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For both viscous and linear viscoelastic fluids, Stokes’ first and second problems are benchmark problems in fluid mechanics and have many applications. They are both exact solutions to the Navier-Stokes equation but have different boundary conditions. In both

problems, there is a semi-infinite space above a flat infinite plane. In Stokes’ 1st problem, the system is initially at rest, and at time t = 0, the plane suddenly begins translating at

a velocity u = U. In Stokes’ 2nd problem, the infinite plane sinusoidally oscillates in

time with a velocityu = U0sin(ωt) whereω is the frequency of the oscillation. For both

problems, no pressure gradient exists, and as the motion of the infinite plate is only in the

xdirection, the fluid velocity is also only in the xdirection and only a function of y. The Navier-Stokes equation (Equation 6.11) in thexdirection reduces to the following for both Stokes’ 1st and 2nd problems with a purely viscous fluid,

∂u

∂t =ν

∂2u

∂y2 (6.18)

whereu is the fluid velocity in the x direction, and ν is the kinematic viscosity η/ρ. For

Stokes’ 1st problem, the boundary conditions are

u(0,t)=      0 t_≤0, U t >0 (6.19) and the condition that the fluid velocity remain finite as we move far away from the plane (asy _{→ ∞}). The initial condition we apply isu(0,0) = 0, indicating the fluid is at rest at

time zero. As I am more interested in Stokes’ 2nd problem with the sinusoidally oscillating plane, I will give the solution to Stokes’ 1st problem:

u(y,t)= U � 1−erf � _y 2√νt �� (6.20) where erf is the error function.

For Stokes’ 2nd problem, the boundary condition of a finite velocity asy_{→ ∞}remains, and in addition, u(0,t)=      0 t _≤0. Usin(ωt) t >0 (6.21) The initial condition is stillu(0,0)=0. We can assume a solution of the form

u(y,t)= ��f(y)eiωt� (6.22)

where� indicates the imaginary part of the solution and f(y) is the amplitude of the os- cillation as a function of the distance from the oscillating plane. Substituting this solution into Equation 6.18 gives

��iωf(y)eiωt� _=ν��d2f(y) dy2 eiωt � =⇒ d 2_f(y) dy2 = iω ν f(y). (6.23)

This second order homogeneous differential equation has a solution of the form

wherec1andc2are constants determined using boundary and initial conditions, andr1, and

r2are roots of the characteristic equation. Immediately, we know f(y) must remain finite as

y_{→ ∞}, which does not happen for the first term. Thusc1 =0, and we are left with

f(y)= c2e−r2y (6.25)

where substitution into the original differential equation, Equation 6.23, gives c2r22e−r2y = iω

ν c2e

−r2y_{. (6.26)}

From this, we can see thatr2= √iω/ν. The solution then becomes

f(y)= c2e− √

iω/νy_{, (6.27)}

and inserting this back into the original solution, Equation 6.22, gives

u(y,t)=��c2e− √

iω/νy_eiωt�_{. (6.28)}

Before taking the imaginary part of this solution, I will substitute √i = (1+ i)/√2 into

the above equation. Making this substitution and taking the imaginary part using Euler’s relationeiθ _{=cosθ+isinθgives}

u(y,t)= c2e− √ω 2ν ysin � ωt₋ � ω 2ν y � . (6.29)

We can use the final set of boundary conditions on the velocity whereu(0,t) = Usin(ωt),

implying thatc2 =U. The solution to Stokes’ 2nd problem for a viscous fluid then becomes

u(y,t)= Ue−√2ων ysin � ωt₋ � ω 2ν y � , (6.30)

a solution which describes an oscillating velocity, maximally valued at the location of the oscillating boundary and decaying as a function of distanceyfrom the boundary.

It is interesting to understand from Equation 6.30 the length scale associated with the penetration of the motion of the lower boundary into the fluid. This length scale is called the penetration depth and is defined as the heighty above the oscillating plate where the velocity is equal to 1/eits original value. Since the value of sine in Equation 6.30 has a

maximum value of one, we can write 1

eU = Ue−

√ω

2ν α (6.31)

whereαis the penetration depth. After some algebra, we have

α=

�

2ν

ω, (6.32)

illustrating that as the viscosity of the fluid above the oscillating boundary increases, the penetration depth increases, and the effects of the boundary are felt farther into the fluid.

This result is depicted in Figure 6.2 with a boundary located at y = 0 oscillating at ω =

1 Hz. Each plot represents a different kinematic viscosity (ν = 1,2,4). In an individual

a velocity equal to the x-intercept. As the plane oscillates, the color of the lines changes from red to green to blue. As the viscosity increases, the coupling between the boundary and the fluid increases as can be seen by the increased penetration depth and the lack of a sharp decrease in velocity with increased height. Conversely, though it is not illustrated in the figure, as the oscillating frequency of the boundary increases, the penetration depth decreases.

direction of plane

Figure 6.2: A simulation of Stokes’ 2nd problem for a purely viscous fluid illustrates the effects of an increasing viscosity on the propagation of stress throughout the fluid. Each

plot represents a different viscosity (ν= 1,2,4), and within a plot, a line indicates an instant

in time after a steady state has been reached. Red lines correspond to earlier times and blue lines correspond to later times. The frequency of the oscillating plane isω =1 Hz. As the