Ockendon J., Howison S., Lacey A., Movchan A. Applied Partial Differential Equations

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298 PARABOLIC EQUATIONS

Verify that (6.68) satisfies these equations and deduce that one possibility

is f = ea/2x, g = eat and h = e"Yau, where ry is an arbitrary constant, and

that

is invariant under the transformation group if

ax+tat+7u

)t;=0.

Finally, show that x/ f and u/t" are both invariant, so that a solution of

the heat equation is u = t7F(x/ f ), where

dn2 +2dFj -ryF=0.

6.22. Show that the wave equation

02u

= c2V2u

ate

is not invariant under the Galilean group of transformations

x'=x-At, y'=y, z'=z, t'=t

(for which the group operation is v = A +,u in the notation of p. 266), where

the constant A, with IAA < c, represents the speed of the (x', y', z') coordinate

system along the x axis. Show that it is, however, invariant under the Lorentz

transformation

x'=7(x-At),

y'=y,

z'=z,

t'=ry(t-LT

where ry = 1/ VF--AY/7, and that these transformations form a group with

group operation v = (A + p)/(1 + Ap/c2).

Remark. Recalling that the components of the electric and magnetic fields in

a vacuum satisfy the wave equation above, this invariance is consistent with

the fact that the speed of light c is the same in all inertial frames, the basic

postulate of the theory of special relativity. It is also possible to show that a

suitable formulation of Maxwell's equations is invariant under Lorentz trans-

formations. The group operation of the Lorentz transformations corresponds

to the relativistic formula for the addition of velocities.

6.23. The equation

=88x2au

forx>0,t>0

models advection in a viscous flow, where x is the distance from a solid

boundary and t is the distance downstream. Show that the equation is in-

variant when we set

x' = eA/3x,

t' = eat, u' = e'rau.

EXERCISES 299

Deduce that t''F(x/t'/3) is a solution as long as

172

dj+3dq-71SF=0,

and that one possibility is u = t-2/3 exp(-x3/9t).

6.24. Show that, if

02U

= 0

is invariant under x' = f (x, y; A), y' = g(x, y; A), then the components U

and V of the infinitesimal generator satisfy

8U 8V 8V 8U

8x 8y ' 8x 8y

6.25. Suppose that V2u = 0, with

10,

8=0, r<1,

r2, 8 = 7r/2, r < 1,

f (8),

r = 1,

in two-dimensional polar coordinates. Show that

u = -? r2(8 cos 28 + log r sin 28) + E bnr2 sin 2n9,

n=1

where

f(8) + ! O cos 29 = E bn sin 2nO for 0 < 8 < 2

n=1

(This question generalises the result of Exercise 5.12.)

* 6.26. Suppose that

with

eV2u+f(u) in n,

8n + eu = 0 on 8f1.

Show that, if you write u = uo + eul + for small a and formally equate

powers of e, you obtain

V2UO

= 0,

= 0

on on,

so that uo = uo(t). Then show that

V2u1 = dt - f (uo)

with

1 = -uo on 8f2,

and deduce that

duo

area(M)

at -

f('°)

vol(52)

300

PARABOLIC EQUATIONS

*6.27. The system

at +

y4u) u)

au 8u ap a2u

"ax)

+ ax

a9,

at+u

+-YP

=e(7-1)(ax)z

with a and y constant, models viscous effects in one-dimensional gas dy-

namics (the right-hand sides of the second and third equations describe the

resistance to extension (as in (2.11)) and the rate of working of this resis-

tance, respectively). Suppose that u, p-po and p- po are all small, where po

and po are constant. Then the flow is localised about waves travelling with

speed dx/dt = ±ao, where ao = 7Po/po. To study a wave near x = aot in

detail, write x = aot + 6y; when terms of order e, e2/6 and 5 are neglected,

show that the perturbations satisfy

8p au au

015

8P 8u

-ao

+ Po

= 0,

-poao 5i + 4 = 0,

-ao

+ 7po

= 0.

The linear dependence of these equations means that we must retain terms

quadratic in the perturbation along with terms involving a and 6. We only

get a sensible answer when e = A62 for some 0(1) constant A. Show that,

with these assumptions, the largest terms to be retained on the right-hand

side of the three equations are

8p 8p au

au 8u 82u 8u

-68t -u8y -p8y, -5po& -pou8y +68y2 +aop8y,

-06

-J!LP -

uay

-1P3i,

respectively. Finally, use cross-differentiation to show that

Of,

02u

7 + 1 _ au

-2po

+ AByz - PO 6 uay 0,

which is (2.50) with a change of notation.

6.28. Suppose that

+U 1 -U)

8t = ,9X2

and u = F(x - Vt), with V > 0. Show that

+VdF +F(1-F)=0,

and that the (F, F') phase-plane has a saddle point at (1, 0), and a stable

node or spiral at (0, 0), depending whether V > 2 or V < 2, respectively.

EXERCISES 301

Fig. 6.5 Phase-plane for travelling waves of the Fisher equation.

Show that the phase-plane is as shown schematically in Fig. 6.5 and deduce

that there is a monotone F in which F -+ 0 as x - V t - +oo and F - 1 as

x - Vt -+ -oo,as long asV>, 2.

6.29. Show that travelling wave solutions of

=a 22+u(1-u)(u-a) for0<a<2

satisfy

+VdF +F(1-F)(F-a)=0.

Using the fact that the phase-plane of (F,dF/4) for a < F < 1 is similar to

that of Fig. 6.5 for 0 < F < 1, show that travelling waves with F(-oo) = a

and F(oo) = 1 exist if

V i Vmin

for some Vmin > 2a(1 - a).

By considering a sequence of phase-planes as V decreases, show that there

is a unique value of V < Vmin such that there is a travelling wave with

F(-oo) = 0 and F(oo) = 1, and that this value is zero when a = z.

6.30. Suppose that

8u 8

1 8u

8t 8x

u2 8x

Define vb to satisfy

1 8u

_u2

8x'

and make the partial hodograph transformation from (x, t) to (0, t) to show

that v = 1/u satisfies the heat equation

8v 02v

8.-j2

302

PARABOLIC EQUATIONS

6.31. Blasius' equation,

d2F d2F

dq3

dqa

is autonomous (i.e. invariant under q' = q + A), so we can lower its order by

writing F = G and dF/di7 = H to give

HdGa + (dG)

Z1 GaG

= 0.

This equation is itself invariant under G' _ (1 + A)G, H' = (1 +.\)2H, for

which the infinitesimal generator is

U = G8G

+2HaH.

Show that UX = 1 and UY = 0 are satisfied by Y = H/G2 and X = log G,

and derive an autonomous second-order equation for Y(X).

6.32. Suppose that

with

02U

8t = 8x2

+ e

for x > 0,

u=0 onx=0, u=0 att=O.

Show that - log(1 - t) is an upper solution and - log(1 - t + h(x, t)) is a

lower solution as long as 8h/8t = 8ah/8xa, with It = t on x = 0 and h = 0

at t = 0. Deduce that u(x,1) -+ oo as x -+ oo.

* 6.33. Suppose that a smooth plane curve r = r(s, t), where s is the are length from

any prescribed point on the curve and t is time, evolves by a curvature flow

in which its velocity v along its normal n is equal to its curvature lc(s, t).

Defining the unit tangent by t = 8r/8s, show that

y=xn+u(s,t)t

for some function u(s, t). Use the Serret-tenet formulae

to show that

8u-aa=0.

57 U38, T

Show that, if the curve is simple and closed and has length L(t), so that

fo x ds = 27r, then

a (L(t), t)

= - [icL]o = -K(L(t), t) [u]o

EXERCISES 303

Deduce that

a2(s, t) ds,

so that the curve grows shorter. If the curve encloses a region ft, show that

the rate of change of area of fl is fan v, ds, and hence that the area of ft

decreases at the rate 2ir.

If two simple dosed curves evolve by curvature flow, and one is initially

inside the other, show that it remains inside (consider what would happen

if they were to touch, the inside curve having larger curvature at the point

of contact). Show that the curve e-(x-l"') = cosay, V > 0, is a travelling

wave solution of (6.88). This curve, known as the Grim Reaper because any

other curves in its path vanish in finite time by the comparison result just

shown, also occurs as the free boundary for a famous Hele-Shaw flow in a

parallel-sided channel, called a Saffman- Taylor finger (see Exercise 7.19).

6.34. Suppose that

V2u + f (U)

in ft

with u = 0 on Oft,

that f > 0, and that f (u)/u is a decreasing function for u > 0. Suppose also

that there is a positive steady state uo and that u - uo = w is small. Show

that, approximately,

V2w

+ f'(uo)w.

Now suppose that AO is the principal eigenvalue of

V20

+ (f'(uo) + \)o = 0,

0=0 on Oft,

and that lAo is the principal eigenvalue of

V2.0 +(f(u0)+p)0,

0 on 011.

Using the appropriate generalisation of the Rayleigh quotient to characterise

AO and po, show that (i) p0 <, AO and (ii) ppo = 0. Deduce that uo is linearly

stable.

6.35. Suppose that

= V2u + Af(u)

in fl, with u = 0 on Oil and u = g at t = 0, where f is convex and positive.

Suppose also that 00 is the principal eigenfunction of V2¢ + µo = 0 in 0,

.0 = 0 on 80, with corresponding eigenvalue po and with f ¢o dx = 1. Use

Jensen's inequality to show that a(t) = fo u4o dx satisfies

> Af (a) - poa

with

a(0) = Oogdx.

304

PARABOLIC EQUATIONS

Deduce that u blows up in finite time if

°O da

(o) )tf (a) - poa

is bounded.

* 6.36. Suppose that

8u 02u

for 0 < x < 1, t > 0

with

u(z, 0) = 0

8x2

and either one boundary condition is prescribed at each end of the interval

0<x<1,

u(0, t) = uo(t),

u(1, t) = u1(t),

or two boundary conditions are prescribed at x = 0 and none at x = 1,

all

u(0,t) = uo(t), (0,t) = vo(t).

Solve formally by taking a Laplace transform in t and, by considering the

convergence of the inverse transform, show that the problem is well posed

with the first conditions but not with the second.

Repeat for the equation 8u/8t = 83u/8x3 with all permutations of boundary

conditions at x = 0 and x = 1. How does your result compare with the case

8u/8t = -83u/8x3?

6.37. Consider the system

82u

+ au + bv,

= D

02v

+ cu + dv,

8t 8x2 8t

8x2

where a, b, c, d and D are constants, and suppose that

a + d < 0 < ad - bc,

so that the solution u = v = 0 is asymptotically stable when diffusion is

neglected (0/8x = 0). Show that there are solutions in which u and v are

proportional to eat sin kx, with

t A > 0, if

Dk4 - (d + aD)k2 + ad - be < 0;

show that this is possible only if ad < 0, be < 0, and, if a (or d) is negative,

D is less than (or greater than) the smaller root of

(az - d)2 + 4bcz = 0.

306 FREE BOUNDARY PROBLEMS

ice

phase bounduy r at , + st

pluue bounduy r at r

Fig. 7.1 The Stefan condition.

discussed in the previous three chapters, so, in each case, we will focus attention

on what happens at the free boundary.

7.1.1

Stefan and related problems

We begin with this, the most famous free boundary problem for parabolic equa-

tions, because of its importance in subjects ranging from metal making to option

pricing. In its simplest guise, the Stefan problem arises as a model for a continu-

ous medium that can transfer heat solely by conduction, so that, as in (6.1), the

temperature u satisfies the heat equation

F= kV2u.

(7.1)

The crucial new ingredient is that we allow the material to change phase, for

example to melt, freeze, vaporise or condense, at a temperature u = u,n which

we assume to be a given constant. Thus we need to solve (7.1), with suitable

initial and fixed boundary conditions, on either side of a free boundary, the phase

boundary. A simple conservation of heat at this boundary is illustrated in Fig. 7.1

for melting. For a small area A of ice, say, this gives

pLAvn 6t = (-kVu n] l

o i A 6t,

(7.2)

sid

where vn is the velocity of the free boundary normal to itself and L is the latent

heat per unit mass that needs to be supplied to the ice at u = u,n - 0 to convert

it to water at u = Urn + 0. Hence, in addition to the Dirichlet condition

U .9 um (7.3)

as we approach the free boundary r from either side, we also have the Stefan

condition on the normal derivatives,133

mnThis `box' argument is analogous to that noted in §1.7 as an alternative way of deriving the

Rankine-Hugoniot conditions.

Free boundary problems

7.1

Introduction and models

This chapter is the most unconventional in the book. Whereas hyperbolic, elliptic

and parabolic problems have been studied over many decades, and many texts

are devoted to each, the subject of free boundary problems has attracted few

specialised publications despite its importance in modern applied mathematics.

In fact, we have already encountered several examples of this type of problem,

although we have not classified them as such. For example, the thin film flows

described in Chapter 1, the shock waves in Chapter 2, the contact problem in

Chapter 5, and the porous medium equation of Chapter 6 could all be posed as

problems involving the solution of partial differential equations in domains that

are unknown a priori; not only the solution of the equations but also the domain

of definition of the equations must be determined. We were able, by judicious

approximation, to reduce the film flow problem to a differential equation for the

shape of the free boundary, which in this case is the surface of the film, but all

the other examples have the distinctive attribute that the free boundary geom-

etry must be calculated and the field equations must be solved simultaneously.

We will adopt this as our working definition of free boundary problems132 and

we note immediately that they are inevitably nonlinear, because the solutions of

partial differential equations almost never depend linearly on the geometry of the

boundaries within which they are to be solved.

We should naturally expect such problems to require more information to be

specified at the free boundary than would be necessary for well-posedness of the

classical initial value or boundary value problems that we have considered hitherto.

Deciding how much extra information is necessary is one of the first challenges for

the development of the theory; in many examples the modelling process suggests

the correct information to be prescribed, but this is not always the case.

The principal reason why this topic has only recently gelled into a coherent

subject can be traced to the spread of applied mathematics into areas of science

which previously lacked detailed quantitative analysis. These applications revealed

many new examples of free boundary problems and we begin by listing some of

them. In most cases we will be dealing with `field' equations identical to those

132The equation of curve shortening (6.88) may be used to describe the evolution of a curve,

but we will not refer to this as a free boundary problem.

305

INTRODUCTION AND MODELS 307

[kVu n]

solid

iiquid = _pLv,,.

(7.4)

Note that, if we write r as f (x, y, z, t) = 0, then n = V f / IV f I and

11V A I.

(7.5)

The condition (7.4) is the `extra' information referred to above; if L = 0 and there

is no phase change, we can simply read off IF as the isotherm u = u,,,.

This model is very simplistic and we need to make several remarks. First, we

have made no comment about whether or not u > u,,, in the liquid or u < u,,, in the

solid. We will see later that the question of whether these inequalities hold or not

has far-reaching implications for the mathematical structure of the model. Second,

the model is a gross simplification of what happens in most practical melting or

freezing situations, because we have neglected many important phenomena such

as convection, radiation and, most of all, the presence of impurities. The latter is

one of the reasons why um should not necessarily be thought of as a prescribed

constant, and it can be shown that even small variations in u,,, can drastically alter

the predictions.134 Nonetheless, the Stefan problem is challenging mathematically

and at once reveals all the nonlinearity inherent in any free boundary problem.

Even though the field equation (7.1) is linear, we can never superimpose two

solutions of a free boundary problem and hope to generate a third because their

domains of definition are different, and this neutralises many of the techniques we

have been describing in previous chapters. In fact, the only obvious simplification

we are ever likely to be able to make to a Stefan problem is to consider its 'one-

phase' specialisation. If, say, the ice is at u =_ u,,, - 0 or the water at u =_ u, + 0,

then clearly we need only solve on one side of IF, which may ease our computational

task. Any free boundary problem in which the solution is trivial on one side of the

free boundary will henceforth be referred to as one-phase.

There are many, many models related to the Stefan problem.135 Suppose, for

example, we neglect the specific heat term, i.e. the left-hand side of (7.1), still of

course retaining the time derivative in (7.5), and also consider a one-phase problem.

This leaves us with the Hele-Shaw free boundary problem in the theory of viscous

flow. The basic idea of its derivation there is as follows. Viscous fluid is forced

either by pumping or suction through the narrow gap between two parallel plates

z = 0 and z = h, as in Fig. 7.2. The equations relating the two 'in-plane' velocity

components (u, v) to the pressure p are quite simple when we neglect inertia and

the in-plane shear force, and only allow for the shear that occurs as we traverse

the gap from z = 0 to z = h. We find

02u 8p 82v 8p

P=;'

I` 8z2 = a

134This is as we might hope, since only a small change in composition

can change iron into

steel, or turn an icy road into a wet one, by the addition of carbon or salt, respectively.

138An interesting example is that of Type I superconductivity, where, if the geometry is right,

the scalar magnetic field H satisfies the parabolic form of Maxwell's equations (as in (6.11)) and

the 'Meissner effect' requires H = 0 in the superconducting region. The Stefan condition follows

from a balance of magnetic flux.