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Numerical Analysis of Differential Equations 215 6 Numerical Solution of Parabolic Equations 6 Numerical Solution of Parabolic Equations TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 215

6 Numerical Solution of Parabolic Equations

6 Numerical Solution of Parabolic Equations TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 216

6.1 The One-Dimensional Model Problem

We consider the following initial boundary value problem (IBVP) modellingheat flow in a thin rod, i.e., in one space dimension:

ut = κuxx, x ∈ (0, 1), t > 0, (6.1a)

u(0, t) = g0(t), t > 0, (6.1b)

u(1, t) = g1(t), t > 0, (6.1c)

u(x, 0) = u0(x), x ∈ [0, 1] (6.1d)

with given (constant) heat conductivity κ (which we set to one in the follo-wing) as well as (possibly time-dependent) Dirichlet boundary values g0, g1

and initial data u0.

6.1 The One-Dimensional Model Problem TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 217

• Steady-state version:

uxx = 0, u(0) = g0, u(1) = g1.

• Analogous IBVP in 2D and 3D:

ut = ∆u + initial and boundary data.

• Related: linear ordinary differential equation

ut = Au , u : t 7→ u(t) ∈ Rn.

Here linear differential operator ∂xx in place of matrix A.

6.1 The One-Dimensional Model Problem TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 218

Series solution by separation of variables: in special cases an ana-lytic representation of the (exact) solution may be constructed using thetechnique of separation of variables. This is helpful for checking numeri-cal approximations and provides important insight into the structure of thesolution.

Inserting the special trial solution u(x, t) = f(x)g(t) into the PDE ut = uxxresults in

fg′ = f ′′g, i.e.g′

g=f ′′

f= const. =: −k2.

For each value of k we obtain a solution

uk(x, t) = e−k2t sin(kx)

of ut = uxx. The boundary condition u(0, t) = u(1, t) = 0 constrains k to thediscrete values

k = km := mπ, m ∈ N.

6.1 The One-Dimensional Model Problem TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 219

Due to the linearity and homogeneity of ut = uxx any linear combination ofthese solutions is also a solution. If we succeed in finding coefficients amin such a way that

u0(x) =∞∑m=1

am sin(mπx),

then the series

u(x, t) :=∞∑m=1

ame−m2π2t sin(mπx)

solves the complete IBVP (6.1).

Since the functions {sin(mπx)}∞m=1 form a complete orthogonal system ofthe function space L2(0, 1), this is possible for all u0 ∈ L2(0, 1).

The coefficients are given by

am = 2

∫ 1

0

u0(x) sin(mπx) dx.

6.1 The One-Dimensional Model Problem TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 220

Example:

u0(x) = 1− 2|x− 12 |.

Here the coefficients are

am =8

m2π2sin

2,

m ∈ N.

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

N=1N=3N=5N=19u

0(x)

Partial sums∑Nm=1 am sin(mπx)

of the Fourier series of u0.

6.1 The One-Dimensional Model Problem TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 221

0 1 2 3 4 5 6 710

−120

10−100

10−80

10−60

10−40

10−20

100

m

t=0t=1/2

Decay of first 4 nonzero Fouriercoefficients of u0.

0

0.5

1

00.2

0.40.6

0.81

0

0.2

0.4

0.6

0.8

1

tx

Fourier series solution of IBVP (6.1) withκ = 1 and u0(x) = 1− 2|x− 1

2 |in domain (x, t) ∈ [0, 1]× [0, 1].

6.1 The One-Dimensional Model Problem TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 222

We first discretize the IBVP (6.1) in the spatial variable x only, leaving timet continuous. To this end we proceed as in the elliptic case and introducethe grid points

0 = x0 < x1 < · · · < xJ < xJ+1 = 1

using a fixed grid spacing ∆x = 1/(J + 1) and approximate

uxx|x=xj ≈ [A∆xu ]j , j = 1, 2, . . . , J,

withA∆x =

1

∆x2tridiag(1,−2, 1). (6.2)

If u = u(t) denotes the vector with components

uj(t) ≈ u(xj , t), t > 0, j = 1, 2, . . . , J,

6.1 The One-Dimensional Model Problem TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 223

then (6.1) is transformed into the semi-discrete system of ODEs

u ′(t) = A∆xu(t) + g(t), (6.3a)

u(0) = u0, (6.3b)

with [u0]j = u0(xj), j = 1, 2, . . . , J as well as

g(t) = 1/(∆x2)[g0(t), 0, . . . , 0, g1(t)]> ∈ RJ .

We can now solve (6.3) with known numerical methods for solving ODEs.Introducing the fixed time step ∆t > 0, we set

Unj ≈ [u(tn)]j ≈ u(xj , tn), tn = n∆t.

The approximation of the solution of a time-dependent PDE as a system ofODEs along the “lines” {(xj , t) : t > 0} is known as the method of lines.

6.1 The One-Dimensional Model Problem TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 224

Applying the explicit Euler method to (6.3) (setting g0(t) = g1(t) = 0 fornow) leads to

Un+1j = Unj +

∆t

∆x2

(Unj−1 − 2Unj + Unj+1

), 1 ≤ j ≤ J, n = 0, 1, 2 . . . .

This corresponds to the finite difference approximation

Un+1j − Unj

∆t︸ ︷︷ ︸≈ut

=Unj−1 − 2Unj + Unj+1

∆x2︸ ︷︷ ︸≈uxx

(6.4)

of the differential equation (6.1a).

In matrix notation:

U n+1 = (I + ∆t A∆x)U n, n = 0, 1, 2, . . . ,

with U n = [Un1 , Un2 , . . . , U

nJ ]>.

6.1 The One-Dimensional Model Problem TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 225

We define the local discretisation error of the difference scheme (6.4) tobe the residual obtained on inserting the exact solution into the differencescheme:

d(x, t) :=u(x, t+ ∆t)− u(x, t)

∆t− u(x−∆x, t)− 2u(x, t) + u(x+ ∆x, t)

∆x2.

Using Taylor expansions in (x, t) one easily obtains:

deE(x, t) =

(∆t

2− ∆x2

12

)uxxxx +O(∆t2) +O(∆x4). (6.5)

Terminology: the explicit Euler method for solving the heat equation isconsistent of first order in time and of second order in space.

6.1 The One-Dimensional Model Problem TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 226

Besides the asymptotic statement (6.5) we also require upper bounds for d.Truncating the Taylor expansions with a remainder term, we obtain in time

u(x, t+ ∆t) = (u+ ∆t ut) |(x,t) +∆t2

2utt(x, τ), τ ∈ (t, t+ ∆t).

Proceeding analogously in x yields

deE(x, t) =∆t

2utt(x, τ)− ∆x2

12uxxxx(ξ, t), ξ ∈ (x−∆x, x+ ∆x),

and we obtain, setting µ := ∆t∆x2 ,

|deE(x, t)| ≤ ∆t

2Mtt −

∆x2

12Mxxxx =

∆t

2

(Mtt +

1

6µMxxxx

), (6.6)

assuming |utt| ≤Mtt und |uxxxx| ≤Mxxxx on [0, 1]× [0, T ].

6.1 The One-Dimensional Model Problem TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 227

Applying the implicit Euler method to (6.3) one obtains, in place of (6.4),the implicit difference scheme

Un+1j − Unj

∆t=Un+1j−1 − 2Un+1

j + Un+1j+1

∆x2. (6.7)

The calculation of un+1 from un is seen to require the solution of a linearsystem of equations with coefficient matrix I −∆tA∆x.

Here Taylor expansion results in a local discretization error of

diE(x, t) = −(

∆t

2+

∆x2

12

)uxxxx +O(∆t2) +O(∆x4).

6.1 The One-Dimensional Model Problem TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 228

Applying instead the trapezoidal rule, which for an ODE y ′(t) = f (t,y(t))

is given by

yn+1 = yn +∆t

2

[f (tn,y

n) + f (tn+1,yn+1)

]yields another implicit scheme

Un+1j − Unj

∆t=

1

2

(Unj−1 − 2Unj + Unj+1

∆x2+Un+1j−1 − 2Un+1

j + Un+1j+1

∆x2

),

which in this context is known as the Crank-Nicolson schemea. This methodis also implicit, requiring in each time step the solution of a linear system ofequations with the coefficient matrix I − ∆t

2 A∆x.

For Crank-Nicolson (CN) there holds

dCN(x, t) = O(∆t2) +O(∆x2).

aJ. CRANK AND P. NICOLSON (1947)

6.1 The One-Dimensional Model Problem TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 229

6.2 Convergence

All three methods considered so far are consistent, i.e., at all points (x, t)

of the domain we have d(x, t)→ 0 as ∆x→ 0 and ∆t→ 0.

To analyze their convergence, we proceed as in the case of numericalmethods for ODEs and consider a finite time interval t ∈ [0, T ], T > 0

as well as a sequence of grids with grid spacings ∆x → 0, ∆t → 0 anddetermine whether at every fixed grid point (xj , tn) also the global erroru(xj , tn)− Unj tends to zero uniformly.

A sequence of grid spacings {(∆x)k, (∆t)k} can approach the point (0, 0)

in different ways. The following figure shows different “refinement curves”in the (∆x,∆t)-plane.

We will see that the explicit Euler method converges only if the refinementsatisfies

µ :=∆t

∆x2≤ 1

2.

6.2 Convergence TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 230

0 0.1 0.2 0.3 0.4 0.5 0.6 0.70

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

∆ x

∆ t

∆ t = ∆ x2/4

∆ t = ∆ x2/2

∆ t = ∆ x2

∆ t = ∆ x∆ t = ∆ x/2

Different refinement curves ∆t,∆x→ 0:red: ∆t/∆x2 = const, blue: ∆t/∆x = const.

6.2 Convergence TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 231

Theorem 6.1 If a sequence of grids satisfies

µk =(∆t)k(∆x)2

k

≤ 1

2for all k sufficiently large,

and if for the corresponding sequences {jk} and {nk} there holds

nk(∆t)k → t ∈ [0, T ], jk(∆x)k → x ∈ [0, 1],

then, under the assumption that |uxxxx| ≤Mxxxx uniformly in [0, 1]× [0, T ],the approximations generated by the explicit Euler method (6.4) convergeto the exact solution u(x, t) as k →∞ uniformly on [0, 1]× [0, T ].

6.2 Convergence TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 232

0 0.2 0.4 0.6 0.8 110

−7

10−6

10−5

10−4

10−3

10−2

t

En

expl. Euler dx=1/10expl. Euler dx=1/20Crank Nicolson, dx = 1/10Crank Nicolson, dx = 1/20

Error E(tn) = maxj |unj − Unj | of numerical solution of model problem(6.1) withg0 = g1 = 0 and u0(x) = x(1− x) obtained with explicit Euler method with

∆t = ∆x2/2 and the Crank-Nicolson method with ∆t = ∆x/2.

6.2 Convergence TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 233

6.3 Stability

In the method of lines, an ODE solver is applied to the semidiscrete system(6.3). Recalling the regions of absolute stability of the explicit and implicitEuler and trapezoidal methods, we are led to investigate the behavior ofthe eigenvalues of the matrix A∆x in (6.2) as ∆x→ 0.

−3 −2 −1 0 1−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

explicit EulerR(λ) = 1 + λ

−1 0 1 2 3−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

implicit EulerR(λ) = 1

1−λ

−2 −1 0 1 2−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

trapezoidalR(λ) = 1+λ/2

1−λ/2

6.3 Stability TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 234

The eigenvalues of A∆x (cf. (5.7)) are given by

λj = − 4

∆x2sin2 jπ∆x

2, j = 1, 2 . . . , J.

The eigenvalue of largest magnitude is

λJ = − 4

∆x2sin2

(π2

(1−∆x))

=−4

∆x2+ π2 +O

(∆x2) .

Therefore, in order to satisfy

|R(∆t λj

)| ≤ 1 for all j = 1, 2, . . . , J

in the explicit Euler method, it is necessary that

∆t

∆x2≤ 1

2. (6.8)

Both implicit Euler and the trapezoidal rule are absolutely stable methods. Therefo-re, as all eigenvalues of A∆x lie in the left half plane, the requirement of absolutestability places no constraints on the time step ∆t.Since the CN scheme (trapezoidal rule) is consisten of order 2 in both x and t, thissuggests a time step ∆t ≈ ∆x.

6.3 Stability TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 235

Absolute stability describes the behavior of a numerical approximation un

at time tn of the solution of an ODE as n → ∞. The appropriate stabilityconcept for the convergence analysis of numerical methods for solvingIVPs for PDEs was developed by Laxa and Richtmyerb.

All three methods considered so far can be written in the formc

U n+1 = B(∆t)U n + gn(∆t), n = 0, 1, . . . (6.9)

with different matrices B(∆t) given in each case by

BeE(∆t) = I + ∆t A∆x, (6.10a)

B iE(∆t) = (I −∆t A∆x)−1, (6.10b)

BCN(∆t) =

(I − ∆t

2A∆x

)−1(I +

∆t

2A∆x

). (6.10c)

aPETER D. LAX (∗ 1906)bROBERT DAVIS RICHTMYER (1910–2003)cWe assume that ∆t is chosen as a fixed given function of ∆x.

6.3 Stability TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 236

A linear finite difference scheme of the form (6.9) is called Lax-Richtmyerstable, if for any fixed stopping time T there exists a constant KT > 0 suchthat

‖B(∆t)n‖ ≤ KT , ∀∆t > 0 and n ∈ N, n∆t ≤ T.

Theorem 6.2 (Lax-Richtmyer equivalence theorem) A consistent linearfinite difference scheme (6.9) is convergent if, and only if, it is Lax-Richtmyerstable.

Examples: For the explicit Euler method applied to (6.1) the resultingmatrix (6.10a) is normal, implying ‖B(∆t)‖2 = ρ(B(∆t)). If condition (6.8)is satisfied, we have ‖B(∆t)‖2 ≤ 1, implying that the explicit Euler methodis Lax-Richtmyer stable and therefore convergent.

The implicit Euler and Crank-Nicolson schemes are also Lax-Richtmyerstable but without constraints on the mesh ratio or time step.

6.3 Stability TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 237

Remark 6.3 In all three schemes considered so far we were able to showthe stronger statement ‖B(∆t)‖ ≤ 1, which is sometimes called strong sta-bility. However, strong stability is not necessary for Lax-Richtmyer stability.

More precisely, for Lax-Richtmyer stability it is necessary that there exist aconstant α such that

‖B(∆t)‖ ≤ 1 + α∆t (6.11)

for all sufficiently small time steps ∆t, which is evident from the inequality

‖B(∆t)n‖ ≤ (1 + α∆t)n ≤ eαT , ∀ n∆t ≤ T.

6.3 Stability TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 238

6.4 Von-Neumann Analysis

A formal procedure based on Fourier analysis for determining stabilitybounds was introduced by von Neumann a.

In general, von Neumann analysis only yields necessary conditions, whichare also sufficient only in special cases, among these linear PDEs with con-stant coefficients, unbounded spatial domain (Cauchy problem) or periodicboundary conditions, uniform grid.

Fundamental fact: in the continuous case (x ∈ R) the functions eiξx, ξ ∈ R,are eigenfunctions of the differential operator ∂x, i.e.,

∂xeiξx = iξ eiξx,

and thus are also eigenfunctions of any linear differential operator withconstant coefficients.

aJOHN VON NEUMANN (1903–1957)

6.4 Von-Neumann Analysis TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 239

On the unbounded spatial grid {xj = j∆x}j∈Z the grid function

vj := eiξxj = eiξj∆x, j ∈ Z,

is an eigenfunction of any difference operator with constant coefficients.

Example: For the central difference operator δ0 defined as(δ0v)j := (vj+1 − vj−1)/(2∆x) there holds

δ0vj := (δ0v)j =eiξ(j+1)∆x − eiξ(j−1)∆x

2∆x=eiξ∆x − e−iξ∆x

2∆xeiξj∆x

=i

∆xsin(ξ∆x) vj

The power series expansion

i

∆xsin(ξ∆x) =

i

∆x

(ξ∆x− (ξ∆x)3

6+O

((ξ∆x)5)) = iξ +O(∆x2ξ3)

shows that the eigenvalue of δ0 approximates the corresponding eigenvalue of ∂x.

6.4 Von-Neumann Analysis TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 240

Any grid function on {xj = j∆x : j ∈ Z} can be represented as the superposition

vj =1

∫ π/∆x

−π/∆xv(ξ)eiξj∆x dξ, j ∈ Z

of “grid waves” eiξj∆x, ξ ∈ [−π∆x, π

∆x] with Fourier coefficients v(ξ) given by

v(ξ) = ∆x

∞∑j=−∞

vje−iξj∆x, ξ ∈

[−π∆x

∆x

].

If the magnitude of grid functions and their Fourier coefficients are measured by

‖v‖ =

(∆x

∞∑j=−∞

|vj |2)1/2

and ‖v‖ =

(∫ π/∆x

−π/∆x|v(ξ)|2 dξ

)1/2

,

then Parseval’s identity states that

‖v‖ =√

2π‖v‖. (6.12)

6.4 Von-Neumann Analysis TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 241

The bound ‖B‖ ≤ 1 + α∆t requires verifying

‖Bu‖ ≤ (1 + α∆t)‖u‖ ∀u.

According to (6.12), an equivalent condition is

‖Bu‖ ≤ (1 + α∆t)‖u‖ ∀u, (6.13)

which is easier to show because the Fourier coefficients are decoupled.

For a one-step method one obtains

un+1(ξ) = g(ξ) un(ξ)

with a so-called amplification factor g(ξ). If we are able to show that

|g(ξ)| ≤ 1 + α∆t.

with a constant α which is independent of ξ, then this implies (6.13).

6.4 Von-Neumann Analysis TU Bergakademie Freiberg, SS 2012

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Numerical Analysis of Differential Equations 242

Examples:

(a) For the explicit Euler scheme applied to (6.3) we obtain the amplificationfactor

g(ξ) = 1− 4∆t

∆x2sin2 ξ∆x

2.

(b) For Crank-Nicolson,

g(ξ) =1 + z(ξ)

1− z(ξ), z(ξ) =

∆t

∆x2

(−1 + cos(ξ∆x)

)≤ 0 ∀ξ.

6.4 Von-Neumann Analysis TU Bergakademie Freiberg, SS 2012