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".


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.

  • $\begingroup$ Thank you, I think I found some references to the results you are mentioning. For example, Goldstein's book Semigroups of linear operators and applications mentions this operator-valued Laplace transform in his Chapter 0 (equation 0.5). $\endgroup$ – Giuseppe Negro Dec 5 '12 at 20:11
  • $\begingroup$ That's a very good reference. $\endgroup$ – Hans Engler Dec 5 '12 at 20:36
  • $\begingroup$ Your post completely answered the given question. Now I'd like to ask you another little thing, if you don't mind. I tried to apply the present method to the Helmholtz equation $(\Delta + 1)u=0$, but ran into a difficulty. (...) $\endgroup$ – Giuseppe Negro Dec 6 '12 at 16:07
  • $\begingroup$ (...) The method doesn't work, because I am not able to find a solution of the evolution equation which is integrable in time. On the contrary, if I try to do the same with the Bessel equation $(-\Delta +1 )u=0$, everything goes fine. Why is that sign so important, in a semigroup-theoretical sense? $\endgroup$ – Giuseppe Negro Dec 6 '12 at 16:51
  • 1
    $\begingroup$ The reason, in a nutshell, is that $\Delta$ is like a diagonal matrix with all negative entries and therefore $e^{(\Delta - 1)t}$ can be expected to converge. On the other hand, $e^{(\Delta + 1)t}$ may contain contributions that behave like $e^{ct}$ with $c>0$, leading to divergence. All this can also be made rigorous, most easily if you consider the case where the domain $\Omega$ is $\mathbb{R}^n$. In that case you can apply the Fourier transform. The equation $u_t = \Delta u - u$ becomes $\hat u_t = (- |\xi|^2 -1)\hat u$ and $u_t = \Delta u + u$ becomes $\hat u_t = (- |\xi|^2 +1)\hat u$. $\endgroup$ – Hans Engler Dec 6 '12 at 21:44

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