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Re: PDE with two variables by NDSolve

  • To: mathgroup at smc.vnet.net
  • Subject: [mg94617] Re: PDE with two variables by NDSolve
  • From: dh <dh at metrohm.com>
  • Date: Wed, 17 Dec 2008 06:36:58 -0500 (EST)
  • References: <gi5jdq$phm$1@smc.vnet.net>


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



Alexei Boulbitch wrote:

> 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




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