day27 pde heat equation implicit

7/27/2019 Day27 PDE Heat Equation Implicit

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Scientific Computing

Partial Differential EquationsImplicit Solution of

Heat Equation


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Explicit Method

We use the centered-difference approximation

for uxx at time step j:

ui1, j

2ui, j

ui1, j

h2

grid point involved with space differencegrid point involved with time difference


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Implicit Method

We use the centered-difference approximation

for uxx at step (j+1) :

ui1, j1

2ui, j1

ui1, j1

h2

grid point involved with space differencegrid point involved with time difference


5/22

We use the forward-difference formula for the time

derivative and the centered-difference formula for the

space derivative at time step (j+1):

Implicit Method

ut(x i,tj ) ui, j1 ui, j

k

cuxx(x i,tj ) cui1, j1 2ui, j1 ui1, j1

h2


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

Or,

Let r = (ck/h2) Solving for ui,j

we get:

Implicit Method

ui, j1 ui, j

k c

ui1, j1 2ui, j1 ui1, j1

h2

ui, j1 uij ck

h2ui1, j1 2ui, j1 ui1, j1

ui, j rui1, j1 (12r)ui, j1 rui1, j1


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Putting in the boundary conditions, we get the

following equations for the (implicit) solution to the

heat equation is:

Implicit Method

ui, j rui1, j1 (12r)ui, j1 rui1, j1 j 0 i 2,K n 2u1, j rg0, j1 (12r)u1, j1 ru2, j1 j 0un1, j run2, j1 (1 2r)un1, j1 rg1, j1 j 0ui,0 fi i 0K n


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Matrix Form of Solution

12r rr 1 2r r

O O O

r 12r r

r 12r

u1, j1

u2, j1

M

un2, j1

un1, j1

rg0, j1

0

M

0

rg1, j1

u1, j

u2, j

M

un2, j

un1, j

ui, j rui1, j1 (12r)ui, j1 rui1, j1 j 0 i 2,K n 2u1, j rg0, j1 (12r)u1, j1 ru2, j1 j 0un1, j run2, j1 (1 2r)un1, j1 rg1, j1 j 0ui,0 fi i 0K n


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This is of the form

To solve for u(:,j) we need to use a linear systems method.Since we have to solve this system repeatedly, a good choice

is the LU decomposition method.

Matrix Form of Solution

Au(:, j 1)b u(:, j)

12r r

r 12r r

O O O

r 12r r

r 12r

u1, j1

u2, j1

M

un2, j1

un1, j1

rg0, j10

M

0

rg1, j1

u1, j

u2, j

M

un2, j

un1, j


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function z = implicitHeat(f, g0, g1, T, n, m, c)

%Simple Implicit solution of heat equation

% Constants

h = 1/n;

k = T/m;

r = c*k/h^2;

% x and t vectors

x = 0:h:1;

t = 0:k:T;% Boundary conditions

u(1:n+1, 1) = f(x)';

u(1, 1:m+1) = g0(t);

u(n+1, 1:m+1) = g1(t);

Matlab Implementation


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% Set up tri-diagonal matrix for interior points of the grid

A = zeros(n-1, n-1); % 2 less than x grid size

for i= 1: n-1

A(i,i)= (1+2*r);

if i ~= n-1

A(i, i+1) = -r;

end

if i ~= 1

A(i, i-1) = -r;end

end

% Find LU decomposition of A

[LL UU] = lu_gauss(A); % function included below



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% Solve for u(:, j+1);

for j = 1:m

% Set up vector b to solve for in Ax=b

b = zeros(n-1,1);

b = u(2:n, j)'; % Make a column vector for LU solver

b(1) = b(1) + r*u(1,j+1);

b(n-1) = b(n-1) + r*u(n+1,j+1);

u(2:n, j+1) = luSolve(LL, UU, b);

endz=u';

% plot solution in 3-d

mesh(x,t,z);

end



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Usage:

f = inline(x.^4);

g0 = inline(0*t);g1 = inline(t.^0);

n=5; m=5; c=1; T=0.5;

z = implicitHeat(f, g0, g1, T, n, m, c);



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Calculated solution appears stable:



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To analyze the stability of the method, we again have to

consider the eigenvalues of A for the equation

We have

Using the Gershgorin Theorem, we see that eigenvalues are

contained in circles centered at (1+2r) with max radius of 2r

Stability

Au(:, j 1)b u(:, j)

A

1 2r r

r 1 2r r

O O O

r 1 2r rr 1 2r


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Thus, if is an eigenvalue of A, we have

Thus, all eigenvalues are at least 1. In solving for u(:,j+1) in

we have

Since the eigenvalues of A-1 are the reciprocal of the

eigenvalues of A, we have that the eigenvalues of A-1 are

all


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Convergence means that as D x and D t approach zero,

the results of the numerical technique approach the

true solution

Stabilitymeans that the errors at any stage of the

computation are attenuated, not amplified, as the

computation progresses

Truncation Errorrefers to the error generated in thesolution by using the finite difference formulas for the

derivatives.

Convergence and Stability


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Example: For the explicit method, it will be stable if r


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Crank-Nicholson Method

We average the centered-differenceapproximation

for uxx at time steps j

and j+1:

ui1, j 2ui, j ui1, jh 2

grid point involved with space difference

grid point involved with time differenceui1, j1 2ui, j1 ui1, j1h2


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We get the following

ui, j1 uij ck

2h

2ui1, j 2ui, j ui1, j

ck

2h2ui1, j1 2ui, j1 ui1, j1


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Putting in the boundary conditions, we get the

following equations for the (C-N) solution to the heat

equation is:


r

2ui1, j1 (1 r)ui, j1

r

2ui1, j1

r

2ui1, j (1 r)ui, j

r

2ui1, j

r

2

g0, j1 (1 r)u1, j1 r

2

u2, j1

r

2

g0, j (1 r)u1, j r

2

u2, j

r

2un2, j1 (1 r)un1, j1

r

2g1, j1

r

2un2, j (1 r)un1, j

r

2g1, j

ui,0 fi


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day27 pde heat equation implicit

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