Consider the Laplace operator defined in the biggest possible subset of $L^2(\mathbb{R}^2)$ and let $z \in \mathbb{C}\backslash\mathbb{R}$. Therefore $z \notin \sigma (\Delta)$ the spectrum of $\Delta$, and the resolvent $R=(-\Delta - zI)^{-1}$ is well defined and bounded in all of $L^2(\mathbb{R}^2)$.

I'm trying to find the integral kernel for this operator, that is a function $K(x,y)$ such that almost everywhere in $\mathbb{R}^2$:

$$(Ru)(x)=\int_{\mathbb{R}^2} u(y)K(x,y) dy$$

Let $f := ( - \Delta - z I)^{- 1} u$ and let $\mathcal{F}$ be the Fourier transform. Now

$$ ( - \Delta - z I) f = u \Rightarrow \mathcal{F} ( ( - \Delta - z I) f) =\mathcal{F} ( u) \Rightarrow ( \xi^2 - z) \hat{f} ( \xi) = \hat{u} ( \xi) . $$

Solving for $\hat{f}$ and applying $\mathcal{F}^{- 1}$ we arrive at (modulo some constant depending on your favourite definition of $\mathcal{F}$)

$$ f ( x) =\mathcal{F}^{- 1} \left( \frac{1}{\xi^2 - z} \hat{u} ( \xi) \right) =\mathcal{F}^{- 1} \left( \frac{1}{\xi^2 - z} \right) \ast u ( x) $$

So I need to calculate

$$ \int_{\mathbb{R}^2} \frac{e^{ix \cdot \xi}}{\xi^2 - z} d \xi $$

and I'm completely stuck. Does anybody have any ideas or references? Thanks.

  • 5
    $\begingroup$ If n equal to 1 or 3,then the inverse fourier transform of $(\xi^{2}-z)^{-1}$ is $c_{n}\frac{e^{-\sqrt{z}|x|}}{|x|}$,for other values of dimension,it can be an expression in terms of bessel functions. For instance,it can be found in stein's book "Singular Integrals and Differentiability Properties of Functions" $\endgroup$
    – user23078
    Nov 23, 2012 at 3:40

3 Answers 3


At least when $\Re z < 0$, and assuming that you're taking $\Delta$ to be a negative operator (i.e., $-\Delta \geq 0$), you can write $(s-z)^{-1} = \int_0^\infty e^{zt} e^{-st} dt$, so that by the functional calculus, you should be able to write your resolvent as $Ru = \int_0^\infty e^{zt} e^{\Delta t}u dt$, and hence your integral kernel as $$ K(x,y) = \int_0^\infty e^{zt} K_H(x,y;t) dt, $$ where $$K_H(x,y;t) = \frac{1}{4\pi t}e^{-|x-y|^2/4t}$$ is the heat kernel for $\mathbb{R}^2$. Even if this sketch is complete nonsense, I still suspect that heat kernel methods might nonetheless be helpful, at least in the regime $\Re z < 0$.

EDIT: Helpful perhaps with regards to any estimates you might want--not so helpful with regards to evaluating your perfectly concrete integral...

  • $\begingroup$ Why complete nonsense? Couldn't this be made to work? However, Evans' technique (see my answer) doesn't need the functional calculus and is simpler to justify. $\endgroup$
    – Mike
    Nov 23, 2012 at 23:19
  • $\begingroup$ Not complete nonsense, but certainly requires more care than what you outlined. On the other hand, if you should ever need to carry out these calculations in the curved case... $\endgroup$ Nov 24, 2012 at 3:35

This is a classical computation' going back to the 19th century. The formula for the integral kernel of the resolvent in any dimension is a function of $r = \mbox{dist}(x,y)$. In three dimensions it is particularly simple: $r^{-1} e^{i\lambda r}$ (up to some constant factors) where $\lambda^2 = z$. The formula in any odd dimension is only a bit harder and less explicit, and in even dimensions (including dimension 2) has a slightly different behaviour so that it must be expressed in terms of Bessel functions. This all appears in almost any good book on classical mathematical physics. One good reference with many explicit formulas, is the book by J.C. Nedelec ``Acoustic and Electromagnetic Equations" (Springer). Anyway, you need to write the integral you have in polar coordinates and do some contour-shifting in the $r$-integral.


I finally found the solution in Evans' book on Partial Differential equations, Chapt. 4.3. Using that

$$ \frac{1}{\xi^2 - z} = \int_0^{\infty} e^{- t ( \xi^2 - z)} d t, $$

and substituting into

$$ \int_{\mathbb{R}^2} \frac{e^{ix \cdot \xi}}{\xi^2 - z} d \xi $$

then using Fubini, completing the square in the exponential function and evaluating a complex integral along a line $ \{ Im = const. \} $, one arrives at the solution.

Thanks everybody for the fine answers.


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