What fails when we try to extend existence and unique for parabolic PDEs for 'PDEs which are 'parabolic in two components''?

What I have in mind is that on $(0,T)\times\Omega$, we have a parabolic pde operator $L$, we have unique solution to

$Lu = f$ when f is Hoelder for some coefficient strictly between 0 and 1.

This is a theorem from Friedman's book on pdes. While i can plough through the proofs which were full of estimates, i don't get the intuitions why they work or why wouldn't they work (Or would they?) for an operator such as

$\bigtriangleup u - \partial_{t_1} - \partial_{t_2} = f$

Where f is a suitable function and the domain is something like $(0,T)\times(0,S)\times\Omega$?

The only answer i found regarding intuition was from User's guide to viscosity solution by crandall-ishii-lions page 52. It says

$u_t+ F(x,u,Du,D^2u)=0$ more or less correspond to $\lambda t+ F = 0$ for large $\lambda$.

I don't quite understand what this means? Does it work if we do $u_t+u_s+F = 0$? What is so nice about having a time evolution that does not (or maybe it does, but i am ignorant) work in higher dimensions?

• The equation you wrote, $$\nabla u + \partial_{t_1} u +\partial_{t_2} u =f$$ is not parabolic, but hyperbolic, in $\tau$ as in Bazin's answer above. So it is simply a linear transport equation, with a parameter $\sigma$ (using again Bazin's notations). Or it is just a typo, and you meant $\Delta u + \partial_{t_1} u +\partial_{t_2} u =f$ ? – username Jan 13 '14 at 9:20
• @AthanagorWurlitzer i mean laplacian... Oh dear – lost1 Jan 13 '14 at 9:24

Change the variables: define $$u(t,s,x)=v(\underbrace{\frac{t+s}{2}}_{\tau},\underbrace{{t-s}}_{\sigma},x).$$ You get $\partial_t u+\partial_s u=\partial_{\tau} v$ and the equation becomes $$\partial_{\tau} v+F(x,v,D_x v,D_x^2 v)=0,$$ where the function $v$ depends on the real parameter $\sigma$.
• Geometrically speaking your parabolic equation is $X+Q$ where $X$ is a real vector field and $Q$ a positive elliptic operator of order 2 (e.g. $-\Delta$). You may note the following equality of vector fields: $$\partial_t+\partial_s=\partial_\tau.$$ – Bazin Jan 13 '14 at 16:39