# Strongly parabolic PDE vs weakly parabolic PDE

In my studies on the Ricci flow, I was faced with a problem. To prove the existence and uniqueness of solutions to the Ricci flow, it is proved that the Ricci flow is a Parabolic PDE type. Then one can find that it is weakly parabolic, so short-time existence does not follow from standard parabolic theory and use the DeTurck trick.

I've sought to understand the difference between strongly parabolic and weakly parabolic equations, But did not get a good result. Please guide me.

Thanks!

-
What is your definition of weakly parabolic? –  timur Apr 4 '13 at 13:03
Could you explain in more detail where the difficulty is? Also, I'm not sure what the difference between weakly and strongly parabolic equations are. I do know that if you use the the DeTurck trick, then the Ricci flow is transformed into a nonlinear parabolic equation that, given a smooth initial Riemannian metric on a closed manifold, has a unique solution. This is explained in a number of places. What book or paper are you using to learn about the Ricci flow? –  Deane Yang Apr 4 '13 at 13:04
I use Hamilton's Ricci Flow by bennett chow. this book is excellent for me. also I have studied "A COMPLETE PROOF OF THE POINCAR´E AND GEOMETRIZATION CONJECTURES – APPLICATION OF THE HAMILTON-PERELMAN THEORY OF THE RICCI FLOW". In these references there exist weakly and strongly parabolic that I don't understand them. –  Sepideh Bakhoda Apr 4 '13 at 13:50
I've looked this up. Indeed, the paper/book by Morgan and Tian call the Ricci flow a "weakly parabolic PDE". The more common term is "degenerate parabolic". Standard PDE theory cannot solve the Ricci flow directly, due to the equation's "gauge invariance" under the action of the group of diffeomorphisms. DeTurck's trick converts the Ricci flow into a strongly parabolic system of PDE's that can be solved using the standard theory of parabolic PDE's. Also, if you are studying the Ricci flow, I suggest just accepting the result on short time existence and focus on more important aspects. –  Deane Yang Apr 4 '13 at 14:20
@Deane: Just to clarify, do you mean by "strongly parabolic" an equation of the form $u_t=-Au$ with $A$ strongly elliptic? –  timur Apr 16 '13 at 14:45

Here is a toy model: Consider a function $u=u(t,x,y,z)$. Then the standard heat equation $\partial_tu=(\partial_x^2+\partial_y^2+\partial_z^2)u$ is strongly parabolic, while e.g. the equation $\partial_tu=(\partial_x^2+\partial_y^2)u$ is only weakly parabolic (more commonly called degenerate parabolic).
Similarly, the Ricci flow it only degenerate parabolic. When you write down the symbol of its linearization you will find some null-directions (this corresponds to the $z$-direction in the above toy example). It is geometrically obvious, that there must be such null-directions, since the equation is invariant under diffeomorphisms. Anyway, after you take care of the diffeomorphisms using DeTurck's trick (how this works is explained in great detail in the books you read) the equation becomes strictly parabolic and you can apply standard theory.