1. Partial Differential Equations
Scientific Computing
Partial Differential Equations
Explicit Solution of
Heat Equation

2. One Dimensional 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

3. One Dimensional Heat Equation
We will solve this equation for x and t values on a grid in x-t space:
Approximate
Solution uij=u(xi, tj) at grid points

4. Finite Difference Approximation
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:

5. Finite Difference Approximation
Then the heat equation (ut =cuxx ) can be approximated as
Or,
Let r = (ck/h2) Solving for ui,j+1 we get:

6. One Dimensional Heat Equation
Note how this formula uses info from the j-th time step to approximate the (j+1)-st time step:

7. One Dimensional Heat Equation
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:

8. Matlab Implementation
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

9. 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);

10. Matlab Implementation
Calculated solution appears correct:

11. Matlab Implementation
Try different T value: T=0.5
Values seem chaotic

12. Matlab Implementation
Why this chaotic behavior?

13. Matrix Form of Solution

14. Matrix Form of Solution
This is of the form
Start: u(0)=u(:,0)=f(:)
Iterate:

15. Matrix Form of Solution
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.

16. Gerschgorin Theorem Let A be a square nxn matrix. Around every element
aii 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.

17. Gerschgorin Theorem 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

18. Gerschgorin Theorem
Example:
Actual eigenvalues in red

19. Gerschgorin Theorem 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

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

21. Matrix Form of Solution
So, the eigenvalue satisfies:
For the error to stay bounded, we need
Thus, we need
For our first example, we had T=0.1 and
so, 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.