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I am interested in knowing whether there is a definition for the symbol of a PDO which is NOT linear. In Wikipedia and in the book I am reading (An Introduction to Partial Differential Equations by Renardy-Rogers) I only found the definition for linear PDOs.

Here is the Wikipedia link:

http://en.wikipedia.org/wiki/Symbol_of_a_differential_operator

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Look like strange to me.... what do you mean by definition of symbol.... If i am denoting velocity by $v$, then what does mean by definition of $v$... –  zapkm Feb 28 '12 at 18:07
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@Pradip: The word "symbol" has a special meaning in this context. –  Hans Lundmark Feb 28 '12 at 21:16

1 Answer 1

The symbol of a nonlinear differential operator is defined as the symbol of its linearization.

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Could you please be more specific about what you mean by linearization. If you can, give an example. Thank you! –  chango Feb 29 '12 at 10:41
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... and in particular depends on the function you are linearising the operator around. (Generally one linearises around an approximate solution.) –  Willie Wong Feb 29 '12 at 10:41
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If your nonlinear PDO is $F:(x,u,\partial u, \partial^2 u) \mapsto 2y$, then its linearisation about a function $v$ is formally $$L_v(x,u,\partial u,\partial^2 u) = \lim_{h\to 0}\frac{1}{h}\left[ F(x,v+hu,\partial v + h\partial u, \partial^2 v + h\partial^2 u) - F(x,v,\partial v,\partial^2 v)\right]$$ –  Willie Wong Feb 29 '12 at 10:50
    
The primary example I'm familiar with is for the Ricci flow; you might be interested in reading Chapters 4 and 5 of Peter Topping's "Lectures on the Ricci Flow" (link), although if you're aiming strictly at PDE instead of geometric PDE you may find it a little obtuse; the PDE in question is $\frac{\partial g}{\partial t}=-2\operatorname{Ric}g$. –  youler Feb 29 '12 at 21:17

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