# Complete Solutions to Extended Stokes' Problems

Authors: Chi-Min Liu
Journal: Mathematical Problems in Engineering
 (2008)

---

## Abstract

The main object of the present study is to theoretically solve the viscous flow of either a finite or infinite depth, which is driven by moving plane(s). Such a viscous flow is usually named as Stokes' first or second problems, which indicates the fluid motion driven by the impulsive or oscillating motion of the boundary, respectively. Traditional Stokes' problems are firstly revisited, and three extended problems are subsequently examined. Using some mathematical techniques and integral transforms, complete solutions which can exactly capture the flow characteristics at any time are derived. The corresponding steady-state and transient solutions are readily determined on the basis of complete solutions. Current results have wide applications in academic researches and are of significance for future studies taking more boundary conditions and non-Newtonian fluids into account.


---

## Body

## 1. Introduction

Stokes’ problems are of great significance
in numerous fields, which include the industry manufacturing, chemical engineering,
geophysical flows, and heat
conduction problems. Early in 1851, Stokes [1] proposed a well-known
paper on pendulums in which the problems of the impulsive and oscillatory
motions of a plane were studied. For a viscous flow motivated by a moving
plane, Stokes’ problems are one-dimensional, initial and boundary value problems.
The momentum equation which is simplified from Navier-Stokes equations is a
partial differential equation with differentiations with respect to only two
variables, time and the coordinate normal to the plane. This simplification
enables the viscous flow to be mathematically solvable. The associated
solutions either for the transient state or the steady state have been well
discussed [2–5].In addition to previous conditions considered
in Stokes’ problems, much more boundary conditions are considered to simulate
the flows in practice. For example, Zeng and Weinbaum [6] theoretically studied
Stokes’ problems for moving half-planes. Their results can be applied to many
practical problems, for example, the flow induced by either earthquakes or fracture
of ice sheets. However, their steady-state solutions cannot capture the transient-state phenomenon
while most of the earthquakes and other specific problems usually occur in a
very short period. Up to now, the transient solutions are still lack and
unclear. Moreover, a flow of a finite depth instead of an infinite depth should
be further investigated as well. These new conditions constitute the extended
Stokes’ problems.Due to above reasons, complete
solutions to extended Stokes’ problems are theoretically derived in the present
paper. The organization of the present paper is as follows. Traditional Stokes’
problems (see Figure1(a)) are revisited in Section 2. Three extended Stokes’ problems
(see Figure 1(b) to Figure 1(d)) are examined in
Sections 3 to 5. According to the
derived solutions, results and conclusions are addressed in Section 6.Traditional Stokes’ problems and three extended Stokes’ problems.
(a)(b)(c)(d)


## 2. Review of Traditional Stokes’ Problems

Traditionally, Stokes’ problems are
classified into two types according to the motions of the rigid boundary below
the fluid. The impulsive and harmonic motions of the boundary lead to the first
and second kinds of problems, respectively (see Figure1(a)). The related
momentum equation, boundary, and initial conditions are
(2.1a)ut=νuyy,(2.1b)u(y=0,t>0)=u0,(2.1c)u(y→∞,t>0)=0,(2.1d)u(y>0,t=0)=0,
for the first problem and
(2.2)ut=νuyy,u(y=0,t>0)=u0cos(σt+θ),u(y→∞,t>0)=0,u(y>0,t=0)=0,
for
the second problem in which u is the velocity, ν the kinematic viscosity, σ the angular frequency, and the subscripts the
differentiations. Solution to the first problem can be obtained through two
main ways, the direct integral transform and the similarity method (for
detailed discussion, see Wang  [7]). The solution is (2.3)u=u0erfc(y2νt),and
the stress on the boundary (wall stress) is (2.4)τb=−u0μπνt.It
is clear that no steady-state solution exists since the velocity profile will
gradually develop by acquiring energy from the moving boundary. As for the
second problem, the dimensionless velocity  profile and wall stress are (2.5)U=12Re{exp(−Y2)⋅exp[i(T−Y2+θ)]⋅erfc(Y2T−iT)+exp(Y2)⋅exp[i(T+Y2+θ)]⋅erfc(Y2T+iT)},(2.6)Tb=−Re{1πTeiθ+2⋅ei(T+θ)[SF(2Tπ)+iCF(2Tπ)]},in
which(2.7)Y=σνy,T=σt,U=uu0,Tb=τbνμu0σ,  and Re denotes the real part of the bracket and SF and CF express the sine and cosine Fresnel integrals
[3]. The steady-state solutions for large times are (2.8)Us=exp(−Y2)cos(T−Y2+θ),Tbs=−cos(T+θ+π4). Erdogan
[2] pointed out that the transient solution (Ut=U−Us) of the cosine oscillation decays more
rapidly than that of the sine case. The wall stress also decays with time.
Furthermore, it is found that there exists a phase difference between the
steady-state velocity and wall stress.


## 3. Stokes’ Problems for a Finite-Depth Fluid

In a mathematical viewpoint, a solution to
an infinite-depth fluid cannot exactly describe the flow of a finite depth, specially
the shallow-water
configuration. Therefore, a viscous flow of a finite depthh shown in Figure 1(b) is considered in this
section.


### 3.1. Solution
to the First Problem

The
governing equation and associated conditions are(3.1a)ut=νuyy,(3.1b)u(y=0,t>0)=u0,(3.1c)uy(y=h,t>0)=0,(3.1d)u(y>0,t=0)=0,
in
which (3.1c) indicates that
no stress is allowed at the free-surface. Applying the Laplace transform to
(3.1a)–(3.1d) leads to (3.2)su^=νu^yy,u^(y=0,s)=u0s,u^y(y=h,s)=0, where u^(y,z,s)=∫0∞u(y,z,t)⋅e−stds. The
solution to (3.2) is (3.3)u^=u0scosh[sν(y−h)]sech[sν⋅h].
With
the help of the inverse transform equations in the first row of Table 1, the inversion of (3.3) is
obtained (3.4)u=u0−u0νh∫0y/νθ2(νχ2h|νth2)dχ,which
can be expressed in a dimensionless form in terms of infinite series (with the
help of (A.2)) (3.5)U=1−∑n=0∞2Kexp(−K2T)sin(KY),where (3.6)Y=yh,T=νth2,U=uu0,K=2n+12π.It is
noted that the above solution excludes the state at T=0.Table 1u^(s)u(t)1scosh(βs)sech(αs),−α≤β≤α1−1α∫0α+βθ2(χ2α|tα2)dχ1sexp(−βs)erfc(β2t)exp(−βs)β2πt3exp(−β24t)


### 3.2. Solution
to the Second Problem

The
momentum equation and conditions for the second problem is the same as (3.1a),
(3.1c), (3.1d)
except that (3.1b) has to be replaced by the oscillating boundary condition.
Hence, the momentum equation and conditions are (3.7)ut=νuyy,u(y=0,t>0)=u0cos(σt+θ),uy(y=h,t>0)=0,u(y>0,t=0)=0. Using the technique of Laplace transform and solving u^ with boundary conditions, we have (3.8)u^=u0[ss2+σ2cosθ−σs2+σ2sinθ]cosh(sν(y−h))sech(sν⋅h).For obtaining the inverse transform of
(3.8), one can rewrite it as (3.9)u^=u0[s2s2+σ2cosθ−sσs2+σ2sinθ]⋅[1scosh(sν(y−h))sech(sν⋅h)].As the inversion of the second bracket is
the same as that of (3.3), the inversion of (3.9) can be expressed by the
convolution integral. After some algebra, the solution is (3.10)U=1−cosθ+cos(T+θ)+2⋅∑n=0∞sinNYN{−exp(−λN2T)cosθ+11+λ2N4⋅[λN2sin(T+θ)−cos(T−θ)+(cosθ−λN2sinθ)exp(−λN2T)]}, where and the
dimensionless variables are defined as (3.11)Y=yh,T=σt,U=uu0,λ=νσh2,N=2n+12π.By ignoring the exponential terms in
(3.10),
one can obtain the steady-state solution.


### 3.3. Results

Figure2 displays the velocity profiles at various values of T for the first problem. It is evident that the
velocity profile will gradually develop, and eventually be consistent with the
wall speed. This result is quite similar to that of the traditional solution
[5]. The significant difference between them is that the similarity phenomenon
only exists in the traditional problem.Figure 2
Velocity profiles of the first
problem for a finite-depth fluid.Results for the second problem are shown in Figures3 and
4. It is
noted that without loss of generality, the initial phase θ in all figures throughout this paper is set to
be zero. The development of the velocity in the first oscillating cycle is
displayed in Figure 3. As the upper boundary is free (no stress exists), the
oscillation of the profile is less obvious than that of the traditional problem
[5]. The comparison between the surface velocity (solid curve) and the wall
velocity (dash curve) in the duration T=0 to T=4π is shown in Figure 4.
It is found that the maximum of the surface velocity is less than that of the
wall velocity according to the viscous dissipation. Moreover, there is a phase difference
between the surface velocity and the wall velocity.Figure 3
Velocity profiles of the second
problem for a finite-depth fluid.Figure 4
Comparison between the surface velocity (solid curve) and the wall velocity
(dash curve) of the second problem for a finite-depth fluid.


## 3.1. Solution
to the First Problem

The
governing equation and associated conditions are(3.1a)ut=νuyy,(3.1b)u(y=0,t>0)=u0,(3.1c)uy(y=h,t>0)=0,(3.1d)u(y>0,t=0)=0,
in
which (3.1c) indicates that
no stress is allowed at the free-surface. Applying the Laplace transform to
(3.1a)–(3.1d) leads to (3.2)su^=νu^yy,u^(y=0,s)=u0s,u^y(y=h,s)=0, where u^(y,z,s)=∫0∞u(y,z,t)⋅e−stds. The
solution to (3.2) is (3.3)u^=u0scosh[sν(y−h)]sech[sν⋅h].
With
the help of the inverse transform equations in the first row of Table 1, the inversion of (3.3) is
obtained (3.4)u=u0−u0νh∫0y/νθ2(νχ2h|νth2)dχ,which
can be expressed in a dimensionless form in terms of infinite series (with the
help of (A.2)) (3.5)U=1−∑n=0∞2Kexp(−K2T)sin(KY),where (3.6)Y=yh,T=νth2,U=uu0,K=2n+12π.It is
noted that the above solution excludes the state at T=0.Table 1u^(s)u(t)1scosh(βs)sech(αs),−α≤β≤α1−1α∫0α+βθ2(χ2α|tα2)dχ1sexp(−βs)erfc(β2t)exp(−βs)β2πt3exp(−β24t)


## 3.2. Solution
to the Second Problem

The
momentum equation and conditions for the second problem is the same as (3.1a),
(3.1c), (3.1d)
except that (3.1b) has to be replaced by the oscillating boundary condition.
Hence, the momentum equation and conditions are (3.7)ut=νuyy,u(y=0,t>0)=u0cos(σt+θ),uy(y=h,t>0)=0,u(y>0,t=0)=0. Using the technique of Laplace transform and solving u^ with boundary conditions, we have (3.8)u^=u0[ss2+σ2cosθ−σs2+σ2sinθ]cosh(sν(y−h))sech(sν⋅h).For obtaining the inverse transform of
(3.8), one can rewrite it as (3.9)u^=u0[s2s2+σ2cosθ−sσs2+σ2sinθ]⋅[1scosh(sν(y−h))sech(sν⋅h)].As the inversion of the second bracket is
the same as that of (3.3), the inversion of (3.9) can be expressed by the
convolution integral. After some algebra, the solution is (3.10)U=1−cosθ+cos(T+θ)+2⋅∑n=0∞sinNYN{−exp(−λN2T)cosθ+11+λ2N4⋅[λN2sin(T+θ)−cos(T−θ)+(cosθ−λN2sinθ)exp(−λN2T)]}, where and the
dimensionless variables are defined as (3.11)Y=yh,T=σt,U=uu0,λ=νσh2,N=2n+12π.By ignoring the exponential terms in
(3.10),
one can obtain the steady-state solution.


## 3.3. Results

Figure2 displays the velocity profiles at various values of T for the first problem. It is evident that the
velocity profile will gradually develop, and eventually be consistent with the
wall speed. This result is quite similar to that of the traditional solution
[5]. The significant difference between them is that the similarity phenomenon
only exists in the traditional problem.Figure 2
Velocity profiles of the first
problem for a finite-depth fluid.Results for the second problem are shown in Figures3 and
4. It is
noted that without loss of generality, the initial phase θ in all figures throughout this paper is set to
be zero. The development of the velocity in the first oscillating cycle is
displayed in Figure 3. As the upper boundary is free (no stress exists), the
oscillation of the profile is less obvious than that of the traditional problem
[5]. The comparison between the surface velocity (solid curve) and the wall
velocity (dash curve) in the duration T=0 to T=4π is shown in Figure 4.
It is found that the maximum of the surface velocity is less than that of the
wall velocity according to the viscous dissipation. Moreover, there is a phase difference
between the surface velocity and the wall velocity.Figure 3
Velocity profiles of the second
problem for a finite-depth fluid.Figure 4
Comparison between the surface velocity (solid curve) and the wall velocity
(dash curve) of the second problem for a finite-depth fluid.


## 4. Stokes’ Problems Motivated by Relatively Moving Planes

A flow generated by relatively moving planes
is studied in this section. Unlike one-dimensional problems investigated in
previous sections, present problem is a two-dimensional problem, as shown in
Figure1(c). Though this problem has been theoretically solved
by Zeng and  Weinbaum [6], it requires more improvements. Firstly, their solution to the
second problem is a steady-state solution in which no decaying terms exist. This may be
applicable for a long-term analysis, but it fails to describe the transient
flow which strongly influences the total flow at the early stage. Secondly, a
more direct transform rather than the semicylindrical transform used in their
derivation seems necessary to standardize the derivation. Above weaknesses will
be improved herein.


### 4.1. Solution
to the First Problem

The momentum equation
and associated conditions are(4.1)ut=ν(uyy+uzz),u(y=0,z>0,t>0)=u0,u(y=0,z<0,t>0)=0,u(y→∞,z,t>0)=0,u(y,z→±∞,t)isfinite,u(y>0,z,t=0)=0. As
(4.1) is purely linear, it can be
decomposed into two subproblems (u=u1+u2) (4.2)u1,t=νu1,yy,u1(y=0,t>0)=u02,u1(y→∞,t>0)=0,u1(y>0,t=0)=0, and
(4.3)u2,t=ν(u2,yy+u2,zz),u2(y=0,z>0,t>0)=u02,u2(y=0,z<0,t>0)=−u02,u2(y→∞,z,t>0)=0,u2(y,z→±∞,t)isfinite,u2(y>0,z,t=0)=0.It is evident that the
former problem governed by (4.2) is the traditional Stokes’ first problem, and
the solution to u1 is a half of (2.3). As for the latter problem,
the flow satisfies the condition (4.4)u2(y,z>0,t)=−u2(y,z<0,t),which
further leads to (4.5)u2(y>0,z=0,t)=0.Since the flow is antisymmetrical with
respect to z=0,
one only needs to solve u2 for the domain of z≥0 only. The resulting equation and conditions are (4.6)u2,t=ν(u2,yy+u2,zz),u2(y=0,z>0,t>0)=u02,u2(y>0,z=0,t>0)=0,u2(y→∞,z≥0,t>0)=0,u2(y,z→+∞,t)isfinite,u2(y>0,z≥0,t=0)=0. Applying the Laplace transform to (4.6) leads to (4.7)su^2=ν(u^2,yy+u^2,zz),u^2(y=0,z>0,s)=u02s,u^2(y>0,z=0,s)=0,u^2(y→0,z≥0,s)=0.Now one further applies
the Fourier sine transform to (4.7). It yields (4.8)u˜zz−(ω2+sν)u˜=−u0ω2s,where u˜(ω,z,s)=∫0∞u^2(y,z,s)sin(ωy)dy.
From the boundedness of u˜ as z approaches infinity and the boundary condition
at z=0,
the solution to u˜ is(4.9)u˜=−u0ων2s(ω2ν+s)exp(−ω2+s/ν⋅z)+u0ων2s(ω2ν+s).The solution of u2 can be obtained by inverting (4.9) twice (4.10)u2=uo2erfc(y2νt)−u0y4πν∫0terfc(z2νt′)(t′)−1.5exp(−y24νt′)dt′, which is equivalent to (4.11)u2=uo2erfc(y2νt)+u02π∫t′=0t′=terfc(z2νt′)exp(−y24νt′)d(yνt′).The dimensionless form of (4.11) is (4.12)U2=12erfc(Y)+1π∫∞Yerfc(ZYY′)exp(−Y′2)dY′,where (4.13)U2=u2u0,Y=y2νt,Z=z2νt.Finally, (4.12) can be further simplified
to be (4.14)U2=1π∫Y∞erf(ZY⋅Y′)exp(−Y′2)dY′.As for the flow in the left domain (Z<0), one can readily derive the solution by following the above processes.


### 4.2. Solution
to the Second Problem

The techniques for
solving the second problem are analogous to those for the first problem. Hence
the following governing equation, boundary and initial conditions,(4.15)ut=ν(uyy+uzz),u(y=0,z>0,t>0)=u0cos(σt+θ),u(y=0,z<0,t>0)=0,u(y→∞,z,t>0)=0,u(y,z→±∞,t)isfinite,u(y>0,z,t=0)=0,are decomposed to be (4.16)u1,t=ν(u1,yy+u1,zz),u1(y=0,z,t>0)=u02cos(σt+θ),u1(y→∞,z,t>0)=0,u1(y>0,z,t=0)=0,
and
(4.17)u2,t=ν(u2,yy+u2,zz),u2(y=0,z>0,t>0)=u02cos(σt+θ),u2(y=0,z<0,t>0)=−u02cos(σt+θ),u2(y→∞,z,t>0)=0,u2(y,z→+∞,t)isfinite,u2(y>0,z,t=0)=0.  The
solution to the former system is a half of (2.5), and the solution to the latter
case is solved to be shown in the dimensionless form (4.18)U2=Y4π∫0Tα−1.5cos(T−α+θ)exp(−Y24α)dα+Z2π3∫0∞∫0Tsin(ωY)⋅G1(ω,T−α)⋅G2(ω,α,Z)dαdω, where (4.19)T=σt,Y=σνy,Z=σνz,U2=u2u0,(4.20)G1(ω,T)=ωω4+1[sinθ(e−ω2T−cosT+ω2sinT)−cosθ(sinT+ω2cosT−ω2e−ω2T)],G2(ω,T,Z)=T−1.5exp(−ω2T−Z24T).


### 4.3. Results

For the first problem, the
velocity profiles for variousZ-sections
at T=0.1 are drawn in Figure 5. At the far end of the
right-hand side (Z≫0), the profile will approach that of the
traditional problem. This implies that the influence of the still half plate
will decay as the value of Z grows. Similarly, the velocity will approach
zero at the other far end (Z≪0) where the effects of the oscillating half plate
are much weaker. Results for the second problem are shown in Figure 6. The phenomena
at the far ends are quite similar to those appearing in the first problem.Figure 5
Velocity profiles of the first problem for an infinite-depth fluid motivated
by relatively moving planes.Figure 6
Velocity profiles of the second problem for an infinite-depth fluid
motivated by relatively moving planes.


## 4.1. Solution
to the First Problem

The momentum equation
and associated conditions are(4.1)ut=ν(uyy+uzz),u(y=0,z>0,t>0)=u0,u(y=0,z<0,t>0)=0,u(y→∞,z,t>0)=0,u(y,z→±∞,t)isfinite,u(y>0,z,t=0)=0. As
(4.1) is purely linear, it can be
decomposed into two subproblems (u=u1+u2) (4.2)u1,t=νu1,yy,u1(y=0,t>0)=u02,u1(y→∞,t>0)=0,u1(y>0,t=0)=0, and
(4.3)u2,t=ν(u2,yy+u2,zz),u2(y=0,z>0,t>0)=u02,u2(y=0,z<0,t>0)=−u02,u2(y→∞,z,t>0)=0,u2(y,z→±∞,t)isfinite,u2(y>0,z,t=0)=0.It is evident that the
former problem governed by (4.2) is the traditional Stokes’ first problem, and
the solution to u1 is a half of (2.3). As for the latter problem,
the flow satisfies the condition (4.4)u2(y,z>0,t)=−u2(y,z<0,t),which
further leads to (4.5)u2(y>0,z=0,t)=0.Since the flow is antisymmetrical with
respect to z=0,
one only needs to solve u2 for the domain of z≥0 only. The resulting equation and conditions are (4.6)u2,t=ν(u2,yy+u2,zz),u2(y=0,z>0,t>0)=u02,u2(y>0,z=0,t>0)=0,u2(y→∞,z≥0,t>0)=0,u2(y,z→+∞,t)isfinite,u2(y>0,z≥0,t=0)=0. Applying the Laplace transform to (4.6) leads to (4.7)su^2=ν(u^2,yy+u^2,zz),u^2(y=0,z>0,s)=u02s,u^2(y>0,z=0,s)=0,u^2(y→0,z≥0,s)=0.Now one further applies
the Fourier sine transform to (4.7). It yields (4.8)u˜zz−(ω2+sν)u˜=−u0ω2s,where u˜(ω,z,s)=∫0∞u^2(y,z,s)sin(ωy)dy.
From the boundedness of u˜ as z approaches infinity and the boundary condition
at z=0,
the solution to u˜ is(4.9)u˜=−u0ων2s(ω2ν+s)exp(−ω2+s/ν⋅z)+u0ων2s(ω2ν+s).The solution of u2 can be obtained by inverting (4.9) twice (4.10)u2=uo2erfc(y2νt)−u0y4πν∫0terfc(z2νt′)(t′)−1.5exp(−y24νt′)dt′, which is equivalent to (4.11)u2=uo2erfc(y2νt)+u02π∫t′=0t′=terfc(z2νt′)exp(−y24νt′)d(yνt′).The dimensionless form of (4.11) is (4.12)U2=12erfc(Y)+1π∫∞Yerfc(ZYY′)exp(−Y′2)dY′,where (4.13)U2=u2u0,Y=y2νt,Z=z2νt.Finally, (4.12) can be further simplified
to be (4.14)U2=1π∫Y∞erf(ZY⋅Y′)exp(−Y′2)dY′.As for the flow in the left domain (Z<0), one can readily derive the solution by following the above processes.


## 4.2. Solution
to the Second Problem

The techniques for
solving the second problem are analogous to those for the first problem. Hence
the following governing equation, boundary and initial conditions,(4.15)ut=ν(uyy+uzz),u(y=0,z>0,t>0)=u0cos(σt+θ),u(y=0,z<0,t>0)=0,u(y→∞,z,t>0)=0,u(y,z→±∞,t)isfinite,u(y>0,z,t=0)=0,are decomposed to be (4.16)u1,t=ν(u1,yy+u1,zz),u1(y=0,z,t>0)=u02cos(σt+θ),u1(y→∞,z,t>0)=0,u1(y>0,z,t=0)=0,
and
(4.17)u2,t=ν(u2,yy+u2,zz),u2(y=0,z>0,t>0)=u02cos(σt+θ),u2(y=0,z<0,t>0)=−u02cos(σt+θ),u2(y→∞,z,t>0)=0,u2(y,z→+∞,t)isfinite,u2(y>0,z,t=0)=0.  The
solution to the former system is a half of (2.5), and the solution to the latter
case is solved to be shown in the dimensionless form (4.18)U2=Y4π∫0Tα−1.5cos(T−α+θ)exp(−Y24α)dα+Z2π3∫0∞∫0Tsin(ωY)⋅G1(ω,T−α)⋅G2(ω,α,Z)dαdω, where (4.19)T=σt,Y=σνy,Z=σνz,U2=u2u0,(4.20)G1(ω,T)=ωω4+1[sinθ(e−ω2T−cosT+ω2sinT)−cosθ(sinT+ω2cosT−ω2e−ω2T)],G2(ω,T,Z)=T−1.5exp(−ω2T−Z24T).


## 4.3. Results

For the first problem, the
velocity profiles for variousZ-sections
at T=0.1 are drawn in Figure 5. At the far end of the
right-hand side (Z≫0), the profile will approach that of the
traditional problem. This implies that the influence of the still half plate
will decay as the value of Z grows. Similarly, the velocity will approach
zero at the other far end (Z≪0) where the effects of the oscillating half plate
are much weaker. Results for the second problem are shown in Figure 6. The phenomena
at the far ends are quite similar to those appearing in the first problem.Figure 5
Velocity profiles of the first problem for an infinite-depth fluid motivated
by relatively moving planes.Figure 6
Velocity profiles of the second problem for an infinite-depth fluid
motivated by relatively moving planes.


## 5. Stokes’ Problems of a Finite-Depth Fluid Motivated by Relatively Moving Planes

A finite-depth
flow motivated by relatively moving planes is studied in this section, as shown
in Figure1(d). In addition to techniques used previously, more mathematical
techniques must be applied to solve current problems. The detailed derivation
is shown below.


### 5.1. Solution to the First
Problem

The
governing equation and associated conditions are expressed as(5.1a)ut=ν(uyy+uzz),(5.1b)u(y=0,z>0,t>0)=u0,(5.1c)u(y=0,z<0,t>0)=0,(5.1d)uy(y=h,z,t>0)=0,(5.1e)u(y,z→±∞,t)isfinite,(5.1f)u(y>0,z,t=0)=0.
The
direct integral transforms cannot work anymore in solving (5.1a),
(5.1b), (5.1c),
(5.1d), (5.1e),
(5.1f) as this flow
system is bounded by Dirichlet and Neumann conditions at
y=0
and
y=h,
respectively. For the sake of eliminating different types of boundary
conditions, the concept of symmetry is employed to generate a symmetrical flow.
Accordingly, the boundary condition (5.1d) is replaced by (5.2)u(y=2h,z,t>0)=u0,where
the flow domain is doubly expanded. Similar to the method introduced in Section
4, the flow system can be decomposed into two subsystems in which U1 is a half of (3.5) and u2 for the domain z>0 is governed by (5.3)u2,t=ν(u2,yy+u2,zz),u2(y=0,z>0,t>0)=u02,u2(y=2h,z>0,t>0)=u02,u2(y>0,z=0,t>0)=0,u2(y,z→+∞,t)isfinite,u2(y>0,z,t=0)=0. By applying the Laplace transform to (5.3), it generates (5.4)su^2=ν(u^2,yy+u^2,zz),u^2(y=0,z>0,s)=u02s,u^2(y=2h,z>0,s)=u02s,u^2(y,z→+∞,t)is  finite,u^2(y>0,z=0,s)=0. The system of u^2 is now shifted to the system of u*(5.5)su*+u02=ν(u2,yy*+u2,zz*),u*(y=0,z>0,s)=0,u*(y=2h,z>0,s)=0,u*(y,z→+∞,t)is  finite,u*(y>0,z=0,s)=−u02s, in
which u*=u^−u0/2s.
After applying the finite sine transform, u˜(s,n,y)=∫02hu*sin(nπy/2h)dy,
to (5.5), it yields (5.6)u˜zz−(sν+n2π24h2)u˜=u0hnπν(1−(−1)n),with
the boundedness condition for z→∞ and u˜(z=0)=−(u0h/nπν)(1−(−1)n).
The solution is (5.7)u˜=(1−(−1)n)⋅[u0hnπνα(exp(−αz)−1)−u0hnπsexp(−αz)],where α=s/ν+n2π2/4h2.
Applying the inverse transforms with the help of the equations in the
second and third rows of Table 1,
the dimensionless solution is solved after some algebra (5.8)U2=12+1π∑n=1∞1−(−1)nnsinnπY2⋅{−exp(−n2π2T4)erf(Z2T)−∫0TZ2πT′3exp(−n2π2T′4−Z24T′)dT′}, in
which Z=z/h and the remaining dimensionless variables are
identical to those shown in (3.6).


### 5.2. Solution
to the Second Problem

Using
the techniques described in previous subsections, the dimensionless solution to
the second problem is the summation of a half of (3.10) and the following
solution (5.9)U2=12cos(T+θ)+1π∑n=1∞1−(−1)nnsinnπY2⋅{cosθ∫0TG1(T−u)H(u)du+sinθ∫0TG2(T−u)H(u)du−G3(T)cosθ+G4(T)sinθ}, where (5.10)H(T)=Z2πλT3exp(−Z24λT−λnT),G1(T)=λn2(e−λnT−cosT)−λnsinT1+λn2,G2(T)=λn(e−λnT−cosT)+λn2sinT1+λn2,G3(T)=λn2e−λnT−λnsinT+cosT1+λn2,G4(T)=λn(cosT−e−λnT)+sinT1+λn2, and λn=n2π2λ/4,Z=z/h  and the remaining variables are identical to those
defined in (3.11).


### 5.3. Results

For
the first problem, the velocity profiles for variousZ-sections for the first and second problems are drawn in
Figure 7 and
Figure 8, respectively. The developments in these two
figures are similar to
those shown in previous section (Figures 5 and  6). The comparison between
the surface velocity (solid curves) and the wall velocity (dash curves) is made
in Figure 9 for Z=1 and Z=−1.
The phase lag and the smaller amplitude for the surface velocity profile at Z=1 are found. Besides, the amplitude of the surface
velocity profile for Z=−1 is much smaller than that for Z=1.
In conclusion, solutions in this section combine the characteristics of problems
in Section 3 (for a finite-depth case) and Section 4 (for relatively moving
planes).Figure 7
Velocity profiles of the first problem for a finite-depth fluid motivated by
relatively moving planes.Figure 8
Velocity profiles of the second problem for a finite-depth fluid motivated
by relatively moving planes.Figure 9
Comparison between the surface velocity (solid curve) and the wall velocity
(dash curve) of the second problem for a finite-depth fluid motivated by
relatively moving planes.


## 5.1. Solution to the First
Problem

The
governing equation and associated conditions are expressed as(5.1a)ut=ν(uyy+uzz),(5.1b)u(y=0,z>0,t>0)=u0,(5.1c)u(y=0,z<0,t>0)=0,(5.1d)uy(y=h,z,t>0)=0,(5.1e)u(y,z→±∞,t)isfinite,(5.1f)u(y>0,z,t=0)=0.
The
direct integral transforms cannot work anymore in solving (5.1a),
(5.1b), (5.1c),
(5.1d), (5.1e),
(5.1f) as this flow
system is bounded by Dirichlet and Neumann conditions at
y=0
and
y=h,
respectively. For the sake of eliminating different types of boundary
conditions, the concept of symmetry is employed to generate a symmetrical flow.
Accordingly, the boundary condition (5.1d) is replaced by (5.2)u(y=2h,z,t>0)=u0,where
the flow domain is doubly expanded. Similar to the method introduced in Section
4, the flow system can be decomposed into two subsystems in which U1 is a half of (3.5) and u2 for the domain z>0 is governed by (5.3)u2,t=ν(u2,yy+u2,zz),u2(y=0,z>0,t>0)=u02,u2(y=2h,z>0,t>0)=u02,u2(y>0,z=0,t>0)=0,u2(y,z→+∞,t)isfinite,u2(y>0,z,t=0)=0. By applying the Laplace transform to (5.3), it generates (5.4)su^2=ν(u^2,yy+u^2,zz),u^2(y=0,z>0,s)=u02s,u^2(y=2h,z>0,s)=u02s,u^2(y,z→+∞,t)is  finite,u^2(y>0,z=0,s)=0. The system of u^2 is now shifted to the system of u*(5.5)su*+u02=ν(u2,yy*+u2,zz*),u*(y=0,z>0,s)=0,u*(y=2h,z>0,s)=0,u*(y,z→+∞,t)is  finite,u*(y>0,z=0,s)=−u02s, in
which u*=u^−u0/2s.
After applying the finite sine transform, u˜(s,n,y)=∫02hu*sin(nπy/2h)dy,
to (5.5), it yields (5.6)u˜zz−(sν+n2π24h2)u˜=u0hnπν(1−(−1)n),with
the boundedness condition for z→∞ and u˜(z=0)=−(u0h/nπν)(1−(−1)n).
The solution is (5.7)u˜=(1−(−1)n)⋅[u0hnπνα(exp(−αz)−1)−u0hnπsexp(−αz)],where α=s/ν+n2π2/4h2.
Applying the inverse transforms with the help of the equations in the
second and third rows of Table 1,
the dimensionless solution is solved after some algebra (5.8)U2=12+1π∑n=1∞1−(−1)nnsinnπY2⋅{−exp(−n2π2T4)erf(Z2T)−∫0TZ2πT′3exp(−n2π2T′4−Z24T′)dT′}, in
which Z=z/h and the remaining dimensionless variables are
identical to those shown in (3.6).


## 5.2. Solution
to the Second Problem

Using
the techniques described in previous subsections, the dimensionless solution to
the second problem is the summation of a half of (3.10) and the following
solution (5.9)U2=12cos(T+θ)+1π∑n=1∞1−(−1)nnsinnπY2⋅{cosθ∫0TG1(T−u)H(u)du+sinθ∫0TG2(T−u)H(u)du−G3(T)cosθ+G4(T)sinθ}, where (5.10)H(T)=Z2πλT3exp(−Z24λT−λnT),G1(T)=λn2(e−λnT−cosT)−λnsinT1+λn2,G2(T)=λn(e−λnT−cosT)+λn2sinT1+λn2,G3(T)=λn2e−λnT−λnsinT+cosT1+λn2,G4(T)=λn(cosT−e−λnT)+sinT1+λn2, and λn=n2π2λ/4,Z=z/h  and the remaining variables are identical to those
defined in (3.11).


## 5.3. Results

For
the first problem, the velocity profiles for variousZ-sections for the first and second problems are drawn in
Figure 7 and
Figure 8, respectively. The developments in these two
figures are similar to
those shown in previous section (Figures 5 and  6). The comparison between
the surface velocity (solid curves) and the wall velocity (dash curves) is made
in Figure 9 for Z=1 and Z=−1.
The phase lag and the smaller amplitude for the surface velocity profile at Z=1 are found. Besides, the amplitude of the surface
velocity profile for Z=−1 is much smaller than that for Z=1.
In conclusion, solutions in this section combine the characteristics of problems
in Section 3 (for a finite-depth case) and Section 4 (for relatively moving
planes).Figure 7
Velocity profiles of the first problem for a finite-depth fluid motivated by
relatively moving planes.Figure 8
Velocity profiles of the second problem for a finite-depth fluid motivated
by relatively moving planes.Figure 9
Comparison between the surface velocity (solid curve) and the wall velocity
(dash curve) of the second problem for a finite-depth fluid motivated by
relatively moving planes.


## 6. Conclusions

One- and two-dimensional
viscous flows of either a finite or infinite depth generated by moving plane(s)
are theoretically examined. Traditional Stokes’ problems are firstly revisited
and three extended problems are subsequently examined. Mathematical techniques
used in the derivation processes include the Laplace transform, the Fourier transform, and transforming the original flow into either
a symmetrical or antisymmetrical flow. Complete solutions, that is, exact
solutions, derived in this study can precisely capture the flow characteristics
at any time. As for the transient solution, it can be obtained by subtracting
the steady-state solution from the complete solution. The shear stress on the
boundary, which indicates external force driving the flow, is also readily
obtained from the velocity profile. Results of Sections3 to
5 are also depicted
and discussed.The field most related
to the present study is the heat-conduction problem. For a traditional problem
of a rod (only one-dimensional dependence in they direction) heated at one side, it owns the
diffusion-type differential equation (referred to (2.1a)), Tt=αTyy,
in which T is the temperature and α the thermal diffusivity. If one end of the rod
is suddenly heated with a constant temperature or harmonically heated, the
solution is referred to the aforementioned Stokes’ first or second problem. Based
on techniques and results given in this paper, the two-dimensional heat-conduction
problem for a plate can be studied through a purely mathematical way.Current solutions are also
valuable to other fields which include chemical engineering, mechanical
manufacturing, and geophysical science. On the basis of the mathematical method
provided in the present study, in the coming future, more conditions on the
boundary could be considered for widely engineering applications.



---

*Source: 754262-2008-10-21.xml*