In my PDE module, the general solution to Laplace's equation $\nabla^2 T=0$ in the plane (in polar coordinates) was shown to be $$T(r,\theta)=A_0+B_0\log r+\sum_{n=1}^\infty(A_nr^n+B_nr^{-n})\cos(n\theta)+(C_nr^n+D_nr^{-n})\sin(n\theta).$$ The lecture notes now claim that, in boundary value problems where solutions have to be bounded at infinity, the $\log r$ and $r^n$ terms are inadmissible, so their coefficients must be zero. For the logarithmic term this makes sense, as $\log r$ is indeed unbounded; but I am struggling to justify the other part of the claim to myself. For example, momentarily letting $\theta=0$, and looking at the boundary value problem $$\lim_{r\to\infty}r\frac{\partial T}{\partial r}=0,$$certain series solutions such as $$T(r)=\frac1r\cos\theta+\sum_{n=1}^\infty(-1)^n\frac{r^{2n-1}}{(2n+1)!}\cos(n\theta)=_{\theta=0}\frac{\sin(r)}{r^2}$$satisfy the boundary condition (in this case, provided $\theta=0$) just fine. Now of course this isn't exactly a counterexample to the lecture notes' claim, since when $\theta=\pi$ I suddenly have a hyperbolic function on my hands; but I can't see how to dismiss the possibility of counterexamples such as these occurring. It certainly does seem believable that, given any choice of coefficients $A_n,B_n,C_n,D_n$ in the general solution, there exists some $\theta$ causing the solution to be unbounded; but I have no idea how to prove this (if it is even true).
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