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

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288 PARABOLIC EQUATIONS

is simply f'(u), remains positive, the maximum principle applies, and the problem

is well posed. But then there is no change of phase. To get different phases the

solution must contain regions where u is near both ±1, and hence, in places, u

takes values for which f' < 0, the so-called `spinodal' region. In this region, the

equation is a backward heat equation, and if the initial data is also spinodal the

problem is ill-posed. We comment that, for times before the blow-up that would

generally occur in the spinodal region, maxima tend to increase and minima to

decrease. Thus u changes so that it abhors the spinodal region and takes values

that increasingly correspond to phase separation.

In any event, (6.102) needs to be regularised, and one way in which this can

be done is by including a higher-order term with the correct sign, and one such

regularised equation is

= V2f (u) -

e2V4tL,

where a is a small constant. This fourth-order parabolic equation is called the

Cahn-Hilliard equation, and the final biharmonic term can be identified with the

energy density contained in the surface of the interface between the phases. The

reason for the choice of sign, and, indeed, a guide to the well-posedness of such an

equation, is given by considering a linear one-dimensional problem with oscillatory

initial data in the spirit of §3.1. Consider

a ax2

e2 8x4

for t > 0

with

u (x, 0) = sin nx. (6.103)

This has the solution u(x,t) = eatsinnx, with J = n2(1 - e2n2). Although low

wavenumbers in which n < 1/e give rise to temporal growth, solutions with high

wavenumbers with n > 1/e decay; the zero solution is unstable, which it was not

in the case of the heat equation, but there is no longer any indication of arbitrarily

rapid growth, as happens with the backward heat equation. Likewise, as long as a is

real, the term -e2V4u in the Cahn-Hilliard equation acts to limit rapid variations

of u, confirming the expectation that the material reduces its interfacial energy.

The lack of stability to low wavenumber perturbations is a clear indication that

there is no maximum principle for this particular equation. Although instability

and the failure of the maximum principle for (6.103) result from the -82u/8x2

term, the maximum principle in fact also fails for 8u/8t = -84u/8x4, as we can

see by taking

and

u=105-24t>0 forx=±3,0<t<2

u=x4+24>0

att=0for-3<, x<3,

so that the fourth-order equation has solution u = x4 + 24(1 - t). This is positive

for 0 S t < 1 and is negative in an expanding interval for 1 < t < 2.

Note that, although Green's functions and transform methods can, in principle,

be applied to parabolic equations of any order, terms like a-p'/4

would appear

when we tried, say, a Laplace transform in time on a fourth-order linear equation.

This makes the inversion more unwieldy than for the heat equation.

HIGHER-ORDER EQUATIONS AND SYSTEMS 289

Another class of problems leading to fourth-order parabolic equations concerns

the flow of thin viscous films, such as paints or other coatings, under the action of

surface tension. As usual, we denote the thickness of the film by h and its pressure

by p; unlike the thin films mentioned in §6.6.1, the pressure, which in the absence

of gravity is approximately equal to its value at the film surface, is now equal to the

surface tension multiplied by the curvature. Hence p is approximately proportional

to -V2h, assuming that the film is thin and that (VhI is small, where V is the

two-dimensional gradient in the plane of the film. We can, however, still use the

lubrication theory approximation to give that the in-plane velocity is proportional

to -h2Vp, so the mass flux in turn is proportional to h3 V2h. Finally, conservation

of mass now means that

at = -V . (h3V(V2h))

(6.104)

with an appropriate scaling, and we have a fourth-order diffusion equation with

cubic `diffusivity'.

Like the Cahn-Hilliard equation, (6.104) is an equation about which a great

deal more could be said. While the former is susceptible to some of the methods

described in §6.6 for semilinear equations, especially the integral estimates ap-

proach, the latter clearly possesses some group invariance that can be exploited

to find similarity solutions. Unfortunately, the degeneracy in (6.104) when h = 0

is much more acute than it was for the porous medium equation, which is a pity

because painters are often interested to know criteria for the occurrence of 'pin-

holing', in which h tends to zero locally in space and time.130

In line with the remarks at the beginning of this section, we conclude with a

cautionary tale about vector equations.

6.7.2

Higher-order systems

When different species of population interact, or when more than one chemical

concentration is of importance in a reaction, or when simultaneous heat conduction

and mass diffusion are considered, models arise in the form of coupled systems of

parabolic equations. In the absence of convection, the basic form of such a system

can sometimes be written as

in S2,

where u and f are now vectors with n components, Vu is the n x m matrix

(8ui/Ox,), and D is an n x n diagonal matrix with positive diagonal elements.131

Each component of u satisfies a boundary condition on Oft.

One particular property of these systems is the occurrence of the so-called Ttrr-

ing or double-diffusive instability. Even though diffusion acts as a stabilising and

smoothing mechanism for a single equation, we now have the following possibil-

ity: u can be a constant steady state such that f(u) = 0 and it can be stable as

13011 can, however, be proved that, if h is strictly positive on the parabolic boundary r of

Fig. 6.1, then h is strictly positive in the interior.

131 in some thermodynamic models D is symmetric rather than diagonal.

290 PARABOLIC EQUATIONS

the solution of the ordinary differential equation du/dt = f(u), so that all the

eigenvalues of the Jacobian matrix (8 f;/8u f) have negative real part, yet u can be

unstable as a solution of the parabolic system. Consider, for example, the linear

system

2 _

_ (0 1)

8x2

+ 2 1

2 )

u for 0 < x < 27r,

u=0 forx=0andx=27r.

The system of ordinary differential equations

dt =

(2 12) u = Au,

say,

has zero as its only steady state. Its stability is determined by the eigenvalues A

of A; these satisfy A2 + A + 2 = 0, so the eigenvalues have negative real parts and

zero is a stable solution.

The boundary value problem has solutions of the form

Cul

) eµt sin kx

for k = 1/2,1,3/2,..., provided that

2 - 14k2 - µ -2u(B-µI)u _

2 1-k2-µ

(U2,) =

The eigenvalue µ now satisfies

µ2 + (1 + 15k2)µ + 4 - (1 - k2)(2 + 14k2) = 0.

The trace of the matrix B remains negative but the determinant is also negative,

and so the eigenvalues are real and of opposite signs, if (1 - k2)(2 + 14k2) >

4. Taking k = 1/2 this inequality is indeed satisfied and the trivial solution is

unstable. The effect of the coupling between the components of u is similar to

that of the coupling terms in the system dx/dt = -z + Ay, dy/dt = -y + Ax for

A>1.

Despite the importance of these systems, for example in oceanic instabilities

where temperature and density interact, their mathematical theory is far less well

advanced than in the scalar case. Particular problems are encountered when the

diffusivity matrix D is not diagonal and depends on u, because of the difficulty of

predicting whether or not D is positive definite and hence whether the system is

strictly parabolic.

EXERCISES 291

Exercises

6.1. Suppose a pollutant concentration c satisfies

-t + vx = DV c,

v = constant,

with

and

=0 on y = ±d

c=co atx=0,

c=c.1

atx=L.

Write x = LX, y = dl' and t = (L/v)r to obtain

ac ac

a2c

(6.105)

8r+al' k(Y}'2+b 8X2

where k = DL/vd2 and d = d/L. Show there is a steady solution c = c(X)

which tends to co as 6 -+ 0 except near X = 1.

If 6 << 1 and the last term in (6.105) is neglected, which boundary conditions

would you expect to be able to satisfy?

6.2. Find the time-periodic solution of

= k

02T

f o r x > 0

with

-kWTx = Q coswt

and T bounded as x -+ oo. Take w = 27r (year)-' to give a reason why the

hottest (coldest) day of the year might be expected to occur about six weeks

after the longest (shortest) day.

6.3. Suppose w is the difference between two solutions of

= V2u+au+ f(x,t)

in f2,

with

and

an + au = g(x, t)

on aft

u = h(x) at t = 0,

where a and a are constants and a > 0. Show that

w2 dx

f (I VwI2 - awe) dx + fan w2 dS = 0.

Deduce that

constant) J w2 dx < 0,

(dt

and thus that w = 0.

292

PARABOLIC EQUATIONS

* 6.4. Suppose

82u

8x2

m-L<x<oo,

with

u(x, 0) = 0,

u(-L, t) = uo = constant, u - 0 as x -> +oo.

Show that there is a solution

u(x, t) = t!

"2'

dtl.

(x+L)/f

Since fx e-"2/4 dq <

e-X2/4

for large positive X, if uo = eKL2 and L >> 1,

then

u <

eKL2-(x+L)'/4i2Wz

unless x is large and negative. Deduce that, if the boundary condition is

replaced by u <

eKxs

for all x sufficiently large and negative, and some

K > 0, then u = 0 for t sufficiently small.

6.5. Show that, if f (t) is infinitely differentiable, then

011

u(x, t)

x2" d°f

(t)

(2n)!

dtn

n=O

satisfies the heat equation. Deduce that there are non-zero solutions of the

heat equation that satisfy u(x, 0) = 0 for all x.

6.6. A particle moves along the real line, starting at the origin and taking a step of

±h with equal probability 1/2 in each time step k. If p(x, t) is the probability

density function of its position at time t, use the argument of §6.1.2 to show

that, in the limit h, k -+ 0 with h2 = k,

8p _ 182p

2 8x2

Explain why the appropriate initial condition is p(x, 0) = 6(x), and show

that p(x, t) = (1/ 27rt)e

x'/2E,

the probability density of a normal distribu-

tion with mean zero and variance t.

Remark. This is the density function of Brownian motion. The discrete ran-

dom walk (before taking the limit h, k -a 0) has a translated binomial distri-

bution, and the normality of the limiting distribution is an example of the

central limit theorem; it can also be established by taking the limit of the

binomial probabilities.

6.7. It is observed numerically that

(e_ta)2

X3.142.

-00

Use a periodic solution of the one-dimensional heat equation to prove that

EXERCISES 293

6.8. Let G(x, t; t) satisfy

with

02G

for0<x<1,t>0,

8t 8x2

8G=0

atx=0,1 and

G=8(x-t) att=0.

Show that the Laplace transform of G in time satisfies

d2G-pG=6(x-.), dG

=0 atx=0,1,

d,2

and hence that

cosh

cosh ((1

- t)f) I (f

sink f) ,

0 <, x < t,

G(x,. ;p) =

cosh ( J) cosh ((1 - x),/) / (f stnh f)

t < x <<, 1.

Deduce that

G(x, t; ) = 1 I

Gept dp

2ri

9tp=constant>0

E_xOC(-1)"e_,,2,2,

cosnrxcosnr(1- (),

x < C

cosnrCcosnr(1 -x),

C s x,

and hence that

G(x, t; ) = I + 2

e-n2r2t

cos nrx cos nvC.

n=1

Remarks.

(i) The Green's function for this problem is G(x, r - t; f ). The function G

here can be identified with G' on p. 251.

(ii) G can be written as

(03

"21)

-93

I x2-,e-"')

where 93 is a theta function. \

(iii) This result can be related to (6.36) by expanding G as a power series

in e-2,/P- and inverting term by term.

6.9. If

= a2

for t > 0, 8 = 0

at x = 0,1, u(x,0) = uo(),

use the result that

(, r) = uo(x)G(x, r; e) dx,

where G is given in Exercise 6.8, to show that u(t, r) -+ fo uo(x) dx. as

r-1o0.

294

PARABOLIC EQUATIONS

6.10. Suppose the Neumann condition in Exercise 6.9 is replaced by the Dirichlet

condition G = 0 at x = 0, 1. Show that G becomes

2 E e-n

z"zt

sin n7rx sin n7r

n=1

\e(x-E-2m)'/4t -e-(x+E-2m)z/4t)

7rt m=-moo ` JJ

6.11. Suppose that f (x) is periodic with period 27r and has the Fourier series

f (x) _ E (an cos nx + bn sin nx)

n=N

with the first N - 1 harmonics absent. Use the large-t behaviour of the

periodic solution of Ou/8t = 8au/8x2 fort > 0, with u(x, 0) = f (x), to show

that, for large t, u(x, t) has at least 2N zeros in each period. Deduce from

the maximum principle that, as t increases, zeros of u can disappear but not

appear, and hence show that f (x) has at least 2N zeros in each period.

6.12. Show that the Green's function for 8u/8t = 82u/8x2 in x > 0, with u

prescribed at x = 0, is, for t < r,

G(x, r - t; f) _

(e-(x-E)'/4(T-t)

-e-(x+E)'/4(T-t)

2 7r(r - t)

Using the result that

02G

OG)

8t) G -

(8x2 + 8

u) dx dt = u(t:, r),

fr 1 o

show that, if u = g(t) on x = 0 and u = h(x) at t = 0, then

r g(t)e-E'/4(r-t)

2/FJ

(r - t)3/2

6.13. Suppose w(x, t) has Laplace transform w = e vim, t i > 0. By deforming

the inversion contour to lie along the negative real axis, show that, for -y > 0,

f +i0o

w(x, t) = /

evt- fx dp

= - fo e-°tsin (x ') ds.

27ri

7-ioo

Deduce that

-2 8

-v'

-x2/4t

w(x, t) =

Wt 8x

e cos dy = 2(at)3/2 e

EXERCISES 295

6.14. By taking Laplace transforms, show that the solution of

82u

inx>0,t>0,

8t - 8x2

with

u=0 att=0

and

u=g(t) onx=0,

is given by the Duhamel formula

where

82v 8v

u(x,t) =J tg(t-r)-(x,r)dr,

v(0 t)=1

t>0 v(x 0)=0

inx>0

, .

, ,

8x2 8t

How is v related to the Green's function for this problem? What is the

physical interpretation of the formula for u?

* 6.15. Show that, if

V2G = a in fl, t > 0,

with

G = 0 on 8f1

and

G = b(x - 4)

at t = 0,

then the eigenvalues Ao, AI, ... of the Helmholtz problem

(V2+A)o=0 in 11,

0=0 on Of?

are such that

e--\^ =

G(x, t; x) dx.

n=o

Assuming you can approximate G for small t by

1 e_k_ I'/4f

2 '(4irt)m/

where m is the dimension of 11, show that, as t -+ 0, En oe-A^t is approx-

imated by vol(f))/(4rt)m/2.

Remark. To make this rigorous, you need to show that the contribution from

the region near the boundary, where the approximation for G is invalid, is

negligibly small as t -+ 0.

296 PARABOLIC EQUATIONS

6.16. Show that the Green's function for

+ (V. V)u = V2u

in 12,

where v is prescribed, and where u is given on Of) and at t = 0, satisfies

5F+ V (Gv) + V2G = b(x

- t;)b(t - r),

with G = 0 on 812 and at t = T > r. Hence show that, if 11 is the real line

-oo < z < oo and v is a constant vector in the x direction, then, for t < T,

G(x, r - tit) =

e-(=-E+IvI(r-t))'/4(r-t)

2 7(r- t)

6.17. Verify that the similarity solution of

_ 02u

02u

8x2+Oyz

inz>0,y>0,t>0,

with u=latt=0,andu=0onx=Dandy=0, is

u=erf(2)erf(2rt

where erf,1 = (2/ f) fo e-'' da.

6.18. Suppose that

=02U

int>0,x>0,y>0,

and u(0,y,t) = u(x,y,0) = 0 and u(x,0,t) = 1. Would you expect u to be

an analytic function of x and t? Show that

(erfc 0 < t < x,

Slerfc

(y/2v' ),

x < t,

where erfc ri = 1 - erf q, satisfies the equations and boundary conditions

except at x = t. What discontinuity occurs at x = t?

* 6.19. Suppose that

with

Da a

-(p-Av(y))0=0 fore>0,

dv'=0 aty=0,1.

For small p, write to = io(o) +p*(1) +p2'(2) +

and A = pa(t) +p2 \(2) +

and equate powers of p to show that

EXERCISES 297

Vi(o) = co = constant,

Am = vo,

where vo = J v(y) dy,

VI(1) =

c-° f (y - y') (vo - v(y')) dy' + constant

Dvo

and

dy2

(1_(I)_A(2)Vo.

Deduce that

_ v

f 1

r _

V(Y)

)

(1)

vg ,

co 0

say, where

=z fi[jV(l_.1)dy']2dyo.

6.20.

(i) Suppose x' = g(x, A) and x" = g(r', µ), where g is a group and g(x, 0) _

x. By differentiating with respect to µ and setting p = 0, show that

ag/aA is the product of a function of g and a function of A.

(ii) Suppose

109

(x, A) = F(g(x, A))

and g(x,0) = x. Show that there is a function G such that g = G(A+

G'' (x)) and hence verify the closure condition for x H 9(x,,\) to form

a group.

*6.21. Let (6.56) be generalised to

x' = f (x, t; A), t' = g(x, t; A),

u' = h(x, t, u; A)

and

U=Ua +v8 +wau;

when u is replaced by u' in (6.59), and the O(A) terms are collected on the

left-hand side, show that there is a new term

8W' au'

_ OW

OW 02u'

02W au'

92W

au'

02W

A -

au at'

at + au

2F-

axl + au2

{ ax')

+ ax72

Deduce that the heat equation is invariant as long as

al'

82V aU av

at = axe

+ 2

ax' ax = °°

au a2u a2w

acs'

02%

02w

at - 5x2

2 axau'

at = axe '

au2

=0.