I have a standard problem of the form:
$ \min_u J[u]=\int_a^b L(x, u, u')dx$
And appart from the boundary conditions $u(a)=u_a, u(b)=u_b$,
I need to impose a condition in a point $c$ such that $ a < c < b$ of the form $u(c)=u_c$.
When I solve the Euler-Lagrange equation:
$\frac{dL}{du}-\frac{d}{dx}\frac{dL}{du'}=0$
I get a general solution with two constants $C_1, C_2$ that I can find with the boundary conditions. But I do not know how to force $u(c)=u_c$.
I know how to include isoperimetric constraints, when they occur in every single point $G(x,u,u')=0$, but I do not know how to impose a constraint in a single point.
Thanks in advance
I probably misunderstood the question, as I believed the point $c$ is not known a priori. In other words, my understanding was that you are looking for a minimiser of $J[u]$ such that $u(c) = u_c$, $u_c$ given, and $c$ is allowed to vary.
A sort of obstacle problem, whereby the minimiser has to attain the value $u_c$ for a $c$, $a<c<b$, and we are looking for the optimal $c$.
In this case the solution could proceed along the following. For fixed $c$, look for an extremal that satisifed the Euler-Lagrange equation in both $[a,c]$ and $[c,b]$, as RRL suggested in his answer. The solution will depend on $c$, so the functional will be parametrised according to c, $J[u] = \hat{J}[u; c]$. The latter is to be interpreted as a function over $c$: given $c$, it computes the functional value, whose argument is the function $u$, extremal for the given $c$.
Minimising over $c$ is then akin to a one-dimensional function minimisation.