Am I doing wrong with the boundary conditions in NDSolveValue?

[wyan (Wen Yan)]

Hi,

I am trying to solve 2D PDEs with NDSolveValue[] (Mathematica 10 windows & linux both), but I always get some errors with the derivative boundary conditions. For example:

uif = NDSolveValue[D[f[x, y], {x, 2}] + D[f[x, y], {y, 2}] == 1 && Derivative[1, 0][f][1, y] == 1 && f[0, y] == 1, f, {x, 0, 1}, {y, 0, 1}]

solves a simple Poisson equation on a square (0,1)x(0,1), with \partial f / \partial x =1 on x=1.
However, the solution uif always gives wrong answer:

Plot[Derivative[1, 0][uif][x, 0.5], {x, 0, 1}]

So the derivative of the solution at x=1 is the negative of the condition specified.

If I specify Derivative[1, 0][f][0, y] == 1 , (same condition at x=0), the answer would be correct. Same thing would happen for all my test PDEs.

Is Mathematica 10 assuming that the boundary condition specified is affected by the normal vector on the boundary? Or how should I specify the boundary conditions appropriately?

Thank you,

[mthiel (Marco Thiel)]

Hi,

the same command does not evaluate inMathematica 9 and gives an error message instead:

NDSolveValue::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. >>

Also, if you invert the sign of your first condition like this:

uif = NDSolveValue[D[f[x, y], {x, 2}] + D[f[x, y], {y, 2}] == 1 && Derivative[1, 0][f][1, y] == -1 && f[0, y] == 1, f, {x, 0, 1}, {y, 0, 1}]

you get a positive 1 derivative just as you wanted.

Plot3D[uif[x, y], {x, 0, 1}, {y, 0, 1}]

I know that this does not solve your problem. I am not really sure what MMA10 does differently from MMA9 here.

Cheers,
Marco