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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.

QuestionIs there some truth in the above reasoning? Can it be formalized somehow?

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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”.

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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.

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