PDEs - Slides (4)

8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite

differences forheat equation

Part IV

20401

Tony Shardlow

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8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


Outline

1 Model parabolic problem

2 Uniqueness

3 Method of lines

4 Euler method

5 Finite differences for heat equation

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8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


Outline


2 Uniqueness

3 Method of lines

4 Euler method


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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


Model parabolic PDE

Consider the heat equation for t > 0 and x [0, 1],

ut =uxx

u(0, t) =u(1, t) = 0,

with initial condition u(x, 0) = u0(x).

We study

1 Uniqueness of soln

2 Finite difference method

3 Stability of time discretisation.

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


Outline


2 Uniqueness

3 Method of lines

4 Euler method


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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


Uniqueness of soln

ut uxx = 0

u(0, t) = u(1, t) = 0u(x, 0) = u0(x)

(H)Multiply the PDE by u and integrate over x [0, 1],

Define the energy : E(t) =

1

0 u2(x, t) dx

and differentiate

E(t) =d

dt

10

u2 dx =

10

t(u2) dx = 2

10

uut dx

=21

0uuxx dx.

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


E(t) =

10

u2 dx; E(t) = 2

10

uuxx dx.

Integrate by parts

E(t) = 2

uux

10

2

10

(ux)2 dx

= 2

10

(ux)2 dx 0 ,

where we apply the boundary conditions u(0, t) = u(1, t) = 0.

We deduce that E(t) E(0) for all t > 0. That is,10

u2(x, t) dx = E(t) E(0) =

10

u20(x) dx.

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


E(t) =

10

u2 dx; E(t) = 2

10

uuxx dx.

Integrate by parts

E(t) = 2

uux

10

2

10

(ux)2 dx

= 2

10

(ux)2 dx 0 ,

where we apply the boundary conditions u(0, t) = u(1, t) = 0.

We deduce that E(t) E(0) for all t > 0. That is,10

u2(x, t) dx = E(t) E(0) =

10

u20(x) dx.

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


Uniqueness of soln

Theorem

The solution to (H) is unique.

Proof.

Let v, u be two solutions. By linearity, w = u v satisfies

wt =wxx, w(0, t) = w(1, t) = 0,

w(x, 0) =0.

E(t) = 1

0 w(x, t)2

dx

is a decreasing function.

But E(0) = 0, so E(t) = 0 for all t. Hence w = 0 for all t andu = v and the solution is unique.

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


A model parabolic PDE problem

HOMEWORK

You can now try Problem 1

Evolution of the temperature profiles for specific initialconditions ..

0 0.5 10

0.2

0.4

0.6

0.8

1

x

temperature

Problem 2.1

0 0.5 10

0.2

0.4

0.6

0.8

1

x

temperature

Problem 2.2

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8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


A model parabolic PDE problem

HOMEWORK


Evolution of the temperature profiles for specific initialconditions ..

0 0.5 10

0.2

0.4

0.6

0.8

1

x

temperature

Problem 2.1

0 0.5 10

0.2

0.4

0.6

0.8

1

x

temperature

Problem 2.2

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8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


Outline


2 Uniqueness

3 Method of lines

4 Euler method


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8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


Heat equation

Consider the Fourier series

u(x, t) =

j=1

uj(t) sin(jx)

Notice that u(0, t) = u(1, t) and boundary conditions aresatisfied.Substitute into the heat equation ut = uxx. Then

ut(x, t) =

j=1uj(t) sin(jx)

uxx(x, t) =

j=1

uj(t) (j22)sin(jx).

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8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


Heat equation

Consider the Fourier series

u(x, t) =

j=1

uj(t) sin(jx)

Notice that u(0, t) = u(1, t) and boundary conditions aresatisfied.Substitute into the heat equation ut = uxx. Then

ut(x, t) =

j=1uj(t) sin(jx)

uxx(x, t) =

j=1

uj(t) (j22)sin(jx).

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


Decoupling the heat equation

After substitution into the heat equation,

j=1

uj(t)j22 sin(jx) =

j=1

uj(t) sin(jx).

and equating coefficients of sin(jx),

ODEs for Fourier coefficients uj(t)

uj = j22uj , j = 1, 2, . . . .

The system is decoupled, as the solution uj can be found inde-pendently of uk for j = k.

From u(0, x) =

j=1 uj(0) sin(jx) = u0(x), the initial data

uj(0) = 2

0u0(x)sinjx dx.

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8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


Method of lines

Replace infinite system of ODEs

u

j = j2

2

uj, j = 1, 2, . . . ,

with finite system of ODEs

uj = j22uj, j = 1, 2, . . . , J

where J is size of system and solve to determine uj(t),j = 1, . . . , J.

Method of lines approximation is

v(x, t) =J

j=1

uj

(t) sin(jx).

For large J, u(x, t) v(x, t).Finite difference methods are often used to solve the ODEs foruj(t). We study finite difference methods for ODEs.

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8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


Outline


2 Uniqueness

3 Method of lines

4 Euler method


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Finite difference method for

8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


Finite difference method forODE

Consider the following initial value problem for the ODE:

dXdt

= f(X), X(0) = X0

For k > 0 (the time step), we can approximate

dX

dt =

X(t+ k) X(t)

k + O(k)

Example: X(t) = et

dX

dt=

det

dt= et =

et+k et

k+ O(k)

As a numerical example, put t = 1 and k = 0.1, then X(t) =et = e = 2.7183 and

et+k et

k =e1.1 e1

0.1 = 2.8588 18/57


8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite




dXdt

= f(X), X(0) = X0


dX

dt=

X(t+ k) X(t)

k+

O(k)

Example: X(t) = et

dX

dt=

det

dt= et =

et+k et

k+ O(k)


et+k et

k =e1.1 e1

0.1 = 2.8588 19/57


8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite




dXdt

= f(X), X(0) = X0


dX

dt=

X(t+ k) X(t)

k+ O(k)

Example: X(t) = et

dX

dt=

det

dt= et =

et+k et

k+ O(k)


et+k et

k =e1.1 e1

0.1 = 2.8588 20/57

8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


Explicit Euler method

Put t = nk and substitute

X(t+ k) X(t)

k= f(X(t)) + O(k)

into ODE.

X((n + 1)k) X(nk)

k= f(X(nk)) + O(k)

Drop the error term and approximate X(nk) by Xn, gives the

Explicit Euler method

Given X0 and time step k, compute Xn X(nk) by

Xn+1 = Xn + kf(Xn)

It is possible to show that |X(nk) Xn| = O(k) for nk = T,under assumptions on f. The method is first order accurate.

We do not prove this. 21/57

8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


Example: explicit Euler method

Consider the ODE

dX

dt= X, X(0) = X0

with X0 = 2 and = 3. In this case, exact solution is

X(t) = e3t2 .

We find explicit Euler approximation for time step k = 0.1,

Xn+1 = Xn + kf(Xn), X0 = 2.

X1 = 2 + 0.1(3 2) = 1.4 ,

X2 = 1.4 + 0.1(3 1.4) = 0.98

The exact solution X1 X(0.1) = 2e0.3 = 1.4816,

X2 1.0976 = X(0.2).

Can get more accurate approximation by taking k smaller 22/57

E l li i E l h d

8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


Example: explicit Euler method

Consider the ODE

dX

dt= X, X(0) = X0

with X0 = 2 and = 3. In this case, exact solution is

X(t) = e3t2 .

We find explicit Euler approximation for time step k = 0.1,

Xn+1 = Xn + kf(Xn), X0 = 2.

X1 = 2 + 0.1(3 2) = 1.4 ,

X2 = 1.4 + 0.1(3 1.4) = 0.98

The exact solution X1 X(0.1) = 2e0.3 = 1.4816,

X2 1.0976 = X(0.2).

Can get more accurate approximation by taking k smaller 23/57

I li i E l h d

8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


Implicit Euler method

Approximating dX/dt by (X(t) X(t k))/k and repeating

gives X(nk) X((n 1)k)

k= f(X(nk)) + O(k)

Drop the error term and approximate X(nk) by Xn, gives the


Given X0 and time step k, compute Xn X(nk) by

Xn = Xn1 + kf(Xn)

A nonlinear system of equations must be solved to determineXn. Hence the method is implicit.

Again possible to show that |X(nk) Xn| = O(k) for nk = T,under assumptions on f. The method is first order accurate.

Not proved. 24/57

Li ODE

8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


Linear test ODE

Linear test equation

dXdt

= X = X, X(0) = X0.

The solution is easily found to be

X(t) = e

t

X0.

Note that X(t) 0 as t if < 0 ( means real part).

What about numerical methods?For what values of , do the numerical approximations Xn 0

as n Hope that the set of such is also { R : < 0}.

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Stability condition for Explicit

8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


Stab ty co d t o o p c tEuler

1 For the case f(X) = X, the explicit Euler method is

Xn+1 = Xn + kXn

2 Rearrange,Xn+1 = (1 + k)Xn.

and the solution Xn = (1 + k)nX0 .

3 Hence as n

Xn 0 |1 + k| < 1

if and only if k belongs tocircle of radius 1 in complex plane with centre at 1.

4 The set of k C such that Xn 0 as n is knownas the region of absolute stability.

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Stability condition for Explicit

8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


y pEuler

1 For the case f(X) = X, the explicit Euler method is

Xn+1 = Xn + kXn

2 Rearrange,Xn+1 = (1 + k)Xn.

and the solution Xn = (1 + k)nX0 .

3 Hence as n

Xn 0 |1 + k| < 1

if and only if k belongs tocircle of radius 1 in complex plane with centre at 1.

4 The set of k C such that Xn 0 as n is knownas the region of absolute stability.

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Time step restriction for explicit

8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


p pEuler

1 Recall for the ODE, then

X(t) 0 as t

if < 0.2 The numerical method decays for all initial data if and

only if |1 + k| < 1 if and only if

timestep k < 2/|| .

That is, there is a restriction on the time step.3 For the method of lines method, = j22 where

j = 1, 2, . . . , J.4 Hence, the time step restriction is

timestep k < 2/J22 .

For large J, the time step condition is very restrictive!

Leads to inefficient method. 28/57


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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


p pEuler


X(t) 0 as t



timestep k < 2/|| .







8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


p pEuler


X(t) 0 as t



timestep k < 2/|| .






Implicit Euler

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


Implicit Euler

Put f(X) = X into the implicit Euler method.


Xn Xn1k

= Xn

It becomes

Xn =1

(1 k)Xn1 .

and the soln Xn = (1 k)nX0 .

HenceXn 0 |1 k| > 1

if and only if k isoutside the circle of radius 1 in complex plane with centre at 1

.

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Implicit Euler

8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


Implicit Euler

Put f(X) = X into the implicit Euler method.


Xn Xn1k

= Xn

It becomes

Xn =1

(1 k)Xn1 .

and the soln Xn = (1 k)nX0 .

HenceXn 0 |1 k| > 1

if and only if k isoutside the circle of radius 1 in complex plane with centre at 1

.

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No stability restriction for

8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


implicit Euler

1 For implicit Euler, the region of absolute stability is

|1 k| > 1 and this holds for all with < 0.2 When the ODE is stable ( < 0), the implicit Euler is

stable too and |Xn| 0 as n .

3 There is no timestep restriction for the implicit Euler

method.This is important for the heat equation, where is large.

4 The implicit Euler is a better method for the timediscretisation of the heat equation.

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No stability restriction for

8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


implicit Euler

1 For implicit Euler, the region of absolute stability is

|1 k| > 1 and this holds for all with < 0.2 When the ODE is stable ( < 0), the implicit Euler is

stable too and |Xn| 0 as n .

3 There is no timestep restriction for the implicit Euler

method.This is important for the heat equation, where is large.

4 The implicit Euler is a better method for the timediscretisation of the heat equation.

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Explicit vs Implicit

8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite



To understand the names implicit and explicit, apply the

methods to dX

dt= f(X), X(0) = X0

for f(x) a nonlinear function. The explicit Euler method is

Xn+1 = Xn + kf(Xn)

so that Xn+1 is determined explicitly from Xn and evaluation off.The implicit Euler method is

Xn = Xn1 + kf(Xn)

so that Xn is determined implicitly by a solution of a nonlinearequation involving f.

HOMEWORK


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8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite



To understand the names implicit and explicit, apply the

methods to dX

dt= f(X), X(0) = X0

for f(x) a nonlinear function. The explicit Euler method is

Xn+1 = Xn + kf(Xn)

so that Xn+1 is determined explicitly from Xn and evaluation off.The implicit Euler method is

Xn = Xn1 + kf(Xn)

so that Xn is determined implicitly by a solution of a nonlinearequation involving f.

HOMEWORK


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Examination

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


Examination

The material from here on is not covered on the exam.

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Outline

8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


Outline


2 Uniqueness

3 Method of lines

4 Euler method


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Discretisation of (x, t) space :m ( t )

8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


um

j u(xj, tm)

Grid spacing in x direction is h.

Grid spacing in t direction is k.

6 6

s s s s s s

c

c

c

c

c

c

c

c

c

c

0 = x0 xj xn = 1j1 j j+1h

h =1

n -

k

mtm

m+1

6

?

umj

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Explicit approximation

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


p pp

ut (xj, tm) = uxx (xj, tm)

Approximate

uxx(xj, tm) 1

h22xu(xj, tm)

= 1h2

u(xj h, tm) 2u(xj, tm) + u(xj + h, tm)

ut(xj, tm) 1

k+t u(xj, tm)

=

1

ku(xj, tm + k) u(xj, tm).

40/57

Finite difference methodexplicit

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method oflines

Euler method

Finite


ut (xj, tm) = uxx (xj, tm)j = 1, 2, . . . , n 1m = 0, 1, 2, . . .

.

Substitute difference approximations to derive the finitedifference method

1

k+t u

mj =

1

h22xu

mj

or1k

um+1j u

mj

= 1

h2

umj1 2um

j + um

j+1

s s s

j 1 j j + 1m

m + 1

Given umj1, um

j , um

j+1, um+1

j is given explicitly by

um+1j = um

j1 + (1 2)um

j + um

j+1, where = k/h2.

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Linear system of equations

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method of

lines

Euler method

Finite


t t t

j 1 j j + 1

m

m + 1

For = k/h2

u

m+1

j = u

m

j1 + (1 2)u

m

j + u

m

j+1,

j = 1, 2, . . . , n 1

m = 0, 1, 2, . . .

This is a matrix-vector multiply at each time level:

um+11...

um+1j...

um+1n1

=

1 2 0.. .

.. .

1 2 . . .

. . .

0 1 2

um1...

umj...

umn1

.

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Consistency I

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method of

lines

Euler method

Finite


Is the explicit approximation consistent?

Theorem (consistency)

The local error at the grid point (xj, tm),

Tmj =1

k

+t u(xj, tm) 1

h2

2xu(xj, tm) j = 1, 2, . . . , n,

is bounded from above:

Tmj

= O(k) + O(h2)

HOMEWORKYou can now try Problem 5

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Consistency I

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method of

lines

Euler method

Finite


Is the explicit approximation consistent?

Theorem (consistency)


Tmj =1

k

+t u(xj, tm) 1

h2

2xu(xj, tm) j = 1, 2, . . . , n,

is bounded from above:

Tmj

= O(k) + O(h2)

HOMEWORKYou can now try Problem 5

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Explicit Approximation II

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method of

lines

Euler method

Finite


1 Is the explicit approximation stable?

It all depends on = r

0 0.5 10

0.2

0.4

0.6

0.8

1Explicit FD Scheme: r is 0.48

x

temperature

0 0.5 10

0.2

0.4

0.6

0.8

1Explicit FD Scheme: r is 0.52

x

temperature

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method of

lines

Euler method

Finite


TheoremIf 1 2 0 1/2 then the explicit approximation

um+1j = um

j1 + (1 2)um

j + um

j+1,

is non-increasing:

maxj

um+1j max

j

umj . ()HOMEWORK


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Part IV

20401

Modelparabolic

problem

Uniqueness

Method of

lines

Euler method

Finite


TheoremIf 1 2 0 1/2 then the explicit approximation

um+1j = um

j1 + (1 2)um

j + um

j+1,

is non-increasing:

maxj

um+1j max

j

umj . ()HOMEWORK


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Part IV

20401

Modelparabolic

problem

Uniqueness

Method of

lines

Euler method

Finite


Proof.Let um = maxj |u

mj |.

|um+1j | = |um

j1 + (1 2)um

j + um

j+1|

|umj1| + |(1 2)um

j | + |um

j+1|

>0

um + (1 2) 0

um + ()>0

um

= um + (1 2)um + u

m = u

m .

This establishes () since |um+1

j | um for all j.

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Implicit Approximation

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method of

lines

Euler method

Finite


ut (xj, tm) = uxx (xj, tm)Approximate

uxx(xj, tm) 1

h22xu(xj, tm)

=1

h2u(xj h, tm) 2u(xj, tm) + u(xj + h, tm)

ut(xj, tm) 1

kt u(xj, tm)

=1

ku(xj, tm) u(xj, tm k).

where

2xu(xj, tm) = u(xj h, tm) 2u(xj, tm) + u(xj + h, tm)

t u(xj, tm) = u(xj, tm) u(xj, tm k).49/57

Implicit Approximation II

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method of

lines

Euler method

Finite


ut (xj, tm) = uxx (xj, tm)j = 1, 2, . . . , n

1

m = 0, 1, 2, . . . .

1

kt u

mj =

1

h22xu

mj

1

kum

j um1

j =1

h2 um

j1

2umj + um

j+1

t

j 1 j j + 1

m 1

m

Thus umj1 + (1 + 2)um

j um

j+1 = um1

j ,

where = k/h2.50/57

Linear system of eqns

8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method of

lines

Euler method

Finite


t

j 1 j j + 1

m 1

m

For = k/h2

umj1 + (1 + 2)um

j um

j+1 = um1

j

j = 1, 2, . . . , n 1m = 0, 1, 2, . . .

This is a tridiagonal matrix solve at each time level:

1 + 2 0. . . . . .

1 + 2 . . .

. . .

0 1 + 2

um1...

umj...

umn1

=

um11...

um1j...

um1n1

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Consistency

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method of

lines

Euler method

Finite


Theorem


Tmj =1

kt u(xj, tm)

1

h22xu(xj, tm) j = 1, 2, . . . , n,

is bounded from above:Tmj = O(k) + O(h2).HOMEWORK


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Consistency

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method of

lines

Euler method

Finite


Theorem


Tmj =1

kt u(xj, tm)

1

h22xu(xj, tm) j = 1, 2, . . . , n,

is bounded from above:Tmj = O(k) + O(h2).HOMEWORK


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Implicit Approximation II

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method of

lines

Euler method

Finite


1 Is the implicit approximation stable?

Yes for any = r

0 0.5 10

0.2

0.4

0.6

0.8

1Implicit FD Scheme: r is 0.48

x

temperature

0 0.5 10

0.2

0.4

0.6

0.8

1Implicit FD Scheme: r is 0.52

x

temperature

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8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method of

lines

Euler method

Finite


Theorem

The implicit approximation

umj1 + (1 + 2)um

j um

j+1 = um1

j

is non-increasing: maxj um+1j maxj umj . ()

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Proof

8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method of

lines

Euler method

Finite


Proof.

Proof. Let um = maxj |um

j |.

| (1 + 2) >0

umj | = |umj1 + um

1j + umj+1|.

(1 + 2) |umj | = |um

j1 + um1

j + um

j+1|

|umj1| + |um1

j | + |um

j+1|

um + |um1

j | + um = 2u

m + |u

m1j |.

(1 + 2) um 2um + |u

m1k |

This establishes () since um |um1k | maxj |u

m1j |.

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Summary

8/3/2019 PDEs - Slides (4)

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Part IV

20401

Modelparabolic

problem

Uniqueness

Method of

lines

Euler method

Finite


1 We have looked at two finite difference approximations for

the heat equation. An explicit and an implicit method.2 The explicit method is easier to implement as it does NOT

involve solving a linear system of equations.

3 Both implicit and explicit methods are consistent with thesame order of accuracy (|Tm

j

| = O(k) + O(h2)). Secondorder accurate in space; first order accurate in time.

4 Stability depends on a step size condition for the explicitmethod ( = k/h2 1/2). The implicit method is alwaysstable.

5 Stable+consistency gives convergence for both methods.The implicit method is preferred in practice due to betterstability.

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