About this Journal Submit a Manuscript Table of Contents
ISRN Applied Mathematics
Volume 2012 (2012), Article ID 274914, 12 pages
http://dx.doi.org/10.5402/2012/274914
Research Article

Another Approach to the Extended Stokes’ Problems for the Oldroyd-B Fluid

Division of Mathematics, General Education Center, Chienkuo Technology University, Changhua City 500, Taiwan

Received 15 March 2012; Accepted 8 May 2012

Academic Editors: Y. Dimakopoulos, S. W. Gong, S. Prudhomme, and X.-S. Yang

Copyright © 2012 Chi-Min Liu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

The extended Stokes problems, which study the flow suddenly driven by relatively moving half-planes, are reexamined for the Oldroyd-B fluid. This topic has been studied (Liu, 2011) by applying the series expansion to calculate the inverse Laplace transform. The derived solution was correct but tough to perform the calculation due to the series expansion of infinite terms. Herein another approach, the contour integration, is applied to calculate the inversion. Moreover, the Heaviside unit step function is included into the boundary condition to ensure the consistence between boundary and initial conditions. Mathematical methods used herein can be applied to other fluids for the extended Stokes’ problems.

1. Introduction

In 1995, Zeng and Weinbaum’s paper studying the viscous flow driven by relatively moving half-planes is a new as well as pioneering work for theoretical fluid mechanics [1]. This problem is later named as the extended Stokes’ problem. Different from the well-known traditional Stokes’ problems in which the flow depends on only one spatial parameter [2, 3], the extended Stokes’ problems possess two spatial dependences. The additional spatial dependence makes the problem more complex than the traditional one. To solve the velocity, one needs to perform two different integral transforms to the momentum equation with the help of boundary and initial conditions given. For the Newtonian fluid, the analytic solutions for a finite-depth and an infinite-depth cases have been provided [4, 5].

As for non-Newtonian fluids, the extended problem for the Oldroyd-B fluid was analyzed [6] by expanding the rheological parameters as well as the spatial and temporal variables in a series form. This expansion provides a way to carry out the inverse transform to pursue the exact velocity profile. As the final solution is also expressed in an infinite series form, it is sometimes time-consuming to perform the calculation. To this end, the contour integration was adopted to perform the inverse calculation for the flow of a second-grade fluid [7]. Although the contour integration is a correct and universal tool for calculating the inverse Laplace transform, it is sometimes laborious, and even impossible, due to the complex mathematical structure of the transformed variable. Different fluid models usually have different transformed types. The variety of transformed types makes a fixed contour path impossible and thus leads to the difficulty of calculating the integration.

From the mathematical viewpoint, it seems easy to reduce the Oldroyd-B model to either the Maxwell fluid or the second-grade fluid by simply setting one parameter to be zero. However, it is not always true for the inversion calculation using the contour integration. For the second-grade fluid, the contour path shown in [7] succeeds in calculating the inversion while the same path cannot be applied to the Oldroyd-B fluid. More singularity points which should be excluded from the contour make the latter case much more complicated. This is why we study the topic again.

The organization of this paper is as follows. In Section 2, the fluid system including the momentum equation, boundary and initial conditions is elucidated. The transform integrals adopted are also introduced. Next, the detailed derivation is presented in Section 3, and concluding remarks are made in Section 4.

2. Problem Description

First, the constitutive equation, boundary and initial conditions considered are shown below (see [6] for details)𝜕1+𝜆𝜕𝑡𝜕𝑢𝜕𝑡=𝜈1+𝜆𝑟𝜕𝜕𝜕𝑡2𝑢𝜕𝑦2+𝜕2𝑢𝜕𝑧2𝑢𝑢,(2.1)(𝑦=0,𝑧>0)=𝑔(𝑡)𝐻(𝑡),(2.2)𝑢(𝑦=0,𝑧<0)=𝑔(𝑡)𝐻(𝑡),(2.3)(𝑦)=0,(2.4)𝑢(𝑧±)isnite,(2.5)𝑢(𝑡=0)=0,(2.6) where 𝑢 represents the velocity along the 𝑥 direction, the fluid constants 𝜈, 𝜆, and 𝜆𝑟 are the kinematic viscosity, the relaxation time, and the retardation times, and 𝑢𝑔(𝑡)=0,𝑢0cos(𝑡𝜎+𝜃)(2.7) are the plate boundary conditions for the first and second problems, respectively. The Heaviside unit step function 𝐻(𝑡) is defined as 𝐻(𝑡)=0,𝑡0,1,𝑡>0.(2.8) The inclusion of the Heaviside unit step function is to avoid the inconsistence between (2.2) and (2.6) at the point (𝑦,𝑡)=(0,0) (also see [8] for reference). Due to the system being the odd function of 𝑧, one only needs to calculate the positive-𝑧 domain, and the boundary condition (2.3) is thus replaced by 𝑢(𝑧=0)=0.(2.9) Now applying the Laplace transform 𝑓(𝑠)0𝑓(𝑡)𝑒𝑠𝑡𝑑𝑡,(2.10) and then the Fourier sine transform 𝑓(𝜔)0𝑓(𝑦)sin(𝑦𝜔)𝑑𝑦,(2.11) to the present PDE system, the transformed velocity is solved to be (see [6] for details) 𝜔̃𝑢=𝜔2+𝑠(1+𝜆𝑠)𝜈1+𝜆𝑟𝑠11exp𝑧𝜔2+𝑠(1+𝜆𝑠)𝜈1+𝜆𝑟𝑠̂𝑔(𝑠),(2.12) where 𝑢̂𝑔(𝑠)=0𝑠,𝑢0𝑠cos𝜃𝑠2+𝜎2𝜎sin𝜃𝑠2+𝜎2,(2.13) for two kinds of problems, respectively. Above methods and results have been elucidated in [6]. The existing paper employed a series expansion to (2.12) to calculate the inverse transformation and then derived the solution in a series form. However, this expansion may result in a possible divergence. In the next section, another approach using the contour integration will be introduced for the sake of overcoming this weakness.

3. Contour Integration

As mentioned above, a different approach for calculating the inverse Laplace transform is adopted to derive the solution. Firstly, the term in the square root of the exponential function of (2.12) is written 𝜔2+𝑠(1+𝜆𝑠)𝜈1+𝜆𝑟𝑠=𝜆𝜈𝜆𝑟𝑠𝑠1𝑠𝑠2𝑠𝑠0,(3.1) where 𝑠01=𝜆𝑟,𝑠(3.2)1=12𝜆1+𝜔2𝜈𝜆𝑟+1+𝜔2𝜈𝜆𝑟24𝜔2𝑠𝜈𝜆,(3.3)2=12𝜆1+𝜔2𝜈𝜆𝑟1+𝜔2𝜈𝜆𝑟24𝜔2𝜈𝜆.(3.4) It is clear that the sign of (1+𝜔2𝜈𝜆𝑟)24𝜔2𝜈𝜆 strongly dominate the pole positions inside the contour path which will be demonstrated later. With the help of (3.1), the inversion of (2.12) is shown 2𝑢=𝜋𝜔=𝜔=01𝑑𝜔𝜔sin(𝑦𝜔)2𝜋𝑖𝑠=𝜅+𝑖𝑠=𝜅𝑖𝐼𝑑𝑠,(3.5) where 𝐼=𝜈𝜆𝑟𝜆𝑠𝑠0̂𝑔(𝑠)𝑠𝑠1𝑠𝑠21exp𝑧𝜆𝜈𝜆𝑟𝑠𝑠1𝑠𝑠2𝑠𝑠0𝑒𝑠𝑡.(3.6) Two cases for calculating (3.5) are discussed below.

3.1. The Case 𝜆𝑟𝜆>0

For the case 𝜆𝑟𝜆>0, the term in the square root of (3.3) and (3.4) can be rewritten as 1𝜔2𝜈𝜆𝑟2+4𝜔2𝜈𝜆𝑟,𝜆(3.7) which possess a nonnegative value implying that both of 𝑠1 and 𝑠2 are negative real numbers for all values of 𝜔. Thus the relation between 𝑠0, 𝑠1, and 𝑠2 is 𝑠1𝑠0𝑠2.(3.8) Now the contour integration shown in Figure 1 is adopted to calculate the inner integral in (3.5). The result is 𝜅+𝑖𝜅𝑖𝐼𝑑𝑠=2𝑖𝑠2𝜈𝜆𝑟𝜆𝑠𝑠0̂𝑔(𝑠)𝑠𝑠1𝑠𝑠2𝑧sin𝜆𝜈𝜆𝑟𝑠𝑠1𝑠𝑠2𝑠0𝑠𝑒𝑠𝑡𝑑𝑠+2𝑖𝑠0𝑠1𝜈𝜆𝑟𝜆𝑠𝑠0̂𝑔(𝑠)𝑠𝑠1𝑠𝑠2𝑧sin𝜆𝜈𝜆𝑟𝑠𝑠1𝑠𝑠2𝑠0𝑠𝑒𝑠𝑡𝑑𝑠+2𝜋𝑖,(3.9) where denotes the contribution from the poles. It is clear that only one pole 𝑠=0 (note ̂𝑔(𝑠)=𝑢0/𝑠) exists for the first problem and the residue is 𝑢=Res(0)=0𝜈𝜆𝑟𝑠0𝜆𝑠1𝑠2exp𝑧𝜆𝜈𝜆𝑟𝑠1𝑠2𝑠0.1(3.10) As for the second problem, there are two poles, 𝑠=𝑖𝜎 and 𝑠=𝑖𝜎, and the result is =𝑢=Res(+𝑖𝜎)+Res(𝑖𝜎)0𝜈𝜆𝑟𝜆𝑟0𝑟1𝑟2cos𝑡𝜎+𝜃+𝜙0𝜙1𝜙2exp𝑧𝜆𝜈𝜆𝑟𝑟1𝑟2𝑟0𝜙cos1+𝜙2𝜙02cos𝑡𝜎+𝜃+𝜙0𝜙1𝜙2𝑧𝜆𝜈𝜆𝑟𝑟1𝑟2𝑟0𝜙sin1+𝜙2𝜙02,(3.11) where 𝑟0=𝜎2+𝜆𝑟2,𝑟1=𝜎2+14𝜆21+𝜔2𝜈𝜆𝑟1+𝜔2𝜈𝜆𝑟24𝜔2𝜈𝜆2,𝑟2=𝜎2+14𝜆21+𝜔2𝜈𝜆𝑟+1+𝜔2𝜈𝜆𝑟24𝜔2𝜈𝜆2,𝜙0=tan1𝜎𝜆𝑟,𝜙1=tan12𝜆𝜎1+𝜔2𝜈𝜆𝑟1+𝜔2𝜈𝜆𝑟24𝜔2,𝜙𝜈𝜆2=tan12𝜆𝜎1+𝜔2𝜈𝜆𝑟+1+𝜔2𝜈𝜆𝑟24𝜔2.𝜈𝜆(3.12)

274914.fig.001
Figure 1: The contour path for the case 𝜆𝑟𝜆>0. Single pole 𝑠=0 is for the first problem and two poles 𝑠=𝑖Ω and 𝑠=𝑖Ω are for the second problem.

The definition sketch for (3.12) is displayed in Figure 2. It is remarked that 𝑟1, 𝑟2, 𝜙1, and 𝜙2 depend on 𝜔. Substituting (3.9) into (3.6) with the help of (3.10) to (3.12), (3.5) can be calculated to obtain the exact solution of flow velocity 𝑢.

274914.fig.002
Figure 2: The relation between poles and 𝑠𝑖 for the second problem for the case 𝜆𝑟𝜆>0.
3.2. The Case 𝜆>𝜆𝑟>0

For this case, it is more complex than the previous case. Firstly the sign of 𝜔 has to be determined by setting the term in the square root of (3.3) and (3.4) to be zero. It reads 1+𝜔2𝜈𝜆𝑟24𝜔2𝜈𝜆>0,for0<𝜔<𝜔1or𝜔>𝜔2,(3.13)1+𝜔2𝜈𝜆𝑟24𝜔2𝜈𝜆<0,for𝜔1<𝜔<𝜔2,(3.14) where 𝜔1=2𝜆𝜆𝑟2𝜆𝜆𝜆𝑟𝜈𝜆2𝑟0.5,𝜔2=2𝜆𝜆𝑟+2𝜆𝜆𝜆𝑟𝜈𝜆2𝑟0.5.(3.15) According to the sign of (3.13) and (3.14), the integration with respect to 𝜔 in (3.5) has to be divided into three parts, as shown below 2𝑢=𝜋𝜔=𝜔1𝜔=01𝑑𝜔𝜔sin(𝑦𝜔)2𝜋𝑖𝑠=𝜅+𝑖𝑠=𝜅𝑖+2𝐼𝑑𝑠𝜋𝜔=𝜔2𝜔=𝜔11𝑑𝜔𝜔sin(𝑦𝜔)2𝜋𝑖𝑠=𝜅+𝑖𝑠=𝜅𝑖+2𝐼𝑑𝑠𝜋𝜔=𝜔=𝜔21𝑑𝜔𝜔sin(𝑦𝜔)2𝜋𝑖𝑠=𝜅+𝑖𝑠=𝜅𝑖𝐼𝑑𝑠.(3.16) For the first and last integrals, the poles 𝑠1 and 𝑠2 are negative real numbers due to the positive sign of (3.13). Therefore, the inner integral can be calculated using the contour path shown in Figure 3: 𝜅+𝑖𝜅𝑖𝐼𝑑𝑠=2𝑖𝑠0𝜈𝜆𝑟𝜆𝑠𝑠0̂𝑔(𝑠)𝑠𝑠1𝑠𝑠2𝑧sin𝜆𝜈𝜆𝑟𝑠𝑠1𝑠𝑠2𝑠0𝑠𝑒𝑠𝑡𝑑𝑠+2𝑖𝑠2𝑠1𝜈𝜆𝑟𝜆𝑠𝑠0̂𝑔(𝑠)𝑠𝑠1𝑠𝑠2𝑧sin𝜆𝜈𝜆𝑟𝑠𝑠1𝑠𝑠2𝑠0𝑠𝑒𝑠𝑡𝑑𝑠+2𝜋𝑖,(3.17) where 𝑠1>𝑠2>𝑠0.(3.18) The residue is evaluated to be equivalent to (3.10) and (3.11) for the first and second problems, respectively.

274914.fig.003
Figure 3: The contour path for the case 𝜆>𝜆𝑟>0 with the conditions 0<𝜔<𝜔1 or 𝜔>𝜔2. Single pole 𝑠=0 is for the first problem and two poles 𝑠=𝑖Ω and 𝑠=𝑖Ω are for the second problem.

As for the second term in (3.16), where 𝑠1 and 𝑠2 are complex conjugates, the contour path shown in Figure 4 is used to calculate the inner integral. It is noted that the contour path goes around the origin instead of 𝑠0. For the first problem, the inner integral is 𝜅+𝑖𝜅𝑖𝐼𝑑𝑠=2𝑖𝑠0𝑢0𝜈𝜆𝑟𝜆𝑠𝑠0𝑠𝑠1𝑠𝑠2𝑧sin𝜆𝜈𝜆𝑟𝑠𝑠1𝑠𝑠2𝑠0𝑠𝑒𝑠𝑡𝑢𝑑𝑠+2𝜋𝑖0𝜈𝜆𝑟𝑠0𝜆𝑠1𝑠2exp𝑧𝜆𝜈𝜆𝑟𝑠1𝑠2𝑠01+2𝜋𝑖,(3.19) where the second term in (3.19) is contributed by the integration around the origin. The residue 𝑠=Res1𝑠+Res2,(3.20) is calculated to be zero.

274914.fig.004
Figure 4: The contour path for the case 𝜆>𝜆𝑟>0 with the condition 𝜔1<𝜔<𝜔2. The poles 𝑠=𝑖Ω and 𝑠=𝑖Ω are for the second problem only.

For the second problem, we have 𝜅+𝑖𝜅𝑖𝐼𝑑𝑠=2𝑖𝑠0𝑢0(𝑠cos𝜃𝜎sin𝜃)𝑠2+𝜎2𝜈𝜆𝑟𝜆𝑠𝑠0𝑠𝑠1𝑠𝑠2𝑧sin𝜆𝜈𝜆𝑟𝑠𝑠1𝑠𝑠2𝑠0𝑠𝑒𝑠𝑡𝑑𝑠+2𝜋𝑖,(3.21) where the residue is contributed from four poles. It reads 𝑠=Res1𝑠+Res2=𝑢+Res(𝑖𝜎)+Res(𝑖𝜎)0𝜈𝜆𝑟𝜆𝑟0𝑟3𝑟4cos𝑡𝜎+𝜃+𝜙0𝜙3𝜙4exp𝑧𝜆𝜈𝜆𝑟𝑟3𝑟4𝑟0𝜙cos3+𝜙4𝜙02cos𝑡𝜎+𝜃+𝜙0𝜙3𝜙4𝑧𝜆𝜈𝜆𝑟𝑟3𝑟4𝑟0𝜙sin3+𝜙4𝜙02,(3.22) where 𝑟3=𝜎2𝜎𝜆4𝜔2𝜈𝜆1+𝜔2𝜈𝜆𝑟2+𝜔2𝜈𝜆,𝑟4=𝜎2+𝜎𝜆4𝜔2𝜈𝜆1+𝜔2𝜈𝜆𝑟2+𝜔2𝜈𝜆,𝜙3=tan12𝜆𝜎4𝜔2𝜈𝜆1+𝜔2𝜈𝜆𝑟21+𝜔2𝜈𝜆𝑟,𝜙4=tan12𝜆𝜎+4𝜔2𝜈𝜆1+𝜔2𝜈𝜆𝑟21+𝜔2𝜈𝜆𝑟,(3.23) where the relation is shown in Figure 5. Now the exact solution of 𝑢 can be acquired by substituting (3.17), and (3.19)–(3.23) into (3.16) to calculate the inverse Fourier transform.

274914.fig.005
Figure 5: The relation between poles and 𝑠𝑖 for the second problem for the case 𝜆>𝜆𝑟>0 with the condition 𝜔1<𝜔<𝜔2.

4. Concluding Remarks

The extended Stokes’ problems for the Oldroyd-B fluid are revisited in this paper by applying the contour integration to calculate the inversion of velocity rather than using the series expansion in the earlier paper. Two cases classified according to the relation between rheological parameters are analyzed. Mathematical techniques used in this paper can be generalized and applied to investigate other fluids for the extended Stokes’ problems.

Acknowledgment

The financial support from the National Science Council of Taiwan with the Grant no. NSC 100-2221-E-270-002 is acknowledged.

References

  1. Y. Zeng and S. Weinbaum, “Stokes problems for moving half-planes,” Journal of Fluid Mechanics, vol. 287, pp. 59–74, 1995. View at Publisher · View at Google Scholar · View at Zentralblatt MATH
  2. H. Schlichting, Boundary Layer Theory, Translated by J. Kestin, McGraw-Hill, New York, NY, USA, 1955.
  3. C.-M. Liu and I.-C. Liu, “A note on the transient solution of Stokes' second problem with arbitrary initial phase,” Journal of Mechanics, vol. 22, no. 4, pp. 349–354, 2006. View at Scopus
  4. C.-M. Liu, “Complete solutions to extended Stokes' problems,” Mathematical Problems in Engineering, vol. 2008, Article ID 754262, 18 pages, 2008. View at Publisher · View at Google Scholar · View at Zentralblatt MATH
  5. G. Riccardi, “Remarks on the solution of extended Stokes problems,” International Journal of Non-Linear Mechanics, vol. 46, no. 7, pp. 958–970, 2011. View at Publisher · View at Google Scholar · View at Scopus
  6. C.-M. Liu, “Extended Stokes' first problem of an Oldroyd-B fluid for relatively moving half-planes,” Mathematical Problems in Engineering, vol. 2011, Article ID 709836, 16 pages, 2011. View at Publisher · View at Google Scholar
  7. C. M. Liu, “Extended Stokes’ problems for second-grade fluids,” in submission.
  8. I. C. Christov, “Stokes' first problem for some non-Newtonian fluids: results and mistakes,” Mechanics Research Communications, vol. 37, no. 8, pp. 717–723, 2010. View at Publisher · View at Google Scholar · View at Scopus