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.

  • $\begingroup$ What is the role of the extra variable $t$? $\endgroup$ – Liviu Nicolaescu Dec 19 '15 at 10:49
  • $\begingroup$ it describe the time dependence $\endgroup$ – Rym Touibi Dec 19 '15 at 10:54
  • $\begingroup$ In this case you have a whole family of operators. $\endgroup$ – Liviu Nicolaescu Dec 19 '15 at 11:18

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 $1$.

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  • $\begingroup$ 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 '15 at 11:49

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