I am curious whether there are any Liouville theorems for the following pde:

$$ - \Delta \phi(x) + a(x) \cdot \nabla \phi(x)=0 \qquad \mbox{in } R^N $$

where $a(x)$ is a smooth bounded vector field with $ |x| |a(x)| \rightarrow 0$ as $ |x| \rightarrow \infty$. As far as the solutions $ \phi$ I am assuming that $ \phi$ is smooth and the gradient satisfies a decay condition like $ | \nabla \phi(x)| \le C |x|^{-\sigma}$ for large $ x$ where $ \sigma>0$ is some small parameter.

As a second question, how about if we additionally assume $a(x)$ is divergence free.

Note if $ \phi$ went to zero at $\infty$ then we can get a Liouville theorem directly from the maximum principle. So one think I tried is taking a derivative of the equation to get a solution which decays... but I am not having much luck.

any comments or counterexamples would be welcomed. Craig


1 Answer 1


If $\Delta u + b(x) \cdot \nabla u = 0$ and $u$ is bounded, then $u$ is constant provided $|b| = O(1/(1+|x|))$. This follows from scaling, the Harnack inequality and the maximum principle.

Indeed, add a constant so that $\inf_{\mathbb{R}^n} u = 0$. Let $v(x) = u(Rx)$ for $R$ large. Then $v$ solves the equation $$\Delta v + Rb(Rx) \cdot \nabla v = 0.$$ By hypothesis, $R|b(Rx)|$ is uniformly bounded in $B_1-B_{1/4}$ for all large $R$, and for $R$ large we may assume by the maximum principle that $\inf_{\partial B_{1/2}}v < \epsilon$ for some $\epsilon$ small. The Harnack inequality, applied to $v$ in overlapping balls in the annulus $B_{3/4}-B_{1/4}$, implies that $v < C(n)\epsilon$ on $\partial B_{1/2}$, so by the maximum principle $u < C\epsilon$ in $B_{R/2}$. Since $\epsilon$ is arbitrarily small and this holds for all $R$ large, $u = 0$.

  • $\begingroup$ Thank you very much for the proof. Do you have any intuition on whether you think the Liouville theorem should hold under this gradient condition? $\endgroup$
    – Craig
    Nov 13, 2014 at 1:52
  • $\begingroup$ Why can we assume that $\inf_{\partial B_{1/2}}<\epsilon$ for $R$ large? I can’t understand how was the maximum principle used in this step. Can you elaborate? Thanks in advance. $\endgroup$
    – Feng
    Apr 14, 2021 at 8:24

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