MaplePrimes - answers and comments on Question, Error: Newton iteration is not converging..

singular pde

Robert Israel — Wed, 30 Nov 2011 01:00:20 Z

I think the singularity in your PDE at r=0 could be causing trouble. You might try a change of variables to remove the singularity. If you take r^2 = s:

> with(PDEtools):
simplify(dchange({r=sqrt(s),Phi(r,t)=B(s,t)},pde1,[s,B(s,t)]));

I'm not sure what you'll do about a boundary condition at s=0, though: the change of variable has made D[1](Phi)(0,t) = 0 automatically true if B is differentiable.

Do I still need a bc at r=0

vish — Wed, 30 Nov 2011 02:33:53 Z

Dear Robert,

That seems like a very good suggestion. I imposed the BC D[1](Phi)(0,t) = 0 in the first place to tackle this singularity at the equation. but it seems that when I make these changes it goes away. I have not tried it yet on maple..

However I have a question. Why does maple ask for boundary conditions even for wellposed initial value problems that only needs initial conditions?

Thanks for your help!

Thanks for the suggestion ! Do I still need a bc at r=0?

vish — Wed, 30 Nov 2011 02:47:08 Z

Dear Robert,

However I have a question. Why does maple ask for boundary conditions even for wellposed initial value problems that only needs initial conditions?

Also if you please can look at the original problem I had, of which this is a branch post, I had to make the spacestep really small for it to match the same function at t=0 which I put in as initial condition. When I try to plot the solution at t=0 for large spacestep it differes significantly from the known function at t=0,, I find it bizzare why maple would change it in the process of solving it

Thanks very much for your attention. Looking forward to your suggestions

Kind regards,

Vish

My code seems to be sensitive to the spacestep command

vish — Wed, 30 Nov 2011 21:55:54 Z

Hi Robert,

After making the variable change you suggested it iterates for longer time.. but the output seems to be sensitive to the spacestep ie the plot is different for the same time but calculated with different spacesteps?

Also does it matter that the spacestep is odd/even or any relationship to range? because when I make say for example spacestep=90, 91 I am able to iterate for much longer time for 91,...

Vish

Boundary conditions

Robert Israel — Wed, 30 Nov 2011 05:59:12 Z

The PDE is second order in r, so Maple says you need 2 boundary conditions. For any problem where "time-based" numerical methods have a chance of working, I think that will in fact be true.