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Let $I$ be the open interval $(0, 1)$. I'm trying to solve a problem in Brezis' Functional Analysis, i.e.,

Exercise 8.18 Check that the mapping $v \mapsto v(0)$ from $H^1(I)$ into $\mathbb{R}$ is a continuous linear functional on $H^1(I)$. Deduce that there exists a unique $u \in H^1(I)$ such that $$ v(0)=\int_I (u' v'+u v) \quad \forall v \in H^1(I) . $$

Show that $u$ is the solution of some differential equation with appropriate boundary conditions. Compute $u$ explicitly.

Below, I prove that

The function $u$ is the unique classical solution of the ODE $$ (1) \quad \left\{\begin{array}{l} -u''+u=f \quad \text { on } I, \\ u'(0)=\alpha, u'(1)=\beta, \end{array}\right. $$

where $f \equiv 0, \alpha=-1, \beta=0$.

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


If $u$ is a classical solution of (1), we have $$ -\int_I u'' v + \int_I u v=\int_I f v, \quad v \in H^1(I). $$

By integration by parts (I.b.P), $$ \begin{align*} \int_I u^{\prime \prime} v &= (u'v)(1)- (u'v)(0) - \int_I u^{\prime} v^{\prime} \\ &= \beta v(1)- \alpha v (0) - \int_I u^{\prime} v^{\prime}, \quad v \in H^1(I), \end{align*} $$

which implies $$ (2) \quad \int_I u^{\prime} v^{\prime}+\int_I u v=\int_I f v + \beta v(1)-\alpha v(0), \quad v \in H^1(I) . $$

By Lax-Milgram theorem, there is a unique $u \in H^1 (I)$ that satisfies $(2)$. In particular, $$ (3) \quad \int_I u^{\prime} v^{\prime}=-\int_I (u-f) v, \quad v \in H^1_0(I), $$

which implies $u \in H^2 (I)$ with $u'' =u-f$. By I.b.P, $$ (4) \quad \int_I u^{\prime \prime} v = (u'v) (1) - (u'v) (0) - \int_I u^{\prime} v^{\prime}, \quad v \in H^1 (I) . $$

We have $(2, 4)$ and the fact that $-u^{\prime \prime} + u=f$ a.e. on $I$ imply $$ (\beta - u' (1)) v (1) +(u'(0)- \alpha) v (0) = 0 \quad v \in H^1 (I), $$ and thus $\beta = u'(1)$ and $\alpha = u' (0)$. Then it follows from (2) that $$ \int_I u^{\prime} v^{\prime}+\int_I u v= v(0), \quad v \in H^1(I) . $$

Notice that $f$ being continuous implies $u''$ being continuous. This completes the proof.

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Clearly $v: H^1(I)\to\mathbb{R}$ by $v\to v(0)$ is linear. First for $\forall v \in H^1(I)$, \begin{eqnarray*} v(0)&=&\int_I (u' v'+u v)\mathrm{d}x\\ &=&\int_I (u' v'+u v)\mathrm{d}x\\ &=&\int_I u'\mathrm{d}v+\int_I uv\mathrm{d}x\\ &=&u'(1)v(1)-u'(0)v(0)+\int_I(-u''+u)v\mathrm{d}x \end{eqnarray*} which implies $$ -u''+u=0, \text{ in }I, \ u'(0)=-1, u'(1)=0. \tag{*}$$ The equation has the solution $$ u=c_1e^t+c_2e^{-t} $$ with $c_1=\frac1{e^2-1},c_2=\frac{e^2}{e^2-1}$.

Conversely for $\forall v \in H^1(I)$, one has, from (*) and by IBP $$ \int_I(-u''+u)v\mathrm{d}x=\int_I(u'v'+uv)\mathrm{d}x-(u'(1)v'(1)-u'(0)v'(0))=\int_I(u'v'+uv)\mathrm{d}x-v(0). $$ Define $$ a(u,v)=\int_I(u'v'+uv)\mathrm{d}x, u, v \in H^1(I)$$ which is continuous and coercive. Since $v: H^1(I)\to\mathbb{R}$ by $v\to v(0)$ is linear, by Lax-Milgram theorem, there is $u\in H^1(I)$ such that $a(u,v)=v(0)$ or $$ \int_I (u' v'+u v)\mathrm{d}x=v(0), \forall v \in H^1(I). $$

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