PDE with two variables by NDSolve

[Alexei Boulbitch]

Dear MathGroup members,

I have a misunderstanding when solving a partial differential equation with M6. It is a rather simple equation for a function z=z(x,y):

pde = \!\(
\*SubscriptBox[\(\[PartialD]\), \({x, 2}\)]\ \(z[x, y]\)\) + \!\(
\*SubscriptBox[\(\[PartialD]\), \({y, 2}\)]\ \(z[x,
y]\)\) - (1 - 3*Exp[-(x^2 + y^2)])*z[x, y] -
Log[(1 + z[x, y])/(1 - z[x, y])] == 0;

with simple boundary conditions fixed on a square 10x10. The function should be there either zero, or something small, but different from zero like 0.01. Let us take zero:

bc = {z[-10, y] == z[10, y] == 0, z[x, -10] == z[x, 10] == 0};

Now application of the NDSolve operator:

sol = NDSolve[{pde, bc}, z, {x, -10, 10}, {y, -10, 10}]

returns the error message:

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. >>

I understand this response, as if Mathematica does not support boundary conditions fixed for both of the variables.
However, in Help/NDSolve/ Scope/Partial Differential Equations/Example Nr 3 shows the following code for SineGordon:

L = 4;
sol = NDSolve[{D[u[t, x, y], t, t] ==
D[u[t, x, y], x, x] + D[u[t, x, y], y, y] + Sin[u[t, x, y]],
u[t, -L, y] == u[t, L, y], u[t, x, -L] == u[t, x, L],
u[0, x, y] == Exp[-(x^2 + y^2)],
Derivative[1, 0, 0][u][0, x, y] == 0},
u, {t, 0, L/2}, {x, -L, L}, {y, -L, L}]

in which the  line u[t, -L, y] == u[t, L, y], u[t, x, -L] == u[t, x, L] fixes very similar type of boundary conditions.

What´s wrong?

Regards, Alexei
--
Alexei BOULBITCH,
Dr. , habil.
IEE S.A.
Department for Material Development
ZAE Weiergewan
11, rue Edmond Reuter
L-5326 Contern, Luxembourg

Tel. ++352-2454-2566
Fax.: ++352 424737201
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E-mail: alexei.b...@iee.lu
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[D. Grady]

It looks like NDSolve is designed to only work with initial value
problems. There's a note in the documentation under NDSolve /
Possible Issues/ Partial Differential Equations that doesn't work with
Laplace's equation, and in the tutorial/NDSolveOverview you can read
more about the methods NDSolve uses; none of them are for elliptic
equations.

One way you could try to work around this is to add a time variable to
your problem, specify some initial condition, and see if the solution
relaxes to a steady state. This is kind of clumsy, but it might be
worth a shot.

-Daniel

[dh]

Hi Alexei,

your PDE is nothing else than "rather simple". It is a non-linear

transcendental PDE.

Further, the boundary conditions from the example you mention:

u[t, -L, y] == u[t, L, y], u[t, x, -L] == u[t, x, L],

only state that the solution is periodic. The other conditions:

u[0, x, y] == Exp[-(x2 + y2)],

Derivative[1, 0, 0][u][0, x, y] == 0

specify boundary conditions just for one point of t.

Daniel

[dh]

Hi Alexei,

I forgot to mention another point about NDSolve. NDSolve is made for

time dependent problems. From the manual:

"all of these functions must depend on a single "independent variable"

t, which is the same for each function."

Daniel