I am trying to find a way to efficiently solve the Laplace equation, by generating a rule for making Ansatz. However, once this ansatz is found, I struggle with finding the coefficients and the structure of the solution.
The problem is:
\begin{equation} u_{xx}+u_{yy}=0 \ \ \ \ \ \ \Omega: \{0<x<1, 0<y<3\} \\ u_x(0,y)=u_x(1,y)=0 \ \ \ \ \ \ \\ u(x,0)=0, u(x,3)=x \end{equation}
From this problem, we can actually see the form of the solution, by looking at the IC:
- $u_x(0,y)=u_x(1,y)=0$
This is a Neumann condition on x, thus the x-part of the solution must be in the form $u(x)=A_n\cos(\frac{n\pi}{a})x$.
- $u(x,0)=0, u(x,3)=x $
This is a Dirichlet condition on y, and note that the second condition suggests that the function is $x$ at the point 3, so this is clearly not a trigonometric behavior, therefore we can conclude that since it is Dirichlet, it must be a sine function, but since it is not trigonometric, it must be hyperbolic sine. So the form must be $u(y)=B_m\sinh\frac{m\pi}{b}y$.
Since the boundary is rectangular, we need an additional constant in front of the Ansatz, hence we can form the Ansatz as
\begin{equation} u(x,y)=C_n + \sum_{n=0}^\infty A_n\cos\frac{n\pi}{a}x \sum_{m=0}^\infty B_m\sinh\frac{m\pi}{b}y \end{equation}
But at this stage, is there an equally rapid way to find the coefficients? Evidently, only the cosine term can be subjected to series expansion, but here we have a zero in the Neumann conditions, so how is that done?
Thanks
The separation of variables technique starts with looking for solutions of the form $X(x)Y(y)$. The sum of such solutions is rarely going to be another separated function of the form $X(x)Y(y)$. Plugging $X(x)Y(x)$ into the original PDE, and separating variables, gives $$ X''(x)Y(x)+X(x)Y''(y)=0 \\ \frac{X''(x)}{X(x)}+\frac{Y''(y)}{Y(y)}=0 \\ \frac{X''(x)}{X(x)}=-\frac{Y''(y)}{Y(y)} $$ The left side does not depend on $y$ and the right side does not depend on $x$. So both sides must be constant in $x$ and in $y$. Therefore, in order for such a solution $X(x)Y(y)$ to exist, it must be the case that there is a constant $\lambda$ such that $$ \frac{X''(x)}{X(x)}=-\lambda,\; \frac{Y''(y)}{Y(y)}=\lambda. $$ If the constant $\lambda$ is positive, you end up with solutions $$ X(x)=A\cos(\sqrt{\lambda}x)+B\sin(\sqrt{\lambda}x), \\ Y(y)=C\cosh(\sqrt{\lambda}y)+D\sinh(\sqrt{\lambda}y) $$ If the constant $\lambda$ is negative, you end up with solutions $$ X(x)=A\cosh(\sqrt{-\lambda}x)+B\sinh(\sqrt{-\lambda}x), \\ Y(y)=C\cos(\sqrt{-\lambda}y)+D\sin(\sqrt{-\lambda}y) $$ The only way to tell which form is needed is to look at the endpoint conditions. If there are two homogenous endpoint conditions in $x$, then the $x$ solutions will involve the ordinary trigonometric functions because the hyperbolic ones cannot satisfy homogeneous endpoint conditions at $x=a$ and at $x=b$, unless you're dealing with the $0$ solution. In your case, you have $a=0$, $b=1$, and $$ X'(0)=0,\;\; X'(1)=0. $$ This dictates solutions $X_n(x)=\cos(\sqrt{\lambda}x)$ where $\lambda=n^2\pi^2$ for $n=0,1,2,3,\cdots$ (the solution $X_0(x)=1$ is a solution.) The corresponding solutions in $Y$ have the form $\sinh(\sqrt{\lambda}x)$ because you need $u(x,0)=0$. So, the general solution becomes $$ u(x,y)=E_0+\sum_{n=1}^{\infty}E_{n}\cos(n\pi x)\sinh(n\pi y), $$ where the $E_n$ are chosen to satisfy $$ x = u(x,3)=E_0+\sum_{n=1}^{\infty}E_n\cos(n\pi x)\sinh(3n\pi) $$ So the problem is reduced to an ordinary Fourier cosine series problem, where the Fourier coefficients are $E_n\sinh(3n\pi)$. And these coefficients are determined by the function $x$ through the usual Fourier orthogonality conditions.