Ngô Quốc Anh

May 3, 2011

A squeezing argument and application to logistic equations

Filed under: Uncategorized — Ngô Quốc Anh @ 1:19

The purpose of this note is to introduce the squeezing method. As an application, we prove some Liouville type theorem for solutions to logistic equations. Here the text is adapted from a paper by Du and Ma published in J. London Math. Soc. in 2001 [here].

Let prove the following theorem.

Theorem. Let $\lambda \in \mathbb R$ and $u \in C^2(\mathbb R^N)$ be a non-negative stationary solution of $u_t - \Delta u = \lambda u - u^p.$

Then $u$ must be a constant.

The basic ingredients in the proof consist of the following three lemmas. But first of all, let us denote by $L$ a (uniformly) second order elliptic operator satisfying the PDE $-L(u)=\lambda a(x)u-b(x) u^p, \quad x\in \mathbb R^N$

where $L(u)=\sum_{ij} (a_{ij}u_{x_i})_{x_j}$

with $a_{ij}$ smooth, $a_{ij}=a_{ji}$ and $\sigma_1 |\xi|^2 \leqslant \sum_{ij}a_{ij}(x)\xi_i\xi_j \leqslant \sigma_2 |\xi|^2$

for some positive constants $\sigma_i$, $i=1,2$.

Lemma 1 (comparision principle). Suppose $\Omega$ is a bounded domain in $\mathbb R^N$, $\alpha (x)$ and $\beta (x)$ are continuous functions on $\Omega$ with $\|\alpha\|_\infty < \infty$ and $\beta (x)$ non-negative and not identically zero. Let $u_1, u_2 \in C^2(\Omega)$ be positive in $\Omega$ and satisfy $Lu_1 + \alpha u_1 - \beta g(u_1) \leqslant L u_2 + \alpha u_2 - \beta g(u_2),\quad x \in \Omega$

and $\displaystyle\limsup_{x \to \partial\Omega} (u_2 - u_1) \leqslant 0$

where $L$ is as above , $g$ is continuous and such that $\frac{g(u)}{u}$ is strictly increasing for $u$ in the range $\inf \{ u_1, u_2\} < u < \sup \{u_1, u_2\}$.

Then $u_2 \leqslant u_1$ in $\Omega$.

Next we obtain

Lemma 2. Let $\Omega \subset \mathbb R^N$ be bounded with smooth boundary and $L$ be as above. Suppose that $\alpha$ and $\beta$ are smooth positive functions on $\overline \Omega$, and let $\mu_1$ denote the first eigenvalue of $-\Delta u=\mu \alpha(x) u$ on $\Omega$ under Dirichlet boundary condition on $\partial \Omega$. Then the problem $-L(u)=\mu u [\alpha (x) - \beta (x) u^{p-1}], \quad x \in \Omega$

with zero Dirichlet boundary $u=0, \quad x\in \partial\Omega$

has a unique positive solution for every $\mu>\mu_1$, and the unique solution $u_\mu$ satisfies $\displaystyle u_\mu \to \left(\frac{\alpha (x)}{\beta (x)}\right)^\frac{1}{p-1}$

uniformly on any compact subset of $\Omega$ as $\mu \to \infty$.

Lastly, we have

Lemma 3. As in Lemma 2, the following problem $-L(u)=\mu u |\alpha (x) - \beta (x) u^{p-1}|, \quad x \in \Omega$

with infinite Dirichlet boundary $u=+\infty, \quad x\in \partial\Omega$

has a unique positive solution for each $\mu>0$, and the unique solution $u_\mu$ satisfies $\displaystyle u_\mu \to \left(\frac{\alpha (x)}{\beta (x)}\right)^\frac{1}{p-1}$

uniformly on any compact subset of $\Omega$ as $\mu \to \infty$. Here by the infinite boundary condition we mean $u \to +\infty$ as $x \to \partial\Omega$.

We are now in a positive to prove the Liouville type theorem.

Due to the Harnack inequality, we shall focus on positive solutions. Assuming $u$ is an entire solution where $\lambda>0$, we prove that $u=\lambda^\frac{1}{p-1}$.

Indeed, let $x_0$ be an arbitrary point in $\mathbb R^N$. For any $\alpha>0$, let us define $u_\alpha (x) = u(x_0 + \alpha (x-x_0)$

It is clear to see that $u_\alpha$ verifies $-\Delta u=\alpha^2(\lambda u-u^p).$

Let $B$ denote the unit ball with center $x_0$. From Lemma 2, for all $\alpha$ sufficiently large, there is some $v_\alpha$ solving $-\Delta v=\alpha^2(\lambda v-v^p), \quad x \in B, \quad v=0, \quad x \in \partial B.$

with property that $\displaystyle v_\alpha \to \lambda^\frac{1}{p-1}$

at $x_0$ as $\alpha \to \infty$. Using Lemma 1, we conclude that $u_\alpha \geqslant v_\alpha$ in $B$ and hence $u(x_0)=u_\alpha(x_0) \geqslant v_\alpha (x_0).$

Let $w_\alpha$ be the unique solution to the following $-\Delta w=\alpha^2(\lambda w-w^p), \quad x \in B, \quad w=+\infty, \quad x \in \partial B.$

Using Lemma 1 and Lemma 3 we can show that $u(x_0)=u_\alpha(x_0) \leqslant w_\alpha (x_0).$

Letting $\alpha \to \infty$ we see that $u(x_0)=\lambda^\frac{1}{p-1}$.

As $x_0$ be arbitrary, we conclude that $u\equiv \lambda^\frac{1}{p-1}$. For the case $\lambda<0$, we refer the reader to the original paper.