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Mar 23, 2018 at 9:50 vote accept Onil90
Mar 23, 2018 at 8:19 comment added Connor Mooney 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 $b(x) = 2x/(1+x^2)$.
Mar 22, 2018 at 13:13 answer added R W timeline score: 2
Mar 22, 2018 at 10:56 history edited Onil90 CC BY-SA 3.0
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Mar 22, 2018 at 10:31 history asked Onil90 CC BY-SA 3.0