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7 PARTIAL DIFFERENTIAL EQUATIONS

Introduction

Elliptic Equations

Parabolic Equations

Hyperbolic Equations

Finite Element Method — An Introduction

Boundary Element Method — An Introduction

Chapter 7 Partial Differential Equations / 2

7.1 Introduction

• Partial Differential Equations (PDE) are differential equations which have at least two independent variables, e.g.

34 2

2

2

2

=+∂∂

+∂∂ u

yuxy

xu

second order & linear

xyxu

xu

=∂∂

∂+⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

2

33

2

2

6 third order & nonlinear

• For a second order linear two-dimensional equation, a general equation is

0,,,,2

22

2

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

+∂∂

+∂∂

∂+

∂∂

yu

xuuyxD

yuC

yxuB

xuA (7.1)

which can be divided into three types.

TABLE 7.1 Types of second order linear PDEs

ACB 42 − Type Examples

< 0 Elliptic 02

2

2

2

=∂∂

+∂∂

yT

xT

Laplace equation

= 0 Parabolic 2

2

xTk

tT

∂∂

=∂∂

Heat conduction equation

> 0 Hyperbolic 2

2

22

2 1x

yct

y∂∂

=∂∂

Wave equation


7.2 Elliptic Equations

• Elliptic PDEs are generally related to steady-state problems with diffusivity having boundary conditions, e.g. the Laplace equation.

• Consider a steady-state 2-D heat conduction problem:

Δz

Δx

Δyq(x)

q(y)

q(x + Δx)

q(y + Δy)

x

y

FIGURE 7.1 Thin plate having a thickness Δz with temperatures at the boundary

Taking q(x) as the heat flux [J/m2⋅s], the heat flow over the element can be written in a time interval of Δt as:

( ) ( ) ( ) ( ) tzxyyqtzyxxqtzxyqtzyxq ΔΔΔΔ++ΔΔΔΔ+=ΔΔΔ+ΔΔΔ

which can be summarised into

( ) ( )( ) ( ) ( )( )( ) ( ) ( ) ( ) 0

0

=Δ

Δ+−+

ΔΔ+−

=ΔΔ+−+ΔΔ+−

yyyqyq

xxxqxq

xyyqyqyxxqxq

For 0, →ΔΔ yx :

0=∂

∂−

∂∂

−y

qxq yx (7.2)


• In conduction heat transfer, the relationship between heat flux and temperature is given by the Fourier Law:

yTCkq

xTCkq yx ∂

∂−=

∂∂

−= ρρ , (7.3)

where k is heat diffusivity [m2/s], ρ is density [kg/m3] and C is heat capasity [J/kg°C]. Combining Eq. (7.2) with Eq. (7.3) gives the Laplace equation:

02

2

2

2

=∂∂

+∂∂

yT

xT

(7.4)

• If a heat source or heat generation of ),( yxQ is present in the domain, a Poisson equation can be formed:

( ) 0,2

2

2

2

=+∂∂

+∂∂ yxQ

yT

xT

(7.5)

• To solve Eq. (7.4), use the central differencing:

( )

( )21,,1,

2

2

2,1,,1

2

2

2

2

y

TTTyT

xTTT

xT

jijiji

jijiji

Δ

+−=

∂∂

Δ

+−=

∂∂

−+

−+

Hence, Eq. (7.4) can be written in an algebraic form:

( ) ( )0

222

1,,1,2

,1,,1 =Δ

+−+

Δ

+− −+−+

yTTT

xTTT jijijijijiji

Taking Δx = Δy gives

04 ,1,1,,1,1 =−+++ −+−+ jijijijiji TTTTT (7.6)


0,0 x

y

m+1,0

0,n+1

i,ji,j+1

i−1,j

i,j−1i+1,j

FIGURE 7.2 Finite difference grid for solving the Laplace equation

• Eq. (7.6) needs boundary conditions (BC), which may be in the form of:

1. Fixed value — Dirichlet (see Fig. 7.3), e.g. Fixed temperature 2. First derivative, or gradient —Neumann, e.g. Fix heat flux or insulated

75°C

T11 T21 T31

T12 T22 T32

T13 T23 T33

100°C

50°C

0°C FIGURE 7.3 Grid/Node representatioan and BC for Fig. 7.1

For node (1,1) — T01 = 75°C and T10 = 0°C:

04 1110120121 =−+++ TTTTT

754 211211 =−− TTT


Hence, the complete system (can be assembled into a matrix equation) is:

15041004175450404

75450404

754

332332

33231322

231312

33322231

2332221221

13221211

323121

22312111

122111

=+−−=−+−−=−+−=−+−−=−−+−−=−−+−=−+−=−−+−=−−

TTTTTTT

TTTTTTT

TTTTTTTTT

TTTTTTT

TTT

• If the Gauss-Seidel method is used, at each node (i,j):

41,1,,1,1

,−+−+ +++

= jijijijiji

TTTTT (7.7)

which can be solve iteratively via relaxation approach:

( ) old,

new,

new, 1 jijiji TTT λλ −+= (7.8)

where λ is the relaxation parameter 21 ≤≤ λ , and is subjected to the termination criterion as followed:

%100new,

old,

new,

,×

−=

ji

jijia T

TTji

ε

Example 7.1

Solve the problem in Fig. 7.3 using the Gauss-Seidel method using the termination criterion %1≤aε dan the relaxation parameter λ = 1.5.

Solution

Let all the initial values be 0°C. For the first iteration:

75.184

007504

1012012111 =

+++=

+++=

TTTTT

( ) ( ) 125.2805.1175.185.1new11 =−+=T

03125.74

00125.2804

2022113121 =

+++=

+++=

TTTTT


( ) ( ) 54688.1005.1103125.75.1new21 =−+=T

13672.154

0054688.10504

3032214131 =

+++=

+++=

TTTTT

( ) ( ) 70508.205.1113672.155.1new31 =−+=T

and, for other nodes:

99554.9618579.34

46900.7445703.18

12696.8067188.38

33

23

23

22

13

12

==

==

==

TT

TT

TT

Error for node (1,1) — for the first iteration, all errors are 100%:

%100100125.28

0125.2811

=×−

=aε

For the second iteration:

68736.6786833.7160108.28

95872.8763333.6135718.22

21973.7595288.5751953.32

33

32

31

23

22

21

13

12

11

===

===

===

TTT

TTT

TTT

The process is repeated until the ninth iteration in which the termination criterion is fulfilled (εa < 1%):

71050.6933999.5288506.33

06402.7611238.5629755.33

58718.7821152.6300061.43

33

32

31

23

22

21

13

12

11

===

===

===

TTT

TTT

TTT


7.3 Parabolic Equations

• Parabolic PDEs are generally related to transient problems with diffusivity, e.g. the 1-D heat conduction equation.

• For a transient problem, there are three approaches:

1. Explicit method, 2. Implicit method, 3. Semi-implicit method — the Crank-Nicolson method.

• Consider a transient 1-D heat conduction problem:

input − output = storage

( ) ( ) TCzyxtzyxxqtzyxq ΔΔΔΔ=ΔΔΔΔ++ΔΔΔ ρ

x

Hot Cold

FIGURE 7.4 A rod with different temperature at its ends

Dividing with the volume zyx ΔΔΔ and the time interval Δt:

( ) ( )tTC

xxxqxq

ΔΔ

=Δ

Δ+− ρ

In the limits of 0, →ΔΔ tx :

tTC

xq

∂∂

=∂∂

− ρ (7.9)

By using the Fourier law, Eq. (7.3), Eq. (7.9) becomes

tT

xTk

∂∂

=∂∂

2

2

(7.10)


0,0 x

t

m+1,0

i,li,l+1

i−1,l

i,l−1i+1,l

FIGURE 7.5 Finite difference grid for the heat conduction equation

• For the explicit method, the right-hand side of Eq. (7.9) can be discretised via central difference as:

( )211

2

2 2x

TTTxT l

il

il

i

Δ+−

=∂∂ −+

and, the left-hand side can be discretised via forward difference as:

tTT

tT l

il

i

Δ−

=∂∂ +1

Hence, Eq. (7.10) can be written in the algebraic form:

( ) tTT

xTTTk

li

li

li

li

li

Δ−

=Δ

+− +−+

1

211 2

(7.11)

( )li

li

li

li

li TTTTT 11

1 2 −++ +−+= λ (7.12)

where 2)( xtk ΔΔ=λ .

xi

Time differencing grid

Space differencing grid

xi−1 xi+1

tl

tl+1

FIGURE 7.6 Computational grid for the explicit method


Example 7.2

Use the explicit method to determine the temperature distribution for a slender rod having a length of 10 cm. At time t = 0, the temperature of the rod is 20°C and the boundary conditions are T(0) = 100°C and T(10 cm) = 50°C. Use the conduction coefficient k = 0.835 cm2/s, the time interval Δt = 0.5 s and the step size Δx = 2 cm. Solution

Calculate λ:

( )( )( ) 104375.0

25.0835.0

22 ==Δ

Δ=

xtkλ

At t = 0:

2004

03

02

01 ==== TTTT

Use Eq. (7.12). At t = 0.5 s:

[ ] 35.28100)20(2201044.02011 =+−+=T

[ ] 2020)20(2201044.02012 =+−+=T

[ ] 2020)20(2201044.02013 =+−+=T

[ ] 1313.2320)20(2501044.02014 =+−+=T

At t = 1 s:

[ ] 9569.34100)352.28(2201044.0352.2821 =+−+=T

[ ] 8715.20352.28)20(2201044.02022 =+−+=T

[ ] 3268.2020)20(2132.231044.02023 =+−+=T

[ ] 6089.2520)132.23(2501044.0132.2324 =+−+=T

The calculation can be continued to produce the result as in Fig. 7.7.


0

20

40

60

80

100

0 2 4 6 8 10 x (cm)

T (°C)

t = 0

t = 15 st = 10 s

t = 6 st = 3 s

t = 20 st → ∞

FIGURE 7.7 Result for Ex. 7.2

• For the explicit method, a more accurate result can be obtained if Δx and Δt approach zeros. However, stability of the results is only obtained if:

( )kxt

2

21

21

or Δ≤Δ

≤λ

If the stability condition is not fulfilled, the results are contaminated by oscillation.

• To prevent oscillation, the implicit method can be used, where the second order spatial derivative is approximated at time l+1:

( )2

11

111

2

2 2x

TTTxT l

il

il

i

Δ+−

=∂∂ +

−++

+

xi



xi−1 xi+1

tl

tl+1

FIGURE 7.8 Computational grid for the implicit method


Hence, Eq. (7.10) becomes

( ) tTT

xTTTk

li

li

li

li

li

Δ−

=Δ

+− ++−

+++

1

2

11

111 2

(7.13)

( ) li

li

li

li TTTT =−++− +

+++

−1

111

1 21 λλλ (7.14)

Eq. (7.14) leads to n simultaneous linear equations having n unknowns.

Example 7.3

Repeat Ex. 7.2 using the implicit method. Solution

From Ex. 7.2, λ = 0.104375. From Eq. (7.14):

( )[ ]( )[ ]

M

44.301044.01044.021

201044.01044.021)100(1044.012

11

12

11

=−+

=−++−

TT

TT

Hence, the following system of linear equations can be formed:

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−

−−−

22.252020

44.30

2088.11044.0001044.02088.11044.00

01044.02088.11044.0001044.02088.1

14

13

12

11

TTTT

[ ]T6152.22,2800.20,6256.20,9634.26=T

• The combination of the explicit and implicit approaches produces the semi-implicit method, and one of its kind is the Crank-Nicolson method:

xi



xi−1 xi+1

tl

tl+1

FIGURE 7.9 Computational grid for the Crank-Nicolson method


In this method, the finite difference term for the spatial derivative is

( ) ( ) ⎥⎦

⎤⎢⎣

⎡

Δ+−

+Δ

+−=

∂∂ +

−++

+−+2

11

111

211

2

2 2221

xTTT

xTTT

xT l

il

il

il

il

il

i (7.15)

Hence, Eq. (7.10) becomes

( ) ( ) li

li

li

li

li

li TTTTTT 11

11

111 1212 +−

++

++− +−+=−++− λλλλλλ (7.16)

Example 7.4

Repeat Ex. 7.2 using the Crank-Nicolson method. Solution

From Ex. 7.2, λ = 0.104375. From Eq. (7.16):

7866.5810437.020874.2)20(10437.020)10437.01(2)100(10437.0

10437.0)10437.01(2)100(10437.0

12

11

12

11

=−

+−+=−++−

TT

TT

By considering other nodes, the following system of linear equations can be formed:

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−

−−−

3496.484040

7866.58

2087.21044.0001044.02087.21044.00

01044.02087.21044.0001044.02087.2

14

13

12

11

TTTT

[ ]T8424.22,1517.20,3652.20,5778.27=T


7.4 Hyperbolic Equations

• Hyperbolic PDEs are generally related to transient problems with convection, e.g. the 1-D wave equation.

• Consider a 1-D wave equation, which is a hyperbolic PDE:

0=∂∂

+∂∂

xua

tu

(7.17)

• One of the methods is the MacCormack’s technique, which is an explicit finite-difference technique and is second-order-accurate in both space and time. By using the Taylor series:

ttuuu t

itt

i Δ⎟⎠⎞

⎜⎝⎛

∂∂

+=Δ+

avg (7.18)

• This method consists of two steps: predictor and corrector. In predictor step, use forward difference in the right-hand side:

⎟⎟⎠

⎞⎜⎜⎝

⎛Δ−

−=⎟⎠⎞

⎜⎝⎛

∂∂ +

xuua

tu t

iti

t

i

1 (7.19)

Thus, from the Taylor series, the predicted value of u is:

ttuuu

t

i

ti

tti Δ⎟

⎠⎞

⎜⎝⎛

∂∂

+=Δ+ (7.20)

• In corrector step, by replacing the spatial derivatives with rearward differences:

⎟⎟⎠

⎞⎜⎜⎝

⎛Δ−

−=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂ Δ+

−Δ+Δ+

xuua

tu tt

itt

i

tt

i

1 (7.21)

The average of time derivative can be obtained using

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

=⎟⎠⎞

⎜⎝⎛

∂∂

Δ+ tt

i

t

i tu

tu

tu

21

avg (7.22)


• Hence the final, “corrected” value at time t+Δt is:

ttuuu t

itt

i Δ⎟⎠⎞

⎜⎝⎛

∂∂

+=Δ+

avg

• The accuracy of the solution for a hyperbolic PDE is dependent on truncation and round off errors, and the term representing it is called artificial viscosity ( )ν−Δ 12

1 xa .

• The effect of artificial viscosity leads to numerical dissipation, which is originated by the even-order derivatives in the truncated term, but it improves stability.

FIGURE 7.10 Numerical dissipation: (a) t = 0, (b) t > 0

• Another opposite effect is known as numerical dispersion, which is originated by the odd-order derivatives in the truncated term and causes ‘wiggles’.

FIGURE 7.11 Numerical dispersion: (a) t = 0, (b) t > 0


7.5 Finite Element Method ⎯ An Introduction

• The Finite Element Method (FEM) is a computer aided mathematical technique used to obtain an approximate numerical solution of a response of a physical system which is subjected to an external loading.

• By using this technique, the computational domain which is theoretically a continuum, is being discretised in form of simple geometries.

• The mesh is the computational domain which is an assembly of discrete elemental blocks known as finite elements, and the vertices defining the elements are called nodes.

• Governing equation is employed at each element to form a set of algebraic equations ⎯ local system.

• Local equations assembled to form a global system which is the solved to yield a vector of variables.

Node & element Sample problem Mesh

y

x

z

x,y

node

element

3-D

2-D

1-D

FIGURE 7.10 Examples of elements and their applications


Geometry, material properties, mesh,BC, IC

Discretisation of structures into elements

Formation of local stiffness matrix [k]

Formation of global stiffness matrix [K]

Boundary conditions as

constraints to the model

Load conditionsto the model

Solving the system of Linear equation

[K]{a}= {F}

Calculaton of strains,stresses and forces

Boundary conditions

Loads

Graphical visualisation

print-out & files

FIGURE 7.11 Algorithm for a force analysis using FEM

• Consider a 1-D steady state heat conduction:

( ) 02

2

=+∂∂ xQ

xTk (7.17)

Eq. (7.17) needs appropriate boundary conditions such that:

)(00

∞=

=

−==

TThqTT

LLx

x (7.18)


T0

L

hT∞

xT0 TL

1 2 3 4

1 2 3

FIGURE 7.12 Boundary condition for a 1D steady state heat conduction

For the first element:

1 2 x1

x x2

1 2

ξ = −1 ξ = +1

ξ

FIGURE 7.13 The first element in the finite element method

The transformation of the coordinate system to a local system is given by an isoparametric coordinate ξ, i.e.

( ) 121

12

−−−

= xxxx

ξ (7.19)

Thus, in order to calculate the temperature at the middle section, a linear interpolation function or a linear shape function can be used:

21)(

21)(

2

1

ξξ

ξξ

+=

−=

N

N (7.20)

Hence, the temperature can be interpolated using the shape function as followed:

eTNTNT NT=+= 2211)(ξ (7.21)


Ni

ξ = -1 ξ = +1

N1 N2

T

ξ = -1 ξ = +1T1

T2

T = N1T1 + N2T2

FIGURE 7.14 Linear shape function

Differentiation of Eq. (7.21) gives

dxxx

d12

2−

=ξ

Using a chain rule:

[ ]e

e

xx

dxd

ddT

dxdT

BT

T

=

−−

=

=

1,11

12

ξξ

where

[ ]1,11

12

−−

=xx

B

• In energy form, the heat conduction problem can be represented by

( ) 02

2

=Ω⎭⎬⎫

⎩⎨⎧

+∂∂

∫ΩdxQ

xTkT

i

( )∫ ∫ ∞−+−⎟⎠⎞

⎜⎝⎛=∏

L

L

L

T TThdxxTQdxdxdTk

0

221

0

2

21 )( (7.22)

Discretisation of Eq. (7.22) gives

( )221

1

1

1

1

T21

22T

∞−−−+⎥⎦

⎤⎢⎣⎡−⎥⎦

⎤⎢⎣⎡=∏ ∑ ∫∑ ∫ TThdlQdlk

Le

eeee

e

eee TNTBBT ξξ (7.23)


Hence, the stiffness matrix for the element is

⎥⎦

⎤⎢⎣

⎡−

−== ∫− 11

112

1

1

T

e

eee

lkdlk ξBBk (7.24)

and, the heat rate vector is

⎭⎬⎫

⎩⎨⎧

== ∫− 11

221

1eeee lQdlQ ξNr (7.25)

The global stiffness matrix can be produced via an assembly of all stiffness matrices for all elements, i.e.

∑←e

ekK (7.26)

Likewise, the global heat rate vector is an assembly by local heat rate vectors:

∑←e

erR (7.27)

To combine with the boundary condition 01 TT = , a penalty method can be used:

( )

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

∞+

+

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

+

+

hTR

RCTR

T

TT

hKKK

KKKKKCK

LLLLLL

L

L

MM

L

MMM

L

L

2

01

2

1

21

22212

11211

(7.28)

or, in a matrix form

K⋅T = R (7.29)

where the penalty parameter C can be estimated as:

410max ×= ijC K (7.30)


Example 7.5

Fig. 7.15 shows plate of three composites, where the temperature at the right-hand side is T0 = 20°C and the left-hand side is subjected to convection with T∞ = 800°C and h = 25 W/m2°C. Obtain a temperature distribution across the plate using the finite element method.

k1 k2 k3

k1 = 20 W/m°Ck2 = 30 W/m°Ck3 = 50 W/m°C

T0 = 20°C

0.3m 0.15m 0.15m

1 2 3 4

1 2 3T4 = 20°CT1

T∞ = 800°C

RAJAH 7.15 Plate of three composites used in Ex. 7.5

Solution

Divide the domain into three elements and construct local stiffness matrices for all elements:

⎥⎦

⎤⎢⎣

⎡−

−=

⎥⎦

⎤⎢⎣

⎡−

−=

⎥⎦

⎤⎢⎣

⎡−

−=

1111

15.050

1111

15.030

1111

3.020

3

2

1

k

k

k

Combine all local stiffness matrices to form a global stiffness matrix:

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−

−−−

=

5500583003410011

7.66K


whereas the heat rate vector is

[ ]T0,0,0,80025×=R

The penalty parameter C can be estimated by 44 1087.6610max ××=×= ijC K

Hence, the finite element system can be solved as followed:

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

×

×

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−

−−−

44

3

2

1

10672,100080025

005,8050058300341001375.1

7.66

TTTT

[ ]T0.20,1.57,0.119,6.304=T


7.6 Boundary Element Method ⎯ An Introduction

• The Boundary Element Method (BEM) is a relatively new for PDE, where it consists only boundary elements, either line (2-D) or surface (3-D) elements.

• Consider a Laplace equation for a steady-state potential flow problems (ψ is the stream function):

02 =∇ ψ

• For a multi-dimensional case, this equation leads to an analytical solution:

r1ln

21π

=∗ψ (7.31)

FIGURE 7.15 Boundary elements

• Eq. (7.31) can be discretised to yield:

⎟⎠⎞⎜

⎝⎛⎟

⎠⎞

⎜⎝⎛

∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂

∂+ ∫∑∫∑ ∗

=

∗

=j

N

j jj

N

jji ds

nds

nψψψψψ

1121 (7.32)

FIGURE 7.16 Example of an analysis using BEM


Exercises

1. A quarter of disc having a radius of 8 cm as shown below has a variation of temperature at the boundary aligned with the principal axes, while the temperature is fixed at 100°C at its outer radius of 8 cm. If the temperature distribution follows the Laplace equation:

02 =∇ T

20°CT1 T2 T3

T4 T5 T6

T7 T8

100°C

0°C 40°C 65°C 85°C 100°C

50°C

75°C

100°C

100°C

By using the grid as shown:

a. Obtain a system of algebraic equations using the finite difference method, b. Solve the system to obtain Ti .

2. By using the implicit technique, solve the following heat conduction problem:

),,(),( 2

2

txTx

txTt ∂

∂=

∂∂ α ,100 << x t > 0

using the following boundary conditions:

.0)0,(,100),10(,0),0( === xTtTtT

Use the constant 10=α , the time step 1.0=Δt , and a model of 10 grid including the grid at the boundary. Compar this solution with the solution obtained by using 3.0=Δt .

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