# Brezis' exercise 8.18: show that $u$ is the solution of some ODE with appropriate boundary conditions

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.

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 $$\int_I (u' v'+u v)\mathrm{d}x=v(0), \forall v \in H^1(I).$$