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nathangrand Feb14-12 01:40 PM

Solve laplace's equation on semi infinite strip

Z= 0 for x=0, y=0
Z= x(1-x) for y=0
Z=0 for y=infinity

Range 0<x<1 and y>0 (suppose strictly speaking should be x=1 and x=0 too)

So all I want to do is solve this

Use separation of variables:

X''/X = a^2 = -Y''/Y

Gives X = Aexp(ax) + Bexp(-ax) and Y=Ccos(ay) + Dsin(ay)

Or completely free to swap these around to give
Y=Aexp(ay)+ Bexp(-ay) and X= Ccos(ax) +Dsin(ax)
which I shall do as get further with boundary conditions

Know that Z=XY

As at y=infinity, Z=0 ==> A=0
At x=0, Z=0 ==>C = 0
At x=1, Z=0 ==> a=n*pi where n is an integer

so have Z=Esin(n*pi*x)exp(-n*pi*y) where E is a new constant

but how on earth do I make this compatible with the remaining boundary condition for y=0

==> Z=x(1-x) = Esin(n*pi*x) ??????

Clearly must have gone wrong somewhere???

kai_sikorski Feb15-12 06:38 AM

Re: Solve laplace's equation on semi infinite strip
The next step is very standard in solving PDEs by separation of variables, any book/article on the topic will explain it. You have to expand [itex]x(x-1)[/itex] as a series using [itex] \sin(\pi n x)[/itex]. You should find that

[tex] x(x-1) = \sum_{n=1}^\infty \frac{4 \left(-1+(-1)^n\right)}{n^3 \pi ^3}\sin(\pi n x)[/tex]

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