Asymptotic Dynamics of Inertial Particles with Memory

Abstract

Recent experimental and numerical observations have shown the significance of the Basset–Boussinesq memory term on the dynamics of small spherical rigid particles (or inertial particles) suspended in an ambient fluid flow. These observations suggest an algebraic decay to an asymptotic state, as opposed to the exponential convergence in the absence of the memory term. Here, we prove that the observed algebraic decay is a universal property of the Maxey–Riley equation. Specifically, the particle velocity decays algebraically in time to a limit that is $\mathcal {O}(\epsilon )$-close to the fluid velocity, where $0<\epsilon \ll 1$ is proportional to the square of the ratio of the particle radius to the fluid characteristic length scale. These results follow from a sharp analytic upper bound that we derive for the particle velocity. For completeness, we also present a first proof of the global existence and uniqueness of mild solutions to the Maxey–Riley equation, a nonlinear system of fractional differential equations.

Appendices

Appendix 1: Proof of Theorem 2

Consider the fractional differential equation

$$\begin{aligned} \frac{d\mathbf {w}}{d\tau } + \kappa \frac{d^{1/2}\mathbf {w}}{d\tau ^{1/2}} + \mathbf {w} = 0, \end{aligned}$$

(31)

with $\mathbf {w}(0) = \mathbf {w}_0$ as initial condition. Let $\mathbf {W}(s) = \left( \mathcal {L}\left[ \mathbf {w}\right] \right) \left( s\right)$ denote the Laplace transform of $\mathbf {w}\left( \tau \right)$. Since

$$\begin{aligned} \left( \mathcal {L}\left[ \frac{d\mathbf {w}}{d\tau }\right] \right) (s) = s\mathbf {W}(s) - \mathbf {w}_0 \end{aligned}$$

and

$$\begin{aligned} \left( \mathcal {L}\left[ \frac{1}{\sqrt{\tau }}\right] \right) (s) = \sqrt{\frac{\pi }{s}}, \end{aligned}$$

the Laplace transform of the Riemann–Liouville derivative in (31) has the expression

$$\begin{aligned} \left( \mathcal {L}\left[ \frac{d^{1/2}\mathbf {w}}{d\tau ^{1/2}}\right] \right) (s)&= \frac{1}{\sqrt{\pi }} \left( \mathcal {L}\left[ \int _{0}^{\tau } \! \frac{d\mathbf {w}}{d\tau }\frac{1}{\sqrt{\tau - \xi }} \ \hbox {d}\xi \right] \right) (s) + \frac{1}{\sqrt{\pi }}\left( \mathcal {L}\left[ \frac{\mathbf {w}_0}{\sqrt{\tau }}\right] \right) (s), \\&= \frac{1}{\sqrt{\pi }}\left( \mathcal {L}\left[ \frac{d\mathbf {w}}{d\tau }\right] \right) (s)\left( \mathcal {L}\left[ \frac{1}{\sqrt{\tau }}\right] \right) (s) + \frac{\mathbf {w}_0}{\sqrt{s}}, \\&= \left( s\mathbf {W}(s) - \mathbf {w}_0\right) \frac{1}{\sqrt{s}} + \frac{\mathbf {w}_0}{\sqrt{s}}, \\&= \sqrt{s}\mathbf {W}(s), \end{aligned}$$

where we used the identity

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}\tau }\int _{0}^{\tau }\frac{\mathbf {w}(s)}{\sqrt{\tau - s}} \ \hbox {d}s = \int _{0}^{\tau }\frac{\dot{\mathbf {w}}(s)}{\sqrt{\tau -s}} \ \hbox {d}s + \frac{\mathbf w(0)}{\sqrt{\tau }}. \end{aligned}$$

Now we use the Laplace transform on (31) and solve for $\mathbf {W}(s)$ to get

$$\begin{aligned} \mathbf {W}\left( s\right) = \frac{\mathbf {w}_0}{s + \kappa \sqrt{s} + 1}. \end{aligned}$$

The denominator can be factorized as

$$\begin{aligned} \mathbf {W}\left( s\right) = \frac{\mathbf {w}_0}{\left( \sqrt{s} + \lambda _{+}\right) \left( \sqrt{s} + \lambda _{-}\right) }, \end{aligned}$$

where

$$\begin{aligned} \lambda _{\pm } = \frac{\left( \kappa \pm \sqrt{\kappa ^2 - 4}\right) }{2}. \end{aligned}$$

Hence the general solution of (31) is

$$\begin{aligned} \mathbf {w}(\tau ; \mathbf {w}_0) = \mathbf {w}_0 \left( \mathcal {L}^{-1} \left[ \frac{1}{\left( \sqrt{s} + \lambda _{+}\right) \left( \sqrt{s} + \lambda _{-}\right) }\right] \right) \left( \tau \right) \end{aligned}$$

(32)

The function $\mathbf {w}(\tau ;\mathbf w_0)$ is proportional to the Mittag-Leffler function of order $1/2$, which is defined as

$$\begin{aligned} E_{1/2}\left( -z\right) = e^{z^2}{{\mathrm{erfc}}}{z} \end{aligned}$$

(33)

for any complex number $z \in \mathbb {C}$ (see, e.g., Bateman et al. 1955, Section 18.1). The Laplace transform of $E_{1/2}$ is given by (see Haubold et al. 2011, Eq. 11.13):

$$\begin{aligned} \left( \mathcal {L}\left[ E_{1/2}\left( -a\sqrt{z}\right) \right] \right) (s) = \frac{1}{\sqrt{s}\left( \sqrt{s} + a\right) } \end{aligned}$$

(34)

for any $a \in \mathbb {C}$.

To study the behavior of $E_{1/2}\left( -z\right)$ as $z \rightarrow \infty$, we will make use of the asymptotic expansion of the complementary error function (Abramowitz and Stegun 1972, Eq. 7.1.23):

$$\begin{aligned} {{\mathrm{erfc}}}{z} \sim \frac{e^{-z^2}}{z\sqrt{\pi }}\left( 1 - \frac{1}{2z^2} + \frac{3}{4z^4} + \mathcal O\left( \frac{1}{z^6}\right) \right) . \end{aligned}$$

(35)

Substituting in (33), we obtain

$$\begin{aligned} E_{1/2}\left( -z\right) \sim \frac{1}{z\sqrt{\pi }}\left( 1 - \frac{1}{2z^2} + \frac{3}{4z^4} + \mathcal O\left( \frac{1}{z^6}\right) \right) . \end{aligned}$$

(36)

The asymptotic expansion of ${{\mathrm{erfc}}}{z}$ is valid only if $|{{\text {arg}}\left( z\right) }| < \frac{3\pi }{4}$ (Abramowitz and Stegun 1972). It also diverges for any finite value of $z$; its sole purpose is to give the rate of decay as $z \rightarrow \infty$.

The general solution will depend on whether the discriminant of $\lambda _{\pm }$, i.e., $\kappa ^2 - 4$, is positive, zero, or negative.

1.1 Case 1: $\kappa > 2$ ($R>16/9$)

We have

$$\begin{aligned} \mathbf {W}\left( s\right) = \frac{\mathbf {w}_0}{\left( \sqrt{s} + \lambda _{+}\right) \left( \sqrt{s} + \lambda _{-}\right) } \end{aligned}$$

or, after some algebra,

$$\begin{aligned} \mathbf {W}\left( s\right) = \frac{\mathbf {w}_0}{\lambda _{+} - \lambda _{-}}\left[ \frac{\lambda _+}{\sqrt{s}\left( \sqrt{s} + \lambda _+\right) } - \frac{\lambda _-}{\sqrt{s}\left( \sqrt{s} + \lambda _-\right) }\right] . \end{aligned}$$

Invert the two terms in the above expression with the rule (34) to get

$$\begin{aligned} \mathbf {w}(\tau ; \mathbf {w}_0) = \frac{\mathbf {w}_0}{\lambda _{+} - \lambda _{-}}\left[ \lambda _+E_{1/2}\left( -\lambda _+\sqrt{\tau }\right) - \lambda _-E_{1/2}\left( -\lambda _-\sqrt{\tau }\right) \right] . \end{aligned}$$

(37)

Since $\kappa - \sqrt{\kappa ^2-4}$ is always greater than zero, we can use the asymptotic expansion (36) to find that in the limit $\tau \rightarrow \infty$,

$$\begin{aligned} \begin{aligned} \mathbf {w}(\tau ; \mathbf {w}_0)&\sim \frac{\mathbf {w}_0}{\lambda _+ - \lambda _-}\left[ \frac{1}{\sqrt{\pi \tau }}\left( 1 - \frac{1}{2\lambda _{+}^{2}\tau }\right) \right. \\&\quad \left. - \frac{1}{\sqrt{\pi \tau }}\left( 1 - \frac{1}{2\lambda _{-}^{2}\tau }\right) + \mathcal O\left( \tau ^{-5/2}\right) \right] , \\&\sim \frac{\mathbf {w}_0}{2\sqrt{\pi }\left( \lambda _+ - \lambda _-\right) } \left( \frac{\lambda _{+}^{2} - \lambda _{-}^{2}}{\lambda _{+}^{2}\lambda _{-}^{2}}\right) \tau ^{-3/2} + \mathcal O\left( \tau ^{-5/2}\right) , \\&\sim \left( \frac{\kappa \mathbf {w}_0}{2\sqrt{\pi }}\right) \tau ^{-3/2} + \mathcal O\left( \tau ^{-5/2}\right) , \end{aligned} \end{aligned}$$

(38)

where we used that $\lambda _+ + \lambda _- = \kappa$ and $\lambda _+\lambda _- = 1$.

1.2 Case 2: $\kappa = 2$ ($R=16/9$)

We have

$$\begin{aligned} \mathbf {W}(s) = \frac{\mathbf {w}_0}{\left( \sqrt{s} + 1\right) ^2} \end{aligned}$$

(39)

or, after a bit of algebra,

$$\begin{aligned} \begin{aligned} \mathbf {W}(s)&= \mathbf {w}_0\left( \frac{1}{\sqrt{s}\left( \sqrt{s} + 1\right) } - \frac{1}{\sqrt{s}\left( \sqrt{s} + 1\right) ^2}\right) \\&= \mathbf {w}_0\left( \frac{1}{\sqrt{s}\left( \sqrt{s} + 1\right) } + 2 \frac{d}{ds}\left( \frac{1}{\sqrt{s} + 1}\right) \right) . \end{aligned} \end{aligned}$$

(40)

We can invert the first term in (40) with (34). The second term can be inverted by using the Laplace transforms (Gorenflo and Mainardi 1997, Equations A.27, A.28 and A.35.)

$$\begin{aligned} \left( \mathcal {L}\left[ \frac{1}{\sqrt{\pi \tau }} - E_{1/2}\left( -\sqrt{\tau }\right) \right] \right) (s) = \frac{1}{\sqrt{s} + 1} \end{aligned}$$

(41)

and

$$\begin{aligned} \left( \mathcal {L}\left[ -\tau f(\tau )\right] \right) (s) = \frac{d}{ds}(\mathcal {L}\left[ f(\tau )\right] )(s). \end{aligned}$$

(42)

Thus the inverse Laplace transform of (39) is

$$\begin{aligned} \mathbf {w}(\tau ; \mathbf {w}_0) = \mathbf {w}_0\left[ E_{1/2}\left( -\sqrt{\tau }\right) \left( 1 + 2 \tau \right) - \frac{2\sqrt{\tau }}{\sqrt{\pi }}\right] . \end{aligned}$$

(43)

With the asymptotic expansion (36), we find that in the limit $\tau \rightarrow \infty$,

$$\begin{aligned} \begin{aligned} \mathbf {w}(\tau ; \mathbf {w}_0)&\sim \mathbf {w}_0 \left[ \frac{1}{\sqrt{\pi \tau }}\left( 1 - \frac{1}{2 \tau } + \frac{3}{4 \tau ^2} + \mathcal O\left( \tau ^{-3}\right) \right) \right. \\&\quad + \left. \frac{2\sqrt{\tau }}{\sqrt{\pi }}\left( 1 - \frac{1}{2 \tau } + \frac{3}{4 \tau ^2} + \mathcal O\left( \tau ^{-3}\right) \right) - \frac{2\sqrt{\tau }}{\sqrt{\pi }}\right] \\&\sim \left( \frac{\mathbf {w}_0}{\sqrt{\pi }}\right) \tau ^{-3/2} + \mathcal O \left( \tau ^{-5/2}\right) \\ \end{aligned} \end{aligned}$$

(44)

1.3 Case 3: $0 < \kappa < 2$ ($R<16/9$)

We have

$$\begin{aligned} \mathbf {W}\left( s\right) = \frac{\mathbf {w}_0}{\left( \sqrt{s} + \lambda _{+}\right) \left( \sqrt{s} + \lambda _{-}\right) } \end{aligned}$$

This is the same Laplace transform as in the case $\kappa > 2$, except that $\lambda _+$ and $\lambda _-$ are now complex conjugate numbers. The inverse Laplace transform is the same as (37):

$$\begin{aligned} \mathbf {w}(\tau ; \mathbf {w}_0) = \frac{\mathbf {w}_0}{\lambda _{+} - \lambda _{-}}\left[ \lambda _+E_{1/2}\left( -\lambda _+\sqrt{\tau }\right) - \lambda _-E_{1/2}\left( -\lambda _-\sqrt{\tau }\right) \right] . \end{aligned}$$

(45)

The quotients

$$\begin{aligned} \frac{\lambda _+}{\lambda _+ - \lambda _-} = \frac{1}{2}\left( 1 - i \frac{\kappa }{\sqrt{4 - \kappa ^2}}\right) \end{aligned}$$

and

$$\begin{aligned} -\frac{\lambda _-}{\lambda _+ - \lambda _-} = \frac{1}{2}\left( 1 + i \frac{\kappa }{\sqrt{4 - \kappa ^2}}\right) \end{aligned}$$

in (45) are also complex conjugates. Since $\left( e^{\overline{w}}\right) = \overline{\left( e^w\right) }$ and ${{\mathrm{erfc}}}{\overline{w}} = \overline{{{\mathrm{erfc}}}{w}}$ for every $w \in \mathbb {C}$, it follows also that $E_{1/2}\left( \overline{w}\right) = \overline{E_{1/2}\left( w\right) }$. Thus

$$\begin{aligned} \mathbf {w}(\tau ; \mathbf {w_0}) = \mathbf {w_0}\left[ \left( \frac{\lambda _+}{\lambda _+ - \lambda _-}E_{1/2}\left( -\lambda _+\sqrt{\tau }\right) \right) + \overline{\left( \frac{\lambda _+}{\lambda _+ - \lambda _-}E_{1/2}\left( -\lambda _+\sqrt{\tau }\right) \right) } \right] \end{aligned}$$

or simply twice the real part of $w(\tau ; w_0)$.

$$\begin{aligned} \begin{aligned} \mathbf {w}(\tau ; \mathbf {w}_0)&= 2\mathbf {w}_0{\text {Re}}\left( \frac{\lambda _+}{\lambda _+ - \lambda _-}E_{1/2}(-\lambda _+\sqrt{\tau })\right) , \\&= 2\mathbf {w}_0\left[ {\text {Re}}\left( \frac{\lambda _+}{\lambda _+ - \lambda _-}\right) {\text {Re}}\left( E_{1/2}\left( -\lambda _+\sqrt{\tau }\right) \right) \right. \\&\quad + \left. {\text {Im}}\left( \frac{\lambda _+}{\lambda _+ - \lambda _-}\right) {\text {Im}}\left( E_{1/2}\left( -\lambda _+\sqrt{\tau }\right) \right) \right] . \end{aligned} \end{aligned}$$

(46)

It is possible to further simplify (45). It turns out that the Mittag-Leffler function $E_{1/2}\left( -z\right)$ may be written as (DLMF, Section 7.19)

$$\begin{aligned} E_{1/2}\left( -z\right) = \sqrt{\frac{4t}{\pi }}\left[ U\left( x, t\right) + i V\left( x, t\right) \right] , \end{aligned}$$

(47)

where

$$\begin{aligned} U\left( x,t\right)= & {} \frac{1}{\sqrt{4\pi t}} \int _{-\infty }^{\infty } \! \frac{e^{-\left( x + s\right) ^2/\left( 4t\right) }}{1+s^2} \ \hbox {d}s, \end{aligned}$$

(48)

$$\begin{aligned} V\left( x,t\right)= & {} \frac{1}{\sqrt{4\pi t}} \int _{-\infty }^{\infty } \! \frac{se^{-\left( x + s\right) ^2/\left( 4t\right) }}{1+s^2} \ \hbox {d}s, \end{aligned}$$

(49)

$z = \frac{1-ix}{2\sqrt{t}}$, $x \in \mathbb {R}$, and $t > 0$. The functions $U\left( x,t\right)$ and $V\left( x,t\right)$ are known as the Voigt functions (DLMF, Section 7.19; Olver et al. 2010). If we set

$$\begin{aligned} z = \frac{1-ix}{2\sqrt{t}} = \lambda _+ \sqrt{\tau } = \left( \frac{\kappa }{2} + i \frac{\sqrt{4 - \kappa ^2}}{2}\right) \sqrt{\tau }, \end{aligned}$$

then we can solve for $x$ and $t$ to get

$$\begin{aligned} t=\frac{1}{\kappa ^2\tau } \end{aligned}$$

and

$$\begin{aligned} x = -\frac{\sqrt{4 - \kappa ^2}}{\kappa }. \end{aligned}$$

Thus

$$\begin{aligned} \begin{aligned} E_{1/2}\left( -\lambda _+\sqrt{\tau }\right)&= \frac{2}{\kappa \sqrt{\pi \tau }}\left[ U\left( -\frac{\sqrt{4-\kappa ^2}}{\kappa },\frac{1}{\kappa ^2\tau }\right) \right. \\&\quad - \left. i V\left( -\frac{\sqrt{4-\kappa ^2}}{\kappa },\frac{1}{\kappa ^2\tau }\right) \right] . \end{aligned} \end{aligned}$$

(50)

Hence (46) can be written as

$$\begin{aligned} \begin{aligned} \mathbf {w}\left( \tau ; \mathbf {w}_0\right)&= \frac{2\mathbf {w}_0}{\kappa \sqrt{\pi \tau }} \left[ U\left( -\frac{\sqrt{4-\kappa ^2}}{a},\frac{1}{\kappa ^2\tau }\right) \right. \\&\quad - \left. \frac{\kappa }{\sqrt{4 - \kappa ^2}}V\left( -\frac{\sqrt{4-\kappa ^2}}{\kappa },\frac{1}{\kappa ^2\tau }\right) \right] . \end{aligned} \end{aligned}$$

(51)

For the asymptotic behavior of $\mathbf {w}(\tau ; \mathbf {w}_0)$ as $\tau \rightarrow \infty$, we can repeat the steps as in the case $\kappa > 2$ and obtain

$$\begin{aligned} \mathbf {w}(\tau ; \mathbf {w}_0) \sim \left( \frac{\kappa \mathbf {w}_0}{2\sqrt{\pi }}\right) \tau ^{-3/2} + \mathcal O\left( \tau ^{-5/2}\right) . \end{aligned}$$

(52)

This asymptotic expansion, however, is justified only if $|{{\text {arg}}\left( \lambda _+\sqrt{\tau }\right) }|$ and $|{{\text {arg}}\left( \lambda _+\sqrt{\tau }\right) }|$ are smaller than $\frac{3\pi }{4}$. Since $\lambda _{\pm } = \left( \kappa \pm i\sqrt{4 - \kappa ^2}\right) /2$, we see that this will be the case whenever $\kappa > 0$, since then $0 < {\text {arg}}\left( \lambda _+\sqrt{\tau }\right) < \frac{\pi }{2}$ and $-\frac{\pi }{2} < {\text {arg}}\left( \lambda _-\sqrt{\tau }\right) < 0$ (to see this, note that the two complex numbers $\lambda _+$ and $\lambda _-$ lie to the right of the imaginary axis, so that the argument cannot be greater than $\pi / 2$). Note that since $\kappa =\sqrt{9R/2}$, the required condition $\kappa >0$ is always satisfied.

Appendix 2: Proof of Theorem 3

We will use the following Gronwall-type inequality.

Lemma 3

(Chu and Metcalf 1967) Let the functions $\alpha , \beta : \mathbb R^+\rightarrow \mathbb R$ be continuous and the function $K(\tau ,s)$ be continuous and nonnegative for $0\le s\le \tau$. If

$$\begin{aligned} \alpha (\tau )\le \beta (\tau )+\int _{0}^\tau K(\tau ,s)\alpha (s) \ \hbox {d}s, \end{aligned}$$

then

$$\begin{aligned} \alpha (\tau )\le \beta (\tau )+\int _{0}^\tau H(\tau ,s)\beta (s)\ \hbox {d}s, \end{aligned}$$

where $H(\tau ,s)=\sum _{j=1}^\infty K_j(\tau ,s)$, $K_1(\tau ,s)=K(\tau ,s)$ and

$$\begin{aligned} K_j(\tau ,s)=\int _{s}^{\tau }K_{j-1}(\tau ,\xi )K(\xi ,s)\ \hbox {d}\xi ,\ \ \ j\ge 2. \end{aligned}$$

Corollary 1

If $K(\tau ,s)=k(\tau -s)$, then one can show that $K_j(\tau ,s)=k_j(\tau -s)$ where

$$\begin{aligned} k_j(\tau )=(k*k*\cdots *k)(\tau ), \end{aligned}$$

where the convolution is $j$-fold. As a result, $H(\tau ,s)=h(\tau -s)$ where

$$\begin{aligned} h(\tau )=\sum _{j=1}^{\infty }k_j(\tau ). \end{aligned}$$

Proof

We prove $K_2(\tau ,s)=k*k(\tau -s)$. The rest follows similarly by induction.

$$\begin{aligned} K_2(\tau ,s)&:=\int _{s}^{\tau }K(\tau ,\xi )K(\xi ,s) \ \hbox {d}\xi \\&=\int _{s}^{\tau }k(\tau -\xi )k(\xi -s)\ \hbox {d}\xi \\&= \int _0^{\tau -s}k(\tau -s-\eta )k(\eta )\ \hbox {d}\eta \\&= k*k(\tau -s)=:k_2(\tau -s), \end{aligned}$$

where we used the change of variable $\eta =\xi -s$. $\square$

Proof of Theorem 3

It follows from the integral equation (17) that

$$\begin{aligned} |{\mathbf {w}(\tau ;\mathbf {y}_0,\mathbf {w}_0)}|\le \psi (\tau )|{\mathbf {w}_0}|+\epsilon L_B\left( 1 - \phi (\tau )\right) +\epsilon L_M\int _{0}^\tau \psi (\tau -s)|{\mathbf {w}(s;\mathbf {y}_0,\mathbf {w}_0)}|\ \hbox {d}s \end{aligned}$$

(53)

where $\tau \in [0, \delta )$. Using Lemma 3 with $\alpha (\tau ) = |{\mathbf {w}(\tau ;\mathbf {y}_0,\mathbf {w}_0)}|$, $\beta (\tau ) = \psi (\tau )|{\mathbf {w}_0}| + \epsilon L_B\left( 1-\phi (\tau )\right)$ and $K(\tau ,s)=\epsilon L_M\psi (\tau -s)$, we get

$$\begin{aligned} |{\mathbf {w}(\tau ;\mathbf {y}_0,\mathbf {w}_0)}|&\le \psi (\tau )|{\mathbf {w}_0}|+\epsilon L_B\left( 1-\phi (\tau )\right) \nonumber \\&\quad + \int _{0}^{\tau }h(\tau -s)\left[ \psi (s)|\mathbf {w}_0|+\epsilon L_B\left( 1-\phi (\tau )\right) \right] \ \hbox {d}s\nonumber \\&= \left[ \psi (\tau )+\int _{0}^\tau \! h(\tau -s)\psi (s) \ \hbox {d}s\right] |{\mathbf {w}_0}|+\epsilon L_B\left( 1-\phi (\tau )\right) \nonumber \\&\quad +\epsilon L_B\int _{0}^{\tau }h(\tau -s)\left( 1-\phi (s)\right) \ \hbox {d}s, \end{aligned}$$

(54)

where $h(\tau ;\epsilon )=\sum _{j=1}^{\infty }k_j(\tau )$ with $k_1=\epsilon L_M\psi$ and $k_j=k_{j-1}*k_1$. Induction on $j$ leads to the expression

$$\begin{aligned} k_j=(\epsilon L_M)^{j}\psi ^{*j}. \end{aligned}$$

Therefore we have the identity

$$\begin{aligned} \psi (\tau )+\int _{0}^\tau h(\tau -s)\psi (s) \ \hbox {d}s&=\frac{k_1(\tau )}{\epsilon L_M}+\int _{0}^{\tau } \!\sum _{j=1}^{\infty }k_j(\tau -s)\frac{k_1(s)}{\epsilon L_M} \ \hbox {d}s\\&= \frac{k_1(\tau )}{\epsilon L_M}+\frac{1}{\epsilon L_M}\sum _{j=1}^{\infty }\int _{0}^{\tau } \! k_j(\tau -s)k_1(s) \ \hbox {d}s\\&= \frac{k_1(\tau )}{\epsilon L_M}+\frac{1}{\epsilon L_M}\sum _{j=1}^{\infty }k_{j+1}(\tau )\\&= \frac{1}{\epsilon L_M}\sum _{j=1}^{\infty }k_{j}(\tau )\\&= \frac{1}{\epsilon L_M}h(\tau ), \end{aligned}$$

where we omitted the dependence of $h$ on the parameter $\epsilon$ for notational simplicity.

This shows that

$$\begin{aligned} |{\mathbf {w}(\tau ;\mathbf {y}_0,\mathbf {w}_0)}| \le \frac{|{\mathbf {w}_0}|}{\epsilon L_M}h(\tau ) + \epsilon L_B \left( 1-\phi (\tau )\right) + \epsilon L_B \int _{0}^{\tau } h(\tau - s)\left( 1-\phi (s)\right) \ \hbox {d}s. \end{aligned}$$

(55)

Since $0 \le \phi (\tau ) \le 1$, we have that $(1-\phi (\tau )) \le 1$ and therefore the inequality can be further simplified to

$$\begin{aligned} |{\mathbf {w}(\tau ;\mathbf {y}_0,\mathbf {w}_0)}| \le \frac{|{\mathbf {w}_0}|}{\epsilon L_M}h(\tau ) + \epsilon L_B\left[ 1-\phi (\tau ) \right] + \epsilon L_B\int _{0}^{\tau } h(s) \ \hbox {d}s. \end{aligned}$$

(56)

So far we have assumed that the series $\sum _{j=1}^{\infty }k_j=\sum _{j=1}^{\infty }(\epsilon L_M)^j\psi ^{*j}$ converges uniformly to a limit $h$. To prove this, we first show that for any $j$ and $\tau \ge 0$, $0\le \psi ^{*j}(\tau )\le 1$. For $j=1$, this property holds since $0\le \psi \le 1$. For $j=2$, we have

$$\begin{aligned} 0\le \psi ^{*2}(\tau ):=\int _{0}^{\tau }\psi (\tau -s)\psi (s) \ \hbox {d}s\le \int _{0}^{\tau }\psi (s) \ \hbox {d}s=1-\phi (\tau )\le 1. \end{aligned}$$

By induction on $j$, we get $0\le \psi ^{*j}(\tau )\le 1$. As a result, $(\epsilon L_M)^j\psi ^{*j}\le (\epsilon L_M)^j$. Since $\epsilon L_M<1$, the series $\sum _{j=1}^{\infty }(\epsilon L_M)^j$ converges. It follows that

$$\begin{aligned} |{h(\tau )}| \le \sum _{j=1}^{\infty }(\epsilon L_M)^j = \frac{\epsilon L_M}{1 - \epsilon L_M} \end{aligned}$$

(57)

by summing up the geometric series. By the dominated convergence theorem, the sequence $\sum _{j=1}^{n}(\epsilon L_M)^j\psi ^{*j}$ converges uniformly to a function $h$ as $n\rightarrow \infty$. Since for any $n$, the series $\sum _{j=1}^{n}(\epsilon L_M)^j \psi ^{*j}$ is continuous, so is the limiting function $h$. This shows that $h:[0,\infty )\rightarrow \mathbb R$ is continuous and $h\ge 0$.

Now, observe that

$$\begin{aligned} \begin{aligned} \int _{0}^\tau h(\xi ) \ \hbox {d}\xi&=\int _{0}^\tau \sum _{j=1}^{\infty } (\epsilon L_M)^j\psi ^{*j}(\xi ) \ \hbox {d}\xi \\&= \sum _{j=1}^{\infty } (\epsilon L_M)^j \int _{0}^\tau \psi ^{*j}(\xi ) \ \hbox {d}\xi \\&\le \sum _{j=1}^{\infty } (\epsilon L_M)^j = \frac{\epsilon L_M}{1 - \epsilon L_M}, \\ \end{aligned} \end{aligned}$$

(58)

where we used the uniform convergence of the series and the fact that, for any $j$,

$$\begin{aligned} 0\le \int _{0}^\tau \psi ^{*j}(\xi )\ \hbox {d}\xi&\le \left( \int _{0}^\tau \psi ^{*(j-1)}(\xi ) \ \hbox {d}\xi \right) \left( \int _{0}^\tau \psi (\xi ) \ \hbox {d}\xi \right) \\&\le \cdots \le \left( \int _{0}^\tau \psi (\xi ) \ \hbox {d}\xi \right) ^j= (1-\phi (\tau ))^j\le 1, \end{aligned}$$

by repeated application of Young’s inequality for convolutions. This also shows that $h(\tau )\rightarrow 0$ as $\tau \rightarrow \infty$, since $|{h}|_1<\infty$ and $h$ is uniformly continuous.

Using inequality (58) in (56) and the definition of $h$, we get

$$\begin{aligned} |{\mathbf {w}(\tau ;\mathbf {y}_0,\mathbf {w}_0)}|&\le \frac{|{\mathbf {w}_0}|}{\epsilon L_M}h(\tau ) + \epsilon L_B\left( 1-\phi (\tau )\right) +\frac{\epsilon ^2 L_M L_B}{1 - \epsilon L_M} \end{aligned}$$

(59)

$$\begin{aligned}&= |{\mathbf {w}_0}|\left[ \sum _{j=1}^{\infty }(\epsilon L_M)^{j-1}\psi ^{*j}(\tau ) \right] + \epsilon L_B\left( 1-\phi (\tau )\right) +\frac{\epsilon ^2 L_M L_B}{1 - \epsilon L_M}. \end{aligned}$$

(60)

This proves part (i) of the theorem.

Taking the sup of $|{\mathbf {w}(\tau ;\mathbf {y}_0,\mathbf {w}_0)}|$ over $[0, \delta )$, we get

$$\begin{aligned} \sup _{0\le \tau <\delta }|{\mathbf {w}(\tau ;\mathbf y_0,\mathbf w_0)}| \le \frac{|{\mathbf {w}_0}| + \epsilon L_B}{1 - \epsilon L_M}, \end{aligned}$$

(61)

which proves part (ii) of Theorem 3.

If $\delta = \infty$, then we can take the limitsup of $|\mathbf w|$. Using inequality (59), we get the asymptotic estimate

$$\begin{aligned} \limsup _{\tau \rightarrow \infty }|\mathbf w(\tau ;\mathbf y_0,\mathbf w_0)|\le \frac{\epsilon L_B}{1-\epsilon L_M}, \end{aligned}$$

(62)

which proves part (iii) of Theorem 3. Here, we used the fact that $\lim _{\tau \rightarrow \infty } h(\tau ) = 0$ and $\lim _{\tau \rightarrow \infty } \phi (\tau ) = 0$. $\square$

Appendix 3: Proof of Lemma 1

Let $\tau _1$, $\tau _2 \in [0, \delta )$. Bound $|{\mathbf {z}_{\mathrm{loc}}(\tau _2) - \mathbf {z}_{\mathrm{loc}}(\tau _1)}|$ by

$$\begin{aligned} \begin{aligned} |{\mathbf {z}_{\mathrm{loc}}(\tau _2) - \mathbf {z}_{\mathrm{loc}}(\tau _1)}|&\le |{\mathbf {y}_{\mathrm{loc}}(\tau _2) - \mathbf {y}_{\mathrm{loc}}(\tau _1)}| + |{\mathbf {w}_{\mathrm{loc}}(\tau _2) - \mathbf {w}_{\mathrm{loc}}(\tau _1)}| \nonumber \\&\le |{\mathbf {w}_0}||{\psi (\tau _2) - \psi (\tau _1)}| + \epsilon \int _{\tau _1}^{\tau _2} \! |{\mathbf {w}_{\mathrm{loc}}(s)}| + |{\mathbf {A_u}(\mathbf {y}_{\mathrm{loc}}(s), s)}| \ \hbox {d}s \nonumber \\&\quad +\, \epsilon \int _{\tau _1}^{\tau _2} \! \psi (\tau _2 - s)\left[ |{\mathbf {M_u}(\mathbf {y}_{\mathrm{loc}}(s), s)}||{\mathbf {w}_{\mathrm{loc}}(s)}|\right. \nonumber \\&\quad \left. +\, |{\mathbf {B_u}(\mathbf {y}_{\mathrm{loc}}(s), s)}|\right] \ \hbox {d}s \nonumber \\&\quad +\, \epsilon \int _{0}^{\tau _1} \! |{\psi (\tau _2 - s) - \psi (\tau _1 - s)}|\left[ |{\mathbf {M_u}(\mathbf {y}_{\mathrm{loc}}(s),s)}||{\mathbf {w}_{\mathrm{loc}}(s)}|\right. \\&\quad \left. +\, |{\mathbf {B_u}(\mathbf {y}_{\mathrm{loc}}(s), s)}|\right] \ \hbox {d}s \nonumber \end{aligned} \end{aligned}$$

Without loss of generality, suppose $\tau _1 \le \tau _2$, so that $|{\psi (\tau _2 - s) - \psi (\tau _1 - s)}| = \psi (\tau _2 - s) - \psi (\tau _1 - s)$. Taking the infinity norm over $[0, \delta )$ to bound $||{\mathbf {M_u}(\mathbf {y}_{\mathrm{loc}}(s),s)}||_\infty$, $||{\mathbf {B_u}(\mathbf {y}_{\mathrm{loc}}(s),s)}||_\infty$, $||{\mathbf {A_u}(\mathbf {y}_{\mathrm{loc}}(s),s)}||_\infty$ and $|{\mathbf {w}_{\mathrm{loc}}(s)}|$ by Theorem 3, we get

$$\begin{aligned} \begin{aligned}&|{\mathbf {z}_{\mathrm{loc}}(\tau _2) - \mathbf {z}_{\mathrm{loc}}(\tau _1)}| \le |{\mathbf {w}_0}||{\psi (\tau _2) - \psi (\tau _1)}| + \epsilon \left( L_A + \frac{|{\mathbf {w}_0}| + \epsilon L_B}{1 - \epsilon L_M}\right) |{\tau _2 - \tau _1}| \\&\quad + \epsilon \left[ \frac{L_M|{\mathbf {w}_0}| + L_B}{1 - \epsilon L_M}\right] \left( |{\tau _2 - \tau _1}| + \int _{0}^{\tau _1} \! \psi (\tau _1 - s) - \psi (\tau _2 - s) \ \hbox {d}s\right) . \end{aligned} \end{aligned}$$

By the results of Theorem 2, $\psi (\tau _1 - s) - \psi (\tau _2 - s) = \phi '(\tau _2 - s) - \phi '(\tau _1 - s) \ge 0$. Integrate and rearrange to obtain

$$\begin{aligned} \begin{aligned} |{\mathbf {z}_{\mathrm{loc}}(\tau _2) - \mathbf {z}_{\mathrm{loc}}(\tau _1)}|&\le |{\mathbf {w}_0}||{\psi (\tau _2) - \psi (\tau _1)}| + \epsilon \left( L_A + \frac{|{\mathbf {w}_0}| + \epsilon L_B}{1 - \epsilon L_M}\right) |{\tau _2 - \tau _1}| \\&\quad + \epsilon \left[ \frac{L_M|{\mathbf {w}_0}| + L_B}{1 - \epsilon L_M}\right] \left( |{\tau _2 - \tau _1}| + \phi (\tau _2) - \phi (\tau _1) \right) \\&\quad + \epsilon \left[ \frac{L_1|{\mathbf {w}_0}| + L_2}{1 - \epsilon L_1}\right] \left( \phi (0) - \phi (\tau _2 - \tau _1)\right) . \end{aligned} \end{aligned}$$

Since both $\psi$ and $\phi$ are uniformly continuous over $[0, \infty )$ by Theorem 2, each of $|{\psi (\tau _2) - \psi (\tau _1)}|$, $|{\phi (\tau _2) - \phi (\tau _1)}|$ and $|{\phi (0) - \phi (\tau _2 - \tau _1)}| \rightarrow 0$ as $|{\tau _2 - \tau _1}| \rightarrow 0$. Hence $|{\mathbf {z}_{\mathrm{loc}}(\tau _2) - \mathbf {z}_{\mathrm{loc}}(\tau _1)}| \rightarrow 0$ as $\tau _1$, $\tau _2 \rightarrow \delta _{-}$.

Now, if we take a sequence $\left. \{t_n\} \right.$ $t_n \in [0, \delta )$ such that $\lim _{n \rightarrow \infty } t_n \rightarrow \delta$, then it follows that $\left. \{\mathbf {z}_{\mathrm{loc}}(t_n)\} \right.$ is a Cauchy sequence. The sequence is convergent in $\mathbb {R}^{2n}$ since $\mathbb {R}^{2n}$ is a complete metric space. The limit is given by the integral equation (16) evaluated at $\tau = \delta$:

$$\begin{aligned} \mathbf {z}_{\mathrm{loc}}(\delta ) = \left( \begin{array}{lr} \mathbf {y}_0 + \epsilon \int _{0}^{\delta } \! \mathbf {w}_{\mathrm{loc}}(s) + \mathbf {A_u}(\mathbf {y}_{\mathrm{loc}}(s), s) \ \hbox {d}s \nonumber \\ \psi (\delta )\mathbf {w}_0 + \epsilon \int _{0}^{\delta } \! \psi (\tau - s)\left[ -\mathbf {M_u}(\mathbf {y}_{\mathrm{loc}}(s), s)\mathbf {w}_{\mathrm{loc}}(s) + \mathbf {B_u}(\mathbf {y}_{\mathrm{loc}}(s), s)\right] \ \hbox {d}s \nonumber \end{array} \right) . \end{aligned}$$

This ends the proof.

Appendix 4: Proof of Lemma 2

Let ${\varvec{\Phi }} = \left( \pmb \xi ,\pmb \eta \right) \in X_K^{\delta ,h}$, and $\tau _1$, $\tau _2 \in [\delta , \delta + h)$. Bound $|{(\mathbf {F}{\varvec{\Phi }})(\tau _2) - (\mathbf {F}{\varvec{\Phi }})(\tau _1)}|$ by

$$\begin{aligned} \begin{aligned} |{(\mathbf {F}{\varvec{\Phi }})(\tau _2) - (\mathbf {F}{\varvec{\Phi }})(\tau _1)}|&\le |{{\varvec{\Phi }}_\mathbf {0}(\tau _2) - {\varvec{\Phi }}_\mathbf {0}(\tau _1)}| + \epsilon \int _{\tau _1}^{\tau _2} \! |{\pmb \eta (s)}| + |{\mathbf {A_u}(\pmb \xi (s), s)}| \ \hbox {d}s \\&\quad + \epsilon \int _{\tau _1}^{\tau _2} \! \psi (\tau _2 - s)\left[ |{\mathbf {M_u}(\pmb \xi (s), s)}||{\pmb \eta (s)}| + |{\mathbf {B_u}(\pmb \xi (s), s)}|\right] \ \hbox {d}s \\&\quad + \epsilon \int _{\delta }^{\tau _1} \! \left( \psi (\tau _2 - s) - \psi (\tau _1 - s)\right) \left[ |{\mathbf {M_u}(\pmb \xi (s), s)}||{\pmb \eta (s)}|\right. \\&\quad \left. +\, |{\mathbf {B_u}(\pmb \xi (s), s)}|\right] \ \hbox {d}s, \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} |{{\varvec{\Phi }}_\mathbf {0}(\tau _2) - {\varvec{\Phi }}_\mathbf {0}(\tau _1)}|&\le |{\mathbf {w}_0}||{\psi (\tau _2) - \psi (\tau _1)}| \\&\quad + \,\epsilon \int _{0}^{\delta } \! |{\psi (\tau _2 - s) - \psi (\tau _1 - s)}|\left[ |{\mathbf {M_u}(\mathbf {y}_{\mathrm{loc}}(s), s)}||{\mathbf {w}_{\mathrm{loc}}(s)}|\right. \\&\quad \left. +\, |{\mathbf {B_u}(\mathbf {y}_{\mathrm{loc}}(s), s)}| \right] \ \hbox {d}s. \end{aligned} \end{aligned}$$

Without loss of generality, suppose $\tau _1 \le \tau _2$, so that $|{\psi (\tau _2 - s) - \psi (\tau _1 - s)}| = \psi (\tau _2 - s) - \psi (\tau _1 - s)$. Taking the infinity norm over $[\delta , \delta + h)$ to bound $||{\mathbf {M_u}(\pmb \xi (s), s)}||_\infty$, $||{\mathbf {B_u}(\pmb \xi (s), s)}||_\infty$, $||{\mathbf {A_u}(\pmb \xi (s), s)}||_\infty$, $||{\pmb \eta (s)}||_\infty$ and $|{\mathbf {w}_{\mathrm{loc}}(s)}|$ by inequality (19), we get

$$\begin{aligned} \begin{aligned} |{(\mathbf {F}{\varvec{\Phi }})(\tau _2) - (\mathbf {F}{\varvec{\Phi }})(\tau _1)}|&\le |{\mathbf {w}_0}||{\psi (\tau _2) - \psi (\tau _1)}| \\&\quad + \epsilon \left( \frac{L_M|{\mathbf {w}_0}| + L_B}{1 - \epsilon L_M}\right) \int _{0}^{\delta } \! \psi (\tau _1 - s) - \psi (\tau _2 - s) \ \hbox {d}s \\&\quad + \epsilon (K + L_A)|{\tau _2 - \tau _1}| + \epsilon \left( L_MK + L_B\right) |{\tau _2 - \tau _1}| \\&\quad + \epsilon \left( L_MK + L_B\right) \int _{\delta }^{\tau _1} \! \psi (\tau _1 - s) - \psi (\tau _2 - s) \ \hbox {d}s. \end{aligned} \end{aligned}$$

By the results of Theorem 2, $\psi (\tau _1 - s) - \psi (\tau _2 - s) = \phi '(\tau _2 - s) - \phi '(\tau _1 - s) \ge 0$. Finally, integrate and rearrange to obtain

$$\begin{aligned} |{(\mathbf {F}{\varvec{\Phi }})(\tau _2) - (\mathbf {F}{\varvec{\Phi }})(\tau _1)}|&\le |{\mathbf {w}_0}||{\psi (\tau _2) - \psi (\tau _1)}| \\&\quad +\, \epsilon \left( \frac{L_M|{\mathbf {w}_0}| + L_B}{1 - \epsilon L_M}\right) \left[ \left( \phi (\tau _1 - \delta ) - \phi (\tau _2 - \delta )\right) \right. \\&\quad \left. +\, \left( \phi (\tau _2) - \phi (\tau _1)\right) \right] \\&\quad + \, \epsilon (K + L_A)|{\tau _2 - \tau _1}| + \epsilon \left( L_MK + L_B\right) |{\tau _2 - \tau _1}| \\&\quad +\, \epsilon \left( L_MK + L_B\right) \left[ \left( \phi (0) - \phi (\tau _2 - \tau _1)\right) \right. \\&\quad \left. +\, \left( \phi (\tau _2 - \delta ) - \phi (\tau _1 - \delta )\right) \right] . \end{aligned}$$

Since both $\psi$ and $\phi$ are uniformly continuous over $[0, \infty )$ by Theorem 2, each of $|{\psi (\tau _2) - \psi (\tau _1)}|$, $|{\phi (\tau _2) - \phi (\tau _1)}|$, $|{\phi (\tau _1 -\delta ) - \phi (\tau _2 - \delta )}|$ and $|{\phi (0) - \phi (\tau _2 - \tau _1)}| \rightarrow 0$ as $|{\tau _2 - \tau _1}| \rightarrow 0$. Hence $|{(\mathbf {F}{\varvec{\Phi }})(\tau _2) - (\mathbf {F}{\varvec{\Phi }})(\tau _1)}| \rightarrow 0$ as $|{\tau _2 - \tau _1}| \rightarrow 0$. This shows that $\mathbf {F}$ maps $X_K^{\delta ,h}$ to a family of uniformly equicontinuous functions in $C(\left[ \delta , \delta + h\right) ; \mathbb {R}^{2n})$.