Consider the equation (with $l_-=0, l_+=l>0)$ \begin{align} \left\{\begin{array}{rclclcl} -\Delta u (s,t) & = & f(s,t) &&&& \text{for } (s,t)\in \mathbb{R}\times (0,l), \\[6pt] u(s,l_\pm) & = & g_\pm(s) &&&& \text{for } s\in \mathbb{R}, \end{array}\right. \end{align} My aim is to show that $u\in H^{2}(\mathbb{R}\times (0,l))$ if $f\in L^{2}(\mathbb{R}\times (0,l))$ and $g_\pm \in H^{\frac 32}(\mathbb{R})$. I follow the ideas in the book "Elliptic problems in domains with piecewise smooth boundaries" by Nazarov and Plamenevsky (Chapter 2.2.2). By density it is sufficient to assume that both $f$ and $g_\pm$ are in $C_c^\infty$ and looking for solutions that are in $C^2$ and continuous up to the boundary. Applying a Fourier transform in $s-$ direction yields the problem \begin{align}\label{eq:DBVP_FT} \left\{\begin{array}{rclclcl} -(\partial_t^2 - \lambda^2) \widehat{u} (\lambda,t) & = & \widehat{f}(\lambda,t) &&&& \text{for } t\in (0,l), \\[6pt] \widehat{u}(\lambda,l_\pm) & = & \widehat{g_\pm}(\lambda) &&&& \text{for } t\in \mathbb{R}. \end{array}\right. \end{align} Now the claim is that \begin{align} ||\widehat{u}(\lambda,\cdot)||_{L^2(0,l)}\lesssim \lambda^{-4}||\widehat{f}(\lambda,\cdot)||_{L^2(0,l)}+ \sum_{\pm} \lambda^{-1} |\widehat{g_\pm}(\lambda)|^2. \end{align} This can be achieved by writing $\widehat{u}(\lambda,\cdot)$ in terms of a Green's function $\Gamma$: \begin{equation}\label{eq:GF} \widehat{u}(\lambda,t)= \int_0^l \Gamma(\lambda, t,\tau)\widehat{f}(\lambda,\tau)d \tau + \frac{\sinh \lambda t}{\sinh \lambda l} \widehat{g_+}(\lambda) - \frac{\sinh \lambda (t-l)}{\sinh \lambda l} \widehat{g_-}(\lambda) \end{equation} with \begin{align} \Gamma(t,\tau) = \frac{1}{2\lambda\sinh\lambda l}\left(\cosh \lambda (t + \tau -l) - \cosh \lambda (l-|t-\tau|)\right). \end{align}
Question 1: The estimate for the boundary values works well but for the estimate involving $\Gamma$ I'm failing. Using Cauchy-Schwarz it suffices to bound $||\Gamma||_{L^2((0,l)^2)}^2$, but when I calculate the integral explicitly (using Mathematica), I can only bound it by $\lambda^{-3}||\widehat{f}(\lambda,\cdot)||_{L^2(0,l)}$. Do you have any ideas how to resolve this problem?
Question 2: I'm not sure why this is sufficient to finally bound the $H^{s+2}$-norm of $u$ as by Plancherel one would need to control terms like $(1+|\lambda|^2+|\lambda|^4)||\widehat{u}(\lambda,\cdot)||_{L^2(0,l)}$.
Question 3: Moreover, I'm interested in a way how to estimate $||\partial_t\widehat{u}(\lambda,\cdot)||_{L^2(0,l)}$ as partial integration would yield again boundary terms (involving the value of $\partial_t\widehat{u}(\lambda,l_\pm)$) which is unkown.
Question 4: According to this answer it is quite reasonable to expect the $\lambda^{-4}$ estimate. As the Green function I gave you above should be correct, is there any other possibility to derive the desired estimate using other inequalities than Cauchy-Schwarz?
Thank you very much in adavance!
Note: I'm aware that there were posted several questions like this here but unfortunately, I haven't found an answer where the inhomogeneous boundary conditions are treated as then arguments involving the standard Poincaré inequality are not available...
I am suggesting a different approach, so this might not be the answer you want.
Let $U=\mathbb{R}\times(0,l)$. By standard trace theory you can find a function $G\in H^{2}(U)$ such that $G(s,l_{\pm})=g_{\pm}(s)$ and $\Vert G\Vert_{H^{2}(U)}\leq C(\Vert g_{-}\Vert_{H^{3/2}(\mathbb{R})}+\Vert g_{+}\Vert_{H^{3/2}(\mathbb{R})})$.
If you define $v=u-G$, you find that $-\Delta v=f+G=:F$ and $v(s,l_{\pm})=0$. So now you are in $H_{0}^{1}(U)$ and Poincare's inequality $\Vert w\Vert_{L^{2}(U)}\leq C\Vert\nabla w\Vert_{L^{2}(U)}$ holds for all $w\in H_{0}^{1}(U)$. Hence, you can take $\Vert\nabla w\Vert_{L^{2}(U)}$ as an equivalent norm in $H_{0}^{1}(U)$. In turn, you can apply Lax Milgram theorem to find a weak solution $v\in H_{0}^{1}(U)$ of $-\Delta v=F$ with $\Vert\nabla v\Vert_{L^{2}(U)}\leq C\Vert F\Vert_{L^{2}(U)}$.
Now, there is a trick to reduce the problem to the entire space. Le $\varphi\in C^{\infty}([0,l])$ be such that $\varphi=1$ in $[0,l/3]$ and $\varphi=0$ in $[2/3l,l]$. Then $v=\varphi v+(1-\varphi)v=:v_{1}+v_{2}$.
Since $v_{1}=0$ in $\mathbb{R}\times\lbrack2/3l,l]$, you can extend it to be zero for $t\geq l$. Thus $v_{1}$ satifies $-\Delta v_{1}=F_{1}$ in $\mathbb{R}\times(0,\infty)$, where $F_{1}=-2\nabla\varphi\cdot\nabla v+\varphi F-v\Delta\varphi\in L^{2}(\mathbb{R}\times(0,\infty))$. By reflecting $v_{1}$ that is $v_{1}(s,t):=-v_{1}(s,-t)$ for $t<0$, you get a solution $v_{1}\in H^{1}(\mathbb{R}^{2})$ of $-\Delta v_{1}=F_{1}$ in $\mathbb{R}^{2}$. Hence, you can use Fourier transforms to show that $\Vert\nabla^{2}v_{1}\Vert_{L^{2}(\mathbb{R}^{2})}\leq C\Vert F_{1}% \Vert_{L^{2}(\mathbb{R}^{2})}$.
For $v_{2}$ you get something similar. Since $v_{2}=0$ in $\mathbb{R}% \times\lbrack0,l/3]$, you can extend it to be zero for $t<0$. Thus $v_{2}$ satifies $-\Delta v_{2}=F_{2}$ in $\mathbb{R}\times(-\infty,l)$ and you can proceed in a similar way.
Finally, observe that \begin{align*} \Vert\nabla^{2}v\Vert_{L^{2}(\mathbb{R}^{2})} & =\Vert\nabla^{2}v_{1}% +\nabla^{2}v_{2}\Vert_{L^{2}(\mathbb{R}^{2})}\leq C\Vert F_{1}\Vert _{L^{2}(\mathbb{R}^{2})}+C\Vert F_{2}\Vert_{L^{2}(\mathbb{R}^{2})}\\ & \leq C\Vert F\Vert_{L^{2}(\mathbb{R}^{2})}+C\Vert v\Vert_{H_{0}% ^{1}(\mathbb{R}^{2})}\leq C\Vert F\Vert_{L^{2}(\mathbb{R}^{2})}\\ & \leq C(\Vert f\Vert_{L^{2}(U)}+\Vert g_{-}\Vert_{H^{3/2}(\mathbb{R})}+\Vert g_{+}\Vert_{H^{3/2}(\mathbb{R})}) \end{align*}