I am trying to derive a variational formulation for the following problem $$\left\{ \begin{array}{ll} \Delta^2u=f, & \Omega \\ \Delta u+\rho \partial_{\nu}u=0, & \partial \Omega \end{array}\right.$$
where $\rho>0$ is constant. I intend to show that the right functional setting is $H^2(\Omega)\cap H^1_0(\Omega)$ and to prove that the resulting problem is well posed.
I am confused as to how to establish the right functional setting, so for a start I choose $C_0^2(\Omega)$ as a space of test functions (functions in $C^2(\Omega)$ compactly supported in $\Omega$) so that the boundary condition makes sense.
Multiplying the equation by $v\in C_0^2(\Omega)$ and integrating over $\Omega$ we obtain $$\int_{\Omega} \Delta^2u\cdot v\,dx=\int_{\Omega} fv\,dx$$
and now integrating by parts (Green's identity) twice on the left hand side we obtain
\begin{eqnarray*}
\int_{\Omega} \Delta^2u\cdot v\,dx &=& \int_{\Omega} div (\nabla \Delta u)\,dx \stackrel{Green}{=} \int_{\partial \Omega} \partial_{\nu}(\Delta u)v\,d\sigma- \int_{\Omega} \nabla \Delta u\cdot \nabla v \,dx
\\
&\stackrel{Green}{=}& \int_{\partial \Omega} \partial_{\nu}(\Delta u)v\,d\sigma – \int_{\partial \Omega} \Delta u\cdot \partial_{\nu}v\,d\sigma + \int_{\Omega} \Delta u\cdot \Delta v\,dx
\\
&=& \int_{\partial \Omega} \partial_{\nu}(\Delta u)v\,d\sigma + \int_{\partial \Omega} \rho \partial_{\nu}u \cdot \partial_{\nu}v\,d\sigma + \int_{\Omega} \Delta u\cdot \Delta v\,dx
\end{eqnarray*}
where in the last equality I use the boundary condition. Now, enlarging the space of test functions by taking the closure of $C_0^2(\Omega)$ in $H^2(\Omega)$, namely $H_0^2(\Omega)\subset H_0^1(\Omega)\cap H^2(\Omega)$ the first integral vanishes ($v$ has zero trace) so our variational formulation is $$\int_{\partial \Omega} \rho \partial_{\nu}u \cdot \partial_{\nu}v\,d\sigma + \int_{\Omega} \Delta u\cdot \Delta v\,dx=\int_{\Omega} fv\,dx$$
How can I rigorously conclude that I need to take the whole $H_0^1(\Omega)\cap H^2(\Omega)$ as my space of test functions?
In order to prove that the problem is well posed I intend to use Lax-Milgram theorem as usual, but I am confused as to how to tackle the integral over $\partial \Omega$. I define a bilinear form $$B(u,v)=\int_{\partial \Omega} \rho \partial_{\nu}u \cdot \partial_{\nu}v\,d\sigma + \int_{\Omega} \Delta u\cdot \Delta v\,dx$$ and I want to check continuity and coercitivity. For the first one I have $$|B(u,v)|\leq \int_{\partial \Omega} \rho |\partial_{\nu}u \cdot \partial_{\nu}v|\,d\sigma + \int_{\Omega} |\Delta u\cdot \Delta v|\,dx$$
For the second integral we have $$\int_{\Omega} |\Delta u\cdot \Delta v|\,dx\leq ||\Delta u||_0||\Delta v||_0\leq ||u||_{H^2(\Omega)} ||v||_{H^2(\Omega)}$$ but what about the first one?
Thanks in advance for any insight.
Best Answer
To begn with, your Boundary-Value Problem (BVP) is under-determined, because it lacks one boundary condition: because the PDE is elliptic and fourth-order, you need two boundary conditions, not only one. Because you insist on working with $H^2\cap H^1_0$, it seems that the hidden BC is $$u=0\qquad\hbox{on }\partial\Omega.$$ So let us assume that your BVP includes this condition. Then your calculation works whenever $v\in C^2(\overline\Omega)$ (you should avoid $C^2_0$, because its elements satisfy $\partial_\nu v=0$ as well).If $v$ vanishes on the boundary (but not necessarily its normal derivative), a solution $u$ of the BVP does satisfy $$B(u,v)=0.$$ If $\Omega$ is bounded with a smooth boundary, then $$|\partial_\nu w|_{L^2(\partial\Omega)}\le C|w|_{H^2}\le C'|\Delta u|_{L^2},$$ where the second inequality is a kind of Poincar\'e inequality. Therefore $B$ is continuous over $H^2(\Omega)$ and $B(u,v)=0$ extends to all elements $v\in H^2\cap H^1_0$ by density.
The existence follows from Lax--Milgram, using the Poincar\'e inequality to show that $B$ is coercive over $H^2\cap H^1_0$.
Conversely, suppose that $u\in H^2\cap H^1_0$ and $B(u,v)=0$ for every $v\in H^2\cap H^1_0$. Then making the calculations backward, you find $$\langle \Delta^2 u,v\rangle_\Omega+\int_{\partial\Omega}(\Delta u+\rho\partial_\nu u)\partial_\nu vds(x)=0,$$ where the first term is duality between $H^{-2}(\Omega)$ and its dual. Taking all $v\in{\mathcal D}(\Omega)$, you first obtain $\Delta^2 u=0$ in the sense of distributions. There remains therefore $$\int_{\partial\Omega}(\Delta u+\rho\partial_\nu u)\partial_\nu vds(x)=0,$$ for every $v\in H^2\cap H^1_0$. Because $v\mapsto\partial_\nu v$ is onto over $H^{1/2}(\partial\Omega)$, this gives you $\Delta u+\rho\partial_\nu u=0$. Finally, $u=0$ on the boundary is ensured because $u\in H^1_0$.