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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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    @Pradip: The word "symbol" has a special meanin$g$ in this conte$x$t.2012-02-28

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

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    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](http://www.warwick.ac.uk/~maseq/topping_RF_mar06.pdf)), although if you're aiming strictly at PDE instead of geometric PDE you may find it a little obtuse; the PDE in question is $\frac{\partial g}{\partial t}=-2\operatorname{Ric}g$.2012-02-29
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See Definition of the principal symbol of a differential operator on a real vector bundle..

For an example, consider the Ricci curvature operator: \begin{align} \mathsf{Ricc}:\Gamma(S^2_+M)&\rightarrow\Gamma(S^2M)\\ g&\mapsto\mathsf{Ricc}(g). \end{align} The linearisation of the Ricci operator at a given metric $g\in\Gamma(S^2_+M)$ is just the directional derivative of the operator at $g$, and is the map \begin{align} D\mathsf{Ricc}|_g:S^2M&\rightarrow S^2M\\ h&\mapsto D\mathsf{Ricc}|_gh=\frac{\text{d}}{\text{d}t}\Big|_{t=0}\mathsf{Ricc}(g+th). \end{align} The hard part is calculating it. Let $(U,\mathsf{x})$ be a chart on $M$ and $\omega\in\Gamma(T^*M)$ be a covector field. Books on the Ricci flow (Topping, Chapter 2 or Chow & Knopf, Chapter 3) show that locally the principal symbol of the Ricci operator is \begin{align} [\hat{\sigma}_\mathsf{Ricc}(\omega)h]_{ij}=\frac{1}{2}g^{st}(\omega_s\omega_ih_{jt}+\omega_s\omega_jh_{it}-\omega_s\omega_th_{ij}-\omega_i\omega_jh_{st}). \end{align} It is then easy to show that the Ricci operator is not elliptic since if we set $h_{ij}=\omega_i\omega_j\neq0$, then $\hat{\sigma}_\mathsf{Ricc}(\omega)h=0$.