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## Wave Equation on a Square Domain

This example shows how to solve the wave equation using the hyperbolic function in the Partial Differential Equation Toolbox™.

We solve the standard second-order wave equation

on a square domain with zero Dirichlet boundary conditions on left and right and zero Neumann boundary conditions on the top and bottom.

Problem Definition

The following variables will define our problem:

• g: A specification function that is used by initmesh. For more information, please see the documentation page for squarereg and pdegeom.

• b: A boundary file used by assempde. For more information, please see the documentation pages for squareb3 and pdebound.

• c, a, f, d: The coefficients of the PDE.

g='squareg';
b='squareb3';
c=1;
a=0;
f=0;
d=1;


Generate Mesh

[p,e,t]=initmesh('squareg');
figure;
pdemesh(p,e,t); axis equal


Generate Initial Conditions

The initial conditions:

• .

• .

This choice avoids putting energy into the higher vibration modes and permits a reasonable time step size.

x=p(1,:)';
y=p(2,:)';
u0=atan(cos(pi/2*x));
ut0=3*sin(pi*x).*exp(sin(pi/2*y));


Define Time-Discretization

We want the solution at 31 points in time between 0 and 5.

n=31;
tlist=linspace(0,5,n);


Find FEM Solution

uu=hyperbolic(u0,ut0,tlist,b,p,e,t,c,a,f,d);

428 successful steps
62 failed attempts
982 function evaluations
1 partial derivatives
142 LU decompositions
981 solutions of linear systems


Animate FEM Solution

To speed up the plotting, we interpolate to a rectangular grid.

figure; set(gcf,'renderer','zbuffer');
delta=-1:0.1:1;
[uxy,tn,a2,a3]=tri2grid(p,t,uu(:,1),delta,delta);
gp=[tn;a2;a3];
newplot;
umax=max(max(uu));
umin=min(min(uu));
for i=1:n
pdeplot(p,e,t,'xydata',uu(:,i),'zdata',uu(:,i),'zstyle','continuous',...
'mesh','off','xygrid','on','gridparam',gp,'colorbar','off');
axis([-1 1 -1 1 umin umax]); caxis([umin umax]);
M(i)=getframe;
end
movie(M,1);