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


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


\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


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.

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Blog at

%d bloggers like this: