## 20 Reputation

11 years, 65 days

## What do the frames represent...

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

Thanks

## Thank you Carl Love 3435 .Also I am look...

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 look...

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

## @yurr  Hello  I wish to s...

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

Thanks for help

## Mathematica code convert to Maple...

Here is the pdf code

Thanks

## Mathematica code convert to Maple...

Here is the pdf code

Thanks

## @Carl Love  Sorry about that, I mea...

they are both equal delta t= delta x

## @Carl Love  Thank you for respond m...

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);

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