Numerical Green-Function Integrals and Cutoff Diagnostics

From Green Functions to Numerical Integrals

A Green function can be finite or infinite, and a numerical integral over a grid can obscure the difference. The main difficulty is local: the Fourier integrand for a simple symmetric walk becomes singular at the origin. In two dimensions the singularity is just strong enough to produce logarithmic divergence, so every grid that avoids the origin returns a finite number even though the Green function is infinite. A useful numerical calculation must therefore inspect the singularity before trusting a cutoff or a quadrature output.

Let \((X_n)_{n\ge0}\) be the simple symmetric random walk on \(S=\mathbb{Z}^d\), defined on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), in discrete time \(n\ge0\). From each \(x\in\mathbb{Z}^d\), the walk moves to each of the \(2d\) nearest neighbors with probability \(1/(2d)\); these probabilities sum to \(2d/(2d)=1\). Write \(\mathbb{P}_0\) for the law with \(X_0=0\). Its Green function at the origin is the expected total number of visits to the origin, including time zero: \(G(0)=\sum_{n\ge0}\mathbb{P}_0(X_n=0)\), with the value \(+\infty\) allowed.

For \(\theta=(\theta_1,\ldots,\theta_d)\in[-\pi,\pi]^d\), define the Fourier symbol \(\phi_d(\theta)=d^{-1}\sum_{\ell=1}^d\cos\theta_\ell\). The Fourier integral formula for the Green function, established by Fourier inversion and summing the return probabilities, is the extended-valued identity

The integrand is interpreted away from its singular point, and the integral is the improper integral of a nonnegative function. Since it is even in each coordinate, the same normalized integral is \(\pi^{-d}\) times the integral over \([0,\pi]^d\). That version is convenient for midpoint grids.

Local Radial Behavior and Dimension

Proof. Taylor’s formula for cosine gives \(\cos u=1-u^2/2+O(u^4)\) near zero. Summing over the finitely many coordinates yields \(1-\phi_d(\theta)=r^2/(2d)+O(r^4)\). Factor out \(r^2/(2d)\): the remaining factor is \(1+O(r^2)\), whose reciprocal is \(1+O(r^2)\) for sufficiently small \(r\). Thus \(f_d(\theta)=2d/r^2+O(1)\). In particular, for some positive constants \(c,C,\delta\), \(c/r^2\le f_d(\theta)\le C/r^2\) whenever \(0<r<\delta\). In \(d\) dimensions the radial volume element is a constant times \(r^{d-1}dr\); comparison therefore reduces local integrability to that of \(r^{d-3}dr\). This integral is finite at zero exactly when \(d-3>-1\), or \(d>2\). For \(d=2\), the leading term is \(4/r^2\), and its normalized integral over the annulus is \((2\pi)^{-2}\int_\epsilon^R (4/r^2)(2\pi r)\,dr=(2/\pi)\log(R/\epsilon)\). The \(O(1)\) remainder has bounded integral on the disk, proving the stated logarithmic form. For \(d=1\), the comparison integral is a constant times \(\int_\epsilon^\delta r^{-2}dr=\epsilon^{-1}-\delta^{-1}\), which has order \(1/\epsilon\). Away from the origin the denominator is positive and continuous, so there are no other singularities on the integration domain. This proves the claim. \(\square\)

This result gives a practical rule: assess the small-\(r\) contribution before choosing a grid. In dimension three, the leading contribution of the ball \(r<\epsilon\) is approximately \((2\pi)^{-3}\int_0^\epsilon (6/r^2)(4\pi r^2)\,dr=3\epsilon/\pi^2\), which tends to zero. In dimension two, the corresponding annular contribution grows like \((2/\pi)\log(1/\epsilon)\), rather than disappearing as the grid is refined.

Worked Grid and Cutoff Calculations

How to Use Quadrature Without Trusting a Cutoff Blindly

A midpoint grid approximates an integral by evaluating the integrand at one point in each cell and multiplying by the cell’s volume. For the normalized integral on \([0,\pi]^d\), a grid with \(m\) equal subintervals per coordinate has cell width \(h=\pi/m\), and each cell has normalized weight \(m^{-d}\). If a grid node lands on the singularity, the integrand is undefined there; assigning it a convenient finite value does not define the improper integral. If the grid avoids the singularity, all its sampled values may be finite even when the integral diverges.

For completeness, suppose a region is partitioned into cells of diameter at most \(\sqrt d\,h\), and \(f\) is Lipschitz there with constant \(L\), meaning \(|f(x)-f(y)|\le L|x-y|\). Let \(x_C\) be the midpoint used in cell \(C\). For every \(x\in C\), \(|f(x)-f(x_C)|\le L\sqrt d\,h\). Integrating this inequality over each cell and summing gives \(|\int f-\sum_C f(x_C)|\le L\sqrt d\,h\,\operatorname{Vol}(D)\), where \(D\) is the region being integrated. Thus the error tends to zero with \(h\). This bound applies only where the Lipschitz condition holds; it cannot be applied across the Green-integrand singularity.

In practice, choose a small radius \(R\), approximate the integral outside the ball \(r<R\) by a grid, and treat the inner region separately. In dimensions three and above, the local contribution is finite and shrinks with \(R\); in two dimensions, report the cutoff dependence and the logarithmic term rather than presenting a finite grid value as the Green function. If greater accuracy is needed, the \(O(1)\) remainder in \(f_d(\theta)=2d/r^2+O(1)\) can be integrated numerically after subtracting the singular term on a neighborhood where that subtraction is defined.

Common Mistakes and Interpretation

Common mistake: treating a finite grid sum as evidence that the integral is finite. In two dimensions, the four-node calculation above gives \(3/2\), but this only says that four sampled values are finite. The origin was not sampled, and the missing contribution grows logarithmically as the cutoff shrinks. A finite grid sum is a numerical approximation only when the relevant integral exists and the quadrature assumptions are satisfied.

A second mistake is losing a factor when pairing symmetric grid nodes. In the three-dimensional calculation, replacing each mixed-sign pair sum \(36/17\) by \(18/17\), while keeping the correct all-equal pair contribution \(4\), gives the incorrect average \((4+3(18/17))/8=61/68\). Alternatively, assigning \(2\) to the all-equal pair while retaining the correct mixed-sign pair sums gives the different incorrect average \((2+3(36/17))/8=71/68\). These are two distinct arithmetic errors, not two ways to obtain the same result. Pairing nodes and writing each pair’s two integrand values before averaging helps prevent both.

Finally, distinguish a cutoff integral from the Green function itself. A cutoff is a diagnostic quantity depending on \(\epsilon\); it becomes useful when its dependence is compared with the local prediction. In dimension two, logarithmic growth is the signal. In dimensions three and higher, the local singularity is integrable, so standard quadrature can be used after the singular neighborhood has been controlled. The numerical method is therefore guided by the dimension-dependent analysis, rather than replacing it.