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 '12 at 18:07
  • 2
    $\begingroup$ @Pradip: The word "symbol" has a special meaning in this context. $\endgroup$ – Hans Lundmark Feb 28 '12 at 21:16

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 '12 at 10:41
  • 1
    $\begingroup$ ... and in particular depends on the function you are linearising the operator around. (Generally one linearises around an approximate solution.) $\endgroup$ – Willie Wong Feb 29 '12 at 10:41
  • 1
    $\begingroup$ If your nonlinear PDO is $F:(x,u,\partial u, \partial^2 u) \mapsto 2y$, then its linearisation about a function $v$ is formally $$L_v(x,u,\partial u,\partial^2 u) = \lim_{h\to 0}\frac{1}{h}\left[ F(x,v+hu,\partial v + h\partial u, \partial^2 v + h\partial^2 u) - F(x,v,\partial v,\partial^2 v)\right]$$ $\endgroup$ – Willie Wong Feb 29 '12 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 '12 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 $\hat{\sigma}_\mathsf{Ricc}(\omega)h=0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.