# Is it true that $u$ is harmonic if and only if $u$ satisfies the mean value property?

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 $$\triangle u = \frac{\partial^2u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2u}{\partial \theta^2} = 0 + \frac{\sin\theta}{r} - \frac{r\sin\theta}{r^2} = 0.$$
• 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? – maths54321 May 29 '20 at 19:32
• 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 $u(1,\theta) = \sin\theta$. – Ivo Terek May 29 '20 at 19:33
• 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 $u=0$. – Ivo Terek May 29 '20 at 19:45