Fractional evolution and convolution quadrature — Examples

For $\alpha =1\mathrm{/}2$, the first-step discrete equation is $h^{-1\mathrm{/}2}(U^1-1)\mathrm{/}\mathrm{\Gamma }(3\mathrm{/}2)+U^1=0$ by [F1]. Multiplying by $\mathrm{\Gamma }(3\mathrm{/}2)\sqrt{h}$ and solving gives $U^1=(1+\mathrm{\Gamma }(3\mathrm{/}2)\sqrt{h}{)}^{-1}$.

For CQ, [F2] gives $(h^{-1\mathrm{/}2}+1)V^1=-1$, so $V^1=-\sqrt{h}\mathrm{/}(1+\sqrt{h})$ and $U^1=1+V^1=(1+\sqrt{h}{)}^{-1}$. Both values are positive and less than $1$ because $h>0$ and $\mathrm{\Gamma }(3\mathrm{/}2)>0$.

Fix any $\epsilon >0$. For any finite disjoint intervals, the sum of absolute endpoint increments of $u(t)=t$ equals the sum of their lengths. Thus [F1]'s defining condition holds with $\delta =\epsilon$, and $u$ is absolutely continuous.

By [F5], $u^{\mathrm{\prime }}=1$ on $(0,T)$, so $u^{\mathrm{\prime }}\in L^1(0,T)$. Using the pointwise representative from [F2], the kernel integral [F6] gives $D_C^{\alpha }u(t)=\frac{1}{\mathrm{\Gamma }(1-\alpha )}{\int }_0^t(t-s{)}^{-\alpha }ds=\frac{t^{1-\alpha }}{(1-\alpha )\mathrm{\Gamma }(1-\alpha )}=\frac{t^{1-\alpha }}{\mathrm{\Gamma }(2-\alpha )},$ using [F4]. This integral is finite at each $t\in [0,T]$.

The direct full-history step $n$ uses one term for each increment $k=1,\dots ,n$. Let $S={\sum }_{n=1}^{N}n$. Reindexing by $n\mapsto N+1-n$ using [F8] also gives $S={\sum }_{n=1}^N(N+1-n)$. Adding these equalities termwise gives $2S={\sum }_{n=1}^N(N+1)=N(N+1)$ by [F7], so the full-history term count is $N(N+1)\mathrm{/}2$. Retaining the sequence requires $N+1$ nodal values.

For $n>m$, the truncated version of [F3] retains exactly the $m$ terms with index lag $j=0,\dots ,m-1$. Since $u^k-u^{k-1}=h$, ${\tilde{\delta }}_h^{\alpha }{u}^n=\frac{h^{1-\alpha }}{\mathrm{\Gamma }(2-\alpha )}{\sum }_{j=0}^{m-1}{a}_j=\frac{h^{1-\alpha }{m}^{1-\alpha }}{\mathrm{\Gamma }(2-\alpha )},$ where the sum telescopes. The full sum through $j=n-1$ likewise telescopes to $n^{1-\alpha }$ and equals the exact derivative from step 1.2. This gives the bias stated.

At step $n$ the truncated evaluation uses $\min (n,m)$ terms, so [F7] gives its total count as ${\sum }_{n=1}^N\min (n,m)$. It can retain only the $m$ most recent increments, equivalently $m+1$ adjacent nodal values. For $n\le m$ no term is omitted, and at $m=N$ the bias is zero at every step.

Fix an arbitrary $\epsilon >0$ and set $\eta =\min (T,{\epsilon }^2\mathrm{/}16)$ and $\delta ={\epsilon }^2\mathrm{/}8$. By [F14] and $u^{\mathrm{\prime }}(s)>0$ for $s>0$, $u$ is increasing. Reorder any finite family of intervals with pairwise disjoint interiors by left endpoints; since $u$ is increasing, absolute endpoint increments equal endpoint increments, and monotonicity across the gaps shows their sum on a subinterval $[a,b]$ is at most $u(b)-u(a)$. For $t>0$, $u^{\mathrm{\prime }}(t)=1\mathrm{/}(2\sqrt{t})$ and $u(t)={\int }_0^t(2\sqrt{s}{)}^{-1}ds$ by [F9]. If $T\le {\epsilon }^2\mathrm{/}16$, the family contributes at most $u(T)-u(0)=\sqrt{T}\le \epsilon \mathrm{/}4<\epsilon$. If $T>{\epsilon }^2\mathrm{/}16$, split each interval at $\eta ={\epsilon }^2\mathrm{/}16$. The pieces below $\eta$ contribute at most $u(\eta )-u(0)=\sqrt{\eta }=\epsilon \mathrm{/}4$. On pieces above $\eta$, $u^{\mathrm{\prime }}(s)\le 1\mathrm{/}(2\sqrt{\eta })=2\mathrm{/}\epsilon$, so [F14] bounds their total increments by $(2\mathrm{/}\epsilon )\delta =\epsilon \mathrm{/}4$. Thus [F1]'s defining condition holds. But $u^{\mathrm{\prime }\mathrm{\prime }}(t)=-1\mathrm{/}(4t^{3\mathrm{/}2})$ is unbounded as $t\downarrow 0$, so $u\mathrm{\notin }{C}^2([0,T])$.

Fix $t>0$. By step 1.1 and [A1], [F10] gives $u^{\mathrm{\prime }}\in L^1(0,T)$. On $(0,t\mathrm{/}2)$, $(t-s{)}^{-1\mathrm{/}2}\le (t\mathrm{/}2{)}^{-1\mathrm{/}2}$; on $(t\mathrm{/}2,t)$, $u^{\mathrm{\prime }}(s)\le 1\mathrm{/}(2\sqrt{t\mathrm{/}2})$. The fractional-integral clause of [F2] makes $(t-s{)}^{-1\mathrm{/}2}$ integrable on $(0,t)$; [F11] bounds the product on each half, and [F12] adds the two finite integrals. Hence $(t-s{)}^{-1\mathrm{/}2}{u}^{\mathrm{\prime }}(s)\in L^1(0,t)$ for every $t>0$.

For the pointwise representative in [F2], apply [F13] to the integrable kernel in step 2.1 with $S(y)=ty$. This is a $C^1$ diffeomorphism with $S^{\mathrm{\prime }}(y)=t$, and the Jacobian cancels the factor $t^{-1}$ in $(ty{)}^{-1\mathrm{/}2}(t-ty{)}^{-1\mathrm{/}2}$. Therefore $D_C^{1\mathrm{/}2}u(t)=\frac{1}{2\mathrm{\Gamma }(1\mathrm{/}2)}{\int }_0^1(1-y{)}^{-1\mathrm{/}2}{y}^{-1\mathrm{/}2}dy=\frac{B(1\mathrm{/}2,1\mathrm{/}2)}{2\mathrm{\Gamma }(1\mathrm{/}2)}=\frac{1}{2}\mathrm{\Gamma }(1\mathrm{/}2)=\mathrm{\Gamma }(3\mathrm{/}2),$ using the Beta–Gamma identity [F4] and $\mathrm{\Gamma }(1)=1$ plus the functional equation [F5]. This holds for every $t>0$.

For the uniform grid, [F6] and [F15] give the equation ${\delta }_h^{1\mathrm{/}2}{U}^1=f(h)$, hence $U^1=\mathrm{\Gamma }(3\mathrm{/}2{)}^2\sqrt{h}$. Also $\mathrm{\Gamma }(3\mathrm{/}2)={\int }_0^{\mathrm{\infty }}\sqrt{s}{e}^{-s}ds<\frac{1}{2}{\int }_0^{\mathrm{\infty }}(s+1)e^{-s}ds=1,$ because $(\sqrt{s}-1{)}^2\ge 0$ with strict inequality except at $s=1$, and [F3]–[F5] evaluate the last integrals. Hence the stated error is positive. Its ratio to $h^{3\mathrm{/}2}$ is $(1-\mathrm{\Gamma }(3\mathrm{/}2{)}^2)\mathrm{/}h$, which is unbounded as $h\downarrow 0$; the smooth-data theorem [F7] cannot be extended to this example.

On the power-graded grid, $t_1=T\mathrm{/}{N}^2$, so [F6] gives $b_{1,1}=N\mathrm{/}(\mathrm{\Gamma }(3\mathrm{/}2)\sqrt{T})$. The first discrete equation, justified by [F15], yields $U^1=\mathrm{\Gamma }(3\mathrm{/}2{)}^2\sqrt{T}\mathrm{/}N$, and subtracting it from $u(t_1)=\sqrt{T}\mathrm{/}N$ gives the displayed graded first-step error. For $N=4$ the nodes follow by substituting $k=0,1,2,3,4$ into $t_k=T(k\mathrm{/}4{)}^2$.