I know that $u$ harmonic $\implies u$ satisfies the mean value property, but does this work the other way around?

Also, if we have Laplace's equation defined inside a disc of radius $1$ with boundary condition $u(1,\theta)=\sin(\theta)$, is it enough to show that if $u(r,\theta)=r\sin(\theta)$ satisfies the mean value property, then $u(r,\theta)=r\sin(\theta)$ is the solution to Laplace's equation satisfying the boundary condition?


Provided $u$ is $C^2$ (so that the Laplacian makes sense), yes. Essentially, if $$u(x) = \frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)} u(y)\,{\rm d}S(y),$$make a change of variables $y = x+rz$ with $z \in \partial B(0,1)$, differentiate under the integral sign and change back to the original variable to obtain that $$\triangle u(x) = \frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)} \triangle u(y)\,{\rm d}S(y).$$Since $\partial B(0,1)$ has no boundary, Stokes' Theorem says that the above integral vanishes and so $\triangle u = 0$.

For the second query, yes, it would suffice in view of the above, but it is much simpler to just verify that $\triangle u =0$ directly by using the polar coordinate expression $$\triangle u = \frac{\partial^2u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2u}{\partial \theta^2} = 0 + \frac{\sin\theta}{r} - \frac{r\sin\theta}{r^2} = 0.$$

  • $\begingroup$ Thanks for this. So it is simply enough to show that $u$ satisfies Laplace's equation and the boundary condition in order to conclude that $u$ is the general solution? $\endgroup$ – maths54321 May 29 '20 at 19:32
  • $\begingroup$ Not the general solution. It is one solution for $\triangle u = 0$ on the disk. But it is the one that satisfies your boundary condition $u(1,\theta) = \sin\theta$. $\endgroup$ – Ivo Terek May 29 '20 at 19:33
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    $\begingroup$ The converse is true even when the function is only continuous. If it satisfies the mean value property then it is smooth. $\endgroup$ – Arctic Char May 29 '20 at 19:39
  • $\begingroup$ Can we be certain that this is the only solution that satisfies the boundary condition? (I haven't covered any additional theorems in this area apart from the maximum principle) $\endgroup$ – maths54321 May 29 '20 at 19:43
  • $\begingroup$ Uniqueness follows from the maximum principle. It suffices to show that if $u$ is harmonic on the disk and vanishes at the boundary, then $u=0$. The maximum principle implies $u\leq 0$. Replacing $u$ by $-u$ and repeating the argument gives $u\geq 0$. So $u=0$. $\endgroup$ – Ivo Terek May 29 '20 at 19:45

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