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I am trying to prove the existence of a weak solution of the problem: $$ -\Delta^2 u = f \in L^2(U)\\ \\ u|_{\partial U}=\Delta u|_{\partial U} = 0 $$ on the bounded open set $U\subset\mathbb{R}^n$ which has smooth boundary. The weak formulation follows from multiplying by the test function $v$ and integration by parts:

$$ \int_U fv\,dx= \int_U \, (\Delta^2 u) v\, dx= \int_U \Delta u \Delta v \, dx + \int_{\partial U} (v \frac{\partial \Delta u}{\partial n} - \frac{\partial v}{\partial n}\Delta u)dS. $$ The second boundary vanishes for $\Delta u|_{\partial U} = 0$ and to make the first boundary term vanish we require $v$ to be in the Sobolev space $H^1_0(U)\cap H^2(U)$, so that $v=0$ on the boundary.

Therefore $u\in H^1_0(U)\cap H^2(U)$ is a weak solution if: $$ \int_U \Delta u \Delta v \, dx = \int_U fv\,dx\,\,\,\, \forall v\in H^1_0(U)\cap H^2(U). $$

Now I want to use the Lax-Milgram theorem on the bilinear form $B[u,v]=\int_U \Delta u \Delta v \, dx$. My problem is: what norm on $H^1_0(U)\cap H^2(U)$ should I use? At first I thought I could use either one of the norms of $H^1_0(U)$ or $H^2(U)$, since clearly $H^1_0(U)\cap H^2(U)$ is a closed subspace of both spaces. However I realized that this argument must be wrong, otherwise I might as well use the norm of the subspace $H^{8}(U)$ and conclude there exists a weak solution in that space.

Or should I use Lax Milgram for both $H^1_0(U)$ and $H^2(U)$ and conclude that the weak solution is in their intersection?

My second problem: when proving the coercivity of $B[u,v]$, I think I need an inequality like $\int_U |\Delta u|^2 dx \geq C\int_U u^2 dx$ for some constant $C$. I know this holds for $u\in H^2_0(U)$ but I don't see why this should hold in $H^1_0(U)\cap H^2(U)$?

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    $\begingroup$ These are called Navier boundary conditions, and the usual way of dealing with them is to treat them as a system of equations. $\endgroup$ – Ray Yang Apr 7 '13 at 19:54
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    $\begingroup$ You mean using Lax-Milgram on $-\Delta u = w$ and $-\Delta w = f$, with $u =0$ and $w=0$ on the boundary? I then get two weak solutions $u$ and $w=-\Delta u$ in $H^1_0(U)$. Wouldn't that mean that $u$ is in $H^1_0(U)\cap H^3(U)$ rather than $H^1_0(U)\cap H^2(U)$? $\endgroup$ – ScroogeMcDuck Apr 7 '13 at 21:32
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    $\begingroup$ Um. Yes, but $H^3(U) \subset H^2(U)$, so isn't that okay? $\endgroup$ – Ray Yang Apr 8 '13 at 6:50
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    $\begingroup$ It is, but I'm not sure the weak solution is supposed to be in $H^3(U)$. Also, $w = -\Delta u$ implies that $w$ is a strong solution which I didn't prove, so I can really only conclude that $u \in H^1_0(U)$ if I use the method of the system of equations. The problem is the $u\in H^2 (U)$ part. $\endgroup$ – ScroogeMcDuck Apr 8 '13 at 14:03
  • $\begingroup$ Just a quick comment (I might expand this into an answer if I find time): I am not sure what you mean by "using a norm". Where exactly do you need a norm in Lax-Milgram? As for the second problem, I am pretty sure your bilinear form is coercive in the sense that you have $B(u,u)+\gamma\|u\|_{L^2}^2>c\|u\|_{H^2}^2$ in $H^1_0\cap H^2$ for some (large) constant $\gamma$. It might even be true with $\gamma=0$. Spend some energy on this please. $\endgroup$ – timur Apr 8 '13 at 15:44
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Let $H=H_0^1\cap H^2$. A good norm to work with is $$\|u\|_H=\|\Delta u\|_2$$

As you can see here, this norm is equivalently to the usual one. Also, the inequality you are looking for is true, in fact, you have that $$\int |\Delta u|^2\geq\lambda_1^2\int|u|^2,\ \forall\ u\in H$$

where $\lambda_1$ is the firt eigenvalue associated with $(-\Delta, H_0^1)$. To prove this inequality, note that

\begin{eqnarray} \int |\nabla u|^2 &=& -\int u\Delta u \nonumber \\ &\leq& \|u\|_2\|\Delta u\|_2 \nonumber \end{eqnarray}

From the last inequality, you can conclude by using Poincare inequality. Now you can easily apply Lax-Milgram.

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  • $\begingroup$ Just a observation. You have to take off the sign of minus from your equation or change the weak formulation, because they are different. $\endgroup$ – Tomás Apr 9 '13 at 12:08
  • $\begingroup$ Is it true that the second displayed formula is not exactly coercivity. One would want an $H^2$-norm in the right hand side. $\endgroup$ – timur Apr 10 '13 at 6:41
  • $\begingroup$ @timur, I do not understood your comment. What formula is not coercive? $\endgroup$ – Tomás Apr 10 '13 at 11:11
  • $\begingroup$ The second formula that is displayed. The left hand side is the $L^1$-norm of $\Delta u$, and the right hand side is the $L^1$-norm of $u$, multiplied by $\lambda_1^2$. My understanding is that in coercivity you need $H^2$-norm of $u$ in the right hand side. $\endgroup$ – timur Apr 10 '13 at 16:03
  • $\begingroup$ You are right, there is an error there, as you can see in the proof a few lines down. I am gonna fixt it, thank you @timur. $\endgroup$ – Tomás Apr 10 '13 at 17:40

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