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I posted this question on Math.SE, but I could not get any help.

The eigenvalue $\lambda(t)$ is characterised as the minimum of the Rayleigh quotient (where $t$ is a scalar variable) $$R(u,\Omega_t)= \frac{\int_{\Omega_t} |\nabla u|^2 dy }{\int_{\Omega_t} u^2 dy}$$ i.e. $\lambda(\Omega_t) = \mathbf{min} \{R(v, \Omega_t) : v \in H^{1,2}(\Omega_t)\}$ 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

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

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

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

$$R(u,\Omega_t)= \frac{\int_{\Omega_t} |\nabla u|^2 dy + \alpha \oint_{\partial \Omega_t} u^2 dS}{\left(\int_{\Omega_t} u^q dy \right)^{2/q}}$$

My attempt: we have $$\lambda(x, \Omega_t) = \frac{\int_{\Omega_t} {\mid \nabla u(x) \mid }^2 \, dx + \alpha \oint_{\partial \Omega_t} u^2\, ds} { \left( \int_{\Omega_t} u^q \, dx\right )^{2/q}} $$

The corresponding eigenvalue equation is given by the following computation.

Let $t \in \mathbb{R} , \varphi \in H^{1,2}\Omega$. If $\lambda(x, \Omega_t) $ is the eigenvalue 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,

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

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

and thus, we get the equation

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

together with the boundary condition

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

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 sufficient conditions on $q \in \mathbb R$ which guarantee the existence of a minimizer $u^*$.

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  • $\begingroup$ I would have two things to point out : first, if you set $\alpha = 0$ and $q=2$, your final equations are the same as in the first case, apart from this curious $\mu(u)$, which probably shouldn't be there. Secondly, you forgot to add orthogonality conditions in both cases, in the spirit of $\int_{\Omega_t} u = 0$. Otherwise, non zero constant functions are obvious global minimizers. $\endgroup$ – Hachino Mar 15 '15 at 9:33
  • $\begingroup$ @Hachino Sir, The curious factor $\mu(u)$ does appear because of the differentiation of $\left(\int_{\Omega_t }(u+s\varphi)^q\right)^{2/q}$ w.r.t $s$ . I do not understand what orthogonality condition you mean . I would be glad if you could elaborate it for me . $\endgroup$ – Learner Mar 15 '15 at 9:46
  • $\begingroup$ Your definition of $\lambda(\Omega_t) := \mathbf{min} \{R(v, \Omega_t) : v \in H^{1,2}(\Omega_t)\}$ implies that 1) $\lambda$ is nonnegative 2) any constant, nonzero function achieves a zero value for the Rayleigh quotient, which means that $\lambda = 0$. But under an additional condition, for instance $\lambda(\Omega_t) := \mathbf{min} \{R(v, \Omega_t) : v \in H^{1,2}(\Omega_t), \int_{\Omega_t} v = 0\}$ (or more generally, that your $v$'s are orthogonal to some subset of eigenfunctions of the Neumann laplacian on $\Omega_t$), $\lambda$ acquires a nontrivial value. $\endgroup$ – Hachino Mar 15 '15 at 9:51
  • $\begingroup$ Btw, your notations and computations suggest that you are trying to understand variations of eigenvalues/eigenfunctions w.r.t. the underlying domain and shape optimization. Is it so ? $\endgroup$ – Hachino Mar 15 '15 at 9:53
  • $\begingroup$ @Hachino Yes, my main objective is to understand the variations of EV/EF's wrt the domain and shape optimisation . $\endgroup$ – Learner Mar 15 '15 at 10:00
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Okay, I checked and you got it mostly right, except the factor $\mu(u)$ in front of the power of $u$ at the very end, in the main equation.

Let's simplify slightly your notations and write $R(\Omega, u)$ and $R(\Omega, u + \varphi)$ for the Rayleigh quotients. I do not compute using a real parameter ; rather, I perform some generalized Taylor exapnsions to get the full derivative of the quotient, which means that $\varphi$ is to be thought of as small itself. Denote by $\|u\|_q^2$ the denominator of $R$.

The numerator of $R(\Omega, u + \varphi)$ expands as
\begin{equation} \int_{\Omega} |\nabla u |^2 + \alpha \int_{\partial \Omega} |u|^2 + 2 \int_{\Omega} \nabla u \cdot \nabla \varphi + 2 \alpha \int_{\partial \Omega} u \varphi + \mathcal{O}(\varphi^2). \end{equation}

Its denominator writes

\begin{align} & \|u\|_q^2 \left( 1 + \frac{q}{\|u\|_q^q} \int_{\Omega} |u|^{q-2} u \varphi + \mathcal{O}(\varphi^2) \right)^{\frac{2}{q}} \\ & = \|u\|_q^2 \left( 1 + \frac{2}{\|u\|_q^q} \int_{\Omega} |u|^{q-2} u \varphi + \mathcal{O}(\varphi^2) \right) . \end{align}

Thanks to the usual $(1+x)^{-1} = 1-x + \mathcal{O}(x^2)$, we get :

\begin{equation} R(\Omega, u + \varphi) = R(\Omega, u) - \frac{2}{\|u\|_q^q} R(\Omega,u) \int_{\Omega} |u|^{q-2} u \varphi + \frac{2}{\|u\|_q^2} \left( \int_{\Omega} \nabla u \cdot \nabla \varphi + \alpha \int_{\Omega} u \varphi \right) + \mathcal{O}(\varphi^2). \end{equation}

Using integration by parts, this rewrites

\begin{equation} R(\Omega, u + \varphi) = \lambda - \frac{2}{\|u\|_q^q} \lambda \int_{\Omega} |u|^{q-2}u \varphi + \frac{2}{\|u\|_q^2} \left( - \int_{\Omega} \Delta u \varphi + \int_{\partial \Omega} (\partial_{\nu} u + \alpha u) \varphi\right). \end{equation}

Finally we are done. The boundary condition that you got was the right one, whereas the main equation is now

\begin{equation} - \frac{\Delta u}{\|u\|_q^2} = \lambda \frac{|u|^{q-2}u}{\|u\|_q^q}, \end{equation}

i.e.

\begin{equation} - {\Delta u} = \lambda \frac{|u|^{q-2}u}{\|u\|_q^{q-2}}. \end{equation} Notice that if we set $q=2$ and $\alpha = 0$, we indeed get back to the usual eigenvalue problem for the Neumann laplacian, which is your first case.


I will attempt to elaborate a bit more on why $q^*$ is important. Recall that the Sobolev embedding theorem tells you in particular that $H^1$ embeds into $L^{q^*}$, where $1 = d(\frac 12 - \frac {1}{q^*})$ and $d$ is the dimension of the ambient space. This embedding means in particular that $L^{q^*}$ scales like $H^1$, in that, for a non zero $f \in H^1$ and $n \in \mathbb{N}$, $\frac{\|f(n \cdot)\|_{H^1}}{\|f(n \cdot)\|_{L^{q^*}}}$ does not depend on $n$, whereas for a general exponent $q$ instead of $q^*$, it will. This means we have three cases to look at :

Case 1 : $q > q^*$.

This is the easiest one and it relies solely on the scaling. Up to a translation, we can assume that $0$ is in the interior of $\Omega$. Let $\rho$ be your favorite function in $\mathcal{C}^{\infty}_c(\Omega)$ localised around $0$ and assume for simplicity that its $L^q$ norm is equal to $1$. Define now $\rho_n$ by $\rho_n(x) := n^{\frac dq} \rho (nx)$. (This defines a family of ultra-thin, ultra-high functions around $0$.)

For any $n \in \mathbb{N}$, we have :

  • $\|\rho_n \|_{L^q(\Omega)} = 1$ (thanks to the $n^{\frac dq}$ factor),
  • $\|\rho_n \|_{L^2(\partial \Omega)} = 0$ (compact support condition),
  • $\|\nabla \rho_n\|_{L^2(\Omega)} = \|\nabla \rho\|_{L^2(\Omega)} n^{d(\frac 1q - \frac{1}{q^*})}$.

Gathering all the equalities above, one obtains hat $R(\Omega, \rho_n) \to 0$ as $n \to \infty$. But the only function cancelling the numerator of $R$ is the zero function, which is not admissible. Thus, in this case, $\lambda = 0$ and there is no minimizer.

Case 2 : $2 \leq q < q^*$.

Here, opposite to what happened in Case 1, you will get both a nontrivial $\lambda$ and the existence of a minimizer.

Let $(u_n)$ be a sequence of functions with the following properties :

  • $\|u_n\|_{L^q} \equiv 1$ for all $n$,
  • $R(\Omega, u_n) \to \lambda$ as $n \to \infty$.

Because $q \geq 2$ and $\Omega$ has finite measure, the first bullet implies that $\|u_n\|_{L^2} \lesssim 1$. From the second bullet, one has in particular that $(\nabla u_n)$ is bounded in $L^2$. Thus, there exists some $v \in H^1(\Omega)$ such that, up to extraction, $u_n \to v$ strongly in $L^2(\Omega)$ and weakly in $H^1(\Omega)$. By interpolation between $L^2$ and $L^{q^*}$, one also has $u_n \to v$ strongly in $L^q(\Omega)$. In particular, $\|v\|_{L^q(\Omega)} = 1$.

On the other hand, by interpolation between $L^2$ and $H^1$, $u_n \to v$ strongly in $\dot{H}^{\frac34}(\Omega)$ (say). Because the trace operator is continous from $\dot{H}^{\frac 34}(\Omega)$ to $\dot{H}^{\frac 14}(\partial \Omega)$, it follows that $u_n \to v$ strongly in $\dot{H}^{\frac14}(\partial \Omega)$. Also, because $(u_n)$ is bounded in $L^2(\partial \Omega)$, up to extraction, $u_n$ converges weakly in $L^2(\partial \Omega)$ to some function $g$. Identifying distributional limits, one obtains that $g$ equals (the trace on $\partial \Omega$ of) $v$.

Eventually, using Fatou lemma, one has

\begin{equation} R(\Omega, v) \leq \liminf_{n \to + \infty} R(\Omega, u_n) = \lambda. \end{equation}

Thus, $v$ is a minimizer. In particular, since $v$ cannot be $0$ (having for instance unit $L^q$ norm), $\lambda$ is nonzero.

The third case (namely $q = q^*$) seems more involved, I'll try later on. If you want any clarification, please go ahead.

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  • $\begingroup$ @Learner : regarding guidance, I strongly suggest you to use such Taylor-like expansions instead of directional derivatives. I believe they are less misleading, especially with intricate computations - think that this case was an introductory exercise. $\endgroup$ – Hachino Mar 15 '15 at 11:09
  • $\begingroup$ But , does the solution(i.e. minimum) really exist for all $q \in \mathbb R$? I have been thinking for quite sometime but couldn't really prove . $\endgroup$ – Learner Mar 16 '15 at 15:27
  • $\begingroup$ Not necessarily, because of Sobolev embeddings. Consider, for simplicity, functions belonging to $\mathcal{C}^{\infty}_c(\Omega)$, thus cancelling part of the numerator. If $q$ is too high (above the exponent given by the Sobolev embedding), then you can build up Dirac-like functions with $H^1$ norm equal to $1$ and big $L^q$ norm. If $d$ is the ambient dimension and $\rho$ some smooth function with compact support, consider $\rho_n(x) := n^{\frac{d}{2}-1} \rho(nx)$. $\endgroup$ – Hachino Mar 16 '15 at 15:44
  • $\begingroup$ OTOH, if $q < q^*$ where $1 = d\left(\frac{1}{2} - \frac{1}{q^*} \right)$, then minimizing sequences will be compact in $L^q$, thus giving strong convergence of the denominator. The existence of a minimizer in $H^1$ follows readily by applying Fatou lemma. If $q = q^*$ I'm not quite sure right now, but profile decomposition could help you - this is precisely studying defects of compactness of bounded sequences of $H^s$ functions in Sobolev-critical $L^p$ spaces. $\endgroup$ – Hachino Mar 16 '15 at 16:02
  • $\begingroup$ So you think there is a bound on $q$ ? Could you tell me how to go about to find the bound on $q$ ? $\endgroup$ – Learner Mar 16 '15 at 16:02

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