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When we are given a differential operator of the form $Lf : = \sum_\alpha a_{\alpha}(x) D^ \alpha f(x) $ , we can define the symbol associated with it to be the function: $a(x,y) := \sum_\alpha a_\alpha (iy)^\alpha $ .

We then can consider an operator (in divergence form) on $L^2 (\Omega) $, given formally by: $ Hf := -b(x)^{-1} \sum_{i,j=1}^{N} \frac{\partial}{\partial x_i} \{a_{i,j} (x) \frac{\partial f}{\partial x_j} \} $ .

Davies' book - "Spectral Theory of differential operators" says : "The reader will observe that the operator $H$ defined above is not that associated with the symbol $a(x,y) := b(x)^{-1} \sum_{i.j=1}^{N} a_{i,j} (x) y_i y_j $ according to the procedure above and then gives the following exercise: Write down the symbol, in the first sense, associated with the operator $H$ , assuming the coefficients $a_{i,j} (x) $ are $C^1 $ functions of $x$ .

Can someone explain me what does he mean by that? What should I do in this question and how can I find this form?

Any detailed answer might help ! Thanks in advance !

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Why do you have $i$ in there? Shouldn't it be: $a(x,y) := \sum_\alpha a_\alpha (y)^\alpha $. – copper.hat Jul 12 '12 at 17:54
No .... This is the definition the book has... – joshua Jul 12 '12 at 18:35

2 Answers 2

Hint: Expand $\dfrac{\partial}{\partial x_i} \left(a_{i,j}(x) \dfrac{\partial f}{\partial x_j}\right) $

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Your point is that I need to transform the divergence form into the reguler one? – joshua Jul 12 '12 at 18:35

$H$ is given in divergence form, $L$ is not.

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