What I understand from your question is , what are the compatibility conditions of the zero and first order of a finite difference approximation for a parabolic equation. If it is the question then answer is

From the finite difference scheme, you first try to find truncation error of the scheme by taking its Taylor's series expansion and then satisfying the given parabolic equation exactly in the expansion.Now what ever left in the expansion is error in the scheme and dominating terms in time and space derivatives are nothing but truncation error. For compatible or consistent finite difference scheme TE must go to zero as time step Dt and space step Dx tend to zero.In some cases it does not happen then we say the scheme is not compatible with the given parabolic equation. For example, duFort and Frankel scheme is inconsistent and it is made consistent by taking Dt/Dx also tend to zero(meaning Dt is going to zero faster than Dx). The order of TE is nothing but the orders of dominating terms.

Dear R.C. Mittal

Thank you for your answer I mean the compatibility condition form the theoretical point of view. If the domain of computation is M=[0,L]x[o,T] for one dimensional heat equation for example in relation to proof of uniqueness and existence of the solution.

@Moh Sabah Hussein. Thanks! My answer is related to finite difference approximation.

For the heat equation, u_t=u_xx with initial data u(x,0)=f(x), u(0,t)=g(t) (ignoring the right boundary) the first compatibility condition is clearly f(0)=g(0). Higher order conditions should be possible to derive using the equations. However, compatibility plays a less significant role for parabolic than for hyperbolic problems.

Consider an initial-value problem with a few discontinuities in the initial data at t=0. The solution of the heat equation is still smooth and unique for t>0.

In the same way, if you take an initial-boundary value problem where the initial data is smooth but incompatible with boundary data at t=0, you'd still have a smooth solution for t>0 provided that g(t) is smooth.

Thank you Magnus.

what about the first order compatibility condition it should be f'(0)=g'(0) also satisfied.

No, that cannot be it. g'=f'' should be another.

Thank you

First order compatibility condition applies to Neumann boundary conditions In the case of Dichlet you go directly from zero order to second order. If, just as a simple example, you have: u_t=u_xx in (0,1)x(0,T); u(x,0)=h(x) and boundary conditions u(0,t)=f(t), u(1,t)=g(t), then zero order are h(0)=f(0), h(1)=g(0), if you want second order you must add the extra conditions f'(0)=h''(0), g'(0)=h''(1).

If the boundary conditions are u_x(0,t)=f(t) and u_x(1,t)=g(t) then the first order compatibility conditions are h'(0)=f(0) and h'(1)=g(0).

However these conditions are needed only to have the (corresponding) regularity of the solution up to the lower corners (0,0) and (1,0)

For zero Dirichlet problems associated to parabolic partial equations, defined on a bounded domain, it is important to assume that the problem satisfies the compatibility conditions on the boundary, to guarantee that the problem has a non-negative local classical solution.

Dear Maan A. Rasheed

Thank you.