A dynamic viscoelastic analogy for fluid-filled elastic tubes

Abstract

In this paper we evaluate the dynamic effects of the fluid viscosity for fluid filled elastic tubes in the framework of a linear uni-axial theory. Because of the linear approximation, the effects on the fluid inside the elastic tube are taken into account according to the Womersley theory for a pulsatile flow in a rigid tube. The evolution equations for the response variables are derived by means of the Laplace transform technique and they all turn out to be the very same integro-differential equation of the convolution type. This equation has the same structure as the one describing uni-axial waves in linear viscoelastic solids characterized by a relaxation modulus or by a creep compliance. In our case, the analogy is connected with a peculiar viscoelastic solid which exhibits creep properties similar to those of a fractional Maxwell model (of order 1 / 2) for short times, and of a standard Maxwell model for long times. The present analysis could find applications in biophysics concerning the propagation of pressure waves within large arteries.

Appendix: Mathematical discussions

We provide the details for the proof of some statements found in the text.

1.1 Proof of the expressions (24) and (31)

Consider the Laplace Transform of the Relaxation-memory function $\widetilde{{\varPhi }} (s)$

$$\begin{aligned} \widetilde{{\varPhi }} (s ) = \frac{2}{\sqrt{s \tau }} \frac{I_1 (\sqrt{s \tau })}{I_0 (\sqrt{s \tau })} \end{aligned}$$

(43)

where we will consider $\tau = 1$ for sake of simplicity (it can be restored by make the substitution $s \, \leftrightarrow \, s \tau$). Then, the Relaxation-memory function is given by

$$\begin{aligned} {\varPhi }(t) = 4 \sum _{n=1} ^\infty \exp \left( -\, \lambda _n ^2 t \right) \end{aligned}$$

(44)

where $\lambda _n \in {\mathbb {R}}$ are such that $J_0 (\lambda _n) = 0, \,\, \forall n = 1, 2, 3, \ldots$ and $t>0$.

Proof

Firstly, consider the power series representation for the Modified Bessel functions of the First Kind:

$$\begin{aligned} \begin{aligned} I_{0} (\sqrt{s})&= 1 + \frac{s}{4} + \frac{s^2}{64} + O(s^3) , \\ I_{1} (\sqrt{s})&= \sqrt{s} \left( \frac{1}{2} + \frac{s}{16} + \frac{s^2}{384} \right) + O(s^{7/2}) , \\ \end{aligned} \end{aligned}$$

(45)

one can eventually deduce that

$$\begin{aligned} \widetilde{{\varPhi }} (s) = \frac{2}{\sqrt{s}} \frac{I_{1} (\sqrt{s})}{I_{0} (\sqrt{s})} = 2 \,\frac{\frac{1}{2} + \frac{s}{16} + \frac{s^2}{384} + \cdots }{1 + \frac{s}{4} + \frac{s^2}{64} + \cdots } \end{aligned}$$

(46)

which means that the function of our concern is regular in $s=0$ and it does not have any branch cuts. Then we can obtain the required function by means of the Bromwich Integral:

$$\begin{aligned} {\varPhi }(t) = \frac{1}{2 \pi i} \int _{Br} \widetilde{{\varPhi }} (s) \, e{{^{st}}} \, ds . \end{aligned}$$

(47)

In particular, $\widetilde{{\varPhi }} (s)$ has simple poles such that: $I_0 (\sqrt{s}) = 0$. Now, if we rename $\sqrt{s}$ as $\sqrt{s} = - i \lambda$, then

$$\begin{aligned} I_{0} (\sqrt{s}) = 0 \quad \Longleftrightarrow \quad J_0 (\lambda ) = 0 . \end{aligned}$$

Moreover,

$$\begin{aligned} \sqrt{s} _n = - i \lambda _n \quad \Leftrightarrow \quad s_n = - \lambda ^2 _n, \quad n \in {\mathbb {N}}. \end{aligned}$$

(48)

From the previous statements, we can then conclude that:

$$\begin{aligned} \begin{aligned} {\varPhi }(t)&= \sum _{s_n} \mathcal {R}es \,\left\{ \widetilde{{\varPhi }} (s) \, e^{st} \right\} _{s_n} \\&= \sum _{n=1} ^\infty \mathcal {R}es \,\left\{ \frac{2}{\sqrt{s}} \, \frac{I_1 (\sqrt{s})}{I_0 (\sqrt{s})} \, e^{st} \right\} _{s = - \lambda ^2 _n} . \end{aligned} \end{aligned}$$

(49)

It is quite straightforward that

$$\begin{aligned} \begin{aligned} \mathcal {R}es \,&\left\{ \frac{2}{\sqrt{s}} \frac{I_{1} (\sqrt{s}) e^{st}}{I_{0} (\sqrt{s})} \right\} _{s = s_n} \\&= \lim _{s \rightarrow s_n} (s - s_n) \, \frac{2}{\sqrt{s}} \frac{I_{1} (\sqrt{s}) e^{st}}{I_{0} (\sqrt{s})} \\&= 4 \exp \left( s_n t \right) . \end{aligned} \end{aligned}$$

(50)

Thus,

$$\begin{aligned} {\varPhi }(t) = \sum _{n=1} ^\infty \mathcal {R}es \,\left\{ \widetilde{{\varPhi }} (s) \, e^{st} \right\} _{s = - \lambda ^2 _n} = 4 \sum _{n=1} ^\infty e^{- \lambda ^2 _n \, t} . \end{aligned}$$

(51)

$\square$

Let us now consider the Laplace Transform of the Creep-memory function $\widetilde{{\varPsi }} (s)$

$$\begin{aligned} \widetilde{{\varPsi }} (s) = \frac{2}{\sqrt{s}} \frac{I_1 (\sqrt{s})}{I_2 (\sqrt{s})} . \end{aligned}$$

(52)

Then ${\varPsi }(t)$ is given by

$$\begin{aligned} {\varPsi }(t) = 8 + 4 \sum _{n=1} ^\infty \exp \left( - \,\mu ^2 _n \, t \right) \end{aligned}$$

(53)

where $\mu _n$ are such that $J_2 (\mu _n) = 0$ and $\mu _n \ne 0$, for $n \in \mathbb {N}$.

Proof

By means of the same procedure shown before we are able to point out that

$$\begin{aligned} \begin{aligned} {\varPsi }(t)&= \sum _{s_n} \mathcal {R}es \,\left\{ \widetilde{{\varPsi }} (s) \, e^{st} \right\} _{s_n} \\&= \sum _{n = 0} ^\infty \mathcal {R}es \,\left\{ \frac{2}{\sqrt{s}} \, \frac{I_1 (\sqrt{s})}{I_2 (\sqrt{s})} \, e^{st} \right\} _{s = - \mu ^2 _n} \end{aligned} \end{aligned}$$

(54)

with $\mu _n$ such that $J_2 (\mu _n) = 0$, $\mu _n \ne 0$, f or $n \in \mathbb {N}$ and $\mu _0 \equiv 0$.

Now, we have to distinguish two cases:

If $s_n \ne 0$, then

$$\begin{aligned} \mathcal {R}es \,\left\{ \widetilde{{\varPsi }} (s) \, e^{st} \right\} _{s_n} = 4 \, \exp \left( s_n t \right) . \end{aligned}$$

(55)

Otherwise, if $s_n = \mu _0 = 0$,

$$\begin{aligned} \mathcal {R}es \,\left\{ \widetilde{{\varPsi }} (s) \, e^{st} \right\} _{s_n = 0} = \lim _{s \rightarrow 0} s \, \widetilde{{\varPsi }} (s) \, e^{st} = 8 . \end{aligned}$$

(56)

Thus,

$$\begin{aligned} \begin{aligned} {\varPsi }(t)&= \sum _{n=0} ^\infty \mathcal {R}es \,\left\{ \widetilde{{\varPsi }} (s) \, e^{st} \right\} _{s = - \mu ^2 _n} \\&= 8 + 4 \sum _{n=1} ^\infty \exp \left( - \,\mu ^2 _n \, t \right) . \end{aligned} \end{aligned}$$

(57)

$\square$

From the above results we are able to conclude that representation of both memory function ${\varPhi }(t)$ and ${\varPsi }(t)$ are given by Dirichlet series (44) and (53), whose convergence is proved in the following.

1.2 On the convergence of the Dirichlet series (44) and (53)

The series (44) is convergent for $t>0$.

Proof

Consider a Generalized Dirichlet Series:

$$\begin{aligned} f(z) = \sum _{n=1} ^\infty a_n \, \exp \left( - \,\alpha _n z \right) , \quad z \in {\mathbb {C}}. \end{aligned}$$

(58)

In general, we have that the abscissa of convergence and the abscissa of absolute convergence would be different, i.e. $\sigma _c \ne \sigma _a$, but they will satisfy the following condition:

$$\begin{aligned} 0 \le \sigma _a - \sigma _c \le d = \limsup _{n \rightarrow \infty } \frac{\ln n}{\alpha _n} . \end{aligned}$$

(59)

If $d=0$, then

$$\begin{aligned} \sigma \equiv \sigma _c = \sigma _a = \limsup _{n \rightarrow \infty } \frac{\ln \left| a_n \right| }{\alpha _n} . \end{aligned}$$

(60)

In our case $a_n = 1$ and $\alpha _n = \lambda ^2 _n \ne 0$. Then, we have to understand the behavior of the coefficients $\lambda _n$ for $n \gg 1$, where $J_0 (\lambda _n) = 0$, $\forall n \in \mathbb {N}$.

Considering the asymptotic representation

$$\begin{aligned} J_{0} (x) \overset{x \gg 1}{\sim } \sqrt{\frac{2}{\pi x}} \, \cos \left( x - \frac{\pi }{4} \right) \end{aligned}$$

(61)

we get to the following conclusion:

$$\begin{aligned} J_0 (\lambda _n) = 0 , \,\,\, \text{ for } \,\, n \gg 1 \, \Longrightarrow \, \lambda _n \propto n , \,\,\, \text{ for } \,\, n \gg 1 . \end{aligned}$$

(62)

Thus,

$$\begin{aligned} \frac{\ln n}{\alpha _n} = \frac{\ln n}{\lambda ^2 _n} \,\, \overset{n \gg 1}{\sim } \,\, \frac{\ln n}{n^2} \,\, \overset{n \rightarrow \infty }{\longrightarrow } \,\, 0 \end{aligned}$$

(63)

which tells us that $d=0$. Finally,

$$\begin{aligned} \sigma \equiv \sigma _c = \sigma _a = \limsup _{n \rightarrow \infty } \frac{\ln \left| a_n \right| }{\alpha _n} = 0 \end{aligned}$$

(64)

being $a_n = 1$.

This result implies that the series (44), with $a_n = 1$ and $\alpha _n = \lambda ^2 _n \ne 0$, converges for $\texttt {Re} \{ z\} = t > 0$. $\square$

In a similar way a we can prove that the series (53) is convergent for $t>0$.

1.3 On the asymptotic representations

Finally, we derive the asymptotic representations for ${\varPhi }(t)$ and ${\varPsi }(t)$ as $t\rightarrow 0^+$ applying the Tauberian theorems to the corresponding Laplace transforms:

$$\begin{aligned} \widetilde{{\varPhi }} (s ) = \frac{2}{\sqrt{s \tau }} \frac{I_1 (\sqrt{s \tau })}{I_0 (\sqrt{s \tau })} , \quad \widetilde{{\varPsi }} (s ) = \frac{2}{\sqrt{s \tau }} \, \frac{I_1 (\sqrt{s \tau })}{I_2 (\sqrt{s \tau })}. \end{aligned}$$

(65)

Then, in view of the asymptotic representation of the modified Bessel functions as $z\rightarrow \infty$ with $|{\mathrm {arg}}(z) | <\pi /2$ and for any $\nu$

$$\begin{aligned} I_\nu (z) \sim \frac{1}{\sqrt{2\pi }} z^{-1/2} \exp (z), \end{aligned}$$

see e.g. [1], we conclude that for $z=\sqrt{s\tau }\rightarrow \infty$ ($s>0$) we get

$$\begin{aligned} \widetilde{\varPhi }(s)\sim \frac{2}{\sqrt{s\tau }}, \quad \widetilde{\varPsi }(s)\sim \frac{2}{\sqrt{s\tau }}, \end{aligned}$$

so that

$$\begin{aligned} \begin{aligned} {\varPhi }(t)&\sim \frac{2}{ \sqrt{\pi \tau }} \, t^{-1/2} , \quad t \rightarrow 0^+, \\ {\varPsi }(t)&\sim \frac{2}{ \sqrt{\pi \tau }} \, t^{-1/2} , \quad t \rightarrow 0^+ . \end{aligned} \end{aligned}$$

(66)