝐵(𝑥𝑖)

2ℎ; 𝑑𝑖 = −𝐷(𝑥𝑖) − 𝐸 ;

Also, dN−1 was changed as d′N−1 = dN−1 − cN−1YN ∶ last row �i = N − 1�

⇒ Let us consider another example where the BVP equation is the same. At boundary, 𝑦(0) =𝑦0(also same as before). However, at the other boundary, gradient is specified instead of functional value: 𝑦𝑁′ = 0. The boundary condition 𝑦𝑁′ can be discretized as 𝑌𝑁+1−𝑌𝑁−1

2ℎ= 0 �0(ℎ2)� 𝑜𝑟 𝑌𝑁+1 = 𝑌𝑁−1

Considering that YN is NOT known in this case, the coefficients (𝑎𝑖 , 𝑏𝑖, 𝑐𝑖,𝑑𝑖) are extended to the rows: 𝑖 = 1 … . .𝑁, 𝑖. 𝑒., the matrix A will have ‘N’ rows and the 1st & last row will be modified as 𝑑1 = 𝑑1 − 𝑎1𝑌0 (same as before) and 𝑎𝑁 = 𝑎𝑁 + 𝑐𝑁; why ? because the Nth equation aNYN−1 + bNYN + cNYN+1 = dN is modified as aNYN−1 + bNYN + cNYN−1 = dN on substitution of the bc. Finally, call Tridiagonal (𝑁,𝑎, 𝑏, 𝑐,𝑑,𝑌); i = 1, N

⇒ It is recommended that you should do your best to keep consistency between the order of methods used to discretize the equation and boundary conditions. Also note that, 2nd order discretization method is reasonably good/accurate for engineering calculations. If you use a higher order method, the coefficient matrix A will no more be tridiagonal that is easy and fast to invert.

Ex: Consider the steady-state one dimensional heat conduction in a fine (length L, cross-section A, perimeter P), whose one end is at a constant temperature 𝑇𝑓 and the other end is insulated. The ambient

temperature is 𝑇𝑎 and the convective heat transfer coefficient at the fix surface is ℎ.

Solve the steady-state temperature profiles in the fin.

L

An energy balance over ′𝐴Δ𝑥′ control volume will result in the following differential equation:

ρCp �∂T∂t

+ V.∇T� = k∇2T + S ∶ cals − m3�

𝑃 𝐴

𝑥

𝑇𝑓

Δ𝑥 𝑇𝑎 ℎ𝑓

𝑇(𝑥) =?

SS Solid −h(T − Ta)Cal

s − m2� . (PΔx)(A Δx)

𝑚3

𝑚2

4

Note that there is no source or sink of heat as such in the fin. However, considering 1D axial heat transfer, heat dissipated from the fin surface is homogenously distributed within the 𝐶𝑉: 𝑆 =−ℎ(𝑇 − 𝑇𝑎)𝑎𝑠 where 𝑎𝑠 is the surface area per unit volume of the fin.

⇒ 𝑘 𝑑2𝑇𝑑𝑥2

− ℎ(𝑇−𝑇𝑎)𝑃𝐴

= 0

or d2Tdx2

+ BT + C = 0 where B = −hP kA

; C = hPkA

Ta and choose Δh = L10�

BC: 𝑇(𝑥 = 0) = 𝑇𝑓 ; −𝑘∇𝑇 (𝑥 = 𝐿) = 0 𝑜𝑟 𝑑𝑇𝑑𝑥

= 0

Discretize the equation using the 2nd order method.

Soln: �𝑑2𝑇

𝑑𝑥2+ 𝐵𝑇 + 𝐶 = 0�

𝑇𝑖−1−2𝑇𝑖+𝑇𝑖+1ℎ2

+ 𝐵𝑇𝑖 + 𝐶 = 0

𝑇𝑖−1 + �𝐵 − 2ℎ2�𝑇𝑖 + 𝑇𝑖+1 = −𝐶 𝑖 = 1,𝑁

• Prepare the tridiagonal matrix

𝑎𝑖 = 1, 𝑏𝑖 = �𝐵 −2

ℎ2 � , 𝑐𝑖 = 1, 𝑑𝑖 = −𝐶, 𝑖 = 1,𝑁

Apply BC/discretize BC 𝑇0 = 𝑇𝑓(𝑖 = 0), 𝑇𝑁+1 = 𝑇𝑁−1 (𝑖 = 𝑁) Therefore, 1st & last row will be modified as 𝑑1 = 𝑑1 − 𝑎1𝑇𝑓

𝑎𝑁 = 𝑎𝑁 + 𝑐𝑁

How? 1st row is modified as �𝐵 − 2 ℎ2 �𝑇1 + 𝑇2 = −𝐶 − 𝑇𝑓

Last row is modified as 𝑇𝑁−1 + �𝐵 − 2 ℎ2 �𝑇𝑁 + 𝑇𝑁−1 = −𝐶

Therefore, you have the tridiagonal coefficient matrix A. Call the subroutine as:

Tridiagonal (𝑁,𝑎, 𝑏, 𝑐,𝑑,𝑦) (Note that the total number of equations is N)

As an example, take L = 1 cm, width of the �in = 0.1 cm, 𝑇𝑓 = 1000 𝐶, 𝑇𝑎 = 300 𝐶, ℎ =100𝑤 𝑚2𝐾, 𝑘 = 20𝑤 𝑚 − 𝐾.⁄ ⁄ Take Δℎ = 0.1 𝑐𝑚, write a programming code to obtain the solution and plot the temperature profiles:

𝐿 = 1 𝑐𝑚 𝑥 30𝑜𝐶

100𝑜𝐶

𝑇

1 𝑖 + 1 𝑖 0 𝑖 − 1 𝑁 − 1 𝑁

(𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑖𝑜𝑛)

5

Ex: Consider the catalytic oxidation of 𝑆𝑂2 into 𝑆𝑂3 under the constant temperature and pressure conditions, over a spherical (diameter 𝑑𝑝 = 1 mm) pt-dispersed porous carbon catalyst. 𝑆𝑂2 in air (1%)

diffuses (𝐷 = 1 × 10−9 m2 s⁄ ) into the pores of the catalyst and is oxidized to 𝑆𝑂3 by a pseudo-zeroth

order reaction, 𝑟 = 0.01moless−m3

. The convective film mass transfer coefficient is 𝑘𝑚 = 0.1 m s⁄ . Determine the steady-state concentration profiles of 𝑆𝑂2 within the pellet.

Soln:

𝜕𝐶𝐴𝜕𝑡

+ 𝑉.∇𝐶𝐴 = 𝐷𝑝𝑜𝑟𝑒∇2𝐶𝐴 − 𝑘

𝑜𝑟 Dporer2

ddr�r2 dCA

dr� + k = 0 (0 ≤ r ≤ rp)

continue in the next lecture………

𝑆𝑜𝑙𝑖𝑑 , 𝑆𝑆

𝐶(𝑟) =?

𝑟𝑝

𝑘𝑚

𝑆𝑂2(1%)𝐶𝑏 = 1%

𝑃,𝑇 = constant

1

Lecture #21

(continued…..from the previous lecture)

𝑜𝑟 𝐷𝑝𝑑2𝐶𝐴𝑑𝑟2

+ 2𝐷𝑝𝑟

𝑑𝐶𝐴𝑑𝑟

− 𝑘 = 0

𝑜𝑟 𝑑2𝑦𝑑𝑟2

+ 2𝑟𝑑𝑦𝑑𝑟

+ 𝐴 = 0 ; 𝐴 = �−𝑘𝐷𝑝�

�h = rpN�

BCs. 𝑟 = 0 𝑑𝐶𝐴

𝑑𝑟= 0 (symmetric)

𝑟 = 𝑟𝑝 − 𝐷𝑝𝑑𝐶𝐴𝑑𝑟

= 𝑘𝑚(𝐶𝐴 − 𝐶𝑏)

Discretized eqn:

𝑦𝑖+1−2𝑦𝑖+𝑦𝑖−1ℎ2

+ 2𝑟𝑖

𝑦𝑖+1−𝑦𝑖−12ℎ

+ 𝐴 = 0 − (1)

(𝑟𝑖 = 𝑖ℎ)

Discretized BCs:

𝑟𝑖 = 0 (𝑖 = 0)

𝑦1−𝑦−12ℎ

= 0

𝑦−1 = 𝑦1

𝑟𝑖 = 𝑟𝑝 − 𝐷𝑝𝑦𝑁+1−𝑦𝑁−1

2ℎ= 𝑘𝑚(𝑦𝑛 − 𝐶𝑏)

𝑦𝑁+1 = �𝑦𝑁−1 −2ℎ𝐷𝑝𝑘𝑚𝑦𝑛 + 2ℎ𝑘𝑚

𝐷𝑝𝐶𝑏�

Before you re-arrange the terms, you should realize that there is a clear discontinuity at 𝑟𝑖 = 0 (or 1st node). Therefore, you cannot proceed without removing discontinuity. One way

of doing this is to approximate 1𝑟�𝑑𝑦𝑑𝑟�𝑖=0

as �𝑑2𝑦𝑑𝑟2

�𝑖=0

because ∇y = 0 at r = 0

or 𝑦+1−2𝑦0+𝑦−1ℎ2

+ 2 𝑦1−2𝑦0+𝑦−1ℎ2

+ 𝐴 = 0

or �3 𝑑2𝑦𝑑𝑟2

+ 𝐴� = 0

or 3 𝑦1−2𝑦0+𝑦−1ℎ2

+ 𝐴 = 0 at 𝑖 = 0

Eq. (1) is applied over the next nodes, i.e., 𝑖 = 1⋯𝑁 without any discontinuity

0 1 2 𝑖 − 1 𝑖 𝑖 + 1 𝑁 − 1 𝑁

(𝑖 = 𝑁)

2

Arrange the terms now,

⎩⎪⎪⎪⎨

⎪⎪⎪⎧ 𝑦−1 + �𝐴 ℎ

23� − 2�𝑦0 + 𝑦1 = −𝐴 𝑟 = 0

� 1ℎ2− 1

𝑟1ℎ� 𝑦0 −

2ℎ2𝑦1 + � 1

ℎ2+ 1

𝑟1ℎ� 𝑦2 = −𝐴 𝑟 = 1

� 1ℎ2− 1

𝑟2ℎ� 𝑦1 −

2ℎ2𝑦2 + � 1

ℎ2+ 1

𝑟2ℎ� 𝑦3 = −𝐴 𝑟 = 2

⋮ � 1ℎ2− 1

𝑟𝑁ℎ� 𝑦𝑁−1 −

2ℎ2𝑦𝑁 + � 1

ℎ2+ 1

𝑟𝑁ℎ� 𝑦𝑁+1 = −𝐴 𝑟 = 𝑁

�

Apply the BCs (𝑖 = 0 and 𝑖 = 𝑁)

1𝑠𝑡 𝑟𝑜𝑤: �𝐴ℎ2

3� − 2�𝑦0 + 2𝑦1 = −𝐴

𝐿𝑎𝑠𝑡 𝑟𝑜𝑤: 2ℎ2𝑦𝑁−1 − � 2

ℎ2+ 2ℎ𝑘𝑚

𝐷𝑝× � 1

ℎ2+ 1

𝑟𝑁ℎ��𝑌𝑁 = −𝐴 − 2ℎ𝑘𝑚

𝐷𝑝� 1ℎ2

+ 1𝑟𝑁ℎ

� 𝐶𝑏

• From the ‘preparation of tridiagonal matrix’ point of view,

𝑎0 = 1, 𝑏0 = �𝐴 ℎ2

3� − 2� , 𝑐0 = 1, 𝑑0 = −𝐴

and

𝑎𝑖 = 1ℎ2� − 1

𝑟𝑖ℎ , 𝑏𝑖 = −2

ℎ2 , 𝑐𝑖 = 1

ℎ2+ 1

𝑟𝑖2ℎ

;𝑑𝑖 = −𝐴 ; 𝑖 = 1 − 𝑁

However, on the substitution of BCs,

𝑐0 = 𝑐0 + 𝑎0 (1st row @ i = 0)

� 𝑎𝑁 = 𝑎𝑁 + 𝑐𝑁, 𝑏𝑁 = 𝑏𝑁 − 𝑐𝑁

2ℎ𝑘𝑚𝐷𝑝

,

𝑑𝑁 = 𝑑𝑁 − 𝑐𝑁 . 2ℎ𝑘𝑚𝐷𝑝

.𝐶𝑏� 𝑙𝑎𝑠𝑡 𝑟𝑜𝑤

Therefore, 𝐴𝑋� = 𝐵. There are (𝑁 + 1) equations to solve (𝑁 + 1) unknown (𝑌0⋯𝑌𝑁) and A is the tridiagonal matrix.

call Tridiagonal (𝑵 + 𝟏,𝒂,𝒃, 𝒄,𝒅,𝒚) �𝑏0 𝑐0 0

0 𝑏𝑁 𝑐𝑁� 𝑖 = 0⋯𝑁

3

Write a programming code to solve the SS profiles of CA(r) for different rp = 0.25,0.5 and 1 mm, km = 0.01, 0.1, 0.5 m s⁄ and Dp = 10−7, 10−9, 10−11 m

2s� .

Plot the pro�iles and interpret the results.

Ex. Consider the steady-state 1D reactive flow (Re > 5000) of a gaseous species A in a 1ong quartz tubular reactor (1D = D, length = L) radiated by UV light. Velocity in the tube is 𝑉.

The species A is converted into B by the 1st order homogeneous reaction (𝑟 = 𝑘𝐶𝐴),𝑎𝑠 it flows in the tube. The inlet concentration of A is 𝐶𝐴𝑜 . Determine the axial concentration profiles 𝐶𝐴(𝑋) =?. The reaction takes place under isothermal conditions and Δ𝐻 ≈ 0.

Soln: 𝜕𝐶𝐴𝜕𝑡

+ 𝑉.∇𝐶𝐴 = 𝐷∇2𝐶𝐴 + (𝑟𝐴)

𝑉� 𝑑𝐶𝐴𝑑𝑋

= 𝐷 𝑑2𝐶𝐴𝑑𝑋2

− 𝑘𝐶𝐴

BC. 𝑋 = 0 𝐶𝐴 = 𝐶𝐴𝑜 , 𝑋 = 𝐿 ∇𝐶𝐴 = 0 or 𝑑𝐶𝐴𝑑𝑋

= 0 (long tube approximation)

𝑑𝐶𝐴𝑑𝑋

+ 𝐵 𝑑2𝐶𝐴𝑑𝑋2

+ 𝐶𝐶𝐴 = 0, 𝐵 = 𝐷𝑉�� ; 𝐶 = 𝑘

𝑉�, Δℎ = 𝐿

𝑁�

𝑦𝑖+1 − 𝑦𝑖−12ℎ

+ 𝐵𝑦𝑖+1 − 2𝑦𝑖 + 𝑦𝑖−1

ℎ2 + 𝐶𝑦𝑖 = 0, 𝑖 = 1,𝑁

for increasingDp

𝐶𝐴𝑜 𝑉� 𝑋

𝐿 𝑋

Δ𝑋

CA(𝑋) =? (r = kCA)

𝑆𝑆

0 1 2 𝑖 − 1 𝑖 𝑖 + 1 𝑁 − 1 𝑁

4

Discretize BCs 𝑦0 = 𝐶𝐴𝑜; 𝑦𝑁+1−𝑦𝑁−12ℎ

= 0 ⇒ 𝑦𝑁+1 = 𝑦𝑁−1

Collect the terms

� 𝐵ℎ2− 1

2ℎ� 𝑦𝑖−1 − �2𝐵

ℎ− 𝐶� 𝑦𝑖 + � 𝐵

ℎ2+ 1

2ℎ� 𝑦𝑖+1 = 0, 𝑖 = 1,𝑁

Prepare the tridiagonal matrix:

𝑎𝑖 = � 𝐵ℎ2− 1

2ℎ�, , 𝑏𝑖 = �2𝐵

ℎ− 𝐶� , 𝑐𝑖 = � 𝐵

ℎ2+ 1

2ℎ� , 𝑑𝑖 = 0; 𝑖 = 1,𝑁

On substitution of BCs,

𝑑1 = 𝑑1 − 𝑎1𝐶𝐴𝑜 (i = 1, 1st row) 𝑎𝑁 = 𝑎𝑁 + 𝑐𝑁 (i = N, last row)

You have 𝐴𝑦� = 𝑏�, where A is the tridiagonal matrix.

𝑦 = 1,𝑁

Plot (numerical solutions using reasonable values for 𝐷,𝑉� ,𝑘,𝐶𝐴𝑜 ,𝐿)

𝐏𝐚𝐫𝐭𝐢𝐚𝐥 𝐃𝐢𝐟𝐟𝐞𝐫𝐞𝐧𝐭𝐢𝐚𝐥 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧𝐬 (𝐏𝐃𝐄𝐬)

- Parabolic equations

1D ∶ A𝜕𝑦𝜕𝑡

+ B𝜕2𝑦𝜕𝑋2

�or 𝐵𝑟𝜕𝜕𝑟�𝑟𝜕𝑦𝜕𝑟� or

𝐵𝑟2

𝜕𝜕𝑟�𝑟2

𝜕𝑦𝜕𝑟�� + C

𝜕𝑦𝜕𝑋

+ Dy + E = 0

Such time dependent 1D (spatial) equations are common in heat/mass/momentum transport. The PDE representing conservation of transport variables usually contains a convective term (𝑉.∇𝑇 or 𝑉.∇𝐶 or 𝑉.∇𝑉) and a diffusion term, �𝛼∇2𝑇 or 𝐷∇2𝐶 or 𝜈∇2𝑉�⃗ �, besides the unsteady-state or transient term

�𝜌𝐶𝑝𝜕𝑇𝜕𝑡

or 𝜕𝐶𝜕𝑡

or 𝜌 𝜕𝑉��⃗

𝜕𝑡�. Therefore,

𝐿 𝑋 0

𝐶𝐴𝑜

5

�

𝜕𝐶𝐴𝜕𝑡

+ 𝑉𝑋𝜕𝐶𝐴𝜕𝑋

= 𝐷𝜕2𝐶𝐴𝜕𝑋2

− 𝑘𝐶𝐴

or 𝜌𝐶𝑝 �𝜕𝑇𝜕𝑡

+ 𝑉𝑋𝜕𝑇𝜕𝑋� = 𝑘

𝜕2𝑇𝜕𝑋2

+ (𝑘𝐶𝐴)(−Δ𝐻)

or 𝜌 �𝜕𝑉�⃗𝜕𝑡

+ 𝑉𝑋𝜕𝑉𝑋𝜕𝑋

� = 𝜇𝜕2𝑉𝑋𝜕𝑋2

− �𝑑𝑝𝑑𝑋

� + 𝜌𝑔

or 𝜌𝐶𝑝𝜕𝑇𝜕𝑡

=𝑘𝑟2

𝜕𝜕𝑟�𝑟2

𝜕𝑇𝜕𝑟�

or 𝜕𝐶𝐴𝜕𝑡

=𝐷𝑟𝜕𝜕𝑟�𝑟𝜕𝐶𝐴𝜕𝑟

� − 𝑘𝐶𝐴 ⎭⎪⎪⎪⎪⎬

⎪⎪⎪⎪⎫

1 DParabolic

PDEequation.

are all examples of the time-dependent 1D parabolic PDE. You may refer the book by Ferziger for the exact mathematical definition of different types of PDE including parabolic elliptic or hyperbolic.

⇒ Why ‘Parabolic’? Because we march on time ‘endlessly’ covering the space.

⇒ Note: If you drop the time-dependent equation (or the unsteady-state or transient term), what do you get? A boundary value problem (BVP), which we covered in the previous lectures.

⇒ Message is clear. Solving 1D parabolic PDE is not very different from solving BVP. The procedure for discretization in the space remains the same as before; it is to be repeated on every time step as you march by refreshing the numerical values from the previous step. In fact, you may like to copy and paste the programming code for BVP and nest it within the time loop. The graphical representation of the numerical calculation is as follows.

𝑡 = 2

𝑡 = 1

𝑡 = 0 0 1

𝑋

𝑖 𝑖 − 1 𝑖 + 1

𝑖 + 1

𝑖 + 1

𝑖 − 1

𝑖 − 1

𝑖

𝑖

march on time

6

Take a general case:

𝐴 𝜕𝑦𝜕𝑡

+ 𝐵 𝜕𝑦𝜕𝑋

= 𝐶 𝜕2𝑦𝜕𝑋2

+ 𝐷 �or 𝐴 𝜕𝑦𝜕𝑡

= 𝑓(𝑦𝑋 ,𝑦𝑋𝑋 ,𝐷)�

�𝑡 = 0 𝑦 = 0 for all 𝐿 > 𝑋 > 0

𝑡 = 0+ 𝑋 = 0 𝑦 = 𝑦0 ; 𝑋 = 𝐿 𝜕𝑦𝜕𝑋

= 0�

Steps

(1) Discretize ‘X’ -terms as before. On time, apply trapezoidal rule for integration:

𝐴𝑦𝑖𝑡+1 − 𝑦𝑖𝑡

Δ𝑡=

12�−𝐵 �

𝑦𝑖+1 − 𝑦𝑖−12Δℎ

�𝑡− 𝐵 �

𝑦𝑖+1 − 𝑦𝑖−12Δℎ

�𝑡+1

�

�+𝐶 �𝑦𝑖+1 − 2𝑦𝑖 + 𝑦𝑖−1

ℎ2�𝑡

+ 𝐶 �𝑦𝑖+1 − 2𝑦𝑖 + 𝑦𝑖−1

ℎ2�𝑡+1

� + 𝐷

(Note: ‘D’ is a const. 𝐷𝑡+𝐷𝑡+1

2= 𝐷!) – This method is better known as Crank-Nicolson.

Also note that you have used a 2nd order accurate scheme on both time and X.

(2) Collect all terms on (𝑡 + 1) on the LHS and 𝑡 on the RHS as 𝐴𝑦(𝑡+1) = 𝑑(𝑡)

12�−

𝐶ℎ2

−𝐵

2Δℎ�𝑦𝑖−1

𝑡+1 + �𝐴Δ𝑡

+2𝐶ℎ2�𝑦𝑖 𝑡+1 +

12�−

𝐶ℎ2

+𝐵

2Δℎ�𝑦𝑖+1

𝑡+1

=12�𝐶ℎ2

+𝐵

2Δℎ�𝑦𝑖−1

𝑡 + �𝐴Δ𝑡−𝐶ℎ2�𝑦𝑖 𝑡 +

12�𝐶ℎ2

−𝐵

2Δℎ�𝑦𝑖+1

𝑡 + 𝐷 𝑖 = 1,𝑁

Note: All terms on RHS are known/calculated from the previous time step.

0

𝑖 − 1

𝑁 − 1 𝑁

𝑖 + 1

𝑖 + 1 𝑖 − 1 𝑖

𝑖 𝑁 − 1 𝑁

𝑡 + 1

𝑡 0 1

1 𝑀𝑎𝑟𝑐ℎ

𝑒𝑥𝑝𝑙𝑖𝑐𝑖𝑡 𝑖𝑚𝑝𝑙𝑖𝑐𝑖𝑡

𝑖𝑚𝑝𝑙𝑖𝑐𝑖𝑡 𝑒𝑥𝑝𝑙𝑖𝑐𝑖𝑡

1

Lecture #22

(… . . 𝒄𝒐𝒏𝒕𝒊𝒏𝒖𝒆)

(3) Apply BCs on both sides (𝑡 + 1) 𝑎𝑛𝑑 (𝑡):

𝑖 = 0 𝑦(𝑜) = 𝑦𝑜 for all 𝑡𝑠 (𝑡 + 1 𝑎𝑛𝑑 𝑡)

𝑖 = 𝑁 𝑦𝑁+1 = 𝑦𝑁−1 for all 𝑡𝑠 (𝑡 + 1 𝑎𝑛𝑑 𝑡)

1st row will change as:

�� 𝐴Δ𝑡

+ 𝐶ℎ2� 𝑦1𝑡+1 + 1

2�− 𝐶

ℎ2+ 𝐵

2Δℎ� 𝑦2𝑡+1 =

� 𝐶ℎ2

+ 𝐵2Δℎ

� 𝑦0 + � 𝐴Δ𝑡− 𝐶

ℎ2� 𝑦1𝑡 + 1

2� 𝐶ℎ2− 𝐵

2Δℎ� 𝑦2𝑡 + 𝐷

� (𝑖 = 1)

Last row will change as:

− 𝐶ℎ2𝑦𝑁−1𝑡+1 + � 𝐴

Δ𝑡+ 𝐶

ℎ2� 𝑦𝑁𝑡+1 = 𝐶

ℎ2𝑦𝑁−1𝑡 + � 𝐴

Δ𝑡− 𝐶

ℎ2� 𝑦𝑁𝑡 + 𝐷 (i = N)

The other rows (2⋯𝑁 − 1) will remain the same. Again the coefficient matrix is a tridiagonal one:

𝐴𝑦(𝑡+1) = 𝑑(𝑡)

You have prepared the tridiagonal matrix as follows:

𝑎𝑖 = −12� 𝐶ℎ2

+ 𝐵2Δℎ

� , 𝑏𝑖 = � 𝐴Δ𝑡

+ 𝐶ℎ2� , 𝑐𝑖 = 1

2�− 𝐶

ℎ2+ 𝐵

2Δℎ� ,𝑑𝑖 = 1

2� 𝐶ℎ2

+ 𝐵2Δℎ

� 𝑦𝑖−1 𝑡 +

� 𝐴Δ𝑡− 𝐶

ℎ2� 𝑦𝑖 𝑡 + 1

2� 𝐶ℎ2− 𝐵

2Δℎ� 𝑦𝑖+1

𝑡 + 𝐷; 𝑖 = 1,𝑁

on application of BCs 𝑑1 = 𝑑1 − 𝑎1𝑦0 (1st row , 𝑖 = 1) Note that y0is substituted for yi−1t before applying BC.

𝑎𝑁 = 𝑎𝑁 + 𝑐𝑁 (last row, i = N)

𝑑𝑛 = 𝐶ℎ2𝑦𝑁−1 𝑡 + � 𝐴

Δ𝑡− 𝐶

ℎ2� 𝑦𝑁𝑡 + 𝐷

∗

2

(4) Start calculations: RHS is known from 𝑡 = 0 (initial conditio

March on timeand call

𝐴𝑦(1) = 𝑏(0) 𝐴𝑦(2) = 𝑏(1)

⋮ 𝐴𝑦(𝑡+1) = 𝑏(𝑡)

Tridiagonal subroutine at every time step, & update 𝑦0 → 𝑦(1) → 𝑦(2) → ⋯

Tridiagonal (𝑁,𝑎, 𝑏, 𝑐,𝑑𝑡,𝑦𝑡+1)

Convince yourself that the SS values will be the same as solving the corresponding BVP by dropping the transient or unsteady-state term.

Ex:

Consider the SS flow of a liquid through a long tube. Reynolds number is 5000 and a radially flat velocity profile may be assumed. At certain time a tracer is injected into the liquid at the inlet. The dispersion coefficient of the tracer in the liquid may be assumed to be 𝐷 𝑐𝑚2 𝑠.⁄ Radial peclet number is large. Determine unsteady-state or transient axial concentration profiles, 𝐶𝐴(𝑡,𝑋) = ?

Soln:

𝜕𝐶𝐴𝜕𝑡

+ 𝑉.∇𝐶𝐴 = 𝐷∇2𝐶𝐴 + 𝑟

𝜕𝐶𝐴𝜕𝑡

+ 𝑉𝑋 . 𝜕𝐶𝐴𝜕𝑋

= 𝐷 𝜕2𝐶𝐴𝜕𝑋2

; (neglecting radial term)

𝑡 = 0, 𝐶𝐴 = 0 for all 𝐿 > 𝑥 > 0

0+, 𝑋 = 0; 𝐶𝐴 = 𝐶𝐴,𝑖𝑛 (Inlet concentration of tracer)

𝑖 = 1,𝑁

1 𝐷 𝑃𝑎𝑟𝑎𝑏𝑜𝑙𝑖𝑐

𝐵𝑉𝑃

𝑡

𝑦2 𝑦�0

Δ𝑋

CAo

t = 0 L

X V�

(no− reaction)

(Pure liquid)

3

𝑋 = 𝐿, ∇𝐶𝐴 = 0 (long − tube approximation) 𝑜𝑟 𝜕𝐶𝐴𝜕𝑋

= 0

𝜕𝐶𝐴𝜕𝑡

+ 𝑉𝑋 . 𝜕𝐶𝐴𝜕𝑋

= 𝐷 𝜕2𝐶𝐴𝜕𝑋2

; 𝑉𝑋 = 𝑉�

𝐶𝐴𝑖𝑡+1 − 𝐶𝐴𝑖

𝑡

Δ𝑡=

12�𝐷 �

𝐶𝐴,𝑖+1 − 2𝐶𝐴,𝑖 + 𝐶𝐴,𝑖−1

ℎ2 �𝑡+1

+ 𝐷 �𝐶𝐴,𝑖+1 − 2𝐶𝐴,𝑖 + 𝐶𝐴,𝑖−1

ℎ2 �𝑡

− 𝑉� �𝐶𝐴,𝑖+1 − 𝐶𝐴,𝑖−1

2ℎ�𝑡+1

− 𝑉� �𝐶𝐴,𝑖+1 − 𝐶𝐴,𝑖−1

2ℎ�𝑡�

�𝐷

2ℎ2+𝑉�

4ℎ�𝐶𝐴,𝑖−1

𝑡+1 − �1Δ𝑡−𝐷ℎ2� 𝐶𝐴,𝑖

𝑡+1 + �𝐷

2ℎ2−𝑉�

4ℎ�𝐶𝐴,𝑖+1

𝑡+1

= −�𝐷

2ℎ2+𝑉�

4ℎ�𝐶𝐴,𝑖−1

𝑡 − �1Δ𝑡−𝐷ℎ2� 𝐶𝐴,𝑖

𝑡 − �𝐷

2ℎ2−𝑉�

4ℎ�𝐶𝐴,𝑖+1

𝑡

𝑖 = 1,𝑁

𝑖 = 0: 𝐶𝐴,0𝑡 = 𝐶𝐴,0

𝑡+1 = 𝐶𝐴,𝑖𝑛 (Inlet BC)

𝑖 = 𝑁: 𝐶𝐴,𝑁+1

𝑡+1 = 𝐶𝐴,𝑁−1 𝑡+1 and 𝐶𝐴,𝑁+1

𝑡 = 𝐶𝐴,𝑁−1𝑡

* Prepare the tridiagonal matrix

𝑎𝑖 = �𝐷

2ℎ2+𝑉�

4ℎ� , 𝑏𝑖 = �

𝐷ℎ2

−1Δ𝑡� , 𝑐𝑖 = �

𝐷2ℎ2

−𝑉

4ℎ� , 𝑑𝑖

= −�𝐷

2ℎ2+𝑉

4ℎ�𝐶𝐴,𝑖−1

𝑡 − �1Δ𝑡−𝐷ℎ2� 𝐶𝐴,𝑖

𝑡 − �𝐷

2ℎ2−𝑉�

4ℎ�𝐶𝐴,𝑖+1

𝑡

Apply BCs:

𝑑1 = 𝑑1 − 𝑎1𝐶𝐴,𝑖𝑛 : Note that 𝐶𝐴,𝑖𝑛 is substituted for 𝐶𝐴,𝑖−1𝑡 in 𝑑1.

𝑎𝑁 = 𝑎𝑁 + 𝑐𝑁

𝑑𝑁 = −� 1Δ𝑡− 𝐷

ℎ2� 𝐶𝐴,𝑁

𝑡 − �𝐷ℎ2� 𝐶𝐴,𝑁−1

𝑡

You have 𝐴𝑦(𝑡+1) = 𝑑(𝑡) ; A ≡ Tridiagonal matrix with i = 1, N rows

𝑖 = 0

𝑖 = 0

𝑖 − 1 𝑖 𝑖 + 1 ℎ

𝑁

𝑁 𝑡 + 1

𝑡 ′Δ𝑡′

4

Call Tridiagonal (N, a, b, c, d, y)

Start with 𝑡 = 0: 𝐶𝐴(𝑖 = 0,𝑁) = 0 (pure solvent) or use a small concentration,𝐶𝐴𝑖 ≪ 𝐶𝐴,𝑖𝑛𝑙𝑒𝑡.

Solve for 𝑦(1) 𝑜𝑟 𝐶𝐴(1) 𝑎𝑡 𝑡 = 1, recalculate RHS 𝑑(2) and keep on marching on time till a

steady-state solution or at 𝑡 = 𝑡𝑓𝑖𝑛𝑎𝑙 is reached.

Plot (use L = 10 cm, CA,i = 0.001, CA,inlet = 1.0, V� = 0.1 cm s⁄ , D = 0.01 cm2 s⁄ )

� � ⇒ �𝜕𝐶𝜕𝑋

= 0�𝛼=𝐿

Note that

• Tridiagonal matrix is called at every time-step. • 𝑏� vector is updated with the recent most 𝐶𝐴 values. • gradient at the exit must be flat (consistent with BC) • There are two steps (ℎ & Δ𝑡). How do you choose them?

No one knows! In general Δ𝑡 < Δ𝑋𝑉

, Δ𝑋2

𝐷. Why? ?

One should choose a fixed Δ𝑋, and then refine ′Δ𝑡′ till there is a convergence in the solution.

Again, use Δ𝑋 = Δ𝑋2� , adjust ′Δ𝑡′ till the solution converges, and so forth!

Ex. Repeat the previous problem, assuming that some solute is irreversibly adsorbed at the tube-wall @ 𝑘𝐶𝐴 rate.

Species balance equation:

𝜕𝐶𝐴𝜕𝑡

+ 𝑉.∇𝐶𝐴 = 𝐷∇2𝐶𝐴 − (𝑘𝐶𝐴)𝑎, where a is the specific surface area per unit

volume of the tube.

or 𝜕𝐶𝐴𝜕𝑡

+ 𝑉� 𝜕𝐶𝐴𝜕𝑋

= 𝐷 𝜕2𝐶𝐴𝜕𝑋2

− 𝑘′𝐶𝐴;

1.0

0.001

0 𝑥 𝐿

𝑡

𝐶𝐴,𝑖𝑛 𝑉� 𝐶𝐴(𝑡,𝑋) =?

r = -kCA

Δ𝑋 L

X

5

𝐼𝐶: 𝑡 = 0 𝐶𝐴 = 𝐶𝐴,𝑖 ,

𝐵𝐶 0+, 𝑋 = 0; 𝐶𝐴 = 𝐶𝐴,𝑖𝑛, 𝑋 = 𝐿, ∇𝐶𝐴 = 0 ⇒ 𝜕𝐶𝐴𝜕𝑋

= 0

𝐶𝐴𝑖𝑡+1−𝐶𝐴𝑖

𝑡

Δ𝑡= 1

2�𝐷 �𝐶𝐴,𝑖+1−2𝐶𝐴,𝑖+𝐶𝐴,𝑖−1

ℎ2 �𝑡+1

+ 𝐷 �𝐶𝐴,𝑖+1−2𝐶𝐴,𝑖+𝐶𝐴,𝑖−1ℎ2

�𝑡− 𝑉� �𝐶𝐴,𝑖+1−𝐶𝐴,𝑖−1

2ℎ�𝑡+1

−

𝑉� �𝐶𝐴,𝑖+1−𝐶𝐴,𝑖−12ℎ

�𝑡− �𝑘′𝐶𝐴,𝑖

𝑡+1 + 𝑘′𝐶𝐴,𝑖𝑡 �� Note: This is the extra term.

Re-arrange,

�𝐷

2ℎ2+𝑉�

4ℎ�𝐶𝐴,𝑖−1

𝑡+1 − �1Δ𝑡−𝐷ℎ2

+𝑘′2�𝐶𝐴,𝑖

𝑡+1 + �𝐷

2ℎ2−𝑉�

4ℎ�𝐶𝐴,𝑖+1

𝑡+1

= −�𝐷

2ℎ2+𝑉�

4ℎ�𝐶𝐴,𝑖−1

𝑡 − �1Δ𝑡−𝐷ℎ2

+𝑘′2�𝐶𝐴,𝑖

𝑡 − �𝐷

2ℎ2−𝑉�

4ℎ�𝐶𝐴,𝑖+1

𝑡

𝑖 = 1,𝑁

𝐵𝐶𝑠: �𝑖 = 0, 𝐶𝐴,𝑜

𝑡 = 𝐶𝐴,𝑜𝑡+1 = 𝐶𝐴,𝑖𝑛 (for all time steps)

𝑖 = 𝑁, 𝐶𝐴,𝑁+1𝑡+1 = 𝐶𝐴,𝑁−1

𝑡+1 𝑎𝑛𝑑 𝐶𝐴,𝑁+1𝑡 = 𝐶𝐴,𝑁−1

𝑡 �

Prepare tridiagonal matrix:

𝑎𝑖 = �𝐷

2ℎ2+𝑉�

4ℎ� , 𝑏𝑖 = −�

1Δ𝑡−𝐷ℎ2

+𝑘′2� , 𝑐𝑖 = �

𝐷2ℎ2

−𝑉�

4ℎ� ,𝑑𝑖

= −�𝐷

2ℎ2+𝑉�

4ℎ�𝐶𝐴,𝑖−1

𝑡 − �1Δ𝑡−𝐷ℎ2

+𝑘′2�𝐶𝐴,𝑖

𝑡 − �𝐷

2ℎ2−𝑉�

4ℎ�𝐶𝐴,𝑖+1

𝑡

Apply BCs.

𝑑1 = −�𝐷ℎ2

+ 𝑉�

2ℎ� 𝐶𝐴,𝑖𝑛 − � 1

Δ𝑡− 𝐷

ℎ2+ 𝑘′

2� 𝐶𝐴,1

𝑡 − � 𝐷2ℎ2

− 𝑉�

4ℎ� 𝐶𝐴,2

𝑡 : i =1 (1st row)

(Note that this step is the same as 𝑑1 = 𝑑1 − 𝑎1𝐶𝐴,𝑖𝑛)

�𝑎𝑁 = 𝑎𝑁 + 𝑐𝑁

𝑑𝑁 = − 𝐷ℎ2𝐶𝐴,𝑁−1𝑡 − � 1

Δ𝑡− 𝐷

ℎ2+ 𝑘′

2� 𝐶𝐴,𝑁

𝑡 � 𝑖 = 𝑁 (𝑙𝑎𝑠𝑡 𝑟𝑜𝑤)

𝐴𝑦(𝑡+1) = 𝑑(𝑡); A is the tridiagonal matrix.

𝑖 = 0

0

𝑖 − 1 𝑖 𝑖 + 1 𝑁 − 1 𝑁

𝑡 + 1

𝑡 𝑖 − 1 𝑖 + 1 𝑁

1

6

Proceed as before with initial condition CA(i = 1, N) = CA,i. Evaluate d(t), solve for

y(t+1), evaluate d(t+1), solve for y(t+2) ⋯ etc. Subroutine Tridiagonal (N, a, b, c, d, y) is called at every time step. Plot qualitatively for the same conditions as before, use k = 0.1 s−1.

The axial concentration profiles for different ′ts′ will be the similar as before; however increase in concentration with ‘t’ will be less than before because of the reactive/adsorptive wall.

𝐶𝐴,𝑖𝑛

𝐶𝐴 𝑡2 𝑡1

𝑡1

𝑡2

(non − reactive wall)

(reactive wall)

𝑋

1

Lecture #23 Example: Consider a spherical (𝑑𝑝 = 0.1 cm) pellet made of steel (𝑘 = 40 W m − K ⁄ , ρ =8000 kg m3 , Cp⁄ = 400 J kg − K.⁄ The initial temperature of the pellet is 300 𝐾. It is immersed in a large oil tank at 400 K. The corrective heat transfer coefficient, ℎ𝑓 at the sphere

surface is 3000 𝑊 𝑚2𝐾.⁄ Solve for unsteady- state temperature profiles in the sphere. Choose Δ𝑡 = 1 𝑠. Determine the average temperature in the sphere and the rate of which the surface temperature decreases at 𝑡 = 20 𝑠.

* Recall: There is a discontinuity at 𝑖 = 0 (𝑟 = 𝑖ℎ)

However, the grad ∇𝑇 = 0 at 𝑟 = 0. Therefore, 1𝑟𝜕𝑇𝜕𝑟≈ 𝜕2𝑇

𝜕𝑟2

Energy balance equation:

𝜌𝐶𝑝𝜕𝑇𝜕𝑡

= 𝑘∇2𝑇 + 𝑆 (𝛼 = 𝑘𝜌𝐶𝑝� )

𝜕𝑇𝜕𝑡

= 𝛼𝑟2

𝜕𝜕𝑟�𝑟2 𝜕𝑇

𝜕𝑟� 𝑟𝑝 > 𝑟 ≥ 0

1𝐶: 𝑡 = 0 𝑇 = 300 𝐾(𝑇𝑜) for all 𝑟𝑝 ≥ 𝑟 ≥ 0

𝐵𝐶: 0+ 𝑟 = 0 ∇𝑇 = 0 𝑜𝑟 𝜕𝑇𝜕𝑟

= 0

𝑟 = 𝑟𝑝 − 𝑘 𝜕𝑇𝜕𝑟

= ℎ(𝑇 − 𝑇𝑓)

or 𝑘 𝜕𝑇𝜕𝑟

= ℎ(𝑇𝑓 − 𝑇)

𝑡

𝑡 = 0 300 𝐾

400 (𝑇𝑓) 𝑟

Δ𝑟 hf

rp

𝑑𝑞

2

The equation is modified as 𝜕𝑇𝜕𝑡

= 𝛼 𝜕2𝑇𝜕𝑟2

+ 2𝛼𝑟𝜕𝑇𝜕𝑟

Apply Crank-Nicholson discretization scheme:

𝑇𝑖𝑡+1 − 𝑇𝑖𝑡

Δ𝑡=

12�𝛼 �

𝑇𝑖+1 − 2𝑇𝑖 + 𝑇𝑖−1ℎ2

�𝑡+1

+ 𝛼 �𝑇𝑖+1 − 2𝑇𝑖 + 𝑇𝑖−1

ℎ2 �𝑡

+2𝛼𝑟𝑖�𝑇𝑖+1 − 𝑇𝑖−1

2ℎ�𝑡+1

+2𝛼𝑟𝑖�𝑇𝑖+1 − 𝑇𝑖−1

2ℎ�𝑡�

Arrange: � 𝛼2ℎ2

− 𝛼2𝑟𝑖ℎ

� 𝑇𝑖−1𝑡+1 − � 1Δ𝑡− 𝛼

ℎ2� 𝑇𝑖𝑡+1 + � 𝛼

2ℎ2+ 𝛼

2𝑟𝑖ℎ� 𝑇𝑖+1𝑡+1 = � 𝛼

2𝑟𝑖ℎ− 𝛼

2ℎ2� 𝑇𝑖−1𝑡 +

� 𝛼ℎ2− 1

Δ𝑡� 𝑇𝑖𝑡 − � 𝛼

2ℎ2− 𝛼

2𝑟𝑖ℎ� 𝑇𝑖+1𝑡 𝑖 = 1,𝑁

Note 𝑖 = 0 requires a modified equation as (Why?): 𝜕𝑇𝜕𝑡

= 3𝛼 𝜕2𝑇𝜕𝑟2

The modified equation is discretized:

𝑇0𝑡+1 −𝑇0𝑡

Δ𝑡= 3𝛼

2��𝑇1−2𝑇0+𝑇−1

ℎ2 �𝑡+1

+ �𝑇1−2𝑇0+𝑇−1 ℎ2

�𝑡�: i = 0

�3𝛼2ℎ2

�𝑇−1𝑡+1 − �3𝛼ℎ2

+1Δ𝑡� 𝑇0𝑡+1 + �

3𝛼2ℎ2

� 𝑇1𝑡+1 = −�3𝛼2ℎ2

� 𝑇−1𝑡 + �3𝛼ℎ2

−1Δ𝑡� 𝑇0𝑡 − �

3𝛼2ℎ2

� 𝑇1𝑡

Prepare tridiagonal matrix:

�𝑎0 =

3𝛼2ℎ2

, 𝑏0 = −�3𝛼ℎ2

+1Δ𝑡� , 𝑐0 =

3𝛼2ℎ2

𝑑0 = −�3𝛼2ℎ2

� 𝑇−1𝑡 + �3𝛼ℎ2

−1Δ𝑡� 𝑇0𝑡 − �

3𝛼2ℎ2

� 𝑇1𝑡� 𝑖 = 0

𝑎𝑖 = �𝛼

2ℎ2−

𝛼2𝑟𝑖ℎ

� , 𝑏𝑖 = �𝛼ℎ2−

1Δ𝑡� , 𝑐𝑖 = �

𝛼2ℎ2

+𝛼

2𝑟𝑖ℎ� ,𝑑𝑖

= −�𝛼

2𝑟𝑖ℎ−

𝛼2ℎ2

� 𝑇𝑖−1𝑡 + �𝛼ℎ2

−1Δ𝑡� 𝑇𝑖𝑡 − �

𝛼2ℎ2

+𝛼

2𝑟𝑖ℎ�𝑇𝑖+1𝑡

𝑖 = 1,𝑁

Discretize BCs (1) 𝑟 = 0 𝜕𝑇𝜕𝑟

= 0 ⇒ 𝑖 = 0; 𝑇−1 = 𝑇+1

0 1 𝑖 + 1

𝑖

𝑁 − 1 𝑁 (𝑟𝑖 = 𝑖ℎ) 0 1

𝑖 − 1

𝑁 𝑁 − 1

𝑡 + 1

𝑖 − 1

𝑖 𝑖 + 1

3

(2) 𝑟 = 𝑟𝑝 𝑘 𝜕𝑇𝜕𝑟

= ℎ𝑓�𝑇𝑓 − 𝑇� 𝑖 = 𝑁;

𝑇𝑁+1 −𝑇𝑁−12h

= ℎ𝑓𝑘�𝑇𝑓 − 𝑇𝑁�

or, 𝑇𝑁+1 = 𝑇𝑁−1 + 2ℎ𝑓ℎ𝐾

�𝑇𝑓 − 𝑇𝑁�

Apply BC

�𝑐0 = 𝑐0 + 𝑎0

𝑑0 = �3𝛼ℎ2− 1

Δ𝑡� 𝑇0𝑡 −

3𝛼ℎ2𝑇1𝑡� 𝑖 = 0

𝑎𝑁 = 𝑎𝑁 + 𝑐𝑁

𝑑𝑁 = − 𝛼ℎ2𝑇𝑁−1𝑡 + � 𝛼

ℎ2− 1

Δ𝑡� 𝑇𝑁𝑡 + � 𝛼

2ℎ2− 𝛼

2𝑟𝑖ℎ� 2ℎ𝑓ℎ

𝑘𝑇𝑁𝑡 − � 𝛼

2ℎ2− 𝛼

2𝑟𝑖ℎ� 2ℎ𝑓ℎ

𝑘𝑇𝑓

Solve 𝐴𝑦(𝑡+1) = 𝑑(𝑡) , 𝑖 = 0,𝑁 with initial condition: 𝑇 = 𝑇𝑖𝑛𝑖 = 300

𝐴𝑦(1) = 𝑑(0) ⇒ 𝐴𝑦(2) = 𝑑(1) ⇒ 𝐴𝑦(3) = 𝑑(2), 𝑒𝑡𝑐.

Example: Consider the catalytic oxidation of 𝑆𝑂2 into 𝑆𝑂3 over a spherical Pt-dispersed porous

carbon catalyst �radius = 𝑟𝑝�. The atmospheric concentration of 𝑆𝑂2 is 1%. The rate of reaction is 𝑟 = 𝑘 (= 0.1) 𝐶𝐴 mole s − m3⁄ . The film mass transfer coefficient is 𝑘𝑚(𝑚 𝑠).⁄ Determine the time-development of concentration profiles within the catalyst. The pore diffusion coefficient of 𝑆𝑂2 is estimated to be 𝐷𝑚2 𝑠⁄ .

𝜕𝐶𝐴𝜕𝑡

= 𝐷𝑟2

𝜕𝜕𝑟�𝑟2 𝜕𝐶𝐴

𝜕𝑟� − 𝑘𝐶𝐴 (1)

𝑡 = 0 𝐶𝐴 = 𝐶𝐴𝑖 ≪< 𝐶𝑏 for all 𝑟𝑝 ≥ 𝑟 ≥ 0

𝐶𝑏 = 0.01 𝐶(𝑡, 𝑟) =?

𝑆𝑆

𝑘𝑚

𝑟𝑝

4

0+ ∇𝐶𝐴 = 0 @ 𝑟 = 0

−𝐷 𝜕𝐶𝐴𝜕𝑟

= 𝑘𝑚(𝐶𝐴 − 𝐶𝑏) @ 𝑟 = 𝑟𝑝

Before you solve, two clarifications follow:

(1) The steady-state solution of eq(1) is the same as that of the equation solved in the previous lectures for BVP. Therefore, one way of checking the present code for this unsteady-state 1D parabolic equation is to compare its SS solution with that from the BVP-code. In the latter lectures you will see that very often it is better to artificially

introduce the transient term �𝑣𝑖𝑧. 𝜕𝑇𝜕𝑡

, 𝜕𝐶𝜕𝑡

, 𝜕𝑉𝜕𝑡� in the BVPs and solve the entire time-

profiles, even when one is asked to solve the SS solution only. This strategy is often followed for solving elliptic PDE.

(2) You should also compare this mass transport example with the previous example on heat transfer. But for the non-homogenous term (𝑘𝐶𝐴), two equations are identically the same, including the IC and BCs.

Recall:

𝜕𝑇𝜕𝑡

= 𝛼𝑟2

𝜕𝜕𝑟�𝑟2 𝜕𝑇

𝜕𝑟�

𝐼𝐶 ∶ 𝑡 = 0 𝑇 = 𝑇𝑜𝐵𝐶 1 ∶ 𝑟 = 0 𝜕𝑇

𝜕𝑟= 0

2 ∶ 𝑟 = 𝑟𝑝 − 𝑘 𝜕𝑇𝜕𝑟

= ℎ𝑓(𝑇 − 𝑇𝑓)

𝜕𝐶𝐴𝜕𝑡

= 𝐷𝑟2

𝜕𝜕𝑟�𝑟2 𝜕𝐶𝐴

𝜕𝑟�

𝐶𝐴 = 𝐶𝐴𝑖 for all 𝑟𝑝 > 𝑟 > 0𝜕𝐶𝐴𝜕𝑟

= 0

−𝐷 𝜕𝐶𝐴𝜕𝑟

= 𝑘𝑚(𝐶𝐴 − 𝐶𝑏)

All it means is that their non-dimensionalized forms are identically the same:

𝜕𝜃𝜕𝜏

=1𝜉2

𝜕𝜕𝜉�𝜉2

𝜕𝜃𝜕𝜉�

where,

⎩⎪⎨

⎪⎧𝜃 =

𝑇 − 𝑇𝑜𝑇𝑓 − 𝑇𝑜

, 𝐶𝐴 − 𝐶𝐴𝑖𝐶𝐴,𝑏 − 𝐶𝐴𝑖

𝜉 = 𝑟 𝑟𝑝�

𝜏 = 𝑡 𝑡𝑐 � ; 𝑡𝑐 =𝑟𝑝2

𝛼 ,

𝑟𝑝2

𝐷

�

𝐼𝐶. 𝜏 = 0 𝜃 = 0

𝐵𝐶 1. 𝜉 = 0 𝜕𝜃𝜕𝜉

= 0

2. 𝜉 = 1 − 𝜕𝜃𝜕𝜉

= 𝐴(𝜃 − 1) , where A = Nu or Sh

5

Therefore, it is clear that their (non-dimensional) solutions will also be the same.

It also follows that one should non-dimensionalize the transport concentration equation, as a good practice, before solving the equation. Apart from learning about transport phenomena from the analogy between heat, mass & momentum, there is every possibility that the non-dimensional forms of the conservation equations & bcs being the same, the non-dimensionalized solutions will also be the same. In such cases or similar cases, the programming code written for one problem will be the same or similar, requiring small modifications.

Let us continue with the non-dimensionalized form of the present mass transport problem:

𝜕𝜃𝜕𝜏

= 1𝜉2

𝜕𝜕𝜉�𝜉2 𝜕𝜃

𝜕𝜉� − 𝐵𝜃 − 𝐶 , where 𝐵 = 𝑘𝑡𝑐, 𝐶 = 𝑘𝑡𝑐

𝐶𝐴𝑖𝐶𝐴,𝑏−𝐶𝐴𝑖

discretize:

𝜃𝑖𝜏+1 − 𝜃𝑖𝜏

Δ𝜏=

12��𝜃𝑖+1 − 2𝜃𝑖 + 𝜃𝑖−1

Δ𝜉2 �𝜏+1

+ �𝜃𝑖+1 − 2𝜃𝑖 + 𝜃𝑖−1

Δ𝜉2 �𝜏

+2𝜉𝑖�𝜃𝑖+1 − 𝜃𝑖−1

2Δ𝜉 �𝜏+1

+2𝜉𝑖�𝜃𝑖+1 − 𝜃𝑖−1

2Δ𝜉 �𝜏

− 𝐵(𝜃𝑖𝜏+1 + 𝜃𝑖𝜏)� − 𝐶

𝑖 = 0,𝑁

@𝑖 = 0 there is a discontinuity in the equation. Therefore,

𝜃0𝜏+1 −𝜃0𝜏

Δ𝜏= 3

2��𝜃1−2𝜃0+𝜃−1

Δ𝜉2 �𝜏+1

+ �𝜃1−2𝜃0+𝜃−1Δ𝜉2

�𝜏

+ 𝐵2

(𝜃0𝜏+1 + 𝜃0𝜏)� − 𝐶 for i = 0

Apply BCs

�𝑖 = 0 𝜃−1 = 𝜃1

𝑖 = 𝑁 − 𝜃𝑁+1−𝜃𝑁−12Δ𝜉

= 𝐴(𝜃𝑁 − 1)

𝑜𝑟 𝜃𝑁+1 = 𝜃𝑁−1 − 2𝐴Δ𝜉(𝜃𝑁 − 1)�

𝑖 = 0: �1Δ𝜏

+3Δ𝜉2

−𝐵2� 𝜃0𝜏+1 −

3Δ𝜉2

𝜃1𝜏+1 = �1Δ𝜏

−3Δ𝜉2

+𝐵2� 𝜃0𝜏 +

32Δ𝜉2

𝜃1𝜏 − 𝐶

𝑖 = 0 𝑖 − 1 𝑖 + 1 𝑖 𝑁

𝑁 𝜏 + 1 𝜏

6

𝑖 = 𝑁: �1Δ𝜉2

� 𝜃𝑁−1𝜏+1 − �1Δ𝜉2

+𝐵2

+1Δ𝜏� 𝜃𝑁𝜏+1 = −�

1Δ𝜉2

� 𝜃𝑁−1𝜏 − �1Δ𝜏

−1Δ𝜉2

−𝐵2�𝜃𝑁𝜏 + 𝐶

𝑖 = 1,𝑁 − 1 �1

2Δ𝜉2−

12Δ𝜉𝜉𝑖

� 𝜃𝑖−1𝜏+1 − �1Δ𝜏

+1Δ𝜉2

−𝐵2�𝜃𝑖𝜏+1 + �

12Δ𝜉2

−1

2𝜉Δ𝜉� 𝜃𝑖+1𝜏+1

= �−1

2Δ𝜉2+

12Δ𝜉𝜉

� 𝜃𝑖−1𝜏 − �1Δ𝜏

−1Δ𝜉2

−𝐵2�𝜃𝑖𝜏 + �−

12Δ𝜉2

−1

2Δ𝜉𝜉� 𝜃𝑖+1 + 𝐶

d

Solve 𝐴𝜃𝜏+1 = 𝑑𝜏 , 𝑖 = 0, 𝑁 with 𝐼𝐶 𝜃 = 0 @ 𝜏 = 0 (same as before,𝐴 ≡ Tridiagonal matrix)

Here, we skipped the steps for preparing tridiagonal matrix and directly substituted discretized BCs into the discretized equations! As an exercise, prepare tridiagonal matrix and see if you get the same discretized equations post substitution of BCs, as before.

Quiz III

1

Lecture #24-25 Elliptic PDE (Method of Lines)

The model equations may be recognized from the following examples:

(1) 𝑘 �𝜕2𝑇

𝜕𝑋2+ 𝜕2𝑇

𝜕𝑦2� + 𝑆 = 0 ; 𝑆 ≡ 𝐼2𝑅

∀� 𝑐𝑎𝑙

𝑠 − 𝑚3�

(2) 𝑘 �𝜕2𝑇

𝜕𝑋2+ 1

𝑟𝜕𝜕𝑟�𝑟 𝜕𝑇

𝜕𝑟�� + 𝑆 = 0

----- SS 2D temperature profiles in the uniform heating of a rectangular plate (1) and a cylindrical wire (2)

(3) 𝑉𝑋𝜕𝐶𝜕𝑋

= 𝐷 𝜕2𝐶𝜕𝑋2

+ 𝐷𝑟� 𝜕𝜕𝑟�𝑟 𝜕𝐶

𝜕𝑟�� − 𝑘𝑐

----- SS 2D concentration distributions of a solute in a tabular reactor flow

(4) 𝑉𝑋𝜕𝑇𝜕𝑋

= 𝑘 𝜕2𝑇𝜕𝑋2

+ 𝑘 𝜕2𝑇𝜕𝑦2

- SS 2D temperature distributions in a flow through rectangular channel.

(5) 𝜌 𝜕𝑉𝑋𝜕𝑋

= 𝜇 �𝜕2𝑉𝑋𝜕𝑋2

+ 1𝑟𝜕𝜕𝑟�𝑟 𝜕𝑉𝑋

𝜕𝑟�� − 𝑑𝑝

𝑑𝑥

- SS 2D axial velocity profiles of an incompressible NF in a pressure-driven horizontal flow.

By now, you must have realized that we are referring to an elliptic PDE which describes a SS 2D (no time-dependent term) heat/mass/momentum transport problem. Why is the equation called elliptic? Because one has to solve over the entire 2D space. Considering that there is the SS consideration, there is no as such ‘marching’ on time and updating of the solution from the previous time-step, as we earlier discussed for the parabolic PDEs, and the solution in this case must be sought one-time only under SS conditions. You will see later that it is more computational extensive requiring iterations.

- The best way of understanding the numerical technique for solving an elliptic PDE is to directly take the example (1) above.

2

Ex: A rectangular plate (𝐿 × 𝑤 × 𝑡ℎ) fabricated from stainless steel (𝑘 = 40 𝑊 𝑚− 𝑘⁄ , 𝜌 =8000 𝑘𝑔 𝑚3,⁄ 𝐶𝑝 = 400 𝐽 𝑘𝑔 𝐾)⁄ is uniformly heated using an electric power source (100 𝑊). The top and bottom ends are insulated, whereas the side surfaces are exposed to atmosphere (𝑇𝑎 = 300 𝐶,ℎ = 100𝑊 𝑚2𝑘⁄ ). Determine the SS temperature profiles in the plate.

2D energy balance over ′Δ𝑥Δ𝑦𝑡ℎ′𝐶𝑉 under SS:

𝜌𝐶𝑝 �𝜕𝑇𝜕𝑡

+ 𝑉.∇𝑇� = 𝑘∇2𝑇 + 𝑆: 𝑊 𝑚3⁄

𝑘 �𝜕2𝑇𝜕𝑥2

+𝜕2𝑇𝜕𝑦2

� + 𝑆 = 0

BCs: 𝑥 = 0 𝑓or 𝑤 > 𝑦 > 0 ; −𝑘 𝜕𝑇𝜕𝑥

= −ℎ𝑓(𝑇 − 𝑇𝑎)

= 𝐿 for 𝑤 > 𝑦 > 0 ; −𝑘 𝜕𝑇𝜕𝑥

= ℎ𝑓(𝑇 − 𝑇𝑎)

𝑦 = 0 and 𝑤 for 𝐿 > 𝑥 > 0, − 𝑘 𝜕𝑇𝜕𝑦

= 0(insulation)

𝑦

𝑀

𝑤 ℎ𝑓 (𝑇𝑎)300𝐶

𝑂

300𝐶(𝑇𝑎)

ℎ𝑓 𝐿

−𝑘𝑑𝑇𝑑𝑦

= 0

ℎ𝑓

−𝑘𝑑𝑇𝑑𝑦

= 0

(𝑖, 𝑗 + 1)

(𝑖 + 1, 𝑗) (𝑖, 𝑗) (𝑖 − 1, 𝑗)

(𝑖, 𝑗 − 1)

Δ𝑥

Δ𝑦

N X

S (𝑊 𝑚3)⁄ = �100 𝑊

𝐿 × 𝑊 × 𝑡ℎ�

(ℎ𝑒𝑎𝑡 𝑠𝑜𝑢𝑟𝑐𝑒)

𝑆𝑆 0, 𝑠𝑜𝑙𝑖𝑑

3

Computational molecule:

Discretize equations over (𝑖, 𝑗). Note that it is a 2D problem. Therefore, one will have to discretize in both directions (𝑥,𝑦) using steps Δ𝑥 and Δ𝑦, respectively. They need not be equal.

𝑘 ��𝜕2𝑇

𝜕𝑋2�𝑖,𝑗

+ �𝜕2𝑇

𝜕𝑦2�𝑖,𝑗� + 𝑆𝑖,𝑗 = 0

�𝑇𝑖−1,𝑗 − 2𝑇𝑖,𝑗 + 𝑇𝑖+1,𝑗

Δ𝑥2� + �

𝑇𝑖,𝑗−1 − 2𝑇𝑖,𝑗 + 𝑇𝑖,𝑗+1Δ𝑦2

� +𝑆𝑖,𝑗𝑘

= 0

(As expected, when discretizing 𝜕2𝑇

𝜕𝑋2, 𝑗 is constant, and when discretizing 𝜕

2𝑇𝜕𝑦2

, 𝑖 is constant)

There are two ways to arrange the discretized terms into

𝐴𝑦� = 𝑏 � form ∶

(1) 1Δ𝑥2

�𝑇𝑖−1,𝑗 − 2𝑇𝑖,𝑗 + 𝑇𝑖+1,𝑗� = − 1Δ𝑦2

�𝑇𝑖,𝑗−1 − 2𝑇𝑖,𝑗 + 𝑇𝑖,𝑗+1� −𝑆𝑖,𝑗𝑘

(2) 1Δ𝑦2

�𝑇𝑖,𝑗−1 − 2𝑇𝑖,𝑗 + 𝑇𝑖,𝑗+1� = − 1Δ𝑦2

�𝑇𝑖−1,𝑗 − 2𝑇𝑖,𝑗 + 𝑇𝑖+1,𝑗� −𝑆𝑖,𝑗𝑘

Notes: 1. Although the linear algebraic equations have taken the form of 𝐴𝑦� = 𝑏 � in both cases, 𝑏 � or RHS terms are not known. Therefore, in principle 𝑦� 𝑜𝑟 𝑇� cannot be solved by inverting the ‘matrix’ formed on LHS, the way we did in the previous examples. In other words, there is no way one can march in 𝑥 𝑜𝑟 𝑦 direction solving unknown variables, because marching in either direction (𝑖 𝑜𝑟 𝑗) creates unknown variables in the other direction (𝑗 𝑜𝑟 𝑖). In fact, one has to solve the ‘entire space’ at one time (without marching). This is the problem in solving an elliptic PDE!

(2) The only way to solve an elliptic PDE or 𝐴𝑦� = 𝑏 � , where 𝐴 is the tridiagonal matrix is by making guess for all variables (𝑇𝑖,𝑗) to start with, so that 𝑏 � (RHS term) is known. Then, 𝑦� can be solved. Next, compare the new/calculated values with guess values –-- iterate till there is

the convergence: 𝐴𝑦𝑘+1 = 𝑏𝑘 b0 = initial guess for �Ti,j�

k = # of iterations.

(𝑖 − 1)

(𝑗 + 1)

(𝑖 + 1)

(𝑗 − 1)

(𝑖, 𝑗)

Δ𝑥 = 𝐿𝑁�

Δ𝑦 = 𝑊𝑀�

� 𝑖 = 0 … 𝑁𝑗 = 0 … 𝑀�

4

(3) Either of the two discretized schemes (1) & (2) can be used to solve 𝑇𝑖,𝑗. In general, one should sweep the discretized set of equations in the direction the expected solution (functional value) is lesser stiff than in the other direction. In the present example, 𝑤 ≪ 𝐿 and the y-ends are insulated. One should use scheme (1). The convergence will be relatively faster.

1Δ𝑥2

�𝑇𝑖−1,𝑗 − 2𝑇𝑖,𝑗 + 𝑇𝑖+1,𝑗� = − 1Δ𝑦2

�𝑇𝑖,𝑗−1 − 2𝑇𝑖,𝑗 + 𝑇𝑖,𝑗+1�𝑔− 𝑆𝑖,𝑗

𝐾 (1)

Prepare tridiagonal matrix:

� 𝑎(𝑖, 𝑗) =

1Δ𝑥2

; 𝑏(𝑖, 𝑗) = −2Δ𝑥2

, 𝑐(𝑖, 𝑗) = 1Δ𝑥2

𝑑(𝑖, 𝑗) = −1Δ𝑦2

�𝑇𝑖,𝑗−1 − 2𝑇𝑖,𝑗 + 𝑇𝑖,𝑗+1�𝑔−𝑆𝑖,𝑗𝑘 ⎭

⎬

⎫ 𝑖 = 0,𝑁𝑗 = 0,𝑀

The superscript "g" stands for the guess values. Once the initial guess is made, one can now

march along ‘𝑗’ direction, solving 𝑇𝑖,𝑗 at every ‘𝑗𝑡ℎ ’ step as 𝐴𝑦�𝑗 = 𝑏, where 𝐴 is the

tridiagonal matrix containing the discretized ‘𝑦’ along ‘𝑖’ direction for a fixed 𝑗, and 𝑏 � is known from the guess.

Discretize BCs

⎩⎪⎨

⎪⎧ (𝑖 = 0), 𝑗: − 𝑘

𝑇1,𝑗 − 𝑇−1,𝑗

2Δ𝑥= −ℎ𝑓(𝑇0,𝑗 − 𝑇𝑎)

or 𝑇−1,𝑗 = 𝑇1,𝑗 −2ℎ𝑓Δ𝑥𝑘

(𝑇0,𝑗 − 𝑇𝑎)

(𝑖 = 𝑁), 𝑗: 𝑇𝑁+1,𝑗 = 𝑇𝑁−1,𝑗 −2ℎ𝑓Δ𝑥𝑘

(𝑇𝑁,𝑗 − 𝑇𝑎)

�

(𝑗 = 0), 𝑖: 𝑇𝑖,−1 = 𝑇𝑖,1

(𝑗 = 𝑀), 𝑖: 𝑇𝑖,𝑀+1 = 𝑇𝑖,𝑀−1

Apply BCs

𝑗 = 0

�

𝑐(0,0) = 𝑐(0,0) + 𝑎(0,0)

𝑏(0,0) = 𝑏(0,0) −2ℎ𝑓Δ𝑥𝑘

𝑎(0,0)

𝑑(0,0) =2Δ𝑦2

�𝑇0,0 − 𝑇0,1�𝑔−𝑆0,0

𝑘−

2ℎ𝑓Δ𝑥𝑘Δ𝑦2

.𝑇𝑎.𝑎(0,0)⎭⎪⎬

⎪⎫

𝑖 = 0

𝑖 = 0 1 𝑖 − 1 𝑖 𝑖 + 1 𝑁 − 1 𝑁

5

𝑑(𝑖, 0) = 2Δ𝑦2

�𝑇𝑖,0 − 𝑇𝑖,1�𝑔− 𝑆𝑖,0

𝑘; 𝑖 = 1,𝑁 − 1

�

𝑎(𝑁, 0) = 𝑎(𝑁, 0) + 𝑐(𝑁, 0)

𝑏(𝑁, 0) = 𝑏(𝑁, 0) −2ℎ𝑓Δ𝑥𝑘

𝑐(𝑁, 0)

𝑑(𝑁, 0) =2Δ𝑦2

�𝑇𝑁,0 − 𝑇𝑁,1�𝑔−𝑆𝑁,0

𝑘−

2ℎ𝑓Δ𝑥𝑘Δ𝑦2

𝑇𝑎 𝑎(𝑁, 0)⎭⎪⎬

⎪⎫

𝑖 = 𝑁

Tridiag�𝑁 + 1,𝑎, 𝑏, 𝑐,𝑑,𝑇𝑖,0� (𝑖 = 0,𝑁)

𝑗 = 1,𝑀 − 1

�

𝑐(0, 𝑗) = 𝑐(0, 𝑗) + 𝑎(0, 𝑗)

𝑏(0, 𝑗) = 𝑏(0, 𝑗) −2ℎ𝑓Δ𝑥𝑘

𝑎(0, 𝑗)

𝑑(0, 𝑗) = 𝑑(0, 𝑗) − �2ℎ𝑓Δ𝑥𝑘Δ𝑦2

�𝑇𝑎𝑎(0, 𝑗)⎭⎪⎬

⎪⎫

𝑖 = 0

⇒ No coefficients will change for the rows 𝑖 = 1,𝑁 − 1, because all of these correspond to interior nodes:

𝑀 − 1 𝑀

𝐽

𝑁 − 1

𝑁 𝑖 0,0

6

�

𝑎(𝑁, 𝑗) = 𝑎(𝑁, 𝑗) + 𝑐(𝑁, 𝑗)

𝑏(𝑁, 𝑗) = 𝑏(𝑁, 𝑗) − �2ℎ𝑓Δ𝑥𝑘

� 𝑐(𝑁, 𝑗)

𝑑(𝑁, 𝑗) = 𝑑(𝑁, 𝑗) − �2ℎ𝑓Δ𝑥𝑘Δ𝑦2

�𝑇𝑎 𝑎(𝑁, 𝑗)⎭⎪⎬

⎪⎫

𝑖 = 𝑁

Tridiag�𝑁 + 1,𝑎, 𝑏, 𝑐,𝑑,𝑇𝑖,𝑗� (𝑖 = 0,𝑁)

𝑗 = 𝑀

�

𝑐(0, 𝑗) = 𝑐(0, 𝑗) + 𝑎(0, 𝑗)

𝑏(0, 𝑗) = 𝑏(0, 𝑗) −2ℎ𝑓Δ𝑥𝑘

𝑎(0, 𝑗)

𝑑(0, 𝑗) =2Δ𝑦2

�−𝑇0,𝑀−1 + 𝑇0,𝑀�𝑔−𝑆0,𝑀

𝑘− �

2ℎ𝑓Δ𝑥𝑘Δ𝑦2

�𝑇𝑎⎭⎪⎬

⎪⎫

𝑖 = 0

𝑑(𝑖, 𝑗) = − 2Δ𝑦2

�−𝑇𝑖,𝑀−1 + 𝑇𝑖,𝑀�𝑔− 𝑆𝑖,𝑀

𝑘; 𝑖 = 1,𝑁 − 1

�

𝑎(𝑁, 𝑗) = 𝑎(𝑁, 𝑗) + 𝑐(𝑁, 𝑗)

𝑏(𝑁, 𝑗) = 𝑏(𝑁, 𝑗) −2ℎ𝑓Δ𝑥𝑘

𝑐(𝑁, 𝑗)

𝑑(𝑁, 𝑗) =2Δ𝑦2

�−𝑇𝑁,𝑀−1 + 𝑇𝑁,𝑀�𝑔−𝑆𝑁,𝑀

𝑘−

2ℎ𝑓Δ𝑥𝑘Δ𝑦2

𝑇𝑎⎭⎪⎬

⎪⎫

𝑖 = 𝑁

Tridiag�𝑁 + 1,𝑎, 𝑏, 𝑐,𝑑,𝑇𝑖,𝑗� (𝑖 = 0,𝑁)

You have, therefore, called Tridiagonal subroutine ‘𝑀 + 1’ times as you marched along ‘𝑗’ direction. At the end, entire ‘𝑥 − 𝑦’ grids have been solved for 𝑇𝑖,𝑗(𝑖 = 0,𝑁; 𝑗 = 0,𝑀). Compare solved values with the guess values and keep iterating till there is a convergence.

⇒ Solving an elliptic PDE is indeed computational extensive because there is a convergence issue using the guesses.

⇒ There is another twist in solving an elliptic PDE. Do you update your guess values of 𝑏� (RHS terms) at ‘𝑗’ with the ones solved at ‘𝑗 − 1’ as you march in y-direction, or do you wait till you have solved till ‘𝑀’ (the last boundary)? This question should remind you of the G-S and Jacobi iterations. Choice is yours. A fast convergence is the criterion. Also, relaxation factor ‘𝑤’ can be used to update the functional values in either case:

𝑇𝑖,𝑗 = 𝑤𝑇𝑖,𝑗 + (1 − 𝑤)𝑇𝑖,𝑗

1

Lecture #25-26 Ex 2: Consider the fully developed SS flow (Re = 200) of an incompressible NF in a long horizontal tube (L, D). The inlet temperature of the liquid is 𝑇𝑜 . The heat is supplied to the flowing liquid at constant flux 𝑞𝑤(𝑊 𝑚2⁄ ) through the tube walls. Determine the 2D (𝑟, 𝑥) SS temperature profiles in the tube.

Energy balance over ′2𝜋𝑟Δ𝑟Δ𝑋′ 𝐶𝑉:

𝜌𝐶𝑝𝑉𝑋.∇𝑇 = 𝐾∇2𝑇 + 𝑆 ( 𝐽 𝑜𝑟 𝑐𝑎𝑙 𝑠 − 𝑚3⁄ )

𝑉(𝑟) 𝜕𝑇𝜕𝑋

= 𝛼 �𝜕2𝑇

𝜕𝑋2+ 1

𝑟𝜕𝜕𝑟�𝑟 𝜕𝑇

𝜕𝑟�� 𝛼 = 𝑘�

𝜌𝐶𝑝 ;

BC. 𝑋 = 0 𝑇(𝑟) = 𝑇𝑜 𝑅 ≥ 𝑟 ≥ 0

𝑋 = 𝐿 + 𝜕𝑇𝜕𝑋

= 0 (long tube approximation)

�𝑟 = 0 𝜕𝑇

𝜕𝑟= 0 (symmetric BC)

𝑟 = 𝑅 − 𝑘 𝜕𝑇𝜕𝑟

= −𝑞𝑊 (𝑊 𝑚2)⁄

𝑜𝑟 𝑘 𝜕𝑇𝜕𝑟

= 𝑞𝑊 ⎭⎪⎬

⎪⎫

𝐿 ≥ 𝑧 ≥ 0

Let us non-dimensionalize the equation & BC

𝜃 = 𝑇To

, 𝑧 = 𝑋Lc

𝑤here Lc = RPe where Pe(radial) = Umax Rα

, ξ = rR

𝑈𝑚𝑎𝑥(1 − 𝜉2)Lc

𝜕𝜃∂z

= α�1Lc2𝜕2𝜃∂z2

+1

R2𝜕2𝜃∂ξ2

+1

R21ξ𝜕𝜃∂ξ�

𝑟

Δ𝑟

𝑋

𝑄 𝑊 𝑚2⁄

𝑟 Δ𝑟

Δ𝑋 𝑄 𝑊 𝑚2⁄

𝐿

𝑅 𝑇𝑜

𝑢(𝑟) = 𝑈𝑚𝑎𝑥 �1 −𝑟2

𝑅2� (𝑅𝑒 = 200); 𝜌 ,𝐶𝑝,𝑘 ≡ 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡

T(r, X)=?

2

or

(1 − 𝜉2)𝜕𝜃∂z

=1𝑃𝑒2

𝜕2𝜃∂z2

+𝜕2𝜃∂ξ2

+1ξ𝜕𝜃∂ξ

BCs:

�𝑧 = 0 𝜃 = 1

𝑧 = 𝐿𝐿𝑐� + 𝜕𝜃

𝜕𝑧= 0�1 > 𝜉 > 0

𝜉 = 0 𝜕𝜃𝜕𝜉

= 0

𝜉 = 1 𝜕𝜃𝜕𝜉

= �𝑞𝑊𝑅 𝑘𝑇𝑜

� = 𝐻(constant) for 𝐿 > 𝑧 > 0

Before solving, let us consider another example on mass transport.

Ex. Consider the fully developed SS flow (𝑅𝑒 = 200) of an incompressible NF in a long horizontal tube (𝐿,𝐷). The inlet concentration of the species A in the liquid is 𝐶𝐴𝑜 .The species are catalytically destroyed at the tube walls by the zeroth order chemical reaction (𝑘 𝑚𝑜𝑙𝑒 𝑠 − 𝑚3).⁄ Determine the 2D (𝑟, 𝑥)𝑆𝑆 concentration profiles in the tube.

Species balance over (2𝜋𝑟Δ𝑟Δ𝑋) 𝐶,𝑉.

𝑉𝑋𝜕𝐶𝐴𝜕𝑋

= 𝐷 �𝜕2𝐶𝐴𝜕𝑋2

+1𝑟𝜕𝜕𝑟�𝑟𝜕𝐶𝐴𝜕𝑟

�� moles s − m3⁄

𝑈𝑚𝑎𝑥 �1 −𝑟2

𝑅2�𝜕𝐶𝐴𝜕𝑋

= 𝐷 �𝜕2𝐶𝐴𝜕𝑋2

+1𝑟𝜕𝜕𝑟�𝑟𝜕𝐶𝐴𝜕𝑟

��

= 𝐷 �𝜕2𝐶𝐴𝜕𝑋2

+ 𝜕2𝐶𝐴𝜕𝑟2

+ 1𝑟𝜕𝐶𝐴𝜕𝑟�

BCs. 𝑋 = 0 𝐶𝐴 = 𝐶𝐴𝑜 , 𝑋 = 𝐿 𝜕𝐶𝐴𝜕𝑥

= 0 (𝑅 > 𝑟 > 0)

𝑟 = 0 𝜕𝐶𝐴𝜕𝑟

= 0 𝑟 = 𝑅 , − 𝐷 𝜕𝐶𝐴𝜕𝑟

= 𝑘 (𝐿 > 𝑧 > 0)

Let us non-dimensionalize the equation & BCs:

𝑋 𝐶𝐴𝑜 𝐶(𝑟,𝑋) 𝑟 = −𝑘 Δ𝑟

𝑟

Δ𝑋

𝑅

𝑢(𝑟) = 𝑈𝑚𝑎𝑥 �1 −𝑟2

𝑅2�

3

𝜃 = 𝐶𝐴𝐶𝐴𝑜� , 𝑧 =

𝑥𝐿𝑐

, 𝜉 =𝑟𝑅

; 𝐿𝑐 = 𝑅𝑃𝑒 where Pe(radial) =𝑈𝑚𝑎𝑥𝐷

(Note that this ′𝑃𝑒′ is based on mass transport)

(1 − 𝜉2)𝜕𝜃𝜕𝑧

=1𝑃𝑒2

𝜕2𝜃𝜕𝑧2

+𝜕2𝜃𝜕𝜉2

+1𝜉𝜕𝜃𝜕𝜉

BC. 𝑧 = 0, 𝜃 = 1 ; 𝑧 = 𝐿𝐿𝑐� , 𝜕𝜃

𝜕𝑧= 0

𝜉 = 0, 𝜕𝜃𝜕𝜉

= 0 ; 𝜉 = 1 − 𝜕𝜃𝜕𝜉

= � 𝑘𝑅𝐷𝐶𝐴𝑜

� = 𝑀 (constant)

⇒ It is clear that the non-dimensionalized equations and BCs of this (mass transport) and previous (heat transport) examples are the same. Therefore, you need to solve only one of the two. The dimensionless solutions will be the same. This situation should remind you of the recommendation that one should non-dimensionalize the equation before solving it. Several such analogous heat, mass, and momentum transport scenarios exist in chemical engineering applications.

⇒In most cases, radial Pe(mass or heat) number in laminar flow regime is of the order of 10 or higher. In such cases, the axial diffusion term can be neglected. In other words,

𝛼 𝜕2𝑇𝜕𝑋2

<< 𝑉𝑋𝜕𝑇𝜕𝑋

𝑜𝑟 𝐷 𝜕2𝐶𝐴𝜕𝑋2

<< 𝑉𝑋𝜕𝐶𝐴𝜕𝑋

. (See BSL book)

If you neglect the axial diffusion term, the 2D elliptic PDE is modified/simplified to 1D parabolic PDE:

𝑉𝑋𝜕𝑇𝜕𝑋

= 𝛼𝑟𝜕𝜕𝑟�𝑟 𝜕𝑇

𝜕𝑟� 𝑜𝑟 𝑉𝑋

𝜕𝐶𝐴𝜕𝑋

= 𝐷𝑟𝜕𝜕𝑟�𝑟 𝜕𝐶𝐴

𝜕𝑟�

This simplified PDE can be solved using the Crank-Nicholson technique described in the preceding lecture. In other words, one can numerically march in ‘X’ direction and solve in ‘r’ direction using Thomas Algorithm for the tridiagonal matrix built from the discretized ‘r’ terms. For now, let us revert to the original (full) non-dimensionalized elliptic PDE:

(1 − 𝜉2)𝜕𝜃𝜕𝑧

=1𝑃𝑒2

𝜕2𝜃𝜕𝑧2

+𝜕2𝜃𝜕𝜉2

+1𝜉𝜕𝜃𝜕𝜉

At 𝜉 = 0, there is a discontinuity and you are solving an approximation:

(1 − 𝜉2)𝜕𝜃𝜕𝑧

=1𝑃𝑒2

𝜕2𝜃𝜕𝑧2

+2𝜕2𝜃𝜕𝜉2

4

Discretize the main conservation equation:

�1 − 𝜉𝑗2� 𝜃𝑖+1,𝑗 −𝜃𝑖−1,𝑗

2Δ𝑧= 1

𝑃𝑒2𝜃𝑖+1,𝑗−2𝜃𝑖,𝑗+𝜃𝑖−1,𝑗

Δ𝑧2 + 𝜃𝑖,𝑗+1−2𝜃𝑖,𝑗+𝜃𝑖,𝑗−1

Δ𝜉2 + 1

𝜉𝑗

𝜃𝑖,𝑗+1−𝜃𝑖,𝑗−12Δ𝜉

or

��1−𝜉𝑗

2�2Δ𝑧

+ 1𝑃𝑒2Δ𝑧2

� 𝜃𝑖−1,𝑗 −2𝜃𝑖,𝑗𝑃𝑒2Δ𝑧2

− ��1−𝜉𝑗

2�2Δ𝑧

+ 1𝑃𝑒2Δ𝑧2

� 𝜃𝑖+1,𝑗 = � 12Δ𝜉𝜉𝑗

− 1Δ𝜉2

� 𝜃𝑖,𝑗−1 +

2𝜃𝑖,𝑗Δ𝜉2

− � 12Δ𝜉𝜉𝑗

+ 1Δ𝜉2

� 𝜃𝑖,𝑗+1

- (1)

Discretize the approximated conservation equation at the line of symmetry:

��1−𝜉𝑗

2�2Δ𝑧

+ 1𝑃𝑒2Δ𝑧2

� 𝜃𝑖−1,0 −2𝜃𝑖,0𝑃𝑒2Δ𝑧2

− ��1−𝜉𝑗

2�2Δ𝑧

− 1𝑃𝑒2Δ𝑧2

� 𝜃𝑖+1,0 = −2𝜃𝑖,−1Δ𝜉2

+ 4𝜃𝑖,0Δ𝜉2

− 2𝜃𝑖,+1Δ𝜉2

Note that RHS terms for all rows contain guess values for all 𝜃𝑖,𝑗to begin with

Discretize BCs:

𝑖 = 0, 𝑗 = 0,𝑀; 𝜃0,𝑗 = 1; 𝑖 = 𝑁, 𝑗 = 0,𝑀; 𝜃𝑁+1,𝑗 = 𝜃𝑁−1,𝑗

𝑗 = 0, 𝑖 = 0,𝑁; 𝜃𝑖,−1 = 𝜃𝑖,1 ; 𝑗 = 𝑀, 𝑖 = 0,𝑁; 𝜃𝑖,𝑀+1 = 𝜃𝑖,𝑀−1 − 2𝑀Δ𝜉

𝑗 = 0

1st row of Tridiagonal matrix,

𝑖 = 1 −2𝜃1,0

𝑃𝑒2Δ𝑧2− �

1 − 𝜉02

2Δ𝑧−

1𝑃𝑒2Δ𝑧2

�𝜃2,0 = −4𝜃1,0

Δ𝜉2+

4𝜃1,0

Δ𝜉2− �

1 − 𝜉02

2Δ𝑧+

1𝑃𝑒2Δ𝑧2

� 𝜃0,0

=1

0 Δ𝑧2 i − 1 i + 1 i N 𝑖 = 1,𝑁𝑗 = 1,𝑀

𝑖 = 1,𝑁𝑗 = 0

Δ𝑧 = 1𝑁� , Δ𝜉 =

𝐿 𝐿𝑐⁄𝑀

i − 1 i + 1 i N − 1 N 1 0

5

middle rows,

𝑖 = 2,𝑁 − 1: ��1 − 𝜉0

2�2Δ𝑧

+1

𝑃𝑒2Δ𝑧2� 𝜃𝑖−1,0 −

2𝜃𝑖,0𝑃𝑒2Δ𝑧2

− �(1 − 𝜉0

2)2Δ𝑧

−1

𝑃𝑒2Δ𝑧2�𝜃𝑖+1,0

= −4𝜃𝑖,0Δ𝜉2

+4𝜃𝑖,0Δ𝜉2

last row, 𝑖 = 𝑁: 2𝑃𝑒2Δ𝑧2

𝜃𝑁−1,0 −2𝜃𝑁,0𝑃𝑒2Δ𝑧2

= −4𝜃𝑁,1Δ𝜉2

+ 4𝜃𝑁,0Δ𝜉2

You have a tridiagonal matrix A in (𝐴𝜃𝑖,0𝑘+1 = 𝑏𝑖,0𝑘 ) to invert

𝑗 = 1,𝑀 − 1

1st row,

𝑖 = 1: −2𝜃1,𝑗

𝑃𝑒2Δ𝑧2− �

1 − 𝜉𝑗2

2Δ𝑧−

1𝑃𝑒2Δ𝑧2

� 𝜃2,𝑗 = −�1 − 𝜉𝑗2

2Δ𝑧−

1𝑃𝑒2Δ𝑧2

�𝜃0,𝑗

� 12Δ𝜉𝜉𝑗

− 1Δ𝜉2

� 𝜃1,𝑗−1 + 2𝜃1,𝑗

Δ𝜉2− � 1

2Δ𝜉𝜉𝑗+ 1

Δ𝜉2� 𝜃1,𝑗+1

middle rows:

𝑖 = 2,𝑁 − 1 : These are middle grids unaffected by BCs.

last row

𝑖 = 𝑁 2

𝑃𝑒2Δ𝑧2𝜃𝑁−1,𝑗 −

2𝜃𝑁,𝑗

𝑃𝑒2Δ𝑧2= �

12Δ𝜉𝜉𝑗

−1Δ𝜉2

� 𝜃𝑁,𝑗−1 +2𝜃𝑁,𝑗

Δ𝜉2− �

12Δ𝜉𝜉𝑗

+1Δ𝜉2

� 𝜃𝑁,𝑗+1

Again, 𝐴𝜃𝑖,𝑗𝐾+1 = 𝑏𝑖,𝑗𝐾 ; 𝑖 = 1,𝑁𝑗 = 1,𝑀 − 1

𝑗 = 𝑀

1st row,

𝑖 = 1 −2𝜃1,𝑀

𝑃𝑒2Δ𝑧2− �

(1 − 𝜉𝑀2)

2Δ𝑧−

1𝑃𝑒2Δ𝑧2

� 𝜃2,𝑀 = �(1 − 𝜉𝑀

2)2Δ𝑧

+1

𝑃𝑒2Δ𝑧2�𝜃0,𝑀

− 2Δ𝜉2

𝜃1,𝑀−1 + 2𝜃1,𝑀Δ𝜉2

− 2𝑀Δ𝜉 � 12Δ𝜉𝜉𝑀

− 1Δ𝜉2

�

middle rows, LHS (same as that of eq1), unaffected by BC =

i=2, N-1 − 2Δ𝜉2

𝜃𝑖,𝑀−1 + 2𝜃𝑖,𝑀Δ𝜉2

+ 2𝑀Δ𝜉 � 12Δ𝜉𝜉𝑀

+ 1Δ𝜉2

�

1

1

6

last row,

𝑖 = 𝑁 2𝜃𝑁−1,𝑀

𝑃𝑒2Δ𝑧2+

2𝜃𝑁,𝑀

Δ𝜉2= −

2Δ𝜉2

𝜃𝑁,𝑀−1 +2𝜃𝑁,𝑀

Δ𝜉2+ 2𝑀Δ𝜉 �

12Δ𝜉𝜉𝑀

+1Δ𝜉2

�

Again, 𝐴𝜃𝑖,𝑀𝑘+1 = 𝑏𝑖,𝑀𝑘

You have called the Thomas Algorithm (M+1) times to invert the tridiagonal A matrix to solve

�𝜃𝑖,𝑗 , 𝑖 = 1,𝑁 𝑗 = 1,𝑀� using the guess values on the RHS of the tridiagonal matrix .

You must have noted we did not prepare the tridiagonal matrix, and instead directly substituted the BCs in the discretized equation! As an exercise, prepare the tridiagonal matrix by defining individuals elements of rows for every ‘j’ viz a(i, j), b(i, j)⋯⋯, substitute BCs and see if you get the same equations or not for different ′𝑗𝑠′.

Now, this is the time to compare the solved 𝜃𝑖,𝑗 values with the guess values you made at the beginning of the iterations before you start the second iteration.

⇒ From programming point of view, you should be able to choose reasonable values of all variables including, D, Vmax, M and Pe,mass transport, or H and Pe,heat transport .

⇒ As earlier noted, solving elliptic PDE is computational extensive, requiring guess values, iterations & convergence. Very often, one artificially inserts a transient term

�𝜕𝐶𝐴𝜕𝑡

or 𝜌𝐶𝑝𝜕𝑇𝜕𝑡

or 𝜌 𝜕𝑉��⃗

𝜕𝑡� and seek SS solutions, which is the focus of the last two lectures.

1

Lecture #26-27 𝑇𝑖𝑚𝑒 − 𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 2𝐷 𝑝𝑎𝑟𝑎𝑏𝑜𝑙𝑖𝑐 𝑃𝐷𝐸 (𝐴𝐷𝐼) 𝑀𝑒𝑡ℎ𝑜𝑑

We are solving, for examples,

𝜕𝐶𝐴𝜕𝑡

+ 𝑉.∇𝐶𝐴 = 𝐷∇2𝐶𝐴 + (−𝑟𝐴) ; 𝐶𝐴(𝑡, 𝑟, 𝑥)

or

𝜌𝐶𝑝 �𝜕𝑇𝜕𝑡

+ 𝑉.∇𝑇� = 𝑘∇2𝑇 + (−𝑟𝐴)(Δ𝐻) ; 𝑇(𝑡, 𝑟, 𝑥)

Note: SS solution must be the same as that of the converged solution of the analogous

2D elliptic PDE �𝜕𝐶𝐴𝜕𝑡

= 𝜕𝑇𝜕𝑡

= 0� discussed in the previous lectures.

Let us take a general case of the time-dependent 2D PDE:

𝜕𝜙𝜕𝑡

= 𝜙𝑋𝑋 + 𝜙𝑋 + 𝜙𝑌𝑌 + 𝜙𝑌 ;𝜙(𝑡, 𝑥,𝑦)

with necessary IC and BCs.

Apply Crank-Nicholson method/scheme to discretize 𝜙𝑋𝑋 ,𝜙𝑌𝑌,𝜙𝑋 ,𝑎𝑛𝑑 𝜙𝑌 terms, in the similar fashion we used to solve time-dependent 1D PDE on 𝜙(𝑡, 𝑥):

�𝜙𝑡+1 − 𝜙𝑡

Δ𝑡�𝑖,𝑗

=12��𝜙𝑖+1,𝑗 − 2𝜙𝑖,𝑗 + 𝜙𝑖−1,𝑗

Δ𝑋2 �𝑡+1

+ �𝜙𝑖+1,𝑗 − 𝜙𝑖−1,𝑗

2ΔX �𝑡+1

+ �𝜙𝑖,𝑗+1 − 2𝜙𝑖,𝑗 + 𝜙𝑖,𝑗−1

Δ𝑌2 �𝑡

+ �𝜙𝑖,𝑗+1 − 𝜙𝑖,𝑗−1

2ΔY �𝑡�

- This way X-derivatives have been discretized implicitly on time, whereas Y- derivatives have been discretized explicitly.

- Alternatively, one can discretize Y-derivatives implicitly on time, whereas X-derivatives can be discretized explicitly.

- Re-arranging the terms as 𝐴∅𝑦𝑡+1 = ∅𝑥

𝑡 𝑜𝑟 𝐴∅𝑥𝑡+1 = ∅𝑦

𝑡 from either scheme, it is clear that ‘A’ will be a tridiagonal matrix and one can proceed on time-step by solving 𝜙 on X-Y plane:

2

�1

2Δ𝑋2−

14Δ𝑋

�𝜙𝑖−1,𝑗𝑡+1 − �

1Δ𝑡

+1

2Δ𝑋2�𝜙𝑖,𝑗𝑡+1 + �

12Δ𝑋2

+1

4Δ𝑋�𝜙𝑖+1,𝑗

𝑡+1

= �1

4Δ𝑌−

12Δ𝑌2

�𝜙𝑖,𝑗−1𝑡 − �1Δ𝑡−

1Δ𝑌2

�𝜙𝑖,𝑗𝑡 − �1

4Δ𝑌+

12Δ𝑌2

�𝜙𝑖,𝑗+1𝑡

- This way you are ‘marching’ in ‘𝑗’ direction and ‘sweeping’ in ‘i’ direction.

By the second scheme, if you ‘march’ in ‘i’ direction and ‘sweep’ in ‘𝑗’-direction, you will get the following equation:

� 14Δ𝑌

− 12Δ𝑌2

�𝑡+1

𝜙𝑖,𝑗−1 + � 1Δ𝑡

+ 1Δ𝑌2

�𝜙𝑖,𝑗𝑡+1 − � 14Δ𝑌

+ 12Δ𝑌2

�𝜙𝑖,𝑗+1𝑡+1

= �1

2Δ𝑋2−

14Δ𝑋

�𝑡𝜙𝑖−1,𝑗 + �

1Δ𝑡−

1Δ𝑋2

�𝜙𝑖,𝑗𝑡 + �1

2Δ𝑋2+

14Δ𝑋

�𝜙𝑖+1,𝑗𝑡

Both schemes will work and the Crank-Nicholson method will produce 2nd order accuracy.

- Alternating Direct Implicit (ADI) method is an improved method producing 4th order accuracy without any extra computational cost (for more, refer the book by Ferziger):

First discretize 𝜙𝑋 and 𝜙𝑋𝑋 terms implicitly and 𝜙𝑌 and 𝜙𝑌𝑌 terms explicitly to solve 𝜙 over half step (Δ𝑡 2⁄ ) and then discretize 𝜙𝑌 and 𝜙𝑌𝑌 terms implicitly and 𝜙𝑋 , and 𝜙𝑋𝑋 terms explicitly:

Step1:

�𝜙𝑡+1 2� − 𝜙𝑡

Δ 𝑡 2⁄�𝑖,𝑗

=12�(𝜙𝑋𝑋 + 𝜙𝑋)𝑡+1 2� + (𝜙𝑌𝑌 + 𝜙𝑌)𝑡�

Step2:

�𝜙𝑡+1 −𝜙𝑡+1 2�

Δ 𝑡 2⁄�𝑖,𝑗

=12�(𝜙𝑋𝑋 + 𝜙𝑋)𝑡+1 2� + (𝜙𝑌𝑌 + 𝜙𝑌)𝑡�

or �𝐴𝜙𝑦

𝑡+1 2� = 𝜙𝑥𝑡

and 𝐵𝜙𝑥𝑡+1 = 𝜙𝑦𝑡+1 2�

�

The procedure can also be understood by the following illustration:

𝑖𝑚𝑝𝑙𝑖𝑐𝑖𝑡 𝑒𝑥𝑝𝑙𝑖𝑐𝑖𝑡

𝑒𝑥𝑝𝑙𝑖𝑐𝑖𝑡 𝑖𝑚𝑝𝑙𝑖𝑐𝑖𝑡

3

Ex: Consider 2D diffusion in a rectangular shaped solid porous carbon block. The block is initially soaked with moisture, say at concentration 𝐶𝑜 (𝑚𝑜𝑙𝑒𝑠 𝑚3)⁄ . At 𝑡 = 0+ all four sides of the block is exposed to dry air (moisture concentration = 𝐶𝑖 ≪ 𝐶𝑜). We are interested in calculating the unsteady-state concentration profiles of moisture within the block, ie. 𝐶(𝑡, 𝑥,𝑦) =?. Pore diffusion coefficient for moisture in solid is 𝐷𝑝𝑜𝑟𝑒.

Ans:

A species balance over ′Δ𝑋Δ𝑦. 1′ 𝐶𝑉 will yield the following conservation equation: 𝜕𝐶𝐴𝜕𝑡

+ 𝑉.∇𝐶𝐴 = 𝐷𝑝𝑜𝑟𝑒∇2𝐶𝐴 + (−𝑟𝐴)

𝜕𝐶𝐴𝜕𝑡

= 𝐷𝑝𝑜𝑟𝑒 �𝜕2𝐶𝐴𝜕𝑋2

+ 𝜕2𝐶𝐴𝜕𝑦2

�

𝑡 = 0 𝐶 = 𝐶𝑜 𝑓𝑜𝑟 𝐿 ≥ 𝑥 ≥ 0 ;𝑤 ≥ 𝑦 ≥ 0

� 0+ 𝐶 = 𝐶𝑖∗ @ 𝑥 = 0 & 𝐿 for 𝑤 > 𝑦 > 0 and @ 𝑦 = 0 & 𝑤 for 𝐿 > 𝑥 > 0�

(𝐶𝑖∗ is the solid phase moisture concentration at the surface of the block in equilibrium with 𝐶𝑖 in atmoshphere).

Step1: discretize ‘𝑖’ implicitly and ‘𝑗’ explicitly over (𝑡 & 𝑡 + 12� )

𝑖, 𝑗 + 1

𝑖 + 1, 𝑗 𝑖 − 1, 𝑗

𝑖, 𝑗 − 1

𝑖, 𝑗

𝑡

𝑖 − 1, 𝑗 𝑖 + 1, 𝑗

𝑖, 𝑗 + 1

𝑖, 𝑗 − 1

𝑖, 𝑗

𝑡

𝑡 + 12�

𝑡 + 1

𝑖 − 1, 𝑗

𝑖, 𝑗 + 1

𝑖, 𝑗 − 1 𝑖, 𝑗 𝑖 + 1, 𝑗

𝑡 + 12�

𝑠𝑤𝑒𝑒𝑝 ′𝑖′ 𝑚𝑎𝑟𝑐ℎ ′𝑖′ 𝑠𝑤𝑒𝑒𝑝 ′𝑗′

𝑚𝑎𝑟𝑐ℎ ′𝑗′

𝐶𝑖 𝑚𝑜𝑙𝑒𝑠 𝑚3⁄

𝐿

𝑦

𝑤

Δ𝑋

Δ𝑦

X

C(t, x, y) =?

4

𝐶𝑖,𝑗𝑡+1 2� − 𝐶𝑖,𝑗𝑡

Δ𝑡2�

=𝐷𝑝𝑜𝑟𝑒

2��𝐶𝑖+1,𝑗 − 2𝐶𝑖,𝑗 + 𝐶𝑖−1,𝑗

Δ𝑋2 �𝑡+1 2�

+ �𝐶𝑖,𝑗+1 − 2𝐶𝑖,𝑗 + 𝐶𝑖,𝑗−1

Δ𝑦2 �𝑡

�

(Δ𝑋 = L𝑁� ,Δ𝑦 = w

𝑀� )

Arrange:

𝐷𝑝𝑜𝑟𝑒2Δ𝑋2

𝐶𝑖−1,𝑗𝑡+1 2� − �𝐷𝑝𝑜𝑟𝑒

Δ𝑋2+ 2

Δ𝑡 �𝐶𝑖,𝑗

𝑡+1 2� + 𝐷𝑝𝑜𝑟𝑒2Δ𝑋2

𝐶𝑖+1,𝑗𝑡+1 2� = −𝐷𝑝𝑜𝑟𝑒

2Δ𝑦2 𝐶𝑖,𝑗−1𝑡 + �𝐷𝑝𝑜𝑟𝑒

Δ𝑦2− 2

Δ𝑡 � 𝐶𝑖,𝑗𝑡 −

𝐷𝑝𝑜𝑟𝑒2Δ𝑦2

𝐶𝑖,𝑗+1𝑡

𝑖 = 1,𝑁 − 1𝑗 = 1,𝑀 − 1

Step-2:

Discretize BCs: 𝐶𝑜,𝑗 = 𝐶𝑁,𝑗 = 𝐶∗; 𝑗 = 0,𝑀

𝐶𝑖,0 = 𝐶𝑖,𝑀 = 𝐶∗; 𝑖 = 0,𝑁

Prepare the tridiagonal matrix:

𝑎(𝑖, 𝑗) = 𝐷𝑝𝑜𝑟𝑒2Δ𝑋2

; 𝑏(𝑖, 𝑗) = −�𝐷𝑝𝑜𝑟𝑒Δ𝑋2

+ 2Δ𝑡

� ; 𝑐(𝑖, 𝑗) = 𝐷𝑝𝑜𝑟𝑒2Δ𝑋2

; 𝑑(𝑖, 𝑗) = −𝐷𝑝𝑜𝑟𝑒2Δ𝑦2

𝐶𝑖,𝑗−1𝑡 +

�𝐷𝑝𝑜𝑟𝑒Δ𝑦2

− 2Δ𝑡

�𝐶𝑖,𝑗𝑡 − 𝐷𝑝𝑜𝑟𝑒2Δ𝑦2

𝐶𝑖,𝑗+1𝑡

� 𝑖 = 1,𝑁 − 1𝑗 = 1,𝑀 − 1�

Substitute BCs and the following coefficients will be modified:

M M − 1

(i, j + 1)

0 𝑖 − 1, j 1 (𝑖, j) 𝑖 + 1, j 𝑁 − 1 𝑁

(𝑖, j − 1) 1

0

5

� 𝑑(1, 𝑗) = 𝑑(1, 𝑗) − 𝑎(1, 𝑗)𝐶∗𝑑(𝑁 − 1, 𝑗) = 𝑑(𝑁 − 1, 𝑗) − 𝑐(𝑁 − 1, 𝑗)𝐶∗� 𝑗 = 1,𝑀 − 1

Convince yourself that no other coefficients will change. This was actually an easy problem, when all four boundary conditions were simple, i.e., functional values were prescribed. Problems are complicated when you have ‘flux’/gradient or mixed boundary conditions; for example,

−𝐷 𝜕𝐶𝜕𝑋

= 𝑘𝑚(𝐶 − 𝐶𝑎𝑡𝑚)

𝑜𝑟 = Flux (known)

and/or −𝐷 𝜕𝐶𝜕𝑦

= 𝑘𝐶 𝑜𝑟 𝑘𝑚(𝐶 − 𝐶𝑎𝑡𝑚)

In such cases the other coefficients may also change.

Tridiagonal �𝑁 − 1,𝑎, 𝑏, 𝑐,𝑑,𝑦(𝑖, 𝑗)� 𝑖 = 1,𝑁 − 1

you will be calling the subroutine ′𝑀 − 1′ times as you ‘march’ along ′𝑗′ direction. Now, you

have the values for 𝐶𝑖,𝑗𝑡+1 2� , 𝑖 = 1,𝑁 − 1

𝑗 = 1,𝑀− 1

Step3:

Now march along ′𝑖′ direction and ‘sweep’ along ′𝑗′ direction to solve for

𝐶𝑖,𝑗𝑡+1 from 𝐶𝑖,𝑗𝑡+1 2� .

𝐶𝑖,𝑗𝑡+1 −𝐶𝑖,𝑗

𝑡+1 2�

Δ𝑡2�

= 𝐷𝑝𝑜𝑟𝑒2

��𝐶𝑖−1,𝑗−2𝐶𝑖,𝑗+𝐶𝑖+1,𝑗

Δ𝑋2 �𝑡+1 2� + �𝐶𝑖,𝑗−1−2𝐶𝑖,𝑗+𝐶𝑖,𝑗+1

Δ𝑦2 �𝑡+1

�

Arrange:

𝐷𝑝𝑜𝑟𝑒2Δ𝑦2

𝐶𝑖,𝑗−1𝑡+1 − �𝐷𝑝𝑜𝑟𝑒Δ𝑦2

+ 2Δ𝑡

�𝐶𝑖,𝑗𝑡+1 + 𝐷𝑝𝑜𝑟𝑒2Δ𝑦2

𝐶𝑖,𝑗+1𝑡+1 = −𝐷𝑝𝑜𝑟𝑒2Δ𝑋2

𝐶𝑖−1,𝑗𝑡+1 2� + �𝐷𝑝𝑜𝑟𝑒

Δ𝑋2−

2Δ𝑡

�𝐶𝑖,𝑗𝑡+1 2� − 𝐷𝑝𝑜𝑟𝑒

2Δ𝑋2 𝐶𝑖+1,𝑗

𝑡+1 2� ;

�𝑗 = 1,𝑀 − 1𝑖 = 1,𝑁 − 1�

𝑒𝑥𝑝𝑙𝑖𝑐𝑖𝑡 𝑖𝑚𝑝𝑙𝑖𝑐𝑖𝑡

6

Step-4:

Discretize BCs: Same as before.

Prepare the tridiagonal matrix:

𝑎(𝑖, 𝑗) = 𝐷𝑝𝑜𝑟𝑒2Δ𝑦2

; 𝑏(𝑖, 𝑗) = −�𝐷𝑝𝑜𝑟𝑒Δ𝑦2

+ 2Δ𝑡

� ; 𝑐(𝑖, 𝑗) = 𝐷𝑝𝑜𝑟𝑒2Δ𝑦2

; 𝑑(𝑖, 𝑗) = −𝐷𝑝𝑜𝑟𝑒2Δ𝑋2

𝐶𝑖−1,𝑗𝑡+1 2� +

�𝐷𝑝𝑜𝑟𝑒Δ𝑋2

− 2Δ𝑡

�𝐶𝑖,𝑗𝑡+1 2� − 𝐷𝑝𝑜𝑟𝑒

2Δ𝑋2 𝐶𝑖+1,𝑗

𝑡+1 2� ;

� 𝑖 = 1,𝑁 − 1𝑗 = 1,𝑀 − 1�

Substitute the discretized BCs, and only the following coefficients will be modified:

� 𝑑(𝑖, 1) = 𝑑(𝑖, 1) − 𝑎(𝑖, 1)𝐶∗ 𝑑(𝑖,𝑀 − 1) = 𝑑(𝑖,𝑀− 1) − 𝑐(𝑖,𝑀 − 1)𝐶∗� 𝑖 = 1,𝑁 − 1

Tridiagonal �𝑁 − 1,𝑎, 𝑏, 𝑐,𝑑,𝑦(𝑖, 𝑗)� 𝑗 = 1,𝑀 − 1

you will be calling the subroutine ′𝑁 − 1′ times as you ‘march’ along ′𝑖′ direction. Thus, you

have solved for 𝐶𝑖,𝑗𝑡+1, 𝑖 = 1,𝑁 − 1𝑗 = 1,𝑀 − 1

1

Lecture #28 Example: Consider the SS flow of a liquid through a long tube. Reynolds number is 180. At time 𝑡 = 0, a tracer is injected into the liquid at inlet to the tube. Diffusion coefficient of tracer in the liquid is 𝐷 𝑐𝑚2 𝑠⁄ . Determine the time profiles of the tracer concentrations (𝑟, 𝑥) in the tube, ie. 𝐶(𝑡, 𝑟, 𝑥) =?

Soln.

A species balance over ′2𝜋𝑟Δ𝑟Δ𝑥′ 𝐶𝑉 yields the following eqn

𝜕𝐶𝜕𝑡

+ 𝑉.∇𝐶 = 𝐷∇2𝐶 + (−𝑟𝐴)

or

𝜕𝐶𝜕𝑡

+ 𝑉(𝑟) 𝜕𝐶𝜕𝑋

= 𝐷 �𝜕2𝐶

𝜕𝑋2+ 1

𝑟𝜕𝜕𝑟�𝑟 𝜕𝐶

𝜕𝑟�� 𝐿 > 𝑋 > 0

𝑅 > 𝑟 > 0

or 𝜕𝐶𝜕𝑡

= �−𝑉(𝑟) 𝜕𝐶𝜕𝑋

+ 𝐷 𝜕2𝐶𝜕𝑋2

�+ 𝐷 �1𝑟𝜕𝐶𝜕𝑟

+ 𝜕2𝐶𝜕𝑟2

�

𝑡 = 0 𝐶 = 0 𝐿 ≥ 𝑥 ≥ 0 and 𝑅 ≥ 𝑟 ≥ 0

�0+ 𝐶 = 𝐶𝑜 and 𝑥 = 0

∇𝐶 = 𝜕𝐶𝜕𝑋

@ 𝑋 = 𝐿 � for all 𝑅 > 𝑟 > 0

(long tube approximation)

�𝜕𝐶𝜕𝑟

= 0 @𝑟 = 0 (𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐) = 0 @𝑟 = 𝑅 (𝑛𝑜𝑛 − 𝑟𝑒𝑎𝑐𝑡𝑖𝑣𝑒 𝑤𝑎𝑙𝑙𝑠)

� 𝑓𝑜𝑟 𝑎𝑙𝑙 𝐿 > 𝑋 > 0

𝐶𝑜 𝑅

𝑂

𝐿

Δ𝑟 𝑟

Δ𝑋

𝑅

𝑟

Δ𝑟 X

C(t,r,x)=?

V(r) = Umax �1 −𝑟2

𝑅2�

2

Note: You will encounter discontinuity at 𝑟 = 0 in the radial diffusion terms, while discretizing. Considering ∇𝐶 = 0, you will be solving the approximated equation instead.

𝜕𝐶𝜕𝑡

= �−𝑉(𝑟) 𝜕𝐶𝜕𝑋

+ 𝐷 𝜕2𝐶𝜕𝑋2

�+ 2𝐷 𝜕2𝐶𝜕𝑟2

𝑎𝑡 𝑟 = 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝐿 > 𝑋 > 0

Discretize the approximated equation:

𝐶𝑖,𝑗𝑡+1 2� − 𝐶𝑖,𝑗𝑡

Δ𝑡2�

=12���−𝑉𝑗�

𝐶𝑖+1,𝑗 − 𝐶𝑖−1,𝑗

2Δ𝑋 + 𝐷

𝐶𝑖+1,𝑗 − 2𝐶𝑖,𝑗 + 𝐶𝑖−1,𝑗

Δ𝑋2 �𝑡+1 2�

+ 2𝐷 �𝐶𝑖,𝑗+1 − 2𝐶𝑖,𝑗 + 𝐶𝑖,𝑗−1

Δ𝑟2 �𝑡� �; 𝑗 = 0

𝑖 = 1,𝑁�

Arrange;

�𝑉𝑗

4Δ𝑋+

𝐷2Δ𝑋2

� 𝐶𝑖−1,𝑗𝑡+1 2� − �

2Δ𝑡

+𝐷Δ𝑋2

� 𝐶𝑖,𝑗𝑡+1 2� − �

𝑉𝑗4Δ𝑋

−𝐷

2Δ𝑋2� 𝐶𝑖+1,𝑗

𝑡+1 2�

= −�𝐷Δ𝑟2

�𝐶𝑖,𝑗−1𝑡 − �2Δ𝑡

+2𝐷Δ𝑟2

�𝐶𝑖,𝑗𝑡 − �𝐷Δ𝑟2

�𝐶𝑖,𝑗+1𝑡 ; � 𝑗 = 0𝑖 = 1,𝑁�

(𝑖, M) (𝑖, M − 1)

(i, j + 1)

𝑖 = 0 𝑖 − 1, j 1 (𝑖, j) 𝑖 + 1, j 𝑁 − 1 𝑁

(𝑖, j − 1)

implicit

0 j =

explicit

1

3

Similarly, discretize the main equation:

𝐶𝑖,𝑗𝑡+1 2� − 𝐶𝑖,𝑗𝑡

Δ𝑡2�

=12���−𝑉𝑗�

𝐶𝑖+1,𝑗 − 𝐶𝑖−1,𝑗

2Δ𝑋 + 𝐷

𝐶𝑖+1,𝑗 − 2𝐶𝑖,𝑗 + 𝐶𝑖−1,𝑗

Δ𝑋2 �𝑡+1 2�

+ 𝐷 �1𝑟𝑗𝐶𝑖,𝑗+1 − 𝐶𝑖,𝑗−1

2Δ𝑟 +𝐶𝑖,𝑗+1 − 2𝐶𝑖,𝑗 + 𝐶𝑖,𝑗−1

Δ𝑟2 �𝑡� �𝑗 = 1,𝑀

𝑖 = 1,𝑁�

Arrange,

�𝑉𝑗

4Δ𝑋+

𝐷2Δ𝑋2

� 𝐶𝑖−1,𝑗𝑡+1 2� − �

2Δ𝑡

+𝐷Δ𝑋2

�𝐶𝑖,𝑗𝑡+1 2� − �

𝑉𝑗4Δ𝑋

−𝐷

2Δ𝑋2� 𝐶𝑖+1,𝑗

𝑡+1 2�

= �𝐷

2𝑟𝑗Δ𝑌−

𝐷2Δ𝑟2

�𝐶𝑖,𝑗−1𝑡 − �2Δ𝑡

+𝐷Δ𝑟2

�𝐶𝑖,𝑗𝑡

− �𝐷

2𝑟𝑗Δ𝑟+

𝐷2Δ𝑟2

�𝐶𝑖,𝑗+1𝑡 ; �𝑗 = 1,𝑀𝑖 = 1,𝑁�

Note: 𝑉𝑗 = 𝑈𝑚𝑎𝑥 �1 − 𝑟𝑗2

𝑅2� ; 𝑟𝑗 = 𝑗Δ𝑟

Discretize BCs & IC

𝑡 = 0, 𝐶(𝑖, 𝑗) = 0 𝑜r a small number 𝐶𝑖𝑛𝑡 ≪ 𝐶𝑜 (𝑁 ≥ 𝑖 ≥ 0 𝑎𝑛𝑑 𝑀 ≥ 𝑗 ≥ 0)

�0+ 𝐶(0, 𝑗) = 𝐶𝑜

𝐶(𝑁+1,𝑗)−𝐶(𝑁−1,𝑗)Δ𝑋

= 0𝑜𝑟 𝐶(𝑁 + 1, 𝑗) = 𝐶(𝑁 − 1, 𝑗)

� 𝑀 > 𝑗 > 0

� 𝐶(𝑖,−1) = 𝐶(𝑖, +1)𝐶(𝑖,𝑀 + 1) = 𝐶(𝑖,𝑀 − 1)�𝑁 > 𝑖 > 0

Prepare tridiagonal matrix

𝑗 = 0: 𝑎(𝑖, 0) = � 𝑉04Δ𝑋

+ 𝐷2Δ𝑋2

� ; 𝑏(𝑖, 0) = −� 2Δ𝑡

+ 𝐷Δ𝑋2

� ; 𝑐(𝑖, 0) = −� 𝑉04Δ𝑋

− 𝐷2Δ𝑋2

� ;𝑑(𝑖, 0) =

−� 𝐷Δ𝑟2

� 𝐶𝑖,−1𝑡 − � 2Δ𝑡

+ 2𝐷Δ𝑟2

�𝐶𝑖,0𝑡 − � 𝐷Δ𝑟2

� 𝐶𝑖,+1𝑡 ; 𝑖 = 1,𝑁

Substitute BCs,

𝑑(1,0) = −�2𝐷Δ𝑟2

�𝐶1,1𝑡 − �

2Δ𝑡

+2𝐷Δ𝑟2

� 𝐶1,0𝑡 − 𝑎(1,0)𝐶𝑜

implicit

explicit

4

𝑎(𝑁, 0) = 𝑎(𝑁, 0) + 𝑐(𝑁, 0)

𝑑(𝑁, 0) = −� 2𝐷Δ𝑟2

�𝐶𝑁,1𝑡 − � 2

Δ𝑡+ 2𝐷

Δ𝑟2�𝐶𝑁,0

𝑡

Tridiagonal �𝑁,𝑎, 𝑏, 𝑐, 𝑑,𝑦𝑡+1 2� (𝑖, 0)� ; 𝑖 = 1,𝑁

Preparation of tridiagonal matrix… continue….

𝑎(𝑖, 𝑗) = �𝑉𝑗

4Δ𝑋+

𝐷2Δ𝑋2

� ; 𝑏(𝑖, 𝑗) = −�2Δ𝑡

+𝐷Δ𝑋2

� ; 𝑐(𝑖, 𝑗) = −�𝑉𝑗

4Δ𝑋−

𝐷2Δ𝑋2

� ;𝑑(𝑖, 𝑗)

= �𝐷

2𝑟𝑗Δ𝑌−

𝐷2Δ𝑟2

�𝐶𝑖,𝑗−1𝑡 − �2Δ𝑡

+𝐷Δ𝑟2

�𝐶𝑖,𝑗𝑡

− �𝐷

2𝑟𝑗Δ𝑟+

𝐷2Δ𝑟2

�𝐶𝑖,𝑗+1𝑡 �𝑗 = 1,𝑀𝑖 = 1,𝑁�

Sbustitute BC (𝑗 = 1,𝑀 − 1)

𝑑(1, 𝑗) = 𝑑(1, 𝑗) − 𝑎(1, 𝑗)𝐶𝑜

𝑎(𝑁, 𝑗) = 𝑎(𝑁, 𝑗) + 𝑐(𝑁, 𝑗)

Tridiagonal �𝑁,𝑎, 𝑏, 𝑐, 𝑑,𝑦𝑡+1 2� (𝑖, 𝑗)� ; � 𝑖 = 1,𝑁𝑗 = 1,𝑀− 1�

You have called the tridiagonal subroutine ′𝑀 − 1′ times for the interior nodes.

Substitute BC @ 𝑗 = 𝑀

𝑑(1,𝑀) = −�𝐷Δ𝑟2

�𝐶𝑖,𝑀−1𝑡 − �2Δ𝑡

+𝐷Δ𝑟2

�𝐶𝑖,𝑀𝑡 − a(1, N). Co

𝑎(𝑁,𝑀) = 𝑎(𝑁,𝑀) + 𝑐(𝑁,𝑀)

𝑑(𝑁,𝑀) = −� 𝐷Δ𝑟2

�𝐶𝑁,𝑀−1𝑡 − � 2

Δ𝑡+ 𝐷

Δ𝑟2� 𝐶𝑁,𝑀

𝑡

Tridiagonal �𝑁,𝑎, 𝑏, 𝑐,𝑑,𝑦𝑡+1 2� (𝑖,𝑀)� ; 𝑖 = 1,𝑁

(Therefore, you have called the Thomas Algorithm (𝑀 + 1) times while sweeping 𝑗 =0,𝑀 rows)

For the last time in this course, let us write down complete eqn

5

𝐴𝑦𝑡+1 2� = 𝑑𝑡 𝑓𝑜𝑟 𝑗 = 0, 𝑖 = 1,𝑁

⎣⎢⎢⎢⎢⎢⎡− �

2Δ𝑡

+𝐷Δ𝑥2

� − �𝑉0

4Δ𝑋+

𝐷2Δ𝑋2

� ⋯ ⋯ ⋯

⋯ �𝑉0

4Δ𝑋+

𝐷2Δ𝑋2

� −�2Δ𝑡

+𝐷Δ𝑥2

� −�𝑉0

4Δ𝑋−

𝐷2Δ𝑋2

� ⋯⋮ ⋮ ⋮ ⋮ ⋮

⋯ ⋯ ⋯ �𝐷Δ𝑥2

� −�2Δ𝑡

+𝐷Δ𝑥2

�⎦⎥⎥⎥⎥⎥⎤

⎩⎪⎪⎨

⎪⎪⎧

𝑌𝑖

⎭⎪⎪⎬

⎪⎪⎫𝑡+1 2�

=

⎩⎪⎪⎨

⎪⎪⎧

𝑑(1,0)⋮

𝑑(𝑖, 0)𝑖 = 2,𝑁 − 1

⋮⋮

𝑑(𝑁, 0) ⎭⎪⎪⎬

⎪⎪⎫𝑡

Similarly, you can write a set of equations for 𝑗 = 1,𝑀 − 1 and 𝑗 = 𝑀 using the coefficient as

above:

𝑗 = 1,𝑀 − 1

⎣⎢⎢⎢⎢⎢⎡− �

2Δ𝑡

+4Δ𝑥2

� − �𝑉𝑗

4Δ𝑋+

𝐷2Δ𝑋2

� ⋯ ⋯ ⋯

⋯ �𝑉𝑗

4Δ𝑋+

𝐷2Δ𝑋2

� −�2Δ𝑡

+𝐷Δ𝑥2

� −�𝑉𝑗

4Δ𝑋−

𝐷2Δ𝑋2

� ⋯⋮ ⋮ ⋮ ⋮ ⋮

⋯ ⋯ ⋯ �𝐷Δ𝑥2

� −�2Δ𝑡

+𝐷Δ𝑥2

�⎦⎥⎥⎥⎥⎥⎤

⎩⎪⎪⎨

⎪⎪⎧

𝑌𝑖

⎭⎪⎪⎬

⎪⎪⎫𝑡+1 2�

=

⎩⎪⎪⎨

⎪⎪⎧

𝑑(𝑖, 𝑗)⋮

𝑑(𝑖, 𝑗)𝑖 = 2,𝑁 − 1

⋮⋮

𝑑(𝑁, 𝑗) ⎭⎪⎪⎬

⎪⎪⎫𝑡

6

𝑗 = 𝑀

⎣⎢⎢⎢⎢⎢⎡− �

2Δ𝑡

+4Δ𝑥2

� − �𝑉𝑀

4Δ𝑋−

𝐷2Δ𝑋2

� ⋯ ⋯ ⋯

⋯ �𝑉𝑀

4Δ𝑋+

𝐷2Δ𝑋2

� −�2Δ𝑡

+𝐷Δ𝑥2

� −�𝑉𝑀

4Δ𝑋−

𝐷2Δ𝑋2

� ⋯⋮ ⋮ ⋮ ⋮ ⋮

⋯ ⋯ ⋯ �𝐷Δ𝑥2

� −�2Δ𝑡

+𝐷Δ𝑥2

�⎦⎥⎥⎥⎥⎥⎤

⎩⎪⎪⎨

⎪⎪⎧

𝑌𝑖

⎭⎪⎪⎬

⎪⎪⎫𝑡+1 2�

=

⎩⎪⎪⎨

⎪⎪⎧

𝑑(1,𝑀)⋮

𝑑(𝑖,𝑀)𝑖 = 2,𝑁 − 1

⋮⋮

𝑑(𝑁,𝑀) ⎭⎪⎪⎬

⎪⎪⎫𝑡

Now, sweep in ′𝑗′ direction and march in ′𝑖′ direction for (𝑡 + 1) based on the 𝑦(𝑖, 𝑗) values you

determined at (𝑡 + 12� ) step (above)

𝐴𝑦𝑡+1(𝑖, 𝑗) = 𝑑𝑡+1 2 � � 𝑖 = 1,𝑁𝑗 = 0,𝑀�

Discretize 𝑗 = 0

𝐶𝑖,𝑗𝑡+1 − 𝐶𝑖,𝑗𝑡+1 2�

Δ𝑡2�

=12���−𝑉𝑗�

𝐶𝑖+1,𝑗 − 𝐶𝑖−1,𝑗

2Δ𝑋 + 𝐷

𝐶𝑖+1,𝑗 − 2𝐶𝑖,𝑗 + 𝐶𝑖−1,𝑗

Δ𝑋2 �𝑡+1 2�

+ 2𝐷 �𝐶𝑖,𝑗+1 − 2𝐶𝑖,𝑗 + 𝐶𝑖,𝑗−1

Δ𝑟2 �𝑡+1

� (𝑖 = 1,𝑁)

𝑗 = 1,𝑀

implicit

explicit

7

𝐶𝑖,𝑗𝑡+1 2� − 𝐶𝑖,𝑗

𝑡+1 2�

Δ𝑡2�

=12���−𝑉𝑗�

𝐶𝑖+1,𝑗 − 𝐶𝑖−1,𝑗

2Δ𝑋 + 𝐷

𝐶𝑖+1,𝑗 − 2𝐶𝑖,𝑗 + 𝐶𝑖−1,𝑗

Δ𝑋2 �𝑡+1 2�

+ 𝐷 �1𝑟𝑗𝐶𝑖,𝑗+1 − 𝐶𝑖,𝑗−1

2Δ𝑟 +𝐶𝑖,𝑗+1 − 2𝐶𝑖,𝑗 + 𝐶𝑖,𝑗−1

Δ𝑟2 �𝑡� (𝑖 = 1,𝑁)

In this last lecture of the course, I leave it here for you to do the remaining part (arranging the terms and applying BCs for 𝑖 = 1, (2⋯𝑁 − 1), 𝑁 rows to invert the matrix) as you sweep in ′𝑗′ direction, as an exercise. This topic on ADI stops here.

Note: (1) The SS solution 𝑦(𝑡, 𝑥, 𝑟) 𝑎𝑠 𝑡 → ∞ of the time-dependent 2D parabolic equation must be the same as that of the corresponding 2D elliptic PDE you have learnt how to solve in the preceding lecture, using ‘method of lines’, ie.

For 𝜕𝐶𝜕𝑡

+ 𝑉.∇𝐶 = 𝐷∇2𝐶 + (−𝑟𝐴) ;

𝐶(𝑡, 𝑥, 𝑟) 𝑎𝑠 𝑡 → ∞ must be the same as that of

𝑉.∇𝐷 = 𝐷∇2𝐶 + (−𝑟𝐴) ⇒ 𝐶(𝑥, 𝑟)

(2) A question arises. When asked to solve the elliptic (2D) PDE, should not or cannot we

artificially insert the transient term 𝜕𝐶𝜕𝑡

and seek the SS solution to the corresponding time-

dependent 2D parabolic equation? Very often, yes. Recall that, solving elliptic PDE requires iterations and there is always a convergence issue. How many iterations? On the other hand, the parabolic equation does not require iterations and you march on ′𝑡′ axis solving 𝑦(𝑥, 𝑟) at every time step without iterations. Therefore, more than often ADI method is preferred over ‘Method of lines’ for solving an elliptic (2D) PDE. Insert the transient term and solve till you have SS solution.

(3) Before closing this chapter, let us answer how we address non-linearity in the differential

term, for example, V𝜕𝑉𝜕𝑥

of the NS equation? Answer is simple. By iterations! Guess velocity

fields (Vo). Discretize the derivative term as before. Solve for velocity fields as before. Iterate till there is convergence.

End – Semester Exam These lectures are usually covered in approximately 42 one-hour or 28 1hr 15min lectures. All comments including suggestions and corrections may be sent to vermanishith@gmail.com .