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Assume $v \in C^{2}(U)$ , where $U=\mathbb{R}^{n} \backslash B_{R}$ (the ball is centered at the origin) solves $$ \begin{cases}\Delta v=0, & \text { on } U \\ v=0, & \text { on } \partial U\end{cases} $$ Then show

(a) For $n=2$ and $$ \lim _{x \rightarrow \infty} \frac{v(x)}{\log x}=0 $$ then $v=0$

(b) For $n \geq 3$ and $$ \lim _{x \rightarrow \infty} v(x)=0 $$ then $v=0$

Why isn't it true that by the minimum principle & maximum principle $v$ is identically 0? should I deal with the special cases of the fundamental solution?

Arctic Char
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mat mat
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  • Well yes it follows from maximum principle for (b). (and maximum principle is usually not directly applicable for non-compact domain). – Arctic Char Jul 13 '21 at 14:58
  • The solution is unique, so if you come up with some function that satisfies the equation and the boundary conditions you are done. The boundary is infinity and the circle, so for any $n\geq 2$ it seems to me that $v=0$ is the solution. (Or maybe I don't understand the notation: $\mathbb{R}^n \backslash B_R$ means the exterior of the circle of radius of $R$ ?) – DanielKatzner Jul 13 '21 at 15:36
  • Why wouldn't it be the fundamental solution restricted to $U$? maybe I should split the entire space with respect to singularity – mat mat Jul 13 '21 at 15:40
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    @DanielKatzner Uniqueness of the solution depends on the maximum principle, and as Arctic Char pointed out the maximum principle depends on $U$ being bounded. – Ali Jul 13 '21 at 15:46
  • Without a constraint at infinity you can furnish two distinct solutions to these BVPs. In particular, in (a), one solution is $\ln(|x|)-\ln(R)$ and another is just $0$. In (b), one solution is $\frac{1}{|x|}-\frac{1}{R}$ and again another is just zero. So the maximum principle is not working in the way you're used to, you have to treat infinity as being on the boundary as well. – Ian Jul 13 '21 at 15:53
  • @Ali To be honest I've never seen a proof of uniqueness. This is a classroom exercise in electromagnetism, so what I can show you is an explicit solution: $v(r,\theta)=\sum_{n\geq1}(a_n \cos n\theta+b_n \sin n\theta)/r^n$, with $a_n$ and $b_n$ constants and $(r,\theta)$ the usual polar coordinates. If you demand $v=0$ on the inner boundary (circle) you are forced to take $a_n=b_n=0$ and the solution dies. – DanielKatzner Jul 13 '21 at 16:01
  • Thank you very much guys. Does the next proof allow to use the maximum principle if we'll send the domain to bounded one?https://math.stackexchange.com/questions/961596/maximum-principle-for-harmonic-functions-on-unbounded-domain – mat mat Jul 13 '21 at 16:02
  • Will it be solvable if we take the limit of $\lim _{x \rightarrow \infty} \frac{u(x)}{\log |x|}=0$? – mat mat Jul 13 '21 at 16:15
  • The maximum principle requires a compact domain. However harmonic functions are preserved by conformal maps, so you can stereographically unproject $v$ to a sphere with two holes, and then, apply the maximum principle. – user7530 Jul 13 '21 at 16:28
  • maybe Kelvin transformation will help that my domain will become the ball? $\bar{u}(x):=u(\bar{x})|\bar{x}|^{n-2}=u(x /|x|)|x|^{2-n}$ – mat mat Jul 13 '21 at 22:59
  • @DanielKatzner Your general solution is just not correct. As I said, in 3D two solutions to this problem with no BC at infinity are $\frac{1}{r}-\frac{1}{R}$ and $0$. – Ian Jul 14 '21 at 02:20
  • @Ian The problem is that I thought that infinity was included as part of the boundary (as is usually the case in mathematical physics). Once you set a boundary condition for all boundaries (connected or not, finite or not), the solution is unique. – DanielKatzner Jul 14 '21 at 10:42

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