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I would like to solve the following nonlinear coupled PDE with a mix of initial conditions and boundary values:

rMax = 0.01;
sol = First@NDSolve[{
Derivative[2, 0][g][r, z] + Derivative[0, 2][g][r, z] == u[r, z]^2,
Derivative[2, 0][u][r, z] + Derivative[0, 1][u][r, z] == -g[r, z],
Derivative[1, 0][u][0, z] == 0.0,
Derivative[1, 0][u][rMax, z] == 0.0,
u[rMax, z] == 0.0,
u[r, 0] == g[r, 0] == Sin[\[Pi] r/rMax],
Derivative[1, 0][g][0, z] == g[rMax, z] == 0.0},
{u, g}, {r, 0, rMax}, {z, 0, 0.01}]

but I receive the following error message (in version 10.0.1.0):

NDSolve::femnonlinear: Nonlinear coefficients are not supported in this version of NDSolve.

The offender is the square term u[r, z]^2 in the first equation; without the square NSolve[] executes without errors. NDSolve seems to apply the FEM method by default to such problems. I'm wondering why NDSolve[] doesn't switch back to another (propagation-type) algorithm? When I add the option Method -> "MethodOfLines", the error message changes to

NDSolve::ivone: Boundary values may only be specified for one independent variable. Initial values may only be specified at one value of the other independent variable.

and I don't quite understand why this is because my time-like variable is z and I'm setting initial conditions only for z=0 and then boundary conditions at r=0 and r=rMax which should be OK?

Any ideas how to solve my problem? Another post suggested calling low-level FEM routines directly, is this a solution? What's the advantage of using FEM on an initial condition/boundary value problem over other methods: speed, accuracy, robustness?

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    $\begingroup$ Try NDSolve[..., Method -> {"MethodOfLines", "TemporalVariable" -> z}] -- You'll get another error message, but perhaps that will give you a clue. $\endgroup$
    – Michael E2
    Commented Sep 26, 2014 at 14:51
  • $\begingroup$ OK, the new error message is NDSolve::tvic: "z cannot be used as the temporal independent variable because the conditions {u[r,0]==Sin[6283.19\ r],g[r,0]==Sin[6283.19\ r]} for that dimension do not constitute sufficient initial conditions given at only one value of z." Why would the condition at z=0 not be sufficient, being at the beginning of the requested solution range? $\endgroup$
    – RonH
    Commented Sep 26, 2014 at 15:07
  • $\begingroup$ Perhaps a time derivative needs to be specified at z == 0? $\endgroup$
    – Michael E2
    Commented Sep 26, 2014 at 15:08
  • $\begingroup$ I may be wrong but won't your Derivative[1, 0][u][0, z] == 0 condition be inconsistent with u[r, 0] == Sin[π r/rMax]? $\endgroup$
    – Silvia
    Commented Sep 27, 2014 at 4:03
  • $\begingroup$ @Silvia: Yes, you're correct, so let's replace the sine term by Cos[Pi/2 r/rMax]. $\endgroup$
    – RonH
    Commented Sep 29, 2014 at 11:06

1 Answer 1

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In version 12.0 you can solve this:

rMax = 0.01;
{usol, gsol} = NDSolveValue[{
   Derivative[2, 0][g][r, z] + Derivative[0, 2][g][r, z] == 
    u[r, z]^2,
   Derivative[2, 0][u][r, z] + Derivative[0, 1][u][r, z] == -g[r, z],
   u[rMax, z] == 0.0, u[r, 0] == g[r, 0] == Sin[\[Pi] r/rMax], 
   g[rMax, z] == 0.0}, {u, g}, {r, 0, rMax}, {z, 0, 0.01}]

I have removed the Neumann zero BCs.

Plot the solution for g:

Plot3D[gsol[r, z], {r, 0, rMax}, {z, 0, 0.01}]

enter image description here

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