Nystrom product integration and fredholm error — Examples

By [A1], the continuous integrands may be integrated as Riemann integrals. Since $Ku(x)=\alpha m_0+\beta x{m}_1$, any solution is affine and has the form $u(x)=1+\alpha m_0+\beta x{m}_1$. Integrating this identity against $1$ and against $y$, using [F8] for the moments and [F9] for linearity, gives $\left(\begin{array}{cc}1-\alpha & -\beta \mathrm{/}2\\ -\alpha \mathrm{/}2& 1-\beta \mathrm{/}3\end{array}\right)\left(\begin{array}{c}{m}_0\\ m_1\end{array}\right)=\left(\begin{array}{c}1\\ 1\mathrm{/}2\end{array}\right)\mathrm{.}$ At the specified parameters the determinant is $(3\mathrm{/}4)(5\mathrm{/}6)-(1\mathrm{/}4)(1\mathrm{/}8)=19\mathrm{/}32\mathrm{\ne }0$, and solving gives $m_0=92\mathrm{/}57$, $m_1=16\mathrm{/}19$, hence $u(x)=(80+24x)\mathrm{/}57$.

The datum conditions in [F1] hold by [F6] and [F7]. The function $u$ is affine and $yu(y)$ is quadratic, so [F3] gives $Q_2(u)=m_0$ and $Q_2(yu)=m_1$. Substituting these two exact quadrature moments into the Nyström formula [F4] yields $1+\alpha m_0+\beta x{m}_1=u(x)$ for every $x$; thus the continuous solution is also a Nyström solution.

Any solution $v$ of the inhomogeneous two-node Nyström equation has the form $v(x)=1+\alpha Q_2(v)+\beta x{Q}_2(yv)$, hence is affine. Exactness [F3] then makes its two Lebesgue moments satisfy the same nonsingular system in step 1.1, so those moments and the reconstructed function are unique; step 2.1 supplies existence. A nonzero vector in the grid-matrix kernel would, by the nullspace correspondence [F5], give a nonzero homogeneous solution, which could be added to this solution and contradict uniqueness. Thus the grid matrix is nonsingular, and its reconstruction is the displayed solution.

Substituting $u=1+cx$ into $u=1+Ku$ gives $c=\lambda (1\mathrm{/}2+c\mathrm{/}3)$, so $c(1-\lambda \mathrm{/}3)=\lambda \mathrm{/}2$. Since $1-\lambda \mathrm{/}3=\delta >0$, the unique coefficient is $c=\lambda \mathrm{/}(2\delta )$, proving the exact formula.

On each midpoint panel, direct integration gives ${\int }_{jh}^{(j+1)h}ydy=h(j+\frac{1}{2})h$ and ${\int }_{jh}^{(j+1)h}{y}^{2}dy-h((j+\frac{1}{2})h{)}^2=h^3\mathrm{/}12.$ Summing the $N$ panels yields $Q_N(y)=1\mathrm{/}2$ and $Q_N(y^2)=1\mathrm{/}3-h^2\mathrm{/}12$.

Any discrete solution has the form $u_N(x)=1+c_{N}x$ because the range of $K_N$ is contained in the span of $x$. Its coefficient satisfies $c_N=\lambda (1\mathrm{/}2+c_N(1\mathrm{/}3-h^2\mathrm{/}12))$, hence $c_N=\lambda \mathrm{/}(2(\delta +\lambda h^2\mathrm{/}12))$; the denominator is positive, so the discrete solution is unique.

On the direction $x$, $Kx=(\lambda \mathrm{/}3)x$, so $I-K$ scales that direction by $\delta$. Also $K_{N}x=\lambda (1\mathrm{/}3-h^2\mathrm{/}12)x$, so $I-K_N$ scales it by $\delta +\lambda h^2\mathrm{/}12$. Directly using the exact and discrete coefficients from steps 1.1 and 2.1 gives $\mathrm{\parallel }{K}_{N}u-Ku{\mathrm{\parallel }}_{\mathrm{\infty }}=\frac{{\lambda }^2{h}^2}{24\delta },\mathrm{\parallel }{u}_N-u{\mathrm{\parallel }}_{\mathrm{\infty }}=\frac{{\lambda }^2{h}^2}{24\delta (\delta +\lambda h^2\mathrm{/}12)}\mathrm{.}$ The ratio of solution error to quadrature residual is exactly $(\delta +\lambda h^2\mathrm{/}12{)}^{-1}$, and the stated lower bound follows when $\lambda h^2\mathrm{/}12\le \delta$.

Let $m_j:=(j+\frac{1}{2})h$. Define $u_{\rho }$ to equal $-1+2\mathrm{\mid }y-m_j\mathrm{\mid }\mathrm{/}\rho$ on each interval $\mathrm{\mid }y-m_j\mathrm{\mid }\le \rho$ and $1$ elsewhere. The windows are disjoint and lie inside $[0,1]$, so $u_{\rho }$ is continuous and $\mathrm{\parallel }{u}_{\rho }{\mathrm{\parallel }}_{\mathrm{\infty }}=1$. Its integral over each window is zero, hence ${\int }_0^1{u}_{\rho }(y)dy=1-2N\rho$.

Every midpoint value is $u_{\rho }(m_j)=-1$, so $K_N{u}_{\rho }=-1$ by [F1] and [F2], whereas $K{u}_{\rho }=1-2N\rho$. Therefore $\mathrm{\parallel }(K-K_N)u_{\rho }{\mathrm{\parallel }}_{\mathrm{\infty }}=2-2N\rho$.

The operators $K$ and $K_N$ both have norm at most $1$, because their defining integrals or positive quadrature weights have total mass $1$. Thus $\mathrm{\parallel }K-K_N\mathrm{\parallel }\le 2$. Letting $\rho \downarrow 0$ in step 2.1 gives $\mathrm{\parallel }K-K_N\mathrm{\parallel }\ge 2$, so $\mathrm{\parallel }K-K_N\mathrm{\parallel }=2$ for every $N$. This disproves the claimed operator-norm convergence.

Since every $m_j>0$, the admissible call $Q_M(\tilde{g};0,1,N)$ equals $h{\sum }_{j=0}^{N-1}g(m_j)$ by [F3]. On the first panel, the substitution $y=ht$ gives ${\int }_0^h{y}^{-1\mathrm{/}2}dy=\sqrt{h}{\int }_0^1{t}^{-1\mathrm{/}2}dt=2\sqrt{h}$ by [F1]; this substitution follows by scaling tagged sums on each truncated interval and then taking the improper limit. The midpoint contribution is $hg(h\mathrm{/}2)=\sqrt{2}\sqrt{h}$. Their difference is $(2-\sqrt{2})\sqrt{h}$.

For every later panel $[jh,(j+1)h]$, $j\ge 1$, the function $g$ is $C^2$ by [F4]. Its second derivative is positive because $y^{-5\mathrm{/}2}=\exp ((-5\mathrm{/}2)\log y)>0$ for $y>0$, using [F4]. The mean value theorem [F5] applied to $g^{\mathrm{\prime }}$ shows that $g^{\mathrm{\prime }}$ is increasing there. For $0\le s\le h\mathrm{/}2$, put $D(s):=g(m_j-s)+g(m_j+s)-2g(m_j)$. By [F9], $D^{\mathrm{\prime }}(s)=-g^{\mathrm{\prime }}(m_j-s)+g^{\mathrm{\prime }}(m_j+s)\ge 0$ in the interior; another application of [F5] on $[0,s]$ gives $D(s)\ge D(0)=0$. Splitting at $m_j$ and substituting $y=m_j-s$ and $y=m_j+s$ gives ${\int }_{jh}^{(j+1)h}g(y)dy-hg(m_j)={\int }_0^{h\mathrm{/}2}D(s)ds$: the left substitution reverses the limits, while the right preserves them, and each has Jacobian magnitude $1$. Thus the integral identity follows directly from the tagged Riemann sums under these affine maps. The integrals exist by [F6], the split at $m_j$ is additive by [F7], and the nonnegative difference integrates to at least zero by [F8]. Thus no later panel cancels the positive first-panel error, and the total midpoint error is at least $(2-\sqrt{2})\sqrt{h}$.

For the product rule, $a\equiv 1$ and $u\equiv 1$ give $g_0\equiv 1$. On each panel the two adjacent hat formulas in [F2] sum to $1$, so $I_n{g}_0=1$. By [F1] and [F2], $K_{n}u(0)=S_0(1)=2=Ku(0)$. Hence this case is exact under product integration although direct midpoint sampling has only the lower-bound rate above.