I'm working on a homework assignment in PDE, and I'm required to use the maximum principle to demonstrate that when $\Delta u(x)=0$ and periodic boundary conditions are applied, $u(x)$ is a constant.

The EXACT wording of the question is: "Let u be harmonic with periodic boundary conditions. Use the maximum principle to show that u is constant."

The maximum principle, as written in my textbook, comes in three parts:

1) Strong max: Let $u$ be harmonic in $\Omega$. If there exists $x_0$ $\epsilon$ $\Omega$ with $u(x_0)=\sup(u(x):x$ $\epsilon$ $\Omega)$ or $u(x_0)=\inf(u(x):x$ $\epsilon$ $\Omega)$, then $u$ is constant on $\Omega$.

Alternatively, using the ball mean property, $$u(x)=constant$$ iff $$u(x_o)=\frac{1}{\omega_d r^d}\int_{B(x_o,r)}u(x)dx = sup(u(x)),x\in \Omega$$

Where B is the ball: $$B(x,r):={y\in R^d:|x-y|\le r}$$

2) Weak max: Let $\Omega$ be bounded and $u$ $\epsilon$ $C^0(\Omega \cup \partial\Omega)$ be harmonic. Then for all $x$ $\epsilon$ $\Omega$, $\min(u(y):y$ $\epsilon$ $\partial\Omega) \le u(x)\le \max(u(y):y$ $\epsilon$ $\partial\Omega)$

3) Translational Corollary: Let $x_0$ $\epsilon$ $\Omega\subset R^d(d\ge 2),$ $u:\Omega\backslash {x_0}\rightarrow R$ be harmonic and bounded. Then u can be extended as a harmonic function on all of $\Omega$; i.e., there exists a harmonic function $\tilde{u}:\Omega\rightarrow R$ that coincides with u on $\Omega\backslash {x_0}$

Periodic boundary conditions are defined as follows:

$$\Omega=(0,L_1)\times ...\times (0,L_n)\subset R^n$$ and, for $$u:\bar{\Omega}\rightarrow R$$ that: $$u(x_1,...,x_{i-1},L_i,x_{i+1},...,x_n)=u(x_1,...,x_{i-1},0,x_{i+1},...,x_n)$$ for all $$x=(x_1,...x_n)\in\Omega,i=1,...,n$$

So far, I have written the following "true" (as best as I can tell) statements...but I can't see why they require $u(x)$ to be constant:

i) $\Delta u(x)=0$ iff $u(x_0)=\frac{1}{\omega_d r^r}\int_{B(x_0,r)}u(x)dx$

ii) $u(x)=constant$ iff $u(x_0)=\sup_{\Omega}(u(x))$

iii) if $\frac{1}{\omega_d r^d}\int_{B(x_0,r)}u(x)dx=\sup_{\Omega}(u(x))$ then $u(x)=constant$

iv) By periodic boundary conditions, (and using the domain for the un-extended $\Omega$ from earlier), $$u(x_0)=\frac{1}{\omega_d r^d}\int_{B(x_0 + nL,r)}u(x+nL)dx$$

Where $n\in Z^d$, and $nL=(n_1*L_1,...,n_d*L_d)$

**Note: $\omega_d$ is the volume of the unit sphere in $d$-dimensions

  • $\begingroup$ Could you maybe give the exact exercise statement and definitions for what is involved? It seems the claim you have written down above is simply wrong: Let $u(x,y) = (e^x + e^{-x})\sin(y)$. Then clearly $u(1,y) = u(-1,y)$, $u(x,0) = u(x,2\pi)$, and also $\Delta u = 0$. But $u$ is not constant... (Take $\Omega = (-1,1)\times (0,2\pi)$) $\endgroup$ – Sam Oct 16 '12 at 0:58
  • $\begingroup$ The exact wording of the question is: Let u be harmonic with periodic boundary conditions. Use the maximum principle to show that u is constant. Jurgen Jost: Partial Differential Equations, 2nd Edition, pg 31. $\endgroup$ – Chris Donlan Oct 16 '12 at 1:02
  • $\begingroup$ Also, I provided the definitions. $\endgroup$ – Chris Donlan Oct 16 '12 at 1:06
  • $\begingroup$ Also, ran into some stuff on the internet that suggest that the only periodic solution to the Laplace equation ($\Delta=\bigtriangledown^2$) is $\phi(x)=constant$...which would be exactly what this thing is asking me to demonstrate...but I don't see how the maximum principle is getting me there... $\endgroup$ – Chris Donlan Oct 16 '12 at 1:25
  • $\begingroup$ I'm thinking your definition of 'periodic boundary conditions' might be wrong. $\endgroup$ – Sam Oct 16 '12 at 3:19

The way I see it, the book asks you to prove the following statement:

Suppose $u$ is a harmonic function on $\mathbb R^d$, satisfying $u(x+z) = u(x)$ for all $x\in \mathbb R^d$ and $z\in \mathbb Z^d$. Then $u$ is constant.

This statement is true. The 'periodic boundary condition' seems to implicitly assume that $u$ can be periodically extended to $\mathbb R^d$, preserving a certain amount of regularity. And not only continuity, but such that the extended function is still $C^2$, I would guess.

The function $u(x,y) = (e^x + e^{-x})\sin(y)$ on $\Omega = (-1,1)\times (0,2\pi)$ admits a continuous extension to $\mathbb R^2$ and is harmonic on $\Omega$, but is not constant. So the extension needs to be smooth enough for the statement to hold. This is where my guess regarding the precise meaning of 'periodic boundary conditions' comes from (also this would describe the situation on a torus).

  • $\begingroup$ OK, awesome, Thanks. I will factor that in and see what the professor says...It sounds like what he said, actually...and I asked one of the guys in my class, and he said something about being confused about the "extension" part. $\endgroup$ – Chris Donlan Oct 16 '12 at 4:34
  • $\begingroup$ Can't vote it up though, till I get to +15 rep. $\endgroup$ – Chris Donlan Oct 16 '12 at 4:41
  • $\begingroup$ @ChristopherDonlan, you really ought to work very hard on solving this a number of different ways in one dimension, then see what you can do in $\mathbb R^2.$ Can you show that a function $f(x)$ in one real variable, continuous on a closed interval with the same beginning and ending value, but with $f''(x) = 0$ within the open interval, must be constant? $\endgroup$ – Will Jagy Oct 16 '12 at 4:49
  • $\begingroup$ LOL. Getting there. There must be a interior maximum or minimum such that $f'(x)=0$ in order for $f(a)=f(b)$...and, in order for $f''(x)=0$, the rest of the values in between the critical point $f'(x)=0$ and $a$ and $b$, they must be either constant or zero... But if they are non-zero constants, then continuity is broken at the point $f'(x)=0$ and therefore they must be zero, making $f(x)=constant$. $\endgroup$ – Chris Donlan Oct 16 '12 at 5:03
  • $\begingroup$ @ChristopherDonlan: The 1-dimensional case might be misleading. The space of "harmonic" functions (I don't even feel comfortable describing them by the same name as their higher-dimensional cousins) on the real line is very simple, but this dramatically changes in dimensions $>1$. $\endgroup$ – Sam Oct 16 '12 at 5:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.