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latex代写-A1

时间：2021-11-11

Project A1

Modelling the flow of a thin film of fluid over a

horizontal substrate

Introduction

The flow of thin films of viscous fluids, such as honey or syrup, can be effectively modelled

by identifying the dominant forces driving the flow and applying the approximations of

lubrication theory, which hold as long as the thickness of the layer of fluid is much smaller

than its horizontal extent. A schematic diagram of a typical thin film of viscous fluid,

of thickness H, spreading under gravity over a horizontal substrate is shown in figure 1.

Such flows are also referred to as viscous gravity currents, and are common in the world

around us, ranging from the spread of honey over morning toast, to lava flows and the

flow of glacial ice sheets [1, 5, 6, 7, for example].

Assuming that vertical shear stress provides the dominant resistance to the flow,

and that inertial effects and the effects of surface tension are negligible, the flow can be

shown to be governed by the following momentum balance equation

0 = −∇p− ρgez + µ

∂2u

∂z2

, (1)

where the velocity field u = (u, 0, 0) is horizontal, and ez = (0, 0, 1) is the unit basis

vector in the z-direction. Here, the pressure p and velocity u may, in general, depend

on all spatial coordinates (x, y, z) as well as on time t, whereas the density ρ and dy-

namic viscosity µ are constants which describe the material properties of the fluid. The

gravitational acceleration is denoted by g. Axes are chosen so that the fluid flows along

the x-direction, z = 0 describes the horizontal substrate, and z = H describes the upper

surface of the thin film of viscous fluid as shown in figure 1. The viscous fluid is supplied

at a specified flux at x = 0, and its leading edge is given by x = xN (t). The setup is

fully two-dimensional, with no variation along the y-direction.

Equation (1) is a version of the so-called Stokes equations, used to describe the flow

of viscous fluids.

xN(t)

H(x,t)

x

z

inflow

Figure 1: Schematic diagram illustrating the side profile of a thin film of viscous fluid

spreading under gravity over a rigid horizontal substrate.

1

Problem 1: hydrostatic pressure

Solve for the pressure by considering the z-component of (1) and applying the boundary

condition that the pressure attains its atmospheric value p0 at the upper surface of the

viscous fluid, that is

p = p0 at z = H. (2)

Problem 2: equation for the horizontal velocity

By substituting your expression for the pressure into the x-component of (1) and as-

suming that the atmospheric pressure p0 is constant, show that the horizontal velocity

u satisfies

∂2u

∂z2

=

ρg

µ

∂H

∂x

. (3)

Problem 3: solving the Stokes equations

Noting that the right-hand side of (3) is independent of z, integrate both side of (3)

twice in z in order to obtain an expression for u in terms of two arbitrary constants, A

and B.

Problem 4: velocity field

Solve for A and B by applying the boundary condition that the upper surface is stress

free, that is,

∂u

∂z

= 0 at z = H, (4)

and that the fluid does not slip at the contact line with the substrate, that is,

u = 0 at z = 0. (5)

The latter is also known as the no-slip condition.

Problem 5: depth-integrated flux

The depth-integrated flux of fluid per unit width is given by

q =

∫ H

0

u dz. (6)

After substituting in your determined values of A and B into your expression for the

velocity, show that the depth integrated flux is given by

q = −CHk

∂H

∂x

. (7)

where C and k are constants that you should determine.

2

Problem 6: Similarity solution

Along with the equation (7) specifying the flux q, the deformation of the upper surface

is governed by the mass conservation equation

∂H

∂t

= −

∂q

∂x

, (8)

along with the source flux boundary condition

q = Q at x = 0, (9)

the vanishing frontal thickness condition

H = 0 at x = xN (t), (10)

and the global mass conservation condition

∫ xN

0

H dx = Qt. (11)

By rescaling the surface height H as

H(x, t) = αtaF (η) (12)

where

η =

x

βtb

and xN = ηNβt

b, (13)

for some constant ηN , find the values of a, b, α and β so that the governing equations

(7) and (8) reduce to the single ordinary differential equation

1

5

F −

4

5

η

dF

dη

=

1

3

d

dη

(

F 3

dF

dη

)

, (14)

and recast the boundary and integral conditions (9)–(11) in terms of only F and its

derivatives. No physical constants, such as µ, ρ, Q, g, and ηN

1 and no time variable t

should appear in these reduced boundary and integral conditions. If any of these appear

in your boundary and integral conditions, that means that your values of a, b, α and

β should be adjusted. A similar, though differently scaled, derivation can be found in

[4] for Newtonian viscous fluids and in [8] for non-Newtonian, power-law fluids. A more

general treatment of self-similar problems, and ways of obtaining similarity solutions,

can be found in [2].

Note: the solution F is known as a similarity solution and the variable η is known

as a similarity variable. Note also that the front x = xN corresponds to η = ηN under

the transformation (13).

1

ηN will actually appear, but only in your transformed version of (10) and (11).

3

Problem 7: Asymptotic solution near the front

In the neighbourhood of the front η = ηN (or, equivalently, x = xN ), the thickness of the

liquid layer is small, while its slope is large. Find an asymptotic solution by following

the steps outlined below.

(a) By changing variables so that

η = ηN − ǫX, (15)

where ǫ is a small parameter, and rescaling F as F = ǫ1/3G, determine an ordinary

differential equation forG in terms ofX only (no η should appear in your equation).

(b) Identify any terms that are small in comparison to the remaining terms, when

ǫ→ 0.

(c) Formulate an approximate equation for G by deleting the small terms identified

above. Your approximate equation should have no ǫ’s and no η’s in it; just ηN and

G as a function of X. If you have any ǫ’s in your equation, that means that you

either did not delete all the small terms, or you did not divide both sides of the

equation by the right power of ǫ.

(d) Show that the solution to this approximate equation is of the form

G ∼ γXr (16)

where γ and r are constants that you should determine.

(e) Recast your approximate solution for G in terms of F and η. This is known as an

asymptotic solution, and it is valid near the front η = ηN .

(f) For what value of ηN does your asymptotic solution for F satisfy the source flux

boundary condition (the version of (9) recast in similarity coordinates for Problem

5)?

(g) Plot the asymptotic solution in Mathematica, or another program, as a function

of η over the domain [0, ηN ] for the value of ηN found in part (f), above.

A more general overview of asymptotics methods, and ways of obtaining asymptotic

solutions to differential equations, can be found in [3].

References

[1] N. J. Balmforth and R. V. Craster. Dynamics of cooling domes of viscoplastic fluid.

J. Fluid Mech., 422:225–248, 2000.

[2] G. I. Barenblatt. Scaling. Cambridge University Press, 2003.

4

[3] E. J. Hinch. Perturbation Methods. Cambridge University Press, 1991.

[4] H. E. Huppert. The propagation of two-dimensional and axisymmetric viscous gravity

currents over a rigid horizontal surface. J. Fluid Mech., 121:43–58, 1982.

[5] H. E. Huppert. Gravity currents: a personal perspective. J. Fluid Mech., 554:299–

322, 2006.

[6] K. N. Kowal and S. S. Pegler M. G. Worster. Dynamics of laterally confined marine

ice sheets. J. Fluid Mech., 790, 2016.

[7] R. A. V. Robison, H. E. Huppert, and M. G. Worster. Dynamics of viscous grounding

lines. J. Fluid Mech., 648:363–380, 2010.

[8] R. Sayag and M. G. Worster. Axisymmetric gravity currents of power-law fluids over

a rigid horizontal surface. J. Fluid Mech., 716, 2013. R5.

5

Project A2

A numerical treatment of similarity solutions

in the real world: thin film flows

Introduction

Many physical phenomena exhibit so-called self-similar behaviour, in which the physical

system grows or decays in time while retaining its general shape and structure, albeit

in rescaled (self-similar) coordinates. For such systems, any initial irregularities (as

formulated via initial conditions) eventually evolve towards a self-similar regime, in which

these initial irregularies become ‘forgotten’ with time. This project will lead you through

an example of such a physical system by examining the fluid mechanics of a thin film

of viscous fluid spreading under gravity over a horizontal, rigid substrate. Such flows

are also referred to as viscous gravity currents, and are common in the world around

us, ranging from the spread of honey over morning toast, to lava flows and the flow of

glacial ice sheets [1, 5, 6, 7, for example].

The key feature behind such flows is that they involve very viscous fluids, such as

honey, or syrup, and as such, viscous forces determine the flow, and any inertial effects

are negligible. Under appropriate assumptions, including that the flow is long and thin,

and resisted dominantly by vertical shear stresses, and that inertial effects and the effects

of surface tension are negligible, the depth-integrated flux of viscous fluid per unit width

can be shown to be given by

q = −

ρg

3µ

H3

∂H

∂x

, (1)

where ρ, µ and H = H(x, t) are the density, dynamic viscosity, and thickness of the

viscous fluid, respectively, and g is the acceleration due to gravity. Here, the flow is fully

two-dimensional, with axes chosen so that the fluid flows along the x-direction, z = 0

describes the horizontal substrate, z = H describes the upper surface of the thin film

of viscous fluid, and there is no variation along the y-direction. The viscous fluid is

supplied at a specified flux at x = 0, and its leading edge, or frontal position, is given

by x = xN (t) as shown in figure 1.

xN(t)

H(x,t)

x

z

inflow

Figure 1: Schematic diagram illustrating the side profile of a thin film of viscous fluid

spreading under gravity over a rigid horizontal substrate.

1

Along with the equation (1) specifying the flux q, the deformation of the upper

surface is governed by the mass conservation equation

∂H

∂t

= −

∂q

∂x

, (2)

along with the source flux boundary condition

q = Q at x = 0, (3)

the vanishing frontal thickness condition

H = 0 at x = xN (t), (4)

and the global mass conservation condition

∫ xN

0

H dx = Qt. (5)

Problem 1: Similarity solution

By rescaling the surface height H as

H(x, t) = αtaF (η) (6)

where the similarity variable η is given by

η =

x

βtb

and xN = ηNβt

b, (7)

for some constant ηN , find the values of a, b, α and β so that the governing equations

(1) and (2) reduce to the single ordinary differential equation

1

5

F −

4

5

η

dF

dη

=

1

3

d

dη

(

F 3

dF

dη

)

, (8)

and recast the boundary and integral conditions (3)–(5) in terms of only F and its

derivatives. No physical constants, such as µ, ρ, Q, g, and ηN

1 and no time variable t

should appear in these reduced boundary and integral conditions. If any of these appear

in your boundary and integral conditions, that means that your values of a, b, α and

β should be adjusted. A similar, though differently scaled, derivation can be found in

[4] for Newtonian viscous fluids and in [8] for non-Newtonian, power-law fluids. A more

general treatment of self-similar problems, and ways of obtaining similarity solutions,

can be found in [2].

Note: the solution F is known as a similarity solution and the variable η is known

as a similarity variable. Note also that the front x = xN corresponds to η = ηN under

the transformation (7).

1

ηN will actually appear, but only in your transformed version of (4) and (5).

2

Problem 2: Asymptotic solution near the front

In the neighbourhood of the front η = ηN (or, equivalently, x = xN ), the thickness of the

liquid layer is small, while its slope is large. Find an asymptotic solution by following

the steps outlined below.

(a) By changing variables so that

η = ηN − ǫX, (9)

where ǫ is a small parameter, and rescaling F as F = ǫ1/3G, determine an ordinary

differential equation forG in terms ofX only (no η should appear in your equation).

(b) Identify any terms that are small in comparison to the remaining terms, when

ǫ→ 0.

(c) By deleting the small terms identified above, show that G satisfies the following

approximate equation

4

5

ηNG

′ =

1

3

(G3G′)′. (10)

(d) Show, by substitution, that this approximate equation has a solution of the form

G ∼ γXr (11)

where γ and r are constants that you should determine.

(e) Recast your approximate solution for G in terms of F and η, and call it Fasym.

You should have no ǫ’s in your formula for Fasym. This is known as an asymptotic

solution, and it is valid near the front η = ηN .

A more general overview of asymptotics methods, and ways of obtaining asymptotic

solutions to differential equations, can be found in [3].

Problem 3: Numerical solution – free boundary problem

Follow the steps outlined below to solve the ordinary differential equation (8), subject

to appropriate boundary conditions, in Mathematica.

(a) Near the front of the viscous fluid (that is, near η = ηN ), the slope of the solution

is negative and large, approaching −∞ as η → ηN . All numerical solvers, no

matter how good, run into problems when the desired solution becomes infinite at

any point in the domain. To eliminate such problems, instead of solving over the

entire interval [0, ηN ], you should instead solve over a shorter interval [0, ηN − ǫ],

where ǫ is a small number. You should use the asymptotic solution Fasym found

in Problem 2(e), as the solution for the interval [ηN − ǫ, ηN ] – the remaining part

of the interval. The asymptotic solution will also be useful in order to formulate

appropriate boundary conditions at η = ηN − ǫ, in order for the numerical solver

3

to be able to give you the correct numerical solution. To do so, substitute your

formula for the asymptotic solution Fasym into the boundary conditions formulated

below:

F (ηN − ǫ) = Fasym(ηN − ǫ) (12)

F ′(ηN − ǫ) = F

′

asym(ηN − ǫ) (13)

(b) Use the Mathematica built-in numerical differential equation solver NDSolve to

solve the ordinary differential equation (8) subject to the boundary conditions

(12) and (13) over the interval [0, ηN − ǫ], assuming that ηN = 1 and ǫ = 0.001,

for now. You will want to be able to change the parameters ηN and ǫ later on,

so you should keep these parameters general when typing in (8), (12) and (13)

into Mathematica, and only assign a specific value to them in NDSolve using a

substitution rule – see

https://reference.wolfram.com/language/ref/Rule.html.

Documentation on NDSolve can be found here:

https://reference.wolfram.com/language/ref/NDSolve.html

(c) Once you obtain the solution, plot the surface height F (η) as a function of η over

the interval [0, ηN − ǫ], save it (right-click the plot in Mathematica, and select Save

Graphic As) and insert your plot into your write-up.

(d) Use your numerical solution obtained in part (c) above to evaluate the flux

qsim = −

1

3

F 3(0)F ′(0) (14)

at η = 0.

(e) According to one of your boundary conditions obtained for Problem 1, this flux,

given in (14), should be 1. As the answer you obtained in part (d), above, is not

equal to 1, this means that the parameter ηN should be adjusted. Recall we have

just arbitrarily chosen ηN = 1 in part (c), above, for the sake of obtaining just

some numerical solution. You should now manually adjust (or write a short script

to programatically adjust) the value of ηN so that the boundary condition

qsim = −

1

3

F 3(0)F ′(0) = 1 (15)

is satisfied at η = 0, to within a tolerance of 10−3. State the value of ηN that

you have found. Plot the surface height F (η) as a function of η over the interval

[0, ηN − ǫ], save it and insert your plot into your write-up.

(f) Overlay on top of your previous plot a plot of the asymptotic solution Fasym over

the whole interval [0, ηN ]. Save this plot and add it to your write-up.

(g) Comment on how well the asymptotic solution approximates the numerical solution

over the entire interval. In which part of the interval is the fit best?

4

References

[1] N. J. Balmforth and R. V. Craster. Dynamics of cooling domes of viscoplastic fluid.

J. Fluid Mech., 422:225–248, 2000.

[2] G. I. Barenblatt. Scaling. Cambridge University Press, 2003.

[3] E. J. Hinch. Perturbation Methods. Cambridge University Press, 1991.

[4] H. E. Huppert. The propagation of two-dimensional and axisymmetric viscous gravity

currents over a rigid horizontal surface. J. Fluid Mech., 121:43–58, 1982.

[5] H. E. Huppert. Gravity currents: a personal perspective. J. Fluid Mech., 554:299–

322, 2006.

[6] K. N. Kowal and S. S. Pegler M. G. Worster. Dynamics of laterally confined marine

ice sheets. J. Fluid Mech., 790, 2016.

[7] R. A. V. Robison, H. E. Huppert, and M. G. Worster. Dynamics of viscous grounding

lines. J. Fluid Mech., 648:363–380, 2010.

[8] R. Sayag and M. G. Worster. Axisymmetric gravity currents of power-law fluids over

a rigid horizontal surface. J. Fluid Mech., 716, 2013. R5.

5

学霸联盟

Modelling the flow of a thin film of fluid over a

horizontal substrate

Introduction

The flow of thin films of viscous fluids, such as honey or syrup, can be effectively modelled

by identifying the dominant forces driving the flow and applying the approximations of

lubrication theory, which hold as long as the thickness of the layer of fluid is much smaller

than its horizontal extent. A schematic diagram of a typical thin film of viscous fluid,

of thickness H, spreading under gravity over a horizontal substrate is shown in figure 1.

Such flows are also referred to as viscous gravity currents, and are common in the world

around us, ranging from the spread of honey over morning toast, to lava flows and the

flow of glacial ice sheets [1, 5, 6, 7, for example].

Assuming that vertical shear stress provides the dominant resistance to the flow,

and that inertial effects and the effects of surface tension are negligible, the flow can be

shown to be governed by the following momentum balance equation

0 = −∇p− ρgez + µ

∂2u

∂z2

, (1)

where the velocity field u = (u, 0, 0) is horizontal, and ez = (0, 0, 1) is the unit basis

vector in the z-direction. Here, the pressure p and velocity u may, in general, depend

on all spatial coordinates (x, y, z) as well as on time t, whereas the density ρ and dy-

namic viscosity µ are constants which describe the material properties of the fluid. The

gravitational acceleration is denoted by g. Axes are chosen so that the fluid flows along

the x-direction, z = 0 describes the horizontal substrate, and z = H describes the upper

surface of the thin film of viscous fluid as shown in figure 1. The viscous fluid is supplied

at a specified flux at x = 0, and its leading edge is given by x = xN (t). The setup is

fully two-dimensional, with no variation along the y-direction.

Equation (1) is a version of the so-called Stokes equations, used to describe the flow

of viscous fluids.

xN(t)

H(x,t)

x

z

inflow

Figure 1: Schematic diagram illustrating the side profile of a thin film of viscous fluid

spreading under gravity over a rigid horizontal substrate.

1

Problem 1: hydrostatic pressure

Solve for the pressure by considering the z-component of (1) and applying the boundary

condition that the pressure attains its atmospheric value p0 at the upper surface of the

viscous fluid, that is

p = p0 at z = H. (2)

Problem 2: equation for the horizontal velocity

By substituting your expression for the pressure into the x-component of (1) and as-

suming that the atmospheric pressure p0 is constant, show that the horizontal velocity

u satisfies

∂2u

∂z2

=

ρg

µ

∂H

∂x

. (3)

Problem 3: solving the Stokes equations

Noting that the right-hand side of (3) is independent of z, integrate both side of (3)

twice in z in order to obtain an expression for u in terms of two arbitrary constants, A

and B.

Problem 4: velocity field

Solve for A and B by applying the boundary condition that the upper surface is stress

free, that is,

∂u

∂z

= 0 at z = H, (4)

and that the fluid does not slip at the contact line with the substrate, that is,

u = 0 at z = 0. (5)

The latter is also known as the no-slip condition.

Problem 5: depth-integrated flux

The depth-integrated flux of fluid per unit width is given by

q =

∫ H

0

u dz. (6)

After substituting in your determined values of A and B into your expression for the

velocity, show that the depth integrated flux is given by

q = −CHk

∂H

∂x

. (7)

where C and k are constants that you should determine.

2

Problem 6: Similarity solution

Along with the equation (7) specifying the flux q, the deformation of the upper surface

is governed by the mass conservation equation

∂H

∂t

= −

∂q

∂x

, (8)

along with the source flux boundary condition

q = Q at x = 0, (9)

the vanishing frontal thickness condition

H = 0 at x = xN (t), (10)

and the global mass conservation condition

∫ xN

0

H dx = Qt. (11)

By rescaling the surface height H as

H(x, t) = αtaF (η) (12)

where

η =

x

βtb

and xN = ηNβt

b, (13)

for some constant ηN , find the values of a, b, α and β so that the governing equations

(7) and (8) reduce to the single ordinary differential equation

1

5

F −

4

5

η

dF

dη

=

1

3

d

dη

(

F 3

dF

dη

)

, (14)

and recast the boundary and integral conditions (9)–(11) in terms of only F and its

derivatives. No physical constants, such as µ, ρ, Q, g, and ηN

1 and no time variable t

should appear in these reduced boundary and integral conditions. If any of these appear

in your boundary and integral conditions, that means that your values of a, b, α and

β should be adjusted. A similar, though differently scaled, derivation can be found in

[4] for Newtonian viscous fluids and in [8] for non-Newtonian, power-law fluids. A more

general treatment of self-similar problems, and ways of obtaining similarity solutions,

can be found in [2].

Note: the solution F is known as a similarity solution and the variable η is known

as a similarity variable. Note also that the front x = xN corresponds to η = ηN under

the transformation (13).

1

ηN will actually appear, but only in your transformed version of (10) and (11).

3

Problem 7: Asymptotic solution near the front

In the neighbourhood of the front η = ηN (or, equivalently, x = xN ), the thickness of the

liquid layer is small, while its slope is large. Find an asymptotic solution by following

the steps outlined below.

(a) By changing variables so that

η = ηN − ǫX, (15)

where ǫ is a small parameter, and rescaling F as F = ǫ1/3G, determine an ordinary

differential equation forG in terms ofX only (no η should appear in your equation).

(b) Identify any terms that are small in comparison to the remaining terms, when

ǫ→ 0.

(c) Formulate an approximate equation for G by deleting the small terms identified

above. Your approximate equation should have no ǫ’s and no η’s in it; just ηN and

G as a function of X. If you have any ǫ’s in your equation, that means that you

either did not delete all the small terms, or you did not divide both sides of the

equation by the right power of ǫ.

(d) Show that the solution to this approximate equation is of the form

G ∼ γXr (16)

where γ and r are constants that you should determine.

(e) Recast your approximate solution for G in terms of F and η. This is known as an

asymptotic solution, and it is valid near the front η = ηN .

(f) For what value of ηN does your asymptotic solution for F satisfy the source flux

boundary condition (the version of (9) recast in similarity coordinates for Problem

5)?

(g) Plot the asymptotic solution in Mathematica, or another program, as a function

of η over the domain [0, ηN ] for the value of ηN found in part (f), above.

A more general overview of asymptotics methods, and ways of obtaining asymptotic

solutions to differential equations, can be found in [3].

References

[1] N. J. Balmforth and R. V. Craster. Dynamics of cooling domes of viscoplastic fluid.

J. Fluid Mech., 422:225–248, 2000.

[2] G. I. Barenblatt. Scaling. Cambridge University Press, 2003.

4

[3] E. J. Hinch. Perturbation Methods. Cambridge University Press, 1991.

[4] H. E. Huppert. The propagation of two-dimensional and axisymmetric viscous gravity

currents over a rigid horizontal surface. J. Fluid Mech., 121:43–58, 1982.

[5] H. E. Huppert. Gravity currents: a personal perspective. J. Fluid Mech., 554:299–

322, 2006.

[6] K. N. Kowal and S. S. Pegler M. G. Worster. Dynamics of laterally confined marine

ice sheets. J. Fluid Mech., 790, 2016.

[7] R. A. V. Robison, H. E. Huppert, and M. G. Worster. Dynamics of viscous grounding

lines. J. Fluid Mech., 648:363–380, 2010.

[8] R. Sayag and M. G. Worster. Axisymmetric gravity currents of power-law fluids over

a rigid horizontal surface. J. Fluid Mech., 716, 2013. R5.

5

Project A2

A numerical treatment of similarity solutions

in the real world: thin film flows

Introduction

Many physical phenomena exhibit so-called self-similar behaviour, in which the physical

system grows or decays in time while retaining its general shape and structure, albeit

in rescaled (self-similar) coordinates. For such systems, any initial irregularities (as

formulated via initial conditions) eventually evolve towards a self-similar regime, in which

these initial irregularies become ‘forgotten’ with time. This project will lead you through

an example of such a physical system by examining the fluid mechanics of a thin film

of viscous fluid spreading under gravity over a horizontal, rigid substrate. Such flows

are also referred to as viscous gravity currents, and are common in the world around

us, ranging from the spread of honey over morning toast, to lava flows and the flow of

glacial ice sheets [1, 5, 6, 7, for example].

The key feature behind such flows is that they involve very viscous fluids, such as

honey, or syrup, and as such, viscous forces determine the flow, and any inertial effects

are negligible. Under appropriate assumptions, including that the flow is long and thin,

and resisted dominantly by vertical shear stresses, and that inertial effects and the effects

of surface tension are negligible, the depth-integrated flux of viscous fluid per unit width

can be shown to be given by

q = −

ρg

3µ

H3

∂H

∂x

, (1)

where ρ, µ and H = H(x, t) are the density, dynamic viscosity, and thickness of the

viscous fluid, respectively, and g is the acceleration due to gravity. Here, the flow is fully

two-dimensional, with axes chosen so that the fluid flows along the x-direction, z = 0

describes the horizontal substrate, z = H describes the upper surface of the thin film

of viscous fluid, and there is no variation along the y-direction. The viscous fluid is

supplied at a specified flux at x = 0, and its leading edge, or frontal position, is given

by x = xN (t) as shown in figure 1.

xN(t)

H(x,t)

x

z

inflow

Figure 1: Schematic diagram illustrating the side profile of a thin film of viscous fluid

spreading under gravity over a rigid horizontal substrate.

1

Along with the equation (1) specifying the flux q, the deformation of the upper

surface is governed by the mass conservation equation

∂H

∂t

= −

∂q

∂x

, (2)

along with the source flux boundary condition

q = Q at x = 0, (3)

the vanishing frontal thickness condition

H = 0 at x = xN (t), (4)

and the global mass conservation condition

∫ xN

0

H dx = Qt. (5)

Problem 1: Similarity solution

By rescaling the surface height H as

H(x, t) = αtaF (η) (6)

where the similarity variable η is given by

η =

x

βtb

and xN = ηNβt

b, (7)

for some constant ηN , find the values of a, b, α and β so that the governing equations

(1) and (2) reduce to the single ordinary differential equation

1

5

F −

4

5

η

dF

dη

=

1

3

d

dη

(

F 3

dF

dη

)

, (8)

and recast the boundary and integral conditions (3)–(5) in terms of only F and its

derivatives. No physical constants, such as µ, ρ, Q, g, and ηN

1 and no time variable t

should appear in these reduced boundary and integral conditions. If any of these appear

in your boundary and integral conditions, that means that your values of a, b, α and

β should be adjusted. A similar, though differently scaled, derivation can be found in

[4] for Newtonian viscous fluids and in [8] for non-Newtonian, power-law fluids. A more

general treatment of self-similar problems, and ways of obtaining similarity solutions,

can be found in [2].

Note: the solution F is known as a similarity solution and the variable η is known

as a similarity variable. Note also that the front x = xN corresponds to η = ηN under

the transformation (7).

1

ηN will actually appear, but only in your transformed version of (4) and (5).

2

Problem 2: Asymptotic solution near the front

In the neighbourhood of the front η = ηN (or, equivalently, x = xN ), the thickness of the

liquid layer is small, while its slope is large. Find an asymptotic solution by following

the steps outlined below.

(a) By changing variables so that

η = ηN − ǫX, (9)

where ǫ is a small parameter, and rescaling F as F = ǫ1/3G, determine an ordinary

differential equation forG in terms ofX only (no η should appear in your equation).

(b) Identify any terms that are small in comparison to the remaining terms, when

ǫ→ 0.

(c) By deleting the small terms identified above, show that G satisfies the following

approximate equation

4

5

ηNG

′ =

1

3

(G3G′)′. (10)

(d) Show, by substitution, that this approximate equation has a solution of the form

G ∼ γXr (11)

where γ and r are constants that you should determine.

(e) Recast your approximate solution for G in terms of F and η, and call it Fasym.

You should have no ǫ’s in your formula for Fasym. This is known as an asymptotic

solution, and it is valid near the front η = ηN .

A more general overview of asymptotics methods, and ways of obtaining asymptotic

solutions to differential equations, can be found in [3].

Problem 3: Numerical solution – free boundary problem

Follow the steps outlined below to solve the ordinary differential equation (8), subject

to appropriate boundary conditions, in Mathematica.

(a) Near the front of the viscous fluid (that is, near η = ηN ), the slope of the solution

is negative and large, approaching −∞ as η → ηN . All numerical solvers, no

matter how good, run into problems when the desired solution becomes infinite at

any point in the domain. To eliminate such problems, instead of solving over the

entire interval [0, ηN ], you should instead solve over a shorter interval [0, ηN − ǫ],

where ǫ is a small number. You should use the asymptotic solution Fasym found

in Problem 2(e), as the solution for the interval [ηN − ǫ, ηN ] – the remaining part

of the interval. The asymptotic solution will also be useful in order to formulate

appropriate boundary conditions at η = ηN − ǫ, in order for the numerical solver

3

to be able to give you the correct numerical solution. To do so, substitute your

formula for the asymptotic solution Fasym into the boundary conditions formulated

below:

F (ηN − ǫ) = Fasym(ηN − ǫ) (12)

F ′(ηN − ǫ) = F

′

asym(ηN − ǫ) (13)

(b) Use the Mathematica built-in numerical differential equation solver NDSolve to

solve the ordinary differential equation (8) subject to the boundary conditions

(12) and (13) over the interval [0, ηN − ǫ], assuming that ηN = 1 and ǫ = 0.001,

for now. You will want to be able to change the parameters ηN and ǫ later on,

so you should keep these parameters general when typing in (8), (12) and (13)

into Mathematica, and only assign a specific value to them in NDSolve using a

substitution rule – see

https://reference.wolfram.com/language/ref/Rule.html.

Documentation on NDSolve can be found here:

https://reference.wolfram.com/language/ref/NDSolve.html

(c) Once you obtain the solution, plot the surface height F (η) as a function of η over

the interval [0, ηN − ǫ], save it (right-click the plot in Mathematica, and select Save

Graphic As) and insert your plot into your write-up.

(d) Use your numerical solution obtained in part (c) above to evaluate the flux

qsim = −

1

3

F 3(0)F ′(0) (14)

at η = 0.

(e) According to one of your boundary conditions obtained for Problem 1, this flux,

given in (14), should be 1. As the answer you obtained in part (d), above, is not

equal to 1, this means that the parameter ηN should be adjusted. Recall we have

just arbitrarily chosen ηN = 1 in part (c), above, for the sake of obtaining just

some numerical solution. You should now manually adjust (or write a short script

to programatically adjust) the value of ηN so that the boundary condition

qsim = −

1

3

F 3(0)F ′(0) = 1 (15)

is satisfied at η = 0, to within a tolerance of 10−3. State the value of ηN that

you have found. Plot the surface height F (η) as a function of η over the interval

[0, ηN − ǫ], save it and insert your plot into your write-up.

(f) Overlay on top of your previous plot a plot of the asymptotic solution Fasym over

the whole interval [0, ηN ]. Save this plot and add it to your write-up.

(g) Comment on how well the asymptotic solution approximates the numerical solution

over the entire interval. In which part of the interval is the fit best?

4

References

[1] N. J. Balmforth and R. V. Craster. Dynamics of cooling domes of viscoplastic fluid.

J. Fluid Mech., 422:225–248, 2000.

[2] G. I. Barenblatt. Scaling. Cambridge University Press, 2003.

[3] E. J. Hinch. Perturbation Methods. Cambridge University Press, 1991.

[4] H. E. Huppert. The propagation of two-dimensional and axisymmetric viscous gravity

currents over a rigid horizontal surface. J. Fluid Mech., 121:43–58, 1982.

[5] H. E. Huppert. Gravity currents: a personal perspective. J. Fluid Mech., 554:299–

322, 2006.

[6] K. N. Kowal and S. S. Pegler M. G. Worster. Dynamics of laterally confined marine

ice sheets. J. Fluid Mech., 790, 2016.

[7] R. A. V. Robison, H. E. Huppert, and M. G. Worster. Dynamics of viscous grounding

lines. J. Fluid Mech., 648:363–380, 2010.

[8] R. Sayag and M. G. Worster. Axisymmetric gravity currents of power-law fluids over

a rigid horizontal surface. J. Fluid Mech., 716, 2013. R5.

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