# Fourier series for the solution of $\frac{\partial^2 u}{\partial x^2} + y^2 \frac{\partial^2 u}{\partial y^2} = 0$

Consider the following problem:

Let $$\Omega = (0, \pi) \times (0, 1)$$ and let $$u$$ be a solution of $$\begin{cases} \frac{\partial^2 u}{\partial x^2} + y^2 \frac{\partial^2 u}{\partial y^2} = 0 \quad \text{ in } \Omega \\ u(0, y) = u(\pi, y) = 0 \quad \text{ for } 0 \leq y \leq 1 \end{cases}$$ Expanding $$u$$ in a Fourier series, $$u(x, y) = \sum_1^\infty c_k(y) \sin(kx)$$ Determine the coefficients $$c_k(y)$$ and show that every bounded solution $$u$$ satisfies $$u(x, 0) = 0 \quad \text{ for } 0 \leq x \leq \pi.$$

I don't know much about Fourier series, so I tried to follow Strauss' PDE book. I began by separating variables: $$u(x, y) = f(x)g(y),$$ which gives us the differential equations $$f''(x) + \lambda f(x) = 0, \quad y^2 g''(y) - \lambda g(y) = 0.$$ From the first equation and the given boundary conditions, it is easy to see where the factors $$\sin(kx)$$ come from. Now, Wolfram Alpha gives $$g(y) = c_1 y^{\frac 12 - \frac 12 \sqrt{4k^2 + 1}} + c_2 y^{\frac 12 + \frac 12 \sqrt{4k^2 + 1}}$$ for the second equation (already using that $$\lambda = k^2$$). The problem is that I am out of ideas about how to compute the constants $$c_1$$ and $$c_2$$ (which is a necessary step for the second part of the problem, I believe), since there are no other boundary conditions given.

Please, what am I missing? Any hints will be appreciated.

Denoting

$$\displaystyle f_k(x)=a_k\sin kx,\quad g_k(x)=b_{1,k}y^{\frac{1+\sqrt{1+4k²}}{2}}+b_{2,k}y^{\frac{1-\sqrt{1+4k²}}{2}},$$

and

$$u_k(x,y)=f_k(x)g_k(y),$$

by the superposition principle, the solution is

$$u(x,y)=\sum_{k=1}^{\infty}u_k(x,y)=\sum_{k=1}^{\infty}a_k\sin kx\cdot g_k(y)=\sum_{k=1}^{\infty}c_k(y)\sin kx,$$

where $$c_k(y)=a_kg_k(y)$$

So,

$$u(x,0)=\sum_{k=1}^{\infty}c_k(0)\sin kx..$$

Since $$c_k(0)=a_k g_k(0)=0$$

we can conclude that

$$u(x,0)=0$$

-----------EDIT-------------

What I wrote about $$c_k(0)$$ is wrong, because you would be dividing by zero.

First, you need bounded solutions. Since the sine terms are bounded, we actually need to look $$c_k(y).$$

For a expression like $$y^r$$ be bounded in the the interval $$(0,1)$$ we need $$r\geq 0$$, so I will apply this idea to the two exponents terms in $$c_k(y)$$

The first one:

$$\frac{1+\sqrt{1+4k^2}}{2}\geq 0 \iff \sqrt{1+4k^2}\geq -1$$

This is satisfied for all $$k$$, so, these terms are bounded.

Now, I will look at the second term. We need

$$\frac{1-\sqrt{1+4k²}}{2}\geq 0 \implies \textrm{some manipulation} \implies k^2\leq 0 \implies k=0$$

But this is a contradiction, because $$k$$ is a positive integer. So none of these therms are bounded, therefore we need $$b_{2,k}=0$$

Conclusion:

$$c_k(y)=a_k b_{1,k}y^{\frac{1+\sqrt{1+4k^2}}{2}}\implies c_k(0)=0.$$

Now we can conclude $$u(x,0)=0$$

• It seems strange, since $\frac 12 - \frac12 \sqrt{4k^2 + 1} < 0$, so by plugging $0$ into $c_k$ you would be dividing by zero... Dec 16 '20 at 15:11
• maybe the bounded solutions are exactly those for which $b_{2, k} = 0$? Dec 16 '20 at 15:12
• I think I got it. I will edit my answer to explain what I thought Dec 16 '20 at 15:22
• Nice. But we are supposed to find the unbounded solutions as well, so I believe that in general the term $b_{2, k}$ should be kept. What do you think? Dec 16 '20 at 16:05
• In general, the solutions include both $b_{1,k}$ and $b_{2,k}$, and since there are no other boundary condition, we can't determine it. Just write as I wrote the expression for $u(x,y).$ Restricting to the bounded solutions, we must have $b_{2,k}=0$ Dec 16 '20 at 16:10