5
$\begingroup$

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.

$\endgroup$
7
$\begingroup$

$\Delta u = \alpha u$ 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 $\beta$).

$\endgroup$
6
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.