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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.

Thanks in advance.

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1 Answer 1

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

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  • $\begingroup$ 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... $\endgroup$ Dec 16, 2020 at 15:11
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    $\begingroup$ maybe the bounded solutions are exactly those for which $b_{2, k} = 0$? $\endgroup$ Dec 16, 2020 at 15:12
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    $\begingroup$ I think I got it. I will edit my answer to explain what I thought $\endgroup$ Dec 16, 2020 at 15:22
  • $\begingroup$ 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? $\endgroup$ Dec 16, 2020 at 16:05
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    $\begingroup$ 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 $b_{2,k}=0$ $\endgroup$ Dec 16, 2020 at 16:10

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