I have following boundary value problem:
$-(au')'=0$ in $\Omega=(0,1)$
$u(0)=1$
$u(1)=0$
with \begin{equation} a(x) = \begin{cases} \kappa \alpha & x \in (0,\epsilon) \\ \alpha & x\in (\epsilon,1) \end{cases} \end{equation}
and $\kappa , \alpha >0$
I wanna find the weak solution $u\in H^1(0,1)$ and show uniqueness. Also im interested if this is a classical solution.
My idea is to determine a weak formulation of the problem and then maybe use following fundamental lemma of calculus of variations:
If $\omega \in L^2((a,b))$ satisfies $$\int ^b_a \omega v'=0 $$ for all $v\in C^{\infty}_0 ((a,b))$ then $\omega$ is constant on $(a,b)$
Unfortunately im totally stuck. Anyone can help and know how to handle it?
Start by multiplying by a test function $v$ and integrating by parts. This initially reads
$$-\int_0^1 (au')' v dx = 0.$$
Integrating by parts gives
$$\left. -(au')v \right |_{0}^1 + \int_0^1 au' v' dx = 0.$$
By choosing test functions with homogeneous Dirichlet boundary conditions, the boundary terms vanish, giving
$$\int_0^1 au'v' dx = 0 \quad \\ u(0)=1,u(1)=0.$$
This choice is necessary, without it the problem does not have a solution (because the test functions that vanish on the boundary already determine $u$ uniquely).
This means $au'$ is constant almost everywhere, which means $u$ is piecewise linear, with the break being at $x=\epsilon$. The boundary conditions plus continuity (which is automatic from the assumption $u \in H^1$, by Sobolev embedding considerations) gives you three equations for the four unknown coefficients. How can you get a fourth equation? (Hint: it is a relationship between the slopes.)