I'm having trouble understanding how to find $u(x,y)$ when the laplace boundary conditions are non zero. The problem below I've attached as a reference….I understand what is going on but when the boundary changes I get lost What I mean is
What steps would I need to take if the boundary conditions were to be $u(x, 0) = 1$ and $u(x, \pi) = \frac{x}{2} (\pi − x)$?
To give some background I've included the following
Given $\Delta u = 0$ and $0 < x$ , $y < \pi$
$u_x(0, y) = u_x(\pi, y) = 0$
$u(x, 0) = 0$ and $u(x, \pi) = \frac{x}{2}(\pi − x)$
We find u(x,y) by applying Fourier separation of variables.
$$X''(x)Y (y) + X(x)Y''(y)=0 \implies \frac{X''(x)}{X(x)}= -\frac{Y''(y)}
{Y (y)} = \lambda$$
then we have
$$Y'' + \lambda Y =0$$
$$X'' − \lambda X =0$$
By the initial conditions and Sturm Liouville $λ_n=n^2$ and $X(x)=\cos(nx)$
Also, $u(x,0)=0 \implies A_n=-B_n$
Then the equation will be
$$u(x,y)=\sum^{\infty}_{n=1} A_n(e^{ny}+e^{-ny})\cos(nx)$$
Also the initial condition $u(x,\pi)=\frac{x}2(\pi-x)$ Then
$$A_n=\dfrac{2}{\pi(e^{n\pi}-e^{-n\pi})}\int_0^{\pi}\frac{x}2(\pi-x)\cos (nx)dx $$
QUESTION:
What steps would I need to take if the boundary conditions were to be $u(x, 0) = 1$ and $u(x, \pi) = \frac{x}{2} (\pi − x)$ Also in this case would the sum start at $0$? since we should have the same eigen-functions and values. Given that $X(x)=\cos (nx)$ where $n=0,1,2,\dots$
I'm trying to understand this so I can solve when $u(x, 0) = 3$ and $u(x, \pi) = 1+\cos x$
Thanks
Best Answer
In order to solve \begin{align} &\Delta u =0,\\ &u_x(0,y)=0, \;\;u_x(\pi,y)=0 \\ &u(x,0)=1, \;\;u(x,\pi)=\frac{x}{2}(\pi-x) \end{align} you can solve two separate problems, and add the solutions: \begin{align} &\Delta u = 0,\\ &u_x(0,y)=0,\;\; u_x(\pi,y)=0 \\ &u(x,0)=1,\;\; u(x,\pi)=0 \\ \\ &\Delta u = 0,\\ &u_x(0,y)=0,\;\; u_x(\pi,y)=0 \\ &u(x,0)=0,\;\; u(x,\pi)=\frac{x}{2}(\pi-x). \end{align} The first of the two problems is solved by $$ u(x,y) = 1-\frac{y}{\pi} $$ The second of the two problems is solved by $$ u(x,y) = A_0y+\sum_{n=1}^{\infty}A_n\cos(nx)\sinh(ny) $$ where the $A_n$ are chosen to satisfy $$ u(x,\pi)=\frac{x}{2}(\pi -x)=A_0\pi+\sum_{n=1}^{\infty}A_n\cos(nx)\sinh(n\pi) $$