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.



Variational formulation for bilaplacian
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 $u\in H^1\_0$.


