Here is a heuristic reasoning.
Suppose that the function $u(x, t)$ solves
$$\partial_t u = \Delta u.$$
Integrating in $t$ we can define a new function $v$:
$$v(x)=\int_0^\infty u(x, t)\, dt.$$
Applying the operator $-\Delta$ to $v$ we get
$$-\Delta v(x)=\int_0^\infty -\partial_t u (x, t)\, dt = u(x, 0).$$
In particular, if $u_0=\delta$, that is if $u(x, t)$ is a fundamental solution for the heat equation, then $v$ is a fundamental solution for the Laplace equation.
Question Is there some truth in the above reasoning? Can it be formalized somehow?
Thank you.
EDIT: I asked the owner of the local course in PDE. He replied that there is some truth in this and suggested to look for the keywords "subordination principle".
Best Answer
This is indeed correct and can be made rigorous, assuming that the integral converges sufficiently well for all $u_0$, which in turn depends on the boundary conditions that are imposed for the Laplacian.
Assume that $\int_0^\infty \Vert u(\cdot,t) \Vert dt < \infty$ for all $u_0$, for a suitable norm (e.g. the $L^2$ norm). By a theorem of Datko and Pazy, this implies that the spectrum of $\Delta$ is contained in the left half plane and bounded away from the imaginary axis. Now write formally $A = \Delta$ and $u(\cdot,t) = e^{At}u_0$. You are then computing $$ \int_0^\infty e^{At} u_0 dt = (-A)^{-1} u_0 = (-\Delta)^{-1} u_0 \, . $$ More generally, for $\lambda$ in a suitable right half plane,
$$ \int_0^\infty e^{At} e^{-\lambda t} dt = (\lambda I - A )^{-1} $$ that is, Laplace transforms of the operator semigroup $\left( e^{At} \right)_{t \ge 0}$ are resolvents of the generator $A$ of the semigroup.
All this can be made rigorous using semigroup theory.