# Ngô Quốc Anh

## April 28, 2015

### On the simplicity of the first eigenvalue of elliptic systems with locally integrable weight

Filed under: Uncategorized — Ngô Quốc Anh @ 0:52

Of interest in this note is the simplicity of the first eigenvalue of the following problem $\begin{array}{rcl}-\text{div}(h_1 |\nabla u|^{p-2}\nabla u) &=& \lambda |u|^{\alpha-1} |v|^{\beta-1}v \quad \text{ in } \Omega\\-\text{div}(h_2 |\nabla v|^{q-2}\nabla v) &=& \lambda |u|^{\alpha-1} |v|^{\beta-1}u \quad \text{ in } \Omega\\u &=&0 \quad \text{ on } \partial\Omega\\v &=&0 \quad \text{ on } \partial\Omega\end{array}$

where $1 \leqslant h_1, h_2 \in L_{\rm loc}^1 (\Omega)$ and $\alpha, \beta>0$ satisfy $\displaystyle \frac \alpha p + \frac \beta q = 1$

with $p,q >1$. A simple variational argument shows that $\lambda$ exists and can be characterized by $\lambda = \inf_{\Lambda} J(u,v)$

where $\displaystyle J(u,v)=\frac \alpha p \int_\Omega h_1 |\nabla u|^p dx + \frac \beta q \int_\Omega h_2 |\nabla v|^q dx$

and $\Lambda = \{(u,v) \in W_0^{1,p} (\Omega) \times W_0^{1,q} (\Omega) : \Lambda (u,v) = 1\}$

with $\displaystyle \Lambda (u,v)= \int_\Omega |u|^{\alpha-1}|v|^{\beta-1} uv dx.$

Further arguments show that $\lambda$ is achieved by some positive pair $(u_0, v_0)\in \Lambda$ in the sense that for any test pair $(\varphi, \psi) \in C_0^\infty(\Omega) \times C_0^\infty (\Omega)$, there holds $\begin{array}{rcl}\displaystyle \int_\Omega h_1 |\nabla u|^{p-2}\nabla u \nabla \varphi &=& \displaystyle \lambda \int_\Omega |u|^{\alpha-1} |v|^{\beta-1}v \varphi\\ \displaystyle \int_\Omega h_2 |\nabla v|^{p-2}\nabla u \nabla \psi &=& \displaystyle \lambda \int_\Omega |u|^{\alpha-1} |v|^{\beta-1}u \psi.\end{array}$

By the simplicity of $\lambda$ we mean the eigenspace associated with $\lambda$ is of dimension one in the sense that whenever $J(u,v)=\lambda \Lambda (u,v)$ we must have $u=k_1 u_0$ and $v=k_2 v_0$ for some constants $k_1, k_2$.

To prove the simplicity of $\lambda$, we first recall the following Picone identity for $p$-Laplacian $\displaystyle {\left| {\nabla u} \right|^p} - \nabla \left( {\frac{u^p}{\phi^{p - 1}}} \right) \cdot \nabla \phi{\left| {\nabla \phi} \right|^{p - 2}} \geqslant 0$

with the equality occurs whenever $u/\phi$ is constant. Here by the density, we only require that $(u, \phi) \in W_0^{1,p} (\Omega) \times W_0^{1,p} (\Omega)$ and $\phi>0$. Upon multiplying both sides by $h_1$ and integrating by parts, we arrive at $\displaystyle \int_\Omega h_1 \left| {\nabla u} \right|^p dx \geqslant -\int_\Omega \left( {\frac{u^p}{\phi^{p - 1}}} \right) \text{div}(h_1 \nabla \phi{\left| {\nabla \phi} \right|^{p - 2}} ) dx.$

In a similar way, for any $(v, \phi) \in W_0^{1,q} (\Omega) \times W_0^{1,q} (\Omega)$ with $\phi>0$, we also obtain $\displaystyle \int_\Omega h_2 \left| {\nabla v} \right|^q dx \geqslant -\int_\Omega \left( {\frac{v^q}{\phi^{q - 1}}} \right) \text{div}(h_1 \nabla \phi{\left| {\nabla \phi} \right|^{q - 2}} ) dx.$

Suppose that $(u,v)$ is also an eigenpair associated with $\lambda$, we first obtain from the definition of $(u_0, v_0)$ the following $\begin{array}{rcl}J(u_0, v_0) &=& \lambda \Lambda (u_0, v_0)\\&=&\displaystyle \lambda \int_\Omega u^\alpha v^\beta \frac{|u_0|^\alpha |v_0|^\beta}{u^\alpha v^\beta} dx\\& \leqslant & \displaystyle \lambda \int_\Omega u^\alpha v^\beta \left( \frac \alpha p \frac{|u_0|^p}{u^p} + \frac \beta q \frac{|v_0|^q}{v^q} \right) dx\\&=& \displaystyle \lambda \int_\Omega \left( \frac \alpha p \frac{u^{\alpha -1 }v^\beta }{u^{p-1}}|u_0|^p + \frac \beta q \frac{u^\alpha v^{\beta-1}}{v^{q-1}} |v_0|^q\right) dx\\&=& \displaystyle \frac \alpha p\int_\Omega \left( \frac{-\text{div}(h_1 |\nabla u|^{p-2}\nabla u)}{u^{p-1}}|u_0|^p \right) dx + \frac \beta q\int_\Omega \left( \frac{-\text{div}(h_2 |\nabla v|^{p-2}\nabla v)}{v^{q-1}} |v_0|^q\right) dx\\&\leqslant & \displaystyle \frac \alpha p\int_\Omega h_1 |\nabla u_0|^p dx + \frac \beta q\int_\Omega h_2 |\nabla v_0|^q dx\\&=&J(u_0,v_0).\end{array}$

The above estimates show that $u/u_0$ and $v/v_0$ are constants as claimed.

## 9 Comments »

1. What is the reason to consider the rhs of the form $|u|^{\alpha-1} |v|^{\beta-1} v$, but not $|u|^{\alpha-2} |v|^\beta u$? Surely, the first eigenvalues and eigenfunctions will be the same for both cases, due to positivity of eigenfunctions. But higher eigenfunctions can be different…

Comment by Vladimir Bobkov — November 7, 2015 @ 15:19

• Thanks Vladimir for this interesting observation.

In this note, I simply adopt standard arguments to this particular example to show the simplicity of the first eigenvalue. You may look at this paper http://dx.doi.org/10.1007/s13398-015-0217-7.

I did not consider or want to compare different nonlinearities as you indicate above.

Comment by Ngô Quốc Anh — November 7, 2015 @ 15:33

• Thank you for the aritcle!

Actually, regardless to the different types of resonance nonlinearity, it seems to me that such type of problems should be considered with two different spectral parameters $\lambda$ and $\mu$. It is in some sense more natural, since there are two equations). Then there arises an eigencurve on the $(\lambda, \mu)$-plane, which visually looks similar to the eigencurve of the Fucik spectral problem for the p-Laplacian. And I think it is possible to adopt the above agrumetns with Picone’s identity to such problems, as well.

In any case, thank you for nice post)

Comment by Vladimir Bobkov — November 7, 2015 @ 16:35

• Thanks Vladimir for your informative comments. If possible, please advise us some materials to look at.

Comment by Ngô Quốc Anh — November 7, 2015 @ 23:39

• I found one very relevant aritcle: http://www.sciencedirect.com/science/article/pii/S0022039606000088
The authors consider the resonant nonlinearity of the form $|u|^{\alpha-2}|v|^\beta u$, and with two spectral parameters $(\lambda, \mu)$.

The global idea of their approach is the following: for each $(\lambda, \mu)$ there exists $t$, such that $(\lambda, \mu) = (\lambda, t \lambda)$. ( $t$ characterizes a slope of the ray in $(\lambda,\mu)$-plane which crosses $(0,0)$ and $(\lambda, \mu)$).
Then they fix this $t$ and consider a one-parameter eigenvalue problem for the same system with respect to $\lambda$. Obviously, the corresponding first eigenvalue $\lambda_1$ will depend on $t$, i.e. $\lambda_1 = \lambda_1(t)$. The set of points $(\lambda_1(t), t \lambda_1(t))$ is, in fact, a parametric representation of the spectral curve in $(\lambda,\mu)$-plane. This curve is continuous and monotone (Lemma 2.3).

I myself worked on related superlinear problems, i.e., with $\frac{\alpha}{p}+\frac{\beta}{q}>1$, where similar curves were also found, see http://www.tandfonline.com/doi/full/10.1080/17476933.2015.1107905#.Vj47grfhCt8

Comment by Vladimir Bobkov — November 8, 2015 @ 1:02

• Very interesting, I will definitely take a look at those mentioned papers :).

Thanks Vladimir :).

Comment by Ngô Quốc Anh — November 8, 2015 @ 1:19

2. Hello Vladimir, me again, what did you mean when you wrote “And I think it is possible to adopt the above agrumetns with Picone’s identity to such problems, as well?”

Comment by Ngô Quốc Anh — November 9, 2015 @ 16:27

• Hi! (Sorry, I don’t know how to correctly call you by name – Quốc or Anh?)

I meant the following: Probably, it is possible to generalize the trick with the Picone identity, which you used in this post, to show that the whole first eigencurve is simple.

Actually, it is definitely doable) Just make the parametrization $(\lambda, \mu) = (\lambda, r \lambda)$, $r > 0$, and consider, as in your post, the minimizer of the energy functional $\displaystyle J_r(u,v) = \frac{\alpha}{p} \int_\Omega |\nabla u|^p \, dx + \frac{1}{r} \frac{\beta}{q} \int_\Omega |\nabla v|^q \, dx$

over the same set $\Lambda$. Then instead of the last chain of inequalities, we get $\displaystyle J_r(u_0, v_0) = \lambda \Lambda(u_0, v_0) = \dots \leq \frac{\alpha}{p} \int_\Omega |\nabla u_0|^p \, dx + \frac{1}{r} \frac{\beta}{q} \int_\Omega |\nabla v_0|^q \, dx = J_r(u_0, v_0).$

Comment by Vladimir Bobkov — November 9, 2015 @ 23:42

• OK OK I got it, thanks :).

Comment by Ngô Quốc Anh — November 10, 2015 @ 0:00

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