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An elliptic operator $L$ is called uniformly elliptic if $a^{ij}(x)\xi _i\xi_j \ge \theta|\xi|^2$. What does this notation mean? All I can find about it is that it is some kind of summation notation or something. Can anyone explain me what it is?

Thank you very much.

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Satisfied with the answer below? –  Did Jul 15 '12 at 17:16
    
@did : understood the definition but i am not able to appreciate ,why it should be defined that way ? –  Theorem Jul 15 '12 at 17:35
    
Well, for one thing, for the LHS to be a number, starting from a vector $(\xi_i)_i$ and a square matrix $(a_{ij}(x))_{ij}$. In the end, I do not know why you leave the question open. –  Did Jul 15 '12 at 18:31
    
@did : Sir , i didn't understand what u meant by leaving the question open . –  Theorem Jul 15 '12 at 18:56
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1 Answer

$$\forall \xi=(\xi_i)_i\qquad\sum\limits_{i,j}a^{ij}(x)\xi_i\xi_j \geqslant \theta\cdot|\xi|^2=\theta\cdot\sum\limits_i\xi_i^2$$

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What is $\xi$ here actually thats my main problem to understand . can u help me ? –  Theorem Jul 10 '12 at 19:48
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@Theorem an arbitrary vector with real components $\xi_i$ –  user31373 Jul 10 '12 at 19:49
    
@Leonid Thanks. –  Did Jul 10 '12 at 19:50
    
@LeonidKovalev : Is that all sir ?? Thank you very much . –  Theorem Jul 10 '12 at 19:51
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In other words, the matrix $A - \theta I$ is positive semidefinite, where $A_{ij} = a^{ij}(x)$. For the operator to be uniformly elliptic on domain $\Omega$, you want there to be some $\theta > 0$ that makes this true for all $x \in \Omega$. –  Robert Israel Jul 10 '12 at 20:27
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