# variational problem-exercice

Let $\Omega$ an open bounded connexe and regular, and let $f \in L^2(\Omega)$ We consider the variational problem: find $u \in H^1(\Omega)$ such: $$\displaystyle\int_{\Omega} A \nabla u \cdot \nabla v dx + \left(\displaystyle\int_{\Omega} u dx\right)\left(\displaystyle\int_{\Omega} v dx\right) = \displaystyle\int_{\Omega} f v dx, \forall v \in H^1(\Omega)$$ with: $$\exists \alpha > 0, A(x) \xi \xi \geq \alpha |\xi|^2, \forall \xi \in \mathbb{R}^n, \exists \beta > 0, |A(x) \xi| \leq \beta |\xi|, \forall \xi \in \mathbb{R}^n$$

-The question is: prove that this variational problem admits a unique solution $u \in H^1(\Omega)$.

My solution is: we put $$a(u,v) = \displaystyle\int_{\Omega} A \nabla u \cdot \nabla v dx + \left(\displaystyle\int_{\Omega} u dx\right)\left(\int_{\Omega} v dx\right)$$ $$L(v) = \displaystyle\int_{\Omega} f v dx$$

We begin to prouve that $a$ is coecive. For this, we have to prouve that $$\exists \lambda > 0, a(v,v) \geq \lambda ||v||^2_{H^1(\Omega)}$$ We have: $a(v,v)= \displaystyle\int_{\Omega} A \nabla v \cdot \nabla v dx + (\displaystyle\int_{\Omega} v dx)\left(\int_{\Omega} v dx\right)$. we prove the coercivity of $a$ by absurd. We suppose that $a$ isn't coercitive, then: $$\forall \lambda > 0, \exists v \in H^1(\Omega): a(v,v) < \lambda ||v||^2_{H^1}$$ then, in particular, for $\lambda = \dfrac{1}{n}$, we have: $$\forall n \in \mathbb{N}, \exists v_n \in H^1: a(v_n,v_n) < \dfrac{1}{n} ||v_n||^2_{H^1}$$ and we can choice $v_n$^such that $||v_n||=1,$ then, we obtain: $$\forall n \in \mathbb{N}, \exists v_n \in H^1(\Omega): ||v_n||_{H^1}=1 , a(v_n,v_n) < \dfrac{1}{n}$$ as $\Omega$ is bounded and régular, we have by Rellich theorem that there exist a subsequence $v_n$ who converge to $v \in L^2(\Omega)$. Then, $(v_n)$ is Cauchy sequence in $L^2(\Omega)$ and as $a(v_n,v_n)=\displaystyle\int_{\Omega} A \nabla v_n \cdot \nabla v_n dx + (\displaystyle\int_{\Omega}v_n dx)^2$ converges to 0, and as $a(v_n,v_n)$ is a sum of positif termes, we deduce that $\displaystyle\int_{\Omega} A |\nabla v_n|^2 dx$ converge to 0 ans $\displaystyle\int_{\Omega} v_n dx$ converge to 0.Then $\nabla v_n$ converge to 0 in $L^2(\Omega)$. So, $v_n$ is a Cauchy sequence in $H^1(\Omega)$ and because of $H^1(\Omega)$ is an Hilbert space, he is a complet space, i.e. All Cauchy sequence converge, then $v_n$ converge in $H^1(\Omega)$ to $v$.

Further, $||\nabla v||_{L^2} = \lim_n ||\nabla v_n||_{L^2} = 0$^and $||v_n||_{L^2}$ converge to 0. and $||v||_{L^2} = \lim_n ||v_n||_{L^2} = 1$

Then, $v=0$ and $||v||_{L^2}=1$ and there is a contradiction. So $a$ is coercitive.

My solution is it correct? please

• Why $\|v_n\|_2\to 0$? – Tomás Jul 2 '13 at 18:32
• because two $a(v_n,v_n)$ is an sum of positifs termes, and $a(v_n,v_n)$ converge to . There exist another method to prouve the coercivity of $a$? – jijiii Jul 2 '13 at 18:39
• Your argument is not right. This does not implies that $\|v_n\|_2\to 0$. – Tomás Jul 2 '13 at 18:40
• so how we can prouve that $a$ is coercive? – jijiii Jul 2 '13 at 18:54
• @jijiii, please do stop posting questions and editing to questions in the part reserved for answers. – DonAntonio Jul 3 '13 at 13:06

I do agree with you, until the part where you prove that $$\tag{1}\|\nabla v\|_2=\lim \|\nabla v_n\|_2=0$$

Then you conclude that $\|v_n\|_2\to 0$ without proof. This part seem to me to be wrong, but we can fix it.

Also, as you have concluded, we have that $$\tag{2}\|v_n\|_{1,2}=\|v_n\|_2+\|\nabla v_n\|_2=1$$

From $(1)$ and $(2)$ we conclude that $\|v_n\|_2\to \|v\|_2=1$. Moreover, from $(1)$ we have that $v$ is a constant function, $\|v\|_2=1$ implies that $v=c$ where $c\neq 0$.

To conclude first your have to prove that $a$ is continuous. After this, you have that $a(v_n,v_n)\to a(v,v)$. But $a(v_n,v_n)\to 0$ and $a(v,v)\neq 0$, because $v$ is constant and not zero.

Okay, so i hope to prouve the coercivity of $a$ in a logical way. We process by absurde, so, we suppose that that $$\forall \nu > 0, \exists v \in H^1(\Omega), a(v,v) \geq \nu ||v||^2_V$$ In partical, for $_nu=\dfrac{1}{n},$ we have $$\forall n \in \mathbb{N}, \exists v_n \in H^1(\Omega):a(v_n,v_n) < \dfrac{1}{n} ||v_n||^2_V$$ from this point I do not know how to think logically and naturally to find the right arguments to prouve the coercivity.

• You to note two things: 1- Because $a$ is bilinear, we can pass $\|v_n\|_V^2$ to the left hand side to conclude that $$a\left(\frac{v_n}{\|v_n\|_V},\frac{v_n}{\|v_n\|_V}\right)<\frac{1}{n}$$ Therefore, we can assume that $v_n$ has norm $1$ for all $n$. 2- The second thing is: because $\|v_n\|_V$ is bounded, we have that $v_n$ has a subsequence (not relabeled) such that $v_n$ converge weak to a element $v\in H^1(\Omega)$, hence, Rellich theorem implies that $v_n$ does converge srtrongly in $L^2$. – Tomás Jul 31 '13 at 23:00
• We have assumed that $||v_n||_V=1$ i.e. $||v_n||_V$ is bounded, so there exist an subsequence $v_n$ such that $v_n$ converge weak to an element $v\in H^1(\Omega).$ The Rellich theorem says that "if $\Omega$ is an open bounded, $\mathcal{C}^1,$ then for all sequence bounded in $H^1,$ we can extract an subsequence who converge in $L^2.$ But here, why the weak convergence of $v_n$ to an element $v\in H^1$ implies the strong convergence of $v_n$ in $L^2(\Omega)?$ – jijii Aug 1 '13 at 11:15
• This is a property of compact operators: if $T$ is a compact operator, then for all $u_n$ such that $u_n\to u$ weakly, we have that $Tu_n\to Tu$ strongly. Rellich theorem implies that $H_0^1$ is compactly embedded in $L^2$, or equivalently the operator $i:H_0^1\to L^2$ defined by $i(u)=u$ is compact, hence, if $v_n\to v$ weakly in $H_0^1$, we conclude that $i(v_n)\to i(v)$ strongly in $L^2$, or $v_n\to v$ strongly in $L^2$. – Tomás Aug 1 '13 at 12:14
• here, $i(v)=v \in L^2(\Omega)$ not in $H^1.$ So, $v_n$ converge weakly to $v$ in $H^1(\Omega)$ where $v\in H^1,$ then by Rellich theorem, $v_n$ converge strongly to $v$ in $L^2(\Omega)$ but where is $v$? in $H^1$ or in $L^2?$ – jijii Aug 1 '13 at 13:02
• $v$ is in $H^1$ – Tomás Aug 1 '13 at 13:07