1. Time Dependent Problem
The Heat Equation
time step
Example:

2. Time Dependent Problem
The Heat Equation
Weak Formulation ( variational formulation)
Fix a time t
Multiply equation (1) by and then integrate over the domain
Green’s theorem gives

3. Variational Formulation
BVP
Spatial Discretization

4. Spatial Discretization
Crank-Nicolson Method

5. Time Discretization
case
case

6. Time Discretization
case
case
case
case
case
case

7. Time Discretization Solve the linear system of equations to find

8. Time Discretization
BVP

9. Example

10. Computer Implementation
function [u1]=HeatSolver2D(p,e,t)
k = 0.001; % time step
T = 0.07; % final time
n = T/k;
np = size(p,2); % number of nodes
x = p(1,:)'; y = p(2,:)'; % node coordinates
u1 = sparse(np,n); % set zero IC
ix=find(sqrt(p(1,:).^2+p(2,:).^2)<0.4);
u1(ix,1)=ones(size(ix));
M = MassMatrix(p,t); %primary mass matrix
A = StiffMatrix(p,t); %primary stiff matrix
[A,M] = boundary(A,M,p,t,e);
% M,A after imbosing boundary
bound = unique([e(1,:) e(2,:)]); % boundary nodes
inter = setdiff([1:np],bound); % interior nodes
bold = Vec_b(0,p,e,t,bound,inter);
for l = 2:n % time loop
time = l*k;
bnew = Vec_b(time,p,e,t,bound,inter);
LHS = (M + 0.5*k*A); % Crank-Nicholson
rhs = (M - 0.5*k*A)*u1(inter,l-1) + 0.5*k*(bnew + bold);
bold = bnew;
u1(inter,l) = LHS\rhs;
end
% draw.m
%To speed up the plotting, we interpolate to a rectangular grid.
x=linspace(-1,1,31);y=x;
[unused,tn,a2,a3]=tri2grid(p,t,u1(:,1),x,y);
% Make the animation
newplot;
nframes = size(u1,2);
Mv = moviein(nframes);
umax=max(max(u1));
umin=min(min(u1));
for j=1:nframes,...
u=tri2grid(p,t,u1(:,j),tn,a2,a3);i=find(isnan(u));u(i)=zeros(size(i));...
surf(x,y,u);caxis([umin umax]);colormap(cool),...
axis([ ]);...
%contourf(x,y,u);colormap(cool)
Mv(:,j) = getframe;...
pause(0.01)
end
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11. Computer Implementation
function M = MassMatrix(p,t)
np = size(p,2);
nt = size(t,2);
M = sparse(np,np);
for K = 1:nt
loc2glb = t(1:3,K); % local-to-global map
x = p(1,loc2glb); % node x-coordinates
y = p(2,loc2glb); % node y-
area=polyarea(x,y);
MK = [2 1 1;
1 2 1;
1 1 2]/12*area; % element mass matrix
M(loc2glb,loc2glb) = M(loc2glb,loc2glb)+ MK;
end
function [A,M] = boundary(A,M,p,t,e);
np = size(p,2); % total number of nodes
bound = unique([e(1,:) e(2,:)]); % boundary nodes
inter = setdiff([1:np],bound); % interior nodes
A = A(inter,inter); % modify stiffness matrix
M = M(inter,inter); % Mass matrix matrix
function [area,b,c] = Gradients(x,y)
area=polyarea(x,y);
b=[y(2)-y(3); y(3)-y(1); y(1)-y(2)]/2/area;
c=[x(3)-x(2); x(1)-x(3); x(2)-x(1)]/2/area;
function b = Vec_b(time,p,e,t,bound,inter);
f = inline('time.*sin(x).*cos(y)','time','x','y');
np = size(p,2);
nt = size(t,2);
b = zeros(np,1);
for K = 1:nt
loc2glb = t(1:3,K);
x = p(1,loc2glb);
y = p(2,loc2glb);
area = polyarea(x,y);
bK = [f(time,x(1),y(1));
f(time,x(2),y(2));
f(time,x(3),y(3))]/3*area; % element load vector
b(loc2glb) = b(loc2glb)+ bK; % add element loads to b
end
b = b(inter); % modify load vector
function A = StiffMatrix(p,t)
np = size(p,2);
nt = size(t,2);
A = sparse(np,np);
for K = 1:nt
loc2glb = t(1:3,K); % local-to-global map
x = p(1,loc2glb); % node x-coordinates
y = p(2,loc2glb); % node y-
[area,b,c] = Gradients(x,y);
AK = (b*b'+c*c')*area; % element stiff mat
A(loc2glb,loc2glb) = A(loc2glb,loc2glb)+ AK;
end

12. T=0
One time step
After 10 time steps
Steady state solution

16. spatial semi-discretization equation
Heat Equation
spatial semi-discretization equation
Fully discrete Equation

17. 1 Consider the heat equation given the primary mass matrix M
5/24 1/ /24 1/ / /24
1/ / / / /24 1/
1/ /12 1/ /48
1/ /24 1/48 1/ /
/ /48 1/ /
/ / / /
1/ / / / /48
1/ / / /12
and the primary stiffness matrix A
For the numerical solution, we use the following triangulation
a) Write down the semidiscrete equation.
b) Apply the forward Euler method with k = 1/3 and then write down the fully discrete equation.
c) Use part (b) to estimate u(1, 3/2, 3/2)