remarks on the solution of extended stokes' problems

International Journal of Non-Linear Mechanics 46 (2011) 958–970

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International Journal of Non-Linear Mechanics

0020-74

doi:10.1

� Tel.

E-m

journal homepage: www.elsevier.com/locate/nlm

Remarks on the solution of extended Stokes’ problems

Giorgio Riccardi �

Department of Aerospace and Mechanical Engineering, Second University of Naples, via Roma, 29-81031 Aversa (CE), Italy

a r t i c l e i n f o

Article history:

Received 10 August 2010

Received in revised form

19 March 2011

Accepted 5 April 2011Available online 1 May 2011

Keywords:

Stokes’ flow

First and second Stokes’ problems

Wall stress

Newtonian fluid

Analytical solution

62/$ - see front matter & 2011 Elsevier Ltd. A

016/j.ijnonlinmec.2011.04.010

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a b s t r a c t

The analytical solutions of first and second Stokes’ problems are discussed, for infinite and finite-depth

flows of a Newtonian fluid in planar geometries. Problems arising from the motion of the wall as a

whole (one-dimensional flows) as well as of only one half of the wall (two-dimensional) are solved and

the wall stresses are evaluated.

The solutions are written in real form. In many cases, they improve the ones in literature, leading to

simpler mathematical forms of velocities and stresses. The numerical computation of the solutions is

performed by using recurrence relations and elementary integrals, in order to avoid the evaluation of

integrals of rapidly oscillating functions.

The main physical features of the solutions are also discussed. In particular, the steady-state

solutions of the second Stokes’ problems are analyzed by separating their ‘‘in phase’’ and ‘‘in quadrature’’

components, with respect to the wall motion. By using this approach, stagnation points have been

found in infinite-depth flows.

& 2011 Elsevier Ltd. All rights reserved.

1. Introduction

The analytical solution of Stokes problems for a Newtonian fluidin a planar geometry is here revised, by following the seminal paperof Liu [1]. A fluid region is bounded by a rigid wall, which moveswith a prescribed velocity having fixed direction, parallel to the wall.The fluid and the wall are at rest at the initial time. By following theliterature, wall velocities constant (first Strokes’ problem) or period-ical (second) in time will be assumed. Moreover, flows in which thewall moves as a whole (one-dimensional) and half wall moves,while the other one is kept fixed, (two-dimensional) will also beinvestigated. Finally, the depth of the fluid region will be assumedinfinite or finite. In these latter kinds of flow, a free surface isassumed to bound the fluid region.

The solution of the first Stokes’ problem in an infinite-depthflow has a well known analytical structure, related to thecomplementary (real) error function. Solutions of the secondproblem in an infinite-depth flow have been discussed in [2–4]and in many other papers. They are usually written in terms oferror functions of complex arguments, because in the correspond-ing real forms integrands containing oscillatory functions appear,the numerical integration of which can lead to severe errors [5].

Recently, these results have been reconsidered in the frameworkof two-dimensional flows. In the paper [6] the steady states havebeen found, while Liu [1] generalizes these solutions, by giving alsothe transient contributions. The effects of side walls on the Stokes

ll rights reserved.

flow on a planar wall have been recently investigated in [7]. Besidesthe first and second Stokes problems, the flows induced by a constantaccelerating plate and by a plate that applies a constant stress are alsoinvestigated. This important paper opens the way to the comparisonwith experiments, where effects of side walls are rarely negligible.

Despite the subject is a quite old one [8], many issues aboutanalytical solutions and their numerical computation appear to beimproved, in particular for two-dimensional flows. The presentpaper is an attempt to fill some of these lacks. It is organized asfollows. In Section 2, the solutions of one-dimensional first andsecond problems are briefly discussed, then they are extended tothe finite-depth case in Section 3. The solution of two-dimensionalproblems is then faced, for infinite (Section 4) and finite-depth(Section 5) flows. Finally, conclusions are offered in Section 6.

2. One dimensional infinite-depth flows

A Newtonian fluid having kinematical viscosity n fills the halfspace y40, bounded by a solid wall at y¼0. Initially (tr0), fluid andwall are at rest. The wall starts to move at time t¼0þ with a givenvelocity (say q), directed along the axis x. The resulting fluid velocity(u) is assumed to be directed along x and to depend on y and t, only.As well known, this flow is described by Stokes’ problem:

@tu¼ n@2yyu,

uð0,tÞ ¼ qðtÞ, uðþ1,tÞ � 0,

uðy,0Þ � 0,

8><>:the solution of which is easily found in terms of Laplace transformin time (qðLÞ and uðLÞ are the transformed functions of q and u,

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G. Riccardi / International Journal of Non-Linear Mechanics 46 (2011) 958–970 959

respectively):

uðLÞðy,sÞ ¼ expð�byÞqðLÞðsÞ: ð1Þ

Here, the complex variable s has a positive real part and b¼ffiffiffiffiffiffiffis=n

p(the principal branch of the root is used). In the following, twodifferent wall velocities will be considered: constant, i.e. qðtÞ � 0 asto0 and qðtÞ � u0 as t40, which leads to the first Stokes’ problemand periodical, i.e. qðtÞ ¼ u0cosðotþyÞ as t40, corresponding to thesecond Stokes’ problem.

2.1. First Stokes’ problem

The solution of this classical problem is here summarized, forlater convenience. The Laplace transform of the wall velocity is

qðLÞðsÞ ¼u0

s, ð2Þ

so that the time derivative of the non-dimensional velocityU1 ¼ u1=u0 (non-dimensional quantities will be indicated bycapital symbols, while the subscript 1 refers to the first solutionof the present paper) is obtained through a Laplace antitransformof the general solution (1):

@tU1 ¼1

2pi

Z mþ i1

m�i1ds expðts�ybÞ ¼: F1, ð3Þ

m being a suitable positive real number. The function F1 iscalculated by applying Cauchy’s theorem to the integral ofexpðts�ybÞ=ð2piÞ on the path of Fig. 1a and then by performingthe limit as M-þ1. The two resulting integrals are evaluatedalong the lower and upper paths of Fig. 1b: it is found that theirsum gives

ffiffiffiffipp

. As a consequence, F1 assumes the following form:

F1ðy,tÞ ¼1

2ffiffiffiffiffiffipnp yt�3=2exp �

y2

4nt

� �: ð4Þ

Once it is inserted in Eq. (3), an integration in time leads to theclassical solution:

U1ðY ,TÞ ¼2ffiffiffiffipp

Z þ1Y=ð2

ffiffiTpÞ

dZ e�Z2

¼ erfcY

2ffiffiffiTp

� �, ð5Þ

in which lengths and times are non-dimensionalized with n=u0

and n=u20, respectively. It can be observed that the velocity (5)

depends on Y and T through the time-rescaled variableY 0 ¼ Y=ð2

ffiffiffiTpÞ: written in terms of a function of Y 0, the above

velocity will be indicated hereafter by U01ðY0Þ. The wall stress w1

follows in non-dimensional form as W1 ¼w1=ðru20Þ, r being the

+iM

−iM

s = (−x) e+i�

s = (−x) e−i�

�

Fig. 1. Integration paths in the plane of s: for the evalu

fluid density. By using the solution (5), one obtains:

W1ðTÞ ¼ �1=ffiffiffiffiffiffipTp

: ð6Þ

2.2. Second Stokes problem

In the second Stokes’ problem, the Laplace transform of thewall velocity is

qðLÞðsÞ ¼u0

2

e�iy

sþ io þeþ iy

s�io

� �, ð7Þ

so that the general solution (1) is specified in the following one:

U2 ¼1

2

8>><>>:e�iðotþyÞ 1

2pi

Z mþ i1

m�i1ds

exp½tðsþ ioÞ�yb�sþ io|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

H þ2

þeþ iðotþyÞ 1

2pi

Z mþ i1

m�i1ds

exp½tðs�ioÞ�yb�s�io|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

H�2

9>>=>>;: ð8Þ

The time derivatives of the functions H27 are easily evaluated

in terms of F1, indeed: @tH72 ¼ expð7 iotÞF1. Once the proper

form of the function F1 (4) is inserted into the above relations andthey are integrated in time, one obtains:

H72 ðy,tÞ ¼H7

2 ðy,0Þþy

2ffiffiffiffiffiffipnp

Z t

0dtt�3=2exp 7 iot� y2

4nt

� �: ð9Þ

Notice that, in order to have u2ðy,0Þ � 0 for any initial phase y,H7

2 ðy,0Þmust vanish, as it can be also proved by integrating alongthe path of Fig. 1a their definitions (8) evaluated in t¼0. Thefunctions H2

7 (9) with H72 ðy,0Þ � 0 are then inserted into the

formula (8) and the non-dimensional quantities T ¼ot andY ¼ yðo=nÞ1=2 are used, according to [4]. In this way, the solution:

U2ðY ,TÞ ¼2ffiffiffiffipp

Z þ1Y=ð2

ffiffiTpÞ

dZ e�Z2

cos Tþy�Y2

4Z2

� �ð10Þ

follows. This solution is the real form of the one in [3] for y¼ 0and p=2 and of the solution in [4].

The numerical evaluation of the solution (10) is not a trivialtask, due to the presence of 1=Z2 in the argument of thetrigonometric function. Numerical integration schemes lead to

+iH

−iH

ation of the function F (a) and of the integrals I1,2.

0

10

20

30

40

50

0 2 4 6 8 10 12Y

T

0

10

20

30

40

50

0 2 4 6 8 10 12Y

Fig. 2. In (a), the level lines (from �3 to 0, step 0.1, black) of log10 m2 are superimposed to the line (green) j2 ¼ 03 , while in (b) the level lines of j2 are drawn (step 451,

red: negative, green: 01, blue: positive). Notice the presence of stagnation points: in (a) they appear as sinks, while in (b) level lines converge on each of them. (For

interpretation of the references to colour in this figure legend, the reader is referred to the web version of this article.)

G. Riccardi / International Journal of Non-Linear Mechanics 46 (2011) 958–970960

a quite poor accuracy, unless huge computational efforts aremade. Here, an analytical procedure is proposed, which enablesus to easily evaluate the velocity (10). First of all, in a finiteprecision calculation of the integral:

Ic,s2 ðxÞ :¼

2ffiffiffiffipp

Z þ1x

dx e�x2 cos

sin

ax2

(a being a positive constant) the integration range can beextended up to a rather small value M of x (e.g., M� 6), due tothe presence of the factor expð�x2

Þ in the integrand function.Then, the interval (x,M) is divided in N sub-intervals ðxn�1,xnÞ

(n¼1,2,y,N), with x0¼x and xN¼M. By using the above decom-position, Ic,s

2 are rewritten as:

Ic,s2 ¼

XN

n ¼ 1

e�x2n�1

X1k ¼ 0

Jc,s2 ,

in which the integrals:

Jc,s2 ðkÞ :¼

ð�1Þk

k!

2ffiffiffiffipp

Z xn

xn�1

dxðx2�x2

n�1Þk cos

sin

ax2

are evaluated through the recurrence formula:

Jc,s2 ðkÞ ¼

1

2kþ1

ð�1Þk

k!

2ffiffiffiffipp x3

nðx2n�x2

n�1Þk�1 cos

sin

ax2

n

(

þ1

k½ð4k�1Þx2

n�1Jc,s2 ðk�1Þ72aJs,c

2 ðk�1Þ�2x4n�1Jc,s

2 ðk�2Þ�

):

Finally, Jc,s2 ð0,1Þ are easily computed in terms of Fresnel’s integrals

C1 and S1 (see [9, p. 300], formulae (7.3.3) and (7.3.4), respec-tively. About their numerical evaluation, see also Appendix A).

In order to highlight the physical properties of the abovesolution, the velocity U2 (10) is rewritten as sum of a part, sayU2

c , in phase with respect to the wall motion and of another one,U2

s , which is in quadrature: U2 ¼Uc2cosðTþyÞþUs

2sinðTþyÞ. More-over, U2

c and U2s are assumed as real and imaginary parts of the

following complex number: Uc2þ iUs

2 ¼m2expð�ij2Þ (the branch�prj2oþp is chosen). Hence, the velocity (10) is also given bythe formula:

U2 ¼m2cosðTþyþj2Þ, ð11Þ

in which m2 and j2 depend on Y and T, but not on y. The use ofthe representation (11) enables us to investigate the solution (10)in terms of its amplitude (m2) and phase relative to the wallmotion (j2), without regard to the initial phase y.

Level lines of log10 m2 and of j2 in the plane (Y,T) are drawn inFig. 2a and b, respectively. At a fixed T, the modulus m2 increasesas Y decreases: it reaches its maximum (1) as Y-0þ . However,this behaviour appears to be not monotonic, due also to thepresence of several points on which m2 vanishes: they appear assinks in the logarithmic scale used in Fig. 2a. At a fixed Y, m2

initially grows with time, afterwards it behaves in an almostperiodic way, unless for a discrete set of abscissae Y, whichcorrespond to the abovementioned stagnation points. Thepresence of such points is also confirmed by an inspection tothe level lines of the phase j2 (Fig. 2b) that converge on each ofthem. Fig. 2b shows that, at Y fixed, two phase behaviours arepossible: if the point is quite close to the wall (Yo4), the phasedelay reaches an asymptotic value for increasing times, whilefurthermost points accumulate delays monotonically growingin time.

The wall stress is calculated by noticing that the functions H27 (9)

can be rewritten in terms of the complementary error function as:

H72 ¼ e7 ioterfc

y

2ffiffiffiffiffintp

� �8 io

Z t

0dt e7 iot erfc

y

2ffiffiffiffiffintp

� �:

By deriving in y the above functions and evaluating the resultingderivatives at the wall, the non-dimensional wall stress W2 ¼

w2=ðru0

ffiffiffiffiffiffiffinopÞ follows:

W2ðTÞ ¼W1ðTÞcosyþffiffiffi2p½C1ð

ffiffiffiTpÞsinðTþyÞ�S1ð

ffiffiffiTpÞcosðTþyÞ�: ð12Þ

The stress (12) is the real form of the one obtained in [4] (seeEq. (13) of that paper). The first term in the right hand side ofEq. (12) is due to the initial non-vanishing value (cosy) of the wallvelocity and it is the only one which is singular as T-0þ . The otherterms can be rearranged in the form: m2wcosðTþyþj2wÞ, m2w andj2w being drawn vs. time in Fig. 3a and b, respectively. Notice theinitial overshoot of the modulus m2w (about 40% of the asymptoticvalue 1) and the asymptotic behaviour of the relative phase j2w,which goes to �3p=2 in an oscillatory way as T-1.

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1Y

T

0

0.2

0.4

0.6

0.8

1

0 0.5 1 1.5 2 2.5 3 3.5 4T

U3

-4

-3

-2

-1

0

0 0.5 1 1.5 2 2.5 3 3.5 4T

W3

Fig. 4. Level lines of the velocity (17) in the (Y,T) plane (a, from the top to the bottom: 0.95, 0.90, y, 0.05), free surface velocity (b) and wall stress (c) vs. time.

0.2

0.4

0.6

0.8

1

1.2

1.4

0 10 20 30 40 50 60 70 80 90 100T

m2w

-165

-150

-135

-120

-105

-90

0 10 20 30 40 50 60 70 80 90 100T

�2w

Fig. 3. Modulus m2w (a) and phase j2w (degrees) of the wall stress vs. the time T for vanishing initial wall velocity. Asymptotic values are drawn with dashed lines.


3. One dimensional finite-depth flows

As discussed in [1], for a finite-depth flow the Stokes’ problemis posed in the following way:

@tu¼ n@2yyu,

uð0,tÞ ¼ qðtÞ, @yuðh,tÞ � 0,

uðy,0Þ � 0,

8><>:h being the height of the fluid. The second boundary conditionenforces vanishing viscous stresses at the interface. Moreover,free surface motion is not considered.

The Laplace transform of the solution is

uðLÞðy,sÞ ¼cosh½ðh�yÞb�

coshðhbÞqðLÞðsÞ: ð13Þ

Notice that the kernel cosh½ðh�yÞb�=coshðhbÞ in Eq. (13) is an evenfunction of b, having a countable set of real and negative poles ofthe first order. They are placed on the points sð1Þk ¼�K2n=h2, K

being ðkþ1=2Þp for any non-negative integer k.


By inserting the Laplace transform of the wall velocity (2) intothe general form of the Laplace transform of the solution (13), thetime derivative of the velocity is written as:

@tu3 ¼u0

2pi

Z mþ i1

m�i1ds est cosh½ðh�yÞb�

coshðhbÞ¼: u0F3 ð14Þ

and the integral in Eq. (14) is evaluated by applying the residuetheorem on the path in Fig. 1a:

F3 ¼ 2n

h2

X1k ¼ 0

K e�K2nt=h2

sinðKy=hÞ: ð15Þ

By following [1], the non-dimensional variables T ¼ nt=h2 andY¼y/h are introduced and by accounting for the result:

X1k ¼ 0

sinðKYÞ

K�

1

2, ð16Þ

which holds for 0oYr1, the solution of the problem becomes:

U3ðY ,TÞ ¼ 1�2X1k ¼ 0

sinðKYÞ

Kexpð�K2TÞ ¼ 2

X1k ¼ 0

KsinðKYÞ

Z T

0dT 0 e�K2T0:

ð17Þ

The velocity (17) has been obtained in [1]. Here it is also rewrittenin an integral form, for later convenience. Notice that the state atT¼0 is included in the above form of the velocity, despite thecontrary is stated at page 5 of the abovementioned paper. Indeed,Eq. (17) gives a velocity U3(Y,0)¼0 for any Y40, by using theFourier series (16). The level lines of the velocity (17) in the plane(Y,T) are drawn in Fig. 4a: the upper line corresponds to the valueU3¼0.95, while the lower one to U3¼0.05. It is shown that a largefluid region near the wall (small Y) accelerates quickly up tovelocities near the wall one (1), while the free surface motion isslower: it employs three units of time to reach an almost unitaryvelocity (see Fig. 4b).

Finally, the non-dimensional wall stress W3 ¼w=ðrnu0=hÞ ¼

@Y U3jY ¼ 0 follows from the velocity (17) as:

W3ðTÞ ¼ �2X1k ¼ 0

expð�K2TÞ: ð18Þ

Its behaviour vs. time is shown in Fig. 4c. It is singular as T-0þ

and becomes negligibly small just after three units of time.


3.2. Second Stokes’ problem

Also the second Stokes’ problem can be easily extended tofinite-depth flows. Indeed, by inserting the Laplace transform ofthe wall velocity (7) inside the general form of the Laplacetransform of the solution (13), the velocity follows through aLaplace antitransform:

u4 ¼u0

2


2pi

Z mþ i1

m�i1ds

eðsþ ioÞt

sþ iocosh½ðh�yÞb�

coshðhbÞ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}H þ

4

þeþ iðotþyÞ 1

2pi

Z mþ i1

m�i1ds

eðs�ioÞt

s�iocosh½ðh�yÞb�

coshðhbÞ|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}H�

4

9>>=>>;: ð19Þ

Introduced the non-dimensional frequency root O :¼ffiffiffiffiffiop

h=ffiffiffinp

(O0 :¼ O=ffiffiffi2p

will be also used), the initial values of the functionsH4

7 follow by applying the residue theorem to the integral on thepath of Fig. 1a:

H74 ðy,0Þ ¼�2

X1k ¼ 0

KsinðKy=hÞ

K28 iO2þ

cosh½O0ð1�y=hÞð18 iÞ�

cosh½O0ð18 iÞ�:

Once the initial values are known, the functions H74 ðy,tÞ are

calculated by observing that their time derivatives are related tothe function F3 (14) through the formula: @tH

74 ¼ e7 iotF3, in a

complete analogy with the second case of Section 2. The use of the

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

Y

m4

Fig. 5. Profiles of the modulus (a) and of the phase (b, degrees) relative to the wall of the

from 1 to 20 with an unitary step. O grows from the right to the left in both figures.

0

0.2

0.4

0.6

0.8

1

1.2

0 1 2 3 4 5

Ω

m4w

Fig. 6. Modulus m4w (a) and relative phase j4w (degrees) of the steady-state wall stress

form (15) of F3 enables us to calculate H47 , for any t. As a

consequence, in terms of the non-dimensional variables Y¼y/hand T ¼ot, the non-dimensional velocity can be written as:

U4ðY ,TÞ ¼ cðYÞcosðTþyÞþsðYÞsinðTþyÞ

�2X1k ¼ 0

KðK2cosyþO2sinyÞK4þO4

e�K2T=O2

sinðKYÞ

¼2

O2

X1k ¼ 0

KsinðKYÞ

Z T

0dT 0e�K2T 0=O2

cosðTþy�T 0Þ, ð20Þ

in which the two functions c and s are given by the followingformulae:

cðYÞ ¼cosðO0YÞcosh½O0ð2�YÞ�þcoshðO0YÞcos½O0ð2�YÞ�

coshð2O0Þþcosð2O0Þ,

sðYÞ ¼sinðO0YÞsinh½O0ð2�YÞ�þsinhðO0YÞsin½O0ð2�YÞ�

coshð2O0Þþcosð2O0Þ:

The form (20) of the solution appears quite satisfactory, thesteady-state part being separated by the transient one. Thewriting of the steady-state part in terms of a Fourier series doesnot have practical interest, due to its slow convergence velocity,but it enables us to recover the form of the solution given in [1].

As discussed in Section 2, the steady-state velocity is writtenas the sum of a component in phase with the wall motion andanother one which is in quadrature, so that U4 ¼m4cosðTþyþj4Þ. The behaviours of the modulus m4 and of the phase j4 vs. Y

are shown in Fig. 5, for different values of the non-dimensionalfrequency O. As expected, the modulus is a decreasing function of

0

0.2

0.4

0.6

0.8

1

-1080 -900 -720 -540 -360 -180 0

�4

stationary part of the velocity (20). The curves are drawn for different values of O,

-140

-130

-120

-110

-100

-90

0 1 2 3 4 5

Ω

�4w

(21) vs. the frequency root O0 . Asymptotic values are also drawn with dashed lines.


the frequency, as the relative phase j4: fluid particles move lessand less for increasing frequency. Moreover, they accumulatelarger and larger delays.

The non-dimensional wall stress W4 ¼w4=ðru0

ffiffiffiffiffiffiffinopÞ¼

@Y U4jY ¼ 0=O follows from the solution (20) as:

W4ðTÞ ¼W3ðT=O

2Þ

Ocosy�2O

X1k ¼ 0

K2siny�O2cosyK4þO4

e�K2T=O2

�sinhð2O0Þ�sinð2O0Þffiffiffi

2p½coshð2O0Þþcosð2O0Þ�

cosðTþyÞ

þsinhð2O0Þþsinð2O0Þffiffiffi

2p½coshð2O0Þþcosð2O0Þ�

sinðTþyÞ:

The first term is due to the non-vanishing initial velocity of thewall (which is just cosy in non-dimensional form), while thesecond one gives the unsteady contribution. The third and fourthterms lead to the steady-state stress, which can be rewritten inthe form m4wcosðTþyþj4wÞ with:

m4w ¼coshð2O0Þ�cosð2O0Þcoshð2O0Þþcosð2O0Þ

� �1=2

,

j4w ¼�pþarctgsinhð2O0Þþsinð2O0Þsinhð2O0Þ�sinð2O0Þ

� �: ð21Þ

The quantities m4w and j4w are drawn in Fig. 6 vs. O0. Asexpected, the modulus vanishes as O-0 and there is a criticalfrequency such that m4w reaches its maximum. Moreover, itsasymptotic value is unitary. Notice also that the relative phaseholds �p=2 as O-0 and reaches its asymptotic value (�3p=4) asO-þ1.

4. Two dimensional infinite-depth flows

A two-dimensional Stokes flow is here considered: half wall(z40) moves, while half is kept fixed. As a consequence, the fluidvelocity u depends on the two spatial variables y and z. It isassumed [1] as the sum of the two velocities uod(y,t) and utd(y,z,t).They satisfy the following one and two-dimensional problems:

problem for uod problem for utd

@tuod ¼ n@2yyuod,

uodð0,tÞ ¼ qðtÞ=2,

uodðþ1,tÞ � 0,

uodðy,0Þ � 0,

8>>>><>>>>:

@tutd ¼ nð@2yyutdþ@

2zzutdÞ,

utdð0,z,tÞ ¼ signðzÞqðtÞ=2,

utdðþ1,z,tÞ � 0,@yutdðþ1,z,tÞ � 0,

utdðy,71,tÞ finite,

utdðy,z,0Þ � 0:

8>>>>>><>>>>>>:

ð22Þ

The problem for uod has been solved above (with q in place of q/2),while the one for utd will be solved below.

First of all, the solution utd is an odd function of z, so that theproblem can be only posed for z40, by accounting for thatutdðy,0,tÞ � 0. As before, the Laplace transform in time of theequation of motion leads to the equation: suðLÞtd ¼ nð@

2yyuðLÞtd þ@

2zzuðLÞtd Þ.

Furthermore, in order to eliminate the derivative in y, a Fouriersine transform in y (F s, indicated with the apex ðFÞ) is also applied,once the following transform:

F s½@2yyuðLÞtd �ðZ,z,sÞ ¼ Z qðLÞðsÞ

2�Z2uðFLÞ

td ðZ,z,sÞ,

has been evaluated. Here, uðFLÞtd is the Fourier sine transform of uðLÞtd .

It follows the differential problem:

@2zzuðFLÞ

td �ðZ2þb2

ÞuðFLÞtd ¼�ZqðLÞ=2,

uðFLÞtd ðZ,0,sÞ � 0,

uðFLÞtd ðZ,þ1,sÞ finite,

8>>><>>>:

which leads to the general form of the solution:

uðFLÞtd ðZ,z,sÞ ¼

Z 1�expð�zffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib2þZ2

qÞ

� �b2þZ2

qðLÞðsÞ

2: ð23Þ

The solution (23) will be particularized below for the twoaforementioned wall velocities (2) and (7).


The first Stokes’ problem has Laplace transform in time of thewall velocity given by Eq. (2), so that from the general form of thesolution (23) the Fourier sine transform uðFÞtd ðZ,z,tÞ follows:

uðFÞtd ¼u0nZ

2

1

2pi

Z mþ i1

m�i1ds

1�expð�zffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib2þZ2

qÞ

sðsþnZ2Þest ¼:

u0nZ2

F5, ð24Þ

m being an arbitrary real and positive number. By integrating onthe path of Fig. 1a, the time derivative of the new function F5 iswritten as:

@tF5 ¼ e�nZ2t erf

z

2ffiffiffiffiffintp

� �, ð25Þ

to be integrated by starting from the initial value F5ðZ,z,0Þ � 0. F5

is then inserted into the form (24) of the velocity uðFÞtd and theinverse Fourier sine transform is applied:

utd ¼u0

4ffiffiffiffiffiffipnp y

Z t

0dtt�3=2 erf

z

2ffiffiffiffiffintp

� �exp �

y2

4nt

� �:

This velocity is non-dimensionalized as in Eq. (5), by using aslength and time scales n=u0 and n=u2

0, respectively. The solutionfollows by adding half velocity (5):

U5ðY ,Z,TÞ ¼2ffiffiffiffipp

Z þ1Y=ð2

ffiffiTpÞ

dZ e�Z2F

Z

YZ

� �, ð26Þ

FðxÞ being ½1þerfðxÞ�=2. It corresponds to the solution in [1],unless a different choice of the length scale (here it has beenpreferred to keep an explicit dependence on time). Notice alsothat U5 (26) goes to U1 for Z-þ1, while it vanishes for Z-�1.Its numerical evaluation is not a trivial task, it will brieflydescribed below.

In the solution (26), the integral:

I5ðxjaÞ :¼2ffiffiffiffipp

Z þ1x

dx erfðaxÞe�x2

,

has to be calculated, a and x being real numbers. Notice thatI5ðxj71Þ ¼ 7 ½1�erf2

ðxÞ�=2, while if jaja1, the modulus of a canbe always assumed smaller than 1. Indeed, if jaj41 an integration byparts gives the relation: I5ðxjaÞ ¼ signðaÞ�erfðxÞ erfðaxÞ�Iðaxj1=aÞ.Assumed jajo1, the error function in the integral I5 is written in apower series and the formula:

I5 ¼4

pX1k ¼ 0

ð�1ÞkJ5ðkÞ

2kþ1

is found, the integrals J5(k) for kZ1 being evaluated through therecurrence relation:

J5ðkÞ :¼a2kþ1

k!

Z þ1x

dxx2kþ1e�x2

¼a2

ðaxÞ2k

k!e�x2

þa2J5ðk�1Þ,

with J5ð0Þ ¼ ae�x2=2.

As in the case of the velocity U1 (5), the solution (26) can berewritten in a self-similar form: once the time-rescaled variablesY 0 :¼ Y=ð2

ffiffiffiTpÞ and Z0 :¼ Z=ð2

ffiffiffiTpÞ have been introduced, the above

velocity is given by the new function U05 ¼U05ðY0,Z0Þ. In Fig. 7a the

level lines of U05 in the plane ðY 0,Z0Þ are drawn. As expected, U05vanishes as Z-�1, while it reaches quickly the value U01ðY

0Þ,

0.001

0.01

0.1

1

10

-1 -0.5 0 0.5 1 1.5 2

Z

Y

-10

-5

0

5

10

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

Z

ψ

Fig. 7. In (a) the level lines of the velocity (26) (from 0.05 on the upper line to 0.95 on the lower one, with step 0.05) are drawn in the plane of the time-rescaled variables

Y 0 ¼ Y=ð2ffiffiffiTpÞ (log scale) and Z0 ¼ Z=ð2

ffiffiffiTpÞ. In (b), C (27) is drawn vs. Z0 (solid line), together with its asymptotes (dashed): 1 as Z0-þ1 and 0 as Z0-�1.


independent of Z0, as Z-þ1. Moreover, a more and more abruptchange from 0 (Zo0) to 1 (Z40) is found as Y-0.

In order to calculate the wall stress, the velocity is written indimensional variables as:

u5 ¼u0y


Z t

0dtt�3=2 1þerf

z

2ffiffiffiffiffintp

� �� exp �

y2

4nt

� �:

It is derived with respect to y and the change of variable from t toz¼ 1=ð2

ffiffiffiffiffintpÞ into the integrals is performed. If zo0, the limit as

y-0þ is directly evaluated by rewriting 1þerfðzzÞ as erfcðjzjzÞ.The non-dimensional stress W5 ¼w5=ðru2

0Þ:

W5ðZ,TÞ ¼W1ðTÞ FðZ0Þþexpð�Z02Þ

2ffiffiffiffipp

Z0

� �¼: W1ðTÞCðZ0Þ ð27Þ

follows. On the contrary, the limit as y-0þ cannot be performedon the above form of the derivative when z is positive. In thiscase the identity: 1þerfðzzÞ � �erfcðzzÞþ2 is used and the aboveform (27) of the stress is recovered. The function C is drawn vs. Z0

in Fig. 7b: as expected, the resulting W5 is negative for Z40 andpositive for Zo0. Moreover, it diverges for vanishing Z0. Notice alsothat the stress on the moving half plate (Z40) is everywhere largerthan the corresponding stress W1 of the one-dimensional case.


The Fourier sine transform of the solution for the secondproblem follows by inserting qðLÞ (7) into the general form of thesolution (23) and by Laplace antitransforming:

uðFÞtd ¼u0nZ

4


2pi

Z mþ i1

m�i1ds


qÞ

sþnZ2

eðsþ ioÞt

sþ io|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}Hþ

6

þeþ iðotþyÞ 1

2pi

Z mþ i1

m�i1ds


qÞ

sþnZ2

eðs�ioÞt

s�io|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}H�

6

9>>=>>;: ð28Þ

It can be shown through integrations on paths like the one in Fig. 1athat H7

6 ðZ,z,0Þ ¼ 0. Hence, equations @tH76 ¼ expð7 iotÞ@tF5 and

(25) enable us to evaluate the functions H76 :

H76 ðZ,z,tÞ ¼

Z t

0dt eð�nZ

2 7 ioÞt erfz

2ffiffiffiffiffintp

� �:

They are used in uðFÞtd (28) and the Fourier sine antitransform isapplied:

utd ¼u0np

Z t

0dt erf

z

2ffiffiffiffiffintp

� �cos½oðt�tÞþy�

Z þ10

dZZ sinðZyÞe�ntZ2

,

in which the internal integral holds:ffiffiffiffipp

y exp½�y2=ð4ntÞ�=½4ðntÞ3=2�.

Finally, half velocity (10) is added to utd/u0, in order toobtain the non-dimensional solution of the present Stokes’problem:

U6ðY ,Z,TÞ ¼2ffiffiffiffipp

Z þ1Y=ð2

ffiffiTpÞ

dZ e�Z2

cos Tþy�Y2

4Z2

� �F

Z

YZ

� �, ð29Þ

in terms of the non-dimensional variables T ¼ot, Y ¼ yffiffiffiffiffiffiffiffiffio=n

pand Z ¼ z

ffiffiffiffiffiffiffiffiffio=n

p. As expected, the solution (29) goes to U2 as

Z-þ1 and vanishes as Z-�1. The important issue of itsnumerically computing will be now discussed.

In the solution (29), the calculation of the following integrals:

Ic,s6 ðxja,bÞ :¼

2ffiffiffiffipp

Z þ1x

dx e�x2

erfðbxÞcos

sin

ax2

with a40 and any real b is needed. As before, due to the presenceof the factor expð�x2

Þ the integration range can be assumed finite,i.e. from x to M (M¼6 is used in the present calculations). Theintegration range is then decomposed in N intervals ðxn�1,xnÞ

(n¼1,2,y,N) with x0¼x and xN¼M. Hence, the following approx-imation is considered:

Ic,s6 C

XN

n ¼ 1

e�x2n�1 erfðbxn�1ÞJ

c,s6 ð0Þþ

X1k ¼ 1

qkðxn�1ÞJc,s6 ðkÞ

" #,

qk(x) being the k-th derivative of the function expð�x2Þ erfðbxÞ

divided by k! and calculated in x¼ x. Moreover, the integrals

Jc,s6 ðkÞ ðkZ3Þ are evaluated through the following recurrence

relation:

Jc,s6 ðkÞ :¼

2ffiffiffiffipp

Z xn

xn�1

dxðx�xn�1Þk cos

sin

ax2

¼2ffiffiffiffipp x3

n

ðxn�xn�1Þk�2

kþ1

cos

sin

ax2

n

�3k

kþ1xn�1Jc,s

6 ðk�1Þ

�3ðk�1Þ

kþ1x2

n�1Jc,s6 ðk�2Þ�

k�2

kþ1x3

n�1Jc,s6 ðk�3Þ

þ2a

kþ1J�s,c6 ðk�2Þ,

while the first three integrals, say Jc,s6 ð0,1,2Þ, are evaluated in

terms of sine and cosine integral functions.Level lines for the modulus m6 and the relative phase j6 in the

(Y,T) plane are shown in Fig. 8 for three values of Z: �4, 0 and þ4.Relevant differences appear between the fields at Z¼0 and �4,while the results at Z¼0 and þ4 are quite similar, unless theamplitude of m6 which has maximum 1/2 at Z¼0 and 1 at Z¼þ4,in both cases at the wall (Y¼0). On the contrary, the field at

0

10

20

30

40

50

0 2 4 6 8 10 12

Y

T

0

10

20

30

40

50

0 2 4 6 8 10 12Y

0

10

20

30

40

50

0 2 4 6 8 10 12Y

0

10

20

30

40

50

0 2 4 6 8 10 12Y

T

0

10

20

30

40

50

0 2 4 6 8 10 12Y

0

10

20

30

40

50

0 2 4 6 8 10 12Y

Fig. 8. Level lines of the decimal logarithm of the modulus m6 (first row) and of the phase j6 (second) are drawn in the (Y,T)-plane. The coordinate Z holds �4 (a), 0

(b) and þ4 (c). The levels of log10 m6 are chosen from �4 to 0 with step 0.1, while the levels of j6 are �1801, �1351,y,þ1801. The corresponding lines are drawn with

three colours: blue for positive levels, green for the zero one and red for negative j6. (For interpretation of the references to colour in this figure legend, the reader is

referred to the web version of this article.)

0

10

20

30

40

50

-4 -2 0 2 4Z

T

0

10

20

30

40

50

-4 -2 0 2 4Z

Fig. 9. For vanishing initial wall velocity (cosy¼ 0) in (a) the level lines (from �4 to 1.4, step 0.2, black) of log10 m6w are superimposed to the line (green) j6w ¼ 03 , while in

(b) the level lines of j6w are drawn (step 201, red: negative, green: 01, blue: positive). (For interpretation of the references to colour in this figure legend, the reader is

referred to the web version of this article.)


Z¼�4 has maximum about 10�2 well inside the field, due to theboundary condition of vanishing velocity enforced at the wall.Notice also that the stagnation points are in different positions,nearer to the wall, with respect to the corresponding ones forZZ0.

The non-dimensional stress W6 ¼w6=ðru0

ffiffiffiffiffiffiffinopÞ reduces

to @Y U6jY ¼ 0 and is evaluated through repeated integrations by

parts as:

W6ðZ,TÞ ¼W2ðTÞCðZ0Þþ1

Z

ffiffiffiffi2

p

rcosðTþyÞ

Z ffiffiTp

0dxS1ðxÞe�Z2=ð4x2

Þ

"

�sinðTþyÞZ ffiffi

Tp

0dxC1ðxÞe�Z2=ð4x2

Þ

#: ð30Þ


In the numerical computation of the above stress, the mainproblem lies in evaluating the integrals:

Pc,s6 ðxÞ :¼

Z x

x0

dxC1

S1ðxÞexp �

ax2

!,

a being a positive constant. The lower bound of integration x0 isnot vanishing: in the case of the stress (30) it can be set to jZj=S,where S¼12 in double precision calculations. Then the interval(x0,x) is divided in N sub-intervals ðxn�1,xnÞ (n¼1,2,y,N) of equalamplitudes and the Fresnel functions are expanded in Taylorseries around the point xn�1 (see Appendix A). As a consequence,the above integrals are approximated as:

Pc,s6 C

XN

n ¼ 1

X1k ¼ 0

qc,sk ðxn�1ÞQ6ðkÞ,

in which qc,sk is the k-th coefficient of the series for C1 or S1 and the

integral Q6(k) is calculated by the following recurrence formula(Z¼

ffiffiffiap

=x and kZ3):

Q6ðkÞ :¼

Z xn

xn�1

dxðx�xn�1Þkexp �

ax2

!

¼x3

nðxn�xn�1Þk�2

kþ1e�Z

2n�

3k

kþ1xn�1Q6ðk�1Þ

�3ðk�1Þx2

n�1þ2akþ1

Q6ðk�2Þ�k�2

kþ1x3

n�1Q6ðk�3Þ:

Finally, the first three integrals Q6(0,1,2) can be easily evaluatedin terms of exponential integral and error functions.

In Fig. 9 the level lines of the modulus m6w and of the relativephase j6w are drawn in the (Z,T)-plane in the case of vanishinginitial wall velocity (cosy¼ 0). Notice that W6 is not continuousacross the line Z¼0: the modulus diverges on that line, while thephase change sign. For Z40, the stress reaches quickly theasymptotic value W2 (independent of Z), while it vanishes forZ-�1. In time (at Z fixed) the stress has a nearly periodicbehaviour, for Z positive as well as for Z negative, but the phasedelay is bounded for Z40, while it grows monotonically for Zo0.

5. Two dimensional finite-depth flows

A two-dimensional Stokes flow (half plane y¼0 moves, whilehalf is kept fixed) in a finite-depth fluid is here considered. Asbefore, the solution u is taken as the sum of two velocities:uodðy,tÞ, which solves a one-dimensional problem, and utdðy,z,tÞ,that is solution of a two-dimensional one. These problems arestated below:

problem for uod problem for utd

@tuod ¼ n@2yyuod,

uodð0,tÞ ¼ qðtÞ=2,

@yuodðh,tÞ � 0,

uodðy,0Þ � 0,

8>>>><>>>>:

@tutd ¼ nð@2yyutdþ@

2zzutdÞ,

utdð0,z,tÞ ¼ signðzÞqðtÞ=2,

@yutdðh,z,tÞ � 0,

utdðy,71,tÞ finite,

utdðy,z,0Þ � 0:

8>>>>>><>>>>>>:

ð31Þ

The problem for uod has been solved above, while the one for utd

will be solved below.The velocity utd is an odd function of z, so that only the

problem with z40 has to be solved, by accounting for the newboundary condition utdðy,0,tÞ � 0. Besides the Laplace transformin time, the Fourier sine transform along z will be used: for thisreason it will be preferred to work in the difference uod�utd ¼: v,rather than in utd. Indeed, v vanishes as z-þ1, while the same isnot true for utd. To this regard, the Fourier transform of @2

zzv:

F s½@2zzv� ¼ zuod�z

2vðFÞ,

will be employed. By accounting for the above result, as well asthe problem (31) for uod, it is found that vðFLÞ satisfies the problem:

@2yyvðFLÞ�ðb2

þz2ÞvðFLÞ ¼ �zuðLÞod ,

vðFLÞð0,z,sÞ � 0,

@yvðFLÞðh,z,sÞ � 0:

8>><>>:By introducing the new function gðs,zÞ :¼ ½b2

ðsÞþz2�1=2, the solu-

tion of the above problem is written as:

vðFLÞðy,z,sÞ ¼zg

sinhðygÞcoshðhgÞ

Z h

0dZcosh½ðh�ZÞg�uðLÞod ðZ,sÞ

(

�

Z y

0dZsinh½ðy�ZÞg�uðLÞod ðZ,sÞ

�:

On the other hand, the velocity uðLÞod is given by Eq. (13) with qðLÞ=2in place of qðLÞ, so that the above solution becomes:

vðFLÞðy,z,sÞ ¼1

zcosh½ðh�yÞb�

coshðhbÞ�

cosh½ðh�yÞg�coshðhgÞ

�qðLÞðsÞ

2: ð32Þ


The Laplace transform in time of the wall velocity assumes inthis case the form (2). From the general form (32) of the solutionit follows:

vðFLÞ

u0ðy,z,sÞ ¼

1


coshðhbÞ�


�1

2s, ð33Þ

the right hand side of which possesses in the s-plane thefollowing singularities: (I) a branch cut along the interval

ð�1,�nz2Þ, across which g jumps from (assume x as a point of

such an interval) g¼ þg0 ¼ iffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�x=n�z2

q, value that is reached from

the above (s¼ xþ iy and y-0þ ), to g¼�g0 from the below; (II) abranch cut along the negative real semi-axis, across which the

function b jumps from b¼ þb0 ¼ iffiffiffiffiffiffiffiffiffiffiffi�x=n

pfrom the above, to �b0

from the below; (III) a simple pole in s¼0; (IV) a countable set of

simple poles, which are the zeros of coshðhbÞ : sð1Þk ¼�K2n=h2;

(V) a countable set of simple poles, which are the zeros of

coshðhgÞ : sð2Þk ¼�K2n=h2�nz2.

The Laplace antitransform in s is performed by integrating onthe path in Fig. 1a (in the limit for M going to infinity), while theevaluation of the residues and of their Fourier sine antitransformslead to the velocity utd:

utd

u0ðY ,Z,TÞ ¼

X1k ¼ 0

KsinðKYÞ

Z T

0dT 0 e�K2T0 erf

Z

2ffiffiffiffiffiT 0p

� �,

written in terms of the non-dimensional quantities T ¼ nt=h2,Y¼y/h and Z¼z/h. Notice that utd/u0 is odd in Z and it goes to U3/2for Z-þ1. The complete non-dimensional solution of theproblem is obtained as:

U7ðY ,Z,TÞ ¼1

2þX1k ¼ 0

sinðKYÞ

KK2

Z T

0dT 0 e�K2T0 erf

Z


� ��e�K2T

� �

¼ 2X1k ¼ 0

KsinðKYÞ

Z T

0dT 0 e�K2T0F

Z


� �: ð34Þ

The above solution possesses the required asymptotic propertiesin Z: it vanishes as Z-�1, while it goes to U3 (17) as Z-þ1.Furthermore, the integral can be evaluated by using the formula(7.4.33) at page 304 of [9]. It follows the solution:

U7ðY ,Z,TÞ ¼1þsignðZÞ

2þX1k ¼ 0

sinðKYÞ

KF7ðKjZ,TÞ, ð35Þ


where F7 is the following function of K, Z and T:

F7ðKjZ,TÞ ¼�e�K2T 1þerfZ

2ffiffiffiTp

� ��

�eþKZ

2signðZÞ�erf K

ffiffiffiTpþ

Z

2ffiffiffiTp

� ��

�e�KZ

2signðZÞþerf K

ffiffiffiTp�

Z

2ffiffiffiTp

� �� , ð36Þ

signðZÞ being 0 for Z¼0, þ1 for Z40 and �1 otherwise. Levellines of U7 in the plane (Y,T) are drawn in Fig. 10 at different Z. Inparticular, in (a) Z is negative, in (b) vanishes and finally in (c) ispositive. In the first case, the velocity vanishes on the axes (T¼0,Y¼0) and grows for increasing T and Y, even if it reaches onlyquite small values. In correspondence to the plane Z¼0, the abovebehaviour changes abruptly, because the fluid in a neighbourhoodof the wall (small Y) moves with velocity about 1/2. For thisreason, the level values decrease from the left to the right, as italso occurs at positive Z (c), where larger velocities have beenfound being U7ð0,Z,TÞ � 1. In Fig. 10e, level lines of the free surfacevelocity in the (Z,T)-plane are also drawn: as expected, U7 growsfor increasing times, but in a faster way for positive Z.

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1Y

T

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1Y

T

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1Y

T

0

0

0

0

T

0

0

0

0

T

Fig. 10. In (a, b, c) the level lines of U7 as a function of (Y,T) at Z¼�0.5, 0, þ0.5 are draw

drawn: positive levels use blue lines, while red lines are employed for negative levels.

while the opposite values are used for negative levels. Finally, in (e) the level lines of t

1 with step 0.02. (For interpretation of the references to colour in this figure legend, t

The non-dimensional wall stress W7 ¼w7=ðrnu0=hÞ ¼ @Y U7jY ¼ 0

follows from the velocity (35) as:

W7ðZ,TÞ ¼X1k ¼ 0

F7ðKjZ,TÞ: ð37Þ

Level lines of the wall stress (37) in the (Z,T)-plane are drawn inFig. 10d: it is negative for Z40 and positive for Zo0. Moreover, itdiverges in correspondence to Z¼0 and at T¼0, but only for positivevalues of Z (for negative Z, it vanishes).


The use of the Laplace transform qðLÞ (7) of the wall velocity inthe general form of the solution (32) leads to the followingsolution:

vðFLÞ

u0ðy,z,sÞ ¼

1


coshðhbÞ�


�

�1

4

e�iy

sþ ioþ

eþ iy

s�io

� �: ð38Þ

The transformed solution (38) possesses in the s-plane thesingularities I, II, IV and V of the first Stokes’ problem, togetherwith two simple poles in the points s¼ 7 io.

0

.2

.4

.6

.8

1

-1 -0.5 0 0.5 1

Z

0

.2

.4

.6

.8

1

-1 -0.5 0 0.5 1

Z

n. Levels go from 0.02 to 1 with step 0.02. In (d) the level lines of the wall stress are

Positive levels are in logarithmic scale from 0.01 to 102 with logarithmic step 0.2,

he free surface velocity as a function of Z and T are drawn. Levels go from 0.02 to

he reader is referred to the web version of this article.)


As before, the Laplace antitransform in s is performed byintegrating on the path in Fig. 1a (in the limit for M going toinfinity), while the calculations of the residues and of their Fouriersine antitransforms lead to the following velocity utd:

utd

u0ðY ,Z,TÞ ¼

1

O2

X1k ¼ 0

KsinðKYÞ

Z T

0dT 0 e�K2T 0=O2

cosðTþy�T 0Þ erfOZ


� �,

ð39Þ

which is odd in Z. Note also that it goes to U4/2 for Z-þ1. Byadding U4/2 to the non-dimensional velocity (39), the followingsolution is finally obtained:

U8ðY ,Z,TÞ ¼1

2cðYÞcosðTþyÞþsðYÞsinðTþyÞ ��X1k ¼ 0

KðK2cosyþO2sinyÞK4þO4

sinðKYÞexpð�K2T=O2Þ

þ1

O2

X1k ¼ 0

KsinðKYÞ

Z T

0dT 0e�K2T 0=O2

cosðTþy�T 0Þ erfOZ


� �

¼2

O2

X1k ¼ 0

KsinðKYÞ

Z T

0dT 0e�K2T 0=O2

cosðTþy�T 0ÞFOZ


� �:

ð40Þ

This solution behaves in the required way for asymptotic valuesof Z. Indeed, it vanished as Z-�1, while it holds U4 as Z-þ1.

Before discussing the numerical computation of the abovesolution, it is convenient to integrate by parts inside the secondseries of the velocity (40) which can be written as follows:

�X1k ¼ 0

sinðKYÞ

K

Z T

0de�a

2T 0cosðTþy�T 0Þ erfbffiffiffiffiffiT 0p

� �

¼1

2cosðTþyÞ signðZÞþ

X1k ¼ 0

sinðKYÞ

KF8ðKjZ,TÞ, ð41Þ

where the new function F8:

F8ðKjZ,TÞ ¼

Z T

0dT 0 e�a

2T 0sinðTþy�T 0Þ erfbffiffiffiffiffiT 0p

� �

�½Ic8cosðTþyÞþ Is

8sinðTþyÞ��e�a2T cosy erf

bffiffiffiTp

� �, ð42Þ

has been introduced. The new quantities used inside the functionF8 (42) are the two constants a :¼ K=O, b :¼ OZ=2 and theintegrals:

Ic,s8 ðTÞ :¼

bffiffiffiffipp

Z T

0dT 0 e�ða

2T 0 þb2=T 0 ÞT 0�3=2 cos

sinT 0: ð43Þ

Notice that the integrand function, say Gc,s8 ðT

0Þ, vanishes whena2T 0 or b2=T 0 are larger than a certain threshold M2 (e.g., M¼6). Asa consequence, Gc,s

8 ðT0ÞC0 as T 04Ta :¼ ðM=aÞ2 (which becomes

smaller and smaller as k-1) and as T 0oTb :¼ ðb=MÞ2. Moreover,

0

0.2

0.4

0.6

0.8

1

-1 -0.5 0 0.5 1

Y

Z

Fig. 11. Level lines of the modulus m8 (a) and of the phase j8 (b) of the steady-state s

one can also replace erfðb=ffiffiffiffiffiT 0pÞ with signðZÞ as T 0oTb. For this

reason, the integral containing erfðb=ffiffiffiffiffiT 0pÞ becomes an elementary

one as TrTb.Assume Ta4Tb, otherwise the above integrals are elementary

ones. In this condition, the first integral in the right hand side ofEq. (41) is splitted in the sum of an integral on ð0,TbÞ, which iselementary, and an integral on ðTb,TÞ. This latter is reduced to alinear combination of the integrals (43) through an integration byparts:Z T

Tb

dT 0 e�a2T 0sinðTþy�T 0Þerf

bffiffiffiffiffiT 0p

� �

¼1

a4þ1cosy�a2siny�

e�a2T erf

bffiffiffiTp

� ��½cosðTþy�TbÞ�a2sinðTþy�TbÞ�e

�a2TbsignðZÞ

þðIc8þa

2Is8ÞcosðTþyÞþðIs

8�a2Ic

8ÞsinðTþyÞ�:

The integrals (43) as T4Tb (g :¼ ab, x :¼ jbj=ffiffiffiTpÞ:

Ic,s8 ¼ signðbÞ

2ffiffiffiffipp

Z M

xdxe�x

2

e�g2=x2 cos

sin

b2

x2ð44Þ

will be now evaluated. The integration domain is decomposed in acertain number, say N, of intervals ðxn�1,xnÞ and the exponentialexpð�x2

Þ is expanded in Taylor series around the point x¼ xn�1.The resulting integrals are evaluated through the recurrencerelation (kZ3Þ:

Jc,s8 ðkÞ :¼

Z xn

xn�1

dxðx�xn�1Þke�g

2=x2 cos

sin

b2

x2

¼x3

nðxn�xn�1Þk�2

kþ1e�a

2Zncos

sinZn

�3k

kþ1xn�1Jc,s

8 ðk�1Þ�3ðk�1Þx2

n�1þ2g2

kþ1Jc,s8 ðk�2Þ

þ2b2

kþ1J�s,c8 ðk�2Þ�

k�2

kþ1x3

n�1Jc,s8 ðk�3Þ,

Z being b2=x2, while Jc,s8 ð0,1,2Þ are evaluated in terms of expo-

nential integral or sine and cosine integrals.Once the problem of numerical computing of the solution (40)

has been solved, the behaviour of the steady-state part of thevelocity U8 is investigated. This part is obtained by setting to0 terms containing expð�K2T=O2

Þ, i.e. the first series in thevelocity (40), and by using ð0,TaÞ as integration range. In Fig. 11level lines of the modulus m8 (a) and of the phase j8 (b) aredrawn in the (Y,Z)-plane. The modulus reaches its maximum 1 onthe moving wall (Y-0, Z40) and its minimum 0 on the fixed one(Y-0, Zo0). m8(Y;Z) is a monotonically growing function of Z atY fixed, while it is a decreasing (increasing) function of Y at Z40(Zo0) fixed. The phase delay with respect to the wall motion j8

0

0.2

0.4

0.6

0.8

1

-1 -0.5 0 0.5 1

Y

Z

olution U8 in the (Z,Y)-plane. O¼ 1 is assumed. Steps are 0.05 in (a) and 13 in (b).

0

1

2

3

4

5

-10 -5 0 5 10Z

mw

8

-180

-150

-120

-90

-60

-30

0

-10 -5 0 5 10Z

�w

8

Fig. 12. Modulus mw8 (a) and phase jw8 (b) of the steady-state wall stress W8 vs. Z. O¼ 1 is assumed. Asymptotic values (21) are also drawn with dashed lines.


reaches its maximum 0 at the moving wall and it is a mono-tonically decreasing function of Y, at Z fixed. On the contrary, it isa monotonically growing function of Z at a fixed Y, even if itsgrowth becomes smaller and smaller once the free surface isreached.

The non-dimensional wall stress W8 ¼w8=ðru0

ffiffiffiffiffiffiffinopÞ¼

@Y U8jY ¼ 0=O is evaluated from the velocity (40) as:

W8 ¼W4ðTÞ

2þ

1

O

X1k ¼ 0

F8ðKjZ,TÞ: ð45Þ

In Fig. 12 the modulus mw8 (a) and the phase jw8 (b) of thesteady-state stress (45) are drawn vs. Z, having assumed O¼ 1. Asexpected, the modulus vanishes as Z-�1, diverges as Z-0 andreaches its asymptotic value (21) as Z-þ1. Notice also thatmw8 is larger than its corresponding one-dimensional value,unless in a rather small interval around Z¼1. The phase delayjw8 with respect to the wall motion experiences a jump ofamplitude �p when Z crosses the value 0. It is not monotonicfor negative Z, while it monotonically reaches its asymptotic value(21) for positive Z. The wall stress is in quadrature with respectto the wall motion as Z-�1, in phase as Z-0� and opposite

as Z-0þ .

6. Conclusions

The analytical solutions of first and second Stokes’ problems inplanar geometries have been investigated and the wall stressesare evaluated. Velocity and stress are given in real form. Analy-tical approaches to the numerical computation of these solutionsare also discussed.

In order to investigate the physical properties of thesteady-state velocity and wall stress in the case of an oscillat-ing wall (second Stokes’ problem), the splitting of the solutionin its ‘‘in phase’’ and ‘‘in quadrature’’ components, with respectto the wall motion, is proposed. In this way, amplitude andphase delay of the solution are directly evaluated. Solutions ininfinite-depth flows exhibit (see Fig. 2 for the one-dimensionalproblem and Fig. 8 for the two-dimensional one) the presenceof stagnation points. It can be argued that a countable set ofsuch points exists, if the entire time interval ð0,þ1Þ isconsidered. On the contrary, stagnation points are not foundin finite-depth flows.

In many cases, the present solutions have simpler formsthan the ones in literature [1]. Indeed, all the solutions ofinfinite-depth problems can be posed in the (dimensional)form:

u¼ u0y


Z t

0dtt�3=2exp �

y2

4nt

� �P1ðt,tÞP2ðz,tÞ,

the functions P1,2 being defined as:

P1 :¼1 first problem,

cos½oðt�tÞþy� second problem,

(

P2 :¼1 1D flow,

F½z=ð2ffiffiffiffiffintpÞ� 2D flow,

(

Also in finite-depth flows, the same rule appears to be valid.The present solutions have just the form:

u¼ 2u0nh2

X1k ¼ 0

KsinðKy=hÞ

Z t

0dt e�K2nt=h2

P1ðt,tÞP2ðz,tÞ:

In the author opinion, this ‘‘unification’’ of the solutionsjustifies the use of their real forms. The same is not true forthe wall stresses, the analytical forms of which can be hardlyrelated. For example, the presence of integrals of Fresnel’sfunctions C1 and S1 in W6 (30) breaks the symmetry withrespect to W5 (27), leading to a much more complicatedbehaviour of the solution.

As stressed by many authors, the use of the real forms of thesolutions leads to severe numerical problems when they arecomputed in practice, due to the presence of integrals of oscillatoryfunctions. For this reason, complex forms with tabulated functionsare often preferred, even if they are much more complicated. In thepresent paper, integration by series and recurrence formulae areadopted to compute the real forms of the solutions in an accurateand efficient way. With this approach, discretization errors are notintroduced. The only error sources are due to the propagation oftruncation errors along recurrence formulae and to the approxima-tion of series with corresponding finite sums. Both kinds of errorscan be handled without difficulties, opening the way to thequantitative use of the real forms of the solutions.

Appendix A. Computation of some special functions by series

Fresnel’s integrals can be evaluated by series, once they are treatedin order to avoid numerical problems. Indeed, they can be written as:

C1

S1ðxÞ ¼

ffiffiffiffi2

p

r Z x

0dx

cos

sinx2¼

1ffiffiffiffiffiffi2pp

Z x2

0

dZffiffiffiZp cos

sinZ:

Assume that x2 lies into the interval: ½2pM,2pðMþ1ÞÞ for some non-negative integer M. By separating the contributions of each periodone obtains:

C1

S1ðxÞ ¼

1ffiffiffiffiffiffi2pp

XMm ¼ 1

Z 2p

0

dzffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffizþ2pðm�1Þ

p cos

sinz

"


þ

Z x2�2pM

0

dzffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffizþ2pM

p cos

sinz

#,

where the first sum must be omitted if M¼0. The above integralsare easily evaluated by series, z being not larger than 2p. Formoderate m (mo5), the series representations of the trigono-metric functions are used, so that the above integrals are evaluatedthrough the recurrence ones (kZ1, a¼ 2pðm�1Þ or 2pM):

Hkðt;aÞ ¼1

k!

Z t

0dx

xkffiffiffiffiffiffiffiffiffiffiffixþa

p ¼1

kþ1=2

tk

k!

ffiffiffiffiffiffiffiffiffiffitþap

�aHk�1ðt;aÞ� �

,

with H0ðt;aÞ ¼ 2ðffiffiffiffiffiffiffiffiffiffitþap

�ffiffiffiapÞ. For large m, the integrals are rewrit-

ten as:Z t

0

dzffiffiffiffiffiffiffiffiffiffizþa

p cos

sinz¼

1ffiffiffiap

X1k ¼ 0

�1=2

k

� �1

ak

Z t

0dzzk cos

sinz,

where the trigonometric integrals are easily evaluated. The aboveprocedure has been successfully tested up to x¼103 by finding anexcellent agreement with the theoretical asymptotic behaviour[10]. Analogous approaches are used in computing the sine Si(x)and cosine Ci(x) integral functions ([9, p. 231], formulae (5.2.1) and(5.2.2), respectively).

The exponential integral:

E1ðxÞ ¼

Z þ1x

dxe�x

x

(see [9], p. 228 formula (5.1.1)) is also computed by expanding in

McLaurin series the exponential function in the range xr1, whilein the range x41 the following approximation:

E1ðxÞ ¼

Z þ1a

dxe�x

xCZ X

adx

e�x

x¼XM

m ¼ 1

e�xm�1

Z xm

xm�1

dxe�ðx�xm�1Þ

x

is used (a¼1 if xr1, a¼x if x41). The value X¼33 is fixed, whilea decomposition having constant step (xm�xm�1) of the order ofthe unity is used, with x0¼a and xM¼X. The integrals are thenevaluated by expanding the exponential functions.

References

[1] C.M. Liu, Complete solutions to extended Stokes’ problems, Math. Probl. Eng.(2008), Art. ID 754262, 18 pp.

[2] R. Panton, The transient for Stokes oscillating plate: a solution in terms oftabulated functions, J. Fluid Mech. 31 (4) (1968) 819–825.

[3] M.E. Erdogan, A note on an unsteady flow of a viscous fluid due to anoscillating plane wall, Int. J. Non-linear Mech. 35 (2000) 1–6.

[4] C.M. Liu, I.C. Liu, A note on the transient solution of Stokes’ second problemwith an arbitrary initial phase, J. Mech. 22 (2006) 349–354.

[5] M.E. Erdogan, C.E. Imrak, On the comparison of the solutions obtained byusing two different transform methods for the second problem of Stokes forNewtonian fluids, Int. J. Non-linear Mech. 44 (2009) 27–30.

[6] Y. Zeng, S. Weinbaum, Stokes problems for moving half-planes, J. Fluid Mech.287 (1995) 59–74.

[7] M.E. Erdogan, C.E. Imrak, Some effects of side walls on unsteady flow of aviscous fluid over a plane wall, Math. Probl. Eng. (2009), Article ID 725196.

[8] H. Schlichting, Boundary Layer Theory, McGraw-Hill, New York, 1979.[9] M. Abramowitz, I.A. Stegun, Handbook of Mathematical Functions with

Formulas, Graph and Mathematical Tables, Dover, New York, 1965.[10] E. Kreyszig, On the zeros of the Fresnel integrals, Can. J. Math. 9 (1957)

118–131.

remarks on the solution of extended stokes' problems

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