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

  • $\begingroup$ Why $\|v_n\|_2\to 0$? $\endgroup$ – Tomás Jul 2 '13 at 18:32
  • $\begingroup$ 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 $a$? $\endgroup$ – jijiii Jul 2 '13 at 18:39
  • $\begingroup$ Your argument is not right. This does not implies that $\|v_n\|_2\to 0$. $\endgroup$ – Tomás Jul 2 '13 at 18:40
  • $\begingroup$ so how we can prouve that $a$ is coercive? $\endgroup$ – jijiii Jul 2 '13 at 18:54
  • 2
    $\begingroup$ @jijiii, please do stop posting questions and editing to questions in the part reserved for answers. $\endgroup$ – 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.

  • 1
    $\begingroup$ 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 $L^2$. $\endgroup$ – Tomás Jul 31 '13 at 23:00
  • $\begingroup$ 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 $L^2(\Omega)?$ $\endgroup$ – jijii Aug 1 '13 at 11:15
  • $\begingroup$ 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 $L^2$. $\endgroup$ – Tomás Aug 1 '13 at 12:14
  • $\begingroup$ 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 $L^2?$ $\endgroup$ – jijii Aug 1 '13 at 13:02
  • $\begingroup$ $v$ is in $H^1$ $\endgroup$ – Tomás Aug 1 '13 at 13:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.