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Now I have known that Fredholm alternative result is valid for the strong elliptic system. But I'm not sure that is it still valid for the general elliptic system, in which the second-order heading coefficient matrix $A(x)$ is only positive definite, rather than strong elliptic type $$A(x)\xi\cdot\xi\geq\lambda|\xi|^2,\quad\forall\xi\in\mathbb{R}^{mn},\quad a.e. x\in\Omega$$

Any answer and reference will be appreciated!

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Check volume 3 of Hormander's four-volume monograph. (I assume that you impose elliptic boundary value conditions, and $\Omega$ is compact. –  Liviu Nicolaescu Jun 6 '13 at 10:00
    
@Liviu Nicolaescu: Please forgive my carelessness, I consider the boundary value problem! –  A.Hoo Jun 6 '13 at 10:28
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@A.Hoo: This book seems to be a great reference on elliptic systems: springer.com/birkhauser/mathematics/scuola+normale+superiore/… –  András Bátkai Aug 7 '13 at 15:38

2 Answers 2

up vote 5 down vote accepted

In 1948 A.V. Bitsadze, “On unique solvability of the Dirichlet problem for elliptic partial differential equations,”Uspekhi Mat. Nauk [Russian Math. Surveys],3, No. 6, 211–212,

constructed an elliptic equation with complex coefficients $$ Lu=\frac{\partial^2 u}{\partial x^2}+2i \frac{\partial^2 u}{\partial x\partial y}+\frac{\partial^2 u}{\partial x^2}=0 $$ for which the Dirichlet problem in the unit circle $D=\{x^2+y^2<1\}$, $$ Lu=0 \text{ in } D,\quad u|_{\partial D}=0, $$ is neither Fredholm, nor Noetherian. Namely, there are infinitely many solutions of this problem of the form $u(z)=f(z)(1-|z|^2)\,$ where $f$ is an analytic function in $\bar D$.

In the real form it is a uniformly elliptic system $$ \frac{\partial^2 u_1}{\partial x^2}-2 \frac{\partial^2 u_2}{\partial x\partial y}-\frac{\partial^2 u_1}{\partial y^2}=0, $$ $$ \frac{\partial^2 u_2}{\partial x^2}+2 \frac{\partial^2 u_1}{\partial x\partial y}-\frac{\partial^2 u_2}{\partial y^2}=0. $$ The notion of a strong elliptic system was introduced exactly to get the case where the corresponding operators are still Noetherian.

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@Andrew: What is the Noetherian operators? In my knowledge, the uniformly elliptic system is just the strong elliptic system, so the fredholm alternative result is valid. What does the conunterexample you posted say? I can't understand it well, can you explain the deep meaning for me? Thanks! :) –  A.Hoo Jun 6 '13 at 10:53
    
A [Noetherian operator][1] has a finite index in particular. And this system is elliptic because the characteristic polynomial has imaginary roots and uniformly elliptic since the coefficients don't depend on $x$. But it is not strongly elliptic. May be you are using some other notion of uniform ellipticity? [1]: A%20Noetherian%20operators –  Andrew Jun 6 '13 at 14:30
    
Here is the reference encyclopediaofmath.org/index.php/Noetherian_operator since the previous was a mess. –  Andrew Jun 6 '13 at 14:32
    
Thanks again!@Andrew :) –  A.Hoo Jun 6 '13 at 23:09

To complement Andrew's answer, Fredholm alternative holds for properly elliptic systems (with complementing boundary conditions). In 3 or more dimensions, any elliptic system in the sense of Douglis-Nirenberg is properly elliptic. Note that this is much more general than strongly elliptic systems. Strongly elliptic systems (with appropriate boundary conditions) have index 0, while a general elliptic system can have a nonzero index. Please see Agmon-Douglis-Nirenberg '64.

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@timur: Thank you for your answer! It furthers my understanding on elliptic systems. :) –  A.Hoo Jun 6 '13 at 23:14

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