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".


1 Answer 1


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). $\endgroup$ Dec 5, 2012 at 20:11
  • $\begingroup$ That's a very good reference. $\endgroup$ Dec 5, 2012 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. (...) $\endgroup$ Dec 6, 2012 at 16:07
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    $\begingroup$ (...) 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? $\endgroup$ Dec 6, 2012 at 16:51
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    $\begingroup$ 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 $\hat u_t = (- |\xi|^2 +1)\hat u$. $\endgroup$ Dec 6, 2012 at 21:44

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