0

Ok so I have a question with tutorial solutions but there are a few points I don't understand.

Q: Find the solution to Laplace's equation $$(∇^2)u=0$$ in the unit square $d=\{(x,y):0≤x≤1,0≤y≤1\}$ subject to the boundary conditions \begin{align*} u&=0&\text{ on }x=0,x=1, \\ u&=\sinπx&\text{ on } y=0, y=1. \end{align*}

If someone could explain in detail all the steps that would be great! Mainly I don't understand how you know whether to use:

  1. $λ=k^2>0$, $F=Ae^{(kx)} + Be^{(-kx)}$
  2. $λ=0$, $F=Ax+B$
  3. $λ=-k^2<0$, $F=A\cos(kx) + B\sin(kx)$

and then why $k=nπ$ making $λ=-n^2π^2$

then why we use form (1.) for G eqn

and then just generally solving for the solution

P. Siehr
  • 3,672
SFL
  • 171
  • I just edited some mathjax to write your formulas readable. Still the question is hard to understand, since not everything you mention is defined (e.g. $λ, k, n$). And the last part is not a full sentence. – P. Siehr Aug 15 '17 at 10:02
  • 1
    Do u know what is Laplace's equation? – Nosrati Aug 15 '17 at 10:02
  • that is how it is in the solutions. my notes give 1.-3. as possible forms to work out the solutions. By the last sentence i mean once you have put in the boundary conditions how do you solve for the solution to the equation.. – SFL Aug 15 '17 at 10:10
  • and yes i know Laplace's. I understand separating into a function of x and a function of y, it is from the point of knowing which formula to use for F & G i do not understand – SFL Aug 15 '17 at 10:11
  • 1
    Is these conditions are true? $u(0,y)=0$, $u(1,y)=0$, $u(x,0)=\sin\pi x$, $u(x,1)=\sin\pi x$. – Nosrati Aug 15 '17 at 10:21
  • yes that is correct – SFL Aug 15 '17 at 10:23
  • If $λ=0$, so $F=Ax+B$ and $F(0)=F(1)=0$ shows $A=B=0$ which is impossible! – Nosrati Aug 15 '17 at 10:30
  • 1
    In the case $λ=k^2>0$, then the assumption $F=Ae^{kx} + Be^{-kx}$ is a non-periodic function but $u(x,0)=F(x)G(0)=\sin\pi x$ says $F(x)$ is periodic, so this case in impossible also! – Nosrati Aug 15 '17 at 10:35
  • no no, we are using number 3 for F so λ=−k^2 <0 – SFL Aug 15 '17 at 10:36
  • and number 1 for G. but i don't understand how you know what one to use. for if other questions like this come up – SFL Aug 15 '17 at 10:36
  • Am I right?.... – Nosrati Aug 15 '17 at 10:45
  • and in the case $λ=-k^2<0$, then $F=A\cos kx + B\sin kx$ and from $u(0,y)=F(0)G(y)=0$ shows $A=0$ and $u(1,y)=F(1)G(y)=0$ shows $B\sin k=0$ says $k=n\pi$ (since $B\neq0$). – Nosrati Aug 15 '17 at 10:46

0 Answers0