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:


  • $\begingroup$ 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 $v$... $\endgroup$
    – zapkm
    Feb 28, 2012 at 18:07
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    $\begingroup$ @Pradip: The word "symbol" has a special meaning in this context. $\endgroup$ Feb 28, 2012 at 21:16

2 Answers 2


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

  • $\begingroup$ Could you please be more specific about what you mean by linearization. If you can, give an example. Thank you! $\endgroup$
    – chango
    Feb 29, 2012 at 10:41
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    $\begingroup$ ... and in particular depends on the function you are linearising the operator around. (Generally one linearises around an approximate solution.) $\endgroup$ Feb 29, 2012 at 10:41
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    $\begingroup$ If your nonlinear PDO is $F:(x,u,\partial u, \partial^2 u) \mapsto 2y$, then its linearisation about a function $v$ is formally $$L_v(x,u,\partial u,\partial^2 u) = \lim_{h\to 0}\frac{1}{h}\left[ F(x,v+hu,\partial v + h\partial u, \partial^2 v + h\partial^2 u) - F(x,v,\partial v,\partial^2 v)\right]$$ $\endgroup$ Feb 29, 2012 at 10:50
  • $\begingroup$ 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 $\frac{\partial g}{\partial t}=-2\operatorname{Ric}g$. $\endgroup$
    – youler
    Feb 29, 2012 at 21:17

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 $\hat{\sigma}_\mathsf{Ricc}(\omega)h=0$.


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