Scientific Computing Partial Differential Equations Explicit Solution of Heat Equation.

Scientific Computing

Partial Differential EquationsExplicit Solution of

Heat Equation

• The one-dimensional Heat Equation for the temperature u(x,t) in a rod at time t is given by:

where c >0, T>0 are constants

10 )()0,(

0 (t)),1(

0 )(),0(

010

1

0

xxfxu

Ttgtu

Tttgtu

Ttxcuu xxt

One Dimensional Heat Equation

• We will solve this equation for x and t values on a grid in x-t space:


t

x00 x ix

nxh /1

jt

00 t

mTtk /

mtT

1nx

ApproximateSolution uij=u(xi, tj) at grid points

• To approximate the solution we use the finite difference approximation for the two derivatives in the heat equation. We use the forward-difference formula for the time derivative and the centered-difference formula for the space derivative:

Finite Difference Approximation

211

1

),(),(2),(),(

),(),(),(

h

txutxutxutxu

k

txutxutxu

jijijijixx

jijijit

• Then the heat equation (ut =cuxx ) can be approximated as

Or,

Let r = (ck/h2) Solving for ui,j+1 we get:

Finite Difference Approximation

€

u(x i, t j+1) − u(x i, t j )

k= cu(x i−1, t j ) − 2u(x i, t j ) + u(x i+1, t j )

h2

jiijjiijji uuuh

ckuu ,1,121, 2

jiijjiji ruurruu ,1,11, )21(

• Note how this formula uses info from the j-th time step to approximate the (j+1)-st time step:


t

x

1ix ix 1ix

1jt

jt

mtT

00 t

00 x 1nx

jiijjiji ruurruu ,1,11, )21(

• On the left edge of this grid, u0,j = g0,j = g0(tj).

• On the right edge of this grid, un,j = g1,j = g1(tj).

• On the bottom row of the grid, ui,0 = fi = f(xi).

• Thus, the algorithm for finding the (explicit) solution to the heat equation is:


nifu

jrgurruu

jruurrgu

nijruurruu

ii

jjnjnjn

jjjj

jiijjiji

0

0)21(

0)21(

2,20)21(

0,

,1,1,21,1

,2,1,01,1

,1,11,

function z = simpleHeat(f, g0, g1, T, n, m, c) %Simple Explicit solution of heat equation h = 1/n; k = T/m; r = c*k/h^2; % Constants x = 0:h:1; t = 0:k:T; % x and t vectors % Boundary conditions u(1:n+1, 1) = f(x)'; % Transpose, since it’s a row vector u(1, 1:m+1) = g0(t); u(n+1, 1:m+1) = g1(t); % compute solution forward in time for j = 1:m u(2:n,j+1) = r*u(1:n-1,j) + (1-2*r)*u(2:n,j) + r*u(3:n+1,j); end z=u'; mesh(x,t,z); % plot solution in 3-d end

Matlab Implementation

Usage: f = inline(‘x.^4’); g0 = inline(‘0*t’); g1 = inline(‘t.^0’); n=5; m=5; c=1; T=0.1; z = simpleHeat(f, g0, g1, T, n, m, c);


Calculated solution appears correct:


Try different T value: T=0.5

Values seem chaotic


Why this chaotic behavior?


Matrix Form of Solution

nifu

jrgurruu

jruurrgu

nijruurruu

ii

jjnjnjn

jjjj

jiijjiji

0

0)21(

0)21(

2,20)21(

0,

,1,1,21,1

,2,1,01,1

,1,11,

1,1

1,2

1,2

1,1

,1

,0

,1

,2

,2

,1

0

0

21

21

21

21

jn

jn

j

j

j

j

jn

jn

j

j

u

u

u

u

rg

rg

u

u

u

u

rr

rrr

rrr

rr

This is of the formStart: u(0)=u(:,0)=f(:) Iterate:


1,1

1,2

1,2

1,1

,1

,0

,1

,2

,2

,1

0

0

21

21

21

21

jn

jn

j

j

j

j

jn

jn

j

j

u

u

u

u

rg

rg

u

u

u

u

rr

rrr

rrr

rr

)1(:,)(:, jubjAu

bIAAuAu

bIAuAbbAuAbAuu

bAuu

mmm )(

)()(

1)0()(

)0(2)0()1()2(

)0()1(

Now, suppose that we had an error in the initial value for u(0), say the initial u value was u(0)+e, where e is a small error.

Then, under the iteration, the error will grow like Ame.

For the error to stay bounded, we must have

Thus, the largest eigenvalue of A must be <= 1.


eAemeeAm

Let A be a square nxn matrix. Around every elementaii on the diagonal of the matrix draw a circle with

radius

Such circles are known as Gerschgorin disks.

Theorem: Every eigenvalue of A lies in one of these Gerschgorin disks.

Gerschgorin Theorem

1,

j N

i ijj j i

r a

Example:

The circles that bound the Eigenvalues are:

C1: Center point (4,0) with radius r1 = |2|+|3|=5

C2: Center point (-5,0) with radius r2=|-2|+|8|=10

C3: Center Point (3,0) with radius r3=|1|+|0|=1

Gerschgorin Theorem

301

852

324

M

Example:

Actual eigenvalues in red

Gerschgorin Theorem

301

852

324

M

-15 -10 -5 5 10

-15

-10

-5

5

10

Example:

C1: Center point (1,0)

radius r1 = |0|+|7|=7

C2: Center point (-5,0)

radius r2=|2|+|0|=2

C3: Center Point (-3,0)

radius r3=|4|+|4|=8

Gerschgorin Theorem

344

052

701

M

-15 -10 -5 5 10

-15

-10

-5

5

10

Circles: center=(1-2r), radius = r or 2r. Take largest radius 2r. Then, if λ is an eigenvalue of A, we have


rr

rrr

rrr

rr

A

21

21

21

21

141

221221

r

rrrr

0 1-2r

So, the eigenvalue satisfies: For the error to stay bounded, we needThus, we need

For our first example, we had T=0.1 andso, we expect the solution to be stable.For the second example, T=0.5, we have r =2.5, so the error in the iterates will grow in an unbounded

fashion, which we could see in the numerical solution.


141 r

11

2

1411 rr

5.0

5

1

5

1.01

2

r

Scientific Computing Partial Differential Equations Explicit Solution of Heat Equation.

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