4

SIGURD ANGENENT AND JOOST HULSHOF

Since the convergence of

u(y, t)

to

U(y)

follows from more general theory, we

will concentrate here on proving the asymptotic formula for

Rep ( t).

1.2. Proof of lemma 1.1.

Both 'P and

rp(r, t)

=

n

/2 are solutions of (1.2),

so that their difference satisfies a linear parabolic equation to which we can apply

the Sturmian theorem: the number of zeroes of

r

r--+

tp( r, t) -

11"

/2 does not increase

with time. Since it starts out being 1, and since the boundary conditions in

tp

force

tp(r, t)-

n/2 to change sign at least once between

r

= 0 and

r

= 1, we conclude

that for each

t

0 the function

r ,__.. tp(r, t) - n

/2 must vanish exactly once.

2. Constructing Formal Solutions

In this section we recall how in

[BHK]

a formal solution for (1.2) is derived.

We consider the function

u(y, t)

=

tp(yR(t), t),

where

R(t)

will be an approximation of

Rq,(t).

This function satisfies

2

Uy f(u)

(2.1) R(t) Ut

=

Uyy

+--

-2-

+

RRtYUy,

y y

Since we expect

u(y, t)

--

U(y)

as

t /

oo, we write

u

=

U(y)

+

v(y, t).

Assuming

v

is small compared with U, at least for y

«

R(t)-

1

,

one obtains the

following equation for

v

28v f"(U;v)

2

(2.2)

R -

8

=

J\1

[v] -

2

v

+

RRtyUy

+

RRtYVy

t

y

Here

J\1

is the differential operator

(2.3)

J\1 _ (_?__)2

+

~__?___

_

f'(U(y)) _ (_?__)2

+

~__?___

_

2._

+

8

- 8y y 8y

y2 -

8y y 8y y2 (1

+

y2)2

0

Also, we use the following notation:

f(n)(u; v)

=

1n

1

(1- Tt-l f(n)(u +TV) dT,

so that the Taylor-Maclaurin formula with remainder can be written as

f "(u) f(n-l)(u) f(n)(u· v)

f(u

+

v)

=

f(u)

+

f'(u)v

+

--v2

+ ... +

vn-l

+ '

vn.

2! (n- 1)! n!

We set

1

v(y, t)

=

a(t)'l/h(y),

where

'l/Jl(y)

is a solution of

(2.4)

J\1~1

=

~o(Y) ~f

yU'(y)

=

1

!yy2

which satisfies

~ 1

(0)

=

0,

and a(t) is determined by the boundary condition at

r

=

1, i.e. at

y

=

R-

1.

Namely, we require

U(R-

1

)

+

a(t)~ 1

(R-1)

=

n,

i.e.

()

_ 2arctanR(t)

(2.5) at -

~ 1 (

1

/R(t))

·

10f

course one does not expect the method of separation of variables to yield solutions for

the nonlinear equation which

v

must satisfy. However, the method may produce functions which

are close to solutions. As the subsequent analysis in this paper shows, this is indeed the case here.