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

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


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

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 $u^*$.

  • $\begingroup$ 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, 2015 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, 2015 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, 2015 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, 2015 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, 2015 at 10:00

1 Answer 1


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}


\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.

  • $\begingroup$ @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, 2015 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, 2015 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, 2015 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, 2015 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, 2015 at 16:02

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.