2.1. Dimensional formulation

We will concern ourselves only with an incompressible fluid at low Reynolds number under the lubrication approximation. Since the radial velocity is always small, the pressure is only a function of the axial coordinate $x$, and the velocity profile is assumed to remain parabolic. Thus, it is sufficient to work only in terms of the flow $Q(x,t)$, since the axial velocity $u_x(x,r,t)$ can be recovered by using the following relations:

(2.1)\begin{gather} Q(x,t) \equiv \int_0^{R(x,t)} u_x(x,r,t) 2{\rm \pi} r \,{\rm d}r , \end{gather}

(2.2)\begin{gather}u_x (x,r,t) = \frac{2Q(x,t)}{{\rm \pi} R(x,t)^2} \left(1 - \frac{r^2}{R(x,t)^2} \right). \end{gather}

Here, $R(x,t)$ is the radius of the tube, and $r$ is the distance from the midline of the vessel. Under our approximations, the equations governing mass continuity and momentum conservation reduce to

(2.3)\begin{gather} \frac{\partial Q}{\partial x} + {\rm \pi}\frac{\partial R^2}{\partial t} = 0, \end{gather}

(2.4)\begin{gather}\frac{\partial P}{\partial x} + \frac{8 \mu}{{\rm \pi} R^4} Q = 0, \end{gather}

where $\mu$ is the dynamic viscosity. Finally, the pressure and radius are coupled via a linear elasticity equation (Timoshenko & Woinowsky-Krieger Reference Timoshenko and Woinowsky-Krieger1959; Takagi & Balmforth Reference Takagi and Balmforth2011)

(2.5)\begin{equation} P-P_a\, f(x,t) = \frac{Eh}{(1-\nu^2)R_0} \left(\frac{R}{R_0}-1\right), \end{equation}

where $E$ is Young's modulus, $h$ is the thickness of the tube, $\nu$ is Poisson's ratio and $R_0$ is the rest radius of the tube, which in this work is assumed to be independent of $x$. A generalization of (2.5) was considered by Macdonald et al. (Reference Macdonald, Arkill, Tabor, McHale and Winlove2008) to represent the elastic response of lymphangions; the form is still a linear relationship between pressure and radial deformation, but with a different coefficient. The term $P_a\, f$ is the prescribed peristaltic force with characteristic amplitude $P_a$ and functional form $f$ whose mean value is zero. Throughout the paper, most of our numerical results for a forward-propagating wave will use $f(x,t) = \cos (2{\rm \pi} (x-ct))$, and our results for a backward-propagating wave will use $f(x,t) = -\cos (2{\rm \pi} (x+ct))$.

Consequences of adding a small bending term to (2.5) will be discussed in Appendix A.

2.2. Dimensionless formulation

We will now work with convenient dimensionless quantities

(2.6a–c)\begin{gather} \bar{R}\equiv \frac{R}{R_0}, \quad \bar{Q} \equiv \frac{Q}{c{\rm \pi} R_0^2}, \quad\bar{P} \equiv \frac{P}{(c{\rm \pi} R_0^2)(8\mu \lambda /{\rm \pi} R_0^4)}, \end{gather}

(2.7a,b)\begin{gather}\bar{x} \equiv \frac{x}{\lambda} ,\quad \bar{t} \equiv \frac{t}{T}, \end{gather}

(2.8a–c)\begin{gather}\kappa \equiv \frac{Eh/(1-\nu^2)R_0}{(c{\rm \pi} R_0^2)(8\mu \lambda /{\rm \pi} R_0^4)}, \quad \eta_P \equiv \frac{P_a}{(c{\rm \pi} R_0^2)(8\mu \lambda /{\rm \pi} R_0^4)} , \quad \eta_R \equiv \frac{\eta_P}{\kappa} = \frac{P_a}{Eh/(1-\nu^2)R_0} . \end{gather}

The radius, flow and pressure non-dimensionalization is similar to that used in previous papers describing peristalsis, such as Shapiro et al. (Reference Shapiro, Jaffrin and Weinberg1969) and Provost & Schwarz (Reference Provost and Schwarz1994). Space is scaled by the peristaltic wavelength $\lambda$, and time is scaled by the peristaltic period $T$. In the above, $\kappa$ is the ratio of the stiffness to the characteristic pressure of peristalsis in a viscous tube, $\eta _P$ is the ratio of the applied peristaltic force to the characteristic pressure of peristalsis in a viscous tube and $\eta _R$ (which can be constructed from the other two parameters) gives the characteristic radial deformation of a stiff vessel.

Using these dimensionless variables, our model for peristaltic pumping in an elastic tube becomes

(2.9)\begin{gather} \frac{\partial \bar{Q}}{\partial \bar{x}} + \frac{\partial \bar{R}^2}{\partial \bar{t}} = 0 , \end{gather}

(2.10)\begin{gather}\bar{Q} ={-}\bar{R}^4 \frac{\partial \bar{P}}{\partial \bar{x}} , \end{gather}

(2.11a,b)\begin{gather}\bar{P}=\kappa (\bar{R}-1 ) + \eta_P f, \quad \bar{R} = 1-\eta_R\, f + \frac{1}{\kappa} \bar{P}. \end{gather}

When periodic boundary conditions are applied, the integral of (2.9) implies that the total volume is conserved. By convention (or by an appropriate definition of $R_0$), we will enforce this dimensionless volume to remain one

(2.12)\begin{equation} \int_0^1 \bar{R}^2 \,{\rm d}\kern 0.06em \bar{x} = 1. \end{equation}

This simple form of the volume constraint was also imposed in Takagi & Balmforth (Reference Takagi and Balmforth2011). Keeping only linear terms in the radial deformation allows one to combine (2.9), (2.10) and (2.11a,b) into a single driven heat equation. Formally, one expands the pressure, flow and radius in powers of $\eta _P$: $\bar {P} = \eta _P \bar {P}_1 + \cdots$, $\bar {Q} = \eta _P \bar {Q}_1 + \cdots$, $\bar {R} = 1 + \eta _P \bar {R}_1 + \cdots$. Then, the terms linear in $\eta _P$ satisfy

(2.13)\begin{equation} \frac{\partial \bar{P}_1}{\partial \bar{t}} - \frac{\kappa}{2} \frac{\partial^2 \bar{P}_1}{\partial \bar{x}^2} \approx \frac{\partial f}{\partial \bar{t}} .\end{equation}

This equation allows us to better understand the role of the parameter $\kappa$. One can think of $\kappa /2$ as a diffusion coefficient or $2/\kappa$ as an elastic relaxation time per peristaltic period. When $\kappa$ is large, the pressure diffuses to an equilibrium configuration of nearly uniform pressure. When $\kappa$ is small, the relaxation time is longer than the peristaltic period, and the pressure distribution closely resembles the external pressure applied to the tube.

3.1. Closely spaced closed valves suppress flow

The flow between two closed valves spaced a distance $x_v$ apart is identically the flow in a flexible pipe of length $x_v$ capped at both ends. When $x_v \ll \lambda$, peristalsis is nearly synchronous across the entire pipe, and $\epsilon$ can be used as an expansion parameter. We change to using spatial coordinate $\bar {y}\equiv \bar {x}/\epsilon$ such that the fluid between valves $i$ and $i+1$ is confined to an interval of length one, $\bar {y} \in [\bar {y}_v^i, \bar {y}_v^{i+1}]$. Since $f$ is nearly constant in space between two closed valves, we Taylor expand $f$ about the midpoint between two valves $\bar {x}_m^i \equiv \frac {1}{2}(\bar {x}_v^i +\bar {x}_v^{i+1})$

(3.6)\begin{equation} f(\bar{y}, \bar{t}) = f (\bar{x}_m^i,\bar{t}) + \epsilon f'(\bar{x}_m^i,\bar{t}) (\bar{y}- \bar{y}_m^i) + \epsilon^2 \tfrac{1}{2}f'' (\bar{x}_m^i,\bar{t}) (\bar{y}-\bar{y}_m^i)^2+\cdots. \end{equation}

Here, the primes denote derivatives with respect to $\bar {x}$. Equations (2.9) and (2.10) become

(3.7)\begin{equation} \left.\begin{gathered} \frac{\partial \bar{Q}}{\partial \bar{y}} +\epsilon \frac{\partial \bar{R}^2}{\partial \bar{t}} = 0 ,\\ \epsilon \bar{Q} ={-}\bar{R}^4 \frac{\partial \bar{P}}{\partial \bar{y}}, \end{gathered}\right\} \end{equation}

which, along with (2.11a,b), can be solved perturbatively with zero flow boundary conditions. We expand $\bar {P}(\bar {y},\bar {t}) = \bar {P}_0(\bar {y},\bar {t}) + \epsilon \bar {P}_1(\bar {y},\bar {t})+ \epsilon ^2 \bar {P}_2(\bar {y},\bar {t})+\cdots$, and similarly for $\bar {Q}(\bar {y},\bar {t})$ and $\bar {R}(\bar {y},\bar {t})$. To lowest order, we have

(3.8)\begin{equation} \left.\begin{gathered} \frac{\partial \bar{Q}_0(\bar{y},\bar{t})}{\partial \bar{y}} = 0,\\ 0 ={-}\bar{R}_0(\bar{y},\bar{t})^4 \frac{\partial\bar{P}_0(\bar{y},\bar{t})}{\partial \bar{y}},\\ \bar{P}_0(\bar{y},\bar{t})= \kappa (\bar{R}_0(\bar{y},\bar{t})-1) + \eta_P f (\bar{x}_m^i,\bar{t}) . \end{gathered}\right\} \end{equation}

From the first equation, the flow is constant in $\bar {y}$, and the boundary conditions fix that constant to zero. From the second equation, the pressure is independent of $\bar {y}$, and the last equation implies that the radius must also be independent of $\bar {y}$. The first-order equations read

(3.9)\begin{equation} \left.\begin{gathered} \frac{\partial \bar{Q}_1(\bar{y},\bar{t})}{\partial \bar{y}} +\frac{\partial \bar{R}_0^2(\bar{t})}{\partial \bar{t}} = 0 ,\\ 0 ={-}\bar{R}_0(\bar{t})^4 \frac{\partial \bar{P}_1(\bar{y},\bar{t})}{\partial \bar{y}} ,\\ \bar{P}_1(\bar{y},\bar{t}) = \kappa \bar{R}_1 (\bar{y},\bar{t}) + \eta_P f'(\bar{x}_m^i,\bar{t}) (\bar{y}- \bar{y}_m^i). \end{gathered}\right\} \end{equation}

The first equation permits flow which is linear in $\bar {y}$, but our boundary conditions forbid this unless $\bar {Q}_1 = 0$, which implies $\bar {R}_0$ is also independent of time. The second equation implies $\bar {P}_1$ is independent of $\bar {y}$, and its time dependence is related to $\bar {R}_1$ by the third equation. The second-order equations read

(3.10)\begin{equation} \left.\begin{gathered} \frac{\partial \bar{Q}_2(\bar{y},\bar{t})}{\partial \bar{y}} + 2\bar{R}_0 \frac{\partial \bar{R}_1(\bar{y},\bar{t})}{\partial \bar{t}} = 0,\\ 0 ={-}\bar{R}_0^4 \frac{\partial \bar{P}_2(\bar{y},\bar{t})}{\partial \bar{y}} ,\\ \bar{P}_2 (\bar{y},\bar{t}) = \kappa \bar{R}_2 (\bar{y},\bar{t}) + \frac{1}{2}\eta \, f'' (\bar{x}_m^i,\bar{t}) (\bar{y}-\bar{y}_m^i)^2. \end{gathered}\right\} \end{equation}

The second equation tells us that $\bar {P}_2$ is independent of $\bar {y}$, and its time dependence is related to $\bar {R}_2$ by the third equation. Integrating the first equation from $\bar {y}_m-{1}/{2}$ to $\bar {y}_m+{1}/{2}$ gives the constraint

(3.11)\begin{equation} \int_{\bar{y}_m-{1}/{2}}^{\bar{y}_m+{1}/{2}} \frac{\partial \bar{R}_1(\bar{y},\bar{t})}{\partial \bar{t}} \,{\rm d}\bar{y} = \frac{1}{\kappa}\frac{\partial \bar{P}_1(\bar{t})}{\partial \bar{t}} = 0. \end{equation}

So, in fact, $\bar {P}_1$ is a constant. The continuity equation and boundary conditions are satisfied by

(3.12)\begin{equation} \bar{Q}_2(\bar{y}, \bar{t}) - \bar{R}_0 \frac{\eta_P}{\kappa} \frac{\partial f'(\bar{x}_m, \bar{t})}{\partial \bar{t}}(\bar{y}-\bar{y}_v^i)(\bar{y}-\bar{y}_v^{i+1})= 0. \end{equation}

Assuming the valves are equally spaced, we can make the replacement $\bar {x}_m^i = \epsilon \lfloor \bar {x}/\epsilon \rfloor +\epsilon /2$, and write an expression for the pressure, flow and radius

(3.13)\begin{gather} \bar{P}(\bar{x}, \bar{t}) = \kappa(\bar{R}_0 - 1) +\eta_P f\left(\epsilon \lfloor \bar{x}/\epsilon \rfloor+\epsilon/2,\bar{t}\right) + \epsilon \bar{P}_1 + \epsilon^2 \bar{P}_2(\bar{t}) + O(\epsilon^3) ,\end{gather}

(3.14)\begin{gather} \bar{Q}(\bar{x}, \bar{t})= \frac{\eta_P }{\kappa}\bar{R}_0 \frac{\partial f\left(\epsilon \lfloor \bar{x}/\epsilon \rfloor+\epsilon/2,\bar{t}\right)}{\partial \bar{t}} \left(\bar{x}- \epsilon \left\lfloor \frac{\bar{x}}{\epsilon} \right\rfloor\right)\left(\bar{x}- \epsilon \left\lfloor \frac{\bar{x}}{\epsilon} \right\rfloor-\epsilon\right)+O(\epsilon^3) ,\end{gather}

(3.15)\begin{align} \bar{R}(\bar{x}, \bar{t}) &= \bar{R}_0 + \frac{\epsilon}{\kappa}\bar{P}_1 - \frac{\eta_P}{\kappa} f'\left(\epsilon \lfloor \bar{x}/\epsilon \rfloor+\epsilon/2,\bar{t}\right)\left( \bar{x}- \epsilon \left\lfloor \frac{\bar{x}}{\epsilon} \right\rfloor-\frac{\epsilon}{2} \right) \nonumber\\ &\quad + \frac{\epsilon^2}{\kappa}\bar{P}_2(\bar{t}) - \frac{1}{2}\frac{\eta_P}{\kappa} f''\left(\epsilon \lfloor \bar{x}/\epsilon \rfloor+\epsilon/2,\bar{t}\right) \left( \bar{x}- \epsilon \left\lfloor \frac{\bar{x}}{\epsilon} \right\rfloor-\frac{\epsilon}{2} \right)^2 +O(\epsilon^3). \end{align}

Here, $\bar {R}_0$ and $\bar {P}_1$ are undetermined constants and $\bar {P}_2(\bar {t})$ is an undetermined function of time. These approximate expressions for a confined fluid agree well with the exact solutions, as demonstrated in figure 3. The pressure is nearly constant in space in between two valves, and the time dependence is dominated by the value of $f(\bar {x}_v^i,\bar {t})$, explaining the step-like profiles observed in figure 3(b). The flow is everywhere continuous, but cusps can be seen at valves, as shown in figure 3(c). Crucially, the flow between the valves is suppressed by $\epsilon ^2$, so the induced flow decreases rapidly as the valve spacing decreases, but may be noticeable in a highly compliant vessel with small $\kappa$. The radius displays rapid oscillations due to the $\bar {y}$-dependent term at $O(\epsilon ^1)$, as seen in figure 3(a). These oscillations describe how slightly different forces applied between two valves can cause a large gradient in the radius.

Figure 3. Exact numerical solutions to (2.9), (2.10), (2.11a,b) and (3.4) with $f(\bar {x},\bar {t})=\cos (2{\rm \pi} (\bar {x}-\bar {t}))$ in a region of many closed valves are shown with solid lines. The analytical predictions (3.13), (3.14), (3.15) are shown with dash-dotted lines, and appear to agree well with the exact solutions. The valve continuum solutions (3.16), (3.17) and (3.18) are shown with black dashed lines. Parameters used for this simulation are $n_v=20$, $\bar {r}_v=0$, $\kappa =0.5$ and $\eta _P=0.5$. Subtracting (or in the case of the flow, dividing) by a constant value eliminates the unknown parameters $\bar {R}_0$ and $\bar {P}_1$. In the case of the analytical solutions, the pressure and radius should be evaluated at the limit as $\bar {x}$ approaches one from below.

Continuous equations can be recovered by taking the limit $\epsilon \rightarrow 0$

(3.16)\begin{gather} \lim_{\epsilon \rightarrow 0}\bar{P}(\bar{x}, \bar{t}) = \kappa(\bar{R}_0 - 1) +\eta_P f(\bar{x},\bar{t}) , \end{gather}

(3.17)\begin{gather}\lim_{\epsilon \rightarrow 0} \bar{Q}(\bar{x}, \bar{t}) =0 , \end{gather}

(3.18)\begin{gather}\lim_{\epsilon \rightarrow 0} \bar{R}(\bar{x}, \bar{t}) = \bar{R}_0. \end{gather}

These equations describe the valve continuum in a region of closed valves and can be understood as a simple consequence of the zero flow boundary conditions. Not only does this enforce $\bar {Q}=0$, but also the radius must be kept constant to prevent induced flow. With this idea in mind, note that we cannot apply an arbitrary radius-imposed peristalsis in between two closed valves because doing so would violate the zero flow condition. The model of Farina et al. (Reference Farina, Fusi, Fasano, Ceretani and Rosso2016) utilizing ideal valves and radially imposed peristalsis considers the case of only two valves, and they find that at least one of the valves must be open at any time. We will later see that it is possible to have radius-imposed peristalsis through many valves, but only if precisely one valve is closed per wavelength.

3.2. Flow through many open valves

Next, we seek a simplified model for a region with many open valves. Notice that we can incorporate open valves into our one-dimensional momentum equation by introducing an additional resistance $\bar {r}_v$ at the location of each valve

(3.19)\begin{equation} \frac{\partial \bar{P}}{\partial \bar{x}} ={-}\bar{R}^{{-}4} \left[1+ \bar{r}_v \sum_i \delta(\bar{x}-\bar{x}_v^i) \right] \bar{Q}. \end{equation}

The calculation will be similar to the previous section, but here there will be two length scales which are important. The valves change the fluidic resistance over a small length scale $\bar {y}\equiv \bar {x}/\epsilon$ while the channel changes the resistance over a longer length scale $\bar {x}$, so we can apply the tools of homogenization theory to derive an appropriate effective resistance (Holmes Reference Holmes2013). We will expand each of the functions $\bar {P}, \bar {Q}, \bar {R}$ in powers of $\epsilon$, and introduce $n_v = \epsilon ^{-1}$, which counts the number of valves per wavelength, with $n_v \bar {r}_v \sim O(\epsilon ^0)$. In a region of open valves

(3.20)\begin{align} &\epsilon \left( \frac{\partial}{\partial \bar{x}} + \frac{1}{\epsilon} \frac{\partial}{\partial \bar{y}} \right) \left(\bar{Q}_0 + \epsilon \bar{Q}_1 + \cdots\right) + \epsilon \frac{\partial}{\partial \bar{t}} \left(\bar{R}_0^2 + \cdots\right) = 0, \end{align}

(3.21) \begin{align}&\epsilon \left(\frac{\partial}{\partial \bar{x}} + \frac{1}{\epsilon} \frac{\partial}{\partial \bar{y}} \right)\left(\bar{P}_0 + \epsilon \bar{P}_1 + \cdots\right)\nonumber\\ &\quad ={-} \epsilon \left(\bar{R}_0^{{-}4}+\cdots\right) \left[1+n_v \bar{r}_v \sum_i \delta(\bar{y}-\bar{y}_v^i) \right] \left(\bar{Q}_0 + \cdots\right), \end{align}

(3.22)\begin{align}&\left(\bar{P}_0 + \epsilon \bar{P}_1 + \cdots\right) = \kappa \left[\left(\bar{R}_0 + \epsilon \bar{R}_1 + \cdots\right) - 1 \right]+ \eta_P f . \end{align}

From the $O(\epsilon ^{0})$ terms in (3.20) and (3.21), we immediately learn that $\bar {Q}_0$ and $\bar {P}_0$ are independent of $\bar {y}$. From (3.22), we can also see that $\bar {R}_0$ is independent of $\bar {y}$. At $O(\epsilon ^1)$ in (3.22), we have

(3.23)\begin{equation} \bar{P}_0 = \kappa(\bar{R}_0 - 1) + \eta_P f. \end{equation}

At $O(\epsilon ^1)$ in (3.20), we have

(3.24)\begin{equation} \frac{\partial \bar{Q}_0(\bar{x},\bar{t})}{\partial \bar{x}} + \frac{\partial \bar{Q}_1(\bar{x},\bar{y},\bar{t})}{\partial \bar{y}} + \frac{\partial \bar{R}_0(\bar{x},\bar{t})^2}{\partial \bar{t}} = 0. \end{equation}

We see that $\bar {Q}_1$ is linear in $\bar {y}$, but since $\bar {y}$ describes effects localized to the valves, we must have that $\bar {Q}_1$ does not grow far from the valves, and thus ${\partial \bar {Q}_1(\bar {x},\bar {y},\bar {t})}/{\partial \bar {y}} = 0$. This leaves us with a continuity equation purely in terms of $\bar {x}$

(3.25)\begin{equation} \frac{\partial \bar{Q}_0}{\partial \bar{x}} +\frac{\partial \bar{R}_0^2}{\partial \bar{t}} = 0 .\end{equation}

Looking at $O(\epsilon ^1)$ in (3.21), we have

(3.26)\begin{equation} \frac{\partial \bar{P}_0(\bar{x},\bar{t})}{\partial \bar{x}} + \frac{\partial \bar{P}_1(\bar{x},\bar{y},\bar{t})}{\partial \bar{y}} ={-}\bar{R}_0(\bar{x},\bar{t})^{{-}4} \left[1+n_v \bar{r}_v \sum_i \delta(\bar{y}-\bar{y}_v^i) \right] \bar{Q}_0(\bar{x},\bar{t}). \end{equation}

Integrating over $\bar {y}$ from an arbitrary point $\bar {y}_0$ to $\bar {y}$ gives

(3.27)\begin{align} &\bar{P}_1(\bar{x},\bar{y},\bar{t}) - \bar{P}_1(\bar{x},\bar{y}_0,\bar{t})\nonumber\\ &\quad ={-} \int_{\bar{y}_0}^{\bar{y}}\left[\frac{\partial \bar{P}_0}{\partial \bar{x}}+\bar{R}_0^{{-}4} \left[1+ n_v \bar{r}_v \sum_i \delta(\bar{y}'-\bar{y}_v^i) \right] \bar{Q}_0\right]{\rm d}\bar{y}'\nonumber\\ &\quad ={-} \left( \frac{\partial \bar{P}_0}{\partial \bar{x}}+\bar{R}_0^{{-}4} \bar{Q}_0\right) (\bar{y}-\bar{y}_0) - n_v \bar{r}_v \bar{R}_0^{{-}4} \bar{Q}_0 \int_{\bar{y}_0}^{\bar{y}}\sum_i \delta(\bar{y}'-\bar{y}_v^i)\,{\rm d}\bar{y}'. \end{align}

Each of the terms on the right-hand side diverges for large $\bar {y}$, so the only way for $\bar {P}_1$ to remain finite is to have these terms cancel as $\bar {y}$ increases

(3.28)\begin{align} \lim_{\bar{y}\rightarrow \infty} \frac{1}{\bar{y}-\bar{y}_0} \left[ - \left( \frac{\partial \bar{P}_0}{\partial \bar{x}}+\bar{R}_0^{{-}4} \bar{Q}_0\right) (\bar{y}-\bar{y}_0) - n_v \bar{r}_v \bar{R}_0^{{-}4} \bar{Q}_0 \int_{\bar{y}_0}^{\bar{y}} \sum_i \delta(\bar{y}'-\bar{y}_v^i)\,{\rm d}\bar{y}\right] = 0. \end{align}

The integral in the second term grows like $\bar {y}-\bar {y}_0$ for large $\bar {y}$, so our homogenized momentum equation simply becomes

(3.29)\begin{equation} \frac{\partial \bar{P}_0}{\partial \bar{x}} ={-}\bar{R}_0^{{-}4} \left[ 1+n_v \bar{r}_v \right]\bar{Q}_0 . \end{equation}

This equation could have been guessed by adding the resistance of our tube and $n_v$ valves in series. In the limit $\epsilon \rightarrow 0$, (3.23), (3.25) and (3.29) are exact. These are the expressions for the valve continuum in a region of open valves. While $\bar {P}_1$ and $\bar {R}_1$ are both discontinuous at the valves, $\bar {Q}_1$ is independent of the microscopic coordinate $\bar {y}$, so much like the theory for regions of closed valves, the leading-order term in our expansion for regions of open valves works particularly well at describing the flow which is the fundamental quantity of interest.

The dependence on valve parameters can be eliminated entirely by rescaling the amplitude and stiffness according to $\eta _P \rightarrow \eta _P/(1+n_v \bar {r}_v)$ and $\kappa \rightarrow \kappa /(1+n_v \bar {r}_v)$. This will give the correct radius and flow, but the result for the pressure will then need to be multiplied by a factor of $(1+n_v \bar {r}_v)$ to get the correct homogenized pressure. This procedure was done to obtain the valve continuum solutions (solid lines) in figure 2. For notational simplicity, we will simply set $\bar {r}_v=0$ until we are interested in exploring specific features related to the number of valves.

Throughout the next section, all analytical results are presented in the limit $\epsilon \rightarrow 0$, and the subscript zero will be re-purposed for a new perturbative expansion.

3.3. Matching conditions

So far, we have derived the equations for a continuum of open valves and a continuum of closed valves. What remains is to match these regions. Matching occurs at coordinates where the valves are closing or opening. From (3.1) and (3.2), we know that the pressure must be continuous at these coordinates, and (2.5) implies the radius is also continuous. Just as in the discrete valve system, the flow is always continuous and is identically zero during opening or closing. By (3.29), the pressure gradient in an open region adjacent to a closed region must be zero to ensure zero flow at the transition, but the pressure gradient in a closed region adjacent to an open region only needs to be non-negative, so the pressure gradient need not be continuous. Thus, although $P, Q$ and $R$ are continuous in the valve continuum, the functions may not be smooth at the closing and opening coordinates. This behaviour can be seen in the valve continuum predictions in figure 2, where cusps can be seen where the shaded and unshaded regions match. Later, we will show that for the case of travelling waves, the pressure gradient is continuous at the closing coordinates but not at the opening coordinates.

To summarize, a fluid in an elastic pipe containing many valves with arbitrary imposed force $f$ satisfies (3.16), (3.17) and (3.18) in a region containing many closed valves, and (3.23), (3.25) and (3.29) in a region containing many open valves. These regions are matched by continuity of $P$, $Q$ and $R$. This is the general form of the valve continuum. Additional boundary conditions could be imposed on a finite tube. We will only consider $f$ in the form of forward-propagating and backward-propagating peristaltic waves imposed on an infinitely long tube with zero net pressure drop such that periodic boundary conditions in $P$, $Q$ and $R$ apply.

4.1. Forward-propagating peristalsis in a stiff tube (radius-imposed peristalsis)

Taking the limit of an infinitely stiff vessel $\kappa \rightarrow \infty$ at fixed $\eta _R$ in (2.11a,b) suggests $\bar {R}(\xi ) = 1 - \eta _R \,f(\xi )$. However, a constant term may be added to this expression corresponding to a constant shift in pressure, and this constant is chosen to ensure that $\langle \bar {R}^2 \rangle =1$ for any $f$ satisfying $\langle\, f \rangle = 0$

(4.8)\begin{equation} \bar{R}(\xi) = \sqrt{1-\eta_R^2 \langle\, f^2 \rangle} - \eta_R\, f(\xi). \end{equation}

Since $f(\xi )$ is known, we have recovered radius-imposed peristalsis as a special case of force-imposed peristalsis where the tube is stiff, but the amplitude of the applied forces $\eta _P$ scales with the stiffness $\kappa$ to induce finite radial deformations of amplitude $\eta _R=\eta _P/\kappa$. This reduction from force-imposed peristalsis to radius-imposed peristalsis was studied in a system without valves by Takagi & Balmforth (Reference Takagi and Balmforth2011).

In a system with ideal valves, radius-imposed peristalsis appears problematic, even before considering travelling waves or a valve continuum. The continuity equation (2.9) with known $\bar {R}(\bar {x},\bar {t})$ is a first-order differential equation in space, but we wish to satisfy two zero flow boundary conditions at the closing and opening coordinates $\xi = 0$ and $\xi = \tilde {\xi }$. The resolution is to have only a single closed valve per wavelength, reducing the number of boundary conditions to one. Indeed, for the current problem of forward-propagating peristaltic waves, we see that the only way to simultaneously satisfy $\bar {R}(\xi ) = \bar {R}(0)$ for the closed valves and to satisfy (4.8) is to have $f(\xi ) = f(0)$, which is the unique maximum of $f$ by (4.3). By our definition of the opening coordinate, $\tilde {\xi }=1$. This is the coordinate of the only closed valve, while all other valves remain open. The pressure will be discontinuous across the closed valve, but the flow can easily be obtained by plugging (4.8) into (4.2)

(4.9)\begin{equation} \bar{Q}(\xi) ={-}2 \eta_R \sqrt{1-\eta_R^2 \langle\, f^2 \rangle} \left(f(\xi) -\max (f) \right)+\eta_R^2 \left(f(\xi)^2 - (\max (f) )^2\right), \end{equation}

and its time average is

(4.10)\begin{equation} \langle \bar{Q} \rangle = 2 \eta_R \max (f) \sqrt{1-\eta_R^2 \langle\, f^2 \rangle} -\eta_R^2 \left((\max (f) )^2 - \langle\, f^2 \rangle \right) .\end{equation}

This solution is valid only if the radius remains positive which is true provided $\eta _R$ is sufficiently small

(4.11)\begin{equation} \eta_R < \frac{1}{\sqrt{(\max f)^2 + \langle\, f^2 \rangle} }. \end{equation}

For larger $\eta _R$, the tube is completely occluded with all of the fluid volume transported in one period, which in dimensionless units is $\langle \bar {Q} \rangle =1$. Figure 5 compares the numerical results of the fraction of valves open and the mean flow with the analytic predictions in the case of sinusoidal peristalsis. The fraction of valves open approaches a number close to one, which will be quantified in the next section. The flow for $\eta _R < \sqrt {2/3}$ is correctly predicted by (4.10), and is equal to one for larger values of $\eta _R$.

Figure 5. Results for forward-propagating peristalsis in a stiff tube with a continuum of valves, $f(\bar {x},\bar {t}) = \cos (2{\rm \pi} (\bar {x}-\bar {t}))$. $(a)$ Fraction of valves open in a stiff vessel for two different choices of large stiffness and varying radial amplitude $\eta _R$. The dashed line is the small-amplitude analytic result (4.30). $(b)$ Mean flow for two different choices of large stiffness and varying radial amplitude $\eta _R$. The dashed line is the analytic result (4.10).

The flow in this limit is drastically different from that of radius-imposed peristalsis without valves. Perhaps the most striking feature is that the leading-order time-averaged flow for small-amplitude peristalsis scales with $\eta _R$ in the valve continuum, as opposed to $\eta _R^2$ for the case without valves. The intuition is that the flow at $O(\eta _R)$ describes inflow at locations along the vessel undergoing expansion and outflow at locations undergoing contraction, a result of the continuity equation (2.9). In the absence of valves, an expansion will pull fluid from the left and the right and a contraction will push fluid to the left and the right. Since the deformation is periodic, the vessel must undergo equal amounts of expansion and contraction at each location of the vessel, and this leading-order term in the flow (which makes no reference to the momentum equation) is zero on average. Non-zero flow results from higher-order terms where the continuity equation couples to the nonlinear term $\bar {R}^{-4}\bar {Q}$ in the momentum equation (2.10). Integrating $\bar {R}^{-4}\bar {Q}$ over one period and applying continuity of pressure gives

(4.12)\begin{equation} \langle \bar{Q}^{nv} \rangle = 1 - \frac{\langle \bar{R}^{{-}2}\rangle}{\langle \bar{R}^{{-}4}\rangle}. \end{equation}

The notation ‘nv’ will be used to denote ‘no valves’. In the valve continuum, expansions may only pull fluid from the left, and contractions may only push fluid to the right, so the leading-order term describing mass continuity with rectification leads to some net fluid flow to the right even without considering the nonlinear factor of $\bar {R}^{-4}$ in the momentum equation. In fact, the valve continuum solution in the special case of radius-imposed peristalsis does not even rely on the precise form of the momentum equation, only that there is precisely one closed valve, across which the pressure is allowed to be discontinuous.

Sample solutions in this regime are given in the left columns of figures 2(a) and 2(b). For a finite $\kappa$, the fraction of valves open is less than one, and a continuous pressure is observed in the closed region which interpolates between that in the open regions. Here, a value of $\kappa = 16$ was used, but the area of the shaded region will get smaller for larger values of $\kappa$, collapsing to $\xi = 1$ in the limit $\kappa \rightarrow \infty$. The pressure for the valveless case is small and nearly averages to zero, but a large negative pressure gradient is established for the case with valves. The flow curve has a similar shape as that of the valveless solution but is shifted upward.

4.2. Forward-propagating, small-amplitude peristalsis

The opposite limit of vanishing stiffness ends up being a hard problem to solve analytically, but we have already discovered interesting solutions at order $\eta _P$, so we will now investigate the leading-order small-amplitude response for arbitrary $\kappa$. We attempt to solve (4.5) up to order $\eta _P$ by expanding $\bar {R}(\xi ) = 1 + \eta _P \bar {R}_1(\xi )+\cdots$

(4.13)\begin{equation} \left.\begin{array}{@{}ll@{}} \dfrac{{\rm d}\bar{R}_1}{{\rm d}\xi} ={-}\dfrac{2}{\kappa} \left(\bar{R}_1-\bar{R}_1(0)\right) - \dfrac{1}{\kappa}\dfrac{{\rm d}f}{{\rm d}\xi} & 0\leqslant \xi \leqslant \tilde{\xi}_1 \\ \bar{R}_1(\xi) = \bar{R}_1(0) & \tilde{\xi}_1 \leqslant \xi \leqslant 1. \end{array}\right\} \end{equation}

Note that $\tilde {\xi }_1$ is defined as the coordinate at which the open and closed radial solutions at order $\eta _P$ match. The constraint (4.6) becomes

(4.14)\begin{align} 0 &= \int_0^{\tilde{\xi}_1} \bar{R}_1(\xi) \,{\rm d}\xi + (1-\tilde{\xi}_1) \bar{R}_1(0)\nonumber\\ &= \int_0^{\tilde{\xi}_1} \left[-\frac{\kappa}{2} \frac{{\rm d}\bar{R}_1}{{\rm d}\xi} + \bar{R}_1(0)-\frac{1}{2}\frac{{\rm d}f}{{\rm d}\xi} \right] {\rm d}\xi + (1-\tilde{\xi}_1) \bar{R}_1(0)\nonumber\\ &= \bar{R}_1(0) - \frac{1}{2}[f(\tilde{\xi}_1)-f(0)]\nonumber\\ & \Rightarrow \bar{R}_1(0) ={-}\frac{1}{2}[f(0)-f(\tilde{\xi}_1)]. \end{align}

The solution to (4.5) with this initial condition is

(4.15)\begin{equation} \bar{R}_1(\xi) ={-}\frac{1}{2}\left[ f(0) - f(\tilde{\xi}_1) \right] - \frac{1}{\kappa} \int_0^\xi {\rm d}\xi' \,\frac{{\rm d}f(\xi')}{{\rm d}\xi} \exp\left({\frac{2}{\kappa} (\xi'-\xi)}\right), \end{equation}

where $\tilde {\xi }_1$ satisfies

(4.16)\begin{equation} \int_0^{\tilde{\xi}_1} {\rm d}\xi' \,\frac{{\rm d}f(\xi')}{{\rm d}\xi} \exp\left({\frac{2}{\kappa} (\xi'-\tilde{\xi}_1)}\right) = 0. \end{equation}

We now have $\bar {R}_1$ and can easily write down $\bar {P}_1$ and $\bar {Q}_1$

(4.17)\begin{gather} \bar{R}_1(\xi)= \begin{cases} -\dfrac{1}{2} [f(0)-f(\tilde{\xi}_1)] - \dfrac{1}{\kappa} \displaystyle\int\nolimits_0^\xi {\rm d}\xi' \,\dfrac{{\rm d}f}{{\rm d}\xi} \exp\left({\dfrac{2}{\kappa} (\xi'-\xi)}\right) & \text{for } 0\leqslant \xi \leqslant \tilde{\xi}_1 \\ -\dfrac{1}{2} [f(0)-f(\tilde{\xi}_1)] & \text{for } \tilde{\xi}_1\leqslant \xi \leqslant 1 , \end{cases} \end{gather}

(4.18)\begin{gather} \bar{P}_1(\xi)= \begin{cases} -\dfrac{ \kappa}{2} [f(0)-f(\tilde{\xi}_1)] + f(\xi) - \displaystyle\int\nolimits_0^\xi {\rm d}\xi' \, \dfrac{{\rm d}f}{{\rm d}\xi} \exp\left({\dfrac{2}{\kappa} (\xi'-\xi)}\right) & \text{for } 0\leqslant \xi \leqslant \tilde{\xi}_1 \\ -\dfrac{\kappa}{2} [f(0)-f(\tilde{\xi}_1)] + f(\xi) & \text{for } \tilde{\xi}_1\leqslant \xi \leqslant 1 , \end{cases} \end{gather}

(4.19)\begin{gather} \bar{Q}_1(\xi)= \begin{cases} - \dfrac{2 }{\kappa} \displaystyle\int\nolimits_0^\xi {\rm d}\xi' \,\dfrac{{\rm d}f}{{\rm d}\xi} \exp\left({\dfrac{2}{\kappa}(\xi'-\xi)}\right) & \text{for } 0\leqslant \xi \leqslant \tilde{\xi}_1 \\ 0 & \text{for } \tilde{\xi}_1\leqslant \xi \leqslant 1 . \end{cases} \end{gather}

Clearly $\kappa$ plays an important role in the shape of these curves, but this will be easier to understand once an explicit solution is given in the next section.

Additionally, we can calculate the mean flow in terms of $\tilde {\xi }_1$

(4.20)\begin{equation} \langle \bar{Q}_1 \rangle ={-}2 \bar{R}_1(0) = f(0)-f(\tilde{\xi}_1) .\end{equation}

Evaluating this simple-looking equation requires knowing $\tilde {\xi }_1$ which depends on $\kappa$. We will now show these results for the special case of a sine wave $f(\xi ) = \cos (2{\rm \pi} \xi )$

(4.21)\begin{gather} {\rm e}^{-({2}/{\kappa})\tilde{\xi}_1} - \cos (2{\rm \pi} \tilde{\xi}_1) + \frac{1}{{\rm \pi} \kappa} \sin (2{\rm \pi} \tilde{\xi}_1) = 0 .\end{gather}

(4.22)\begin{gather} \bar{R}_1(\xi)= \begin{cases} -\dfrac{1}{2} \left[1-\cos(2{\rm \pi} \tilde{\xi}_1)\right] & {}\\ \quad +\,{\rm \pi} \left[ \dfrac{{\rm \pi} \kappa {\rm e}^{-({2}/{\kappa})\xi} + \sin (2{\rm \pi} \xi)- {\rm \pi}\kappa \cos (2{\rm \pi} \xi) }{1+({\rm \pi} \kappa)^{2}}\right] , & \text{for } 0\leqslant \xi \leqslant \tilde{\xi}_1 \\ -\dfrac{1}{2} \left[1-\cos(2{\rm \pi} \tilde{\xi}_1)\right] , & \text{for } \tilde{\xi}_1\leqslant \xi \leqslant 1 \end{cases}\end{gather}

(4.23)\begin{gather} \bar{P}_1(\xi)= \begin{cases} -\dfrac{\kappa}{2} \left(1-\cos(2{\rm \pi} \tilde{\xi}_1)\right) & {}\\ \quad +\, \dfrac{({\rm \pi} \kappa)^2 {\rm e}^{-({2}/{\kappa})\xi} + \cos (2{\rm \pi} \xi) + {\rm \pi}\kappa \sin (2{\rm \pi} \xi)}{1+({\rm \pi} \kappa)^{2}} , & \text{for } 0\leqslant \xi \leqslant \tilde{\xi}_1 \\ -\dfrac{\kappa}{2} \left(1-\cos(2{\rm \pi} \tilde{\xi}_1)\right)+ \cos (2{\rm \pi} \xi), & \text{for } \tilde{\xi}_1 \leqslant \xi \leqslant 1 \end{cases}\end{gather}

(4.24)\begin{gather} \bar{Q}_1(\xi) = \begin{cases} 2{\rm \pi} \left[\dfrac{{\rm \pi} \kappa {\rm e}^{-({2}/{\kappa}) \xi} + \sin (2{\rm \pi} \xi) - {\rm \pi}\kappa \cos (2{\rm \pi} \xi) }{1+({\rm \pi} \kappa)^{2}}\right], & \text{for } 0\leqslant \xi \leqslant \tilde{\xi} \\ 0, & \text{for } \tilde{\xi}_1 \leqslant \xi \leqslant 1 \end{cases}\end{gather}

(4.25)\begin{gather} \langle \bar{Q}_1 \rangle= 1 - \cos (2{\rm \pi} \tilde{\xi}_1) .\end{gather}

Two cases can be solved exactly: when $\kappa \rightarrow \infty$, $\tilde {\xi }_1 \rightarrow 1$ and $\langle \bar {Q}_1 \rangle \rightarrow 0$; when $\kappa \rightarrow 0$, $\tilde {\xi }_1 \rightarrow \frac {1}{2}$ and $\langle \bar {Q}_1 \rangle \rightarrow 2$. For all finite $\kappa$, $\langle \bar {Q}_1 \rangle$ is positive. To quantify the effect a valve has on peristaltic flow, we review the leading-order results for small-amplitude peristalsis without valves. For a more complete study of valveless force-driven peristalsis, see Takagi & Balmforth (Reference Takagi and Balmforth2011) and Elbaz & Gat (Reference Elbaz and Gat2014). The equations for the radius, pressure and flow read

(4.26)\begin{gather} \bar{R}_1^{nv} (\xi)= {\rm \pi}\left[\frac{\sin (2{\rm \pi} \xi) - {\rm \pi}\kappa \cos (2{\rm \pi} \xi)}{1+({\rm \pi} \kappa)^2}\right] , \end{gather}

(4.27)\begin{gather}\bar{P}_1^{nv} (\xi)= \left[\frac{\cos (2{\rm \pi} \xi) + {\rm \pi}\kappa \sin (2{\rm \pi} \xi)}{1+({\rm \pi} \kappa)^2}\right], \end{gather}

(4.28)\begin{gather}\bar{Q}_1^{nv} (\xi)= 2{\rm \pi} \left[\frac{\sin (2{\rm \pi} \xi) - {\rm \pi}\kappa \cos (2{\rm \pi} \xi)}{1+({\rm \pi} \kappa)^2}\right], \end{gather}

(4.29a,b)\begin{gather}\langle \bar{Q}_1^{nv} \rangle = 0, \quad\langle \bar{Q}_2^{nv} \rangle ={-}4 \int_0^1 \bar{R}_1^{nv}(\xi') \frac{{\rm d}f}{{\rm d}\xi}(\xi')\,{\rm d}\xi' = \frac{4{\rm \pi}^2}{1+({\rm \pi} \kappa)^{2}} . \end{gather}

Note that the leading-order contribution to the mean flow is at order $\eta _P^2$, and the direction of the flow is always in the direction of peristalsis. We contrast these intuitive results with those predicted in the valve continuum where the leading-order contribution to the mean flow is order $\eta _P$, and we will soon find that the magnitude of the flow is independent of peristalsis direction, but always in the direction of the valve. The lowest-order term in the valveless problem is pure fluctuations that average to zero, but in the valve problem, these fluctuations are rectified. Note that even for the valveless problem, net flow is only permitted due to nonlinearity in the momentum equation, but the nonlinearity arises from the factor of $\bar {R}^{-4}$ in the resistance rather than from valves. This geometric nonlinearity is weaker than the step-function nonlinearity imposed by valves, and it does not show up in the calculations until order $\eta _P^2$. This quantitatively explains the different scalings of the net flow with $\eta _P$ for the valveless and valve continuum problems.

There is another interesting observation we can make by comparing the solutions with and without valves. The solutions in between open valves with zero valve resistance (4.22), (4.23) and (4.24) seem to closely resemble the solutions for the valveless problem (4.26), (4.27) and (4.28). Other than some constant terms which enforce volume conservation, the only difference between the two solutions is a term proportional to $\textrm {e}^{-({2}/{\kappa })\xi }$. It is best to think of ${\kappa }/{2}$ as the diffusion coefficient as in (2.13). Following a disturbance, the system relaxes on a time scale of ${2}/{\kappa }$. For a valveless system, this term is transient, but for a system with valves, this becomes part of the steady state. This term is particularly important when $\kappa$ is large such that the relaxation time is small compared with the period of peristaltic pumping. The extreme case $\kappa \rightarrow \infty$ corresponds to a system driven quasistatically, in which case the pressure diffuses completely such that $\bar {P}_1=0$, much like how a gas compressed quasistatically with a piston maintains a spatially uniform pressure throughout the process.

It is difficult to continue our analysis to higher order in $\eta _P$ analytically due to the challenge in finding higher-order corrections to $\tilde {\xi }$, so even for moderate-amplitude peristalsis, one must resort to numerically solving (4.2), (4.4) and (2.11a,b). In the next couple subsections, we will further simplify the sinusoidal solutions to study the limits of large and small $\kappa$.

4.2.1. Forward-propagating, small-amplitude peristalsis, large $\kappa$

Even at small amplitude, the condition for valve closure (4.21) is transcendental. We know that, at $\kappa = \infty$, all but one valve is open, and $\tilde {\xi }_1=1$. Perturbing $\tilde {\xi }_1 \approx 1 - \delta \tilde {\xi }_1$ and keeping the lowest-order terms in $\delta \tilde {\xi }_1$ and $\kappa ^{-1}$ in (4.21) gives

(4.30)\begin{equation} \tilde{\xi}_{1} \approx 1-\frac{1}{\rm \pi} \kappa^{{-}1/2} + \frac{1}{3{\rm \pi} } \kappa^{{-}3/2} . \end{equation}

The $\kappa ^{-3/2}$ correction is only necessary for getting the correct pressure scaling. Even for large values of $\kappa$ (such as $\kappa =40$ in figure 5), the fraction of valves open appears noticeably far from one due to the slowly varying $\kappa ^{-1/2}$ term. The radius, pressure, flow and mean flow can now easily be calculated

(4.31a) \begin{gather} \bar{R}_1(\xi)= \begin{cases} -\dfrac{1}{\kappa} \cos(2{\rm \pi} \xi), & \text{for } 0\leqslant \xi \leqslant \tilde{\xi}_1 \\ -\dfrac{1}{ \kappa}, & \text{for } \tilde{\xi}_1 \leqslant \xi \leqslant 1, \end{cases} \end{gather}

(4.31b) \begin{gather} \bar{P}_1(\xi)= \begin{cases} -\dfrac{2}{\kappa}\left[(\xi - \dfrac{1}{2})-\dfrac{1}{2{\rm \pi} } \sin (2{\rm \pi} \xi)\right], & \text{for } 0\leqslant \xi \leqslant \tilde{\xi}_1 \\ -1+\cos(2{\rm \pi} \xi)+\dfrac{1}{ \kappa}, & \text{for } \tilde{\xi}_1 \leqslant \xi \leqslant 1, \end{cases} \end{gather}

(4.31c) \begin{gather} \bar{Q}_1(\xi) = \begin{cases} \dfrac{2}{\kappa}\left[1- \cos (2{\rm \pi} \xi) \right], & \text{for } 0\leqslant \xi \leqslant \tilde{\xi}_1 \\ 0, & \text{for } \tilde{\xi}_1 \leqslant \xi \leqslant 1 , \end{cases} \end{gather}

(4.31d) \begin{gather} \langle \bar{Q}_1 \rangle= \frac{2}{\kappa}.\end{gather}

These solutions for $\bar {R}_1$ and $\bar {Q}_1$ agree with the $O(\eta _R)$ solutions in § 4.1, but now we also have estimates for the fraction of open valves and the pressure in the region of closed valves. The pressure becomes discontinuous for infinite stiffness, but we now see that, for any finite stiffness, the pressure in the small region of closed valves continuously interpolates to the solutions in the open regions. Although we were able to explain the mean flow in § 4.1 purely as a result of the continuity equation (and constraints placed by the valves), we can now also understand the mean flow as a consequence of the linear pressure drop established across the open region of valves. Physically, this pressure drop arises from the rectification of the diffusive motion described by (2.13) for the valveless case.

4.2.2. Forward-propagating, small-amplitude peristalsis, small $\kappa$

At small $\kappa$, we can neglect the exponential term in (4.21). We will keep the exponential term in (4.22), (4.23) and (4.24) to ensure the boundary condition at $\xi =0$ is satisfied, and write (4.21), (4.22), (4.23), (4.24) and (4.25) as

(4.32a) \begin{gather} \tilde{\xi} = \frac{1}{2} + \frac{\kappa}{2}, \end{gather}

(4.32b) \begin{gather} \bar{R}_1(\xi)= \begin{cases} -1 + {\rm \pi}\sin(2{\rm \pi} \xi)+{\rm \pi}^2 \kappa \left({\rm e}^{-({2}/{\kappa})\xi} -\cos(2{\rm \pi} \xi) \right), & \text{for } 0\leqslant \xi \leqslant \tilde{\xi}_1 \\ -1, & \text{for } \tilde{\xi}_1 \leqslant \xi \leqslant 1 , \end{cases}\end{gather}

(4.32c) \begin{gather} \bar{P}_1(\xi)= \begin{cases} \cos(2{\rm \pi} \xi) -\kappa + {\rm \pi}\kappa \sin(2{\rm \pi} \xi) + {\rm \pi}^2 \kappa^2 {\rm e}^{-({2}/{\kappa} )\xi }, & \text{for } 0\leqslant \xi \leqslant \tilde{\xi}_1 \\ \cos (2{\rm \pi} \xi)-\kappa, & \text{for } \tilde{\xi}_1 \leqslant \xi \leqslant 1 , \end{cases}\end{gather}

(4.32d) \begin{gather} \bar{Q}_1(\xi) = \begin{cases} 2{\rm \pi} \sin(2{\rm \pi} \xi)+2{\rm \pi}^2 \kappa \left({\rm e}^{-({2}/{\kappa})\xi} -\cos(2{\rm \pi} \xi) \right), & \text{for } 0\leqslant \xi \leqslant \tilde{\xi}_1 \\ 0, & \text{for } \tilde{\xi}_1 \leqslant \xi \leqslant 1 , \end{cases}\end{gather}

(4.32e) \begin{gather} \langle \bar{Q}_1 \rangle = 2. \end{gather}

The peristaltic period is much shorter than the elastic response time, so the fluid pressure is dominated by the external pressure, and the radial deformation is out of phase with $f$. Since $f$ has negative slope half of the time, $\bar {P}_1$ also has negative slope approximately half of the time, suggesting approximately half of the valves should be open. Up to some constants fixed by matching conditions and the exponentially small term, the solutions in the open region exactly match those for the valveless problem. The diffusion mechanism discussed in the previous subsection is suppressed.

An example solution in this regime is given in the left column of figure 2(c). Because the vessel is highly compliant, even though $\eta _P$ is small, the radial deformation is large. The pressure for the cases with and without valves is similar, so the flow with valves is just the rectification of the flow without valves

(4.33)\begin{equation} \bar{Q}_1 \approx \bar{Q}_1^{nv} \varTheta (\bar{Q}_1^{nv}). \end{equation}

4.3. Forward-propagating, large-amplitude peristalsis

Large-amplitude peristalsis is already capable of pushing fluid in one direction, so if the peristalsis direction agrees with the valve direction, the flow response of the valve problem will essentially be that of the valveless problem. This is confirmed in the left column of figure 2(d). The radius is narrow throughout most of the tube, but is very large in a small region which traps and pushes the fluid in the direction of peristalsis. Since the fluid is completely trapped, the mean flow approaches one in the limit $\eta _P \rightarrow \infty$. The deviation from one can be found by considering the fluid in the occluded region where both $\bar {R}^{-4}$ and $\eta _P$ are large such that the $\bar {R}'$ term in (4.5) is negligible. Then $\bar {R}^2$ satisfies an algebraic equation whose solution is real if and only if $\bar {R}^2(0) \leqslant {1}/{8{\rm \pi} \eta _P}$. This leads to the scaling

(4.34)\begin{equation} 1-\langle \bar{Q} \rangle \approx \frac{1}{8{\rm \pi} \eta_P}. \end{equation}

A thorough analysis of large-amplitude force-imposed peristalsis can be found in Takagi & Balmforth (Reference Takagi and Balmforth2011). Although it is mathematically unsurprising that the valve system behaves like a valveless system when driven by large-amplitude peristalsis, it is worth stressing the implication of this. If a biological or engineered system is reliably driven by large-amplitude peristalsis, there is no reason for it to contain valves. We have already seen one reason why a system may utilize valves. Given a small perturbation, the valveful system can transport volume proportional to the amplitude of the perturbation (as opposed to the amplitude squared as is the case for valveless peristalsis). We will see an even more apparent difference when we consider backward-propagating peristaltic waves in the next section.

A complete summary of the fraction of valves open and the mean flow for the case of a forward-propagating sinusoidal wave is shown in figure 6. The fraction of valves open is well approximated by the small-amplitude result even for $\eta _P=1$, as shown in figure 6(a). However, for larger values of $\eta _P$, this is a less useful metric for quantifying the system since the behaviour is nearly that of the valveless system. In figure 6(b), it is shown that, for a fixed $\eta _P$, the flow decreases with $\kappa$, indicating a larger flow response in more compliant tubes subject to forces of the same magnitude. In figure 6(c), it is clear that the flow for small $\eta _P$ is linear in $\eta _P$, a unique feature to the system with ideal valves, but for $\eta _P$ close to or greater than one, the flow approaches the fully occluded limit $\langle \bar {Q} \rangle = 1$.

Figure 6. Results for forward-propagating peristalsis with a continuum of valves, $f(\bar {x},\bar {t}) = \cos (2{\rm \pi} (\bar {x}-\bar {t}))$. $(a)$ The fraction of valves per wavelength which are open at any given time is calculated numerically (points) as a function of $\kappa$ and compared with the small-amplitude result (4.21) (dashed line). $(b)$ The mean flow divided by $\eta _P$ is calculated numerically as a function of $\kappa$ and compared with the small-amplitude result (4.25). $(c)$ The mean flow is calculated as a function of $\eta _P$. For small $\eta _P$, the scaling is linear for each $\kappa$, as demonstrated by the dashed line.

5.1. Backward-propagating peristalsis in a stiff tube (radius-imposed peristalsis)

In the limit $\kappa \rightarrow \infty$ at fixed small $\eta _R$, we again recover radius-imposed peristalsis with $\bar {R}$ given by (4.8), but in this case, the flow is given by

(5.8)\begin{gather} \bar{Q}(\xi) = 2 \eta_R \sqrt{1-\eta_R^2 \langle\, f^2 \rangle} \left(f(\xi) -\min (f) \right)-\eta_R^2 \left(f(\xi)^2 - (\min (f) )^2\right) \end{gather}

(5.9)\begin{gather}\langle \bar{Q} \rangle ={-}2 \eta_R \min (f) \sqrt{1-\eta_R^2 \langle\, f^2 \rangle} + \eta_R^2 \left((\min (f) )^2 - \langle\, f^2 \rangle \right) . \end{gather}

This looks similar to the forward peristalsis solution. In fact, (5.9) suggests that, for a sine wave, the flow induced by reverse peristalsis is larger than that induced by forward peristalsis satisfying (4.10). The caveat is that (5.9) is valid over a smaller range of values than (4.10). To see this, note that in addition to satisfying (4.11), we also need to satisfy (5.7) which gives an additional constraint

(5.10)\begin{equation} \eta_R < \frac{1}{\sqrt{2}|\min f|+\sqrt{(\min f)^2 - \langle\, f^2 \rangle }}. \end{equation}

For the case of a sine wave, (4.11) gives a bound of $\eta _R<\sqrt {2/3}$, but (5.10) places a stronger bound of $\eta _R < \sqrt {2/9}$. To be clear, the bound only tells us when the analytic solution is guaranteed to give an unphysical solution, it does not tell us when the radius-imposed solutions will work. Although we found good agreement with the analytic result for all physical values in the case of a forward-propagating wave in a stiff tube, the same is not true here. In fact, it appears the analytic solution (5.8) only works for values much smaller than (5.10). In that limit, only the linear term in (5.9) contributes, and for the case of a sine wave, the mean flow is identical to that of a forward-propagating wave. Indeed, comparing the left and right columns of figures 2(a) and 2(b), the radius and flow look nearly identical. For an $f$ with minimum value larger in magnitude than its maximum value, the small-amplitude backward-propagating wave will produce more forward flow than the small-amplitude forward-propagating wave. This effect is demonstrated using Gaussian wave forms in Appendix B.

The qualitative explanation of the $O(\eta _R)$ term in (5.9) is the same as that for the forward-propagating wave. Expansions and contractions induce oscillatory flow which is rectified by the valves, and the pressure in the vessels is irrelevant for determining the flow.

5.2. Backward-propagating, small-amplitude peristalsis

Proceeding as in § 4.2, we wish to solve

(5.11) \begin{equation} \left.\begin{array}{@{}ll@{}} \bar{R}_1(\xi) = \bar{R}_1(0) & 0\leqslant \xi \leqslant \tilde{\xi}_1 \\ \dfrac{{\rm d}\bar{R}_1}{{\rm d}\xi} = \dfrac{2}{\kappa} \left(\bar{R}_1-\bar{R}_1(0)\right) - \dfrac{1}{\kappa}\dfrac{{\rm d}f}{{\rm d}\xi} & \tilde{\xi}_1 \leqslant \xi \leqslant 1. \end{array}\right\} \end{equation}

In this case, $\tilde {\xi }_1$ satisfies

(5.12)\begin{equation} \int_{\tilde{\xi}_1}^1 {\rm d}\xi' \,\frac{{\rm d}f(\xi')}{{\rm d}\xi} \exp\left({-\frac{2}{\kappa} (\xi'-\tilde{\xi}_1)}\right) = 0, \end{equation}

and the remaining quantities of interest are

(5.13)\begin{gather} \bar{R}_1(\xi)= \begin{cases} \dfrac{1}{2}[f(\tilde{\xi}_1)-f(0)] & \text{for } 0\leqslant \xi \leqslant \tilde{\xi}_1 \\ \dfrac{1}{2}[f(\tilde{\xi}_1)-f(0)] - \dfrac{1}{\kappa} \displaystyle\int\nolimits_{\tilde{\xi}_1}^{\xi} {\rm d}\xi' \, \dfrac{{\rm d}f(\xi')}{{\rm d}\xi} \exp\left({-\dfrac{2}{\kappa} (\xi'-\xi)}\right) & \text{for } \tilde{\xi}_1\leqslant \xi \leqslant 1 , \end{cases} \end{gather}

(5.14)\begin{gather} \bar{P}_1(\xi)= \begin{cases} \dfrac{\kappa}{2}[f(\tilde{\xi}_1)-f(0)] +f(\xi) & \text{for } 0\leqslant \xi \leqslant \tilde{\xi}_1 \\ \dfrac{\kappa}{2}[f(\tilde{\xi}_1)-f(0)] +f(\xi) - \displaystyle\int\nolimits_{\tilde{\xi}_1}^{\xi} {\rm d}\xi' \,\dfrac{{\rm d}f(\xi')}{{\rm d}\xi} \exp\left({-\dfrac{2}{\kappa} (\xi'-\xi)}\right) & \text{for } \tilde{\xi}_1\!\leqslant\! \xi \!\leqslant\! 1 , \end{cases} \end{gather}

(5.15)\begin{gather} \bar{Q}_1(\xi)= \begin{cases} 0 & \text{for } 0\leqslant \xi \leqslant \tilde{\xi}_1 \\ \dfrac{2 }{\kappa} \displaystyle\int\nolimits_{\tilde{\xi}_1}^\xi {\rm d}\xi' \, \dfrac{{\rm d}f}{{\rm d}\xi} \exp\left({-\dfrac{2}{\kappa}(\xi'-\xi)}\right) & \text{for } \tilde{\xi}_1\leqslant \xi \leqslant 1 , \end{cases} \end{gather}

(5.16)\begin{gather} \langle \bar{Q}_1 \rangle = 2 \bar{R}_1(0) = f(\tilde{\xi}_1)-f(0) . \end{gather}

For the special case of a sine wave $f(\xi ) = -\cos (2{\rm \pi} \xi )$, the fraction of time open satisfies the same equation as that of the forward-propagating wave

(5.17)\begin{equation} \exp\left({-\frac{2}{\kappa} (1-\tilde{\xi}_1)}\right) - \cos \left(2{\rm \pi} (1-\tilde{\xi}_1) \right)+ \frac{1}{{\rm \pi} \kappa} \sin \left(2{\rm \pi} (1-\tilde{\xi}_1)\right) = 0 . \end{equation}

The mean flow is also identical

(5.18)\begin{equation} \langle \bar{Q}_1 \rangle= 1 - \cos (2{\rm \pi} (1-\tilde{\xi}_1)).\end{equation}

Figures 7(a) and 7(b) confirm this claim numerically. For $\eta _P\ll 1$, the numerical results agree with the predictions for $1-\tilde {\xi }_1$ and $\langle \bar {Q}_1\rangle$. It is not, in general, true that an arbitrary low-amplitude peristaltic wave in a valve-filled tube produces the same flow regardless of pumping direction.

Figure 7. Results for backward-propagating peristalsis with a continuum of valves, $f(\bar {x},\bar {t}) = -\cos (2{\rm \pi} (\bar {x}+\bar {t}))$. The dashed lines in (a,b) correspond to the small-amplitude analytic expressions (5.17) and (5.18), which are identical to (4.21) and (4.25) for the forward-propagating wave.

5.3. Backward-propagating, large-amplitude peristalsis

For the amplitudes considered in figure 7(c), the flow appears to reach 1 at a slower rate as compared with figure 6(c). Unlike forward-propagating peristaltic waves, large-amplitude backward-propagating peristaltic waves produce solutions which are qualitatively different from the valveless problem. For backward-propagating peristaltic waves, the tube becomes constricted in the region of open valves, and the radius approaches the maximum value (5.7) in the closed regions. For $\kappa$ not too large, we expect the terms on the right side of (5.4) to dominate since $\bar {R}^{-4}$ and $\eta _P$ are both large. Setting $R'(\xi )=0$, we are left solving an algebraic equation which is independent of $\kappa$

(5.19)\begin{equation} {-} \eta_P \frac{{\rm d}f}{{\rm d}\xi} \bar{R}^4 + \bar{R}^{2} - \bar{R}(0)^2 = 0 . \end{equation}

Just like the $\kappa \rightarrow 0$ limit, the solution to this equation is to have $f'(\tilde {\xi }) = 0$, so the open region of valves will correspond to where $f$ is decreasing. With this in mind, we can uniquely pick the sign to this quadratic equation that gives a positive area

(5.20)\begin{equation} \bar{R}(\xi)^2 = \frac{-1+ \sqrt{1+4\bar{R}(0)^2 \eta_P \big|\,f'(\xi)\big|}}{2\eta_P \big|\,f'(\xi)\big|}. \end{equation}

In order to fix $\bar {R}(0)$, we would need to solve (5.5). This is difficult to solve in general, but we can approximately solve this problem for the case of a sine wave where $\tilde {\xi }=\frac {1}{2}$. We will assume that the integral is dominated by the largest values of the radius close to the closed regions, where $\sin (2{\rm \pi} \xi )$ is approximately linear

(5.21)\begin{align} 1 &= \frac{\bar{R}(0)^2}{2} + \int_{{1}/{2}}^1 \frac{-1+ \sqrt{1+8{\rm \pi} \bar{R}(0)^2 \eta_P |\sin (2{\rm \pi} \xi)|}}{4{\rm \pi} \eta_P |\sin (2{\rm \pi} \xi)|} {\rm d}\xi \nonumber\\ &= \frac{\bar{R}(0)^2}{2} + 2 \int_{0}^{{1}/{4}} \frac{-1+ \sqrt{1+8{\rm \pi} \bar{R}(0)^2 \eta_P \sin (2{\rm \pi} \xi)}}{4{\rm \pi} \eta_P \sin (2{\rm \pi} \xi)} {\rm d}\xi \nonumber\\ &\approx \frac{\bar{R}(0)^2}{2} + 2 \int_{0}^{{1}/{4}} \frac{-1+ \sqrt{1+8{\rm \pi} \bar{R}(0)^2 \eta_P (2{\rm \pi} \xi)}}{4{\rm \pi} \eta_P (2{\rm \pi} \xi)} {\rm d}\xi \nonumber\\ &= \frac{\bar{R}(0)^2}{2} + \frac{1}{{\rm \pi} \eta_P} \left[ \sqrt{1+2{\rm \pi} \bar{R}(0)^2 \eta_P } - 1 - \log \left(\frac{1+\sqrt{1+2{\rm \pi} \bar{R}(0)^2 \eta_P}}{2} \right) \right] \nonumber\\ &= \frac{\bar{R}(0)^2}{2} + \sqrt{\frac{2}{{\rm \pi} \eta_P}} \bar{R}(0) + O\left( \frac{\log \eta_P}{\eta_P}\right). \end{align}

We can now write down the maximum radius and mean flow

(5.22)$$\begin{gather} \implies (\max \bar{R})^2\approx 2 - \sqrt{\frac{2}{{\rm \pi} \eta_P}} , \end{gather}$$

(5.23)$$\begin{gather}1 - \langle \bar{Q} \rangle \approx \sqrt{\frac{2}{{\rm \pi} \eta_P}}. \end{gather}$$

This is a distinct power law from the forward case which approaches 1 as $\eta _P^{-1}$. It is not surprising that much larger amplitudes are needed in order to achieve maximum flow for a backward-propagating wave as compared with a forward-propagating wave. Perhaps it is more surprising that this limit is ever achieved. As a technical note, one can also derive an $\eta _P^{-1/2}$ power law without making the approximation on the third line but instead assuming that the $4\bar {R}(0)^2 \eta _P f'(\xi )$ term dominates the square root; however, this gives a slightly smaller prefactor which does not agree as well with our numerical results. The analytical results are confirmed in figure 8. Although the radius is small in the open region, the large imposed force produces a pressure gradient which drives flow. This is entirely different from large-amplitude peristalsis without valves where the flow is confined to a region of large radius but small pressure gradient.

Figure 8. Large-amplitude peristalsis against a continuum of valves with $f(\bar {x}, \bar {t}) = - \cos (2{\rm \pi} (\bar {x}+\bar {t}))$. $(a)$ Numerical solutions for the cross-sectional area and flow using $\eta _P = 25, \kappa = 0.5$ are shown with solid lines and compared with the analytic solution (5.20) with $\bar {R}(0)^2$ given by (5.22) shown with dotted lines. $(b)$ Deviation of the mean flow from the fully occluded limit. The dots show numerical solutions for two different choices of $\kappa$, and the dashed line is the analytical large-amplitude prediction (5.23).