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Let $D$ be the open bounded subset in $\mathbb{R}^{n}$ with smooth boundary, $\alpha$ and $\beta$ be different non-null real numbers, and $u$ and $v$ be in $W_0^{1,2}(D)\setminus\left\{ 0\right\} $ such that $\Delta u=\alpha u$ and $\Delta v=\beta v$ in weak solutions sense. Prove that $$\int_{D}\nabla u\cdot\nabla v\,dx=\int_{D}uv\,dx=0$$
I don't understand what "in weak solutions sense" means. Can anyone tell me what it means so I can solve the problem.
Thanks in advanced.
$\Delta u = \alpha u$ in the sense of weak solutions means, that for each $w \in W^{1,2}_0(D)$ we have $$ - \int_D \nabla u \cdot \nabla w \, dx = \alpha \int_D uw\, dx $$ (same for $v$ and $\beta$).
Continue martini 's hints, I will solve this problem.
$\Delta u = \alpha u$ in the sense of weak solutions means, that for each $w \in W^{1,2}_0(D)$ we have $$ - \int_D \nabla u \cdot \nabla w \, dx = \alpha \int_D uw\, dx \; \; \; \; \; \; \; \; \; \; \; \; (1)$$ $\Delta v = \beta v$ in the sense of weak solutions means, that for each $w \in W^{1,2}_0(D)$ we have $$ - \int_D \nabla v \cdot \nabla w \, dx = \beta \int_D vw\, dx \; \; \; \; \; \; \; \; \; \; \; (2)$$
Choose $w=v$ in $(1)$ and $w=u$ in $(2)$ we have $$ - \int_D \nabla u \cdot \nabla v \, dx = \alpha \int_D uv\, dx$$ and
$$ - \int_D \nabla u \cdot \nabla v \, dx = \beta \int_D uv\, dx$$ But $\alpha \neq \beta$ so we are done.
I hope I don't have mistakes.
