I have this problem: \begin{equation} \begin{cases} \Delta\:g+ \lambda \:g=0\quad {\rm in}\;D \\ g=0\quad {\rm on} \; \partial D.\end{cases} \end{equation} The domain $D$ is made by two triangles ${T}_1$ and ${T}_1$ with a common edge $L$.
I have found a function $u$ which is $0$ on every edge of ${T}_1$ different from $L$ and such that $\Delta\:u+ \lambda \:u=0$ in ${T}_1$. Then, similarly, I have found a function $v$ which is $0$ on every edge of ${T}_2$ different from $L$ and such that $\Delta\:v+ \lambda \:v=0$ in ${T}_2$.
Then I prove that the eigenfunctions $u$ and $v$ fit toghether smoothly (${C}^1$) along their common interface $L$ and I call $w$ the function defined on $D$, which is $u$ on ${T}_1$ and $v$ on ${T}_2$.
I have now to prove that $w$ is a weak solution of the starting problem. I've just begun studying PDE and I don't kwow well how to proceed. Is this the condition that I have to verify: \begin{equation*} \int_{D}{\bigtriangledown w\cdot\bigtriangledown \phi\: dx}=\lambda\:\int_{D}{w\:\phi dx} \quad{\rm for\; all\;\phi\in{C}_0^\infty }\quad? \end{equation*} Does this condition come directly from the definition of weak solution?
In order to prove that $w$ is a weak solution I have thought to write the integral at the first member in this way: $\int_{{T}_1\bigcap supp(\phi)}{\dots}+\int_{{T}_2\bigcap supp(\phi)}{\dots}$ then to use the following identity: $\bigtriangledown w\cdot \bigtriangledown \phi=div(\phi\:\bigtriangledown w)-\phi\:\bigtriangleup w$ and the divergence theorem, but I can't prove the condition of weak solution. How can I prove this? Thanks for the help!