Compact Numerical Quadrature Formulas for Hypersingular Integrals and Integral Equations

Abstract

In the first part of this work, we derive compact numerical quadrature formulas for finite-range integrals $I[f]=\int^{b}_{a}f(x)\,dx$, where f(x)=g(x)|x−t|^β, β being real. Depending on the value of β, these integrals are defined either in the regular sense or in the sense of Hadamard finite part. Assuming that g∈C ^∞[a,b], or g∈C ^∞(a,b) but can have arbitrary algebraic singularities at x=a and/or x=b, and letting h=(b−a)/n, n an integer, we derive asymptotic expansions for ${T}^{*}_{n}[f]=h\sum_{1\leq j\leq n-1,\ x_{j}\neq t}f(x_{j})$, where x _j =a+jh and t∈{x ₁,…,x _n−1}. These asymptotic expansions are based on some recent generalizations of the Euler–Maclaurin expansion due to the author (A. Sidi, Euler–Maclaurin expansions for integrals with arbitrary algebraic endpoint singularities, in Math. Comput., 2012), and are used to construct our quadrature formulas, whose accuracies are then increased at will by applying to them the Richardson extrapolation process. We pay particular attention to the case in which β=−2 and f(x) is T-periodic with T=b−a and $f\in C^{\infty}(-\infty,\infty)\setminus\{t+kT\}^{\infty}_{k=-\infty}$, which arises in the context of periodic hypersingular integral equations. For this case, we propose the remarkably simple and compact quadrature formula $\widehat{Q}_{n}[f]=h\sum^{n}_{j=1}f(t+jh-h/2)-\pi^{2} g(t)h^{-1}$, and show that $\widehat{Q}_{n}[f]-I[f]=O(h^{\mu})$ as h→0 ∀μ>0, and that it is exact for a class of singular integrals involving trigonometric polynomials of degree at most n. We show how $\widehat{Q}_{n}[f]$ can be used for solving hypersingular integral equations in an efficient manner. In the second part of this work, we derive the Euler–Maclaurin expansion for integrals $I[f]=\int^{b}_{a} f(x)dx$, where f(x)=g(x)(x−t)^β, with g(x) as before and β=−1,−3,−5,…, from which suitable quadrature formulas can be obtained. We revisit the case of β=−1, for which the known quadrature formula $\widetilde{Q}_{n}[f]=h\sum^{n}_{j=1}f(t+jh-h/2)$ satisfies $\widetilde{Q}_{n}[f]-I[f]=O(h^{\mu})$ as h→0 ∀μ>0, when f(x) is T-periodic with T=b−a and $f\in C^{\infty}(-\infty,\infty)\setminus\{t+kT\}^{\infty}_{k=-\infty}$. We show that this formula too is exact for a class of singular integrals involving trigonometric polynomials of degree at most n−1. We provide numerical examples involving periodic integrands that confirm the theoretical results.

Appendices

Appendix A: Proof of Theorem 5.1

Proof of Part 1

We start by noting that

Making the variable transformation y=2π(x−t)/T in the integral inside the square brackets, and using the fact that the integrand is T-periodic, we obtain

$$ I[f_m]=\frac{T}{2\pi}L_{m} e^{\mathrm{i}\,2m\pi t/T}; \quad L_{m}=\int^{\pi}_{-\pi}\frac{e^{\mathrm{i}my}}{\sin^2(\frac{1}{2}y)}\,dy. $$

(A.1)

Next,

and since $\int^{\pi}_{-\pi}\frac{\sin(my)}{\sin^{2}(\frac{1}{2}y)}\,dy=0$ due to its integrand being odd, it follows that

$$ L_{m}=\int^{\pi}_{-\pi}\frac{\cos(my)}{\sin^2(\frac{1}{2}y)}\,dy \quad \Rightarrow \quad L_{-m}=L_m. $$

(A.2)

Thus, it suffices to treat only the case m=0,1,…. Now,

which, upon dividing both sides by $\sin^{2}(\frac{1}{2}y)$, gives the identity

$$ \frac{\cos[(m+1)y]}{\sin^2(\frac{1}{2}y)}-2\frac{\cos(my)}{\sin^2(\frac{1}{2}y)} +\frac{\cos[(m-1)y]}{\sin^2(\frac{1}{2}y)}=-4\cos(my). $$

(A.3)

Integrating both sides of this identity over [−π,π], and invoking (A.2), we obtain

$$ L_{m+1}-2L_{m}+L_{m-1}=-4\int^\pi_{-\pi}\cos(my)\,dy. $$

(A.4)

Upon setting m=0, and invoking the fact that L ₋₁=L ₁, (A.4) gives

$$ 2L_1-2L_0=-8\pi\quad \Rightarrow \quad L_1=L_0-4\pi, $$

(A.5)

while for m≥1, because $\int^{\pi}_{-\pi}\cos(my)\,dy=0$, (A.4) gives the recurrence relation

whose general solution is of the form L _m =Am+B. We can determine A and B by invoking the values of L ₀ and L ₁. Now, applying to L ₀ the known result

we obtain

As for L ₁, by (A.5), we have

Consequently, A=−4π and B=0. Hence, also by the fact that L _−m =L _m ,

$$ L_{m}=-4\pi|m|, \quad m=0,\pm1,\pm2,\ldots . $$

(A.6)

Substituting this in (A.1), we obtain (5.2).

Proof of Part 2

To prove (5.3) and (5.4), we proceed similarly. First, by (4.5),

(A.7)

(A.8)

Next,

By the fact that

and since $\sum^{n}_{j=1}=\sum^{n}_{j=1}w_{n-j+1}$, we have $\sum^{n}_{j=1} \frac{\sin (my_{j})}{\sin^{2}(\frac{1}{2}y_{j})}=0$. Consequently,

$$ W_{m,n}=\sum^n_{j=1} \frac{\cos (my_j)}{\sin^2(\frac{1}{2}y_j)} \quad \Rightarrow \quad W_{-m,n}=W_{m,n}. $$

(A.9)

Thus, it suffices to treat only the case m=0,1,….

Next, for every m≥0, there exist unique integers k,r≥0, r≤n−1, such that m=kn+r. (Thus, k=0 and r=0 for m=0, while k=1 and r=0 for m=n.) By the fact that

we realize that

$$ W_{m,n}=W_{kn+r,n}=(-1)^k W_{r,n}. $$

(A.10)

Thus, we need to concern ourselves only with 0≤m≤n−1. (Note that this also implies that |W _m,n |≤max_{0≤i≤n−1}|W _i,n | for every m, hence $\{W_{m,n}\}^{\infty}_{m=-\infty}$ is a bounded sequence for fixed n.)

Setting y=y _j in (A.3) and summing over j from 1 to n, and invoking (A.9), we obtain

$$ W_{m+1,n}-2W_{m,n}+W_{m-1,n} =-4\sum^n_{j=1}\cos(my_j)=-4\bigg[ \operatorname{Re} \sum^n_{j=1}e^{\mathrm{i}my_j}\bigg]. $$

(A.11)

Upon setting m=0 in (A.11), and invoking the fact that W _−1,0=W _1,0, we first obtain

$$ 2W_{1,n}-2W_{0,n}=-4n\quad \Rightarrow\quad W_{1,n}=W_{0,n}-2n. $$

(A.12)

Let us now consider 1≤m≤n−1 in (A.11). For this, we need to compute $\sum^{n}_{j=1}\cos(my_{j})$. Invoking (A.8), we have

since 0<2mπ/n<2π for 1≤m≤n−1. As a result, $\sum^{n}_{j=1}\cos(my_{j})=0$, and (A.11) becomes

$$ W_{m+1,n}-2W_{m,n}+W_{m-1,n}=0,\quad m=1,2,\ldots,n-1, $$

(A.13)

which enables us to compute W _m,n for 2≤m≤n when W _0,n and W _1,n are known. The general solution of (A.13) is of the form W _m,n =Am+B for some constants A and B, which we determine by invoking two initial values, namely, those of W _0,n and W _1,n . To compute W _0,n , let

$$ \theta_j=\frac{1}{2}y_j\quad \text{and}\quad \xi_j=\cos\theta_j, \quad j=1,\ldots,n, $$

(A.14)

and note that ξ ₁,…,ξ _n are the zeros of T _n (ξ), the nth Chebyshev polynomial of the first kind. (For Chebyshev polynomials and their properties, see Rivlin [16], for example.) By (A.9) and (A.14),

and since ξ _n−j+1=−ξ _j , j=1,…,n, we have

Now, since ξ _j are the zeros of T _n (ξ), we have the identity

Letting ξ=1 in this identity, and making use of the facts that T _n (1)=1, $T_{n}'(1)=n^{2}$, we have

As for W _1,n , using (A.12), we have

Consequently, A=−2n and B=n ², and hence

Recalling also the fact that W _−m,n =W _m,n , we replace m by |m| everywhere, thus obtaining

$$ W_{m,n}=n^2-2|m|n,\quad m=0,\pm1,\ldots,\pm n. $$

(A.15)

Finally, substituting (A.15) in (A.7), we obtain (5.3).

The result in (5.4) is obtained by substituting (A.10), with W _r,n =n ²−2rn, in (A.8). The result in (5.5) is obtained by subtracting (5.2) from (5.4).

Proof of Part 3

This follows by invoking Part 2 of the theorem in

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Appendix B: Proof of Theorem 10.1

Proof of Part 1

We start by noting that

Making the variable transformation y=2π(x−t)/T in the integral inside the square brackets, and using the fact that the integrand is T-periodic, we obtain

$$ I[f_m]=\frac{T}{2\pi}K_{m} e^{\mathrm{i}\,2m\pi t/T}; \quad K_{m}=\int^{\pi}_{-\pi}{\cot\biggl(\frac{1}{2}y\biggr)}\,{e^{\mathrm{i}my}}\,dy. $$

(B.1)

Next,

and since $\int^{\pi}_{-\pi}{\cot(\frac{1}{2}y)}\,{\cos(my)}\,dy=0$ due to its integrand being odd, it follows that

$$ K_{m}=\mathrm{i}\int^{\pi}_{-\pi}\cot\biggl(\frac{1}{2}y\biggr)\sin(my)\,dy\quad \Rightarrow\quad K_0=0,\qquad K_{-m}=-K_m. $$

(B.2)

Thus, it suffices to treat only the case m=1,2,…. For m≥1, the integral in (B.2) representing K _m is regular. By making the variable transformation $z=\frac{1}{2}y$ in this integral, and invoking Gradshteyn and Ryzhik [4, p. 366, formula 3.612(7)], we obtain

$$ K_{m}=4\mathrm{i}\int^{\pi/2}_0{\cos z}\,\frac{\sin(2mz)}{\sin z}\,dz =2\pi\mathrm{i}, \quad m=1,2,\ldots . $$

(B.3)

Substituting this in (B.1), we obtain (10.2), with sgn(0)=0.

Proof of Part 2

By (9.10), we have first

(B.4)

Next,

By the fact that

and since $\sum^{n}_{j=1}w_{j}=\sum^{n}_{j=1}w_{n-j+1}$, we have $\sum^{n}_{j=1} \cot(\frac{1}{2}y_{j})\cos (my_{j})=0$. Consequently,

$$ U_{m,n}=\mathrm{i}\sum^n_{j=1} \cot\biggl(\frac{1}{2}y_j\biggr)\sin (my_j)\quad \Rightarrow \quad U_{0,n}=0,\qquad U_{-m,n}=-U_{m,n}. $$

(B.5)

Thus, it suffices to treat only the case m=1,2,….

Next, for every m≥0, there exist unique integers k,r≥0, r≤n−1, such that m=kn+r. (Thus, k=0 and r=0 for m=0, while k=1 and r=0 for m=n, and, therefore, U _±kn,n =0 for k=0,1,….) By the fact that

we realize that

$$ U_{m,n}=U_{kn+r,n}=(-1)^k U_{r,n}. $$

(B.6)

Thus, we need to concern ourselves only with 1≤m≤n−1. (Note that this also implies that |U _m,n |≤max_{0≤i≤n−1}|U _i,n | for every m, hence $\{U_{m,n}\}^{\infty}_{m=-\infty}$ is a bounded sequence for fixed n.)

Now,

$$ \sin\biggl(my\pm \frac{1}{2}y\biggr) =\sin(my)\cos\biggl(\frac{1}{2}y\biggr)\pm \cos(my)\sin\biggl(\frac{1}{2}y\biggr). $$

(B.7)

Dividing both sides of this identity by $\sin(\frac{1}{2}y)$, we obtain

$$ \frac{\sin(my\pm \frac{1}{2}y)}{\sin(\frac{1}{2}y)} =\cot\biggl(\frac{1}{2}y\biggr)\sin(my)\pm \cos(my). $$

(B.8)

Let us now set y=y _j in this identity and sum over j from 1 to n. Invoking (B.5), we obtain

$$ \sum^n_{j=1}\frac{\sin(my_j\pm \frac{1}{2}y_j)}{\sin(\frac{1}{2}y_j)}= -\mathrm{i}U_{m,n}\pm \sum^n_{j=1}\cos(my_j). $$

(B.9)

Note that the y _j in (B.4) are the same as the y _j in (A.8), and hence $\sum^{n}_{j=1}\cos(my_{j})=0$ for 1≤m≤n−1, as shown in the proof of Theorem 5.1 given in Appendix A. Consequently, (B.9) becomes

$$ D_{m+1}=-{\mathrm{i}}U_{m,n}=D_m,\quad 1\leq m\leq n-1;\qquad D_k=\sum^n_{j=1}\frac{\sin(ky_j-\frac{1}{2}y_j)}{\sin(\frac{1}{2}y_j)}. $$

(B.10)

Therefore, D _n =D _n−1=⋯=D ₂=D ₁=n, so that

Recalling also (B.5), we have

$$ U_{m,n}=\mathrm{i}\,\text{sgn}(m) n,\quad m=0,\pm1,\ldots,\pm(n-1); \qquad \text{sign}(0)=0. $$

(B.11)

Substituting (B.11) in (B.4), we obtain (10.3). Combining (B.11) with (B.5) and (B.6), we obtain (10.4) with (10.5). The result in (10.6) is obtained by subtracting (10.2) from (10.4).

Proof of Part 3

This follows by invoking Part 2 of the theorem in

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