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Question: How do I express bc for PDE

Question:How do I express bc for PDE

Thomas Dean 322 Maple 2016
pde pdsolve pdetools
January 18 2017
0

I have a PDE with boundary conditions, from a NASA paper.  I always seem to have problems expressing the bc.

PDE := diff(u(x, t), t) = (1/4)*exp(2)*exp(-u(x, t))*(diff(diff(u(x, t), x), x))/(x^2+2);

The initial/boundary conditions are

@t=0, u(x, t) = 2-2*ln(-x^2+2)

@x=0, diff(u(x,t),x)=0  ## this is the bc I have problem expressing

@x=1, u(x,t) = 2+ln(1+t)

The exact solution given in the paper:

2 + ln(1+t) - 2*ln(2-x^2)

I tried

ics := u(x, 0) = 2*(1-ln(2-x^2));

bcs := D[1](u(0,t))=0, u(1,t)=2+ln(1+t);

PDEtools[Solve]([PDE,ics,bcs]); ## no solution

How do I do this?

Tom Dean
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