Solving $\nabla^2 \phi(x,y) +\lambda \phi(x,y) = 0$

Question

I wish to obtain the eigenfunctions/eigenvalues of $\nabla^2 \phi(x,y) +\lambda \phi(x,y) = 0$ with boundary conditions $\phi(x,0) = \phi(x,\pi) = \phi(0,y) = \phi(\pi,y) = 0$. To do this, I consider the method of Separation of Variables.

Ansatz: $\phi = X(x)Y(y)$. Then substituting appropriately I obtain: $$ \frac{X''}{X} = -\frac{(Y'' + \lambda Y)}{Y} = -\mu $$ which yields the two equations $$\begin{cases}X'' + \mu X &= 0 \\ Y'' + (\lambda - \mu)Y &=0\end{cases}$$ and corresponding boundary conditions: $$X(x)Y(0) = X(x)Y(\pi) = X(0)Y(y) = X(\pi)Y(y) = 0$$

Questions:

1. In the equation for $X$ the variable $-\mu$ is called the eigenvalue of the $X$ equation because we can write $DX = -\mu X$ where $D := \partial_x^2$ and so $-\mu$ is an eigenvalue of this equation. For the $Y$ equation the eigenvalue is $-(\lambda - \mu )$. If I find the values for $\lambda, \mu$ then...what does this mean for the original problem. Is the product of the eigenfunctions of the respective equations again an eigenfunction for the original problem?
2. I have issue actually solving this system of equations with the boundary conditions. From my observation it does not seem possible to solve for the four coefficients the result from solving the equations. See below:

Solving the $X$ equation I obtain: $X = c_1 \cos(\sqrt(\mu)x)+c_2\sin(\sqrt{\mu}x)$

Solving the $Y$ equation I obtain $Y = c_3\cos(\sqrt{(\lambda - \mu)}y) + c_4\sin(\sqrt{(\lambda-\mu)}y)$

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I obtain $c_1 = c_3 = 0$. But now I have the equations $$\begin{cases}c_2\sin(\sqrt{\mu}x)c_4\sin(\sqrt{(\lambda-\mu)}\pi) &=0 \\ c_2\sin(\sqrt{\mu}\pi)c_4 \sin(\sqrt{(\lambda-\mu)}y) &= 0\end{cases}$$

Answer 1

This method of jamming your separated variables together is not a good way to go. Instead, you should recognize, since each of these holds for all $(x,y) \in [0,\pi]^2$, \begin{align*} X(x)Y(0) &= 0 &&\implies& Y(0) &= 0 \text{,} \\ X(x)Y(\pi) &= 0 &&\implies& Y(\pi) &= 0 \text{,} \\ X(0)Y(y) &= 0 &&\implies& X(0) &= 0 \text{, and} \\ X(\pi)Y(y) &= 0 &&\implies& X(\pi) &= 0 \text{.} \end{align*} That is, you've separated the variables, which is predicated on being able to separate the boundary/initial conditions...

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(This answer is written for the version of the two conditions at the end of the Question as it was when the answer was written. I hope that some square roots have appeared since then.)

Nevertheless, we can still answer your question.

Let's look at your first equation, which is to hold for all $x \in [0,\pi]$: $$ c_2 \sin(\sqrt{\mu}x) c_4 \sin((\lambda + \mu)\pi) = 0 \text{.} $$ A product is zero if one (or more) of its multiplicands is zero. This gives four possibilities:

• $c_2 = 0$,
• $\sin(\sqrt{\mu}x) = 0$, which requires $\mu = 0$ (for any other choice of $\mu$, we can choose an $x$ giving a nonzero left-hand side; recall that $\lambda$ and $\mu$ are constants, so do not depend on $x$ or $y$),
• $c_4 = 0$, or
• $\sin((\lambda + \mu)\pi) = 0$, requiring $\lambda + \mu \in \Bbb{Z}$.

Your second equation gives the options

• $c_2 = 0$,
• $\sqrt{\mu} \in \Bbb{Z}$,
• $c_4 = 0$, or
• $\lambda + \mu = 0$.

The intersection of these two lists of options (the conjuction of these two disjunctions followed by some simplification) gives the options

• $\mu = \lambda = 0$,
• $\sqrt{\mu}, \lambda + \mu \in \Bbb{Z}$, that is, $\mu$ is the square of an integer and $\lambda$ is an integer,
• $c_2 = 0$,
• $c_4 = 0$,
• $\lambda = -\mu$, or
• $\mu = 0$.

We now know that every solution (compatible with separated variables) satisfies one or more of those six constraints.