1
$\begingroup$

Let $\Omega$ be a bounded smooth domain in $\mathbb{R}^d$ and let $L$ be a uniformly elliptic second order partial differential operator: $$Lu(x,t)=-\sum_{i,j=1}^{d}{a_{ij}(x,t)u_{x_{i}x_{j}}(x,t)}+\sum_{i=1}^{d}{b_{i}(x,t)u_{x_i}(x,t)}$$ where $x\in\Omega$ and $t>0.$ My question is: Does exist a positive eigenfunction of the corresponding homogeneous Dirichlet problem and an associated positive eigenvalue?under which conditions? Could you provied me by some references? Thanks.

$\endgroup$
3
  • $\begingroup$ What is the role of the extra variable $t$? $\endgroup$ Dec 19, 2015 at 10:49
  • $\begingroup$ it describe the time dependence $\endgroup$
    – Rym Touibi
    Dec 19, 2015 at 10:54
  • $\begingroup$ In this case you have a whole family of operators. $\endgroup$ Dec 19, 2015 at 11:18

1 Answer 1

1
$\begingroup$

This happens for operators ```in divergence form'', i.e., Laplace operators associated to a Riemann metric on $\Omega$. If the Riemann metric is described by the tensor $(g_{ij}(x))_{1\leq i,j \leq d}$ satisfying the positivity condition: $\exists c>0$ $\newcommand{\bR}{\mathbb{R}}$

$$ \sum_{ij}g(ij)(x)\xi_i\xi_j\geq c\vert \xi\vert^2,\;\;\forall \xi\in\bR^d. $$

If $(g^{ij}(x))$ denotes the inverse matrix

$$\bigl(\; g^{ij}(x)\;\bigr):=\bigl(\; g_{ij}(x)\;\bigr)^{-1},\;\;\forall x\in\Omega, $$

and $|g|$ the determinant

$$ |g|=\det \bigl(\; g_{ij}(x)\;\bigr), $$

then the associated Laplacian is $\newcommand{\pa}{\partial}$

$$ Lu=-\frac{1}{\sqrt{|g|}}\sum_{i,j}\pa_{x_i}\left(\sqrt{|g|}g^{ij}\pa_{x_j} u\right). $$

The first nonzero eigenvalue of this operator is given by

$$\lambda_1=\min\int_\Omega u(x)Lu(x)\sqrt{|g|} dx,\;\mbox{subject to the constraint}\;\int_\Omega u(x)^2\sqrt{|g|} dx=1,\;\;u\bigl|_{\pa \Omega}=0.$$

The minimizers of this problem are eigenfunctions of $L$ corresponding to $\lambda_1$. It is not difficult to see that if $u$ is a minimizer of this problem, the so is $|u|$. To prove that $|u|>0$ in the interior of $\Omega$ use the maximum principle. In particular, this also show that $\lambda_1$ has multiplicity $1$.

$\endgroup$
1
  • $\begingroup$ Thank you for your answer. In this kind of operator "divergence ooerator" we cannot make depent the coefficients on a time variable? $\endgroup$
    – Rym Touibi
    Dec 19, 2015 at 11:49

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.