Compatible PDEs If we have an overdetermined system of pdes what does one have to check to be sure that they are compatible? Suppose each pde is derived from a different Hamiltonian and we have that the Poisson brackets of any pair of these Hamiltonians vanish (i.e. the Hamiltonians are in involution) then how does that information help?

An example I have in mind is:

Suppose we have some function $u=u(t,x)$ -- assume it to be very well-behaved. And some Hamiltonians as functionals of $u$ say $H_1, H_2,...$ such that each pair are in involution. And they lead to PDEs $$u_{t_1}=f_1(x, u, u_x,...)\\ u_{t_2}=f_2(x, u, u_x,...)\\ \vdots$$

 A: Let us consider for simplicity two times, $t_1$ and $t_2$. We can evolve from the point $(0,0)$ in the 2D "time"-space to another point $(T_1,T_2)$ by different means. 


*

*One way, for example, is to use the 1st evolution equation to arrive to the point $(T_1,0)$ and then use the 2nd evolution equation to go from $(T_1,0)$ to $(T_1,T_2)$ (with fixed $T_1$!). 

*But we could also first use the 2nd equation to go to the point $(0,T_2)$ and then the 1st to arrive to $(T_1,T_2)$. Compatibility means that we have to obtain the same result for $u(T_1,T_2,x)$ in the end.


Now, infinitesimally (when $T_1,T_2\rightarrow0$) this amounts to checking the equality of mixed derivatives $u_{t_1t_2}$ calculated in two different ways: as $\partial_{t_2}f_1$ with $u_{t_2}$ computed using the 2nd equation and as $\partial_{t_1}f_2$ with $u_{t_1}$ computed from the 1st equation. When both equations are hamiltonian, it is then sufficient for the Hamiltonians to be in involution (this is an easy consequence of the Jacobi identity for the Poisson bracket).
You can find a very pedagogical exposition of these matters in this book (Section 1.2). I guess it may also be extremely helpful to you in future.
