Systematic way to determine the ansatz and sign in eigenvalue problem resulting form separation of variables fo Laplace's equation

Question

To solve

$$ \nabla^2 u = 0$$

in a rectangle $(0 \leq x \leq L; 0 \leq y \leq H)$ using separation of variables, we assume the solution is of the form

$$u(x,y) = h(x)\phi(y)$$

or

$$u(x,y) = h(y)\phi(x)$$

to obtain two ODEs (one of which is an eigenvalue problem in for $\phi$).

$$\dfrac{1}{h} \dfrac{d^2h}{dx^2} = - \dfrac{1}{\phi} \dfrac{d^2\phi}{dy^2} = \pm \lambda$$

or

$$\dfrac{1}{h} \dfrac{d^2h}{dy^2} = - \dfrac{1}{\phi} \dfrac{d^2\phi}{dx^2} = \pm \lambda$$

Apparently, the imposed boundary conditions determine which 1- ansatz to use from the above, and 2- the sign of the separation constant $\lambda$. Examples for specific choices that I came across are for the follwoing sets of boundary conditions:

a) $u(0,y) = f(y); u(L,y)=u(x,0)= u(x,H) = 0$

$$u(x,y) = h(x) \phi(y)$$ $$\dfrac{1}{h} \dfrac{d^2h}{dy^2} = - \dfrac{1}{\phi} \dfrac{d^2\phi}{dx^2} = -\lambda$$

b) $u(x,H) = f(y); u_x(0,y) = u_x(L,y)=u(x,0) = 0$

$$u(x,y) = h(x) \phi(y)$$ $$\dfrac{1}{h} \dfrac{d^2h}{dx^2} = - \dfrac{1}{\phi} \dfrac{d^2\phi}{dy^2} = -\lambda$$

c) $u(L,y) = f(y); u_x(0,y) =u(x,0)= u(x,H) = 0$

$$u(x,y) = h(x) \phi(y)$$ $$\dfrac{1}{h} \dfrac{d^2h}{dx^2} = - \dfrac{1}{\phi} \dfrac{d^2\phi}{dy^2} = \lambda$$

My questions how choose the appropriate ansatz and sign of $\lambda$. What is the reasoning and systematic way behind it?

Thanks

Answer 1

So the "different" types of ansatz you showed are exactly the same, since you do not know the $h,\phi$ or $\lambda$ at this point. For this reason you do not need to worry about the sign on front of the $\lambda$, that comes later. So your separation of variables ansatz is simply $$ u(x,y)=X(x)Y(y)\Rightarrow X''Y+XY''=0\Rightarrow \frac{X''}{X}=-\frac{Y''}{Y}=\lambda $$ now the introduction of boundary conditions is straightforward, we just plug in the ansatz. Lets start with homogenous Dirichlet $u(x,0)=u(x,H)=u(0,y)=u(L,y)=0$: $$ u(x,0)=X(x)Y(0)=0\Rightarrow Y(0)=0\\ u(x,H)=X(x)Y(H)=0\Rightarrow Y(H)=0\\ u(0,y)=X(0)Y(y)=0\Rightarrow X(0)=0\\ u(L,y)=X(L)Y(y)=0\Rightarrow X(L)=0 $$ Here I have omited the trivial solution $X(x)=0$ or $Y(y)=0$ which both yield $u(x,y)=0$. If we now solve the ODEs $$ X''=\lambda X, X(0)=X(L)=0 $$ we find that $$ X(x)=c\sin(\sqrt{-\lambda}x) \text{ for } \lambda=-\frac{\pi^2 n^2}{L^2}, n\in\mathbb{N} $$ And therefore $\lambda<0$. Unfortunately the other ODE leads the exact opposite result $\lambda=\frac{\pi^2 n^2}{H^2}$ and therefore in this case the only solution is the trivial solution $u(x,y)=0$.

Looking at your other boundary conditions, the problem is that depending on $f$ a simple separation ansatz might not be sufficient to solve the problem: Indeed if we plug the first set into the ansatz: $$ u(0,y)=X(0)Y(y)=f(y)\Rightarrow Y(y)=f(y)/X(0)\\ u(L,y)=X(L)Y(y)=0\Rightarrow X(L)=0\\ u(x,0)=X(x)Y(0)=X(x)f(0)/X(0)\Rightarrow f(0)=0\\ u(x,H)=X(x)Y(H)=X(x)f(H)/X(0)\Rightarrow f(H)=0\\ \Rightarrow \begin{cases}f''=-\lambda f, f(0)=f(H)=0\\ X''=\lambda X,X(L)=0\end{cases} $$

so we see there is a strong assumption on $f$ for the ansatz to work.

Answer 2

You're solving $\nabla^2 u = 0$ on a rectangular region $R$ of the plane, which will require boundary conditions on the four edges of $R$. Because of the linearity of $\nabla^2$, you can reduce the problem to solving $\nabla^2=0$ with $3$ of the $4$ boundary conditions being $0$, and the $4$th boundary condition specified as some function which is not identically $0$; then you can add these $4$ solutions in order to obtain the desired solution where all $4$ required edge conditions are satisfied. This approach is justified by the linearity of the Laplacian; non-linear equations cannot generally be reduced in this way.