poisson equation sobolev regularity - explicit computation

Question

Consider the Dirichlet problem on the circular strip $\Omega = \mathbb{T} \times (0,1)$:

$$\Delta u(x,y) = f(x,y) \ \text{ with } \ u(x,0) = u(x,1) = 0$$

It is a known result that for $f \in H^{s-2}$ we should have $u\in H^s$. I have a problem showing this explicitly. My attempt is as follows:

Writing $u(x,y) = \sum_k u_k(y)e^{ikx}$ and $f$ as fourier series in $x$ converts problem to a series of ODES on interval $(0,1)$:

$$u_k''(y) - k^2u_k(y) = f_k(y)$$ with homogeneous boundary conditions. The solutions in terms of Green's function is $$u_k(y) = \int_0^1 G_k(y,t) f_k(t) dt$$ where $$G_k(y,t) = \begin{cases} \dfrac{\sinh[k(t-1)]\sinh(ky)}{k\sinh(k)} , \quad y \leq t\\ \dfrac{\sinh(kt)\sinh(k(y-1))}{k\sinh(k)} , \quad y \geq t \\ \end{cases}$$

Now we want to show $u \in H^s$, in fact show $$ \|u\|_s^2 = \sum_{p+q \leq s} \sum_k (1+k^2)^p \int_0^1 |u_k^{(q)}(y)|^2 dy \leq c\|f\|_{s-2}^2$$ Let's investigate the case when $p=s$ and $q=0$, it is enough to show my problem. In order for the above to work, we expect that $$\|u_k\|_{L^2(0,1)}^2 \approx k^{-4} \|f_k\|_{L^2(0,1)}^2$$ so that $$(1+k^2)^s\|u_k\|_{L^2(0,1)}^2 \approx (1+k^2)^{s-2}\|f_k\|_{L^2(0,1)}^2 $$

My problem is instead of the desired $k^{-4}$ I get only $k^{-3}$:

Note by Cauchy- Schwarz $$\int_0^1 |u_k(y)|^2 dy = \int_0^1 |\int_0^1 G_k(y,t)f_k(t) dt \ |^2 dy \leq \|f_k\|_{L^2}^2 \int_0^1 \int_0^1 |G_k(y,t)|^2 dy dt$$ and $$\int_0^1 |G_k(y,t)|^2 dy dt = $$ $$ = \frac{1}{k^2\ \sinh^2(k)} \int_0^1 ( \sinh^2[k(t-1)] \int_0^t \sinh^2(ky) dy + \sinh^2(kt) \int_t^1 \sinh^2[k(y-1)] dy ) dt$$ $$= \frac{1}{4k^3\ \sinh^2(k)} \int_0^1 \sinh^2[k(t-1)] \{\sinh(2kt)-2kt\} + \sinh^2(kt)\{-\sinh[2k(t-1)] + 2k(t-1)\} dt$$ $$ = \frac{1}{4k^3\ \sinh^2(k)} (k + \frac{1}{k} - \frac{\cosh(2k)}{k} + \sinh(k)\cosh(k) ) $$ $$ \approx \frac{1}{k^3}$$

I've checked this computation many times and always get this $k^{-3}$. I wonder what can be going wrong? Perhaps there is some better way to bound it than with Cauchy-Schwartz right at the beginning of the computation?

Answer 1

Admittedly, what I'm about to write doesn't resolve the issues with your computation, but it does allow you to prove the $L^2$ estimate in a different way.

Starting from $u_k'' -k^2 u_k = f_k$ we multiply by $\bar{u}_k$ and integrate by parts: $$ \int_0^1 |u_k'|^2 + k^2 |u_k|^2 = -\int_0^1 f_k \bar{u}_k. $$ Since $u_k(0) = u_k(1) =0$ we can use Poincare on $(0,1)$ to bound $$ \int_0^1 |u_k|^2 \le \int_0^1 |u_k'|^2. $$ Plugging in and using Cauchy-Schwarz then shows that $$ (1+k^2) \int_0^1 |u_k|^2 \le \left( \int_0^1 |f_k|^2\right)^{1/2} \left( \int_0^1 |u_k|^2 \right)^{1/2} $$ and hence $$ (1+k^2)^s \int_0^1 |u_k|^2 \le (1+k^2)^{s-2}\int_0^1 |f_k|^2. $$

Also, with a bit of work you can extend this method to get control of $u_k^{(q)}$ as well.