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Let $I$ be the open interval $(0, 1)$. Let $k \in \mathbb R \setminus \{1\}$. We consider the space $$ V := \{v \in H^1 (I) : v(0) = kv(1)\}, $$ and the symmetric bilinear form $a$ defined on $V$ by $$ a(u, v) = \int_I [ u'v' + uv ] - \left ( \int_I u \right) \left ( \int_I v \right). $$

I am trying to solve a problem in Brezis' Functional Analysis

Exercise 8.30

  1. Check that $V$ is a closed subspace of $H^1 (I)$.

In what follows, $V$ is equipped with the Hilbert structure induced by the $H^1$ inner product.

  1. Prove that $a$ is a continuous and coercive bilinear form on $V$.
  2. Deduce that for every $f \in L^2 (I)$ there exists a unique solution of the problem $$ (1) \quad u \in V \quad \text{and} \quad a(u, v)=\int_I f v \quad \forall v \in V. $$
  3. Show that the solution $u$ of $(1)$ belongs to $H^2 (I)$ and satisfies $$ (2) \quad \begin{cases} -u'' + u-\int_I u = f \quad \text {on} \quad I, \\ u(0)=k u(1) \text { and } u'(1)=k u'(0). \end{cases} $$
  4. Conversely, prove that any function $u \in H^2(I)$ satisfying $(2)$ is a solution of $(1)$.
  5. Let $k_n \in \mathbb{R} \setminus \{1\}$ for all $n$. Assume $k_n \xrightarrow{n \to \infty} k \neq 1$. Set $$ V_n = \{v \in H^1(I) : v(0)=k_n v(1)\} . $$ Given $f \in L^2 (I)$, let $u_n$ be the solution of $$ (1_n) \quad u_n \in V_n \quad \text { and } \quad a(u_n, v) = \int_I f v \quad \forall v \in V_n . $$ Prove that $u_n \xrightarrow{n \to \infty} u$ in $H^1 (I)$, where $u$ is the solution of $(1)$. Deduce that $u_n \xrightarrow{n \to \infty} u$ in $H^2 (I)$.

There are possibly subtle mistakes that I could not recognize in my below attempt of (6.). Could you please have a check on it?

From (4.) and (5.), we get that $(1)$ is equivalent to $(2)$. I showed that $$ a(v, v) \ge \varepsilon_n \| v \|_{H^1}, \quad \forall v \in V_n, $$ where $\varepsilon_n := \frac{(k_n-1)^2}{2(k_n^2+1)}$. Because $k_n \xrightarrow{n \to \infty} k \neq 1$, WLOG we can assume that there is $\varepsilon >0$ such that $$ a(v, v) \ge \varepsilon \| v \|_{H^1}, \quad \forall v \in \bigcup_n V_n. $$

It follows that $$ \|f\|_{L^2} \|u_n\|_{H^1} \ge \|f\|_{L^2} \|u_n\|_{L^2} \ge \int_I f u_n =a(u_n, u_n) \ge \varepsilon \| u_n \|_{H^1}, $$ which implies $\frac{1}{\varepsilon} \|f\|_{L^2} \ge \| u_n \|_{H^1}$. Then $(u_n)$ is bounded in $H^1 (I)$. It follows from the first equation of $(2)$ that $(u_n'')$ is bounded in $L^2 (I)$. Then $(u_n)$ is bounded in $H^2 (I)$. Notice that $H^2 (I)$ is uniformly convex and thus reflexive. We denoted by $\rightharpoonup$ the weak convergence in the weak topology of $H^2 (I)$.

Let $(n_k)$ be an arbitrary sub-sequence of $\mathbb N$. Then $(u_{n_k})_k$ is bounded in $H^2 (I)$. There is $v \in H^2 (I)$ and a sub-sequence (also denoted by $(u_{n_k})_k$ for simplicity) such that $u_{n_k} \rightharpoonup v$. Then $\| u_{n_k} - v \|_{L^2} \to 0$ and $\| u_{n_k}' - v' \|_{L^2} \to 0$. By Helly’s selection theorem, we can extract a further sub-sequence (also denoted by $(u_{n_k})_k$ for simplicity) such that $u_{n_k} \to v$ everywhere and $u_{n_k}' \to v'$ everywhere.

We have $u_{n_k} \rightharpoonup v$ together with $-u_{n_k}'' + u_{n_k} - \int_I u_{n_k} = f$ implies $-v'' + v - \int_I v = f$. We have $u_{n_k} (0) = k_{n_k} u_{n_k} (1)$ and $u_{n_k}'(1) = k_{n_k} u_{n_k}'(0)$ together with $u_{n_k} \to v$ everywhere and $u_{n_k}' \to v'$ everywhere imply $v (0) = k v (1)$ and $v'(1)=k v'(0)$. To sum up, $v$ is a solution of $$ \begin{cases} -v'' + v-\int_I v = f \quad \text {on} \quad I, \\ v(0)=k v(1) \text { and } v'(1)=k v'(0). \end{cases} $$

By uniqueness of $(1)$, we get $u=v$. This completes the proof.

Akira
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1 Answers1

1

You can use this to show $u_n\to u$. You already showed that $\|u_n\|_{H^1},\|u\|_{H^1}\le \frac1\varepsilon\|f\|_{L^2}$. Now $$ |u_n(1)|\le |u_n(1)-u_n(x)|+|u_n(x)|,|u_n'(0)|\le |u_n'(0)-u'_n(x)|+|u'_n(x)| $$ which implies $$\begin{eqnarray} |u_n(1)|&=&\int_I|u_n(1)|\le \int_I|u_n(1)-u_n(x)|+\int_I|u_n(x)|\\ &\le&\int_I\bigg|\int_x^1u_n'(t)dt\bigg|+\|u_n\|_{L^2}\\ &\le&\|u'_n\|_{L^2}+\|u_n\|_{L^2}\le \frac1\varepsilon\|f\|_{L^2},\\ |u_n'(0)|&=&\int_I|u_n'(0)|\le \int_I|u_n'(0)-u_n'(x)|+\int_I|u_n'(x)|\\ &\le&\int_I\bigg|\int_0^xu_n''(t)dt\bigg|+\|u_n'\|_{L^2}\\ &\le&2\|u_n\|_{L^2}+\|f\|_{L^2}\le (\frac2\varepsilon+1)\|f\|_{L^2},\\ \end{eqnarray}$$ Let $v_n=u_n-u$ and then $v_n$ satisfies $$ -v_n''+v_n-\int_I v_n=0 \tag{*}$$ and the boundary conditions $$ v_n(0)=u_n(0)-u(0)=k_nu_n(1)-ku(1)=(k_n-k)u_n(1)+kv_n(1)=o(1)+kv_n(1) $$ and $$ v_n'(1)=u_n'(0)-u'(0)=k_nu_n'(0)-ku'(0)=(k_n-k)u_n'(0)+kv_n'(0)=o(1)+kv_n'(0). $$ Multiplying (*) by $v_n$ and using IBP, one has $$ \|v_n'\|^2_{L^2}+\|v_n\|^2_{L^2}=\|v_n\|_{L^1}^2+v_n(1)v_n'(1)-v_n(0)v_n'(0)\le\|v_n\|_{L^2}^2+o(1)$$ and hence $$ \|v_n'\|^2_{L^2}\le o(1). $$ So $$ v_n'\to 0 \text{ in }L^2 $$ which also implies $v_n\to0$ in $L^2$.

xpaul
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