Fractional evolution and convolution quadrature

In the improper integral, set $x=(t-s)\mathrm{/}t$. On every compact truncation this is a differentiable decreasing substitution with $ds=-tdx$, so [F5] and passage to the endpoint limits give ${\int }_0^t(t-s{)}^{\beta -1}ds=t^{\beta }{\int }_0^1{x}^{\beta -1}dx=t^{\beta }B(\beta ,1),$ where the last equality is [F1] because $(1-x{)}^0=1$ on $(0,1)$.

By [F2] and [F4], $B(\beta ,1)=\mathrm{\Gamma }(\beta )\mathrm{/}\mathrm{\Gamma }(\beta +1)$. Applying [F3] gives $B(\beta ,1)=1\mathrm{/}\beta$. Substitution into step 1.1 proves the formula and convergence of the stated improper integral.

For each $t$, [F1] and $\beta >0$ give $\mathrm{\mid }{I}^{\beta }g(t)\mathrm{\mid }\le \mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\Gamma }(\beta {)}^{-1}{\int }_0^t(t-r{)}^{\beta -1}dr=\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}{t}^{\beta }\mathrm{/}(\beta \mathrm{\Gamma }(\beta ))$. By [F2] this is the first bound, and it also gives $I^{\beta }g(0)=0$.

Suppose $0\le s\le t\le T$ and put $\delta =t-s$. Since $\beta -1<0$, the kernel $(t-r{)}^{\beta -1}$ is no larger than $(s-r{)}^{\beta -1}$ for $0\le r<s$. Thus [F1] gives $\mathrm{\Gamma }(\beta )\mathrm{\mid }{I}^{\beta }g(t)-I^{\beta }g(s)\mathrm{\mid }\le \mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\left({\int }_0^s\right((s-r{)}^{\beta -1}-(t-r{)}^{\beta -1})dr+{\int }_s^t(t-r{)}^{\beta -1}dr)\mathrm{.}$ The first integral is $(s^{\beta }-t^{\beta }+{\delta }^{\beta })\mathrm{/}\beta$. Its integrand is nonnegative, so the integral is nonnegative; and $t^{\beta }-s^{\beta }\ge 0$ by monotonicity of positive powers in [F3], so the same formula bounds it above by ${\delta }^{\beta }\mathrm{/}\beta$. The second integral is ${\delta }^{\beta }\mathrm{/}\beta$. Therefore the right side is at most $2\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}{\delta }^{\beta }\mathrm{/}\beta$.

Divide step 1.2 by $\mathrm{\Gamma }(\beta )>0$ and use [F2]. This gives the stated Hölder bound when $s\le t$; interchanging $s,t$ gives it for every pair. The bound tends to zero with $\mathrm{\mid }t-s\mathrm{\mid }$, so $I^{\beta }g$ is continuous.

The integrands are measurable. Applying Tonelli [F5] to their absolute values on $0\le r\le s\le t\le T$, for each fixed $r$ set $x=t-s$ and $y=s-r$; the region lies in $[0,T{]}^2$, so the kernel integral is at most ${\int }_0^T{x}^{a-1}dx{\int }_0^T{y}^{b-1}dy=T^{a+b}\mathrm{/}(ab)$ by [F6]. Hence ${\int }_0^T{\int }_0^t{\int }_0^s(t-s{)}^{a-1}(s-r{)}^{b-1}\mathrm{\mid }g(r)\mathrm{\mid }drdsdt\le T^{a+b}\mathrm{\parallel }g{\mathrm{\parallel }}_1\mathrm{/}(ab)<\mathrm{\infty }$, so the double integral is absolutely integrable for almost every fixed $t$ and Tonelli permits swapping $r,s$. The same theorem gives $\mathrm{\parallel }{I}^{a}g{\mathrm{\parallel }}_1\le \mathrm{\Gamma }(a{)}^{-1}{\int }_0^T\mathrm{\mid }g(r)\mathrm{\mid }{\int }_r^T(t-r{)}^{a-1}dtdr\le T^a\mathrm{\parallel }g{\mathrm{\parallel }}_1\mathrm{/}\mathrm{\Gamma }(a+1)$ using [F4]–[F6] and positivity [F7].

After swapping the order, the inner kernel integral is ${\int }_r^t(t-s{)}^{a-1}(s-r{)}^{b-1}ds=(t-r{)}^{a+b-1}{\int }_0^1(1-y{)}^{a-1}{y}^{b-1}dy=(t-r{)}^{a+b-1}B(b,a)$ by the substitution $y=(s-r)\mathrm{/}(t-r)$ using [F8] and the definition [F2]. This applies for almost every $t$ by step 1.1; if $g$ is continuous, Tonelli and this formula bound the nonnegative double integral for each fixed $t$ by $\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}B(b,a){\int }_0^t(t-r{)}^{a+b-1}dr<\mathrm{\infty }$ using [F5]–[F6], so Fubini applies for every $t$.

Substituting [F3] into step 2.1 cancels $\mathrm{\Gamma }(a)\mathrm{\Gamma }(b)$ in [F1], leaving $I^a(I^{b}g)(t)=\mathrm{\Gamma }(a+b{)}^{-1}{\int }_0^t(t-r{)}^{a+b-1}g(r)dr=I^{a+b}g(t)$ almost everywhere, and everywhere for continuous $g$. This proves the semigroup identity and the $L^1$ bound in step 1.1.

For the constant function, [F1] and [F6] give $I^c\mathbf{1}(t)=t^c\mathrm{/}(c\mathrm{\Gamma }(c))$ for $t>0$. Positivity from [F7] makes the quotient well-defined, and [F4] rewrites it as $t^c\mathrm{/}\mathrm{\Gamma }(c+1)$; both sides vanish at $t=0$.

The recurrence [F1] gives $\mathrm{\mid }{\omega }_0\mathrm{\mid }=1$, ${\omega }_1=-\alpha$ with $\mathrm{\mid }{\omega }_1\mathrm{\mid }=\alpha <1$, and for $j\ge 2$ gives ${\omega }_j=((j-1-\alpha )\mathrm{/}j){\omega }_{j-1}$ with $0<(j-1-\alpha )\mathrm{/}j<1$, so induction yields $\mathrm{\mid }{\omega }_j\mathrm{\mid }\le 1$ for all $j$. For each $0\le r<1$, [F2] bounds ${\sum }_{j\ge 0}\mathrm{\mid }{\omega }_j\mathrm{\mid }{r}^j$ by a convergent geometric series; the radius definition [F3] therefore gives radius at least $1$, and the complex series sum $W$ is holomorphic on $\mathbb{D}$ by [F3].

For real $x\in (-1,1)$ the same coefficients define a real power series of radius at least $1$; termwise differentiation [F4] gives $W^{\mathrm{\prime }}(x)={\sum }_{j\ge 0}(j+1){\omega }_{j+1}{x}^j$. By [F1], $(j+1){\omega }_{j+1}=(j-\alpha ){\omega }_j$, hence $W^{\mathrm{\prime }}(x)={\sum }_{j\ge 0}(j-\alpha ){\omega }_j{x}^j=x{W}^{\mathrm{\prime }}(x)-\alpha W(x)$ and $(1-x)W^{\mathrm{\prime }}(x)=-\alpha W(x)$. Since $1-x>0$, [F5] gives $H^{\mathrm{\prime }}(x)=0$ for $H(x):=W(x)(1-x{)}^{-\alpha }$; [F6] makes $H$ constant, and $H(0)={\omega }_0=1$, so $W(x)=(1-x{)}^{\alpha }$ throughout $(-1,1)$.

Fix $z\in \mathbb{D}$. If $1-z$ were real $r\le 0$, then $z=1-r\ge 1$ would be positive real, so [F12] would give $\mathrm{\mid }z\mathrm{\mid }=z\ge 1$, a contradiction. Thus $1-z$ is in the slit plane and $P(z):=(1-z{)}_{\mathrm{Log}}^{}$ is defined; [F8] makes the principal logarithm holomorphic there, and [F9] with the branch definition makes $P$ holomorphic on $\mathbb{D}$.

By [F7] the unit disc is a complex domain. For every real $x\in (-1,1)$, [F10] identifies $P(x)$ with $(1-x{)}^{\alpha }$, so step 1.2 gives $P(x)=W(x)$. The agreement set contains an interval accumulating at $0\in \mathbb{D}$; since $W$ and $P$ are holomorphic by steps 1.1 and 1.3, the identity theorem [F11] gives $W=P$ throughout $\mathbb{D}$, including $z=0$.

Suppose $D_C^{\alpha }u=f$ almost everywhere and put $v=u^{\mathrm{\prime }}\in L^1$ by [F4]. Then $u-u_0=I^{1}v$. Applying $I^{\alpha }$ to the Caputo identity and using [F1] and [F2] gives $u-u_0=I^{\alpha }f$ almost everywhere. Both sides are continuous: $u$ is absolutely continuous and [F3] applies to $f$. If their continuous difference were nonzero at some point, continuity would make its absolute value bounded below by a positive constant on a relative interval of positive length (one-sided at an endpoint), contradicting equality almost everywhere. Thus the Volterra identity holds for every $t\in [0,T]$.

Conversely, suppose $u-u_0=I^{\alpha }f$. Applying $I^{1-\alpha }$ and using [F2] gives $I^{1-\alpha }(u-u_0)=I^{1}f$. Since $u-u_0=I^{1}v$ by [F4], [F2] also gives $I^{1-\alpha }(u-u_0)=I^1(I^{1-\alpha }v)=I^1{D}_C^{\alpha }u$. Both integrands belong to $L^1$ by [F1], [F2], and [F4], so their indefinite integrals are continuous. These primitives agree almost everywhere by the preceding identities, hence everywhere by the continuity argument in step 1.1. Applying [F5] gives $f=D_C^{\alpha }u$ almost everywhere.

For fixed $n$, the function $w(s)=(t_n-s{)}^{-\alpha }$ is positive and increasing on $[0,t_n)$, and $b_{n,k}=\mathrm{\Gamma }(1-\alpha {)}^{-1}{h}_k^{-1}{\int }_{t_{k-1}}^{t_k}w(s)ds$ by [F1] and [F4]. The factor $\mathrm{\Gamma }(1-\alpha {)}^{-1}$ is positive by [F7]. The average of an increasing function on an earlier interval is at most its value at the shared endpoint, which is at most its average on the next interval. Each integral is finite because $\alpha <1$. Thus the weights are positive and nondecreasing in $k$.

Put $\varphi (t)=t^{\alpha }$. By [F5], ${\varphi }^{\mathrm{\prime }}(s)=\alpha s^{\alpha -1}$ for $s>0$, and this derivative is decreasing because its derivative is $\alpha (\alpha -1)s^{\alpha -2}<0$ and [F8] applies. It is in $L^1(0,T)$ by [F9], with integral $T^{\alpha }$. Its indefinite integral is $\varphi (t)-\varphi (0)=t^{\alpha }$, so $\varphi$ is absolutely continuous by [F6] and its Caputo derivative is defined. In this proof, $D_C^{\alpha }\varphi (t_n)$ denotes the pointwise integral representative in [F2]; the displayed Beta-kernel calculation below proves that this value is finite at every grid node. On an interval $[a,b]=[t_{k-1},t_k]$ its secant slope ${\bar{\varphi }}_k^{\mathrm{\prime }}$ is the average of ${\varphi }^{\mathrm{\prime }}$, hence $e_k(s):={\bar{\varphi }}_k^{\mathrm{\prime }}-{\varphi }^{\mathrm{\prime }}(s)$ is increasing and ${\int }_a^b{e}_k(s)ds=0$. Since $w(s)=(t_n-s{)}^{-\alpha }$ is also increasing, their pairwise differences have the same sign. Expanding the double integral and using ${\int }_a^b{e}_k=0$ gives ${\int }_a^b{e}_k(s)w(s)ds=\frac{1}{2(b-a)}{\int }_a^b{\int }_a^b(e_k(s)-e_k(r))(w(s)-w(r))drds\ge 0.$ These integrals are finite: $e_k,w\in L^1(a,b)$, and the only possible singular product ${\varphi }^{\mathrm{\prime }}(s)w(s)$ occurs when $n=k=1$; there it is the integrable Beta kernel $\alpha s^{\alpha -1}(t_1-s{)}^{-\alpha }$ from [F3]. Expanding the absolute value of the double-integral integrand bounds it by the sum of products of the finite integrals $\int \mathrm{\mid }{e}_k\mathrm{\mid }$, $\int w$, and $\int \mathrm{\mid }{e}_k\mathrm{\mid }w$, which justifies the expansion. Summing over intervals and using [F2] yields ${\delta }_{\mathcal{T}}^{\alpha }\varphi (t_n)\ge D_C^{\alpha }\varphi (t_n)$. Direct substitution $s=t_{n}y$ and [F3]–[F4] give $D_C^{\alpha }(t^{\alpha })(t_n)=\frac{\alpha }{\mathrm{\Gamma }(1-\alpha )}{t}_n^{0}B(\alpha ,1-\alpha )=\mathrm{\Gamma }(1+\alpha )\mathrm{.}$ Thus, using $\mathrm{\Gamma }(1+\alpha )>0$ from [F7], $q^n:=t_n^{\alpha }\mathrm{/}\mathrm{\Gamma }(1+\alpha )$ satisfies $q^0=0$ and ${\delta }_{\mathcal{T}}^{\alpha }{q}^n\ge 1$.

Expanding [F1] by collecting coefficients gives ${\delta }_{\mathcal{T}}^{\alpha }{z}^n=b_{n,n}{z}^n+{\sum }_{k=1}^{n-1}(b_{n,k}-b_{n,k+1})z^k-b_{n,1}{z}^0\mathrm{.}$ If some $z^n<0$, let $m$ be the first such index. Then $z^k\ge 0$ for $k<m$, the middle coefficients in this display are nonpositive by step 1.1, and $-b_{m,1}{z}^0\le 0$. The terms $(b_{m,m}+\lambda )z^m$ are strictly negative because $b_{m,m}>0$ and $\lambda \ge 0$. Therefore ${\delta }_{\mathcal{T}}^{\alpha }{z}^m+\lambda z^m<0$, contradicting the hypothesis. This proves comparison.

For the residual equation, set $z_{\pm }^n:=R{q}^n\pm e^n$. Then $z_{\pm }^0=0$ and ${\delta }_{\mathcal{T}}^{\alpha }{z}_{\pm }^n+\lambda z_{\pm }^n=R({\delta }_{\mathcal{T}}^{\alpha }{q}^n+\lambda q^n)\pm r^n\ge R-\mathrm{\mid }{r}^n\mathrm{\mid }\ge 0.$ Comparison from step 2.1 gives $z_{\pm }^n\ge 0$, hence $\mathrm{\mid }{e}^n\mathrm{\mid }\le R{q}^n\le R{T}^{\alpha }\mathrm{/}\mathrm{\Gamma }(1+\alpha )$ for every $n$. This proves residual stability.