Spectral collocation and Galerkin methods — Examples

Substitution of $x_0=0,x_1=\pi \mathrm{/}2,x_2=\pi ,x_3=3\pi \mathrm{/}2$ into $p$ using [F2] gives respectively $-1,8,-5,2$. The frequencies $0,1,2$ lie in [F1]'s interpolation space, so $p$ is the interpolant of these samples.

By [F3] and [F5], $p^{\mathrm{\prime }}(x)=-2\sin x+3\cos x+8\sin (2x)$. At the four nodes, the final term is zero and the first two terms give $3,-2,-3,2$.

Exactness [F4] therefore gives $D_{4}U=(3,-2,-3,2)$, including the vanishing nodal derivative of the Nyquist mode.

The subtraction and addition identities [F2] give $x_j-x_{j+1}=2\sin \left(\frac{(2j+1)\pi }{2N}\right)\sin \left(\frac{\pi }{2N}\right),x_0-x_1=2{\sin }^2\left(\frac{\pi }{2N}\right)\mathrm{.}$ By [F3], $\sin (\pi \mathrm{/}(2N))\le \pi \mathrm{/}(2N)$, hence $x_0-x_1\le {\pi }^2\mathrm{/}(2N^2)$.

If $N$ is odd, choose $j=(N-1)\mathrm{/}2$; then $(2j+1)\pi \mathrm{/}(2N)=\pi \mathrm{/}2$ and [F4] gives sine equal to $1$, so the gap formula in step 1.1 and [F3] give a central gap at least $2\mathrm{/}N\ge \sqrt{2}\mathrm{/}N$. If $N$ is even, choose $j=N\mathrm{/}2-1$ and write $t=\pi \mathrm{/}2-\pi \mathrm{/}(2N)\ge \pi \mathrm{/}4$. The addition and subtraction formulas [F2] give $\sin (\pi \mathrm{/}4)=\cos (\pi \mathrm{/}4)$ and $2\sin (\pi \mathrm{/}4)\cos (\pi \mathrm{/}4)=\sin (\pi \mathrm{/}2)=1$. Thus ${\sin }^2(\pi \mathrm{/}4)=1\mathrm{/}2$; the chord bound [F3] gives $\sin (\pi \mathrm{/}4)\ge 1\mathrm{/}2>0$, so uniqueness of the nonnegative square root [F5] gives $\sin (\pi \mathrm{/}4)=\sqrt{2}\mathrm{/}2$. Monotonicity [F4] now gives $\sin t\ge \sqrt{2}\mathrm{/}2$. In either parity, [F3] also gives $\sin (\pi \mathrm{/}(2N))\ge 1\mathrm{/}N$, so the selected central gap in step 1.1 is at least $\sqrt{2}\mathrm{/}N$.

Put $a=\pi \mathrm{/}(2N)$. Every adjacent gap has the form $2\sin ((2j+1)a)\sin a$, with $a\le (2j+1)a\le \pi -a$. Reflecting about $\pi \mathrm{/}2$ and using monotonicity [F4] gives $\sin ((2j+1)a)\ge \sin a$; equality occurs at $j=0$ and $j=N-1$. Hence the smallest spacing is exactly $2{\sin }^{2}a$, and [F3] bounds it by $2\mathrm{/}{N}^2\le 2{\sin }^{2}a\le {\pi }^2\mathrm{/}(2N^2)$. For every $\epsilon >0$, [F6] supplies an integer $n\ge 1$ with $1\mathrm{/}n<\epsilon$; for every integer $N\ge \max \{2,n\}$, positivity and monotonicity of the canonical naturals and reciprocal order give $0<1\mathrm{/}N\le 1\mathrm{/}n<\epsilon$. Thus $1\mathrm{/}N\to 0$, and the endpoint-to-central-gap ratio is at most ${\pi }^2\mathrm{/}(2N)$, which tends to zero by the fixed-multiple limit law in [F6]. For $N\ge 3$, the selected central angle, reflected into $[0,\pi \mathrm{/}2]$ if necessary, is strictly larger than $a$; strict monotonicity in [F4] makes its gap larger than the endpoint gap. For $N=2$, the nodes are $1,0,-1$ and both gaps equal $1$. This proves the quadratic endpoint scale with its stated nonuniformity exception.

The set $\mathrm{\Omega }=\{z\in \mathbb{C}:\mathrm{\mid }z\mathrm{\mid }<3\mathrm{/}2\}$ is an open neighborhood of $[-1,1]$ and omits $2$. For $z,z_0\in \mathrm{\Omega }$ with $z\mathrm{\ne }{z}_0$, $\frac{f_a(z)-f_a(z_0)}{z-z_0}=\frac{1}{(2-z)(2-z_0)}⟶\frac{1}{(2-z_0{)}^2}(z\to z_0)\mathrm{.}$ Thus [F4] makes $f_a$ holomorphic on a neighborhood of the interval. To verify analyticity in the power-series sense as well, fix a real $c<2$ and put $t=(x-c)\mathrm{/}(2-c)$ for $\mathrm{\mid }x-c\mathrm{\mid }<(2-c)\mathrm{/}2$. Finite geometric summation gives $\frac{1}{2-x}={\sum }_{k=0}^M\frac{(x-c{)}^k}{(2-c{)}^{k+1}}+\frac{t^{M+1}}{(2-c)(1-t)}$. Since $\mathrm{\mid }t\mathrm{\mid }<1\mathrm{/}2$, the remainder has absolute value at most $2^{-M}\mathrm{/}(2-c)$, tending to zero by [F3]. The series therefore represents $f_a$ on a neighborhood of every $c<2$, establishing real analyticity on $(-\mathrm{\infty },2)$ by [F15], in particular near $[-1,1]$.

The function $f_s(x)=\mathrm{\mid }x\mathrm{\mid }$ is continuous by [F14] and is not differentiable at $0$: for $h>0$ its difference quotient is $1$, while for $h<0$ it is $-1$, so the two-sided limit in [F11] does not exist. By [F7], $\mathrm{\mid }\cos \theta \mathrm{\mid }=\cos \theta$ on $[0,\pi \mathrm{/}2]$ and $\mathrm{\mid }\cos \theta \mathrm{\mid }=-\cos \theta$ on $[\pi \mathrm{/}2,\pi ]$. The subtraction identity [F9] gives $\cos (n(\pi -\theta ))=(-1{)}^n\cos (n\theta )$, so reflection makes each odd coefficient integral its own negative. Therefore $a_0(f_s)=\frac{1}{\pi }{\int }_0^{\pi }\mathrm{\mid }\cos \theta \mathrm{\mid }d\theta =\frac{2}{\pi }\mathrm{.}$ For odd $n$, that reflection makes $a_n(f_s)=0$.

Put $q=2-\sqrt{3}$. Then $0<q<1$, $q^{-1}=2+\sqrt{3}$, $q+q^{-1}=4$, and $1-q^2=2\sqrt{3}q$. By [F8], $z=q{e}^{i\theta }$ has modulus $q$. For $m\ge 1$, put $S_m(z)={\sum }_{n=1}^m{z}^n$. Multiplying the finite sum by $1-z$ cancels consecutive terms and gives $(1-z)S_m=z-z^{m+1}$. The triangle inequality in [F3] gives $\mathrm{\mid }1-z\mathrm{\mid }\ge 1-q>0$, so $\mid S_m(z)-\frac{z}{1-z}\mid =\frac{\mathrm{\mid }z{\mathrm{\mid }}^{m+1}}{\mathrm{\mid }1-z\mathrm{\mid }}\le \frac{q^{m+1}}{1-q}\mathrm{.}$ Induction using the addition formulas in [F5] and Euler's identity in [F8] gives $z^n=q^n(\cos (n\theta )+i\sin (n\theta ))$. Thus $C_m(\theta ):=1+2{\sum }_{n=1}^m{q}^n\cos (n\theta )$, a continuous function by [F14], differs from $1+2\mathrm{Re}(z\mathrm{/}(1-z))$ by at most $2q^{m+1}\mathrm{/}(1-q)$, uniformly for $\theta \in [0,\pi ]$.

Multiplying numerator and denominator by $1-q\cos \theta +iq\sin \theta$ and using [F10] gives $\mathrm{Re}\frac{z}{1-z}=\frac{q\cos \theta -q^2}{1-2q\cos \theta +q^2}\mathrm{.}$ Hence $1+2\mathrm{Re}\frac{z}{1-z}=\frac{1-q^2}{1-2q\cos \theta +q^2}=\frac{\sqrt{3}}{2-\cos \theta },$ because the denominator equals $2q(2-\cos \theta )$. By step 1.3, the continuous functions $C_m\mathrm{/}\sqrt{3}$ satisfy ${\sup }_{0\le \theta \le \pi }\mid \frac{C_m(\theta )}{\sqrt{3}}-f_a(\cos \theta )\mid \le {\eta }_m:=\frac{2q^{m+1}}{\sqrt{3}(1-q)}⟶0$ by [F3]. This also establishes the claimed convergent Chebyshev expansion through [F1].

Fix a coefficient index $n\ge 0$. By [F14], the functions $f_a(\cos \theta )$, $C_m\mathrm{/}\sqrt{3}$, and their products with $\cos (n\theta )$ are continuous; [F12] makes them integrable. Since $\mathrm{\mid }\cos (n\theta )\mathrm{\mid }\le 1$ by [F7], [F13] bounds the corresponding coefficient differences by ${\eta }_m$ for index zero and by $2{\eta }_m$ for positive indices. For every $m\ge n$, finite integration and orthogonality [F2] give the coefficients of $C_m\mathrm{/}\sqrt{3}$ as $1\mathrm{/}\sqrt{3}$ at zero and $2q^n\mathrm{/}\sqrt{3}$ at each $1\le n\le m$. Letting $m\to \mathrm{\infty }$ therefore yields exactly the asserted coefficients of $f_a$; no interchange of an unbounded infinite sum and integral is assumed.

For $m\ge 1$, symmetry about $\pi \mathrm{/}2$ and [F5] give ${\int }_0^{\pi }\mathrm{\mid }\cos \theta \mathrm{\mid }\cos (2m\theta )d\theta =2{\int }_0^{\pi \mathrm{/}2}\cos \theta \cos (2m\theta )d\theta =\frac{2(-1{)}^{m-1}}{4m^2-1},$ where the last equality follows by integrating $\frac{1}{2}[\cos ((2m+1)\theta )+\cos ((2m-1)\theta )]$ using [F6]–[F7]. Multiplying by $2\mathrm{/}\pi$ gives $a_{2m}(f_s)=4(-1{)}^{m-1}\mathrm{/}(\pi (4m^2-1))$. This coefficient magnitude is asymptotic to $1\mathrm{/}(\pi m^2)$.

By [F3], $p(x{)}^2={\cos }^{2}x+2\cos x\cos (2x)+{\cos }^2(2x)=1+\cos x+\frac{1}{2}\cos (2x)+\cos (3x)+\frac{1}{2}\cos (4x)\mathrm{.}$

At every grid node [F4] turns this into $p(x_j{)}^2=\frac{3}{2}+2\cos (x_j)+\frac{1}{2}\cos (2x_j)$. The right side belongs to [F1]'s interpolation space.

The modes at frequencies $3$ and $4$ in step 1.1 are identified on the grid by the congruences $3\equiv -1(\mathrm{m}\mathrm{o}\mathrm{d}4)$ and $4\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}4)$, exactly as [F2] states. Thus the displayed interpolant has the same samples as $p^2$, with the two alias contributions combined.

Under [A1], let $\mu$ be Lebesgue measure restricted to the relative Borel sets of $[-1,1]$. It is a positive measure with total mass $2$ by [F10]. Every relative open neighborhood of a point of $[-1,1]$ contains a positive-length interval, including a one-sided interval at either endpoint, so [F10] and [F14] give that neighborhood positive measure. Thus the relative support is all of $[-1,1]$, containing at least three points. Every $\mathrm{\mid }x{\mathrm{\mid }}^k$ for $0\le k\le 3$ is continuous and bounded on $[-1,1]$ by [F14], so [F11] gives its integrability. These are exactly the order-three, $m=2$ moment-data conditions in [F1]. The primitives $x,x^2\mathrm{/}2,x^3\mathrm{/}3,x^4\mathrm{/}4$ give moments ${\mu }_0=2,{\mu }_1=0,{\mu }_2=2\mathrm{/}3,{\mu }_3=0$ by [F11].

By [F9], $g_{00}=1$, $g_{01}=\cos \theta$, and $g_{11}={\cos }^2\theta$. Substitution of the quarter-turn values [F8] into [F5] gives $M_{00}^S=\pi$, $M_{01}^S=0$, and $M_{11}^S=\pi \mathrm{/}3$. The exact mass matrix is $G=\mathrm{diag}(\pi ,\pi \mathrm{/}2)$ by [F7].

The monic polynomial $p_2(x)=x^2-1\mathrm{/}3$ is orthogonal to $1$ and $x$, since ${\mu }_2-{\mu }_0\mathrm{/}3=0$ and ${\mu }_3-{\mu }_1\mathrm{/}3=0$ by step 1.1 and [F12]. Finite linearity extends this to every polynomial of degree below two, so uniqueness in [F1] identifies $p_2$ with the Gaussian orthogonal polynomial. Its roots satisfy $x^2=1\mathrm{/}3$, hence are $x_1=-1\mathrm{/}\sqrt{3}$ and $x_2=1\mathrm{/}\sqrt{3}$ by [F14]. Their cardinal polynomials are ${\mathrm{\ell }}_1(x)=(1-\sqrt{3}x)\mathrm{/}2$ and ${\mathrm{\ell }}_2(x)=(1+\sqrt{3}x)\mathrm{/}2$, so [F12] and the first two moments give $w_1=w_2=1$. The rule Definition and exactness theorem in [F1] therefore give $Q_2(f)=f(-1\mathrm{/}\sqrt{3})+f(1\mathrm{/}\sqrt{3})$, exact for degrees through three.

The Gaussian sum for the basis $1,x$ is ${\left(\begin{array}{cc}1& -1\mathrm{/}\sqrt{3}\\ 1& 1\mathrm{/}\sqrt{3}\end{array}\right)}^{\text{\hspace{0.17em}⁣}T}\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1& -1\mathrm{/}\sqrt{3}\\ 1& 1\mathrm{/}\sqrt{3}\end{array}\right)=\left(\begin{array}{cc}2& 0\\ 0& 2\mathrm{/}3\end{array}\right)\mathrm{.}$ The entries agree with exact integrals by [F2], as also asserted by [F3].

The only nonzero entry error is $\mathrm{\mid }{G}_{11}-M_{11}^S\mathrm{\mid }=\pi \mathrm{/}6$. The Simpson theorem [F6] applies with $M=2$, $h=\pi \mathrm{/}2$, and $i=j=1$, and gives its bound ${\pi }^5\mathrm{/}180$. This bound is at least $\pi \mathrm{/}6$, since [F13] gives ${\pi }^4>81>30$, equivalently ${\pi }^5\mathrm{/}180>\pi \mathrm{/}6$. Therefore the displayed finite calculation matches the general entrywise estimate.