Solve laplace's equation on semi infinite strip

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SUMMARY

The discussion focuses on solving Laplace's equation, ∇²(Z) = 0, on a semi-infinite strip with specific boundary conditions. The solution employs the method of separation of variables, leading to the general forms for X and Y as X = Aexp(ax) + Bexp(-ax) and Y = Ccos(ay) + Dsin(ay). The boundary conditions yield Z = Esin(nπx)exp(-nπy), where E is a constant, and the challenge lies in matching this with the boundary condition Z = x(1-x) at y=0. The resolution involves expanding x(x-1) as a Fourier sine series.

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  • Understanding of Laplace's equation and boundary value problems.
  • Familiarity with the method of separation of variables in partial differential equations.
  • Knowledge of Fourier series, particularly sine series expansions.
  • Basic concepts of exponential and trigonometric functions in mathematical analysis.
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  • Study the method of separation of variables in greater detail.
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Mathematicians, physicists, and engineering students involved in solving partial differential equations, particularly those interested in boundary value problems and Fourier analysis.

nathangrand
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\nabla^2(Z)=0

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?
 
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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 x(x-1) as a series using \sin(\pi n x). You should find that

x(x-1) = \sum_{n=1}^\infty \frac{4 \left(-1+(-1)^n\right)}{n^3 \pi ^3}\sin(\pi n x)
 
Last edited:

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