1 Introduction

Slip at the wall is a phenomenon observed not only with non-Newtonian [1] but also with Newtonian fluids [2]. Several slip laws have been used in the literature, the simplest one being the Navier slip law which states that the velocity of the fluid relative to that to the wall, \(u_{w}\), is linearly proportional to the wall shear stress, \(\tau _{w}\):

$$\begin{aligned} u_{w}=\dfrac{\tau _{w}}{\beta } \end{aligned}$$
(1)

where \(\beta \) is the slip coefficient, also defined as \(\beta \equiv \eta /b\), where \(\eta \) is the viscosity and \(b\) is the slip length, i.e. the distance beyond the wall at which the fluid velocity extrapolates to zero. It should be noted that \(\beta \rightarrow \infty \) is equivalent to no-slip, while \(\beta =0\) corresponds to the full-slip case.

The above slip equation, as well as, its non-linear and other extensions [1], is static, i.e. it implies that the slip velocity adjusts instantaneously to the wall shear stress. This assumption obviously holds for steady-state flows. In transient flows, however, slip relaxation effects can become important, see e.g. the slip of polymer melts [3–5]. A memory, i.e. dynamic, slip model in which the slip velocity depends on the past states of the wall shear stress has been proposed by Pearson and Petrie [6]. Its simplest version is the following extension of (1):

$$\begin{aligned} u_{w}+\varLambda \dfrac{\partial u_{w}}{\partial t}=\dfrac{\tau _{w}}{\beta }, \end{aligned}$$
(2)

where \(\varLambda \) is the slip relaxation time. This model induces a delay on the development of the slip velocity. When \(\varLambda =0\), the static Navier slip model is recovered. As noted by Renardy [7] a relationship of the form (2) may arise from a competition between high stresses, which cause loss of adhesion, and a relaxation mechanism by which adhesion is restored. The experiments of Lim and Schowalter [8] on narrow molecular weight distribution polybutadienes suggested that the transition from slip to stick conditions exhibits the characteristics of a relaxation process. In an attempt to find an explanation for the melt fracture extrusion instability, Renardy [7] showed that the equations governing the two-dimensional inertialess flow of an upper-convected Maxwell fluid can become ill-posed when combined with the slip Eq. (2), leading to short-wave instabilities. However, as pointed out by Graham [9], the addition of any finite solvent viscosity to the model renders it well-posed. Black and Graham [10] suggested that this pathology may stem from the unphysical property that even if the shear stress vanishes, the model allows a finite slip velocity to persist. They thus proposed a modified dynamic slip model involving an evolution equation for the fraction of available polymer segments that are strongly interacting with the solid surface.

Other dynamic slip models have been used in the literature, such as the power-law model by Hatzikiri-akos and Dealy [4], to explain the observed slip in transient shear experiments of molten polymers (high density polyethylenes). Transient observations were made for both exponential and oscillatory shear and the results were fitted with a dynamic slip model in conjunction with a Maxwell model of viscoelasticity. Hatzikiriakos and Dealy [4] pointed out that melt slip is a physicochemical process in which the polymer/wall interface may undergo continuous change with time before attaining steady state. Hatzikiriakos and Kalogerakis [5] extended the above model simulating the behaviour of a polymer/metal interface by means of a network kinetic theory. The calculated time-dependent slip velocities with experimental data on high-density polyethylene. Consequently, Hatzikiriakos [3] extended this single-mode model to a multi-mode one. Aral and Kalyon [11] investigated the time and temperature-dependent development of wall slip in the case of concentrated suspensions with Newtonian binders in steady torsional flow. They showed that for these generalised Newtonian materials there is a critical time for both the wall velocity and the shear stress to reach steady values and that this critical time decreases with increasing apparent shear rate.

Graham [9] used a power-law dynamic slip model with non-zero slip yield stress, as well as, its two-mode generalisation and noted that, both viscoelasticity and a dynamic slip model are necessary in order to explain the instabilities and nonlinear dynamics of polymer melts in oscillatory shear. Further, it was shown rigorously that for linear and weakly nonlinear viscoelasticity, in oscillatory shear with a static power-law slip law there are no non-periodic responses. Lan et al. [12] combined the dynamic slip models of [4] and [10] with different constitutive equations, such as Wagner’s equation and the Liu model, to determine the slip parameters for a linear low-density polyethylene. All models gave good predictions of sliding plate rheometer slip data in steady shear by showed insufficient gap dependence in exponential shear. A multi-mode extension of slip Eq. (2), similar to the generalised Maxwell mechanical model for viscoelasticity, has been employed by Kazatchkov and Hatzikiriakos [13], giving a better fit to experimental data for molten polymers in shear flow. This new dynamic slip model involving multiple slip relaxation times was used along with Wagner’s constitutive equation in solving the transient shear response of a linear low-density polyethylene, yielding improved predictions of the stress response in start-up of steady shear and large-amplitude oscillatory tests. As noted by Hatzikiriakos [1] the effect of dynamic slip has also been taken into account in stochastic simulations of polymer melts (see e.g. [3, 5, 14–16]).

In a recent paper Thalakkottor and Mohseni [17] carried out molecular dynamic simulations to study slip at the fluid-solid boundary in unsteady flow. Interestingly, their numerical experiments revealed that the slip velocity is dependent not only on the shear rate but also on fluid acceleration and follows an equation which is identical to Eq. (2). Although the original derivation was based on some characteristics of gases, it has been verified that an analogous formulation is valid for simple liquid flows.

Several analytical solutions have been reported for steady-state Newtonian, Poiseuille flows with static slip in various geometries using static slip equations, such as the Navier slip equation [18–20], and the slip yield stress model [21, 22]. Using various static slip laws, both linear and non-linear, Ferrás et al. [23] presented solutions of Newtonian and Non-Newtonian Couette and Poiseuille flows. Additionally, analytical solutions for time-dependent, Newtonian flows with the Navier slip model (1), have also been reported in the literature [24–28]. Solutions of various start-up and cessation Newtonian flows with slip yield stress models have been reported in [29, 30]. An analytical solution corresponding to a pressure gradient varying linearly with time was also recently reported by Tang [31].

It is the purpose of this work to derive analytical solutions of the Newtonian Poiseuille flow for the planar and axisymmetric geometries and the Newtonian circular Couette flow using the dynamic slip Eq. (2). As demonstrated below, the time-dependence in the boundary condition leads to a more difficult mathematical problem to solve than its static counterpart. The spatial problem is of Sturm–Liouville type where the eigenvalue parameter appears in the boundary condition and thus, the orthogonality condition of the corresponding spatial eigenfunctions must be considered more carefully. The Sturm–Liouville problem encountered here, is a special case of more general ones which were tackled rigorously, see e.g. Churchill [32], Walter [33] and Fulton [34]. This type of problems, involving time derivatives in their boundary conditions are also common in free and forced vibration problems governed by the wave equation [35] and in problems in diffusion or in heat transfer between a solid and a fluid using the heat equation [36]. More references on these problems, as well as, on electric circuits involving long cables and diffusion processes in probability theory are provided in Fulton [34].

Despite the fact that wall slip of Newtonian fluids is supported by experimental observations [2], it can be argued that there is no justification for using a dynamic slip equation in transient Newtonian flows. The recent numerical dynamics experiments of [17] demonstrated that the dynamic slip equation applies in unsteady flows of simple liquids. Moreover, the analytical solutions derived below may be useful in checking numerical non-Newtonian simulation codes and in start up and cessation of steady shear in MEMS devices.

In Sect. 2, we consider the axisymmetric Poiseuille flow of a Newtonian fluid with dynamic wall slip. Following the solution procedure in [35], the appropriate orthogonality condition for the spatial eigenfunctions is derived, and the full procedure for the solution is given. The implications of the dynamic slip term in the slip equation are then discussed. In Sect. 3, the results for the planar Poiseuille flow are summarised. The circular Couette flow is solved analytically in Sect. 4 and finally, in Sect. 5 the conclusions of this work are provided.

2 Axisymmetric Poiseuille flow

We consider the start-up, pressure-driven flow of a Newtonian fluid in an infinitely long circular tube of radius \(R\). The flow is assumed to be laminar, incompressible and unidirectional and gravity is neglected. In cylindrical polar coordinates, the velocity component in the \(z\)-direction, \(u_{z}(r,t)\), satisfies

$$\begin{aligned} \rho \frac{\partial u_{z}}{\partial t}=G+\eta \left( \frac{\partial ^{2} u_{z}}{\partial r^{2}}+ \frac{1}{r}\frac{\partial u_{z}}{\partial r}\right) , \end{aligned}$$
(3)

where \(\rho \) is the density and \(G\) is the constant pressure gradient. Initially the flow is at rest,

$$\begin{aligned} u_{z}(r,0)=0. \end{aligned}$$
(4)

As for the boundary conditions, \(u_{z}\) is finite along the axis of symmetry and slip is assumed along the wall following the slip law (2). Given that \(\tau _{w}=|\tau _{rz}|_{r=R}\), one gets

$$\begin{aligned} u_{w}=-\dfrac{\eta }{\beta }\dfrac{\partial u_{w}}{\partial r}-\varLambda \dfrac{\partial u_{w}}{\partial t}. \quad \end{aligned}$$
(5)

It is clear that the steady-state solution of the problem (3)–(5), is given by

$$\begin{aligned} u^{s}(r)=\frac{R^{2}G}{4\eta }\left[ 1+2B-\left( \frac{r}{R}\right) ^{2}\right] , \end{aligned}$$
(6)

where

$$\begin{aligned} B\equiv \frac{\eta }{\beta R}, \end{aligned}$$
(7)

is a dimensionless inverse slip number (where \(B=0\) corresponds to no-slip and \(B\rightarrow \infty \) to the full-slip case). In order to homogenise Eq. (3) we look for solutions of the form

$$\begin{aligned} u_{z}(r,t)=u^{s}-\bar{u}_{z}(r,t), \end{aligned}$$
(8)

where \(\bar{u}\) is the solution of the homogeneous problem. Using separation of variables,

$$\begin{aligned} \bar{u}_{z}(r,t)=\sum _{n=1}^{\infty }X_{n}(r)T_{n}(t) \end{aligned}$$
(9)

we obtain the following ordinary differential equations

$$\begin{aligned} T_{n}'(t)+\dfrac{\lambda _{n}^{2}\nu }{R^{2}}T_{n}(t)=0, \end{aligned}$$
(10)

and

$$\begin{aligned}{}[rX'_{n}(r)]'+r \lambda _{n}^{2}/R^{2}X_{n}(r)=0, \end{aligned}$$
(11)

where, \(\nu \equiv \eta /\rho \) is the kinematic viscosity and \(\lambda _{n},\) \(n=1,2,\ldots ,\) are the eigenvalues of the problem to be determined. The boundary conditions dictate that \(X_{n}(0)\) is finite and

$$\begin{aligned} X_{n}(R)=-BRX'_{n}(R)+ \lambda _{n}^{2}\dfrac{\varLambda \nu }{R^{2}}X_{n}(R). \end{aligned}$$
(12)

It is worth noting that the solution to the homogeneous problem, \(\bar{u}_{z}\), corresponds to the solution of the cessation problem. Equation (11) is Bessel’s equation of zero order, which is a special type of Sturm–Liouville problem where the eigenvalues appear in the boundary condition (12). The solution to the homogeneous problem thus becomes

$$\begin{aligned} \bar{u}_{z}(r,t)=\sum _{n=1}^{\infty }X_{n}(r)e^{-\displaystyle {\dfrac{\lambda _{n}^{2}\nu }{R^{2}}t}}, \end{aligned}$$
(13)

and

$$\begin{aligned} X_{n}(r)\equiv A_{n}J_{0}(\lambda _{n}r/R)+ C_{n}Y_{0}(\lambda _{n}r/R). \end{aligned}$$
(14)

Here, \(J_{i}\) and \(Y_{i}\) are the order \(i\) Bessel functions of the first and second kind, respectively. Given that \(X_{n}(0)\) is finite, \(C_{n}=0\) and Eq. (12) leads to the following transcendental equation:

$$\begin{aligned} J_{0}(\lambda _{n})=\dfrac{\lambda _{n}B J_{1}(\lambda _{n})}{1-\lambda _{n}^{2}\varLambda \nu /R^{2}}. \end{aligned}$$
(15)

We still need to determine the coefficients \(A_{n}\) from the initial condition (4), which requires that

$$\begin{aligned} \sum _{n=1}^{\infty }A_{n}J_{0}(\lambda _{n}r/R)=u^{s}(r)=\frac{R^{2}G}{4\eta }\left[ 1+2B-\left( \frac{r}{R}\right) ^{2}\right] . \end{aligned}$$
(16)

In such problems, one multiplies both sides of the equation above with the eigenfunction, integrates, and obtains an expression for the coefficients \(A_{n}\) by means of the orthogonality property of the eigenfunctions. However, since the eigenvalue appears in the boundary condition (12), the orthogonality condition must be considered more carefully.

Now we consider the one-dimensional problem in \(r\) and introduce,

$$\begin{aligned}&[rX'_{m}(r)]'+r\lambda _{m}^{2}/R^{2}X_{m}(r) = 0,\end{aligned}$$
(17)
$$\begin{aligned}&X_{m}(R) = -BRX'_{m}(R)+ \lambda _{m}^{2}\dfrac{\varLambda \nu }{R^{2}}X_{m}(R), \end{aligned}$$
(18)

such that, \(X_{m}\), \(X_{n}\) and \(\lambda _{m}\), \(\lambda _{n}\) are distinct (\(m\ne n\)). Multiplying (11) by \(X_{m}\) and integrating by parts gives

$$\begin{aligned} RX_{n}'(R)X_{m}(R)-\int _{0}^{R}rX_{n}'X_{m}'\,\mathrm {d}r +\lambda _{n}^{2}R^{2}\int _{0}^{R}rX_{n}X_{m}\,\mathrm {d}r=0. \end{aligned}$$
(19)

Multiplying (17) by \(X_{n}\), integrating by parts, and subtracting from (19), we have

$$\begin{aligned} R[X_{n}'(R)X_{m}(R)-X_{n}(R)X_{m}'(R)]+(\lambda _{n}^{2}-\lambda _{m}^{2})/R^{2}\int _{0}^{R}rX_{n}X_{m}\,\mathrm {d}r=0. \end{aligned}$$
(20)

Using now the b.c.s. (12) and (18) we get

$$\begin{aligned} (\lambda _{n}^{2}-\lambda _{m}^{2})\left[ \frac{\varLambda \nu }{B}X_{n}(R)X_{m}(R)+\int _{0}^{R}rX_{n}X_{m}\,\mathrm {d}r \right] =0. \end{aligned}$$
(21)

Since \(\lambda _{m}\) and \(\lambda _{n}\) are distinct

$$\begin{aligned} \frac{\varLambda \nu }{B}X_{n}(R)X_{m}(R)+\int _{0}^{R}rX_{n}X_{m}\,\mathrm {d}r =\delta _{m,n}N_{n} \end{aligned}$$
(22)

where

$$\begin{aligned} N_{n}=\frac{\varLambda \nu }{B}X_{n}^{2}(R)+\int _{0}^{R}rX_{n}^{2}\,\mathrm {d}r, \end{aligned}$$
(23)

and \(\delta _{m,n}\) is the Kronecker delta.

Expression (22) is the appropriate orthogonality condition for this problem, and the RHS of Eq. (23) is the square of the norm of the eigenfunctions \(X_{n}(r)\). This result is a special case, of more general expressions on Sturm–Liouville operators, obtained by Churchill [32], and Langer [36], and similar results also hold if the governing equation is the heat equation, i.e. in diffusion problems [36] and for the wave equation, i.e. in forced vibration problems [35]. Setting \(\varLambda =0\), the orthogonality condition (22) leads to its static counterpart [24–28].

In order to determine the coefficients \(A_{n}\), Eq. (16) must be supplemented by an extra term. Thus, multiplying (16) by \(rJ_{0}(\lambda _{m} r/R)\), and integrating gives

$$\begin{aligned} \int _{0}^{R}ru^{s}(r)J_{0}(\lambda _{m} r/R)\,\mathrm {d}r=\sum _{n=1}^{\infty }A_{n}\int _{0}^{R}rJ_{0}(\lambda _{n} r/R)J_{0}(\lambda _{m} r/R)\,\mathrm {d}r. \end{aligned}$$
(24)

Also, consider (16) times \((\varLambda \nu /B) J_{0}(\lambda _{m} r/R)\) which gives

$$\begin{aligned} \frac{\varLambda \nu }{B}u^{s}(r)J_{0}(\lambda _{m} r/R)=\frac{\varLambda \nu }{B}\sum _{n=1}^{\infty }A_{n}J_{0}(\lambda _{n} r/R)J_{0}(\lambda _{m} r/R), \end{aligned}$$
(25)

and when \(r=R\),

$$\begin{aligned} \frac{\varLambda \nu }{B}u^{s}(R)J_{0}(\lambda _{m})=\frac{\varLambda \nu }{B}\sum _{n=1}^{\infty }A_{n}J_{0}(\lambda _{n}) J_{0}(\lambda _{m}). \end{aligned}$$
(26)

Adding (24) and (26) gives

$$\begin{aligned}&\frac{\varLambda \nu }{B}u^{s}(R)J_{0}(\lambda _{m})+\int _{0}^{R}ru^{s}(r)J_{0}(\lambda _{m})\,\mathrm {d}r \nonumber \\&\quad =\sum _{n=1}^{\infty }A_{n}\left[ \frac{\varLambda \nu }{B}J_{0}(\lambda _{n})J_{0}(\lambda _{m}) +\int _{0}^{R}rJ_{0}(\lambda _{n} r/R)J_{0}(\lambda _{m} r/R)\,\mathrm {d}r\right] =\sum _{n=1}^{\infty }A_{n}N_{n}\delta _{n,m} =A_{m}N_{m}. \end{aligned}$$
(27)

where, the orthogonality condition (22) has been used to obtain the right-hand side of (27). Thus, the coefficients are given by

$$\begin{aligned} A_{n}=\frac{2 R^{2}G}{\eta }\frac{ J_{1}(\lambda _{n})}{\lambda _{n}^{3}\left[ \left( 1+\dfrac{2\varLambda \nu }{BR^{2}}\right) J_{0}^{2}(\lambda _{n})+J_{1}^{2}(\lambda _{n})\right] }, \end{aligned}$$
(28)

and the solution to the problem becomes

$$\begin{aligned} u_{z}(r,t)=\frac{R^{2}G}{4\eta }\left[ 1+2B-\left( \frac{r}{R}\right) ^{2} -8\sum _{n=1}^{\infty } \frac{ J_{1}(\lambda _{n}) J_{0}\left( \lambda _{n}r/R\right) e^{-\displaystyle {\dfrac{\lambda _{n}^{2}\nu }{R^{2}}t}}}{\lambda _{n}^{3}\left[ \left( 1+\dfrac{2\varLambda \nu }{BR^{2}}\right) J_{0}^{2}(\lambda _{n})+J_{1}^{2}(\lambda _{n})\right] }\right] . \end{aligned}$$
(29)

Therefore, the slip velocity is

$$\begin{aligned} u_{w}(t)=\frac{R^{2}G}{2\eta }\left[ B -4\sum _{n=1}^{\infty } \frac{ J_{0}(\lambda _{n})J_{1}(\lambda _{n})}{\lambda _{n}^{3}\left[ \left( 1+\dfrac{2\varLambda \nu }{BR^{2}}\right) J_{0}^{2}(\lambda _{n})+J_{1}^{2}(\lambda _{n})\right] } e^{-\displaystyle {\dfrac{\lambda _{n}^{2}\nu }{R^{2}}t}}\right] \end{aligned}$$
(30)

and the volumetric flow rate is given by

$$\begin{aligned} Q(t)=\frac{\pi R^{4}G}{8\eta }\left[ 1+4B -32\sum _{n=1}^{\infty } \frac{J_{1}^{2}(\lambda _{n})}{\lambda _{n}^{4}\left[ \left( 1+\dfrac{2\varLambda \nu }{BR^{2}}\right) J_{0}^{2}(\lambda _{n})+J_{1}^{2}(\lambda _{n})\right] } e^{-\displaystyle {\dfrac{\lambda _{n}^{2}\nu }{R^{2}}t}}\right] . \end{aligned}$$
(31)

Setting \(\varLambda =0\) in Eq. (29) we obtain the solution for the static slip case

$$\begin{aligned} u_{z}(r,t)=\frac{R^{2}G}{4\eta }\left[ 1+2B-\left( \frac{r}{R}\right) ^{2} -8\sum _{n=1}^{\infty } \frac{ J_{0}\left( \lambda _{n}r/R\right) e^{-\displaystyle {\dfrac{\lambda _{n}^{2}\nu }{R^{2}}t}}}{\lambda _{n}^{3}\left( 1+B^{2}\lambda _{n}^{2}\right) J_{1}(\lambda _{n})}\right] , \end{aligned}$$
(32)

where \(\lambda _{n}\) are the roots of

$$\begin{aligned} J_{0}(\lambda _{n})=B\lambda _{n} J_{1}(\lambda _{n}). \end{aligned}$$
(33)

In the case of no-slip, \(B=\varLambda =0\). It turns out that \(\lambda _{n}\) are the roots of \(J_{0}\) and the standard-textbook solution is obtained.

$$\begin{aligned} u_{z}(r,t)=\frac{R^{2}G}{4\eta }\left[ 1-\left( \frac{r}{R}\right) ^{2} -8\sum _{n=1}^{\infty } \frac{ J_{0}\left( \lambda _{n}r/R\right) e^{-\displaystyle {\dfrac{\lambda _{n}^{2}\nu }{R^{2}}t}}}{\lambda _{n}^{3}J_{1}(\lambda _{n})}\right] . \end{aligned}$$
(34)

To non-dimensionalise the solution, we scale the velocity by the mean steady-state velocity \(U\), lengths by \(R\), pressure by \(\eta U/R\), and time by \(\nu /R^{2}\) and denote all dimensionless variables by stars. Hence, the dimensionless velocity is given by

$$\begin{aligned} u_{z}^{*}(r^{*},t^{*})=\dfrac{2}{1+4B}\left[ 1+2B-{r^*}^{2}-8\sum _{n=1}^{\infty } \frac{ J_{1}(\lambda _{n})J_{0}\left( \lambda _{n}\right) e^{-\displaystyle {\lambda _{n}^{2}t^{*}}}}{\lambda _{n}^{3}\left[ \left( 1+\dfrac{2\varLambda ^{*}}{B}\right) J_{0}^{2}(\lambda _{n})+J_{1}^{2}(\lambda _{n})\right] }\right] , \end{aligned}$$
(35)

where

$$\begin{aligned} \varLambda ^{*}=\dfrac{\varLambda \nu }{R^{2}}, \end{aligned}$$
(36)

is the dimensionless dynamic slip number. The non-dimensional slip velocity at the wall is given by

$$\begin{aligned} u_{w}^{*}(t^{*})=\dfrac{4}{1+4B}\left[ B -4\sum _{n=1}^{\infty } \frac{J_{0}(\lambda _{n})J_{1}(\lambda _{n})}{\lambda _{n}^{3}\left[ \left( 1+\dfrac{2\varLambda ^{*}}{B}\right) J_{0}^{2}(\lambda _{n})+J_{1}^{2}(\lambda _{n})\right] }e^{-\displaystyle {\lambda _{n}^{2}t^{*}}}\right] , \end{aligned}$$
(37)

and the dimensionless volumetric flow rate becomes

$$\begin{aligned} Q^{*}(t^{*})=1 -\dfrac{32}{1+4B}\sum _{n=1}^{\infty } \frac{J_{1}^{2}(\lambda _{n})}{\lambda _{n}^{4}\left[ \left( 1+\dfrac{2\varLambda ^{*}}{B}\right) J_{0}^{2}(\lambda _{n})+J_{1}^{2}(\lambda _{n})\right] } e^{-\displaystyle {\lambda _{n}^{2}t^{*}}}. \end{aligned}$$
(38)
Fig. 1
figure 1

Evolution of the velocity profile in start-up axisymmetric Poiseuille flow with \(B=0.1\): a \(\varLambda ^{*}=0\) (static slip); b \(\varLambda ^{*}=1\); and c \(\varLambda ^{*}=10\)

Full size image
Fig. 2
figure 2

Evolution of the velocity profile in start-up axisymmetric Poiseuille flow with \(B=1\): a \(\varLambda ^{*}=0\) (static slip); b \(\varLambda ^{*}=1\); and c \(\varLambda ^{*}=10\)

Full size image
Fig. 3
figure 3

Evolution of the slip velocity, in start-up axisymmetric Poiseuille flow for different values \(\varLambda ^{*}\): a \(B=0.1\); and b \(B=1\)

Full size image
Fig. 4
figure 4

Evolution of the volumetric flow rate, in start-up axisymmetric Poiseuille flow for different values of \(\varLambda ^{*}\): a \(B=0.1\); b \(B=1\)

Full size image

The evolution of the velocity profile for two slip numbers, \(B=0.1\) and \(1\) (corresponding to moderate and strong slip), and three relaxation times (\(\varLambda ^{*}=0\), 1 and 10) is illustrated in Figs. 1 and 2. Compared to the static slip model \((\varLambda ^{*}=0)\), it takes longer for the flow to develop to steady-state for higher values of the relaxation time. This is due to the fact that the slip boundary condition changes dynamically and the past states of the wall shear stress affect the slip velocity. This is more clearly seen in Fig. 3, which shows the evolution of the slip velocity for different values of \(\varLambda ^{*}\).

Similar conclusions can be drawn for the effect of the relaxation time on the evolution of the dimensionless volumetric flow rate, illustrated in Fig. 4. An interesting observation is that for higher values of \(B\) (i.e. for stronger slip), \(Q^{*}\) needs more time to develop, and thus at any given time, \(Q^{*}\) decreases with increasing \(B\). Actually, there is more flow as \(B\) increases, which cannot be seen due to our non-dimensionalisation.

In the case of cessation i.e. when the flow is initially fully-developed flow and the pressure-gradient is suddenly set to zero, the velocity is given by:

$$\begin{aligned} u_{z}^{*}(r^{*},t^{*})=\dfrac{16}{1+4B}\sum _{n=1}^{\infty } \frac{ J_{1}(\lambda _{n})J_{0}\left( \lambda _{n}\right) e^{-\displaystyle {\lambda _{n}^{2}t^{*}}}}{\lambda _{n}^{3}\left[ \left( 1+\dfrac{2\varLambda ^{*}}{B}\right) J_{0}^{2}(\lambda _{n})+J_{1}^{2}(\lambda _{n})\right] }. \end{aligned}$$
(39)

Similarly, the slip velocity and the volumetric flow rate correspond to the last terms of Eqs. (37) and (38).

3 Planar geometry

3.1 Poiseuille flow

In the case of plane Poiseuille flow in a tube with \(-1\le y^{*}\le 1\), one finds that the dimensionless velocity is given by

$$\begin{aligned} u^{*}_{x}(y^{*},t^{*})=\frac{3}{2(1+3B)}\left[ 1+2B-{y^*}^{2} -4\sum _{n=1}^{\infty } \frac{\sin \lambda _{n}\cos (\lambda _{n}y^{*})e^{-\displaystyle {\lambda _{n}^{2}t^{*}}}}{\lambda _{n}^{2}\left( \dfrac{2\lambda _{n}\varLambda ^{*}}{B}\cos ^{2} \lambda _{n}+\lambda _{n}+\sin \lambda _{n}\cos \lambda _{n}\right) }\right] , \end{aligned}$$
(40)

where similar scales are used for the non-dimensionalisation, \(B\equiv \eta /(\beta H)\) is the slip number, and the eigenvalues, \(\lambda _{n}\), are the solutions of the transcendental equation

$$\begin{aligned} \tan \lambda _{n}=\frac{1}{B \lambda _{n}}(1-\varLambda ^{*}\lambda _{n}^{2}), \end{aligned}$$
(41)

where \(\varLambda ^{*}=\varLambda \nu /H^{2}\). Hence, the non-dimensional slip velocity takes the form

$$\begin{aligned} u^{*}_{w}(t^{*})=\frac{3}{1+3B}\left[ B-2\sum _{n=1}^{\infty } \frac{\sin \lambda _{n}\cos \lambda _{n}}{\lambda _{n}^{2}\left( \dfrac{2\lambda _{n}\varLambda ^{*}}{B}\cos ^{2} \lambda _{n}+\lambda _{n}+\sin \lambda _{n}\cos \lambda _{n}\right) } e^{-\displaystyle {\lambda _{n}^{2}t^{*}}}\right] , \end{aligned}$$
(42)

and the dimensionless volumetric flow rate is given by,

$$\begin{aligned} Q^{*}(t^{*})=1-\dfrac{6}{1+3B}\sum _{n=1}^{\infty } \frac{\sin ^{2} \lambda _{n}}{\lambda _{n}^{3}\left( \dfrac{2\lambda _{n}\varLambda ^{*}}{B}\cos ^{2} \lambda _{n}+\lambda _{n}+\sin \lambda _{n}\cos \lambda _{n}\right) }e^{-\displaystyle {\lambda _{n}^{2}t^{*}}}. \end{aligned}$$
(43)

Obviously, the velocity in the case of cessation is given by

$$\begin{aligned} u^{*}_{x}(y^{*},t^{*})=\frac{6}{2(1+3B)}\sum _{n=1}^{\infty } \frac{\sin \lambda _{n}\cos (\lambda _{n}y^{*})e^{-\displaystyle {\lambda _{n}^{2}t^{*}}}}{\lambda _{n}^{2}\left( \dfrac{2\lambda _{n}\varLambda ^{*}}{B}\cos ^{2} \lambda _{n}+\lambda _{n}+\sin \lambda _{n}\cos \lambda _{n}\right) }. \end{aligned}$$
(44)

3.2 Couette flow

Here we consider the planar Couette flow in a tube with \(0\le y^{*}\le 1\) where the wall at \(y^{*}=1\) moves with speed \(V\). One finds that the non-dimensional velocity (scaled by \(V\)) is given by

$$\begin{aligned} u^{*}_{x}(y^{*},t^{*})=\dfrac{y^{*}+B}{1+2B}-\sum _{n=1}^{\infty } C_{n}\left[ \sin (\lambda _{n}y^{*})+\dfrac{\lambda _{n}}{BI}\cos (\lambda _{n}y^{*})\right] e^{-\displaystyle {\lambda _{n}^{2}t^{*}}}, \end{aligned}$$
(45)

where

$$\begin{aligned} C_{n}=\dfrac{(B^{2}\lambda _{n}^{2}+I^{2})\sin \lambda _{n}}{\lambda _{n}^{2}B(I^{2}+2B+B^{2}\lambda _{n}^{2}+2\varLambda ^{*}\lambda _{n}^{2})},\qquad I=1-\lambda _{n}^{2}\varLambda ^{*}, \end{aligned}$$
(46)

and the eigenvalues, \(\lambda _{n}\), are the solutions of the transcendental equation

$$\begin{aligned} \tan \lambda _{n}=\frac{2BI\lambda _{n}}{B \lambda _{n}}(B^{2}\lambda _{n}^{2}-I^{2}). \end{aligned}$$
(47)

The non-dimensional slip velocities at \(y^{*}=0\), and \(1\) are given by

$$\begin{aligned} u^{*}_{w1}(t^{*})=\dfrac{B}{1+2B}-\sum _{n=1}^{\infty } C_{n}\dfrac{\lambda _{n}}{BI}e^{-\displaystyle {\lambda _{n}^{2}t^{*}}}, \end{aligned}$$
(48)

and

$$\begin{aligned} u^{*}_{w2}(t^{*})=\dfrac{1+B}{1+2B}-\sum _{n=1}^{\infty } C_{n}\left( \sin \lambda _{n}+\dfrac{\lambda _{n}}{BI}\cos \lambda _{n}\right) e^{-\displaystyle {\lambda _{n}^{2}t^{*}}}, \end{aligned}$$
(49)

respectively. The dimensionless volumetric flow rate is given by,

$$\begin{aligned} Q^{*}(t^{*})=1-\dfrac{6}{1+3B}\sum _{n=1}^{\infty } \frac{\sin ^{2} \lambda _{n}}{\lambda _{n}^{3}\left( \dfrac{2\lambda _{n}\varLambda ^{*}}{B}\cos ^{2} \lambda _{n}+\lambda _{n}+\sin \lambda _{n}\cos \lambda _{n}\right) }e^{-\displaystyle {\lambda _{n}^{2}t^{*}}}. \end{aligned}$$
(50)

In the case of cessation, the velocity is given by

$$\begin{aligned} u^{*}_{x}(y^{*},t^{*})=\sum _{n=1}^{\infty } C_{n}\left[ \sin (\lambda _{n}y^{*})+\dfrac{\lambda _{n}}{BI}\cos (\lambda _{n}y^{*})\right] e^{-\displaystyle {\lambda _{n}^{2}t^{*}}}. \end{aligned}$$
(51)

4 Circular Couette flow

In this section we derive the analytical solution for the start-up Newtonian circular Couette flow, i.e. the flow between two vertical coaxial cylinders of infinite length and radii \(\kappa R\) and \(R\), with \(0<\kappa <1\). The outer cylinder is stationary while the inner cylinder rotates about the common axis with angular velocity \(\varOmega \), in the absence of gravity. Hence, the angular velocity component \(u_{\theta }(r,t)\) satisfies,

$$\begin{aligned} \rho \frac{\partial u_{\theta }}{\partial t}=\eta \frac{\partial }{\partial r}\left( \frac{1}{r}\frac{\partial }{\partial r}(r u_{\theta })\right) . \end{aligned}$$
(52)

The fluid is initially at rest, thus

$$\begin{aligned} u_{\theta }(r,0)=0. \end{aligned}$$
(53)

We assume that both walls have the same properties, so the b.c.s. are given by

$$\begin{aligned} u_{\theta }&= \varOmega \kappa R-u_{w1}, \quad {\text {on}} \, r=\kappa R \end{aligned}$$
(54)
$$\begin{aligned} u_{\theta }&= u_{w2}, \qquad \qquad \quad \, {\text {on}} \, r=R , \end{aligned}$$
(55)

where \(u_{w1}\) and \(u_{w2}\) are the slip velocities at the inner and outer walls, respectively. The wall shear stress is defined as

$$\begin{aligned} \tau _{w}=|\tau _{r\theta }|=-\eta r\dfrac{\partial }{\partial r}\left( \dfrac{u_{w}}{r}\right) . \end{aligned}$$
(56)

For this type of flow, the slip velocity at the inner wall, which is rotating, is actually decreasing with time. This is a consequence of the sign convention employed in b.c.s. (54) and (55). In start-up flow, the velocity of the fluid at both walls will be increasing, which implies that \(u_{w1}\) will be decreasing and \(u_{w2}\) will be increasing with time. Thus the appropriate slip equations are

$$\begin{aligned} u_{w1}-\varLambda \dfrac{\partial u_{w1}}{\partial t}&= \dfrac{\tau _{w1}}{\beta }, \end{aligned}$$
(57)
$$\begin{aligned} u_{w2}+\varLambda \dfrac{\partial u_{w2}}{\partial t}&= \dfrac{\tau _{w2}}{\beta }. \end{aligned}$$
(58)

Hence the b.c.s. become

$$\begin{aligned} u_{\theta }&= \varOmega \kappa R+\dfrac{\eta }{\beta }r\dfrac{\partial }{\partial r}\left( \dfrac{u_{w}}{r}\right) -\varLambda \dfrac{\partial u_{w}}{\partial t}, \quad {\text {on}} \, r=\kappa R \end{aligned}$$
(59)
$$\begin{aligned} u_{\theta }&= -\dfrac{\eta }{\beta }r\dfrac{\partial }{\partial r}\left( \dfrac{u_{w}}{r}\right) -\varLambda \dfrac{\partial u_{w}}{\partial t}, \quad {\text {on}} \, r=R. \end{aligned}$$
(60)

The steady-state velocity profile is given by

$$\begin{aligned} u_{\theta }^{s}= \kappa C \varOmega R\left[ (2B-1)\dfrac{r}{R}+\dfrac{R}{r}\right] , \end{aligned}$$
(61)

where \(B\equiv \eta /(\beta R)\) and

$$\begin{aligned} C=\dfrac{ \kappa ^{2}}{\kappa -\kappa ^{3}+2B(1+\kappa ^{3})}. \end{aligned}$$
(62)

Note that the steady-state slip velocities satisfy \(u_{w2}^{s}=\kappa ^{2}u_{w1}^{s}\) and are given by,

$$\begin{aligned} u_{w1}^{s}=2 CB\varOmega R/\kappa \end{aligned}$$
(63)

and

$$\begin{aligned} u_{w2}^{s}=2\kappa CB\varOmega R. \end{aligned}$$
(64)

Since the b.c. (59) is non-homogeneous we follow a similar procedure as in Sect. 2 thus, the solution to the homogeneous problem, \(\bar{u}_{\theta }\) becomes

$$\begin{aligned} \bar{u}_{\theta }(r,t)=\sum _{n=1}^{\infty }D_{n}Z_{1}(\lambda _{n} r/R) e^{-\displaystyle {\dfrac{\lambda _{n}^{2}\nu }{R^{2}}t}}, \end{aligned}$$
(65)

where,

$$\begin{aligned} Z_{1}(\lambda _{n}r/R)\equiv J_{1}(\lambda _{n}r/R)+ \delta _{n}Y_{1}(\lambda _{n}r/R), \end{aligned}$$
(66)

and \(\lambda _{n}\), \(\delta _{n}\) are the \(n\)th pair of the roots of

$$\begin{aligned} B\lambda _{n}Z_{0}(\kappa \lambda _{n})&= \left( 1+ \dfrac{2B}{\kappa }-\lambda _{n}^{2}\dfrac{\varLambda \nu }{R^{2}}\right) Z_{1}(\kappa \lambda _{n}), \end{aligned}$$
(67)
$$\begin{aligned} B\lambda _{n}Z_{0}(\lambda _{n})&= \left( -1+2B+\lambda _{n}^{2}\dfrac{\varLambda \nu }{R^{2}}\right) Z_{1}(\lambda _{n}). \end{aligned}$$
(68)

The appropriate orthogonality condition for the eigenfunctions \(Z_{1}\) can be found as

$$\begin{aligned} \frac{\varLambda \nu }{B}\left[ Z_{1}(\lambda _{n})Z_{1}(\lambda _{m})+ \kappa Z_{1}(\kappa \lambda _{n})Z_{1}(\kappa \lambda _{m})\right] + \int _{\kappa R}^{R}rZ_{1}(\lambda _{n}r/R)Z_{1}(\lambda _{m}r/R)\,\mathrm {d}r =\delta _{m,n}N_{n} \end{aligned}$$
(69)

where

$$\begin{aligned} N_{n}=\frac{\varLambda \nu }{B}[Z_{1}^{2}(\lambda _{n})+\kappa Z_{1}^{2}(\kappa \lambda _{n})]+\int _{\kappa R}^{R}rZ_{1}^{2}(\lambda _{n}r/R)\,\mathrm {d}r, \end{aligned}$$
(70)

and the coefficients in (65) are given by:

$$\begin{aligned} D_{n}=\dfrac{2\varOmega \kappa RC}{\lambda _{n}}\left[ \dfrac{\kappa -\kappa ^{3}+2B(1+\kappa ^{3})}{L}\right] Z_{1}(\kappa \lambda _{n}) \end{aligned}$$
(71)

where

$$\begin{aligned} L&= \dfrac{2\varLambda \nu }{R^{2}}\kappa \lambda _{n}[Z_{1}^{2}(\lambda _{n})+\kappa Z_{1}^{2}(\kappa \lambda _{n})]+ \lambda _{n}[Z_{0}^{2}(\lambda _{n})+Z_{1}^{2}(\lambda _{n}) \nonumber \\&-\kappa ^{2}(Z_{0}^{2}(\kappa \lambda _{n})+ Z_{1}^{2}(\kappa \lambda _{n}))]-2[Z_{0}(\lambda _{n})Z_{1}(\lambda _{n})- \kappa (Z_{0}(\kappa \lambda _{n})Z_{1}(\kappa \lambda _{n}))] \end{aligned}$$
(72)

The solution to the full problem thus becomes,

$$\begin{aligned} u_{\theta }(r,t)=\varOmega \kappa R C\left[ (2B-1)\dfrac{r}{R}+\dfrac{R}{r}-\sum _{n=1}^{\infty }\widetilde{D}_{n}Z_{1}\left( \dfrac{\lambda _{n} r}{R}\right) e^{-\displaystyle {\dfrac{\lambda _{n}^{2}\nu }{R^{2}}t}}\right] , \end{aligned}$$
(73)

where

$$\begin{aligned} \widetilde{D}_{n}=\dfrac{D_{n}}{CR}. \end{aligned}$$
(74)

Scaling the velocity by \(\varOmega \kappa R\) and the rest of the quantities as in Sect. 2, the non-dimensional velocity profile is given by

$$\begin{aligned} u_{\theta }^{*}(r^{*},t^{*})= C\left[ (2B-1)r^{*}+\dfrac{1}{r^{*}}-\sum _{n=1}^{\infty }\widetilde{D}_{n}Z_{1}\left( \lambda _{n} r^{*}\right) e^{-\displaystyle {\lambda _{n}^{2}t^{*}}}\right] \end{aligned}$$
(75)

and the dimensionless slip velocities by

$$\begin{aligned} u_{w1}^{*}(t^{*})=C\left[ \dfrac{2B}{\kappa ^{2}}+\sum _{n=1}^{\infty }\widetilde{D}_{n} Z_{1}\left( \kappa \lambda _{n} \right) e^{-\displaystyle {\lambda _{n}^{2}t^{*}}}\right] \end{aligned}$$
(76)

and

$$\begin{aligned} u_{w2}^{*}(t^{*})=C\left[ 2B-\sum _{n=1}^{\infty }\widetilde{D}_{n}Z_{1}\left( \lambda _{n}\right) e^{-\displaystyle {\lambda _{n}^{2}t^{*}}}\right] . \end{aligned}$$
(77)

In the case of cessation, the velocity is given by

$$\begin{aligned} u_{\theta }^{*}(r^{*},t^{*})= C\left[ (2B-1)r^{*}+\dfrac{1}{r^{*}}-\sum _{n=1}^{\infty }\widetilde{D}_{n}Z_{1}\left( \lambda _{n} r^{*}\right) e^{-\displaystyle {\lambda _{n}^{2}t^{*}}}\right] . \end{aligned}$$
(78)

In Figs. 5 and 6, the evolution of the velocity profile is plotted, for slip numbers, \(B=0.1\), and \(B=1\), respectively, \(\kappa =0.25\) and three relaxation times (\(\varLambda ^{*}=0.1\), 1 and 10). Similarly to the axisymmetric Poiseuille problem, higher values of \(\varLambda ^{*}\) increase the time that the velocity evolves to steady-state. In the no-slip case \((B=0)\), the fluid velocity at the inner wall, which is rotating, is always constant and equals the speed of rotation. However, it can be seen here where slip is present that the fluid velocity at the inner wall evolves gradually to its steady-state value, and it is always less than the speed of rotation. This can also be inferred indirectly from Fig. 7, which shows the evolution of the two slip velocities for two slip numbers (\(B=0.1\) and \(1\)) and two relaxation times (\(\varLambda ^{*}=0\) and \(1\)). The slip velocity at the inner wall initially, equals the speed of rotation and then drops eventually to its steady-state value \(u_{w1}^{*s}\). On the other hand, \(u_{w2}^{*}\) increases with time to reach the steady-state value \(u_{w2}^{*s}=\kappa ^{2}u_{w1}^{*s}\).

Fig. 5
figure 5

Evolution of the velocity profile in start-up circular Couette flow with \(B=0.1\) and \(\kappa =0.25\): a \(\varLambda ^{*}=0\); b \(\varLambda ^{*}=1\); and c \(\varLambda ^{*}=10\)

Full size image
Fig. 6
figure 6

Evolution of the velocity profile in start-up circular Couette flow with \(B=1\) and \(\kappa =0.25\): a \(\varLambda ^{*}=0\); b \(\varLambda ^{*}=1\); and c \(\varLambda ^{*}=10\)

Full size image
Fig. 7
figure 7

Evolution of the slip velocities in start-up circular Couette flow for different values of \(\varLambda ^{*}\) and \(\kappa =0.25\): a \(u^{*}_{w1}\); and b \(u^{*}_{w2}\)

Full size image

5 Conclusions

We have derived analytical solutions for the start-up Newtonian Poiseuille and Couette flows with dynamic wall slip. For this purpose, the appropriate orthogonality condition for the spatial eigenfunctions has been derived. Under a dynamic slip condition, the slip velocity rather than depending on the instantaneous value of the wall shear stress, also depends on its past states. This effect delays the evolution of the slip velocity, and also the flow development.