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The Eigen value $\lambda(t)$ which is characterised by the Rayleigh quotient (where $t$ is a scalar variable): $$R(u,\Omega_t)= \frac{\int_{\Omega_t} |\nabla u|^2 dy }{\int_{\Omega_t} u^2 dy}$$ $\lambda(\Omega_t) = \mathbf{min} \{R(v, \Omega_t) : v \in H^{1,2}(\Omega_t)\}$ The minimiser $u*$ of the Rayleigh quotient satisfies : $$\left. \frac{d}{ds} R(u* + s\varphi) \right \vert_{s=0} =0 $$ Solving it , the conditions i get are

$$\Delta u^* + \lambda^* (\Omega_t) u^* =0 \quad \textrm{in} \quad \Omega_t $$

$$\partial_{\nu} u^* = 0 \quad \textrm{in} \quad \partial \Omega_t$$

$\textbf{Similarly, I want to find the conditions that are}$ $ \textbf{fulfilled by the minimizer $u^*$ for the Rayleigh quotient defined as follows .}$

$$R(u,\Omega_t)= \frac{\int_{\Omega_t} |\nabla u|^2 dy + \alpha \oint_{\partial \Omega_t} u^2 dS}{\left(\int_{\Omega_t} u^q dy \right)^{2/q}}$$ $\textbf{My Attempt: }$ We have $$\lambda(x, \Omega_t) = \frac{\int_{\Omega_t} {\mid \nabla u(x) \mid }^2 \, dx + \alpha \oint_{\partial \Omega_t} u^2\, ds} { \left( \int_{\Omega_t} u^q \, dx\right )^{2/q}} $$ Corresponding eigen value equation is given by , Let $t \in \mathbb{R} , \varphi \in H^{1,2}\Omega$ . If $\lambda(x, \Omega_t) $ is the eigen value then the following holds $\frac{d}{ds} R(u_t + s \varphi , \Omega_t) \rvert_{s=0}$ $$\frac{d}{ds} \left(\frac {\int_{\Omega_t} {\mid \nabla (u+s\varphi) \mid}^2\,dx}{\left(\int_{\Omega_t} {\mid u +s\varphi \mid}^q \, dx \right)^{2/q}} \right) = \frac{d}{ds} \left (\frac{\int_{\Omega_t} {\mid \nabla u \mid}^2 + 2 s \nabla u \nabla \varphi + s^2 {\mid \nabla \varphi \mid}^2 \, dx} {\left( \int_{\Omega_t} (u+s \varphi )^q \, dx\right )^{2/q}} \right)_{s=0} $$

$$=\frac{2 \int_{\Omega_t} \nabla u \nabla \varphi\,dx }{ \left( \int_{\Omega_t} u^q \, dx\right )^{2/q}} - 2 \left( \int_{\Omega_t} {\mid u \mid }^q\right)^{\frac{-2}{q} -1 } \int_{\Omega_t} u^{q-1} \varphi \, dx \int_{\Omega_t} {\mid \nabla u \mid}^2 dx ... (1*) $$

Evaluating next term, $$\frac{d}{ds} \left( \frac{\alpha \oint_{\partial \Omega_t} (u+s \varphi)^2 \, ds}{ \left(\int_{\Omega_t} (u+s\varphi)^q \, dx \right )^{2/q} }\right )_{s= 0} $$

$$= \frac{2 \alpha \int_{\partial \Omega_t} u \varphi \, ds - 2 \alpha \oint_{\partial \Omega_t} u^2 \, ds \left(\int_{\Omega_t} u^q \, dx \right )^{-1} \int_{\Omega_t} u^{q-1} \varphi \, dx }{ \left( \int_{\Omega_t} u^q \, dx \right )^{2/q} } ... (2*) $$

Adding $(1)$ and $(2)$ gives :

$$\frac{2 \int_{\Omega_t} \nabla u \nabla \varphi dx - \int_{\Omega_t} {\mid \nabla u \mid}^2 dx \left( \int_{\Omega_t} u^q \, dx \right )^{-1} \int_{\Omega_t} u^{q-1} \varphi \, dx + \alpha \int_{\partial \Omega_t} u \varphi \, ds - \alpha \int_{\partial \Omega_t } u^2 \, ds \left( \int_{\Omega_t} u^q dx \right)^{-1} \int_{\Omega_t} u^{q-1} \varphi \, dx }{ \left(\int_{\Omega_t} {\mid u\mid }^q\, dx \right)^{2/q}}$$

Using integration by parts yields,

$$\frac{-2 \int_{\Omega_t} \Delta u \varphi dx + 2 \int_{\partial \Omega_t} \partial_{\nu} u \varphi \, ds - 2 \left ( \int_{\Omega_t} u^q \right)^{-1} \int_{\Omega_t} u^{q-1} \varphi \, dx \int_{\Omega_t} {\mid \nabla u \mid}^2 \, dx +2 \alpha \int_{\partial \Omega_t} u \varphi \, ds }{\left(\int_{\Omega_t} {\mid u\mid }^q \, dx \right)^{2/q}}$$

$$ -\frac{2 \alpha \int_{\partial \Omega_t} u^2 \, ds \left(\int_{\Omega_t} {\mid u \mid}^q \right)^{-1} \int_{\Omega_t} u^{q-1} \varphi \, dx }{\left(\int_{\Omega_t} {\mid u\mid }^q \, dx \right)^{2/q}}$$

$$ \Delta u + \lambda \mu (u) u^{q-1} = 0 \mathtt \;on\; \Omega_t $$

$$ \partial_{\nu} u + \alpha u = 0 \mathtt \; on \; \partial \Omega_t $$

Where

$$\mu(u) = \left ( \int_{\Omega_t} u^{q-1} \, dx \right )^{-1} $$

Can someone check if the resulted equations are the right necessary conditions ? I am looking also for the sufficient conditions so as to know for which $q \in \mathbb R$ the minimum even exists .

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    $\begingroup$ You can just use **two asterisks** for bold. $\endgroup$
    – user856
    Mar 9, 2015 at 15:05

1 Answer 1

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Almost. It has to be $$ \left\{ \begin{aligned} \Delta u &+ \lambda \mu(u) |u|^{q-2} u=0 &&\text{in}~ \Omega_t,\\ \partial_\nu u &+ \alpha u = 0 &&\text{on}~ \partial \Omega_t, \end{aligned} \right. $$ where $$ \mu(u) = \left(\int_{\Omega_t} |u|^q \, dx\right)^{-1}. $$

Certainly, to be able to obtain the minimizer of $\lambda(\Omega_t)$, one has to assume that $1 \leq q < +\infty$ if the dimension $N$ of the space is $1$ or $2$, and $1 \leq q < \frac{2N}{N-2}$ if $N \geq 3$. This follows from the Rellich-Kondrachov theorem.

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