Consider the following Cauchy problem 
$$u\_t - \Delta u + f(x,u,u\_x) = 0, \quad (t,x) \in (0,T) \times \mathbb{R}^n,$$
with initial condition $u(0,x) = g(x), \ x \in \mathbb{R}^n.$


Suppose that $u \in C^2((0,T) \times \mathbb{R}^n)$ is a solution of the initial value problem. 


Can we obtain (under some reasonable assumptions) that 
$$u\_t(0,x) - \Delta u(0,x) + f(x,u,u\_x)= 0,$$
that is, that the solution $u$ solves the PDE at time $t=0$ too?



Solution to semilinear heat equation at $t=0$: $u_t(0,x) - \Delta u(0,x) + f(x,u,u_x)= 0 \ ?$
The question is vague, because the OP does not mention in which sense the PDE is to be satisfied at initial time.


In addition, the assumption that $u$ is $C^2$ for $T>0$ is very weak, if not useless. For let me assume that $f$ is a smooth function with $f(\cdot,0,0)\equiv0$, and that the initial data $g$ is $L^1\cap L^\infty({\mathbb R}^n)$. Then it is classical, because of the smoothing property of the heat semi-group, that $u$ is actually $C^\infty$ for $t>0$ ; hence the assumption is automatically fulfilled. Therefore the question asks whether, if $g\in L^1\cap L^\infty({\mathbb R}^n)$, one has
$$u\_t(0)=\Delta g-f(\cdot,g,\nabla g).$$
certainly, this does not hold pointwisely, since the right-hand side may not be a function. It holds in the distributional sense ; just pass to the limit (in ${\cal D}'$) in the equation satisfied at $\{s\}\times{\mathbb R}^n$, as $s\rightarrow0+$. But then $u\_t(0)$ is just defined as the limit (in ${\cal D}'$) of $u\_t(s)$ as $s\rightarrow0+$, and this is a tautology. The answer will be better if we can prove that the distributional limit of $\frac{u(s)-g}s$ equals $\Delta g-f(\cdot,g,\nabla g)$. I guess that this can be done by the usual techniques of semi-groups.


