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Let $\Omega$ a bounded domain, connexe and regular, and let $f \in L^2(\Omega).$ Let the variational problem: Find $u \in H^1(\Omega)$ such $$\int_{\Omega} \nabla u \nabla v dx + (\int_{\Omega} u dx)(\int_{\Omega} v dx) = \int_{\Omega} f v dx, \forall v \in H^1(\Omega)$$ 1- Prouve that this variational problem admits a unique solution in $H^1(\Omega).$

2- Deduce the boundary problem associate to this variational problem ( study the two cases $u \in H^2(\Omega)$ and $u \notin H^2(\Omega)$.

I dont't understand in the question 2, why we mus study the two cases $u \in H^2$ and $u \notin H^2$?

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  • $\begingroup$ Sorry about the question, but what is a limit problem? $\endgroup$ – Tomás Jun 3 '13 at 22:35
  • $\begingroup$ sorry, there is a boundary problem. $\endgroup$ – jijiii Jun 3 '13 at 22:36
  • $\begingroup$ Are you sure that there is this integral product? Is not $\int_\Omega uvdx$? $\endgroup$ – Tomás Jun 3 '13 at 22:38
  • $\begingroup$ Yes, i'm sure for this. $\endgroup$ – jijiii Jun 3 '13 at 22:39
  • $\begingroup$ Could you find the boundary value problem associated to it? $\endgroup$ – Tomás Jun 3 '13 at 22:39
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Case 1: $u\in H^2(\Omega)$

In this case, we have that $$\tag{1}\int_\Omega \nabla u\nabla v=-\int_\Omega v\Delta u+\int_{\partial\Omega}\frac{\partial u}{\partial\nu}v,\ \forall\ v\in H^1(\Omega)$$

From $(1)$, we conclude that $$\tag{2}-\int_\Omega v\Delta u+\int_{\partial\Omega}\frac{\partial u}{\partial\nu}v+\int_\Omega (\int_\Omega u)v=\int f v,\ \forall\ v\in H^1(\Omega)$$

If we take $v\in C_c^\infty(\Omega)$ in $(2)$, we can conclude by the Fundamental Lemma of Calculus of Variation that $$\tag{3}-\Delta u(x)+\int_\Omega u=f(x),\ a.e.\ x\in\Omega$$

By using $(3)$, we conclude from $(2)$ that $$\tag{4}\int_{\partial\Omega} \frac{\partial u}{\partial\nu}v=0,\ \forall\ v\in H^1(\Omega)$$

$(4)$ implies that $\frac{\partial u}{\partial\nu}=0$ in $\partial\Omega$, so your boundary balue problem is

$$ \left\{ \begin{array}{rl} -\Delta u+\int_\Omega u=f, &\mbox{ in $\Omega$} \\ \frac{\partial u}{\partial\nu}=0 &\mbox{ in $\partial\Omega$} \end{array} \right. $$

Case 2: $u\notin H^2(\Omega)$

I dont know how to treat the case. It is worth to note that the solution is unique, so I think that it is possible to prove that $u\in H^2(\Omega)$, by using difference quotient methods and then we are on the first case again, but this is just a guess.

Update: Suppose that $u\in H^1(\Omega)$ satisfies $$\tag{5}\int_\Omega \nabla u\nabla v+\left(\int_\Omega u\right)\left(\int_\Omega v\right)=\int_\Omega fv,\ \forall\ v\in H^1(\Omega)$$

Take $v=1$ in $(5)$ and get $$\tag{6}\int_\Omega u=\frac{1}{|\Omega|}\int_\Omega f$$

From $(5)$ and $(6)$ we concude that $$\tag{7}\int_\Omega\nabla u\nabla v=\int_\Omega\left(\frac{1}{|\Omega|}\int_\Omega f-f\right)v,\ \forall\ v\in H^1(\Omega)$$

As you can verify in Brezis's book chapter 9, equation $(7)$ implies that $u\in H^2(\Omega)$, so the same argument as above can be used.

Remark: After the update, we note that the first part of the proof could be carried without distinguishing two distinct cases.

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  • $\begingroup$ What is Fundamental Lemma of Calculus of Variation? $\endgroup$ – Tony Stark Jun 3 '13 at 23:29
  • $\begingroup$ @TonyStark en.wikipedia.org/wiki/… $\endgroup$ – Tomás Jun 3 '13 at 23:44
  • $\begingroup$ and for the case $u \notin H^2$ any one have an idea? $\endgroup$ – jijiii Jun 4 '13 at 12:15
  • $\begingroup$ non idea for the case 2? $\endgroup$ – jijiii Jun 5 '13 at 9:40
  • $\begingroup$ Did you tried to apply my guess? Take a look in the Brezis book chapter 9. There he does this type of accounte that I have suggested here. $\endgroup$ – Tomás Jun 5 '13 at 10:13

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