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I would like to know if there exists a Liouville theorem for solutions $u : \mathbb{R}^n \to \mathbb{R}$ of uniformly elliptic equations of the kind $$ D_i \left( a_{ij} D_j u \right) + b_i D_i u = 0. $$ I assume the coefficients $a_{ij},b_i \in C^{\infty}(\mathbb R^n) \cap L^{\infty}(\mathbb{R}^n)$.

Any hint/reference would be highly appreciated!

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    $\begingroup$ If $b$ decays like $1/|x|$ and $n \geq 2$ then bounded solutions are constants. One sees this using the scaling invariance of $\|b\|_{L^{\infty}(B_2 \backslash B_1)}$ and the Harnack inequality (see e.g. mathoverflow.net/questions/186856/… ). This is false in the case $n = 1$ because the annulus is not connected; take e.g. $u = \tan^{-1}(x)$ and $b(x) = 2x/(1+x^2)$. $\endgroup$
    – Connor Mooney
    Commented Mar 23, 2018 at 8:19

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What do you mean by the Liouville theorem? If the absence of bounded or positive harmonic functions, then the answer is "no" due to the presence of a vector field $b$. The corresponding counterexample can be constructed already for $n=1$. Take the diffusion coefficient $a=a_{11}$ to be equal identically 1, and let $b=b_1$ be odd and such that it converges to $+1$ at $+\infty$ (and therefore to $-1$ at $-\infty$). Then the space of bounded harmonic functions is 2-dimensional. This can be seen either directly, or from probabilistic considerations: the corresponding diffusion process on $\mathbb R$ converges to one of the two ends of the real line, so that it has a non-trivial behaviour at infinity.

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    $\begingroup$ It might be instructive to add the following concrete example: if we drop the condition that $b$ be smooth, we can choose $b(x) = \operatorname{sgn}(x)$ and consider the equation $u''(x) + \operatorname{sgn}(x) u'(x) = 0$ in the sense of distributions. This equation is solved by all constant functions and by the $H^1$-function $\tilde u$ given by $\tilde u(x) = 1 - e^{-x}$ for $x \ge 0$ and $u(x) = e^x - 1$ for $x < 0$. $\endgroup$
    – Jochen Glueck
    Commented Mar 22, 2018 at 15:18
  • $\begingroup$ @JochenGlueck Nice example! Thanks! $\endgroup$
    – Onil90
    Commented Mar 23, 2018 at 9:43

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