Special functions for heat kernel expansion

Abstract

In this paper, we study an asymptotic expansion of the heat kernel for a Laplace operator on a smooth Riemannian manifold without a boundary at enough small values of the proper time. The Seeley–DeWitt coefficients of this decomposition satisfy a set of recurrence relations, which we use to construct two function families of a special kind. Using these functions, we study the expansion of a local heat kernel for the inverse Laplace operator. We show that the new functions have some important properties. For example, we can consider the Laplace operator on the function set as a shift one. Also, we describe various applications useful in theoretical physics and, in particular, we find a decomposition of local Green’s functions near the diagonal in terms of new functions.

Appendix A

Lemma 11

The integral (85) from Theorem 2 has the following series expansion near zero for $m=0$

$$\begin{aligned} T(\omega ,s)&= 2\int _{{\mathbb {R}}_+} d\rho \,\rho J_0(\rho \sqrt{2\omega })\left( e^{-s/\rho ^2}-1\right) \nonumber \\&=\sum _{k=1}^{+\infty } \frac{s^k[\ln (s\omega /2) -2H_{k-1}-H_{k} +3\gamma ]}{(k-1)!(k-1)!k!}(\omega /2)^{k-1}. \end{aligned}$$

(117)

Proof

First of all we show that the integral satisfies the following differential equation $2\partial _\omega ^{} s\partial _s^2T(\omega ,s)=T(\omega ,s)$. This can be achieved by explicit differentiation, integration by parts, and using the properties of the Bessel functions

$$\begin{aligned} \frac{d}{dx}J_0(x)=-J_1(x),\,\,\,\,\,\,\frac{1}{x}J_1(x)+\frac{d}{dx}J_1(x)=J_0(x). \end{aligned}$$

(118)

According to the above mentioned, let us take an ansatz for (117) in the form

$$\begin{aligned} T(\omega ,s)= g_0(\omega )+\sum _{k=1}^\infty \left( s^k\ln (s)f_k(\omega )+s^k g_k(\omega )\right) , \end{aligned}$$

(119)

where the coefficients $f_k(\omega )$ and $g_k(\omega )$ should be found. Let us apply the operator $2\partial _\omega ^{} s\partial _s^2$ to (119). Then we get the recurrent relations

$$\begin{aligned} {\left\{ \begin{array}{ll} 2\partial _\omega f_1(\omega )=g_0(\omega ); \\ 2k(k+1)\partial _\omega f_{k+1}(\omega )=f_k(\omega ); \\ 2(2k+1)\partial _\omega f_{k+1}(\omega )-2k(k+1)\partial _\omega g_{k+1}(\omega )=g_k(\omega ). \end{array}\right. } \end{aligned}$$

(120)

where $k\geqslant 1$.

To solve them, we have to find the initial conditions, using the integral representation (117). One can note that $T(\omega ,0)=0$, so $g_0(\omega )=0$. It is evident, that we can explicitly integrate the equations from (120). However, we have such arbitrariness as integration constants $\{f_k(0)\}_{k\geqslant 1}$ and $\{g_k(0)\}_{k\geqslant 2}$. Let us find them, using the asymptotic behavior of the construction (85) for quite small values of $\omega$. For those purposes we cut ${\mathbb {R}}_+$ in two intervals [0; 1] and $[1;+\infty )$. So we get

$$\begin{aligned} 2\int _0^1d\rho \, \rho J_0(\rho \sqrt{2\omega })\left( e^{-s/\rho ^2}-1\right)&= \int _0^1 d\rho \, \left( e^{-s/\rho }-1\right) +o(1)\\\nonumber &=s\ln (s)+s\omega -s-\sum _{k=2}^{+\infty } \frac{(-s)^k}{k!(k-1)}+o(1),\end{aligned}$$

(121)

and

$$\begin{aligned} 2\int _1^{+\infty } d\rho \,\rho J_0(\rho \sqrt{2\omega })\left( e^{-s/\rho ^2}-1\pm \frac{s}{\rho ^2}\right) =\sum _{k=2}^{+\infty } \frac{(-s)^k}{k!(k-1)}+s\ln (\omega )+s\left( 2\gamma -\ln (2)\right) +o(1). \end{aligned}$$

(122)

Using the last calculations and the form of ansatz (119), we get

$$\begin{aligned} \sum _{k=1}^{+\infty } s^k\ln (s) f_k(0)+\sum _{k=2}^{+\infty }s^kg_k(0)=s \ln (s). \end{aligned}$$

(123)

Therefore, $f_1(0)=1$ and $f_k(0)=g_k(0)=0$ for $k\geqslant 2$. Finally, we need to find the coefficient $g_1(\omega )$, that corresponds to s. This can be achieved by subtracting the logarithmic part $s\ln s$ and differentiating by the parameter s with a further transition $s\rightarrow 0$. So we get

$$\begin{aligned} \partial _s\big |_{s=0}\bigg [2\int _{{\mathbb {R}}_+} d\rho \, \rho J_0(\rho \sqrt{2\omega })\left( e^{-s/\rho ^2}-1\right) -s\ln (s)\bigg ]=&-2\int _1^{+\infty } \frac{d\rho }{\rho }J_0(\rho \sqrt{2\omega })\nonumber \\&-2\int _0^1\frac{d\rho }{\rho }\left( J_0(\rho \sqrt{2\omega })-1\right) +\gamma -1\\ \nonumber =&\ln (\omega )+3\gamma -\ln 2-1, \end{aligned}$$

(124)

where in the second equality we have used the change of variable $\rho \rightarrow \rho /\sqrt{2\omega }$. This means that $g_1(\omega )=\ln (\omega )+3\gamma -\ln (2)-1$. Solving the recurrent relations, we find

$$\begin{aligned} f_k(\omega )=\frac{(\omega /2)^{k-1}}{k!(k-1)!(k-1)!},\,\,\, g_k(\omega )=\frac{\ln (\omega )-2H_{k-1}-H_k+3\gamma -\ln 2}{k!(k-1)!(k-1)!}(\omega /2)^{k-1}, \end{aligned}$$

(125)

which leads to the statement of the lemma. $\hfill\square$