Classical solvability of the radial viscous fingering problem in a Hele-Shaw cell Текст научной статьи по специальности «Математика»

Математические заметки СВФУ Июль—сентябрь, 2018. Том 25, № 3

YAK 517.9

CLASSICAL SOLVABILITY OF THE RADIAL VISCOUS FINGERING PROBLEM IN A HELE—SHAW CELL A. Tani and H. Tani

Abstract. We discuss a single-phase radial viscous fingering problem in a Hele—Shaw cell, which is a nonlinear problem with a free boundary for an elliptic equation. Unlike the Stefan problem for heat equation Hele—Shaw problem is of hydrodynamic type. In this paper a single-phase Hele—Shaw problem in a radial flow geometry admits a unique classical solution by applying the same method as for Stefan problem and justifying the vanishing the coefficient of the derivative with respect to time in a parabolic equation.

DOI: 10.25587/SVFU.2018.99.16953 Keywords: radial viscous fingering, Hele—Shaw problem, unique classical solution

1. Introduction

Viscous fingering occurs in the flow of two immiscible, viscous fluids between the plates of a Hele-Shaw cell [1]. Due to pressure gradients or gravity, the initially planar interface separating the two fluids undergoes a Saffman-Taylor instability and develops finger-like structure. The Saffman-Taylor problem [2] is a widely studied example of hydrodynamic pattern formation where interfacial instabilities grow and evolve (see [3-5] and the literatures therein). Experiments and theory focus on two basic flow geometries: (i) rectangular [2] and (ii) radial [6]. For both geometries the initial developments of the interface instability were discussed through the linear stability theory in [3]. However, the significant differences have been reported between the dispersion relation derived from the linear analysis and that obtained from the experiments if the surface tension effect is not negligible, which call attention to the validity of Hele-Shaw approximation and the new formulation with physically reasonable effects. Among them one is the wetting layer effect [7-10] and another is viscous normal stress effect [11,12].

For the radial fingering phenomena linear stability was discussed in [9] with the former effect and in [12] with the latter effect. Weakly nonlinear analysis was done first by Miranda and Widom [13] without any effect, and recently by Tani in [10] with the wetting effect and in [14] with the viscous normal stress effect.

Most of mathematical results have been established by applying the complex function theory (see [15] and the literatures therein). There are a few results from the exact mathematical analysis: for example, [14,16] for one-phase problem with surface tension and [17,18] for two-phase problem in the standard Holder spaces; [19] for one-phase problem and [20] for one-phase problem with surface tension in

© 2018 A.Tani and H. Tani

the little Holder spaces; [21-23] for one-phase problem with surface tension in the Sobolev spaces. There are some papers not only [19-22] discussing in the multidimensional case, which are absurd from applied viewpoint because Hele-Shaw flows are inherently two-dimensional.

Such a fingering pattern may be found even at a single fluid interface when an incompressible fluid surrounded by air is located between the parallel plates of a horizontal Hele-Shaw cell. We consider this problem in a radial flow geometry and in the classical framework in this paper.

This paper is organized as follows. In section 2 we formulate the problem discussed in this paper and describe the main theorems. In section 3 the problem in section 2 is reformulated. In section 4 the linear problem of the reformulated one is analyzed by following the arguments due to Bazalii [16,17] and Bizhanova and Solonnikov [24]. The most importance is to get a uniform estimate of the solution with respect to a parameter as in [17,25]. For that it necessitates to modify the discussion used in the Stefan problem since the boundary conditions on the interface for the Hele-Shaw problem cause further difficulty. In sections 5 and 6 the original nonlinear problem and the passing to the limit of the parameter are discussed.

2. Formulation of the problem

We consider a slow quasi-stationary displacement of air by a fluid in the Hele-Shaw cell. Fluid is assumed to be immiscible and incompressible. The motion is quasi-two-dimensional and all characteristics of the flow are averaged over the cell thickness. This approximation, so-called Hele-Shaw approximation, is traditional for problems of this type.

The motion of such a fluid is described by

V • v = 0, v = -MVp in Q(t), t> 0. (2.1)

Here v is the velocity vector field and p is the pressure in the fluid; the second equation in (2.1) means Darcy's law (M = b2/(12^) is mobility; ^ is the fluid viscosity; b is the width of two plates). In discussing the fingering problem it is sufficient to consider (2.1) under the following geometrical situation:

fl{t) = jx G R2 | R* < |x| < R{t) + C

where R* is the radius of the hole through which the displacing fluid is injected at a flow rate Q(t), R(t) is the time-dependent unperturbed radius satisfying

t

nR(t)2 = nR2 + y Q(t) dr, R0 = R(0) > R* o

and Z is the perturbed radius.

In addition, we impose the following boundary conditions: Q(t) 2

on r* = {x e R2 | |x| = R*}, t > 0,

2nR* J' ' (2.2)

v • n = Vn, p = pe on r(t), t> 0,

where

m = jiel2 I \x\ = R(t) + ((^,t

Vn is the normal velocity of the interface T(t), n is the unit outward normal vector on T* or T(t) and pe is the pressure in the displaced air, and the initial conditions are

v|t=o = v0, p = p0 on fi(0) = fi, ( )

Clt=o = C0 on T(0) = T (Z0 > R* - Ro). (2'3)

Our problem is to find (v,p) and Z satisfying (2.1)-(2.3), which is reduced to find p and Z satisfying

Ap = 0 in fi(t), t > 0,

-MVp-n=-^- on r*, i > 0,

2irR* (2.4)

-MVp • n = Vn, p = pe on T(t), t> 0, p|t=0 = p0 on fi, Z|t=0 = Z0 on T.

As the compatibility conditions p0 is assumed to satisfy

Ap0 = 0 in fi,

-MVp°- n=7®- on rt, (2.5)

2nR*

p0 = p0 = pe|t=0 on T.

It is more convenient to rewrite (2.4) in polar coordinates (r, 0):

IA (V^) =0

r dr y dr J r2 d02 (R* < r < R(t) + Z(0, t), 0 < 0 < 2n, t > 0), Mdp = (r = E 0 < t > 0) dr 2nR*

{dp id(dp\_ e (2-6)

p = pe (r = R(t) + Z(0, t), 0 < 0 < 2n, t > 0),

p|t=0 = p0 (R* < r < R0 + Z0(0), 0 < 0 < 2n), Z|t=0 = Z0 (0 < 0< 2n).

Besides problem (2.6) we consider the following problem:

Id/ dp\ ^ 1 d2p dp ^ r dr \ dr J r2 dd2 dt (R* < r < R(t) + Z(e, t), 0 < e < 2n, t > 0),

= -Mr (r = R*, O<0<2tt, t> 0), dr 2nR*

p = pe (r = R(t) + z(e, t), 0 < e < 2n, t > 0), p|t=o = p0 (R* < r < Ro + Z0(e), 0 < e < 2n), Z|t=0= Z0 (0 < e < 2n)

with e > 0 and f being given later.

Now let us transform the free boundary problem (2.7) into the problem on a fixed domain. Introduce the transformation from

fi(t) = {R* < r < R(t) + Z(e, t), 0 < e < 2n}

onto

n = {R* < r' < R0 + z0(e'), 0 < e' < 2n}

by the change of the independent variables

r' = fl° + c° R*(r-R*) + R*, e' = e, t' = t,

lO t t, — n*

R + C-R,

and the unknown functions

p(r, e, t) = p'(r', e', t'), z(e,t) = z'(e', t').

By omitting all the primes for simplicity, problem (2.7) takes the form

= Jif/p - e/ in ii, t> 0, dt z J

dp- Q{t) R + on r; = {r = Rt, 6>G [0,2tt]}, t>0,

dr 2nR* MR0 + Z0 - R*

dt KS)dr ns)de 27ri?'

p = pe on r = {r = R0 + Z0(e), e e [0,2n]}, t > 0,

p|t=0 = p0 on n, Z|t=0 = Z0 on [0, 2n].

Here

e e d d A 1

* ^ r0+(°-R* v

d2 0 / 1 dZ0 1 dZ V „ , d2

+ 2 D I AO D Jfl D I A D flfl H7"

de2 VR0 + Z0 - R* de r + z - R* de; v *'drde

+ [[r*+ R: (r-R«)

+

1

Ro + Z0 - R* dZ0

flp + z° -

R + C-R*

1

dZ

Ro + z0 - R* d9 R + Z - R* <9(9

(r - R*)2

dr2

+ ■

r - R* "

1

Ro + Z0 - R*

1

dZ0

dZ

VR0 + Z0 - R* d( R + Z - R*

1

dZ0

1

dZ

R0 + Z0 - R* d( R + Z - R*

dr'

62(Z) = M

Now we choose

R0+(°-R* R+C-R*

1 +

1

1

dZ dZ0

6i(Z) = -M

(R0 + Z0)2 Vdej J (R0 + Z0)2 d(

1 dZ

(R0 + Z0)2 de'

f= ^p0 d 1 Ro + (°-R* dr 8t[ + U

on O.

t=0

Before describing the main results, we introduce function spaces. For a domain tt in ln (n G N) and any T > 0 let Cl(U) and Cl/J2{QT) (QT = ft x [0,T]) with l = k + a, k £ Z, k > 0, a £ (0,1) be the standard Holder spaces constructed with the use of the following semi-norms:

|u|(c) = |u|(0) + (u)(c), |u

|u|(0) = sup |u(x,t)|, (u)(a) = (u)Xa) + (u)

\ (a/2)

(x,t)eQi

{u)[xa) = sup = sup

x,yen,te[o,T] \x y I t,t'e[o,T]

(see [26]). We also use the semi-norm

|u(x, t) - u(x, t')| li-i'l"

sup + e (0,1),

x,yent,t'e[o,T\

|x - y|a|t -

and define the Banach spaces Ek+a(QT) (k = 0,1,2) that are obtained as the completion of infinitely differential functions in respective norms

I E<*(QT)

= Ea'a/2[u]= sup |u(x,t)| + (u)(a) + [u

](a,a/2)

(x,t)£QT

Da'a[u]= sup |u(x,t)| + (u)Xa) + (u)(c) + 'uHa,a)

] (a,a)

lu\\k+c

(x,t')EQT

\ Ek+a(QT)

k-1

Ec,c/2 [d£u] + E Dc,c [D>]

j=0

2

2

2

1

2

2

t

u

u

c

u

(D£ = Y^ di (j is a multi-index), k = 1, 2),

|j|=k

and the space P2+a(QT) with the norm ||u||P2+c^q ^ = ||e<9tw||a + |M|2+a- We also introduce the spaces

E2+a(QT) = {u | u G E2+a(QT), dtu G E1+a(QT)},

IMI e2+"(Qt) = \\uh+a + ||9tw||i+a, E2+a(QT) = {u | u G E2+a(QT), dtu G Ea(QT)} ,

IMI E2+a(QT) = \\uh+a + \\dtu\\a, E2+a(QT) = \u I \\u\\e1+^t) =J2Ea'a/2№u] < 00 } '

I k=0 J

E2+a{QT) = In | = J2Ea'a/2№u] < ool ,

I k=0 k=0 J

42+"(QT) = \u | IMI^(qt) = +Ea'a'2\dtu] < 00} .

I k=0 )

Denote by E a(QT), Pe2+a(QT), etc. the spaces of the corresponding spaces whose 00

elements are equal to zero at t = 0 together with their admissible derivatives with respect to t.

For a smooth manifold r in Rn C'(r), cX't/2(rT), etc. are defined with the help of partition of unity and of local maps. The following is our main result.

Theorem 2.1. Let T > 0 and a G (0,1). Assume that (p°,C°) G C3+a(TI) x C4+a([0,2n]) satisfy the compatibility conditions, dp0/dr < 0 on r, pe G c3+a'(3+a)/2(rT) and Q G Ca([0,T]). Then there exists T0* > 0 depending on the data of the problem such that problem (2.8) with e = 0 has a unique solution (PjC) e E2+a(QTt) x E2+a(TT*).

Theorem 2.1 is proved in several steps. First we consider problem (2.8) with e > 0 in the classes of smooth functions, and then show that this solution converges as e ^ 0. For that it is necessary to use the function classes where the uniform estimate in e holds.

Theorem 2.2. Let e0 > 0 and the corresponding assumptions in Theorem 2.1 hold. Then there exists T0 > 0 depending on the data of the problem and e0 such that problem (2.8) with a fixed e G (0, e0] has a unique solution (pe,Ze) G P|+"(QTo) x E2+«(TTo).

3. Reformulation of the problem

Let C G c4+"'2+a/2([0, 2n] x [0, T]) be an extension of Z0 such that

; <K dX\ dt' dt2 J

t=o ^'dt'dt2

t=0

where (dZ/dt, d2Z/dt2)|t=0 are obtained from the third equation in (2.8) and its derivative in t at t = 0.

We seek a solution of problem (2.8) in the form

P = P*+P° + J~R\ d-fc, C = C* + C (3.1)

R + C - R* dr

Then (2.8) becomes

e—^— = + in ÎÎ, t> 0,

dt

dp*

onr.)t>0, (32)

% - - bi(C)% = p - d(C)C = *2 on r, t > 0,

p*|t=o = 0 on ft, Z* L=0= 0 on [0, 2n].

Here

**

r, 9

1!l

de2

l2( 1 < 1 gcyr

\r0 + z° - R* de R + C-Rtde)y *'drde ((„ , R + C-R* , D Q\2 fRo + C°-R^ 2

+ + + U + Z-R*

/ 1 dÇ° 1 aç\2

+ \r0 + - r* de R + C-Rtde) {r *')dr2

dp0 d ( Z* A = *«(p*, o = - w -sir- a.) ) - ef,

*.=*.ia =-I R+C~R* Q{t)

dr V R + C - R* dr / R0 + z0 - R* 2nR*M '

dr wdr d(9 dt 2nR'

= Pe - p0;

R0 + Z0 - R* dp0

d(0 = — -

R + c - R* dr

with (p, Z) substituted by (3.1). Note that the assumption dp0/dr < 0 on r implies

d(C) > 0.

4. Linear Problem

In this section we consider the following linear problem: du

e—--Jz?*w = 6 in il, t > 0,

dt

du

onrt,i>0, (41)

ff " ~ = V-i. u- d(C)g = V-2 on r, t > 0,

u|t=0 = 0 on O, p|t=0 = 0 on [0,2n]

for given , ^i, under the conditions b2 > 0, d > 0.

First we study three model problems in the whole- and half-spaces:

du

e— -i?U = / ((x^) gR2, t>0), w|t=o = 0; (4.2)

du

e— - Jfw = / ((xi,x2) G R+ = {(xi,x2) G R2 | x2 > 0}, i > 0)

du dx2

= 0, u|t=0 = 0;

X2=0

(4.3)

du

e— — Jzfw = 0 ((xi,x2) G R+, t > 0)

<9p <9w dt dx2

= gi, u - dp|x2=0 = g2, (4.4)

=0

(u, g)|t=0 = 0.

In the above

y d2

ajkdxjdxk

is the second order partial differential operator with positive real coefficients ajk which constitute positive definite symmetric matrix, and b and d are positive constants.

For problems (4.2)-(4.4) we can assume without loss of generality ajk = j by changing the independent variables (cf. [26]). Then the solutions to problems (4.2) and (4.3) for Jz? = A are given by

t

u(x, t) = J dr y re(x - y, t - t)f (y, t) dy, (4.5)

0 R2

and

t

u(x, t) = J dT j Ne(x - y, t - t)f (y, t) dy, (4.6)

0 R2

respectively, where

re(x,t) = en/2-1(4nt)-n/2 exp

4t

X> — , n ) ,

six I2

Ne(x, t) = Te(x', xn, t) + Te(x', — xn, t), n = 2.

Applying the same arguments as in the estimation of the volume potential for the heat equation to the integrals in (4.5) and (4.6), we have uniform estimates with respect to e for the solutions of the problems (4.2) and (4.3) (see, [17, 26])

du

dt

+ |M|2+a < Ci|

(4.7)

For problem (4.4) by making use of the Fourier-Laplace transformation

J^[u](£,x2,s) = u(£,X2, s) = y e-st dt J e-iu(xi, X2, t) dxi

0 -TO

as in [16,17, 24], the parabolic equation is reduced to an ordinary differential equation in x2, so that we get

û(Ç,x2,s) = ïï(Ç,s)e-r'X2 (x2>0), re = re{s,£) = y?es + Ç2, Rere>0. (4.8) From the boundary conditions in (4.4) it follows that

Solving these, we obtain

Q

breQ + s Q = Q1, Q — d Q = Q2.

■(Qi — breQ2), Q = d Q + g2.

dbr,

(4.9)

The solution of problem (4.4) is given through the inverse Fourier-Laplace transformation

v{x1,t) = {^^')-1\v} = ^ j e1Xli J estv(Z,s)ds

Res=a>0

as follows.

e(xi,t) = (^T1

1

s + dbre v(xi, t) = dp + Q2

* (Qi — 6^)-1[reQ2])

(4.10)

where * means a convolution with respect to xi and t.

By following the arguments in [16], let us derive the explicit representation of

Z (xi,t) = (^i?)-i For simplicity let db =1. Since

1

s + dbre(s, £)

t > 0.

(4.11)

O

(s + ■Ves + Ç2)-1 = J exp[—r(s + Ves + £2)] dr,

e

a

a

oo

1

ö+iw

J exp[—r(s + y/es + £2) + si] ds

1

27ri

ö+iw

ö+iw

I + I +

t2t er2

= S(t - t) * rr3/2 exp v 7 t 2a/7t f

where * means a convolution with respect to t, so that

4t

(a > 0),

Ze(£,t) = j S(t-r)*^=rt-3/2exp 0

t

dr

2a/7T

J T(t - T)-3/2 exp

e 4i ^ , er2

dT

t/e

1 f t — ez

2a/7T J Z3/2 0

■ exp

' 2 (i-ez)2 z ~

Therefore, we have

4z

dz.

Ze(xut) = — / Ze(£,t)cos(£a:i)d£

t/e

1 ft - ez 0

4z

From (4.12) it easily follows Lemma 4.1. Inequalities

|Ze(xi,t)| < C2t

1 + Vei

x2 + i2 '

(t - ez)2

4z

Ze(xi,t)

d2

-Ze(xi,t)

— Z

dx1 e

dxf

dtdxi

hold with a constant C2 independent of e. Lemma 4.1 implies

Ze(xi,t) Ze(xi ,t)

< c2

<c2-

1 + a/Ü

X2 + i2 '

1 +ei (x2 + i2)3/2

w

limi i Ze(xi - £,t)f(£)d£ = / (xi) t^0 J

dz. (4.12)

(4.13)

(4.14)

a-iw

e

w

w

2

x

1

for any bounded continuous function f (xi). Introduce the notation

t to

7(xi ,t) = (Z * q) (xi, t) = y dr y Ze(xi — y,t — t)g(y, t) dy,

0 — to

t—h to

Wh(xi,t)^y dT y Ze(xi — y,t — t )g(y, t ) dy (h > 0).

Je

0 -o

For wh it is clear to hold

t —h to d f f d

■7^wh(x1,t)= / dr / — Ze(xi -y,t-T)(g(y,r) -g(x1,T))dy

0 —o

it—h to to

+ y ff(xi,r)dr y — Ze(xi -y,i-r)dy + y Ze(x1-y,h)g(y,t-h)dy.

Making use of (4.14) and the formula

t/e

d ~ 1 f e(t — ez)

diZ^t) = -^J z^t + ez)***9

0

—C2z —

{t-ezf Az

dz

t/e

1 f C2(t — ez) I" _2 (t — ez)2 ' v 7 exp v 7

J z1/2(i + ez) 0

—C2z —

4z

dz (4.15)

together with the estimates in Lemma 4.1, we have after passing to the limit h ^ 0

t to

d f f d

—w(x1,t)= dT / —Ze(x1-y,t-T)(g(y,T)-g(x1,T))dy

0 -to

t to

d

+ j g(x1,T)dr j ^ze(x!-y,i-r)dy + g(x1,t). (4.16)

dt e

0 —o

Analogously we have

d2 f f d / d d \

—w{x1,t) = jdrj —Ze(x1-y,t-r)^-g(y,r)-—g(x1,r)j dy. (4.17)

We begin by estimating the first term denoted by w^ in the right hand side of (4.16). Following the arguments in [26], we get

w^(xi, t) — W^ (xi, t)

t

y dr y — Ze(xi - y,t - T)(g(y,T)-g(x1,T))dy

d

0 |xi—y|<2|xi—x1

t

d

J dr j ■^Ze(x[-y,t-T)(g(y,T)-g(x'1,T))dy

dt e

0 |xi-y|<2|xi-xi|

t

+ J dr J -y,i-r) - ^Ze{x[-y,t-T)

0 |xi-y|>2|xi-x1|

x (g(y,T) - g(xi,T))dy t 4 + J(g(x[,T)-g(xUT)) dr J -ZE(x'1-y,t-T)dy = YJIJ. (4.18)

0 |xi-y|>2|xi-xi| j=1

Applying Lemma 4.1 to I1, we have

\h\ <cs<*№ / ^-»r^/^^f^Ld,

|xi-y|<2|xi-xi| 0

|xi-y|<2|xi-xi|

< C3(e0,T)(g)xa)|xi - x1|a. /2 is estimated by just the same way as I1. For /3 we do in the same way with the help of the mean value theorem. Finally for /4 we get

t

\h\ <C'4(g)ia)\^i-^i\a fdr j ^-ZE(x'1-y,r)dy

0 |xi-y|>2|xi-xi|

< C4(e0,T)(g)xa) |xi - |a due to (4.15) and Lemma 4.1. Second term in the right hand side of (4.16) is estimated similarly. Hence we obtain

/ dw \ (a)

(-Qt) <C5(e0,T){g)M. (4.19)

Next the difference with respect to t of w^ is expressed for t' < t:

t cc

d

w'[xi,t) — w'[xi,t) = I dr I —Ze{xi—y

\x1,t)-w\x1,t')= J dr j ^_Ze{x1-y,t-T){g{y,T) - g{x1,T))dy

2t'-t -c t' c

J dr J —Ze(x1-y,t'-T)(g(y,T)-g(x1,T))dy

or e

2t'-t -c

2t'-t c

+ / dT / (J^^1 ~ (9(y,T) -g{xi,T))dy

-c -c

3

= £ /j". j=i

Lemma 4.1 yields

t œ

\I[\<C^ I dr I

2t'-t -c

< C6(g)Xa) J

(xi - y)2 + (t - t)2 t

1 + y/e{t - t)

2t'-t

(t - T)

i-a

dT < C6(ec,T)(g)Xa)|t - t'|a.

12 is estimated by just the same way as lj. For I3 we do in the same way with the help of the mean value theorem and the estimate of d2Ze(xi, t)/dt2:

d2

< C7^exp

ex i

IF

t/e

+ C^ z-5/2 exp

ci (t - ez)2

4z 4z

dz.

After some lengthy calculations we get

(wt)ta) < C8(£0,T)(g)ia). Second term in the right hand side of (4.16) is estimated similarly. Hence we obtain

From (4.16) we can derive

(4.20)

d2

dtdx

t cc

■w(x1,t) = JdT J Ze{xi-y,t-T) (J^9(y,T) ~ dy

0 -œ

t œ

f d f d d + J d^9{Xl'T)dT j Q-tUxi-y,t-t)dy + —g{Xl,t).

Then, repeating the above arguments, we obtain the estimates

(4.21)

\dtd^/x 0

We come to the estimate [dw/dt](a,a). Since w^ can be written as

cc cc

(xi,t) = j dr j ^Ze(x1-y,T)(g(y,t-T)-g(x1,t-T)) dy,

c -c

vHXl,t)-wHXl,t-At) f

(At)a -W {Xl,t)

cc cc

J dr j ^_Z£(xi — y, t) {ip{y, t — t) — f(xi, t — r)) dy

c -c

-c

t

w

/ dr / ^j_Z£(xi — y,t — t) {ip{y, t) — if(xi, r)) dy

— œ

g(xi,t) - g(xi,t - At)

0 — œ

(At)c

for At > 0, where

^(xi,t) = Again the above argument implies

^t)W<C-11(e0)T)<v,>(a)=C-ll(e0)T) Sup

Cii(£0,T) sup

xi ,t

i,x1,t |xi x'l|C

|g(xi,t) - g(x1,t) - g(xi,t - At) + g(xi,t - At)|

|xi - xi|a(At)a

and hence

[wt]Ka) < Cn(£0,T)[g](a'a). Second term in the right hand side of (4.16) is estimated similarly. Thus we obtain

dw

dt

< Cn(£0,T)[g]

Similarly we have

d2w

dtdxi

(a,a/2)

< Ci2(£0,T)

dg

dxi

(a,a/2)

(4.22)

(4.23)

Now it is necessary to get the representation of (^Jz?) i[rec/2]:

(^Jf)-1^] = 2^ /dr J TE(x1-y,x2,t-T)g2(y,T)dy

0 — œ

X2=0

t' œ

2^ / dr / Ti(xi -y,x2,t' -t)G2(y,t) d y

0 —œ

X2=0

where t' = t/e, G2(xi;t') = g2(xi;t), ri = re|e=i (see [25]). From this we have

((^-Sf) — i[reg2])(a'a/2) < Ci3(G2)(i+a'(i+a)/2)

= Ci3 ((g2)Xi+a) + e(i+a)/2(g2)t(i+a)/2) ).

Equation (4.9) and estimates (4.19)-(4.23) lead to

< Ci4(e0,T)(||gi||i+a + ||g2||2+a) .

dp ¥

(4.24)

i+a

The same arguments can be applied to (4.17) with the help of Lemma 4.1, so that similar estimates to (4.19), (4.20) and (4.22) hold for d2w/dxi:

d2w dxi

< Ci5(e0,T)

dg

dxi

tœ

a

a

moreover, we have

(a)

< Cie(£0,T)

/-

dx\ /t ibV u' \ dxi

(a)

Lower order derivatives of w are easily estimated. Therefore, from (4.9) we deduce

|M|2+a < Cir(£0,T)(||gi||i+a + ll^lh+a) ,

\dx\/t

(a)

< Ci8 (£0,T)

(a)

+

(I

(i + a,(i+a)/2)

(4.25)

Estimates (4.24) and (4.25) easily give the estimate for v(xi;t) in (4.10):

dv (

Qj + IM|2+a < C19 (j|ffl||l+a + ||ff2||2+a +

For u(x,t) we prepare two representations:

dg2

dt

u(x, t) = -2 j dr j -^—Te(x1-y,x2,t-T)v(y,T)dy,

0 -c

t c

¿(x, i) = -2 J dr J Ve(x1-y,x2,t-T)g(y,T)dy, +

(4.26)

(4.27)

(4.28)

since i is represented as

£ = <72 + dg = -¡—{-gi + sg).

Here we use (4.27) for the estimates of d2u/dx2, d2u/dx2, du/dxi, and (4.28) for the estimates of d2u/dxidx2, du/dx2. Introduce the notation with t' = t/e

u(x, et') = U(x,t'), v(xi;et') = V(xi,t'), g(xi;et') = G(xi;t').

Then, (4.27) and (4.28) are represented as

dr

U{x,t') = - 2—±*V and U{x, t') = -2I\ * G, 0x2

respectively. We can find the following estimates (see [26]):

(S • vp «nr (£ • M(V)ST' + (V>r»).

Simultaneously we have

\dx2/t, + \dx2/v ~ 21\\dxl/v +\dt'/t,

/£PU\{a) /82U\

\ dxi! x \ dxl /

(a)

C2

+ / dV\{a) J dV\

dxL x

\dx2 /

+W

(a/2)

+W„

d2

w

x

1

x

a

t

c

-c

t

x

x

These inequalities yield

dxi/ t

2u \ (a/2)

d*u\w"> /d2u\ \dx2/t

<62

(Pv\{a/2) /dv\

dxi/1

(a/2)

+£KSi/t

dxi

(Pu\{a) /d2u

+

(a)

<62

2„ \ (a/2)

(4.29)

/ dv

(a)

+e

i+a/2

/ cfe

(a/2)

\dt/t

Furthermore, by the same method as that for wt we have

d2u

dxi j

(a,a/2)

< C22

d2V

dxi j

(a,a/2)

+ sup

1

d2V

dxi

xiM' |At'|a/2|s|a/2

d2V, A d2V, . a / x d2V. . A .. —) - TT^-^i'i + Ai - s) + TT^-^i'i + Ai)

(xi; t' — s)

dx

dxi

and hence

d2u dxi j

(a,a/2)

< C23

d2v

dxi j

< C22 (a,a/2)

dxi dxi j

(a,a/2) ,

+

\

+ e'

2v\ (a)

(a)

(4.30)

Analogously we obtain the estimate for [d2u/dx2](a'a/2). By virtue of the expression (4.28) we get

*L = -2 _

dx2 dx2 dx2dxi

d2U _ _2dT\ 9G dx2 dxi

from which it follows

dx2

(a)

< C24(G)t,a),

dxidx2

(a/2)

and hence

(a)

i—2/ t-

d2u \(a/2)

(a/2)

< C25

/

(a/2)

(4.31)

r\ r\ / _I ^ ZD \ r\

t v dxidx2 /1 \ ax^ t

Each term in the right hand side of (4.29)-(4.31) is estimated by virtue of (4.24)-(4.26).

Similar arguments are applicable to other terms appearing in the norm ||u||2+a. And finally from the first equation in (4.4) we derive the uniform estimate e||du/dt||a with respect to e.

Lemma 4.2. Let e0 be a fixed positive number, and b and d are positive constants. Assume that gi G EEi+a(RT) and g2 £ E?2+a(RT) with a G (0,1) and T > 0. Then problem (4.4) with any e G (0,e0] has a unique solution

u g Pe2+a(R+T), ge E 0 ' 0

2+a

t2 = R+ x [0, T], RT = R x [0, T])

x

t

t

t

+

satisfying the inequality du

dt

+ IMk+a + II^Ha+a +

dp ~dt

1+a

< C26 ( IlgiHl+a + IlgaHa+a

dga

dt

(4.32)

where C26 is a positive constant depending on eo, but not on e.

Now we shall solve problem (4.1) by the regularizer method.

In the following (0,r) corresponds to the above (x1;x2). For a suitably small positive number A we introduce two coverings {wk} and {ftk} of ft as follows (cf. [27, 28]): Let k be in if two dimensional squares wk and ftk included completely in ft are mapped by n1 from those with common center (rk, 0k) and with the length of their edges, in parallel directions of axes, equal to A/2 and A, respectively. For k G . //> satisfying ujk fl r ^ 0, u>k and ftk C ft are defined in the local coordinates (z1, z2) with an origin at (rk, 0k) G r by

= n2 {|z2| < A/2, 0 < zi - Fr(Z2) < A} ,

ftk = n2 {|z2| < A, 0 < zi - Fr(Z2) < 2A} ,

where equation z1 = Fr (z2) represents r around (rk, 0k) G r and n2 is a transformation from (z1, z2) to (r, 0) G ftk. By {wk} and {ftk}, k G we denote the coverings of r* in ft and by II3 the transformation from z to (r, 0) G ftk.

Let the partitions of unity {nk} and {n*} subordinated to {wk} and {ftk} be such as

nk (r, 0)

1 for (r, 0) G wk, 0 for (r, 0) G ft \ ftk,

jnk(r, 0)| < C^A-^,

=

Ek (nk (r,0))2'

Obviously, {n* (r, 0)} have properties

nk

(r,0) =0 if (r,0) Gft\ftk, Ei?fe(r,0)^(r,0) = 1, |D^(r,0)| <C28A

First let T* = yA2, where 7 G (0,1) will be specified later. Then the regularizer is defined by

®H = £ £ n*uk + £ n*, H = (^*,^1,^2) .

Here uk = n1uk, where uk is a solution of problem

duk

~dT

- n-1 (i?*^) uk = n-1nk^ in R2 x (0,T), uk |t=0 = 0 on R2 (:S?.,k ^*(rk ,0k; d/dr, d/d0)|t=o);

(4.33)

e

a

a

u = n3uk, where uk is a solution of problem

dgk

— - n^1 (jsf,ifc) uk = II3 W in R* x (0, T), uk|t=0 = 0 on R+ = {(zi, Z2) G R2 | zi > 0}, (4.34)

n3 ink "*;

zi=0

duk dzi

uk = n2uk, gk = n2gk, where (uk, gk) is a solution of problem duk

~dT

(uk ,gk) |t=0 = 0, (4.35)

= n-ink uk — dkgk |zi=0 = n-ink 02

— - n^1 (Jif^fc) uk = II^ W in R* x (0, T),

dgk ^ duk dt dzi

zi=0

with

bfeeZi = n2(b2, bi)(C)|e=efc,t=0, dk = d(C))|e=efc,t=0, e^ = (1,0).

Then it is not difficult to see that MH = (u', g') satisfies du'

e—--Jz?*u = 6 - in n,t> 0,

dt

du'

-T— = -0* - «^2-ff on r*, t > 0, dr

dg' - du' - du' (4.36)

u' — d(g)g' = 02 ^4H on r, t > 0, (u', g')|t=0 = 0. Here the operator = ( i, 2, 3, 4) is defined on

^t = Ea(QT) x E1+a{T*T) x E1+a(TT) x ^2+"(rT) 0 0 0 0

by the formulae

«^H = i?*u' — ^ ^ n*

j=i,2,3 ke.#,-

ke.#a i

ke.#2 i

J4H = —d(g)g' + E n*n2dk gk.

ke.#2

One can find the solution of problem (4.1) in the form (u, g) + & + &2 + ... )H = — 9' )-iH (jf is an identity operator),

for which it is necessary to show that the operator 3" is a contraction on J^T • We first note that r G C2+a implies that Fr G C2+a(B,5) (B = {z2 G R | |z2| < 5}) and Fr(0) = 0, dFr/dz2(0) = 0, ||Fr|| < C with some constants 5 and C

being independent of z2. We take A small enough to satisfy A < 5/2. Clearly,

F (Z2)| < C |z2|1+a,

We estimate each term in <^1H, .i^H, «i^H, <i?4H in such a way that for the lower order terms the interpolation inequalities, for example,

sup |Vu(x)| < c((u)ia))(1+a)/2((u)X2+a))(1-a)/2, (V2u)(a/2) < cAa[V2u](a'a/2) xeo

for u G E2+a(QT) are used, while for the highest order terms the smallness of their 0

coefficients like (4.37) derived from the smallness of A and 7 are used. Then we get the following estimate after some lengthy, but straightforward calculations:

\\&H\u = \\&iH\\Ea(Qt) + \\,%H\\Ei+a{Ttt) + \№H\\Ei+a(Tt) + \\,%H\\E2+a{Tt)

<c29(A,7)(£ E E (4-38)

j=1,2,3 ke.#3- ' ke.#2

for any t G (0, T*), where Qk = x (0, t), rtk = rt n Qk and

C29(A, Y) = C29,1(A)C29,2(7) + C29,3(A) + C29,4(7)

is a positive constant with C29j(•) being dependent non-decreasingly on their argument and C29j(0) = 0 (j = 1, 2, 3,4). By virtue of Lemma 4.2 and (4.7) we see that ,9 is the contraction operator on for suitably small A and 7. Hence problem (4.1) is solvable on [0, yA2]. In the same way one can solve problem (4.1) on [7A2, 2yA2]. Repeating the above argument finite times, we finally obtain the solution on [0,T] with arbitrary T > 0. Summing up the above leads to

Lemma 4.3. For any T > 0 and any £0 > 0 problem (4.1) with a Sxed £ G

(0, £0] has a unique solution (u, g) G Pe2+a(QT) x E (Dp) satisfying

00

< C30IIH||^T, (4.39)

where C30 is a positive constant depending on £0, but not on £.

dFr(z2) dz2

< C|Z2 |C

(4.37)

5. Nonlinear Problem: Proof of Theorem 2.2

In this section we construct the solution to problem (3.2) by the successive approximation method.

Let (pn, Zn) (n =1, 2, 3,...) be a solution of problem (4.1) with

0 = $e(pn-1,C-1), = **(£-!), V>1 = *1(pn-1,C-1), ^2 = ^2

for a given (p^, Q-i) & Pe2+a(QT) x E+a(TT), and (p*,C0) = (0,0).

00

It is easily seen that there exists a constant M > 0 dependent on e0 and T, but not on e such that

||($e(0,0),W,(0),Wi(0,0),^)||^T < M. (5.1)

From the same argument as in (4.38) we derive the following inequality with the help of (5.1):

< ||($e(0, 0),W,(0),Wi(0,0),*2)IU

+ i(^e(p;-i, c:-i) - $e(0,0),^,(c:-i ) - ^w^n-i.c-i) - *i(0,0), 0)u

< M + K\\P*n-l\\E^(Qt) + IIC-lb2+.(rt))

for any t G (0,T) and any ft > 0, where C^ is a positive constant depending on ft non-increasingly, x > 0 is a constant depending on a (possibly) and F(■) is a polynomial in its argument. Thus, Lemma 4.3 yields

\\Pn\\p^(Qt) + HCnllÊ2+a(rt) ^ C30(M + f3(\\P*n-l\\E^(Qt) + IIC-lll^(rt))

+ |K_1||fî2+(,(Qt) + ||C-illB2+«(rt))) (5.3)

for any t G (0, T). Now we take first ft = 1/(4C30), and then

rpf • , T ( M

1 = mm< 1, I -

i/x

V2C^ F (2C30 M)/

Therefore, from (5.3) it follows that

llPnllPc+c"(Qt) + IICII^(rt) ^ 2C30M

for any t G (0, T') if

lbn-lllpe2+"(QT) + IIC-ill^+«(rT) - 2CsoM,

that is, for n = 0,1, 2,...

(p*n,C) e Pe2+a(QT>) x E2+a(YT>), 00

\\Pn\\p^(QTr, + \\CJE^{TT,)<2C^M.

For the convergence of the sequence {(pi, C*)} we consider the difference (pi — Pn-i, C* — C*-i), which satisfies (4.1) with

u=pn—pi-i, g = a — c*-i, 0 = $e(p;-i,C-i)—$e(p;-2,c:-2),

0* = w*(c*-i)—**(c*-2), "i = *i(p;-i,c,-i) — *i(p;-2,C-2), "2 = 0.

Similarly as above, with the help of the interpolation inequalities we can obtain a similar estimate as (5.2):

H^n-i^-i) — $e(p;-2, cu), **(c*-i) — **(C-2),

(5.4)

C-i) - eu), 0)iu

< (/î' + C^'Î^F'dlp^iH^+cfQ ) + ||Pn—2 II E2+a (Qt

+ yen

+ ne: — 2 II _g2+a(rt) Pn — 2 II E2+<*(Qt)

+ yCn—i — Cn — 2y _§2 + a(pt))

for any t G (0,T') and any ft' > 0, where C^' is a positive constant depending on ft' non-increasingly, x' > 0 is a constant depending on a (possibly) and F'(■) is a polynomial in its argument. Therefore, Lemma 4.3 and (5.4) yield

\\p*n~ P*n-l\\p2+"(Qt) + IIC - Cn-lll£2+c(rt)

•Jn Pn— 1IIP2 (Qt) T llSn - Sn

— 1 Sn —2 N _E2 + a(pt)

for any t G (0,T'). Again taking first ft' = 1/(4C30), and then

1 \1/x' ■

we obtain

II Pn ~ Pn-l\\P2+a (Qt) + lie ~~ Cn—1II B2+o(rt)

< ¿(IK-1 -P*n-2llE2^(Qt) + IIC-1 - CUII^r,)) for t G (0, To). (5.5)

Consequently, the sequence {(p* , £*)} converges to (p*, £*) uniformly in Pe2+a(QT ) x

0 0

^ 2+a

F0 (rTo), which satisfies

Wp*Wp^(qt ) + IIClB2+„(rT ) < 2C30M. (5.6)

The uniqueness of such a solution follows from the similar inequality as (5.5) for two solutions to problem (3.2).

6. Passing to the limit e ^ 0: Proof of Theorem 2.1

Theorem 2.2 means that for any £ G (0, £0] there exists a smooth solution (p*, Z*) of problem (3.2), and hence the following integral identity holds for any sufficiently smooth test function ^(r,0,t) with <^(r,0,T*) = 0 (0 < T* < T0):

- J ep*^ drd6 - J ^p*E(f drdOdt = J $e(p*, drd6dt. (6.1

Since the set of (p*, Z*) belongs to the compact subset

{(Pe.C) I M\\e2+^To) + IICIb2+„(rTo) < 2C30M}, one can select a convergent subsequence of it such that

(Pk ,Ck ) ^ (p*,c*) as ^ 0 (6.2)

in E2+a(QTo )xE2+a(TT0 )• Passing to the limit in (6.1) along the selected subsequence, we establish that the function (p*, Z*) is a solution of (3.2) with e = 0.

If problem (3.2) with e = 0 admits a unique solution on [0,Tq] for some T0* G (0, To], then the convergence result (6.2) holds for the full sequence, so that the proof of Theorem 2.1 is completed.

Finally let us prove the uniqueness of solution (3.2) with e = 0. Let (p*,Z*) and (p**,Z**) be solutions of (3.2) with e = 0 satisfying (5.6). Indeed, analogously to the case e > 0, with the help of the interpolation inequalities and (5.6) we can obtain

II(p*,Z*) - (p**,Z**),**(C*) - **(C**),* i (p*,Z*) - * i (p**,Z**),0)11.*;.

< (I3" + C^"F"(4C30M))(\\p*-p**\\El+am + IIZ* -ril^+-(rt))

for any t G (0,T0) and any ft'' > 0, where C^» is a positive constant depending on ft'' non-increasingly, x'' > 0 is a constant depending on a (possibly), F''(•) is a polynomial in its argument and

= E*a(QT) x £,1+a(r,T) x E*1+a(TT) x E^+a{rT). o o o o

Repeating the same arguments as in section 4 with Te, Ne and Ze replaced by

r0(x) = —77- log |x|, N0(x1,x2) = r(xi,x2) +T(xi,-x2) 2n

and

Zo(xi,t) = (^J^)-1 respectively, we have

1

s + MCI

- 1 * td=i ~ 2

\\P* ~P**\\El+°(Qt) + \\C -C*hl+~{rt)

< C30 (ft" + C^"F"(4C30M))(\\p* -p**\\El+am + lie - ril^(rt)) for any t G (0, T0). Again taking first ft'' = 1/(4C3'0), and then

i/x"-

Tq = min I Tq ,

1

,4C30Cß» F ''(4C30M )/ we conclude the uniqueness of a solution to problem (3.2) with e = 0 on [0, Tq ].

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Submitted June 19, 2018 Atusi Tani

Department of Mathematics, Keio University, 3-14-1 Hiyoshi, Yokohama 223-8522, Japan tani@math.keio.ac.jp

Hisasi Tani

Department of Mechanical Engineering, Texas A&M University, TX 77843-3123, USA, htani0926atamu.edu