Integral kernel for the resolvent of the laplace operator - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T20:51:38Z http://mathoverflow.net/feeds/question/114195 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114195/integral-kernel-for-the-resolvent-of-the-laplace-operator Integral kernel for the resolvent of the laplace operator Mike 2012-11-22T23:11:42Z 2012-11-23T23:03:52Z <p>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 $L^2(\mathbb{R}^2)$.</p> <p>I'm trying to find the integral kernel for this operator, that is a function $K(x,y)$ such that almost everywhere in $\mathbb{R}^2$:</p> <p>$$(Ru)(x)=\int_{\mathbb{R}^2} u(y)K(x,y) dy$$</p> <p>Let $f := ( - \Delta - z I)^{- 1} u$ and let $\mathcal{F}$ be the Fourier transform. Now</p> <p>$$( - \Delta - z I) f = u \Rightarrow \mathcal{F} ( ( - \Delta - z I) f) =\mathcal{F} ( u) \Rightarrow ( \xi^2 - z) \hat{f} ( \xi) = \hat{u} ( \xi) .$$</p> <p>Solving for $\hat{f}$ and applying $\mathcal{F}^{- 1}$ we arrive at (modulo some constant depending on your favourite definition of $\mathcal{F}$)</p> <p>$$f ( x) =\mathcal{F}^{- 1} \left( \frac{1}{\xi^2 - z} \hat{u} ( \xi) \right) =\mathcal{F}^{- 1} \left( \frac{1}{\xi^2 - z} \right) \ast u ( x)$$</p> <p>So I need to calculate</p> <p>$$\int_{\mathbb{R}^2} \frac{e^{ix \cdot \xi}}{\xi^2 - z} d \xi$$</p> <p>and I'm completely stuck. Does anybody have any ideas or references? Thanks.</p> http://mathoverflow.net/questions/114195/integral-kernel-for-the-resolvent-of-the-laplace-operator/114202#114202 Answer by Branimir Ćaćić for Integral kernel for the resolvent of the laplace operator Branimir Ćaćić 2012-11-23T01:31:59Z 2012-11-23T08:05:11Z <p>At least when $\Re z &lt; 0$, 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 $\Re z &lt; 0$.</p> <p>EDIT: Helpful perhaps with regards to any estimates you might want--not so helpful with regards to evaluating your perfectly concrete integral...</p> http://mathoverflow.net/questions/114195/integral-kernel-for-the-resolvent-of-the-laplace-operator/114223#114223 Answer by Rafe Mazzeo for Integral kernel for the resolvent of the laplace operator Rafe Mazzeo 2012-11-23T08:44:42Z 2012-11-23T08:44:42Z <p>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. </p> http://mathoverflow.net/questions/114195/integral-kernel-for-the-resolvent-of-the-laplace-operator/114299#114299 Answer by Mike for Integral kernel for the resolvent of the laplace operator Mike 2012-11-23T23:03:52Z 2012-11-23T23:03:52Z <p>I finally found the solution in Evans' book on Partial Differential equations, Chapt. 4.3. Using that</p> <p>$$\frac{1}{\xi^2 - z} = \int_0^{\infty} e^{- t ( \xi^2 - z)} d t,$$</p> <p>and substituting into</p> <p>$$\int_{\mathbb{R}^2} \frac{e^{ix \cdot \xi}}{\xi^2 - z} d \xi$$</p> <p>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.</p> <p>Thanks everybody for the fine answers.</p>