1
$\begingroup$

Let $\Omega \subset \mathbb R^n$ be a smooth, bounded and open. Assume that $u=0$ if $u\in H^1(\Omega )$ and satisfies $$-\Delta u-u=0\quad \text{on }\Omega \quad \text{and}\quad \partial _\nu u=u=0\quad \text{on }\partial \Omega .$$ This property is know as unique continuation principle. Let $f\in L^2(\Omega )$ and assume that $v$ is a solution to $-\Delta v-v=f$ in $\Omega $. Prove $$\|v\|_{H^1(\Omega )}\leq C(\|f\|_{L^2(\Omega )}+\|v\|_{H^{1/2}(\partial \Omega )}+\|\partial _\nu v\|_{H^{-1/2}(\partial \Omega )}).$$

There is just things that I don't understand in the proof.

Proof : Assume by contradiction there exist sequences $v_n\in H^1(\Omega)$ and $f_n\in L^2(\Omega)$ such that $$\|v_n\|_{H^1(\Omega)}=1\quad\text{ and}\quad \|f_n\| _ {L^2(\Omega )}+\|v_n\| _ {H^{1/2}(\Omega )}+\|\partial _\nu v_n\| _ {H^{-1/2}(\partial \Omega )}\to 0.$$ By the Reileigh-Kondrashov compactness theorem there exists a weakly convergent subsequence of $\{v_n\}$ in $H^1(\Omega)$ such that $\{v_n\}$ converges strongly in $L^2(\Omega),$ i.e., $\nabla v_n\rightharpoonup \nabla v$ in $L^2(\Omega)$ and $v_n\to v$ in $L^2(\Omega)$ (after relabeling the subsequence).
Multiplying the equality $-\Delta v_n-v_n=f_n $ by any function $\varphi\in H^1(\Omega)$ and integrating by parts we get

$$\int_\Omega \nabla\varphi\nabla v_ndx=\int_\Omega \varphi v_n+ \int_\Omega \varphi f_n+\int_{\partial\Omega}\partial_\nu v_n\varphi \ \ \ (1).$$ Sending now $n$ to infinity, the second and the third summand will go to zero by our contradiction assumption, thus we get by the weak convergence $v_n\rightharpoonup v,$ $$\int_\Omega\nabla\varphi\nabla v=\int_\Omega\varphi v \ \ \ \ (2).$$ On the on the other hand if we take $\varphi=v_n$ in (1) and let $n$ go to infinity we get $\|\nabla v_n\|^2\leq 2\|v_n\|^2$ for $n$ big enough. Now, (2) implies that $$-\Delta v-v=0$$ in the weak sense. Also, by the definition of $H^{1/2}$ and $H^{-1/2}$ norms and the strong convergence $v_n\to v$ we get $$\|v\| _ {H^{1/2}(\partial \Omega)}=\|\partial_\nu v \| _ {H^{-1/2}(\partial\Omega)}=0,$$

Question 1 : I don't understand how we get $\|v\| _ {H^{1/2}(\partial \Omega)}=\|\partial_\nu v \| _ {H^{-1/2}(\partial\Omega)}=0.$ The thing I know is that $\|v\|_{H^{1/2}(\partial \Omega) }=\int_{\partial \Omega } \hat u(x)(1+|x|)^2dx$ and $\|\partial_\nu v\|_{H^{-1/2}(\partial \Omega )}=\sup_{\|\varphi \|=1}|\left<\partial _\nu v,\varphi\right>|$. So how do I get that those norms are $0$ ?

thus from the given unique continuation principle we get $v=0$ in $\Omega.$ From the strong convergence $v_n\to v$ we get $\|v_n\| _ {L^2(\Omega)}\to 0$, thus the inequality $\|\nabla v_n\|^2\leq 2\|v_n\|^2$ implies $\|v_n\| _ {H^1(\Omega)}\to 0$ which is a contradiction.

Question 2 : Why the inequality $\|\nabla v_n\|^2\leq 2\|v_n\|^2$ implies $\|v_n\| _ {H^1(\Omega)}\to 0$ ?

1 Answers 1

1

For the second question, we have $$\parallel\nabla v_n\parallel^2\leq 2\parallel v_n\parallel^2 $$ and, $$\parallel v_n\parallel_{L^2(\Omega)}\longrightarrow 0$$ then, $$\parallel\nabla v_n\parallel_{L^2(\Omega)}\longrightarrow 0 $$ Thus,$$\parallel v_n\parallel_{H^1(\Omega)}=\left(\parallel \nabla v_n\parallel_{L^2(\Omega)}^2+\parallel v_n\parallel_{L^2(\Omega)}^2\right)^{\frac{1}{2}}\longrightarrow 0 $$

To answer your first question, we already have $$\parallel f_n\parallel_{L^2(\Omega)}+\parallel v_n\parallel_{H^{\frac{1}{2}}(\partial\Omega)}+\parallel\partial_\nu v_n\parallel_{H^{-\frac{1}{2}}(\partial\Omega)}\longrightarrow 0 $$ and knowing that the $\parallel\cdot\parallel$ is always positiv we can deduce that$$\parallel f_n\parallel_{L^2(\Omega)}\longrightarrow 0 $$ $$\parallel v_n\parallel_{H^{\frac{1}{2}}(\partial\Omega)}\longrightarrow 0$$ $$ \parallel\partial_\nu v_n\parallel_{H^{-\frac{1}{2}}(\partial\Omega)}\longrightarrow 0$$ Now we write, $$\parallel v\parallel_{H^{\frac{1}{2}}(\partial\Omega)}=\parallel v-v_n+v_n\parallel_{H^{\frac{1}{2}}(\partial\Omega)} \leq\parallel v-v_n\parallel_{H^{\frac{1}{2}}(\partial\Omega)}+\parallel v_n\parallel_{H^{\frac{1}{2}}(\partial\Omega)}$$ but we have $v_n \rightarrow v$ which means $\parallel v_n-v\parallel_{H^{\frac{1}{2}}(\partial\Omega)}\longrightarrow 0$

Then, we conclude that $$\parallel v\parallel_{H^{\frac{1}{2}}(\partial\Omega)}=0$$ You can follow the same stpes to get to $$\parallel \partial_\nu v\parallel_{H^{-\frac{1}{2}}(\partial\Omega)}=0$$ but do not forget that you can switch the '$\lim\limits_{n\rightarrow\infty}$' and the '$\partial_\nu$'

  • 0
    Something looks strange... when you write $\|v\|_{H^{1/2}}\leq \|v_n-v\|_{H^{1/2}}+\|v_n\|_{H^{1/2}}$, I agree that $\|v_n\|_{H^{1/2}}\to 0$, but to write $\|v-v_n\|_{H^{1/2}}\to 0$ looks strange... how can $v_n$ has 2 limit... after, of course we get $v=0$, but to $\|v_n-v\|_{H^{1/2}}\to 0$ looks wrong a priori since $v\neq 0$ a priori. – 2017-01-24
  • 0
    when we write $v_n\longrightarrow v$ that means the distance between the term of the sequence and it's limits is decreasing as $n$ gets bigger, which is translated in norm form as follows: $\parallel v_n-v\parallel\longrightarrow 0$ This is always true whether the limit of the sequence is 0 or any other constant $l$. – 2017-01-24