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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 $u_n \to u$ in $H^1 (I)$, where $u$ is the solution of $(1)$. Deduce that $u_n \to u$ in $H^2 (I)$.
  6. What happens to the sequence $\left(u_n\right)$ if $k_n$ converges to $1$?

I am trying to solve (7.). I have proved that $$ a(v, v) \ge \varepsilon_n \| v \|_{H^1}, \quad \forall v \in V_n, $$ where $\varepsilon_n := \frac{(k_n-1)^2}{2(k_n^2+1)}$. This implies $\| u_n \|_{H^1} \le \frac{1}{\varepsilon_n} \|f\|_{L^2}$. We have $\varepsilon_n \to 0$, so I guess $(u_n)$ diverges in $H^1 (I)$. However, I could not prove it.

Could you elaborate on (7.)?

Thank you so much for your help!

Akira
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1 Answers1

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Let $u_n, u$ be the solutions of $$ \begin{cases} -u'' + u-\int_I u = f \quad \text {on} \quad I, \\ u(0)=k u(1) \text { and } u'(1)=k u'(0). \end{cases} $$ for $k=k_n,1$, respectively; namely $$ \begin{cases} -u_n'' + u_n-\int_I u_n = f \quad \text {on} \quad I, \\ u_n(0)=k_n u_n(1) \text { and } u_n'(1)=k_n u_n'(0), \end{cases} $$ and $$ \begin{cases} -u'' + u-\int_I u = f \quad \text {on} \quad I, \\ u(0)=u(1) \text { and } u'(1)=u'(0). \end{cases} $$ Let $v_n=u_n-u$ and then $v_n$ satisfies $$ -v_n'' + v_n-\int_I v_n =0 \quad \text {on} \quad I. $$ which has the general solution $$v_n=\beta_n+c_1^{(n)}e^t+c_2^{(n)}e^{-t}\tag{*}$$ where $\beta_n,c_1^{(n)},c_2^{(n)}$ are constants depending on $n$. Note $$ \begin{eqnarray} v_n(0)&=&u_n(0)-u(0)=k_n u_n(1)-u(1)\\ &=&k_nv_n(1)+(k_n-1)u(1)=k_nv_n(1)+o(1),\\ v_n'(1)&=&u_n'(1)-u'(1)=k_nu_n'(0)-u'(0)\\ &=&k_nv_n'(0)+(k_n-1)u'(0)=k_nv_n'(0)+o(1),\\ \end{eqnarray} $$ Now using (*) one has \begin{eqnarray} \beta_n+c_1^{(n)}+c_2^{(n)}&=&k_n(\beta_n+c_1^{(n)}e+c_2^{(n)}e^{-1})+o(1),\\ c_1^{(n)}e-c_2^{(n)}e^{-1}&=&k_n(c_1^{(n)}-c_2^{(n)})+o(1),\\ \end{eqnarray} from which one can get \begin{eqnarray} c_1^{(n)}&=&-\frac{\beta_n(k_n-1) (e k_n-1)}{k_n^2+e^2 \left(k_n^2+1\right)-4 e k_n+1}+o(1),\\ c_2^{(n)}&=& \frac{e\beta_n(e-k_n) (k_n-1)}{k_n^2+e^2 \left(k_n^2+1\right)-4 e k_n+1}+o(1). \end{eqnarray} Clearly if $\{\beta_n\}$ is bounded and $\lim_{n\to\infty}k_n=1$, then $$\lim_{n\to\infty}c_1^{(n)}=\lim_{n\to\infty}c_2^{(n)}=0. $$ From this one can conclude that if $\beta_n\to\beta$, then $v_n\to\beta$ as $n\to\infty$.

xpaul
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