Variational formulation for bilaplacian - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T04:31:14Z http://mathoverflow.net/feeds/question/96854 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/96854/variational-formulation-for-bilaplacian Variational formulation for bilaplacian Marc 2012-05-13T21:31:15Z 2012-05-14T05:45:56Z <p>I am trying to derive a variational formulation for the following problem <code>$$\left\{ \begin{array}{ll} \Delta^2u=f, &amp; \Omega \\ \Delta u+\rho \partial_{\nu}u=0, &amp; \partial \Omega \end{array}\right.$$</code></p> <p>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.</p> <p>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.</p> <p>Multiplying the equation by $v\in C_0^2(\Omega)$ and integrating over $\Omega$ we obtain $$\int_{\Omega} \Delta^2u\cdot v\,dx=\int_{\Omega} fv\,dx$$</p> <p>and now integrating by parts (Green's identity) twice on the left hand side we obtain </p> <p><code>\begin{eqnarray*} \int_{\Omega} \Delta^2u\cdot v\,dx &amp;=&amp; \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 \\ &amp;\stackrel{Green}{=}&amp; \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 \\ &amp;=&amp; \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*}</code></p> <p>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 $$\int_{\partial \Omega} \rho \partial_{\nu}u \cdot \partial_{\nu}v\,d\sigma + \int_{\Omega} \Delta u\cdot \Delta v\,dx=\int_{\Omega} fv\,dx$$</p> <p>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?</p> <p>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 $$|B(u,v)|\leq \int_{\partial \Omega} \rho |\partial_{\nu}u \cdot \partial_{\nu}v|\,d\sigma + \int_{\Omega} |\Delta u\cdot \Delta v|\,dx$$</p> <p>For the second integral we have <code>$$\int_{\Omega} |\Delta u\cdot \Delta v|\,dx\leq ||\Delta u||_0||\Delta v||_0\leq ||u||_{H^2(\Omega)} ||v||_{H^2(\Omega)}$$</code> but what about the first one?</p> <p>Thanks in advance for any insight.</p> http://mathoverflow.net/questions/96854/variational-formulation-for-bilaplacian/96870#96870 Answer by Denis Serre for Variational formulation for bilaplacian Denis Serre 2012-05-14T05:45:56Z 2012-05-14T05:45:56Z <p>To begn with, your Boundary-Value Problem (BVP) is under-determined, because it lacks <em>one</em> boundary condition: because the PDE is elliptic and <em>fourth</em>-order, you need <strong>two</strong> 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 $$|\partial_\nu w|<em>{L^2(\partial\Omega)}\le C|w|</em>{H^2}\le C'|\Delta u|_{L^2},$$ 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.</p> <p>The existence follows from Lax--Milgram, using the Poincar\'e inequality to show that $B$ is coercive over $H^2\cap H^1_0$.</p> <p>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 $u\in H^1_0$.</p>