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I don't have a strong background in partial differential equations so some of these questions might be quite basic. I first want to give some definitions which I am using.

Definition:

We define a second-order elliptic partial differential equation:

$Au_{xx}+2Bu_{xy}+Cu_{yy}+Du_{x}+Eu_{y}+F=0$ where $B^{2}-AC<0$. Assuming $u_{xy} = u_{yx}$.

We define a uniformly elliptic operator in the following way:

Definition:

A partial differential operator $L$ is (uniformly) elliptic if there exists a constant $\theta>0$ such that $\sum_{i,j=1}^{\infty}a^{i,j}(x)\xi_{i}\xi_{j} \geq \theta|\xi|^{2}$

for a.e. $x \in U$ and all $\xi \in \mathbb{R}^{n}$

1.What is the connection between elliptic operators and elliptic partial differential equations? 2.What is the importance of this property $\sum_{i,j=1}^{\infty}a^{i,j}(x)\xi_{i}\xi_{j} \geq \theta|\xi|^{2}$ which is used in the definition of uniformly elliptic operators above? 3.Do uniformly elliptic operators always give you a linear partial differential equation?

Thanks for any assistance.

Nate Eldredge
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Lucio D
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  • Related: https://math.stackexchange.com/q/21525/532409 , https://math.stackexchange.com/q/878746/532409 – Quillo Jun 08 '22 at 13:00

1 Answers1

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  1. The constant matrix $a = (a^{i,j}) = \left[\begin{matrix} A & B \\ B & C \end{matrix}\right]$ satisfies the ellipticity condition if and only if $A>0$ and $AC-B^2>0$. For $n\times n$ matrices, the ellipticity condition is saying that $a$ is positive definite uniformly over $x$. And one criterion for positive definite is that the determinants of the $k\times k$ upper left hand corner matrices all have positive determinant. Note in the $2\times2$ case, the condition $A>0$ is not so important, because if $A<0$ you can simply multiply the whole equation by $-1$.

  2. Ellipticity conditions are important because they guarantee uniqueness of solutions (although usually with other more benign hypotheses). This is a huge subject. I would recommend the book "Maximum Principles in Differential Equations" by Murray H. Protter, Hans F. Weinberger as a place to start.

Stephen Montgomery-Smith
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    Thanks for your response. Could you explain how we show that the elliplicity condition implies that the matrix $a = ((a^{i,j}(x)))$ is positive definite. The definition of positive definite that I am familiar with states that an $nxn$ matrix $M$ is positive definite if $Z^{T}MZ$ is positive for any nonzero column transpose of $Z$. – Lucio D Dec 01 '13 at 12:10
  • This is a uniform bound on the positive definiteness. So what we have is that $M-\theta I$ is positive definite for some $\theta>0$. And this results in $z^T M z \ge \theta z^T z = \theta |z|^2$. – Stephen Montgomery-Smith Dec 01 '13 at 15:17
  • Okay but how does $M-\theta I$ is positive definite follow from the elliplicity condition? Just to confirm, for the rest of the calculation you are saying $Z^{T}(M-\theta I)Z = Z^{T}MZ-Z^{T}\theta Z > 0$ $\therefore$ $Z^{T}MZ \geq \theta Z^{T}Z = \theta |Z|^{2}$ – Lucio D Dec 01 '13 at 15:31
  • Another equivalent condition for a symmetric matrix $M$ to be positive definite is that all its eigenvalues are positive. Let $\theta$ be the minimum of all the eigenvalues. Then $M-\theta I$ is positive semi-definite. – Stephen Montgomery-Smith Dec 01 '13 at 15:36
  • I still don't see how this follows from $\sum_{i,j=1}^{\infty}a^{i,j}(x)\xi_{i}\xi_{j} \geq \theta|\xi|^{2}$? How do you go from this to stating $a = ((a^{i,j}))$ is positive definite? – Lucio D Dec 01 '13 at 15:42
  • I you mean the reverse implication? That condition says $z^T M z \ge \theta|z|^2$. And $\theta|z|^2 > 0$. By the way, I don't think the upper limit in the sum should be infinity - it should be $n$. I think that is a typo. – Stephen Montgomery-Smith Dec 01 '13 at 15:48
  • Yes it's a typo thanks. I'm looking for a line of reasoning which starting with the elliplicity condition $\sum_{i,j=1}^{n}a^{i,j}(x)\xi_{i}\xi_{j} \geq \theta|\xi|^{2}$ and then showing that it follows from this assumption that $a$ is positive definite. Sorry if these questions are basic, I have only recently started studying PDE's. – Lucio D Dec 01 '13 at 15:54
  • If I write $M = (a^{i,j})$, then $z^T M z = \sum_{i,j=1}^n a^{i,j} z_i z_j$. – Stephen Montgomery-Smith Dec 01 '13 at 16:02
  • I understand, thanks for all of your responses. – Lucio D Dec 01 '13 at 16:08