ANALYSIS THE OF COATING FLOWS NEAR THE CONTACT LINE
AVNER FRIEDMAN* and JUAN J.L. VELAZQUEZ**
coat surfaces. Such ows are used for coating photographic lms, magnetic tapes, optical devices and, of course, for painting surfaces. Various techniques are used for depositing the uid on surfaces; the uid may be Newtonian or Non-Newtonian, often being a 2-phase uid. We refer to 9] 11] 13] 15] for a general introduction to the subject. The boundary of the ow region consists of three parts: (i) the part 0 in contact with the container from which the uid emerges; (ii) the part 1 in contact with air, and (iii) the part 2 in contact with the surface which is being coated. In many applications 2 is a part of!a plane, the upper surface of a horizontal substrate moving with a given xed velocity U. The surface 1 is a free boundary; it is one of the unknowns of the problem. The other unknowns are the velocity ! and pressure p in the uid. v Numerical methods based on nite elements have been developed by Kistler and Scriven 10]. One of the di culties is to determine the location of the so called contact line 0 = 2 \ 1 where the free boundary meets the substrate, and to analyze the shape of 1 near 0. There are a few existence results for steady coating ow problems with Newtonian uids. These are usually restricted to \quiescent" situations whereby the ow velocity is \very small," an assumption which does not usually cover industrial applications; see 14] 17] and the references given there. Other free boundary problems for viscous ow are treated in 2] 3] 8], but here again there is a \smallness" assumption. We also refer to 4] for additional articles on free boundary problems for viscous ow, mostly by numerical methods. In this paper we shall consider a 2-dimensional steady Stokesian ow. The free boundary is then a curve y = f(x); 1 < x < 0 with f(0) = 0. The top surface of the moving substrate is given y = 0, and the contact line 0 is identi ed with the origin (0,0); see Figure 1.
0. Introduction. The term \coating ows" refers to any viscous ow which is used to
*University of Minnesota, Institute for Mathematics and its Applications, Minneapolis, Minnesota 55455. This work partially supported by the National Science Foundation Grant DMS{86{12880. **Departamento de Matematica Aplicada, Universidad Complutense, Facultad de Matematicas, 28040 Madrid, SPAIN. Partially supported by CICYT Research Grant PB90{0235 and Fulbright Fellowship.
Figure 1
We wish to determine the existence of solutions and the behavior of the free boundary near the contact point 0 . The free boundary conditions are (0.1)
T ! = (f)! ; ! ! = 0 n n vn
( > 0)
where ! is the outward normal, (f) the curvature, and T is the stress tensor; is the n reciprocal of the capillary number. On the substrate we impose the no-slip condition: (0.2)
! = ! on y = 0 : vU
The no-slip condition causes some di culties near the contact point 0. It was pointed out by several authors (see 6] 9] and the references given there) that if the contact angle at 0 is not equal to 0 or , then the no-slip condition predicts unbounded force about 0; such a force is physically impossible. (In this sense the solutions established in 8] and 17] are not physically acceptable.) This motivated several di erent approaches based on (i) ! molecular theory near 0, (ii) rolling type motion near 0, (III) ! = U near 0, and other v adhoc assumptions (see 6] 9] 12] and the references therein). However the situation where the free boundary is tangent to the substrate at 0 has not been analyzed for coating ow. The purpose of this paper is to show that this situation 2
is consistent with the no-slip condition. More speci cally, we establish the existence of smooth solutions satisfying (0.1) and (0.2), with f 0 (0) = 0. ! v We shall study rst the \linearized problem" about ! = U near (0; 0). This is an eigenvalue problem. It has two sequences of eigensolutions, (0.3)
8 1 > (fn ; n) with fn = ( x) n ; < n =n+ 2 ; >ee : (fn ; n) with fen (x) = ( x)en ; ; en = n
2
1; 1 ; 22 1 2i = 21 i log 1 + 2i ; = U :
where is determined by (0.4)
1 For coating ow U > 0 so that 0 < < 2 (The case U < 0, or 1 < < 0, represents a 2 cavity problem with f(x; 0); x < 0g as the nose or obstacle.) By superposition we can form linearized solutions (f0 ; 0 ) with (0.5) Suppose, for such a pair, (0.6)
f0 =
X
n
(An fn + Bnfen ) :
8 > A( x) + higher order terms near x = 0 ; > 3 < 2 f0 (x) = > : B( x) + lower order terms near x = 1; 0 < < 1
and f0(x) > 0 if 1 < x < 0. Then we prove, and this is the main result of the paper, that there exists a solution to the coating problem (satisfying (0.1) and (0.2)) with free boundary (0.7)
y = "f(x) = "f0 (x) + "2f1 (x; ") ;
for any " > 0 small enough, where (0.8)
jf1 (x; ")j Cjxj jf1 (x; ")j Cjxj
+
near x = 0 ; near x = 1 ; 3
>0; >0:
The class of functions f0 satisfying (0.6) is not at all restrictive: Given any function g0(x), continuous and positive for 1 < x < 0, such that
B0( x) near x = 1 ; we can approximate it by functions f0 having the form (0.5) and satisfying (0.6). (The functions f0 (x)( x) approximate g0(x)( x) near x = 0 and the functions g0(x)( x) approximate g0(x)( x) near 1.) The conclusion of the paper is then that the no-slip condition is consistent with the free boundary conditions provided the contact angle is (as indicated in Figure 1). In x1 we describe the coating problem in detail and state the main results of the paper. In x2 we study the linearized eigenvalue problem. In x3 we transform the coating problem into an equivalent problem in the xed domain fy > 0g; we also outline the proof of existence. The details of the proof are given in xx4{9. In x9 it is also shown that any g0 as in (0.9) can be approximated by functions f0 to which the existence theorem can be applied. In this paper we consider the ow problem in fy > 0g and we do not impose boundary conditions at y = 1; this explains why we can obtain a large family of solutions. In a future work we shall extend the methods of this paper to study ows in domains such as 0 < y < h, where boundary conditions are imposed at y = h; for more details see Remark 9.1. ! x1. Statement of the main result. We are given velocity U = (U; 0) of the substrate surface fy = 0g where U is any real number; for the coating problem, U > 0. Denote by Ca the capillary number and set 1 = Ca ( > 0) : We wish to determine a free boundary
and a velocity function ! and pressure p in the ow region v = f(x; y) ; x 2 R and y > f(x) if such that (1.1) (1.2)
! = rp in v r ! = 0 in v
(0.9)
g0(x) A0 ( x)
near x = 0 ;
y = f(x) > 0 ( 1 < x < 0) with f(0) = 0 ; f 0 (0) = 0
1 < x 0 ; y > 0 if x > 0g ;
; ;
4
and (1.3) (1.4) (1.5) where T is the stress tensor 1 @v @v Tij = p ij + 2 @xi + @xj j i
! ! = ! on vU !!
2 1
= f(x; 0); x > 0g ; = f(x; f(x)); x < 0g ;
1
v n = 0 on
T ! = ! on n n
(x1 = x; x2 = y) ;
x
is the outward normal to 1, and is the curvature xx = (f) = (1 +ff 2 )3=2 : x
(fx 1) n = (1 + ;f 2 )1=2
v = ( y ; x) ; and reformulate problem (C) also in terms of (f; ). Introduce the sequences 1 n = 2 + n (n = 1; 2; . . . ) ; (1.6) en = + n (n = 1; 2; . . . )
where 2 1 ; 1 is uniquely determined by 22 = 21 i log 1 + 2i ; 1 2i
:
We refer to problem (1.1){(1.5) as problem (C). One may introduce the stream function
!
= U:
1 < <0 . 2 !! In x2 we prove that if we linearize problem (C) about v = U near (0,0), then we get a linear homogeneous problem for in fy 0g. This problem and the associated \free e boundary" f have eigensolutions (fn ; n ) and (fn; en ) where Notice that if U > 0 (U < 0) then < 0 ( > 0) and 0 < < 1 2
fn (x) = ( x) n ; fen (x) = ( x)en
5
( 1 < x < 0) :
We need to introduce some notation: B (x; y) = f(x; y); (x x)2 + (y y)2 < 2g ;
B (x) = B (x; 0); B = B (0; 0) ; B+ = B \ fy > 0g ;
+
0
= f(x; y); x > 0; 0 y = f(x; y); x < 0; 0 y =
+
0 0
0 xg
; ;
0
0 jxjg 0
<2: Set X = (x; y); r = jXj. In the sequel will denote any one of the sets + ;; BR and fy > 0gnBR ; where 0 0 < 1 ; R > 0 ; here 0 ( 0 ) denotes the x-axis (the negative x-axis). For X 2 and 2 (0; 1) we de ne e jD '(X)j = sup j'(X) '(X)j e jX Xj re e where the \sup" is taken over X; X in such that jX Xj < 2 ; jX Xj < r . 2 Let and be any real numbers. Let g(r) be a continuous positive function for 0 < r < 1, such that (1.7) g(r) = r if r < 1 ; g(r) = r if r > 2 : For any integer m 0 we de ne the norm X rm+ k X rjkj k jD '(X)j + k'kg ;m+ = sup g(r) g(r) jD '(X)j : jkj=m jkj m
0
where 0 <
0
0
Since the particular choice of g which satis es (1.7) will not be of any consequence to us, we shall use the less precise but more explicit notation (1.8) k'k ;;m+ = k'kg ;m+ ; g as in (1.7) : By superposing eigensolutions of the linearized problem we obtain a linear class of solutions (f0 ; 0 ); we denote this class by A. If (f0 ; 0 ) belongs to A and <1 kf0k ; ;4+ < 1 ; k 0 k ;
0 0 ;4+
for some > 0; > 0; 0 > 0, then we say that (f0 ; 0 ) belongs to A ;;4+ . We can now state the main result of this paper.
0
6
be positive real number such that > 3 ; 0 < < 1, 2 and = n or = ~n for some n. Let and be any positive numbers such that < min( ; 1 ) and < 1. Finally, let (f0 ; 0 ) 2 A ;;4+ for some 0 < 0 < 1 and positive and small. Then, for any su ciently small " > 0 there exists a solution (f" ; " ) of problem (C) with
Theorem 1.1. Let ;
0
(1.9) (1.10) where f1 satis es: (1.11) and
f" = "f0(x) + "2f1 (x; ") ;
"=
Uy + " 0 (x; y) + "2 1 (x; y; ")
kf1 k
+;
0
;4+
C <1
(x < 0) (x 0)
8 < 1 (x; y + "f0 (x) + "2f1 (x; "); ")) e1 (x; y; ") = : 1 (x; y)
k e1 k
+;
+ 0 BR ;4+ 0
(which is de ned in all of fy > 0g) satis es (1.12)
0 C <1 8 0<1; R>0;
the constant C is independent of " Notice that can be chosen such that > 1 . 2 Remark 1.1. The solution is not bounded in general; e may grow exponentially if x2 +y2 ! 1 and jy=xj > 1. Furthermore, we do not make any assertions about uniqueness. The main point of Theorem 1.1 is to exhibit a large class of smooth solutions of problem (C) in a neighborhood of the contact point. In a future work we shall extend the methods of the present paper to ows lying in regions such as 0 < y < h, and impose boundary conditions at y = h; see Remark 9.1 for more details. lem (C) about ! = U in a neighborhood of (0; 0). Set v
! !!!
x2. The eigenvalue problem. In this section we consider the linearization of prob!
v = U + G ; G = (Gx; Gy ) ;
7
`1 = f(x; 0); x < 0g ; l2 = f(x; 0); x > 0g :
Then (2.1) (2.2) (2.3) (2.4) (2.5) and (2.6)
G = rp = ! ; in the ow region ; r G=0
!! ! G n= U ! n !! ! !! !
!
9
nT(G; p)! = (f) n G = 0 on `2 :
T(G; p) n = 0
9 > > = > on the free boundary ; > ;
n Here ! is obtained by rotating ! clockwise by 90 , and 0 1 @Gx + @Gy ! B @Gx p 0 B @x ! 2 @y @x +@ T(G; p) = 1 @Gx + @Gy @Gy 0p 2 @y @x @y !! p0 + (G) 0p
is the stress tensor. On the free boundary
1
1 C C A
!
1
and since f 0 (x) 0 near x = 0;
is approximately `1,
0 n = (1(f ;f 021) ; + )1=2
!! U n=
Uf 0 (x) (1 + f 0 (x)2 )1=2
!
Uf 0 (x) ;
n j = (0; 1) ; ! i = (1; 0) : Hence, for the linearized problem, (2.3) and (2.4) become
(2:30 ) (2:40 )
and
!
!
Gy = Uf 0 on `2 ; @Gx + @Gy = 0 on ` : 2 @y @x
8
Equation (2.5) can be written as
p + ! (G)! = (f) ; n n
and since p is determined only up to an additive constant, this equation is equivalent to
!
dp + d ! (! ! = d (f) ds ds n G) n] ds where s is the length parameter with ds=dx > 0. From (2.1), dp = ( ! ! ( ! ! = G : G) G) i x ds @@ ! . Therefore by (2.4) and @s @x , d! (! ! = ! (! d! = ! (! ! = 0 ; n G) n n G) n G) n ds ds !! @ ! d ! ! !@ n ds (G) n j @x (G) j = @x @Gy : @y
We conclude that
Also d! = n=ds
@ d Gx + @x @Gy = dx (f) @y
provided Thus (2.5) becomes (2:50 )
2 Gx(x; 0) + @ Gy (x; 0) = @x@y
d f 00 (x) : dx >0 :
(x < 0) :
f 0 (x) = 0(jxj ) for some
f (3)(x)
From (2.2) we know that there exists a stream function :
!
G=(
y ; x)
and 2 = 0 in the ow region (by taking the curl of (2.1)). Conditions (2:30 ){(2:50 ), written for , read: (2:300 )
x (x; 0) = Uf 0 (x)
on `1 ;
9
(2:400 ) (2:500 ) (2:5000 ) Set
yy (x; 0)
xx (x; 0) = 0
on `1 ;
2 xxy (x; 0) + yyy (x; 0) =
f (3)(x) on `1 :
Substituting f 0 from (2:300 ) into (2:500 ), we get 2 xxy (x; 0) + yyy (x; 0) = U xxx(x; 0) on `1 : = U:
2
Then the nal version of the linearized problem for is the following: (2.7) (2.8) and (2.9) (2.10)
yy xx = 0
= 0 in fy > 0g ; if x > 0 ;
x (x; 0) = y (x; 0) = 0
at (x; 0); x < 0 ; at (x; 0); x < 0 :
(2 xxy + yyy ) 1 f 0 (x) = U
x (x; 0)
xxx = 0
The corresponding \linearized" free boundary is given by (2.11) (x < 0) with f(0) = 0 :
We may view (2.7){(2.10) as an eigenvalue problem. It has a sequence of solutions of the form = r B( ) when varies over the sequence (1.6) ((r; ) are polar coordinate about the origin). It will be more convenient, however, to use complex notation and write the eigenfunctions in the form (2.12)
b + Bz 1z (z = x + iy) b bb where and the complex coe cients A; B; A; B are to be determined.
(z; z) = Az + Bz
1 z + Az
We note that any biharmonic function in a disc jzj < has the form
r 2 h1 + h2
10
(r = jzj)
where each hj is harmonic function and thus has an expansion (Aj zm + Aj zm ) m m (j = 1; 2)
This suggests that should be a superposition of functions of the form (2.12) where is an integer 0. Since however the stream function is de ned only in fy 0g and may have a singularity at z = 0, we allow more general terms as in (2.12), where is still to be determined. We need to rewrite (2.7){(2.10) in complex notation using 1@ 1@ @ @ @ = 2 @x i @y ; @ = 2 @x + i @y ; or
@ = @ + @ ; @ = i(@ @) : @x @y By direct calculation we nd that (2.7){(2.10) become @ 2 @ 2 = 0 if y > 0 ; @ = 0 ; @ = 0 if x > 0 ; y = 0 ;
(@ 2 + @ 2 ) = 0 if x < 0; y = 0 ;
(2.13) (2.14) (2.15)
(2.16) i (@ 3 + 5@ 2@ 5@@ 2 @ 3 ) (@ 3 + 3@ 2@ + 3@@ 2 + @ 3)] = 0 if x < 0; y = 0 : Substituting (2.12) into (2.14) we obtain after some simple calculation, (2.17) (2.18)
b B + A+(
( 1)ei +(
( 2) A + (
A+(
b 1)B + B = 0 ; b 1)B = 0 :
(
Next, from (2.15) we obtain after noting that z = rei if x < 0, (2.19) 1)e i
(
2) A + (
b
1)( 11
2)ei 1)(
4) B
2)e i
(
4) B
b=0:
Finally (2.16) gives (2.20) ( 1)( 2)ei +( ( ( 1)( 1)( 1)(
(
3) (i
1)A
( 3) (i
2)( 2)e 2)ei
(
3)ei
(
b + 1)A
+ 3) + (
5) (i
1) + ( 1)(
1)( 2)(
2)ei
(
3) (5i
3)]B
bb In case = 1 or = 2 we can compute A; A; B; B directly and discover that 0. Hence we may assume that 6= 1; 6= 2. Factoring out 1 from (2.19) and ( 1)( 2) from (2.20), we arrive at
(2.21) (2.22) where (2.23)
3) (5i
3)e i
(
3) (i
b + 1)]B = 0:
ei A + (
b 2)ei B + e i A + ( b z 1A + e
3)(i
i
ei z1 + ei z2 B + e z1 = i
1 ; z2 = (
b z2 B = 0
b 2)e i B = 0 ;
1) + (5i
3) :
We now concentrate on the systems (2.17), (2.18), (2.21), (2.22) where the coe cients bb A; B; A; B are arbitrary unknown complex numbers. The determinant of the coe cients is 1 1 0 1 0 1 1 1 ei ( 2)ei ei ( 2)e i ei z1 ei z2 e i z1 e i z2 Add to the third row e i times the rst row and ei times the second row. Then add to the fourth row ei z1 times the rst row. The result is 1 1 0 1 0 1 1 1 0 0 2 cos 2( 2) cos 0 ei (z2 ( 1)z1 ) e i z 1 e i z2 ei z1 If we add to the fourth row e i (z2 ( 1)z1 ) times the second row, we nd that the determinant vanishes if and only if either (2.24) cos 12 =0
or ( 2) e i z1 ei (z2 ( 1)z1)] = 0 : Substituting z1 ; z2 from (2.23) we nd that (2.25) is equivalent to 1 2i (2.26) e2i = 1 + 2i ; or (2.27) (2.28) or, (2.29) = +n (n = 1; 2; . . . ) where is uniquely determined by 1 2i 1 e2i = 1 + 2i ; 1 < < 2 2 1 = 21 i log 1 + 2i ; 2 < < 1 : 1 2i 2 (2.25)
e
i
z2 ei z1 (
1)ei (z2 (
1)z1) ;
Note that n 0 gives in nite velocity for ! at (0; 0), and is therefore excluded from (2.27). v The solutions to (2.24) are given by 1 (2.30) = 2 + n (n = 0; 1; 2; . . . ) : Observe that the two sequences (2.27) and (2.30) are mutually disjoint; (2.30) is a limiting case of (2.27) when j j ! 1. Consider rst the case (2.30). We proceed to solve the system (2.17), (2.18), (2.21), (2.22) for this case. From (2.17), (2.18) we get (2.31) (2.32)
A=
b 1B B;
b A=
1B B; b
Since e2 i = 1, equation (2.21) depends linearly on (2.17), (2.18) and may therefore be dropped. Substituting (2.31), (2.32) into (2.22) we obtain a linear relation between B and b B, from which we deduce, after some calculation, that b (2.33) B = 1+ i B : 1i 13
Therefore (2.31) and (2.32) become (2.34) (2.35) 2 i B; 1+ i i b A = 1 1 + ( 1) 1 + i B : 1
A= 1
We nally need to choose B so as to obtain a real valued solution , i.e., a solution with (2.36) Setting (2.37) the second relation in (2.36) becomes (2.38) 1i e2i = 1 + i :
b b A = A and B = B :
B = jBjei ;
An easy calculation shows that with this choice of , also the rst relation in (2.36) is satis ed. Consider next the case (2.26). Equations (2.31), (2.32) are still valid. Next we use (2.26) in (2.23) and obtain, after some simple calculation which make use of (2.31), (2.32), (2.39)
b B= B:
A=
2B: 2 B; A= b 2
Consequently, equations (2.31), (2.32) become (2.40)
We now choose B as in (2.37) with such that (2.41) It follows that so that is real-valued. We summarize: 14
e2i = 1 :
b B = B and A = A ;
having the form (2.42) (z; z) = A z + B z 1z + A z + Bz 1z where either (2.43) A = 1 2i i B ; = 1 + n (n = 0; 1; 2; . . . ) ; + 2 or (2.44) A = ( 2)B ; = + n (n = 1; 2; . . . ) ; here A and B are the complex conjugates of A and B respectively, B has the form (2.37) 1 and is de ned by (2.38) if = 2 + n and by (2.41) if = + n. 1 We shall now superpose these solutions. We begin with the case = 2 + n. From (2.43), @ (A z ) = 2i B z 1 B z 1 : @z 1+i Setting (2.45) we then have (2.46) and, by (2.37), (2.38), (2.47) In case = Setting (2.48) we have (2.49) and, by (2.37), (2.41), (2.50) We summarize:
1 (z) 1 (z)
Theorem 2.1. There exist two sequences of real-valued solutions of (2.13){(2.16)
=
1 X
n=0
1 (z)
An z
1 2
+n
;
1 (z) 1 (z)
=
1 X
n=0
Bn z
1 2
+n
@ @z
=!
= 1 2i i +
1 X
n=0 nzn
z
@ @z
1 (z)
z1=2
;
n 2 R ; !2
1i = 1+ i :
1
+ n we have, by (2.44), @ (A z ) = 2 B z @z z
2 (z)
Bz
2 (z)
:
+n
=
1 X
n=1
Anz
2 (z)
+n
;
=
1 X
n=1
Bn z
;
@ @z
2 (z)
= 2 2z(z)
1 X
n=1 nzn
@ @z ;
2 (z)
= iz
n2R:
15
Theorem 2.2. The system (2.13){(2.16) has solutions of the form
(2.51) where
1; 2
(z; z) = 2Re
2 X
j=1
j (z) + j (z) z
1 (z); 2 (z)
z
are given by (2.47), (2.50) and
are given by (2.46), (2.49).
This solution will be su ciently general for our purposes. Let us compute the linearized free boundary y = f(x) corresponding to . From (2:300 ) we have 2 f(x) = U Re( 1 +
1 2 X ! 2i = U Re 1+i n=1
"
2+
1+
2)
1 2
(z = z = x < 0)
zn+ + 2ie zn nn n 1 n+ 2
#
;
We have taken
0
= 0, since f 0 (x) must vanish at x = 0. Setting r = x and noting that Re ! 1 2i i +
6= 0 ; Re iz
6= 0 if z = x ;
( n 2 R; en 2 R) ;
we deduce that (2.52)
f(x) =
1 Xh
n=1
1 n rn+ 2
en rn+ i +
the coe cients n; en are not the same as before, but they are nevertheless arbitrary. x3. The perturbation problem. Consider problem (C). We shall try to nd a solution with (3.1) velocity ! = U + "G, pressure "p and free boundary v 1 < x < 0g 1 = fy = "f(x);
! != n ! !
where " > 0 is su ciently small and G; p; f depend on ". The outward normal is ("f 0 (x); 1) 1 + "2 f 0 (x)2 ]1=2
and the unit tangent vector is taken to be
!
(1; 0 (x)) = 1 + "2"f0 (x)2 ]1=2 : f 16
Set
ds = 1 + "2 f 0 (x)2 ]1=2dx ;
" = f(x; y); x < 0; y > "f(x)g
f(x; y); x 0; y > 0g :
Then (C) becomes: (3.2) and (3.3) (3.4) (3.5)
! !
G = rp ; r G = 0 in the ow domain G = 0 on
!
!
"
2
;
1
U(1; 0) ("f 0 (x); 1) + "G(x) ("f 0 (x); 1) = 0 on
! pI + (G) ! = n
; :
f 00 (x) ! 2 f 0 (x)2 ]3=2 n on 1+"
1
We wish to simplify (3.5). Taking the scalar product with ! and then di erentiating n along 1, we get (3.6) Also, from (3.2),
dp + d ! (! ! = ds ds n G) n
d f 00 (x) ds 1 + "2 f 0 (x)2 ]3=2 :
dp = ( ! ! : G) ds Since d! = !=R (R = radius of curvature) and n=ds
! !! ! ( f 00 (G) n = !( pI + (G))! = ! 1 + "2 f 02)3=2 ! = 0 ; n ]n
we also have
Hence (3.6) becomes (3.7) ! ! nd n ( G) !+ 1 + "2f1(x)2 ]1=2 ! dx (G) ! = 0 17
d! (! ! = ! (! d! = 0 : n G) n n G) n ds ds d f 00 (x) 1 + "2 f 0 (x)2 ]1=2 dx 1 + "f 0 (x)2 ]3=2 on
1:
From (3.5) we also have (3.8)
! !!
(G) n = 0 on
!
1
:
We shall reformulate this problem in terms of the stream function , where
G=(
y ; x) ;
the existence of is equivalent to the condition r G = 0. ! Equation rG = rp is equivalent to (3.9) Conditions (3.3), (3.4) become (3.10) (3.11) Since
x (x; 0) = y (x; 0) = 0
2
!
= 0 in
"
:
if x > 0 ; if y = "f(x) ; x < 0 : 1( 2 xx
yy ) xx xy yy ) yy ) C
Uf 0 (x)
x = "f 0 (x) y xy
2 ( xx we can easily check that (3.8) reduces to (3.12)
yy xx = 4" xy f 0 (x) yxx + yyy
0 ! (G) = B @1
1 A
1:
"2 f 0 (x)2 (
"Q1 ("; f; ) on
Next, from (3.7) we get by direct calculation
"f 0 (x)(
xxx + xyy )] yyx )
or
+ 1 + "21f 0 (x)2 ] xxy "f 0 (x)( xxx 00 d = dx (1 + "f f(x) 2 ]3=2 ; 2 0 (x) 1 + "2 f 0 (x)2 ] 1
"2 f 0 (x)2
yxx
2 xxy + yyy 2"f 0 (x) xxx + "2 f 0 (x)2 yyy
xxx + xyy )] =
"3 f 0 (x)3 (
d f 00 (x) dx 1 + "2 f 0 (x)2 ]3=2 :
18
Expanding the right-hand side we can rewrite the last equation in a more convenient way: 2 xxy + yyy + f (3)(x) = 2"f 0 (x) xxx "2 f 0 (x)2 yyy + "3f 0 (x)3 ( xxx + xyy )
0 00 (x)2 + 3 "2 1 f (x)f0 (x)2 ]3=2 ; + "2 f where all the terms on the right-hand side are O("). It will be useful to eliminate f (3) from the left-hand side of the last equation. We do this by di erentiating (3.11) twice, then solving for f (3) and substituting into the left-hand side. This results in the equation (3.13) (2 xxy + yyy ) xxx = 2" f 0 (x) xxx + "f (3)(x) y
2 f 0 (x)2 )1=2 f (3) (x) 1 (1(1 +2" 0 (x)2 )1=2 +" f
+ 2"f 00 (x) yx + " f 0 (x) yxx "2 f 0 (x)2 yyy + "3 f 0 (x)3 ( xxx + xyy )
2 0 2 )1=2 f 0 00 (x)2 3U"2 (1 + (x)f0 (x)2 )3=2 "Q2 ("; f; ) on 1 ; + Uf (3) (x) 1 (1(1 +2" 0f (x)1=2 + " f (x)2 ) "2f where = U= . We shall refer to problem (C) in its formulation (3.9){(3.13) also as problem (C"). It is not convenient to work directly with the system (3.9){(3.13) since the ow region is unknown. We shall therefore perform a change of variables which transforms " onto " fy > 0g. Let (t) be a function satisfying:
(3.14)
2 C 5 0; 1]; 0 (t) 0; and
(t) = 1 if 0 t
1 ; (t) = 0 if t 3 : 20 40 We shall be working with positive functions f(x) which approximate A( x) for x near 0 and B( x) for x near 1, where A; B are positive constants and > 1 and 0 < < 1. Therefore, for any given 0 > 0, the free boundary f(x; "f(x)); 1 < x < 0g is contained in =4 if " is small enough. We introduce the function
0
(3.15)
(x; y) =
x; y + "f(x)
19
jxj
y
if x < 0
= (x; y) if x 0
and note that = 1 on the free boundary, so that the free boundary for is mapped onto the negative real axis for . We also note that (x; y) = (x; y) if (x; y) 2 : =
0
We now wish to rewrite the system (3.9){(3.13) in terms of . Take for example equation (3.11). Using the relations (3.16)
@x = @x + "@y @y = @y + "@y Uf 0 (x)
yf f 0 + jxj2 0 ; f0 ( x) yf f 0 + jxj2 0 + f 0 @y :
we can rewrite (3.11) in the form (3.17)
x="
@y
We can use (3.16) to rewrite the expression in braces in (3.17) in terms of x and y : it becomes a linear combination (with bounded coe cients) of products of f 0 or f=x by x or y . The same calculation can be carried out for the other equations, using the relations
@ k = @ (@ )@y ]k :
System (3.9){(3.13) then takes the form (3.18) (3.19) (3.20) (3.21) (3.22)
2
= "H4 ; f] in fy > 0g ; if x > 0 ;
x (x; 0) = y (x; 0) = 0
Uf 0 (x)
yy
x = "H1 xx = "H2
; f] at (x; 0); x < 0 ; ; f] at (x; 0); x < 0 ; ; f] at (x; 0); x < 0
0
(2 xxy + yyy )
xxx = "H3
where Hj ; f] is a linear combination (with bounded coe cients) of products of derivatives of f of order j by derivatives of of order j, and H4 ; f] = 0 outside . The Hj ; f] are linear in the derivatives of . 20
The functions f and Theorem 1.1, we set (3.23) We can then write (3.24) where (3.25) (3.26)
0 (x; y; ") 1 (x; y; ")
will actually depend on " and, to conform to the notation of = 0 (x; y) + " 1 (x; y; ") ;
f = f0(x) + "f1(x; ") :
(x; y) = 0(x; y; ") + " 1 (x; y; ")
= =
0
0 1
x; y + " f0 + "f1 (x; ")] x; y + " f0 (x) + "f1 (x; ")]
jxj
y
; y ;" : jxj
Note that 1 coincides in =2 with the function e1 (de ned in Theorem 1.1). Since (f0 ; 0 ) is a solution to (2.7){(2.11), we can rewrite (3.10){(3.22) as a system of equations for (f1 ; 1) as follows: (3.27) (3.28) (3.29) (3.30) (2 and (3.31)
0 Uf1
1 1;x = H1 2 0 ; f0 + "f1 ] + "H1 0 1 ; f0 + "f1 ] 1;yy 2 1 = "H4 22 0 ; f0 + "f1 ] + " H4 1;x (x1 0) = 1;y (x; 0) 1 1;xx = H2 1 ; f0 + "f1 ]
in fy > 0g ;
= 0 if x > 0 ;
1 ; f0 + "f1 ]
2 0 ; f0 + "f1 ] + "H2
at (x; 0); x < 0 ; at (x; 0); x < 0
1;xxy + 1;yyy )
1 1;xxx = H3
2 0 ; f0 +"f1 ]+"H3
1 ; f0 +"f1 ]
at (x; 0); x < 0 :
22 2 Actually H4 ; H2 and H3 depend also on
and ".
21
Lemma 3.1. If
(3.32) and (3.33)
kf0k kf1 k
;
0
;
0
;4+
C0 ; k 0k
;
0
;
0
;4+
C0
(C0 > 1)
;4+
C ; k 1k
;4+
C
(C > C0) ;
then the functions
1 1 Hm = Hm 0 ; f0 + "f1 ] 2 2 ; Hm = Hm 1 ; f0 + "f1 ]
satisfy: (3.34)
1 kH4 k2
; 0
5;2
5
2 e C(C0 + "C ); kH4 k2
; 0
5;2
5
e C(C ) ; e C(C ) (0 i 3)
i;2 (3.35) kHi1 k2 ;41 i+
0
1i
e C (C0 + "C ) ; kHi2 k2
1 i;2 1 i ;4 i+ 0
e where C(t) is a positive monotone increasing function of t.
1 Proof. consider the relation (3.31). It is obtained from (3.17) by using (3.23). H1 contains the terms
(3.36)
@y
0
0 0 f0 + yf20 0 + f0@y x
0
:
Using (3.32), (3.33), the estimate (3.35) for i = 1; j = 1 readily follows. The proof for 2 0 0 i = 1; j = 2 is similar; H1 contains products of f0 or f1 by rst derivatives of 0 or 1 . The proof of the other estimates in (3.35) and (3.34) can also be established by similar straightforward calculation. We now outline the proof of Theorem 1.1. Outline. Denote by X(C ) the set of pairs f1 (x); 1(x; y) satisfying (3.33). For any such pair, consider the problem (3.37) (3.38)
2 1 = H4 2 0 ; f0 + "f1 ] + "H4 1 ; f0 + "f1 ]
in fy > 0g ;
x (x; 0) = y (x; 0) = 0
if x > 0 ;
22
(3.39)
yy
1 2 xx = H2 0 ; f0 + "f1 ] + "H2 1 ; f0 + "f1 ]
at (x; 0); x < 0 ; at (x; 0); x < 0
(3.40) (2 xxy + yyy ) and (3.41)
2 1 xxx = H3 0 ; f0 +"f1 ]+"H3 1 ; f0 +"f1 ]
UF 0
1 2 x = H1 0 ; f0 + "f1 ] + "H1 1 ; f0 + "f1 ]
at (x; 0); x < 0 :
We shall construct (in xx6{8) a special solution (x; y) of (3.37){(3.40) and then de ne F(x) by (3.41), and a mapping T by
T(f1 ; 1) = (F; ) We shall prove that if C is chosen large enough then T maps X(C ) into itself and is a contraction; hence it has a xed point which is a solution to problem (C"). To construct we rst establish in xx4,5 a useful representation for biharmonic functions in fy > 0g satisfying mixed boundary conditions
(3.42) Consider (3.43)
2
8 < yy xx = g1; (2 xxy + yyy ) : x(x; 0) = y (x; 0) = 0 if x > 0 :
xxx = g2
at (x; 0); x < 0;
= H in fy > 0g
together with (3.42), where H and g1; g2 satisfy estimates similar to those derived in Lemma 3.1 for the right-hand sides of (3.27), (3.29) and (3.30). In x6 we construct a special solution W of (3.43) such that (3.44)
kWk
+;
+ 0 BR ;4+ 0
e C(C )
for any 0 < < 1; 0 < < 1. In x7 we add to it a biharmonic function ' whose boundary data 'x(x; 0), 'y (x; ; 0) ( 1 < x < 1) coincide with Wx(x; 0); Wy (x; 0) respectively. In x8 we add to W + ' another biharmonic function ' so as to obtain a b function = W + ' + ' satisfying (3.43) and (3.42), as well as an estimate such as (3.44). b The function ' consists of the special solution constructed in x5 plus an appropriate nite b linear combination of eigenfunctions of the linearized problem. In x9 we apply the above construction to the special case of (3.37){(3.40) in order to show that the mapping T, which is well de ned by the above construction, maps X(C ) into itself and is a contraction. We nally prove, also in x9, that the class of functions f0 , with (f0 ; 0 ) as in Theorem 1.1, is dense in a natural weighted norm. 23
x4. Mixed boundary conditions. Consider the problem
(4.1) (4.2)
2' 1
= 0 in fy > 0g ;
'1;x(x; 0) = (x); '1;y (x; 0) = 0 for x > 0
where (x) is the Dirac function. Using the Fourier transform in x one can easily derive the following formulas (which are special cases of results obtained in 1]).
Lemma 4.1. There exists a solution '1 of (4.1), (4.2), which is given by
(4.3) (4.4)
y3 Dx'1(x; y) = 2 (x2 + y2 )2 ;
2 Dy '1(x; y) = 2 (x2 xy y2 )2 : +
Consider next the problem (4.5) (4.6)
2' 2
= 0 in y > 0 ;
'2;x(x; 0) = 0; '2;y (x; 0) = (x) for y = 0 :
2 Dx'2(x; y) = 2 (x2 xy y2)2 ; + 2 Dy '2(x; y) = 2 (x2 x y 2)2 : +y
Lemma 4.2. There exists a solution '2 of (4.5), (4.6), given by
(4.7) (4.8)
In this and in the next section we are concerned with the mixed boundary value problem: (4.9) (4.10) (4.11)
2
= 0 in fy > 0g ; if x > 0 ; if x < 0 ;
x (x; 0) = y (x; 0) = 0
( yy
xx )(x; 0) = g1 (x)
24
(4.12)
(2 xxy + yyy )(x; 0)
xxx (x; 0) = g2 (x)
if x < 0 :
Observe that the boundary operators change abruptly from x > 0 to x < 0. Problems of this type have been considered in earlier papers (see 16] and the references therein) where Lp and C estimates were derived. However we shall need much sharper estimates. To derive such estimates (in x8) we shall establish an integral representation for a solution of (4.9){(4.12). Our approach is motivated by 16;x3]. We rst write the boundary conditions in two di erent forms: (4.13) and (4.14)
x (x; 0) = h1 (x); y (x; 0) = h2 (x)
(x 2 R) ;
8 < ( yy xx )(x; 0) = g1(x) (x 2 R) ; : (2 xxy + yyy )(x; 0) xxx(x; 0) = g2(x) (x 2 R) ;
and then try to express the hi in terms of the gi. This step will be carried out in the present section. In x5 we shall use the Green function constructed in Lemmas 4.1, 4.2 in order to represent the solution of (4.9){(4.12) in terms of the functions g1(x); g2 (x) (de ned for x < 0, only). The subsequent calculations are somewhat formal. They can be justi ed however without my di culty if hi and gi have compact support on decay fast enough at in nity. These conditions on hi; gi are satis ed in the actual applications in this paper. The Fourier transform b( ; y) satis es
2 (i )2 + Dy ] b = 0 :
We are interested only in solutions that do not grow exponentially as y ! 1, and therefore take (4.15) (4.16) and (4.17)
b( ; y) = A( )e
j jy + B( )ye j jy :
We can rewrite (4.13), (4.14) in the form
i bx ( ; 0) = b1( ); by ( ; 0) = h2( ) h
8b < ( yy (i )2 b)( ; 0) = b1( ) ; g : 2(i )2 by + byyy ]( ; 0) + (i )3 b( ; 0) = b2( ) : g
25
We want to simplify (4.17) by using (4.15). This can be done along the general procedure used in 1]. Set
e B1( ; ) =
Notice that
2 2
2
e B2( ; ) = (2
;
+ 3)
3
:
e e B2(i ; Dy ) = iB2( ; iDy )
and therefore (4.17) can be written in the form (4.18) Set
e e B1(i ; Dy ) = B1( ; iDy ) ;
e iB2( ; iDy ) b = b2( ) : g
e B1( ; iDy ) b = g1( ) ; b
M+ ( ; ) = ( ij j)2 = ( 2 2 ) 2ij j : e Dividing Bi( ; ), as a polynomial in , by M+, we get
e e0 B2( ; ) = M+( ; )( + 2 ij j) + B2( ; ) ; e0 B2 ( ; ) =
2
e e0 e0 B1( ; ) = M+( ; ) + B1( ; ); B1( ; ) = 2ij j ;
+ 3(2 i sgn 1) :
Since, by (4.15), M+( ; iDy ) b = 0, the boundary conditions in (4.18) simplify to
e0 g B1( ; iDy ) b = b1( ) ; e0 iB2 ( ; iDy ) b = g2( ) b
and then, by (4.16), (4.19)
g b1( ) = 2(i )(i sgn )b2( ) ; h g b2( ) =
26
(i )2 b2( ) + (i )2 2 i sgn h
1]b1( ) : h
We now introduce the Hilbert transform
H'(x) = 1
Z1 '(y) x y dy
1
where the integral is taken in the sense of principal value (P:V:). Then
d H '( ) = (i sgn )b( ) : '
Consequently (4.19) may be written in the form (4.20) (4.21) 2
Z1 Dh2( )
1
x
d = g1(x) ; x
x
2 Dxh1 (x) +
2
Z1 D2 h1( )
1
d=
2 Dxh2(x) g2(x) :
We now assume that (4.22) Then (4.23) 2
h1(x) = h2(x) = 0 if x > 0 :
Z0 Dxh2( ) d = g1(x) ; x
1
(4.24)
2 Dxh1 (x) +
2
Z0 D2 h1( ) x 2 d = Dxh2(x) g2(x) : x
1
To invert the operators on the left-hand sides we shall use the following well known relations (see 7; pp. 159{160 and 187{190]): if 1 (4.25)
Z1 '(y)
0
2 x y dy = f(x); f 2 L (0; 1) then
1 C 1 Z x + y p dy f(y) '(x) = px 2 x y yx
0
27
where any constant C is possible; if '(x) 2 (4.26)
Z1 '(y)
0
dy = f(x) ; f 2 L2(0; 1) and xy
Z1 f(y)y f(x) + 2 x 2 2 ( 1; 1) then '(x) = = 1 + 4 2 (1 + 4 2) x y dy
0
where is the real number de ned in (2.28) or (2.29). Taking in (4.25) y = ; '(x) = 2Dh2( x); f(x) = g1( x), we obtain from (4.23)
0 1 Z x + p g1(p) d : C 2Dxh2(x) = p x x2 x
1
The last integral can be written as
Z0 x
1
x
p g1(p)
xd +
Z0
1
x
2
2
p g1(p)
The rst and last integrals are equal to const =
p x and the middle integral is equal to
Z0 2 d + ( ) p g1(p) x d : x 1
p1
Hence, for a suitable choice of C, (4.27)
Z0 2xg1 ( ) p (x ) d : x1
p x Z0 g ( ) p 1(x ) d : Dxh2(x) = 2 1
2 Dx h1(x) = 1 +14
Similarly, by using (4.26) we obtain from (4.24)
2 2 Dxh2(x) + g2(x)]
(4.28)
2( + (1 + 4x)2)
Z0 ( ) 2 Dxh2( ) + g2( )]d : x
1
28
Our goal in the next section is to establish integral representation for a solution of (4.9){ (4.12). This representation is given in Theorem 5.1. The proof is based on Theorem 4.3 below, whose proof however is valid only if (4.29) 1 0< < 2 :
We shall rst assume that (4.29) is satis ed and prove both Theorem 4.3 and part of 1 Theorem 5.1. We shall then use analytic continuation to extend this part to 2 < < 0, 1 and then proceed to complete the proof of Theorem 5.2 for all 2 < < 1 . 2 is a bounded solution of (4.9){ (4.12) and set h1(x) = x(x; 0); h2(x) = y (x; 0) for x < 0. If h1(0) = h01(0) = 0 and h2(0) = 0 then (4.30)
Theorem 4.3. Assume that (4.29) holds. Suppose
h2(x) =
Z0 p
x
2
d
Z0
1
p g1(( ) ) d ;
(4.31)
h1(x) = 1 +14
Z0 Z
2
x
d
2 (1 + 4 2)
Z0 Z
x
1
2 Dxh2(q) + g2(q)]dq
d
1
Z0 ( ) 2 ( q) dq q Dxh2( ) + g2 ( )]d :
1
Proof. Formula (4.30) follows by integrating (4.27). To prove (4.31) we need to show R0 of the right-hand side of (4.28) vanishes. Equiva(since h01(0) = 0) that the integral 1 lently, if we set Z1 f(y)y f(x) + 2 x '(x) = 1 + 4 2 (1 + 4 2 ) x y dy
0 2 where f(x) = ( Dx h2 + g2)(x), then we need to show that
(4.32)
Z1
0
'(x)dx = 0 :
29
We have (1 + 4 (4.33)
2)
Z1
0
'(x)dx =
Z1
0
f(x)dx + 2
Z1
0
x
Z1 f(y)y
0
x y dy
=
Z1
0
f(x)dx + 2
Z1
0
Z1 x y f(y)dy x y dx
0
provided we may use Fubini's theorem for the P:V: integrals; this will be justi ed later on. 1 Since 0 < < 2 we have (setting x = uy) (4.34) Using contour deformation and the residue theorem we also get Z1 u 2 i) (1 e du = i(1 + e 2 i ) ; u1 0 or Z1 u u 1 du = 2 : Thus the left-hand side of (4.34) is equal to 2 and, therefore, the right-hand side of (4.33) vanishes; this completes the proof of (4.32). It remains to justify the change of order of integration in (4.33). Setting h(x; y) = f(y)x y we have 1 h(x; y) = O x1 O y(3=2) as x ! 1; y ! 1 : We may write
0
Z1 u Z1 x y x y dx = u 1 du : 0 0
(4.35)
Z1 Z1 h(x; y) dx x y dy 0 0 8 9 Z1 > Z h(x; y) Z h(x; y) > < = = dx > dy + lim dy> !0 xy; :jy xj " x y 0 jy xj<" 2 3 Z h(x; y) 7 Z1 Z h(x; y) Z1 6 = dy !0 x y dx + dx 4 lim x y dy5 :
0
jx yj "
0
jy xj<"
30
From the form of h and its smoothness,
Z
jy xj "
h(x; y) dy = xy
Z
jy xj "
h(x; y) h(x; x) dy = ("; x) xy
where ("; x) = O(") and, in fact,
Z
j ("; x)jdx = O(") ; uniformly in
:
Hence the last term in (4.35) is O("). Similarly the rst term on the right-hand side of (4.35) di ers from Z1 Z1 h(x; y) dy x y dx by O("). Taking " ! 0 in (4.35), the change of order of integration in (4.33) is thereby justi ed. Remark 4.1. From (4.30), (4.31) we see that
0 0
h1(x) jxj1 ; h2(x) jxj1=2 as x ! 1 :
x5. Representation of the solution of (4.9){(4.12). Set
(5.1) Then by (4.31). (5.2) (q) = 1 +14
2 2 Dx h2(q) + g2(q)] :
h1(x) = dq1
x
Z0
Zq
1
1
(q2 )dq2
2
Z0
x
dq1
Zq
1
1
Z0 ( ) ( q2 ) dq2 q ( )d : 2
1
Introduce the contour C and a small circle about q2 , as shown in Figure 2.
31
Figure 2
We shall temporarily assume that (5.3) is holomorphic in a domain D containing C and the negative x-axis, and (q) ! 0 if q 2 D; jqj ! 1 :
Then, by Cauchy's theorem,
(5.4)
Z0 Z ( q2 ) ( ) ( )d = P:V: (e2 q2
C
i
Z ( q2 ) ( ) + ( )d q2
1
1) ( q2q) ( ) 2
( )d
where we have taken the branch of ( ) arg( ) = 2 arg( ) = 0
( = x + iy) for which on fx < 0; y = 0+g ; on fx < 0; y = 0 g : 32
When shrinks to q2 the integral over converges to
i(e2
2 (e2 i 1)
i + 1)
( ):
Substituting this into (5.4) and using (2.28), we get
1
Z ( q2 ) ( ) 2 Z0 ( q2 ) ( ) = P:V: ( )d q2 q2 1 C
1
(q2 ):
Substituting this into (5.2) we obtain a simpler expression for h1: (5.5)
h1(x) = 2 (e2 i 1)
Z0
x
dq1
Zq
1
1
Z ( q2 ) ( ) dq2 ( )d : q2
C
We shall now use the Green functions '1; '2 introduced in Lemmas 4.1, 4.2 in order to represent the solution of (4.9){(4.13) in terms of h1; h2 as de ned by (5.5) (4.30) for x < 0 (recall (5.1)). We write (5.6) where
21
= =
22
1+ 2
= 0 and
Dx 1 = h1 ; Dy 1 = 0 at y = 0 ; Dx 2 = 0 ; Dy 2 = h2 at y = 0 :
By Lemma 4.1 (5.7) (5.8) where (5.9)
Dx 1 (x; y) = Dy 1(x; y) =
Z0 Z0
1
K1(x K2(x
; y)h1( )d ; ; y)h2( )d
1
8 > K1 (x; y) = 2 2 y3 2 2 ; > < (x + y ) > > K2 (x; y) = 2 2 xy2 2 2 : :
(x + y ) 33
In the sequel we let K denote either K1 or K2 ; K(x with poles at = x iy. We wish to evaluate
; y) is a meromorphic function in
I
(5.10)
Z1
1
K(x
; y)h1( )d
1
= 2 (e2 i 1)
Z0
1
d K(x
; y) dq1
Z0
Z0
1
Z ( q2 ) ( ) dq2 ( )d : q2
C
We shall use the following relations, valid for any L1 function g,
Z0
=
dq1
Zq
1
Z
1
1
g(q2 )dq2 = dq1
Z0
Z
1
gdq2 + dq1 dq2
Z0
Zq
1
dq2 gdq1 + dq2 gdq1 =
q2
Z0
Z0
Z0
Z
1
g(q2 )dq2
Z0
q2g(q2 )dq2 :
Using this in (5.10) yields
I = 2 (e2
i
1)
1
Z0
1
d K(x
3Z 2Z Z dq2 + q2 dq25 ( q2) ( ) ( )d ; y) 4 q2 1 0 C
1
and, by changing the order of integration, (5.11) where
I = 2 (e2 i 1)
Z0
1
d K(x
; y) d ( )( ) H( ; )
C
Z
(5.12)
Z ( q2 ) dq2 ; H( ; ) = G( ; ) + Q( ; ) ; G( ; ) = q2 Z0 q2( q2 ) Q( ; ) = dq2 : q2
34
1
We note that for xed x; y; ,
; y)j j Cj3 for j j large. Using these estimates and (5.3), we may change the order of integration in (5.11):
(5.13)
jH( ; )j Cj j1 ; jK(x
(5.14)
I = 2 (e2 i 1)
1
Z
C
d ( )( )
Z0
1
K(x
; y)H( ; )d :
We next introduce a contour as shown in Figure 3.
Figure 3
xed the functions G( ) = G( ; ) and Q( ) = Q( ; ) are analytic in in x < 0g. If Im > 0, the integral de ning G( ) can be taken as any contour (from 1 to ) above the x-axis and, if Im < 0, as any contour below the x-axis; the same holds for Q( ). Also, since
R2 nf(x; 0);
For
arg(
2 + i0)
=2
2 i G(
; arg(
2
i0) = 0 for
2i
2
>0; i0)
the functions G( ); Q( ) in (5.12) satisfy
G( + i0) = e
i0); Q( + i0) = e
35
Q(
for 1 < < 0. It follows that
Z
K(x
; y)H( )d =
2 i)
Z i0
= (1 e
Z0
1
1 i0
K(x
; y)H( i0)d :
2 i) 1
i0)d
+i0 Z
1+i0
K(x
; y)H( + i0)d
K(x
; y)H(
Consequently, (5.15)
I = 2 (e2 i 1) 1 (1 e
Z
d K(x
; y) d ( )( ) H( ; ):
C
Z
Consider the integral
J
Z
C
d ( )( ) H( ; ) = d ( )( )
C
Z
Z ( q2 ) dq2 q2
1
Z0 q2 ( q2) dq2 ; + q2
2
and deform C into a contour C 0 as shown in Figure 4.
Figure 4
36
Each time we cross a point q2 (in ) we get an additional term due to the pole of ( )( ) =(q2 ) at = q2 . After changing also the order of integration, we end up with
Z d ( )( ) Z d ( )( ) Z1 + dq2( q2 ) q2 dq2 ( q2 ) J= q2 q2 1 C0 C0 Z
+2 i
Z
Substituting this into (5.15) and collapsing C 0 onto the negative real axis, we obtain
1
(q2 )dq2 + 2 i q2 (q2 )dq2 :
Z0
I = 2 (e2
i
1) 1(1 e 1)
2 i) 1
Z
K(x
2Z 3 Z0 ; y) 42 i (q2 )dq2 + (q2 )q2 dq25
1
+ (e2 i
Z0
1
dq2 ( q2)
Z0 Z0
1 1
( d ( )q ) 2
+ dq2( q2 ) q2
Z0
( d ( )q ) 2
:
Observe that K(x ; y) as well as the expression in the rst brackets are holomorphic in in a domain D0 containing and the negative real axis (the denominator in K is 6= 0) and the product is bounded by C=j j1+ for some > 0, as j j ! 1; 2 D0. (here we use the assumption (5.3)). Hence the integral over is equal to zero. The result is that (5.16)
I = 22 (1 e
2 i) 1
Z0
where H is de ned in (5.12). Finally we deform into two circles !1 : j (x + iy)j = " ;
1
d ( )( )
Z
K(x
; y)H( ; )d
!2 : j (x iy)j = " both in the clockwise direction, so that
(5.17)
Z
K(x y; )H( ; )d =
2 XZ
j=1 !j
K(x y; y)H( ; )d :
37
We shall now use (5.16), (5.17) to evaluate I for K1 . Writing 2 K1 (x ; y) = 2y ( (x1+ iy))2 ( (x + iy))( (x iy)) + ( (x1 iy))2 and H( ; ) = H(x + iy; ) + H 0 (x + iy; )( (x + iy)) + where H 0 ( ; ) = @H( ; )=@ we nd, by the residue theorem, that Z K1 (x ; y)H( ; )d = y iH 0 (x + iy; ) H(x + iy; ) y
!1
and similarly,
Z
!2
K1(x
; y)H( ; )d = y iH 0 (x iy; ) + H(x y iy; ) :
Substituting this into (5.17) and observing that G0 ( ; ) + Q0 ( ; ) = 0 we get (5.18) Z K1(x ; y)H( ; )d = xG(x iy; ) xG(x + iy; ) + Q(x iy; ) Q(x + iy; )] : Substituting this into (5.16) and recalling (5.10), (5.7), we nd that (5.19) where (5.20) (5.21) Similarly, using
Dx 1 (x; y) = 2 (1 e
2 i) 1
Z0
1
d ( )( ) xG(x iy; ) xG(x + iy; )
+ Q(x iy; ) Q(x + iy; )] ;
G(x iy; ) G(x + iy; ) = Q(x iy; ) Q(x + iy; ) =
x iy Z x+iy
( q) q ( q) q
dq ; dq :
x iy Z
x+iy
K2 (x
; y) = 21 (
1 2 2 (x + iy)) ( (x + iy))( ( (x + iy)) + iy] ; 38
(x iy)) + (
1 (x iy))2
we calculate and similarly
Z
!1
K2(x
; y)H( ; )d = yH 0 (x + iy; ) ; y)H( ; )d = yH 0 (x iy; ) :
Z
!2
K2 (x
Substituting this into (5.16) and recalling (5.10), (5.8) and the relation G0 ( ; )+Q0 ( ; ) = 0 we obtain (5.22) Dy 1(x; y) = 2 (1 e
2 i) 1
Z0
1
d ( )( ) y G(x iy; ) G(x + iy; )] :
So far we have established (5.19), (5.22) for 0 < < 1 . To extend these results to 2 1 < < 0, we use analytic continuation. For any a > 0 the function a is an entire 2 holomorphic function in . From (4.27) we have
jD2h2(x)j C=jxj3=2
for jxj large:
1 1 We can therefore extend the de nition of D2 h1(x) to 2 D where D = 2 < Re < 2 ; for such values of the integral in (4.28) is absolutely convergent, uniformly in , thus de ning a holomorphic function in . Since, by assumption, h1(0) = h01(0) = 0, we clearly have 0
h1(x; ) h1(x) = dx0 D2 h1(x00 )dx00
0 0
Zx Zx
and so h1(x; ) is also holomorphic in , 2 D. Clearly 2 jDxh1(x; )j Cjxj for jxj large, so that (5.23)
1 Ref
g
jh1(x; )j Cjxj1
Ref
g:
Denote the right-hand side of (5.7) by k(x; y; ), and the right-hand side of (5.19) by Dx 1 (x; y; ). Since jK1 (x ; y)j C=j j4 39
if j j is large, the function k(x; y; ) is holomorphic in , 2 D. The function Dx 1 (x; y; ) can obviously also be extended as holomorphic function in 2 D. Since we have proved that k(x; y; ) = Dx 1 (x; y; ) if 0 < < 1 , this relation must hold also for 1 < < 0, by analytic continuation. 2 2 Thus the function Dx 1 de ned in (5.7) can be represented in the form (5.19) for all 1 < < 1. 2 2 Similarly we can establish for the function Dy 1 , de ned by (5.8), the representation (5.22) for all 1 < < 1 . 2 2 So far we have established (5.19) and (5.22) under the restriction (5.3). In general what we know about is that it is given by (5.1) and therefore j ( )j C for j j large:
j j3=2
If we can approximate ( ) by a sequence of functions n( ), each satisfying (5.3), such that the integrals in (5.19), (5.22) and (5.7), (5.8) corresponding to n converge to the same integrals corresponding to , then the proof of (5.19) and (5.22) will be completed for general . To carry out this approximation, we rst approximate by functions e n with compact support (using cut-o functions), and then proceed to approximate the e n by functions n satisfying (5.3). Suppose
e n( ) = 0 if < b :
For any " > 0 choose a polynomial p( ) such that (5.24) Introduce the function
jp e n j " in
b 1; 0] :
1 +b+ 2 1 if 0 if
f( ) = 1 1 + tanh A 2
Then jfj 1 and if A ! 1 then f( ) ! Hence, for A large,
:
(
+b >0 ; +b+1< 0 :
jfp e n j < 2" if
40
b< <0:
If b 1 < < b then e n = 0 and so, by (5.24),
jpj < " ; so that also jfpj < " :
Consequently
b 1< < b: Finally if A is chosen large enough, depending on p, then
The function n = fp provides the desired approximation to e n and to ; it satis es (5.3) and, in fact, it decreases exponentially fast at in nity. We now turn to 2 . By Lemma 4.2 and the proof of (4.27)
jfp e nj < " if
jfp e nj = jfpj < " if
< b 1:
(5.25) (5.26) where
Dx 2(x; y) = 2 Dy 2 (x; ky) = 2 h2( ) = 21
Z0
Z0
1
2 h2( ) ((x y (x 2 + ) 2)2 d ; )y
1
2 h2 ( ) ((x y(x )2 +) y2 )2 d
Z0 p
Z0 g1( )d q dq p
1
(q
)
:
Introduce a contour C as above and assume rst that g1( ) is holomorphic in a domain D containing C and the negative x-axis, and that it decreases fast top as 2 D; j j ! p o the negative real axis such that arg zero + i0 = ( > 1. De p a branch of ning 2 2 i0 = 0 ( 2 < 0), we easily nd by deforming contours that 0), arg 2
Z
C
g1( )d = 2P:V: Z p g1( )d : p( ) ( ) 1
0
Hence (5.27) Set (5.28)
h2( ) = 41
Z0 p
Z g1 ( ) d q dq p q :
C
2 2 K3 (x; y) = 2 (x2 xy y2 )2 ; K4(x; y) = 2 (x2 x y 2 )2 : + +y
41
Then the right-hand sides of (5.25), have (5.26) the form (5.29) where (5.30)
I=
Z0
1
d K(x
Z g1 ( ) ; y) p Z( ; )d
C
Z( ; ) = 41 Z( + 0i; ) = Z(
Z p dq qq 0
i0; ) if
0 i0
and K = K3 or K = K4. Note that (5.31)
< 0:
Next we introduce a contour as before. Then, by (5.31),
Z
and thus (5.32)
d K(x
; y)Z( ; ) = 2
Z
1 i0
d K(x
; y)Z(
i0; )
I=1 2
Z
C
g1( ) Z d K(x dp
; y)Z( ; ) :
We next deform C into C 0 as above. The term which arises as we cross the singularity at a point = q is Z 1 Z g (q)dq Z d F( ) d K(x ; y) 4 1 and this is equal to zero by Cauchy's theorem (since F( ) is analytic in D and converges to zero fast enough if j j ! 1; 2 D). We thus get 1 Z d g1 ( ) p I=2 0
C
0
Z
d K(x
; y)Z( ; ) ; y)Z( ; )
(5.33)
Z0 g1( ) Z = dp d K(x
1
after collapsing C 0 onto the negative real axis. 42
Finally we may deform to obtain (5.34)
Z
d K(x
; y)Z( ; ) =
2 XZ
Since the kernel K3 coincides with the kernel K2 , we can proceed as in the case of K2 to derive Z K3(x ; y)Z( ; ) = y Z 0(x iy; ) Z 0 (x + iy; )] where Z 0 ( ; ) = @Z( ; )=@ . Hence (5.35) where
1
j=1 !j
d K(x
; y)Z( ; ) :
Z0 g1( ) Dx 2 (x; y) = d p y Z 0 (x iy; ) Z 0 (x + iy; )]
1 Z p q dq ; Z 0 ( ; ) = 1 p : (5.36) Z( ; ) = 4 q 4 0 Similarly, by the residue theorem, Z K4 (x ; y)Z( ; )d = Z(x + iy; ) + iyZ 0 (x + iy; )] and It follows that (5.37) and (5.38)
!1
Z
!2
K4 (x
; y)Z( ; )d = Z(x iy; ) iyZ 0 (x iy; )] :
Z0 g1( ) Dy 2 (x; y) = d p Z(x iy; ) Z(x + iy; )]
1
iy Z 0(x + iy) + Z 0(x iy)] ; Z(x iy; ) Z(x + iy; ) = 41
x iy Z x+iy
p q dq : q
So far we have assumed that g1( ) is holomorphic in the domain D (D as in (5.3)) and g1( ) ! 0 fast, if j j ! 1; 2 D. Since any g1 with compact support can be approximated by g1;n analytic in D and decaying to zero exponentially fast as j j ! 1 (the proof is the same as for above), we conclude that (5.35) and (5.37) are true for general g1. We summarize: 43
( 1 < x 0), de ne functions h1(x); h2 (x) for 1 < x 0 by (4.27), (4.28) and h1(0) = h01 (0) = 0; h2(0) = 0. De ne also a function (x), for 1 < x < 0, by (5.1). Then a bounded solution of (4.9){(4.12) is given by = 1 + 2 where 1 and 2 are determined by (5.19), (5.22) and (5.35), (5.37).
Theorem 5.1. Given any continuous functions g1 (x); g2 (x) with compact support in
x ! 1.
This result is valid actually whenever g1(x); g2 (x) are decreasing su ciently fast as
2
function satisfying: (6.1) (6.2) where 0 <
0
x6. Special solution of
3 = H. Let > 2 ; 0 < < 1 and let H(x; y) be a
0
H = 0 if (x; y) 2 =
;
kHk2
; 0
5;2
5
N
< 1; 0 < < 1 and N is a positive constant. In this section we shall prove: Theorem 6.1. For any 0 < < 1; 0 < < min( ; 1 ), there exists a solution W to
2W
(6.3) such that (6.4)
= H in fy > 0g
+;
0
kWk
+ 0 BR ;4+
CN
R2 nB5=4
0 for any 0 < 0 < 1 and R > 0, where C is a constant independent of N.
Introduce nonnegative C 1 functions 1; 2 such that supp 1 B4=3 ; supp and 0 1 1 and 1 + 2 1. We rst construct a solution to (6.5)
2
2
= 1H H1 in fy > 0g : (K constant ; = (x; y))
We introduce the fundamental solution
Kj j2 log j j
of
2
in R2 and consider the special solution of (6.5)
(6.6)
'( ) = K j
B2
Z
qj2 log j
44
qjH1 (q)dq :
In order to estimate this function we cover the support of H1 by discs Bj (xj ) = B i (xj ; 0) where j and jxj j decrease exponentially to zero; for instance, if 0 is small enough (which may assume to be the case) we can take (6.7) Write (6.8) By (6.2), (6.9)
xj = jxj j = 3 4 '=
j
;
j 1 =2 3 j 4
(j = 0; 1; 2; . . . ) :
XZ
j B (xj ) j
j
qj2 log j
qjhj (q)dq
X
j
Ij :
jhj (q)j CNjxj j2 5 ; jD hj (q)j CNjxj j2 5 :
Ij = jxj
Substituting q = jxj j in Ij we get
j4 log jx
(6.10)
+ jxj
j4
Z
jj B1=2 ( 1)
Z
jxj j
2
2
hj ( jxj j)d hj ( jxj j)d Ij0 + Ij00 :
B1=2 ( 1)
jxj j
log jx j j
The rst integral can be written as a quadratic polynomial in , log jxj jfaj j j2 + bj where, by (6.9), + cj g
4
jaj j CNjxj j2
(6.11)
3
; jbj j CNjxj j2
; jcj j CNjxj j2
5
:
Summing over j we get a quadratic polynomial
X
j
Ij0 = aj j2 + b + c ; jaj + jbj + jcj CN :
Next we wish to estimate the Ij00 . It will be important to observe that (6.12) if 1 2 B1=2( 1) then j j 2 : 45
We apply Taylor's formula to
g(u) = ju j2 log ju j ( 2 B1=2( 1)) for u near 0 and then substitute u = =jxj j. We get
log jx j PM jx j ; C jx j j j j where PM is a polynomial in =jxj j of degree M and M is such that (6.13)
jxj j
2
+
M <2 1 M +1 ; here we can take any 1. We shall actually need to choose such that < 1; + >2; + >M ; (6.15) which is possible since > 3=2. Applying any k-th derivative to the Taylor expansion of g(u) we also get
(6.14) (6.16) log jx j PM jx j ; jxj jk jxj j j j provided =jxj j is small. If =jxj j is large (or just stays away from zero and ) then (6.16) is still valid since + > 2; + > M. If =jxj j is near then the left-hand side of (6.13) remains bounded and, by (6.12), it is bounded by the right-hand side for a suitably chosen constant C. The same considerations apply to (6.16) with k = 1. If k = 2, however, then we must replace (6.16) by
@k
jxj j
2
C
+
k
PM jx j ; jxj j2 j (6.17) + jxCj2 log jx j j j in order to bound the left-hand side also for =jxj j near . For k = 3; 4 we need to add still larger terms on the right-hand side of (6.16) in order to control the left-hand side for =jxj j near :
@2
jxj j
2
log jx j j
C
jj jxj j
+
2
(6.18)
@ k jx j j
log jx j j 1 + jxCjk k j
xj
2
PM jx j ; j
2
jxj jk
C
jj jxj j
+
k
n
46
jxj j
<1 4
o
(k = 3; 4) :
This estimate, although good enough for our purposes in case k = 3, will not be su ciently 4 precise for k = 4. We note that any fourth order derivative @u of juj2 log juj is a kernel (u) of a singular integral. Using this fact we can write
(6.19)
@ k jx j j
2
log jx j j
PM jx j ; j
jxj j4
1
jxj j
n
jxj j
<1 4
o
jxj j4
C
jj jxj j
+
4
:
We proceed to estimate the Ij00 . Using (6.13) and (6.9) we nd that (6.20) where
I 00 = jxj j4
j
Z
B1=2 ( 1)
PM jx j ; j
hj ( jxj j)d + R00 j
Since 1 is a polynomial in of degree M with coe cients bounded by
j+ jR00 j CNjxj j4 jx jj + jxj j2 5 = CNjxj j 1 j j + : j j 00 > 0 ; jRj j CNj j + . The rst term on the right-hand side of (6.20)
CNjxj j2
1M
(using (6.9)):
Since 2 (6.21)
1 > M, we conclude that
X
j
Ij00 PM ( )
CNj j
+
where PM ( ) is a polynomial in of degree M. Similarly to (6.21) we can derive the estimate (6.22)
3 2 X 00 @ 4 Ij PM ( )5
j
CNj j
1+
:
To estimate @ 2 Ij00 and @ 3 Ij00 we need to use (6.17) and (6.18) (for k = 3) respectively. We shall carry out the details just for @ 3 Ij00 . We have
2 Z 3 I 00 = @ 3 6jx j4 @j 4j
B1=2 ( 1)
PM
47
3 ; hj ( jxj j)d 7 + R00 5j jxj j
where and
2 Z ej j CNjxj j4 6 jR 4
j jR00 j CNjxj j4 jx jj j j
1
B1=2 ( 1) jxj j
3+ 3+
1 jx j2 jx j3 j
j
5 + jR
ej j
5
n
jxj j
1 <4
3 od 7 jx1j3 jxj j2 5j
:
In the last integral the only possible points for which the integrand does not vanish identically are such that j j jxj j. Consequently
e jRj j Cjxj j2
CNj j
Hence and, since (6.23) Similarly (6.24) 1> ,
4 3+
Z
1
1
B1=2 ( 1) xj
n
jxj j
1 <4
od
CNjxj j2
4
jxj j
:
3+
jR00 j CNj j j
@ 3 Ij00 PM ( )]
jxj j
1
CNj j
3+
: :
j@ 2 Ij00 PM ( )]j CNj j
2+
We proceed to evaluate @ 4 Ij00 . By (6.19),
2 Z 4 I 00 = @ 4 6jx j4 @j 4j
B1=2 ( 1)
PM
3 ; hj ( jxj j)d 7 + R00 + Rj 5j jxj j
4+
where and
Rj =
Z
B1=2 ( 1)
jR00 j CNj j j jxj j
jxj j
1
hj ( jxj j) n
jxj j
1 <4
od :
Using (6.9) we can apply the Schauder estimates as in 5] to conclude that
jRj j + jxj j j@ Rj j CNjxj j2
48
5
:
Since j j jxi j if Rj 6= 0, it follows that (6.25)
jRj j + j j j@ Rj j CNj j
4+
jxj j
1+
:
Combining the estimates for R00 and Rj , we get j (6.26)
j@ k Ij00 PM ( )]j + j j j@ k+ Ij00 PM ( )]j
CNj j
4+
(jkj = 4) :
Combining the estimates (6.26) with (6.21){(6.24) and recalling (6.8), (6.10), (6.11), we get the following result: Lemma 6.2. The function ' de ned by (6.8) in a solution to (6.5) in fy > 0g, satisfying: (6.27)
jDk '( ) PM ( ]j CNr
jkj+
4 +
(jkj 4) ; (jkj = 4)
jDk+ '( ) PM ( )]j CNr
in fy > 0g where = (x; y); r = j j, and PM ( ) is a polynomial of degree M with coe cients bounded by CN. We next turn to solving (6.28) (6.29)
2
= 2 H H2 :
Proceeding similarly to (6.6) it seems natural to take
'0( ) = K
Z
fx< 1g
j
j
qj2 log j
qjH2 (q)dq :
We can cover the support of H2 by discs B j (xj ) where 4 jxj j = 3
; =1 4 23
j
(j = 1; 2; . . . )
provided 0 is small enough. Then we split the domain of integration in (6.29) into these discs. However, unlike what we did before, we now must add to the j-th term a special biharmonic function 'j ; otherwise the series will not converge. Thus, we shall replace '0 by a somewhat di erent function (6.30)
' = Ij e
49
where
Ij = jxj
(6.31) + jxj
j4 log jx
j4
Z
jj B1=2 ( 1)
Z
jxj j
2
2
log jxjjj j
hj ( jxj j)d Ij0 + Ij00 ;
B1=2 ( 1)
jxj j
log jxjjj j
hj ( jxj j)d
the harmonic function 'j enter through the factor 1=j j in the \log" terms of both Ij0 and Ij00 . From (6.2) we have (6.32)
jhj (q)j CNjxj j2 5 ; jD hj (q)j CNjxj j2 5 :
3 < 0.
We can estimate the Ij0 as before, simply replacing by and noting that 2 We then get the same result (6.11) as before. It remains to estimate the Ij00 . We expand
u 1 j = log juj + log juj juj = log juj + c0(juj)2 u + O juj2 for juj large and use this expansion to establish the relation, for =jxj j large,
(6.33) log ju
jxj j
(6.34)
2
log jxjjj j
j j2 log j j + c( ) log j j jxj j2 jxj j jxj j jxj j
2 c c + e(jx)jj2j + b(jx) j + j j2 log jxjjj j j j
n
jxj j
<1 4
o
j = O jx jj j
for any 0 < <
where c; e; c depend on ; xj ; and are uniformly bounded. (In fact, the right-hand side cb may be replaced by O(log j j=jxj j). If =jxj j varies in a bounded set away from 0 then (6.34) is still true since the left-hand side remains bounded. Finally, if =jxj j tends to zero then the left-hand side converges linearly to zero (it is here that we need the factor 1=j j in \log"), and therefore (6.34) remains valid. We need to estimate the contributions of the various terms in f g and of the O-term to Ij00 ; 50
We begin with (6.35)
j2 j jxj jx jj2 log jx jj j j j4
Z
B1=2 ( 1)
hj ( jxj j)d
CNjxj j2
The next term contributes (6.36)
3 j j2 (j log j
jj + log jxj j) :
CNjxj j2
2j
j(j log j j j + log jxj j) :
The contributions from the next two terms in f g are bounded as in (6.35), (6.36), and the last term contribution is bounded by
CNjxj j4jxj j2
5
= CNjxj j
1
= CNj j
jxj j
1+
since j j jxj j if the integrand is not identically zero. Finally the O-term in (6.34) contributes to Ij00 a function bounded by
j CNjxj j4 jx jj
j
jxj j2
5
= CNj j
jxj j
1+
:
, we obtain
Summing over j in the above estimates and choosing < 1 (6.37)
j Ij00 j Nfj j2 log j j + j j2 + Cj j + O(j j )g :
Next we wish to estimate derivatives @ k Ij00 . The procedure is similar to that used for near 0. If we apply @ k to the left-hand side of (6.34) we get the error term 1 O j j jkj jxj jk jxj j jkj provided j j=jxj j is large. The same is true if j j=jxj j is near 0 (since jkj < 0). Finally, if =jxj j is near we need to modify the estimate when jkj 2 in the same way as before. It will be enough to consider just the case jkj = 4. The various terms obtained by applying @ k to the sum in f g in (6.34) can be handled essentially as before, with minor changes. Therefore we concentrate only on the new term which involves a singular integral:
jxj jx j4 j
j4
1
Z
B1=2 ( 1)
jxj j
51
n
jxj j
<1 4
o hj ( jxj j)d :
By the Schauder estimates 5] this term and jxj j times its derivative @ are bounded by CNjxj j2 5 = CNj j 4 jxj j 1+ ; where we used the fact that j j jxj j if the integral is 6= 0. Since 1 + < 0, we can k f g, deduce that sum over j and, together with the estimates of the other terms in @
3 2 X 00 X 00 5 Ij ] CNj j @ k 4 Ij + j j @ k+
j j
4
if jkj = 4. We summarize: The function ' de ned by e (6.38) with j j j (6.39) and (6.40)
j
'( ) = K e
satis es:
XZ
j B (xj ) j
j
qj2 log j j jqj H2 (q)dq
j
@k ' e
A1 j j2 log j j + A2j j2 + A3 log j j + A4 + A5 log j j CNr jkj
if jkj 4 ; + A5 log j j
@ k+ ' e
A1 j j2 log j j + A2 j j2 + A3 log j j + A4 CNr
4
where j j = r and (6.41)
if jkj = 4
jAj j CN :
Actually we have only proved so far a cruder result than (6.39), namely,
e j@ k 'j C j j2 j log j j + j j2 + j j j log j j + j j + r jkj : However a little bit more careful analysis shows that the bounds j j log j j and j j actually come from terms A3 log j j and A4 . The same remark applies to (6.40). Fix a point 0 in fy < 0g and consider the function
(6.42)
h
i
'( ) = '( ) b e
B1j
0j
2 log j
0 j + B2 j
0j
2
+ B3 ( 0) log j 0 j + B4 ( 0 ) + B5 log j 0 j] We can choose B1; B2; . . . ; B5 in a unique way so that all the unbounded terms in ' cancel e out. Hence: 52
< min( ; 1 ). Then the function ' de ned by b (6.42), where ' is de ned in (6.38), is a solution to (6.28) satisfying: e
Lemma 6.3. Suppose let 0 <
(6.43)
We return to Lemma 6.2, and prove: Lemma 6.4. PM ( ) is a biharmonic function. Proof. We can write PM as a sum of homogeneous polynomials n of degree n:
jDk '( )j CNr jkj b jDk+ '( )j CNr 4 b
(jkj 4) ; (jkj = 4) :
PM ( ) =
In view of (6.27), the biharmonic functions
M X
n=n0
n(
):
'R ( ) R1 '(R ) n converge to n ( ) as R # 0; hence n ( ) is biharmonic. Proceeding similarly with ' n we discover that n +1 is biharmonic, etc. Any biharmonic function has the form h1(z) + zh2(z) + c:c: where h1; h2 are holomorphic functions and c:c: stands for the complex conjugate. It follows that the biharmonic polynomial PM ( ) which occurs in Lemma 6.1 is a nite sum of expressions (6.44) j = Aj z k + Aj z k + Bj z k 1 z + B j z k 1 z where jAj j; jBj j are bounded by CN and the bar means c:c:. For any such j and for any R > 0 one can construct polynomials Qj (z); Rj (z) such that ej j Qj (z)e z + zRj (z)e z + c:c:] = O(jzjR ) for jzj < 1. Of course we also have 0 ej ! 0 exponentially fast if jzj ! 1; jImzj < 0 j Re zj 0 provided 0 < 1; the convergence is in the (4 + )-norm. Note that the function ' constructed in Lemma 6.2 satis es (6.43) if j j > 1 whereas the function ' constructed in Lemma 6.3 satis es (6.27) if j j < 1. Consequently, the b function i i X hh z + zRj (z)e z + c:c: ( = (x; y)) Qj (z)e (6.45) W( ) = '( ) + '( ) b
0 0 0 0 0 2 2 2 2
satis es the assertions of Theorem 6.1. 53
j
nd a solution to (7.1) (7.2) (7.3)
x7. Solution to auxiliary boundary value problems. In this section we wish to
2' = 0
in fy > 0g ;
'x(x; 0) = Wx(x; 0) ; x 2 R ; 'y (x; 0) = Wy (x; 0) ; x 2 R :
It will be convenient to use integral representation for 'x; 'y rather than directly for '. Lemmas 4.1, 4.2 provide us with the representations (7.4) (7.5) where
2 @K1 (x; y) = 2 (x; y3 ; @K2@y y) = 2 (x2 x y 2 )2 ; @x (x2 + y2)2 +y @K1(x; y) = @K2 (x; y) = 2 xy2 : @y @x (x2 + y2)2 0 0 @' = Z @K1 (x s; y)D W(s; 0)ds + Z @K2 (x s; y)D W(s; 0)ds ; x y @x @x @y 0 0 @' = Z @K1 (x s; y)D W(s; 0)ds + Z @K2 (x s; y)D W(s; 0)ds x y @y @x @y
1
1
1
1
From the estimates in Theorem 6.1 we deduce that the integrals in (7.4), (7.5) are convergent and de ne smooth functions which we tentatively denote by '1 and '2 respectively. One can verify that @'1 = @'2 : @y @x Hence there exists a function ' such that @' = ' ; (7.6) @x 1 By direct computation @'1 + (7.7) @x
@' = ' : 2 @y @'2 = 0 : @y
54
Hence (7.8) We decompose
2' = 0
:
Wx(x; 0) = hj (x; 0); Wy (x; 0) = kj (x; 0)
3 j and length 3 j ; j = where hj ; kj are supported in intervals with center xj = 4 4 0; 1; 2; . . . ; and then split the integral in (7.4) into in nite series, @' = @ j (7.9) @x @x where @ j =@x is an integral corresponding to the integrands hj ; kj . Thus
1 2 y3 @ j = Z ds 2 hj (s; 0) 2 (x (x s)2s)y y2 ]2 kj (s; 0) : @x (x s)2 + y2 ]2 + 1
Substituting s = jxj j , we get
@ j=2 @x
(7.10)
Z b b B1=2 ( 1) B1=2 (1)
jxj j jxj j
x x
d
jxj j jxj j jxj j
y
2
y
3
x
2
+ jxj j
y
h (jx j 22j j ; 0)
; 0)
2
+ jxj j
2 kj (jxj j y2
b where BR ( 1) denotes the interval with center 1 and length 2R. Consider rst the terms with j > 0. By (6.46), if j > 0 then
(7.11)
jhj j + jkj j CNjxj j
1+
; and similar bounds hold for
the 3 + derivatives:
Expanding the function u3= u3 + (v )2]2 about u = v = 0 by Taylor's formula, we b b nd that if 2 B1=2 (+1) B1=2 ( 1) (so that stays away from 0) then (7.12)
jxj j jxj j
x
2
y
3
+ jxj j
y
22
PM
1
x;y; jxj j jxj j
j C jx jj j
1+
55
where = (x; y); PM 1 is polynomial of degree M 1 in =jxj j; M is as in (6.14) and is a su ciently small positive number such that 1+ >M 1 (which is possible by (6.14)). This choice enables us to conclude that (7.12) is valid also for (x=jxj j; y=jxj j) near 1 and, in fact, whenever x ; y 2 B ( 1; 0) B (+1; 0) : = If
jxj j jyj j
1=4
3=4
x ; y 2 B ( 1; 0) B (+1; 0) 3=4 3=4 jxj j jxj j then we need add to the right-hand side the term
jxj j jxj j
x
2
(7.13)
y
3
+ jxyj j Using (7.11) we then easily nd that (7.14)
22
n
jxxj j ;
y jxj j
2 B3=4 (
1;0)
B3=4 (+1;0)
]
o:
Similarly one can establish the corresponding 3 + estimates
b + Cj Njxj j 1+ b where Cj is a constant c times the integral of the function in (7.13). If this integral is nonzero then j j jxj j and therefore we can replace the last term on the right-hand side of (7.14) by CNj j 1+ jxj j : In (7.14) PM 1;j is a polynomial degree M with uniformly bounded coe cients. Summing over all j = 1; 2; . . . , we get X@ j 1+ (7.15) : @x PM 1( ) CNj j j>0 2 X Dk 4 @
j>0
@ j Njx j j @x
1+
PM
1;j (x; y) jxj jM 1
CNj j
jxj j
1+ 1+
jxj j
1+
(7.16)
3 2 X@ j Dk+ 4 @x PM 1( )5
j>0
3 jP M 1 ( )5 @x
CNj j CNj j
56
1
jkj+
(jkj 3) ; (jkj = 3);
4
+
the coe cients of PM 1( ) are bounded by CN. We now proceed with the part of the series (7.9) for which j estimate
jxj j jxj j
x
2
0. Here we use the
y
2 22
+ jxyj j
Cjj
e if jxj j 1 e
1
x ; y 2 B ( 1; 0) B (+1; 0) 1=2 jxj j jxj j = 1=2
where e is any small positive number. Analogously to (7.14) we get, for j 0, (7.17)
@j @x
CNj j
1 1
jxj j
e jxj j
e
1
b + CNjxj j
1
b
where b is equal to the integral of (7.13), so that j j jxj j if the last term in (7.17) is nonzero. We conclude that X@j 1 e ; for any 0 < e < : (7.18) @x Cj j j0 Similarly,
(7.19)
3 2 X jDk 4 @@xj 5
j0
CNj j CNj j
1
jkj
e (jkj 3) ; e (jkj = 3) :
jDk+
3 2 X@ j5 4 @x
j0
4
@x (with another polynomial), since this sum corresponds to integrating over Dx W(x; 0) for jxj > 1. Similarly the estimate (7.19) X near 1 holds for @@xj . Since further and e can be taken arbitrarily small, they can j>0 be absorbed into the de nitions of and , respectively. Thus:
j0
The estimates (7.16) hold near 0 also for
X@
j
(7.20)
8 k @' 1 >D e > < @x P ( ) CNj j > k+ @' e >D : @x P ( ) CNj j
57
jkj+
4 +
(jkj 3) ; (k = 3) if j j < 1 ;
(7.21)
e where P is a polynomial. b Similar estimates can be derived for @'=@y, with another polynomial P. From (7.20) e b and the corresponding estimates for @'=@y we easily deduce that @ P=@y = @ P=@x and therefore there exists a polynomial P such that @P = P ; @P = P : e b @x @y
Lemma 7.1. The function '( ) ( = (x; y)) satis es
8 k @'( ) >D < @x CNj j 1 > k+ @'( ) :D CNj j @x
jkj
4
(jkj 3) ; (jkj = 3) if j j 1
(7.22) and (7.23)
8k < jD (' P)( j CNj j jkj+ (jkj 4) ; : jDk+ (' P)( )j CNj j 4 + (k = 4) if j j 1; y > 0 ; 8k < jD '( )j CNj j jkj (jkj 4) ; : jDk+ '( )j CNj j 4 (jkj = 4) if j j > 1; y > 0
Px(x; 0) = Py (x; 0) = 0 ; x 2 R :
where P is a biharmonic polynomial of degree < + with coe cients bounded by CN, and (7.24) Proof. We have already proved the assertions (7.22), (7.23). It remains to show that P is biharmonic and that (7.24) is satis ed. Write P is a sum of homogeneous polynomials (7.25) Consider the functions
n X
1
n=n0
Qn ; where n1 < + :
Since n0 < + , we deduce from (7.22) that, as R ! 0,
'R(x; y) = R1 '(Rx; Ry) n
0 0
(R # 0) :
'R(x; y) ! Qn (x; y)
58
@'R(x; 0) ! @Qn (x; 0) ; @'R (x; 0) ! @Qn (x; 0) : @x @x @y @y Since by (6.4) and (7.2), (7.3) @'R(x; 0) ! 0 ; @'R(x; 0) ! 0 if R ! 0 ; @x @y it follows that Qn is biharmonic polynomial satisfying (7.24). Applying this proof to ' Qn ; ' (Qn + Qn ), etc., Lemma 7.1 follows. We wish to eliminate P from Lemma 7.1 by modifying the function '. To do this we need to study biharmonic polynomials whose gradient vanishes on y = 0. Lemma 7.2. If ' is a real-valued biharmonic polynomial of degree N with r'(x; 0) = 0 for all x 2 R, then ' has the form
0 0 0 0 0 1
and
(7.26)
N N N X n 1 n X Bn n X n 1 '= n Bnz n z + z Bnz + c:c: n=2 n=2 n=2
where Bn are complex numbers, and Bn is the complex conjugate of Bn. Proof. From the paragraph following (2.12) we know that ' must be a linear combinations of functions ee Azn + Bzn 1 z + Azn + Bz n 1z : Equating @' and @' to zero an z = r and z = rei = r for all r > 0, we get the equations
ee B + (n 1)B = nA = 0 ;
nei ei
(n 1) A + (n
b nA + (n 1)B + B = 0 ;
1)ei 1)e i
(n 3) + e i (n 1) B (n 3) + ne i (n
(n 1) B + (n
e=0; 1) A = 0 : e
We easily compute the solution to this systems: 1 e A = n (n 1)B + B] ; e e1 A = n (n 1)B + B] e e e where B; B are arbitrary. Since ' is real-valued, the conditions B = B, A = A must hold, and this yields the n-th degree homogeneous polynomials in (7.26). 59
By Lemma 7.2, the polynomial P in (7.22) is a linear combination of polynomials (7.27)
k 1 zk k zk + z zk 1 + c:c: : k kk k We wish to construct a biharmonic function which near the origin coincides with (7.27), up to any given order, and, at the same time, decreases exponentially fast to zero in 0 0 jImzj 0 j Re zj; 0 < 1, as jzj ! 1. We can write 1 k e z = X j zj ; j 2 R : z
2
Setting Bn = n n we have
1 X Bn
n=k
j=k
zn = Bn zn = zk e n n=k n d Bnzn = z dz (zk e
z2 ) ;
2
1 X
z2
:
Also
1 X
n=k
z
d Bnzn 1 = z dz (zk e z ) : n=k
1 X
Hence, as easily computed, the real biharmonic function ' = 'k de ned by (7.26) with N = 1 and Bn = n n satis es: (7.28)
'k = k
k1 k
k zk
k z k + z z k 1 + c:c: k k
e + Pk+2
e with k = 1, where Pk+2 is a polynomial of degree k + 2, and 'k ! 0 exponentially fast 0 if jImzj 0j Re zj; jzj ! 1. Similarly, taking 1 X e izk e z = ej zj ; Bn = nen
2
e and forming the corresponding 'k by (7.26) with N = 1 and Bn replaced by Bn, we e obtain (7.28) with 'k replaced by 'k and k = i. By taking a linear combination with real e coe cients of 'k and 'k we obtain a new function, which we denote again by 'k , of the e form (7.26) with N = 1 such that it satis es (7.28) with any given complex coe cients k ; the Bn are also complex.
60
j=k
We now write P as a sum of homogeneous polynomials Qk , as in (7.25). We apply the above construction to obtain a function 'n of the form (7.28) (with N = 1) such that, near the origin,
0
P 'n contains only terms of order
0 0
n0 + 1 ;
0 and 'n (z) ! 0 exponentially fast if jzj ! 1; jImzj 0j Re zj. Repeating the construction, we get similar functions 'n for n = n0 + 1; . . . ; n1 such that, near the origin,
P
n X
1
n=n0
'n contains only terms of order > n0 + 1
0
whereas each 'n decays at 1 in the same way as 'n . De ning (7.29) we obtain the following result.
Theorem 7.3. The function ' is a solution to (7.1){(7.3) satisfying: e
'=' e
n X
1
n=n0
'n
(7.30)
0 for any 0 < 0 < 1 ; R > 0.
k'k
+;
0
+ 0 BR ;4+
CN
tion we wish to nd a special solution to (8.1) (8.2) (8.3) (8.4)
x8. Solution to inhomogeneous mixed boundary value problem. In this sec2' = 0
in fy > 0g ;
'x(x; 0) = 'y (x; 0) = 0 ; x > 0 ;
('yy 'xx)(x; 0) = g1(x) ; x < 0 ; (2'xxy + 'yyy ) 'xxx](x; 0) = g2(x) ; x < 0 61
where g1; g2 satisfy: (8.5) (8.6)
kg1k kg2k
+ +
0 ;2+
2;
2 3
N; N:
0 ;3+
3;
e small, and positive and small ( < e). Then there We chose e < with exists a positive integer M and " 2 (0; 1) such that e 1 (8.7) + e 2 < M +" < + 1 e ; M +">1 : e 2 As in xx4,5 we associate to g1; g2 functions h1(x); h2 (x) de ned by (4.27), (4.28) for x 0 with h1(0) = h01(0) = h2(0) = 0 and a function de ned by (5.1). According to Theorem 5.1, there is a solution to (8.1){(8.4) given by
= (8.8)
1 2 1+ 2
;
is de ned by (5.7), (5.8) or (5.19), (5.22); is de ned by (7.25), (5.26) or (5.35), (5.37):
0 Dh2(x) = Z p g1(s)ds b (x) 2p h x (x s) 2
We wish to evaluate the function and use this to construct the special solution '. We begin with 1 . By (4.27) (8.9)
1
b We split g1 into a series g1j using a sequence of overlapping intervals B j (xj ) with centers xj and length 2 j , where
xj =
b g1j is supported in B j (xj ). Then
(8.10)
3 j ; =1 3 j j 24 4
(j = 0; 1; 2; . . . ) ;
b2(x) = X h
j
Z
b B j (xj )
p g1j (s)ds s) : s (x
We rst consider the case j > 0. Clearly Z g1j (s)ds p s (x s) = jx 11=2 (8.11) jj
Z
B1=2 ( 1)
b B j (xj )
g1jp jxj j) d ( s jxxj j
62
and (8.12)
jDk g1j ( jxj j)j CNjxj j 2 jD2+ g1j ( jxj j)j CNjxj j
k+
4 +
(k = 0; 1; 2) ;
:
(8.13)
m 1 = 1 X x n + x m+1 1 : xa a n=0 a a xa x =b b For any 2 B1=2( 1), if jx j 2 B3=4( 1) then, by (8.13), j
We shall use the expansion
e if x=jxj j is small enough, where PM 2 is a polynomial of degree M 2 in x=jxj j. Since M + " > 1 (see (8.7)), this relation remains true also for x=jxj j large and, in fact, as long e x 2 B ( 1). as jx j = 3=4 j It follows that
jxj j1=2
= jx j1=2
j
jxj j
x
1
e = PM
2
x; jx j
j
x + O jx j M j
2+e "
1
Z
1
Z Z
b B1=2 ( 1)
g1jp jxj j) (
jxj j
2
x
d x; jx j
j M 2+e "
b B1=2 ( 1)
g1jp jxj j) P ( M
d d:
+ jx 11=2 j
j
b B1=2 ( 1)
g1jp jxj j) O x ( xj
Using (8.12) and (8.7) we obtain, upon summing over all j > 0,
PM 2 (x) + NO(jxjm
2+e) "
where PM 2(x) is a polynomial of degree M 2 with coe cients bounded by CN. x Next we consider the integral on the right-hand side of (8.11) in case jx j 2 B3=4 ( 1). j We apply the Schauder estimates to get Z g1j ( jxj j) d CN jx j 2 0+ 1 p
jxj j1=2
B1=2 ( 1)
jxj j
x
jxj j1=2
j
63
for any 0 > 0. Furthermore, if the integral is actually nonzero then necessarily jxj jxj j. Therefore the last bounded may be replaced by
CN jx j jx j1=2 j
j
2
0+
x xj
M 2+e "
;
2+e). "
and using (8.9) we get, upon summing over j > 0, the bound NO(jxjM Thus, altogether, (8.14) Obviously,
Z0 g1 ( ) d p x = PM 2(x) + NO(jxjM
3 2
2+e) "
:
Z3=2 g1( ) d p
1
x
can be expanded near x = 0 using (8.13). Combining this remark with (8.14) and recalling (8.9), we conclude that (8.15)
Dh2(x) = P (x) + NO(jxjM px M2
2+e) "
with another polynomial PM 2. Similarly we can estimate the 2 + derivative the left-hand side of (8.15), from which we deduce, after using (8.7), that (8.16)
jDk h2(x) ( x)3=2 PM 2(x)]j Cjxj 1 jD3+ h2(x) ( x)3=2 PM 2(x)]j Cjxj
k+e
4 +
(0 k 3) ; if 1 x<0:
We now turn to the case where jxj 1. We shall use the estimates (8.17)
jDk g1j ( jxj j)j CNjxj j 2 jD2+ g1j ( jxj j)j CNjxj j
x 2 B ( 1), b jx j 3=4
j
k
4
(k = 0; 1; 2) ;
for j 0. One of the following cases must hold: (i)
64
(ii) (iii)
x 2 B ( 1), and x < 5 b jxj j = 3=4 xj 6 x 2 B ( 1), and x > 6 b jx j = 3=4 x 5
j j
8 2 B1=2 ( 1), 8 2 B1=2 ( 1).
+b "
In case (i) we use the Schauder estimates to get the bound
CNjxj
2
1 2
for any " > 0. In case (ii) we bound the j-th term by b x CNjxj j 2 or CN x jxj j 2 ( < 0) j 5 + ", giving us (upon summation) the same bound as before. and choose = 2b In case (iii) we expand (x=jxj j) ] 1 by (8.13) with m = 2. The error term is bounded by x x C x 3 and therefore also by C x j j where = 2 1 + ". We obtain, upon summation on j > 0, 2b C1N + C2N + NO(jxj 2 +b); C constants: " i x x2 We conclude that Dh2(x) C1N C2N CNjxj...