Hypergeometric Expression for the Resolvent of the Discrete Laplacian in Low Dimensions

Abstract

We present an explicit formula for the resolvent of the discrete Laplacian on the square lattice, and compute its asymptotic expansions around thresholds in low dimensions. As a by-product we obtain a closed formula for the fundamental solution to the discrete Laplacian. For the proofs we express the resolvent in a general dimension in terms of the Appell–Lauricella hypergeometric function of type C outside a disk encircling the spectrum. In low dimensions it reduces to a generalized hypergeometric function, for which certain transformation formulas are available for the desired expansions.

Appendices

Comparison with Previous Results

1.1 Result from the Previous Paper [7]

In [7] we computed the singular parts of the resolvent expansions around all the thresholds in all the dimensions. They should coincide with the results of this paper. For a quick comparison here we review the results from [7] according to the present setting only for the low dimensions.

Theorem A.1

Let $d=1$. There exists ${\mathcal {E}}_0(z,n)$ analytic in $|z|<4$ such that for any $z\in {\mathbb {C}}\setminus [0,4]$ with $|z|<4$ and $n\in {\mathbb {Z}}$

$$\begin{aligned} G(z,n) = {\mathcal {E}}_0(z,n)+\frac{1}{2\sqrt{-z}}F^{(1)}_B\biggl (\genfrac{}{}{0.0pt}{}{\frac{1}{2}+n;\frac{1}{2}-n}{\frac{1}{2}};\frac{z}{4}\biggr ). \end{aligned}$$

Similarly, there exists ${\mathcal {E}}_1(z,n)$ analytic in $|z-4|<4$ such that for any $z\in {\mathbb {C}}\setminus [0,4]$ with $|z-4|<4$ and $n\in {\mathbb {Z}}$

$$\begin{aligned} G(z,n) = {\mathcal {E}}_1(z,n)+\frac{(-1)^{n+1}}{2\sqrt{z-4}} F^{(1)}_B\biggl (\genfrac{}{}{0.0pt}{}{\frac{1}{2}+n;\frac{1}{2}-n}{\frac{1}{2}};\frac{4-z}{4}\biggr ). \end{aligned}$$

Theorem A.2

Let $d=2$. There exists ${\mathcal {E}}_0(z,n)$ analytic in $|z|<4$ such that for any $z\in {\mathbb {C}}\setminus [0,8]$ with $|z|<4$ and $n\in {\mathbb {Z}}^2$

$$\begin{aligned} G(z,n)&= {\mathcal {E}}_0(z,n) \\&\quad -\,\frac{1}{4\pi }[\log (-z)] F^{(2)}_B\biggl (\genfrac{}{}{0.0pt}{}{\frac{1}{2}+n_1,\frac{1}{2}+n_2;\frac{1}{2}-n_1,\frac{1}{2}-n_2}{1}; \frac{z}{4},\frac{z}{4}\biggr ). \end{aligned}$$

There exists ${\mathcal {E}}_1(z,n)$ analytic in $|z-4|<4$ such that for any $|z-4|<4$ with $\mathop {\mathrm {Im}}z>0$ and $n\in {\mathbb {Z}}^2$

$$\begin{aligned} G(z,n)&= {\mathcal {E}}_1(z,n) -\frac{\mathrm {i}}{8\pi }\Bigl [\log \Bigl (-\frac{(z-4)^2}{16}\Bigr )\Bigr ] \\&\quad \times \Bigl [ (-1)^{n_2}F^{(2)}_B \biggl (\genfrac{}{}{0.0pt}{}{\frac{1}{2}+n_1,\frac{1}{2}+n_2;\frac{1}{2}-n_1,\frac{1}{2}-n_2}{1};\frac{z-4}{4},\frac{4-z}{4}\biggr ) \\&\qquad +\, (-1)^{n_1}F^{(2)}_B\biggl (\genfrac{}{}{0.0pt}{}{\frac{1}{2}+n_1,\frac{1}{2}+n_2;\frac{1}{2}-n_1,\frac{1}{2}-n_2}{1};\frac{4-z}{4},\frac{z-4}{4}\biggr ) \Bigr ]. \end{aligned}$$

There exists ${\mathcal {E}}_2(z,n)$ analytic in $|z-8|<4$ such that for any $z\in {\mathbb {C}}\setminus [0,8]$ with $|z-8|<4$ and $n\in {\mathbb {Z}}^2$

$$\begin{aligned} G(z,n)&= {\mathcal {E}}_2(z,n) +\frac{(-1)^{|n|}}{4\pi }[\log (z-8)] \\&\quad \times F^{(2)}_B\biggl (\genfrac{}{}{0.0pt}{}{\frac{1}{2}+n_1,\frac{1}{2}+n_2;\frac{1}{2}-n_1,\frac{1}{2}-n_2}{1};\frac{8-z}{4},\frac{8-z}{4}\biggr ). \end{aligned}$$

1.2 Comparison of the Singular Parts

For $d=1$ the singular parts of Theorem 2.2 and Theorem A.1 clearly coincide due to (2.1). Hence we examine only the case $d=2$. We begin with the embedded threshold $z=4$.

Corollary A.3

The singular parts from Theorems 2.3 and A.2 around $z=4$ coincide, which is equivalent to the following relations: For all $k\in {\mathbb {N}}_0$ and $n\in {\mathbb {Z}}^2$

$$\begin{aligned}&\sum _{|\alpha |=2k} \frac{(-1)^{\alpha _1}}{\alpha _1!\alpha _2!} \Bigl (\frac{1}{2}+n_1\Bigr )_{\alpha _1} \Bigl (\frac{1}{2}+n_2\Bigr )_{\alpha _2} \Bigl (\frac{1}{2}-n_1\Bigr )_{\alpha _1} \Bigl (\frac{1}{2}-n_2\Bigr )_{\alpha _2} \\&= \frac{4^{2k}}{(2k)!}\Bigl (\frac{1+n_1+n_2}{2}\Bigr )_k \Bigl (\frac{1+n_1-n_2}{2}\Bigr )_k \Bigl (\frac{1-n_1+n_2}{2}\Bigr )_k \Bigl (\frac{1-n_1-n_2}{2}\Bigr )_k \end{aligned}$$

and

$$\begin{aligned}&\sum _{|\alpha |=2k+1} \frac{(-1)^{\alpha _1}}{\alpha _1!\alpha _2!}\Bigl (\frac{1}{2}+n_1\Bigr )_{\alpha _1} \Bigl (\frac{1}{2}+n_2\Bigr )_{\alpha _2} \Bigl (\frac{1}{2}-n_1\Bigr )_{\alpha _1} \Bigl (\frac{1}{2}-n_2\Bigr )_{\alpha _2} \\&= \frac{4^{2k+1}}{(2k+1)!}\Bigl (\frac{n_1+n_2}{2}\Bigr )_{k+1} \Bigl (\frac{n_1-n_2}{2}\Bigr )_{k+1} \Bigl (\frac{2-n_1+n_2}{2}\Bigr )_k \Bigl (\frac{2-n_1-n_2}{2}\Bigr )_k . \end{aligned}$$

Proof

Since Theorems 2.3 and A.2 provide asymptotic expansions of the same function, it is clear that their singular parts have to coincide.

Let us verify that this is equivalent to the asserted identities. For that we set

$$\begin{aligned}&s_j[n]=\\&\frac{1}{8} \Bigl [ (-1)^{n_2} \sum _{|\alpha |=j} \frac{ (-1)^{\alpha _2}\left( \frac{1}{2}+n_1\right) _{\alpha _1} \left( \frac{1}{2}+n_2\right) _{\alpha _2} \left( \frac{1}{2}-n_1\right) _{\alpha _1} \left( \frac{1}{2}-n_2\right) _{\alpha _2} }{j!\alpha _1!\alpha _2!} \\&\qquad +\, (-1)^{n_1} \sum _{|\alpha |=j} \frac{ (-1)^{\alpha _1}\left( \frac{1}{2}+n_1\right) _{\alpha _1} \left( \frac{1}{2}+n_2\right) _{\alpha _2} \left( \frac{1}{2}-n_1\right) _{\alpha _1} \left( \frac{1}{2}-n_2\right) _{\alpha _2} }{j!\alpha _1!\alpha _2!} \Bigr ], \end{aligned}$$

which are the coefficients of the singular part from Theorem A.2. First let $j=2k$ be even, and then

$$\begin{aligned} s_{2k}[n]&= \frac{(-1)^{n_1}+(-1)^{n_2}}{8(2k)!} \\&\quad \times \sum _{|\alpha |=2k} \frac{ (-1)^{\alpha _1} }{\alpha _1!\alpha _2!}\Bigl (\frac{1}{2}+n_1\Bigr )_{\alpha _1} \Bigl (\frac{1}{2}+n_2\Bigr )_{\alpha _2} \Bigl (\frac{1}{2}-n_1\Bigr )_{\alpha _1} \Bigl (\frac{1}{2}-n_2\Bigr )_{\alpha _2} . \end{aligned}$$

Then the coincidence of the singular parts implies the first identity of the assertion for |n| even, and hence also for |n| odd. Next let $j=2k+1$ be odd, and we have

$$\begin{aligned} s_{2k+1}[n]&= \frac{(-1)^{n_1}-(-1)^{n_2}}{8(2k+1)!} \\&\quad \times \sum _{|\alpha |=2k+1} \frac{(-1)^{\alpha _1}}{\alpha _1!\alpha _2!} \Bigl (\frac{1}{2}+n_1\Bigr )_{\alpha _1} \Bigl (\frac{1}{2}+n_2\Bigr )_{\alpha _2} \Bigl (\frac{1}{2}-n_1\Bigr )_{\alpha _1} \Bigl (\frac{1}{2}-n_2\Bigr )_{\alpha _2} . \end{aligned}$$

Similarly to the above, the coincidence of the singular parts implies the first asserted identity for |n| odd, and hence also for |n| even. The converse is clear from the above argument, and we are done. $\square$

Remark

We do not try to give direct proofs of the identities from Corollary A.3. The identities in Corollary A.3 have been verified using the computer algebra system Maple.

Similarly we discuss the endpoint thresholds $z=0,8$ as follows.

Corollary A.4

The singular parts from Theorems 2.4 and A.2 around $z=0,8$ coincide, i.e., for all $|w|<1$ and $(m,l)\in {\mathbb {Z}}^2$

$$\begin{aligned}&(-1)^{m+l} F^{(2)}_B\biggl (\genfrac{}{}{0.0pt}{}{\frac{1}{2}+m+l,\frac{1}{2}+m-l;\frac{1}{2}-m-l,\frac{1}{2}-m+l}{1};w,w\biggr ) \\&= {_2F_1}\biggl (\genfrac{}{}{0.0pt}{}{\frac{1}{2},\frac{1}{2}}{1};w(2-w)\biggr ) {_4F_3} \biggl (\genfrac{}{}{0.0pt}{}{m,-m,l,-l}{1,\frac{1}{2},\frac{1}{2}};(w-1)^2\biggr ) \\&\quad +\,\sum _{\mu =1}^{|m|} (-1)^\mu \biggl [{_2F_1}\biggl (\genfrac{}{}{0.0pt}{}{\frac{1}{2}+\mu ,\frac{1}{2}-\mu }{1};w(2-w)\biggr ) \\&\qquad \qquad +\,{_2F_1}\biggl (\genfrac{}{}{0.0pt}{}{\mu -\frac{1}{2},\frac{3}{2}-\mu }{1};w(2-w)\biggr )\biggr ] \\&\qquad \times {_4F_3}\biggl (\genfrac{}{}{0.0pt}{}{1+|m|-\mu ,\mu -|m|,l,-l}{1,\frac{1}{2},\frac{1}{2}};(w-1)^2\biggr ) \\&\quad +\,\sum _{\nu =1}^{|l|} (-1)^\nu \biggl [ {_2F_1} \biggl (\genfrac{}{}{0.0pt}{}{\frac{1}{2}+\nu ,\frac{1}{2}-\nu }{1};w(2-w)\biggr ) \\&\qquad \qquad +\,{_2F_1}\biggl (\genfrac{}{}{0.0pt}{}{\nu -\frac{1}{2},\frac{3}{2}-\nu }{1};w(2-w) \biggr )\biggl ] \\&\qquad \times {_4F_3}\biggl (\genfrac{}{}{0.0pt}{}{m,-m,1+|l|-\nu ,\nu -|l|}{1,\frac{1}{2},\frac{1}{2}};(w-1)^2\biggr ) , \end{aligned}$$

and for all $|w|<1$ and $(m,l)\in {\mathbb {N}}_0^2$

$$\begin{aligned}&(-1)^{m+l} F^{(2)}_B\biggl (\genfrac{}{}{0.0pt}{}{\frac{3}{2}+m+l,\frac{1}{2}+m-l;-\frac{1}{2}-m-l,\frac{1}{2}-m+l}{1};w,w\biggr ) \\&={} (2m+1)(2l+1)(w-1) {_2F_1}\biggl (\genfrac{}{}{0.0pt}{}{\frac{1}{2},\frac{1}{2}}{1};w(2-w)\biggr ) \\&\qquad \times {_4F_3}\biggl (\genfrac{}{}{0.0pt}{}{1+m,-m,1+l,-l}{1,\frac{3}{2},\frac{3}{2}}; (w-1)^2\biggr ) \\&\quad -\, (2l+1)(w-1) \sum _{\mu =-m}^{m}(-1)^\mu {_2F_1}\biggl (\genfrac{}{}{0.0pt}{}{\frac{1}{2}+\mu ,\frac{1}{2}-\mu }{1};w(2-w)\biggr ) \\&\qquad \times {_4F_3}\biggl (\genfrac{}{}{0.0pt}{}{1+m-|\mu |,|\mu |-m,1+l,-l}{1,\frac{1}{2},\frac{3}{2}}; (w-1)^2\biggr ) \\&\quad -\, (2m+1)(w-1) \sum _{\nu =-l}^{l}(-1)^\nu {_2F_1}\biggl (\genfrac{}{}{0.0pt}{}{\frac{1}{2}+\nu ,\frac{1}{2}-\nu }{1};w(2-w)\biggr ) \\&\qquad \times {_4F_3}\biggl (\genfrac{}{}{0.0pt}{}{1+m,-m,1+l-|\nu |,|\nu |-l}{1,\frac{1}{2},\frac{3}{2}}; (w-1)^2\biggr ) . \end{aligned}$$

Proof

We can prove the assertion similarly to Corollary A.3. However, it requires fairly long computations, and we omit it. $\square$

Remark

In contrast to Corollary A.3 we have no idea how we could possibly directly prove the identities from Corollary A.4. However, they can be verified using the computer algebra system Maple, in this case for fixed but arbitrary values of m and l.

Renormalized Expectation of Random Work

Proof of (1.3)

It is elementary to compute $P(X_k=n)$. In fact, we have

$$\begin{aligned} P(X_k=n) =\frac{1}{(2d)^k}\sum _{|\alpha |=(k-|n|)/2}\frac{k!}{\alpha !\prod _{j=1}^d\bigl [(\alpha _j+|n_j|)!\bigr ]} \end{aligned}$$

if $k\ge |n|$ and $k-|n|$ is even, and

$$\begin{aligned} P(X_k=n)=0 \end{aligned}$$

otherwise. Then we obtain

$$\begin{aligned} \sum _{k=0}^\infty \Bigl (\frac{2d}{2d-z}\Bigr )^k&P(X_k=n) \\&= \sum _{\genfrac{}{}{0.0pt}{}{k\ge |n|,}{k-|n|\text {:even}}} \sum _{|\alpha |=(k-|n|)/2} \frac{k!}{\alpha !\prod _{j=1}^d\bigl [(\alpha _j+|n_j|)!\bigr ]} (2d-z)^{-k} \\&= \sum _{\alpha \in {\mathbb {N}}_0^d} \frac{(2|\alpha |+|n|)!}{\alpha !\prod _{j=1}^d\bigl [(\alpha _j+|n_j|)!\bigr ]} (2d-z)^{-2|\alpha |-|n|}, \end{aligned}$$

and hence (1.3) follows by Theorem 2.1. $\square$