Consider, the heat equation on Euclidean space $$\partial_t u - \Delta u = 0, $$ where $u : [0, \infty) \times \mathbb{R}^n \to \mathbb{R}^n$ is a function.
I could phrase the following question in a more general context, where the Laplace operator is replaced by an elliptic operator of even order, but I will keep it in the context of the heat equation for simplicity (and I assume that an answer would generalize). However, I guess that for high order elliptic operators, solutions would be more regular and hence my question would have an easier answer.
This equation can be understood in a "classical sense", that is, the function $u$ is $C^1$ in $t$ and for every $t \geq 0$, we have $u(t, \cdot) \in H^2(\mathbb{R}^n)$ (or, for even more simplicity, we can assume that $u(t, \cdot)$ is $C^2$).
However, there is also another "interpretation" of the heat equation, that of considering a curve $$\gamma : [0, \infty) \to H^2(\mathbb{R}^n) \subset L^2(\mathbb{R}^n) $$ which is differentiable (as a vector-valued map) and which satisfies the "vector-valued ODE" $$\gamma'(t) = \Delta (\gamma(t)) $$ for all $t \geq 0$. Semigroup theory shows that this ODE can have a unique solution $$\gamma \in C^1([0, \infty), H^2(\mathbb{R}^n)). $$ However, with this interpretation, we may then define the function $$u : [0, \infty) \times \mathbb{R^n} \to \mathbb{R^n}, \quad u(t, x) := \gamma(t)(x). $$ To what extent does this function $u$ solve the heat equation in the "classical" sense as above. More precisely, can we conclude that $u$ is $C^1$ (or at least differentiable) in $t$?
I am not exactly sure how to relate these two concepts of "differentiability", one of them being pointwise differentiability of a function and the other being $H^2$-differentiability of a vector-valued curve. If, instead, one would consider a $C^1$ curve $$ \gamma : [0, \infty) \to C^2(\mathbb{R}^n), $$ then it is pretty clear that the function $(t, x) \mapsto \gamma(t)x$ is $C^1$ is $t$ and $C^2$ in $x$. But, in the PDE situation above, the $H^2$-derivative is weaker than the $C^2$-derivative (at least on bounded domains).
Edit.
I could also ask the converse question as well: if $u$ is $C^1$ in $t$, $C^2$ in $x$, and solves the heat equation "classically", then does $u$ solve the heat equation as an $H^2$-valued curve? More precisely, does classical differentiability imply $H^2$-differentiability on $\mathbb{R}^n$ (on bounded domains this is easier to answer)?
The general theory of semigroups gives you, for any Cauchy problem, that
For any generator $A$ in a Banach space $X$: $u_0\in D(A) \Rightarrow u\in C(\mathbb R^+;D(A))\cap C^1(\mathbb R^+;X)$.
For an analytic semigroup generator $A$: $u_0\in X \Rightarrow u\in C((0,\infty);D(A))\cap C^1((0,\infty);X)$.
The heat equation is a particular case of 2, while the wave and Schrodinger equations belong to 1, when they are written in the abstract form.