Thread: Laplace's equation in 2 dimension

1. Laplace's equation in 2 dimension

Hi there, I'm doing some revision for an upcoming test and have all the past papers we have answers for ask us to find separable solutions when dealing with the one dimensional heat equation which I think I can do now but then in other papers we don't have answers for it talks about Laplaces equation in two dimensions which I don't think we even toucjed on in class so other than part a) which I think I can get (elliptic) I'm pretty lost. Any help appreciated!!

2. Since you are looking for separable solutions assume it has the form.

$u(x,y)=X(x)Y(y)$

This gives

$u_{yy}=XY'' \quad u_{xx}=X''Y$

This gives

$X''Y+XY''=0 \iff \frac{X''}{X}=-\frac{Y''}{Y}=-\lambda^2$

With

$\lambda^2 > 0$ (why?)

This gives the system

$X''+\lambda^2X=0 \quad Y''-\lambda^2 Y=0$

The boundary conditions gives

$X(0)=X(\pi)=0$

The solutions to the X equation are

$X =c_1\cos(\lambda x)+c_2\sin(\lambda x)$

This gives

$X(0)=c_1=0$

$X =c_2\sin(\lambda x)$

$X(\pi)=0 =c_2\sin(\lambda \pi) \iff n\pi =\lambda \pi \implies \lambda = n$

with

$n \in \mathbb{Z}^+$

$X(x)=\sin(n x)$

Now if you solve the Y equation you get

$Y(y)=d_1\cosh(ny)+d_2\cosh(ny)$

This gives

$u(x,y)=X(x)Y(y)=\sin(n x)( d_1\cosh(ny)+d_2\cosh(ny))$

For C just use this formula

For D just mimic the process done in b)