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$\textbf{Problem}$ Let $\Omega$ be an open, bounded and connected subset of $\mathbb{R}^n$. Suppose that $\partial \Omega$ is $C^{\infty}$. Consider an eigenvalue problem \begin{align*} \begin{cases} -\Delta u=\lambda u & \textrm{ in } \; \Omega \\ \frac{\partial u}{\partial \nu}=-u & \textrm{ on } \partial \Omega \end{cases} \end{align*} Define a bilinear operater $(\cdot,\cdot)_{H^1}$ by \begin{align*} (u,v)_{H^1}:=\int_{\Omega} \nabla u \cdot \nabla v \;dx + \int_{\partial \Omega} uv \; d\sigma \end{align*} Show that there exists a constant $\theta>0$ independent of $u,v$ such that \begin{align*} (u,u)_{H^1} \geq \theta \Vert u \Vert _{H^1(\Omega)}^2 \end{align*}

$\textbf{Attempt}$

\begin{align*} (u,u)_{H^1}&=\int_{\Omega} \nabla u \cdot \nabla u \;dx + \int_{\partial \Omega} u^2 \; d\sigma \\ &=\int_{\Omega} \nabla \cdot(u\nabla u)-u\Delta u \; dx +\int_{\partial \Omega} u^2 \; d\sigma \\ &=\int_{\partial \Omega} u \frac{\partial u}{\partial \nu} \; d\sigma +\int_{\Omega} \lambda u^2 dx +\int_{\partial \Omega} u^2 \; d\sigma \\ &=-\int_{\partial \Omega} u^2 \; d\sigma +\int_{\Omega} \lambda u^2 dx +\int_{\partial \Omega} u^2 \; d\sigma\\ &=\lambda \Vert u \Vert _{L^2(\Omega)}^2 \end{align*} I don't know how to get $\lambda \Vert u \Vert_{L^2(\Omega)}^2 \geq \theta \Vert u \Vert_{H^1(\Omega)}^2$...

Any help is appreciated..

Thank you!

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  • $\begingroup$ What is the connection between the taks 'Show that...' and the eigenvalue problem? As it is stated now, there is no connection, and it looks like the inequality has to be proven for all $u\in H^1$. $\endgroup$
    – daw
    Dec 10, 2018 at 14:19
  • $\begingroup$ If I prove the problem, then I'll get every eigenvalue is a positive real number. $\endgroup$
    – user453447
    Dec 10, 2018 at 14:22

2 Answers 2

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Here is a standard proof by contradiction. Assume that the inequality does not hold. Then for all $n$ there is $u_n\in H^1(\Omega)$ such that $$ \|\nabla u_n\|_{L^2(\Omega)}^2 + \|u_n\|_{L^2(\partial\Omega)}^2 < n \|u_n\|_{H^1(\Omega)}^2. $$ This implies $u_n\ne0$, and we can assume w.l.o.g. $\|u_n\|_{H^1(\Omega)}=1$. Then $\nabla u_n\to0$ in $L^2(\Omega)$ and $u\to0$ in $L^2(\partial\Omega)$ follows immediately.

Since $(u_n)$ is bounded in $H^1$, after extracting a subsequence (denoted the same for simplicity), we have $u_n\rightharpoonup u$ in $H^1(\Omega)$. Then $\nabla u=0$, and $u$ is a constant. Since $\|u\|_{L^2(\partial\Omega)}=0$, it follows $u=0$. By compact embedding, $u_n\to0$ in $L^2(\Omega)$. This implies $u\to0$ strongly in $H^1(\Omega)$, which is a contradiction to $\|u_n\|_{H^1}=1$.

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If you want to show that for $u$ that are solutions of the eigenvalue problem,$$(u,u)_{H^1}\ge\theta \int_\Omega |\nabla u|^2 + |u|^2 dx,$$ you have already done all the hard work. From your equalities, using the first and last line, $$ (u,u)_{H^1}=\frac12\int_\Omega \nabla u \cdot \nabla u dx + \frac12 \int_{\partial\Omega} |u|^2 d\sigma + \frac\lambda2 \int_\Omega u^2 dx. $$ Dropping the middle term and using $\theta=\min(\frac12,\frac\lambda 2)$ you get what you want.

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  • $\begingroup$ Is $\lambda$ always positive? $\endgroup$
    – user453447
    Dec 10, 2018 at 14:15
  • $\begingroup$ You have $\lambda=\frac{(u,u)_{H_1}}{\|u\|_{L^2}^2}$ (and you assume $u$ nonzero as it is an eigenfunction). That means clearly $\lambda\ge 0$. If $\lambda=0$, you get that $\int_\Omega |\nabla u|^2 dx+\int_{\partial\Omega} |u|^2 d\sigma=0$ so $u$ is constant, and zero on the boundary, so zero everywhere. So $\lambda>0$. $\endgroup$
    – Kusma
    Dec 10, 2018 at 14:29

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