@Carl Love Hi

Yes, you are right I want the intersection of the surface with a horizontal plane .

Thanks 

Thank you Carl Love 3435 .Also I am looking to make it as movie do you have any idea .

Thank you Carl Love 3435 .Also I am looking to make it as movie do you have any idea .

@yurr 

Hello 

 I wish to solve nonlinear PDE equation numerically  in u & v coordinate . where u=t-x, v=t+x . we note x=(v-u)/2.. 

pde := diff(u(t,x),t,t) - diff(u(t,x),x,x) + x^2*u(t,x);  

 We need only initial condition let f(u)=sech(u) & g(v)=sech(v). So u((i,0)=f(ih), u(0,j)=g(jh) 

we summe step size of time j =  step size of spatial i .

the stencil we use , 

u(i,j)=-u(i-1,j-1)+(1-h^2/8*v(h/2*(j-i-1)))*u(i,j-1)+(1-h^2/8*v(h/2*(j-i+1)))*u(i-1,j);

Thank you 

If you have any queastion please contact me my email  mos1401(at)gmail.com  

@yurr 

Thanks for help 

please contact me by email mos1401at gmail.com  

linear_wave2.pdf@Carl Love 

Here is the pdf code

Thanks 

linear_wave2.pdf@Carl Love 

Here is the pdf code

Thanks 

@Carl Love 

Sorry about that, I mean
they are both equal delta t= delta x

@Carl Love 

Thank you for respond my question
Yes, h=delta t
Here is the solution for this equation by use pdsolve, which include the boundary and the initial condition

f:=x-> exp(-2*x^2):
g:= x-> x:

IBC := {psi(0,x)=f(x), D[1](psi)(0,x)=g(x), psi(t,-1)=0, psi(t,1)=0};
pde := diff(psi(t,x),t,t)-diff(psi(t,x),x,x)+x^2*psi(t,x)=0;
pde_sol := pdsolve(pde,IBC,numeric,spacestep=1/2,timestep=1/4):
pde_sol:-animate(t=0..10,axes=boxed,frames=100);