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Question 1: It appears that when studying an elliptic equation $Lu=f$ in $\Omega$ with $u = g$ on $\partial \Omega$ we need to have $g=0$ in order that the inverse operator, $K=L^{-1}$ is linear. Otherwise $K(f_1+f_2) \neq Kf_1 + Kf_2$.

Is this inideed the case? Does it make any sense at all to speak of the "spectrum" of $L$ on $\Omega$ with respect to the boundary condition $g$?

I'm trying to understand when a maximum principle holds for $-\Delta u - \epsilon u$ on a domain $\Omega$. If I fix Dirichlet boundary conditions then for $\epsilon$ small enough I will have only the trivial $u \equiv 0$ solution. In some sense then, a maximum principle holds for the operator $-\Delta - \epsilon I$. However I can't seem to say anything about a general maximum principle here since my $\epsilon$ depended on my Dirichlet boundary conditions.

Question 2: Does a maximum principle still hold for $-\Delta u - \epsilon u$ on some arbitrary domain $\Omega$ when $\epsilon$ is small enough?

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Usually one decomposes the problem to $Lu = f$ with Dirichlet boundary, and the Laplace equation $Lv = 0$ with $v = g$ on the boundary. For fixed boundary function $g$ the solution to the second is unique, and the first can be inverted by General Nonsense (as you alluded to in the title). –  Willie Wong Sep 8 '10 at 23:08
    
For your second question, is $-\Delta$ a positive or negative operator? In the latter case, the maximum principle holds always (see Evans' PDE book for example), for the former, set $u = \prod \sin( \sqrt{\epsilon / n} x_i)$. For most domains $u$ will not attain a maximum value on its boundary. –  Willie Wong Sep 8 '10 at 23:33
    
Right but in general $-\Delta u - Vu$ does not have a maximum principle. You need $-V \geq 0$. However I was wondering if $V$ is negative but very small, will a maximum principle still hold. –  Dorian Sep 8 '10 at 23:46
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3 Answers 3

Here is what is done with spectral theory:

The boundary condition is just encoded in the domain. That is instead of considering the operator $-\Delta$ with domain $H_0^{2}(\Omega)$, one considers other domains $\mathcal{D}$. However, these are usually of the form $$ \mathcal{D} = \{f \in H^2(\Omega):\quad f|_{\partial \Omega} = g\cdot \partial{f}| _{\partial \Omega}\}, $$ where $g$ is some function. Here, I denote by $f| _{\partial \Omega}$ the trace of $f$ on the boundary of $\Omega$ and by $\partial{f}| _{\partial \Omega}$ the trace of the normal derivative. One can consider slightly more general situations by relating $f$ and $\partial f$ through a linear operator (non-local condition).

The obstruction to considering that $f = g$ on $\partial \Omega$ is that, one needs $\mathcal{D}$ to be a subspace of $L^2(\Omega)$.

Proof $\mathcal{D}$ is a subspace .

First $H^2(\Omega)$ is a Hilbert space, so it suffices to check if $f_1, f_2$ satisfy $$ f_j|_{\partial \Omega} = g\cdot \partial{f_j}| _{\partial \Omega},\quad j=1,2 $$ then for $\lambda$ $$ (f_1 + \lambda f_2) |_{\partial \Omega} = g\cdot \partial{ (f_1 + \lambda f_2)}| _{\partial \Omega} $$ Now this is a consequence of the linearity of the trace operators.

About question 2:

Consider $\Omega = \mathbb{R}$. Then $\sigma(-\Delta) = [0, \infty)$ and any non-zero solution of $-\Delta u = \epsilon u$ has infinitely many zeros. This follows from $u(x) = a exp(i k x) + b exp(-ikx)$ with $\epsilon = k^2$.

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But this is where I'm confused. Your domain $D$ is not a linear space. If you hadd two functions in it it falls out of the space. When considering the solution operator to your differential equation, it seems essential that it is linear and that appears to only be true when the boundary conditions are homogeneous. If one considers $-\Delta$ on your domain $\mathcal{D}$, the inverse operator $K = (-\Delta)^{-1}$ is no longer linear so how can we still use spectral theory? –  Dorian Sep 9 '10 at 1:07
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My $\mathcal{D}$ is a linear space. (I'm adding it to the post). –  Helge Sep 9 '10 at 9:04
    
One should be able to show that some form of maximum principle holds for $-\Delta-\epsilon, if $\inf(\sigma(- \Delta)) > \epsilon$. –  Helge Sep 9 '10 at 13:38
    
Ok but $f = g \frac{\partial u}{\partial \nu}$ is a Robin boundary condition not a Dirichlet condition. Secondly I'm just interested in the case of a Bounded domain where one would expect a maximum principle to hold. My question is, if $\Omega$ is bounded and $g$ is given on $\partial \Omega$, will a maximum principle hold for $-\Delta u + \epsilon u$ for $\epsilon$ small enough. ie. $\sup_{x \in \Omega} u(x) \leq \sup_{x \in \partial \Omega} u^+(x)$. –  Dorian Sep 9 '10 at 14:10
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The point is "spectral theory applies to Robin boundary conditions" and not much more (with spectral theory being naive, one can do smart things like adding solutions etc...) –  Helge Sep 9 '10 at 15:06
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On question 1: Given an elliptic operator $L$ on a domain $\Omega$, let $\Gamma$ denote its spectrum with Dirichlet boundary conditions. Then if $g$ is a function on the boundary of $\Omega$, the boundary value problem $Lu = \lambda u$ with $u$ equal to $g$ on the boundary has at most one solution for each $\lambda \notin \Gamma$. If $\lambda \in \Gamma$, then the space of solutions, if nonempty, is a translate of the corresponding eigenspace for the Dirichlet problem.

On question 2: Do you need a sharp maximum principle or just a sup norm bound? The latter can be obtained using Moser iteration, which involves proving an iterative $L_p$ bound and taking a limit $p \rightarrow \infty$.

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I would like just a weak maximum principle so that $\sup_{x \in \Omega} u(x) \leq \sup_{x \in \partial \Omega} u^+(x)$. –  Dorian Sep 9 '10 at 14:12
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  1. Let $Y=L^2(\Omega)\times L^2(\partial\Omega)$ and define $A:H^2(\Omega)\to Y$ by $Au=(-\Delta u,u|_{\\partial\Omega})$. Obviously $A$ is linear, and from elliptic estimates and the maximum principle one can show that $A$ is invertible. It is late at night so the function spaces here may have to be adjusted a bit but I am sure it can be made to work.

  2. I can say something when the Dirichlet boundary condition does not change sign. If the Dirichlet boundary condition is nonpositive then maximum principle will hold, if the boundary condition is nonnegative then the minimum principle will hold. This is from sub- and super-harmonic considerations.

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