Let $u$ be an absolutely continuous function on $(a,b) \subset \mathbb{R}$. Find the unique absolutely continuous function $v$ satisfying,
$$ \int_a^bu(x)v(x)dx + \int_a^bu^\prime(x)v^\prime(x)dx = u(b) $$
I'm completely stumped on how to find a function like this. Initially I thought something simple like $v(x) = x$ would work out after integrating by parts, but unfortunately that did not work. I also tried a function like $v(x) = \int_0^xu^\prime(y)dy$, but this also didn't work.
Are there general techniques one can use to solve integral relations like this? Thanks.
If we do integrate the second integral by parts, we find we want $$ u(b) = u(b)v'(b)-u(a)v'(a) + \int_a^b u(x)(v(x)-v''(x)) \, dx. $$ We can satisfy this for any $u$ if we can find $v$ satisfying $$ v'(b)=1, \qquad v'(a) = 0, \qquad v''-v = 0. $$ Staring at this for a bit, we realise that $$ v(x) = \frac{\cosh{(x-a)}}{\sinh{(b-a)}} $$ will suffice.
I wouldn't go so far as to say that there are general techniques to deal with this type of problem, but it has a certain flavour of the calculus of variations to it, so that may be a source of applicable techniques in some cases.