I'm trying to solve a differential equation (Dirichlet problem):
$\Delta \varphi(x, y) = 0$
$0 \leq x \leq a, \ \ \ 0 \leq y \leq a$.
The boundary values are:
$\varphi(a, y) = \varphi(x, a) = \varphi_0, \ \ \ \varphi(0,y) = \varphi(x, 0) = 0$.
After substitution $\varphi(x, y) = f(x)g(y)$ we get:
$\underbrace{\frac{\partial^2 f}{\partial x^2}\frac{1}{f}}_{\text{constant}} + \underbrace{\frac{\partial^2 g}{\partial x^2}\frac{1}{g}}_{\text{-constant}} = 0 \Longrightarrow \frac{\partial^2 f}{\partial x^2}\frac{1}{f} = -\frac{\partial^2 g}{\partial x^2}\frac{1}{g} = k^2 $
So we have two options for those functions – either sinusoidal or exponential functions.
Now, at lessons we had solved the same equation but when $\varphi(x, a) = \varphi_0$ and all the other boundaries had $\varphi = 0$.
Then, we could guess that $f(x) = \sum_{n=1}^{\infty} sin \frac{n \pi x}{a} $ because $\varphi$ has zeros on the both sides of the $x$ domain, and exponential functions couldn't. For $g(y)$, we guessed that it's $g(y)=Ae^{ky} + Be^{-ky}$ , because it has zero only on the one side of $y$ domain. Then, we solved the whole thing using Fourier series and boundary conditions.
The current problem is similar but I'm reaching a wall. As both boundary conditions are the same for $x$ and $y$, they have one zero – they both fit exponential functions. It looks like both $f(x)$ and $g(x)$ are of exponential kind, which makes it impossible for me to continue with the Fourier series. The teacher says the solution must be a series.
What am I doing wrong? How to get the solution as a series?
Best Answer
You have 2 non-homogeneous boundary conditions here. The way they are arranged does not allow for a simple series solution. Instead, break up the solution to $\varphi = u + v$, such that the boundary conditions are split up
\begin{cases} \nabla^2u = 0, && 0 < x < a, \ 0 < y < a \\ u = 0, && x = 0, x = a, y = 0 \\ u = \varphi_0, && y = a \end{cases} \begin{cases} \nabla^2v = 0, && 0 < x < a, \ 0 < y < a \\ v = 0, && y = 0, y = a, x = 0 \\ v = \varphi_0, && x = a \end{cases}
The first BVP will have a sine series in $x$ and the second BVP will have a sine series in $y$. Since they are symmetric, you can just solve one of them and swap $x$ and $y$ for the other.
The final solution looks something like this
$$ \varphi(x,y) = \sum_{n=1}^\infty a_n \sin\left(\frac{n\pi x}{a}\right)\sinh\left(\frac{n\pi y}{a}\right) + \sum_{n=1}^\infty b_n \sin\left(\frac{n\pi y}{a}\right)\sinh\left(\frac{n\pi x}{a}\right) $$