Show that there exists a unique $v_0 \in H^1(0,1)$ such that $u(0)=\int_0^1(u'v_0'+uv_0), \forall u \in H^1(0,1)$

Question

Show that there exists a unique $v_0 \in H^1(0,1)$ such that $u(0)=\int_0^1(u'v_0'+uv_0), \forall u \in H^1(0,1)$. Further Show that $v_0$ is the solution of some differential equation with appropriate boundary conditions. Compute $v_0$ explicitly.

Let $f: H^1(0,1) \to \mathbb{R}$ be defined by $f(u)=u(0)$. Then I showed that $f$ is linear and continuous. Hence there exists $v_0 \in H^1(0,1)$ such that $u(0)=\int_{0}^1u'v_0'+\int_{0}^1uv_0$

The differential equation is given by $f=-u''+u$ with the conditions $u'(0)=u'(1)=0$.

How do I compute $v_0$ explicitly??

Answer 1

You know that if $$ u(0) = \int_{0}^{1}u'v_0'+uv_0 dx,\;\;\; u\in H^1(0,1), $$ then $$ 0=\int_{0}^{1}\varphi' v_0'+\varphi v_0 dx, \;\; \varphi\in\mathcal{C}_{c}^{\infty}(0,1), $$ which implies that $v_0'$ has a weak derivative and $v_0''=v_0 \in L^2$. That's enough to imply to $v_0$ is twice absolutely continuous, and \begin{align} u(0)&=\int_{0}^{1}u'v_0'+uv_0'' dx \\ &= uv_0'|_{0}^{1}-\int_{0}^{1}(u'v_0'-u'v_0')dx\\ &= u(1)v_0'(1)-u(0)v_0'(0). \end{align} So, what you need is $v_0''=v_0$ and $v_0'(0)=-1$, $v_0'(1)=0$. So the solution must be $$ v_0(x)=\cosh(1-x)/\sinh(1) $$

Answer 2

Your differential equation is wrong. Using integration by parts, we get $$u(0) = \int_0^1 -u \, v_0'' + u \, v_0 \, \mathrm{d}x + u(0) \, v_0'(0) - u(1) \, v_0'(1).$$ This equation has to be satisfied for all $u \in H^1$. Hence...