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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

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    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 $\Omega = (-1,1)\times (0,2\pi)$)2012-10-16
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    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 31.2012-10-16
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    Also, I provided the definitions.2012-10-16
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    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...2012-10-16
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    I'm thinking your definition of 'periodic boundary conditions' might be wrong.2012-10-16
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    Indeed, I just looked at page 13 of your book: Periodic boundary conditions are not boundary conditions at all, but rather the description of a function defined (and in your case $\in C^2(\mathbb R^d)$) on all of $\mathbb R^d$ which is periodic w.r.t. the lattice $\mathbb Z^d$. In this case, the result is true and I think you will be able to prove it yourself. =) This is the same as with elliptic functions in complex analysis.2012-10-16
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    Well, its the definition given by the book, pg 13: "...the so-called periodic boundary condition. This means the following. We consider a domain of the form $\Omega = (0,L_1)\times ... \times (0,L_d)\subset R^d$ and require for $u:\bar{\Omega}\rightarrow R$ that $$u(x_1,...,x_{i-1},L_i,x_{i+1},...,x_d)=u(x_1,...,x_{i-1},0,x_{i+1},...,x_d)$$ for all $x=(x_1,...,x_d)\in \Omega,i=1,...,d.$ This means that u can be periodically extended from $\Omega$ to all of $R^d$. A reader familiar with ... blah blah...[visualize a torus]"2012-10-16

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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).

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    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.2012-10-16
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    Can't vote it up though, till I get to +15 rep.2012-10-16
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    @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?2012-10-16
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    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 $f(x)=constant$.2012-10-16
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    @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 $>1$.2012-10-16
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    Did you manage to prove that any $u\in C^2(\mathbb R^d)$ which is periodic wrt $\mathbb Z^d$ and harmonic must be constant? Because 'periodic boundary condition' really is just a reformulation of this condition on $u$ as I've tried to point out...2012-10-16
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    @Sam, Yeah, I think it will follow from the 1D example. I re-read your post. My mistake on the other two comments. Forgot that you accounted for $C^2$ as versus $C^{\infty}$...2012-10-16
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    I really appreciate the help. I was stuck, big time.2012-10-16
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    @ChristopherDonlan: I don't see how it can follow from the 1D case? Note that if $u(x_1, \dots, x_n)$ is harmonic, then the function obtained by fixing certain components, such as $x_1 \mapsto u(x_1, c_2, \dots, c_n)$ with $c_i$ constant, need not be harmonic itself (was this maybe your idea?). It really is best to make use of the maximum principle here (you are free to choose a suitable bounded domain $\Omega$ to apply it to, you simply need an interior maximum inside of $\Omega$ to prove that $u$ is constant).2012-10-16
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    Ok, thanks, I'll try that.2012-10-16
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    @Christopher: since $u$ is continuous and periodic in $x_1, \ldots, x_d$, it has a maximum value, but since $u$ harmonic, it is constant.2012-10-17