Find $v \in H^1(0,1)$ which satisfies the following Determine the function $v \in H^1(0,1)$ which satisfies the equation $u(0)=\langle u,v \rangle_{H^1}$ for all $u\in H^1(0,1)$ .
It is clear that on $H^1(0,1)$; $u(0)=\int_{0}^{1}(uv+u'v')$.What can be  the function $v$ satisfying the above equation? Please help?

If we assume that $v$ is twice-differentiable, we can integrate the $u'v'$ term by parts, so the equation becomes
$$ u(0) = u(1)v'(1)-u(0)v'(0) + \int_0^1 (-v''(x)+v(x))u(x) \, dx. $$
Therefore a natural thing to do now is find a function that satisfies $v'(0)=-1$, $v'(1)=0$, and $-v''+v=0$ in between. Some elementary differential equation solving later, we find that the unique choice for $v$ that satisfies these is
$$ v(x) = \frac{e^x+e^{2-x}}{e^2-1}, $$
and so this is the only possible twice-differentiable $v$.
