Existence of stationary solutions of the Navier#Stokes ... - Peter Wittwer
Existence of stationary solutions of the Navier-Stokes equations in two dimensions in the presence of a wall Matthieu Hillairet
Peter Wittwer
Université de Toulouse, IMT
Département de Physique Théorique
Toulouse, France
Université de Genève, Switzerland
[email protected]
[email protected]
Abstract We consider the problem of a body moving within an incompressible ‡uid at constant speed parallel to a wall, in an otherwise unbounded domain. This situation is modeled by the incompressible NavierStokes equations in an exterior domain in a half space, with appropriate boundary conditions on the wall, the body, and at in…nity. Here we prove existence of stationary solutions for this problem for the simpli…ed situation where the body is replaced by a source term of compact support.
Contents 1 Introduction
1
2 Reduction to an evolution equation
3
3 Functional framework
7
4 Proof of main lemmas 4.1 Proof of Lemma 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Proof of Lemma 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10 11 12
A Derivation of the integral equations
18
B Basic bounds B.1 Continuity of semi-groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B.2 Convolution with the semi-group e t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B.3 Convolution with the semi-group e jkjt . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20 20 21 23
1
Introduction
The present paper is the main step in an e¤ort to develop the mathematical framework which is necessary for the precise computation of the hydrodynamic forces that act on a body that moves at small constant speed parallel to a wall in an otherwise unbounded space …lled with a ‡uid. A very important practical application of such a situation is the description of the motion of bubbles rising in a liquid parallel to a nearby wall. Interesting recent experimental work is described in [14] and in [6]. Numerical studies can be found in [2], [5], [13], and [15]. The computation of hydrodynamic forces is reviewed in [10]. In what follows we consider the situation of a single bubble of …xed shape which rises with constant velocity in a regime of Reynolds numbers less than about …fty. The resulting ‡uid ‡ow is then laminar. The Stokes equations provide a good quantitative description (forces determined within an error of one Supported in part by the Swiss National Science Foundation.
1
percent) only for Reynolds numbers less than one. For the larger Reynolds numbers under consideration the Navier-Stokes equations need to be solved in order to obtain precise results. The vertical speed of the bubble depends on the drag, and the distance from the wall at which the bubble rises requires one to …nd the position relative to the wall where the transverse force is zero. Since at low Reynolds numbers the transverse forces are orders of magnitude smaller than the forces along the ‡ow, this turns out to be a very delicate problem which needs to be solved numerically with the help of high precision computations. But, if done by brute force, such computations are excessively costly even with today’s computers. In [3] we have developed techniques that lead for similar problems to an overall gain of computational e¢ ciency of typically several orders of magnitude. See also [4] and [10]. These techniques use as an input a precise asymptotic description of the ‡ow. The present work is an important step towards the extension of this technique to the case of motions close to a wall. In what follows we consider the two dimensional case. For convenience later on we place the position of the wall at y = 1. Namely, let x = (x; y), let + = f(x; y) 2 R2 j y > 1g, let B + be a compact set with smooth boundary @B, and let e1 = (1; 0). Then, in a frame comoving with the body, the Navier-Stokes equations are u ru which have to be solved in the domain
=
@x u+ u
+
rp = 0 ; r u=0;
(1) (2)
n B, subject to the boundary conditions
u(x; 1) = 0 ; lim u(x) = 0 ;
x2R;
x!1
uj@B =
e1 :
(3) (4) (5)
The standard technique to solve this problem is to prove the existence of weak solutions. Such solutions are constructed by considering a nested sequence of …nite domains that converge to + . Existence follows by a compactness argument. See for example [17], [8], [9], for the case of = R3 n B, and [12], [11] for the case of = + n B. Weak solutions constructed this way are smooth, the only shortcoming of the method is that only very little information is obtained about the behavior of the solutions at in…nity. In order to obtain such information a classic way is to consider the problem in an appropriately chosen weighted Sobolev space. Such methods are well developed for the case of isotropic weights, but become very technical if, as in the present case, anisotropic weights are needed. See for example [7], [1]. In this paper, we propose a new strategy which takes advantage of this anisotropy. In order to obtain detailed information at in…nity, we construct a classical solution in a function space which is motivated by the theory of dynamical systems. Namely, we chose the coordinate y to play the role of time and rewrite (1)-(5) as a system of evolution equations with respect to this variable. Information on the large time behavior of the dynamical system then naturally provides detailed information at in…nity. In order to get a system of ordinary di¤erential equations we use the Fourier transform in the x coordinate. We then choose the function spaces which are well adapted to the problem. These spaces come up naturally once the problem is formulated in this form. However, to use our techniques based on the Fourier transform we need that the problem be formulated on all of + . This is achieved as follows. Let u ~ be a smooth solution of the above problem and let ~ be the corresponding stream function, i.e., u ~ = ( @y ~; @x ~). One can then always use a smooth cut o¤ ~ function such that the function = is equal to ~ outside a su¢ ciently large disk D + containing B and zero inside some smaller disk D1 (but su¢ ciently large to still contain B). Inside B, is also de…ned equal zero. Let u = ( @y ; @x ). Then, since u = u ~ in the complement of D, we …nd that u satis…es (1), (2) for a certain smooth force term F of compact support. Motivated by these remarks we consider in what follows the equation @x u+ u = F + u ru + rp ;
(6)
in the domain + , subject to the incompressibility conditions (2) and the boundary conditions (3) and (4), and with F a smooth vector …eld with compact support in + , i.e., F 2 Cc1 ( + ). 2
The following theorem is our main result (see Section 3, Theorem 7 and Theorem 8 for a precise formulation): Theorem 1 For all F 2 Cc1 ( + ) with F su¢ ciently small in a sense to be de…ned below, there exist a unique vector …eld u = (u; v) 2 H 1 ( + ) and a function p satisfying the Navier-Stokes equations (6), (2) in + subject to the boundary conditions (3) and (4). Moreover, there exist constants C1 , C2 such that, uniformly in (x; y); ju(x; y)j C1 =y 3=2 and jv(x; y)j C2 =y 3=2 . To our knowledge this is the …rst proof of existence which provides detailed information on the behavior of solutions at in…nity. The proof that the functions F, obtained from the original exterior problem by the truncation procedure, are su¢ ciently small to apply the present theorem will be given in a subsequent paper. See [16]. The rest of this paper is organized as follows. In Section 2 we reduce the equation (6) and (2) to a set of integral equations for an evolution equation for which the coordinate y plays the role of time. In Section 3 we formulate the problem as a functional equation and prove the uniqueness of solutions. Existence of solutions is proved in Section 4.
2
Reduction to an evolution equation
Let u = (u; v) and F = (F1 ; F2 ). Then, the Navier-Stokes equations (6) are equivalent to ! = @y u + @x v ; @x !+ ! = q + ; @x u + @y v = 0 ;
(7) (8) (9)
q = @x (u!) + @y (v!) ; = @y F1 + @x F2 :
(10) (11)
where
The function ! is the vorticity of the ‡uid. Once the equations (7)-(9) are solved, the pressure p can be obtained by solving the equation p = r (F + u ru) in
+,
subject to the Neumann boundary condition
@y p(x; 1) = @y2 v(x; 1) : We now rewrite (7)-(9) as an evolution equation with y playing the role of time. Let q0 = u! ; q1 = v! ;
(12) (13)
Q0 = q0 + F2 ; Q1 = q1 F1 ;
(14) (15)
Q = @x Q0 + @y Q1 :
(16)
and let furthermore
and Then, instead of (7)-(8), we get the system of equations @y ! =
;
(17)
@y = @y u = @y v =
@x2 !
(18) (19) (20)
+ @x ! + Q ; ! + @x v ; @x u : 3
As we will see later on, the equations (17)-(20) have a special algebraic structure which permits to rewrite the solution as a sum of functions with di¤erent asymptotic behavior at in…nity, and this splitting will make the analysis simpler. With this in mind, we set v=!+ u = @x From (20) we then get that we get that
+ @y
= @y v =
@y u =
;
(21)
1
+
:
@x u =
! + @x v =
(22)
, and therefore that @y ! + @x ! + @x
=
:
, and from (19) (23)
Taking the derivative of (23) with respect to x gives, using (22), that @y + @y = @y @x u = @x ! + @x2 ! + @x2 , which, using (18), gives that ( @x2 ! + @x ! + Q) + @y = @x ! + @x2 ! + @x2 , and therefore we get that @y = @x2 + Q. We conclude that, instead of the system of equations (17)-(20), we can solve the system of equations @y ! =
;
@y = @y =
@x2 !
@y =
(24) ;
(25) (26)
+ @x Q0 + @y Q1 ;
(27)
@x2
+ @x ! + @x Q0 + @y Q1 ;
with v given by (21) and u given by (22). We now make a second change of variables which allows to express u in a more direct way. This will lead to additional signi…cant simpli…cations. Namely, we set = @x + Q1 ; = @x + Q1 :
(28) (29)
Substituting (28)-(29) into (24)-(27) we get that @y ! = @x + Q1 ; @x @y = @y = @x @y =
@x2 ! @x @x2
(30)
+ @x ! + @x Q0 ; Q1 ;
(31) (32)
+ @x Q0 :
(33)
All the terms on the right hand side containing only y-derivatives have disappeared and we can therefore instead of (30)-(33) solve the equations @y ! @y @y @y
= @x + Q1 ; = @x ! + ! + Q0 ; = @x Q1 ; = @x + Q0 ;
(34) (35) (36) (37)
with v given by (21) and with u given by u=
+
:
(38)
We now convert (34)-(37) into a system of ordinary di¤erential equations by taking the Fourier transform in the x-direction. De…nition 2 Let f^, g^ be complex valued functions de…ned almost everywhere on the inverse Fourier transform f = F 1 [f^] by Z 1 f (x; y) = F 1 [f^](x; y) = e ikx f^(k; y) dk ; 2 R 4
+.
Then, we de…ne
(39)
and ^ = f^ g^ by ^ (k; y) = (f^ g^)(k; y) =
Z
f^(k
k 0 ; y)^ g (k 0 ; y) dk 0 ;
R
whenever the integrals make sense.
We note that for functions f , g which are smooth and of compact support in f = F 1 [f^], and that f g = F 1 [f^ g^], where Z ^ f (k; y) = F[f ](k; y) = eikx f (x; y) dx ;
+
we have that
(40)
R
and similarly g^ = F[g]. With these de…nitions we formally have in Fourier space, instead of (34)-(37), the equations @y ! ^=
^1 ; ik ^ + Q
(41)
^0 ; @y ^ = (ik + 1)^ !+Q ^1 ; @y ^ = ik ^ Q
(43)
@y ^ =
^0 : ik ^ + Q
(44)
^ 0 = q^0 + F^2 ; Q ^ 1 = q^1 F^1 ; Q
(45)
(42)
From (14), (15) we get
(46)
from (12), (13) we get 1 (^ u ! ^) ; 2 1 (^ v ! ^) ; q^1 = 2
q^0 =
(47) (48)
and instead of (38) and (21) we have the equations ^+ ^ ; v^ = ! ^+ ^:
u ^=
(49) (50)
It is (41)-(50) that we solve in Section 3 in appropriate function spaces. We also show that the constructed solutions correspond via inverse Fourier transform to strong solutions of (2), (3), (6) with …nite Dirichlet integral. We now rewrite (41)-(50) as a system of integral equations (see Appendix A for a detailed derivation). From now we will use s, t 1 instead of y for the time variable, and , 0 for time di¤erences. We set p = k 2 ik ; (51) and de…ne, for k 2 R n f0g and
0, the functions Kn by,
1 e ; for n = 1; 2 ; 2 1 K3 (k; ) = e e ; 2 ik
Kn (k; ) =
(52) (53)
and the functions Gn by, 1 jkj e ; for n 2 1; 2 ; 2 1 jkj jkj G3 (k; ) = e e jkj : 2 ik
Gn (k; ) =
5
(54) (55)
We furthermore de…ne, for t 1, and n = 1, 2, 3, the intervals In by, I1 = [1; t], and In = [t; 1), otherwise. Using this notation, a representation of a classical solution to (41)-(44), which satis…es the boundary condition (3), in the sense that u ^(k; 1) = ^(k; 1) + ^(k; 1) = 0 and v^(k; 1) = ! ^ (k; 1) + ^(k; 1) = 0, is: X X X X ^= ^n;m ; ! ^= ! ^ n;m ; (56) m=0;1 n=1;2;3
^=
X
X
m=0;1 n=1;2;3
^n;m ;
^=
m=0;1 n=1;2;3
X
X
^n;m ;
(57)
m=0;1 n=1;2;3
with ^n;m (k; t) = Kn (k; t
1)
Z
gn;m (k; s
^ m (k; s) ds ; 1) Q
(58)
fn;m (k; s
^ m (k; s) ds ; 1) Q
(59)
kn;m (k; s
^ m (k; s) ds ; 1) Q
(60)
hn;m (k; s
^ m (k; s) ds ; 1) Q
(61)
In
! ^ n;m (k; t) = Kn (k; t
1)
^n;m (k; t) = Gn (k; t
1)
Z
Z
In
In
^n;m (k; t) = Gn (k; t
1)
Z
In
with Gn , Kn and In as de…ned above, with g1;0 (k; ) = g2;0 (k; ) =
ik ik ik ik
g3;0 (k; ) =
2
(jkj + )
e
e
+ 2 (jkj + ) e
(jkj + )
jkj
!
jkj (jkj + ) e ik
jkj
!
jkj (jkj + ) 2 e ik
jkj
)e
+ 2 (jkj + ) e
ik ik
g3;1 (k; ) = e
e
+
(jkj + ) e ik
2
(jkj + ) (1 + )e ik
ik
g3;m (k; ), f2;0 (k; ) =
jkj ik jkj (jkj + )2 jkj e + e ik jkj jkj jkj ik (jkj + )2 k2;0 (k; ) = ( + )e jkj ik jkj jkj ik jkj k3;0 (k; ) = e ; jkj jkj k1;1 (k; ) = ik jkj k2;1 (k; ) = ik
and h2;1 (k; ) =
(63)
!
;
(65)
;
(66) (67)
k1;0 (k; ) =
jkj
2
jkj
e
(jkj + ) e ik
jkj
2
(1
(jkj + ) )e ik
jkj
ik
(g2;0 (k; ) + 2e
), and
2(jkj + )e
;
(68)
2 (jkj + ) e
;
(69) (70)
(jkj + ) +2 e ik (jkj + ) +2 e ik
!
!
;
(71)
;
(72)
;
ik jkj k1;m (k; ), h3;m (k; ik jkj . jkj k2;1 (k; ) + 2e
and with h1;m (k; ) =
;
;
with f1;m (k; ) = ik g1;m (k; ), f3;m (k; ) = f2;1 (k; ) = ik g2;1 (k; ) 2e , with
k3;1 (k; ) = e
(62)
(64)
2
g2;1 (k; ) =
;
; 2
g1;1 (k; ) =
!
2
ik
(
e
jkj
(73) )=
ik jkj k3;m (k;
6
), h2;0 (k; ) =
ik jkj
k2;0 (k; ) + 2e
jkj
,
3
Functional framework
We start by de…ning adequate function spaces. Let , r ;r (k; t)
0 and k 2 R, and let
1 1 + (jkj tr )
=
:
(74)
Let furthermore (k; t) = ~ (k; t) =
;1 (k; t)
; ;2 (k; t) :
De…nition 3 Let R0 = R n f0g. We de…ne, for …xed of functions f 2 C(R0 [1; 1); C), for which the norm kf ; B
;p;q k
is …nite. Furthermore, we set W = B
= sup sup t 1 k2R0
; 27 ; 25
The following properties of the spaces B without mention: - if , - if
0
0, and p, p0 , q, q 0
> 1, p > 0 and q
0, then B
B
;p;q
;p;q
0, B
;p;q
to be the Banach space
jf (k; t)j (k; t) + t1q ~ (k; t)
1 tp
; 72 ; 52 ,
0, and p, q
and V = B
; 25 ;1
B
; 12 ;0
B
; 12 ;1 .
will be important below and will be routinely used \B
0 ;p0 ;q 0
Bminf
0;
;g;minfp0 ;pg;minfq 0 qg .
0, then
(k; t) 7!
1 tp
(k; t) 2 L2 (
+ ),
(k; t) 7!
1 ~ (k; t) 2 L2 ( tq
+)
:
Therefore, and because the Fourier transform is an isometry of L2 (R), we have that f = F L2 ( + ), whenever f^ 2 B ;p;q for some > 1, p > 0, q 0. - if > 1, p 0 and q 0, then f^ 2 B ;p;q is bounded by kf^; B Therefore, the function k 7! sup jf^( : ; t)j is in L1 (R).
;p;q k(1
1
[f^] 2
+ jkj)
, uniformly in t.
W ; (~ u1 ! ~ 2 ) ; 21 (~ v1 ! ~2) ;
(75)
t 1
Next, we rewrite the problem of solving (41)-(50) as a functional equation: Lemma 4 Let
> 1. Then, C :
V V ! ((~ !1 ; u ~1 ; v~1 ); (~ !2 ; u ~2 ; v~2 )) 7 !
1 2
de…nes a continuous bilinear map. Lemma 5 Let
> 1. Then, L :
W ! ~0; Q ~1) 7 ! (Q
V (~ !; u ~; v~) ;
(76)
de…nes a continuous linear map. Here, (~ !; u ~; v~) = (^ ! ; ^ + ^; ! ^ + ^), with (^ ! ; ^; ^; ^) given in terms of ^ ^ ~ ~ the integral equations (56)-(73), with (Q0 ; Q1 ) = (Q0 ; Q1 ). The maps C and L are studied in Section 4.1 and Section 4.2, respectively. Now let F = (F1 ; F2 ) 2 ~ = (F~1 ; F~2 ) = (F[F1 ]; F[F2 ]) be the Fourier transform of F. Note that (F~2 ; F~1 ) 2 W Cc1 ( + ), and let F for all > 1. 7
De…nition 6 Let
> 1. A triple (~ !; u ~; v~) is called an
-solution if:
(i) (~ !; u ~; v~) 2 V ; (ii) (~ !; u ~; v~) = L[C[(~ !; u ~; v~); (~ !; u ~; v~)] + (F~2 ; F~1 )] : With this de…nition at hand we can now give a precise formulation of Theorem 1: ~ = (F~1 ; F~2 ) be the Fourier Theorem 7 (Existence) Let > 1, F = (F1 ; F2 ) 2 Cc1 ( + ), and let F ~ ~ transform of F. If k(F2 ; F1 ); W k is su¢ ciently small, then there exists a unique -solution (~ !; u ~; v~) in V , with k(~ !; u ~; v~); V k C k(F~2 ; F~1 ); W k, for some constant C depending only on the choice of . Proof. Let " := k(F~2 ; F~1 ); W k. Since > 1, we have by Lemma 4 and Lemma 5 that the map N : V ! V , N [x] = L[C[x; x] + (F~2 ; F~1 )] is continuous. We now show that for " small enough there is a constant such that N is a contraction on the ball U = fx 2 V j kx; V k < g. Namely, let x 2 U . Then, by Lemma 4 there exists a constant C1 such that kC[x; x]; W k C1 ( )2 , and therefore kC[x; x] + (F~2 ; F~1 ); W k C1 ( )2 + " . Using now Lemma 5 it follows that there exists a constant C2 such that kN [x]; V k C2 (C1 ( )2 + " ). We set C = 2C2 . Now, we assume that "
1, and r, s 0 and let a, b be continuous functions from R0 satisfying the bounds (see (74) for the de…nition of ;r and ;s , respectively), ja(k; t)j jb(k; t)j
;r (k; t)
; (k; t) : ;s
Then, the convolution a b is a continuous function from R j(a b) (k; t)j uniformly in t
1 tr
const:
[1; 1) to C
[1; 1) to C and we have the bound
;s (k; t)
+
1 ts
;r (k; t)
;
(90)
1, k 2 R.
Proof. We only prove (90). Since the functions ;r and case k 0. Cutting the integral into two parts we have, j(a b) (k; t)j ;s (k=2; t)
const:
1 tr
Z
;s
k=2 ;r (k 1
0
0
; t) dk +
;r (k=2; t)
are even in k, it su¢ ces to consider the
Z
1
k 0 ; t) dk 0
;s (k
k=2
;s (k; t)
+
1 ts
;r (k; t)
;
and (90) follows. Corollary 10 Let, for i = 1; 2,
i
> 1, and pi , qi
0. Let fi 2 B
i ;pi ;qi
, and let
= minf 1 ; 2 g ; p = minfp1 + p2 + 1; p1 + q2 + 2; p2 + q1 + 2g ; q = minfq1 + q2 + 2; p1 + q2 + 1; p2 + q1 + 1g : Then f1 f2 2 B
;p;q
and there exists a constant C, depending only on kf1 f2 ; B
;p;q k
C kf1 ; B
1 ;p1 ;q1
k kf2 ; B
i,
such that
2 ;p2 ;q2
k
Proof. Using Proposition 9 we …nd that 1 1 1 1 (k; t) + q1 ~ 1 (k; t) (k; t) + q2 ~ 2 (k; t) tp1 1 t tp2 2 t const: const: minf 1 ; 2 g (k; t) + q1 +q2 +2 ~ minf 1 ; 2 g (k; t) p +p +1 1 2 t t const: const: const: const: + p1 +q2 +2 1 (k; t) + p1 +q2 +1 ~ 2 (k; t) + q1 +p2 +1 ~ 1 (k; t) + q1 +p2 +2 t t t t and the claim follows after regrouping of the terms involving and ~, respectively. Now let (~ !1 ; u ~1 ; v~1 ), (~ !2 ; u ~2 ; v~2 ) 2 V = B j 2 f1; 2g that u ~i ! ~ j 2 B ; 27 ; 25 , and that k~ ui ! ~j ; B
; 27 ; 52 k
B
; 52 ;1
const:k~ ui ; B
B
; 12 ;0
; 21 ;0 k
; 12 ;1 .
k~ !j ; B
2
(k; t) ;
Using Corollary 10 we …nd for i,
; 52 ;1 k
const:k(~ !i ; u ~i ; v~i ); V k k(~ !j ; u ~j ; v~j ); V k Similarly we …nd that v~i ! ~j 2 B k~ vi ! ~j ; B We conclude that
( 21
u ~1
! ~ 2 ; 21
; 27 ; 52 , ; 27 ; 25 k
and that const:k~ vi ; B
k~ !j ; B
; 52 ;1 k
const:k(~ !i ; u ~i ; v~i ); V k k(~ !j ; u ~j ; v~j ); V k v~1 ! ~2) 2 W = B
1 1 u ~1 ! ~2; v~1 ! ~ 2 ); W k 2 2 This completes the proof of Lemma 4. k(
; 12 ;1 k
; 72 ; 25
B
; 72 ; 25 ,
and that
const:k(~ !1 ; u ~1 ; v~1 ); V k k(~ !2 ; u ~2 ; v~2 ); V k
11
4.2 Let
Proof of Lemma 5 be as de…ned in (51), and de…ne =
by 1 2
Re( ) =
We have that j j = (k 2 + k 4 )1=4
q p 2 k 2 + k 4 + 2k 2 :
jkj1=2 + jkj
23=4 j j
and that jkj
j
Therefore, we have in particular that for
j
p
j j
2j
(91)
23=4 (1 + jkj) ; j:
(92) (93)
0 e
jkj
e
:
(94)
In what follows we prove Lemma 5 by providing bounds for the norms of ~, ! ~ , ~, and ~ in terms of ~ ~ the norms of Q0 and Q1 . We systematically use the notation introduced above, but, for simplicity, we set 1 1 (95) (k; s) = 7 (k; s) + 5 ~ (k; s) ; s2 s2 ~0; Q ~ 1 ); W k with C a constant independent of k and t. This constant may be di¤erent and kQk = C k(Q from instance to instance changing even within the same line. For ~ we have: Proposition 11 Let gi;j be as given in Section 2. Then we have the bounds j)ej
jg1;0 (k; )j
const:(1 + j
jg2;0 (k; )j
const:(1 + jkj)e
jg3;0 (k; )j
const:e
j
const:e const:ej
jg1;1 (k; )j jg2;1 (k; )j jg3;1 (k; )j
jkj
j
j
j
j2
minf1; j
2
g;
;
minf1; j
const:(1 + jkj)e e
j
(97) jg ;
(98)
for jkj 1 j for jkj > 1
jkj
(96)
;
;
(99) (100)
;
(101)
uniformly in 0 and k 2 R0 (and uniformly in k 2 R0 , jkj the case of (99)).
1 and k 2 R0 , jkj > 1, respectively, for
Proof. From (62) we get that jg1;0 (k; )j
const:(1 + j
j)ej
j
:
Expanding the exponential functions in (62) the …rst two terms cancel, so that 2
g1;0 (k; ) = (e +2
1
(jkj + ) e ik
and therefore, since for all z 2 C with Re(z) ez
(jkj + ) ik
)
jkj
e
1 + jkj
1+ ;
(102)
0 and N 2 N0 ,
PN
1 n n=0 n! z z N +1
12
const: ;
(103)
and for all z 2 C with Re(z) > 0
PN
1 n n=0 n! z z N +1
ez
const:eRe(z) ;
(104)
we …nd from (102) using (92) that jg1;0 (k; )j
j2
const:j
2
(1 + j
j)ej
j
:
This completes the proof of (96). The bounds (97) and (98) follow using (92), and (94). We now prove (99). Using (92) we …nd from (65) for k 1 that jg1;1 (k; )j For jkj that
const:(1 + j
j)ej
j
:
1 we use the fact that if we expand the exponential functions in (65) the …rst term cancels, so ! 2 (jkj + ) jkj (jkj + ) e jkj g1;1 (k; ) = (e 1) + e 1 2 1 : (105) ik ik ik
From (105) we …nd, using (92), (103), and (104) that jg1;1 (k; )j
j j2 j e jkj
const:
j
+ j j2
+
const:ej
j
:
To prove (100) we use that 2
(jkj + ) 2k 2 + 2 jkj j j = ; ik ik and the result again follows using (92), and (94). The bound (101) is trivial. 1+
As a consequence of Proposition 11 we have: ~0; Q ~ 1 ) 7! ~ de…nes a continuous linear map from W to B ; 3 ;0 . Proposition 12 Let > 1. Then, (Q 2 ~0; Q ~ 1 ) 7! ~i;j , with ~i;j as given in (56), de…nes continuous linear maps on W , with More precisely, (Q values ~i;0 2 B ; 25 ; 32 , i = 1; 2; 3, ~1;1 2 B ; 32 ;0 , ~2;1 2 B ; 52 ; 23 , and ~3;1 2 B ; 23 ; 12 . Proof. Using (96), Proposition 20, and Proposition 21 we …nd, with the notation (95), that Z t j~1;0 (k; t)j kQk e (t 1) (1 + j j)ej j(s 1) minf1; j j2 (s 1)2 g (k; s) ds 1 Z t 1 1 (t 1) (k; s) + 5 ~ (k; s) = kQk e (1 + j j)ej j(s 1) minf1; j j2 (s 1)2 g 7 2 2 s s 1 1 1 1 1 kQk 2 t 2 ~ (k; t) + 5 (k; s) + 3 ~ (k; t) ; t t2 t2 and therefore ~1;0 2 B
; 25 ; 32 .
Using (97) and Proposition 25 we …nd that Z 1 (t 1) j~2;0 (k; t)j kQke (1 + jkj)e jkj(s 1) (k; s) ds t
(t 1)
kQke and therefore ~2;0 2 B
; 52 ; 32 .
j~3;0 (k; t)j
e
jkj(t 1)
Using (98) with minf1; j kQk kQk
ik 1 5
t2
e
(t 1)
e
1
t
(k; t) +
5 2
jg
(t 1)
(k; t) +
3
t2
~ (k; t)
13
3
t2
~ (k; t)
;
j j and Proposition 22 we …nd that Z 1 e (s 1) j j (k; s) ds t
1
1
;
ds
and therefore ~3;0 2 B
; 52 ; 32 .
Using (99), Proposition 20 and Proposition 21, we …nd for jkj > 1 that
j~1;1 (k; t)j
kQk e kQk
and for jkj
(t 1)
Z
t
ej
j(s 1)
1
j
j (k; s) ds
1 1 ~ (k; t) + 7 t t2
(k; t) +
Z
1) (k; s) ds
1 5
t2
~ (k; t)
;
1 that kQk e
(t 1)
kQk e
(t 1)
+ kQk e
(t 1)
j~1;1 (k; t)j
t
ej
Z
t+1 2
1 Z t
t+1 2
kQk ~ (k; t) + and therefore ~1;1 2 B
; 32 ;0 .
ej
j(s 1)
ej
(s
(k; t) +
3
t2
1) (k; s) ds 1
j(s 1)
1
(t 1)
kQk e
; 52 ; 32 .
Z
1
s 1
(k; s) +
5 2
~ (k; t)
1
t2
1 3
s2
~ (k; s)
ds
;
(t 1)
e
jkj(s 1)
(1 + jkj) e
t
kQke
1
jkj(t 1)
(k; s) ds
(k; t) +
5
t2
1 3
t2
~ (k; t)
;
Finally, using (101) and Proposition 22 we …nd that
j~3;1 (k; t)j
kQk
e
ik 1
kQk and therefore ~3;1 2 B
(s
Using (100) and Proposition 25 we …nd that
j~2;1 (k; t)j
and therefore ~2;1 2 B
j(s 1)
1
t
(t 1)
Z
(t 1)
e
1
(s 1)
e
(k; s) ds
t
3 2
1
(k; t) +
1
t2
~ (k; t)
;
; 23 ; 12 .
For ! ~ we have: Proposition 13 Let fi;j be as given in Section 2. Then we have the bounds jf1;0 (k; )j
const:ej
jf2;0 (k; )j
)e
jf3;0 (k; )j
const:(jkj + jkj
const:e
jf1;1 (k; )j
const:(1 + j
j
jf2;1 (k; )j
const: (1 + jkj) e
jf3;1 (k; )j uniformly in
j
minfj 1=2
const:e
minf1; j j)e
jkj
minf1; j
j3
j; j
jkj
2
g;
(106)
;
(107)
j g;
(108)
2
j
minf1; j
j g;
;
(110)
jg ;
(111)
0 and k 2 R0 .
Proof. Since ik
const:
j j 1+j j
14
(109)
const: minf1; j
jg ;
(106) follows immediately from (96), (107) from (97), (108) from (98), (110) from (100), and (111) from (101). Finally, in order to prove (109), we note that 2
f1;1 (k; ) = e
(jkj + ) e ik
+
and therefore we …nd using (92) that jf1;1 (k; )j functions in (112) we see that
2
jkj (jkj + ) e ik
2
f1;1 (k; ) = (e
(jkj + ) ik
1) +
j)ej
const:(1 + j e
1
2
jkj
j
(112)
. Expanding the exponential
jkj (jkj + ) e ik
and therefore we …nd using (92), (103), and (104) that jf1;1 (k; )j
;
jkj
const:(1 + j
1
;
j)ej
j
j
j .
As a consequence of Proposition 11 we have: ~0; Q ~ 1 ) 7! ! Proposition 14 Let > 1. Then, (Q ~ de…nes a continuous linear map from W to B ; 52 ;1 . ~ ~ More precisely, (Q0 ; Q1 ) 7! ! ~ i;j , with ! ~ i;j as in (56), de…nes continuous linear maps on W , with values ! ~ 1;0 2 B ; 27 ; 52 , ! ~ 2;0 2 B ;3;2 , ! ~ 3;0 2 B ; 52 ; 23 , ! ~ 1;1 2 B ; 52 ;1 , and ! ~ i;1 2 B ; 52 ; 32 , i = 2; 3. Proof. Using (106), Proposition 20, and Proposition 21 we …nd that j~ !1;0 (k; t)j
(t 1)
kQk e
;3;2 .
j~ !3;0 (k; t)j
kQk e
(t 1)
kQk e
(t 1)
j~ !1;1 (k; t)j
; 52 ; 32 .
kQk
kQk e
; 25 ;1 .
j3 (s 1)2 g (k; s) ds !
j; j 1 5
t2
~ (k; s)
;
Z
1
t
1 t3
jkj(t 1)
e
j2 g
(t 1)
e
ik 1
1=2
(jkj + jkj
j
jkj(s 1)
(k; s) ds
1 ~ (k; t) t2
(k; t) +
;
j we get from (108) and Proposition 22 that Z 1 e (s 1) j j (k; s) ds
(t 1)
e
)e
t
(k; t) +
5
t2
1 3
t2
~ (k; t)
;
Using (109) and Proposition 20 we …nd that
kQk and therefore ! ~ 1;1 2 B
minfj
(k; s) +
Using that minf1; j
kQk and therefore ! ~ 3;0 2 B
j(s 1)
Using (107) and Proposition 25 we …nd that
j~ !2;0 (k; t)j
and therefore ! ~ 2;0 2 B
ej
1
1
; 27 ; 52 .
t
t2 1 ~ (k; t) + 7 t3 t2
kQk and therefore ! ~ 1;0 2 B
Z
(t 1)
Z
1
t
j)ej
(1 + j
1 1 ~ (k; t) + 5 t t2
(k; t) +
j(s 1)
1 3
t2
minf1; j
~ (k; t)
j(s
1)g (k; s) ds
;
Using (110) and Proposition 25 we …nd that
j~ !2;1 (k; t)j
kQk
1 e 2
kQk e
(t 1)
Z
t
(t 1)
e
1
(1 + jkj) e
jkj(t 1)
15
1 t
5 2
jkj(s 1)
(k; t) +
(k; s) ds 1 3
t2
~ (k; t)
;
and therefore ! ~ 2;1 2 B j~ !3;1 (k; t)j
; 52 ; 32 .
Finally, using (111) and Proposition 22 we …nd that
kQk kQk
and therefore ! ~ 3;1 2 B
e
ik 1
(t 1)
Z
(t 1)
e
1
(s 1)
e
t
(k; t) +
5
t2
1 3
t2
~ (k; t)
minf1; j
jg (k; s) ds
;
; 25 ; 32 .
For ~ we have: Proposition 15 Let ki;j be as given in Section 2. Then we have the bounds
uniformly in
3=2
jk1;0 (k; )j
const:(1 + jkj)ejkj minf1; jkj
jk2;0 (k; )j
const:(1 + jkj)e
jk3;0 (k; )j
e
jk1;1 (k; )j
const:(1 + jkj)e
jk2;1 (k; )j
const:(1 + jkj)e
jk3;1 (k; )j
e
jkj
jkj
g;
(113)
;
(114)
;
(115) jkj
jkj
2
1=2
jkj
minf1; (1 + jkj
1=2
) jkj
g;
(116)
;
(117)
;
(118)
0 and k 2 R0 . const:(1 + jkj)ejkj . Expanding the exponential functions
Proof. From (68) we get that jk1;0 (k; )j in (68) we …nd that 2
k1;0 (k; ) = ejkj
1
jkj
+
(jkj + ) ik
e
jkj
1 + jkj
2
jkj (jkj + ) e ik
1+
;
and therefore using (104) and (103) that jk1;0 (k; )j
2
const: jkj
2 jkj
e
3=2
const:(1 + jkj) jkj
2
+ jkj
2
2 jkj
e
(1 + jkj) + j
3
j
2
:
This completes the proof of (113). Using (92) and (94) we …nd (114), (115), (117) and (118). From (71) we get that jk1;1 (k; )j const:(1 + jkj)ejkj . Expanding the exponential functions in (71) we …nd that ! 2 jkj (jkj + ) (jkj + ) jkj jkj k1;1 (k; ) = e 1 e 1 +2 e 1 ; ik ik ik and therefore using (104) and (103) that jk1;1 (k; )j
const: jkj ejkj + jkj (1 + jkj) + (1 + jkj) j 1=2
const:(1 + jkj)(1 + jkj1=2 ) jkj
j
ejkj :
As a consequence of Proposition 15 we have: ~0; Q ~ 1 ) 7! ~ de…nes a continuous linear map from W to B ; 1 ; 3 . Proposition 16 Let > 1. Then, (Q 2 2 ~0; Q ~ 1 ) 7! ~i;j , with ~i;j as in (57), de…ne continuous linear maps on W , with values More precisely, (Q ~1;0 2 B 3 , ~1;1 2 B 1 3 , and ~i;j 2 B 5 3 , i = 2; 3, j = 0; 1. ;1; 2
;2;2
;2;2
16
Proof. Using (113), Proposition 23, and Proposition 24 we …nd that Z t 3=2 ~1;0 (k; t) kQk e jkj(t 1) (1 + jkj)ejkj(s 1) minf1; jkj (s
1)2 g (k; s) ds
1
kQk and therefore ~1;0 2 B
1
t3=2
t1=2
(k; t) +
1
t
(k; t) +
5 2
1
3
t2
~ (k; t)
;
;1; 23 .
Using (114) and Proposition 25 we …nd that Z 1 jkj(t 1) ~2;0 (k; t) (1 + jkj)e jkj(s 1) (k; s) ds kQk e t
kQke and therefore ~2;0 2 B
; 52 ; 32 .
~3;0 (k; t)
and therefore ~3;0 2 B
1
2jkj(t 1)
t
(k; t) +
5 2
1
3
t2
~ (k; t)
Using (115) and Proposition 25 we …nd that Z 1 jkj jkj(t 1) jkj(t 1) e jkj(s e kQk e ik t 1 1 kQk (k; t) + 3 ~ (k; t) ; 5 2 t t2
;
1)
(k; s) ds
Using (116), Proposition 23, and Proposition 24 we …nd that Z t 1=2 kQk e jkj(t 1) (1 + jkj)ejkj(s 1) minf1; (1 + jkj1=2 ) jkj (s 1)g (k; s) ds
~1;1 (k; t)
; 32 ; 12 .
1
kQk and therefore ~1;1 2 B
1
t1=2
(k; t) +
1
t
5 2
(k; t) +
1
3
t2
~ (k; t)
;
; 21 ; 32 .
Using (117) and Proposition 25 we …nd that Z 1 jkj(t 1) ~2;1 (k; t) kQk e (1 + jkj)e jkj(s 1) (k; s) ds t
kQke and therefore ~2;1 2 B
; 52 ; 32 .
~3;1 (k; t)
and therefore ~3;1 2 B
1
2jkj(t 1)
t
(k; t) +
5 2
1
3
t2
~ (k; t)
Using (118) and Proposition 25 we …nd that Z 1 jkj jkj(t 1) jkj(t 1) e e kQk e jkj(s ik t 1 1 (k; t) + 3 ~ (k; t) ; kQk 5 2 t t2
;
1)
(k; s) ds
; 52 ; 32 .
For ~ we have: Proposition 17 Let hi;j be as given in Section 2. Then we have the bounds
uniformly in
3=2
jh1;0 (k; )j
const:(1 + jkj)ejkj minf1; jkj
jh2;0 (k; )j
const:(1 + jkj)e
jh3;0 (k; )j
e
jh1;1 (k; )j
const:(1 + jkj)e
jh2;1 (k; )j
const:(1 + jkj)e
jh3;1 (k; )j
e
jkj
jkj
g;
(119)
;
(120)
;
(121) jkj
jkj
2
jkj
;
1=2
minf1; (1 + jkj ;
1=2
) jkj
g;
(122) (123) (124)
0 and k 2 R0 . 17
Proof. The bounds (119)-(124) immediately follow from (113)-(118) using the de…nitions. As a consequence of Proposition 17 we have: ~0; Q ~ 1 ) 7! ~ de…nes a continuous linear map from W to B ; 1 ; 3 . Proposition 18 Let > 1. Then, (Q 2 2 ~ ~ ~ ~ More precisely, (Q0 ; Q1 ) 7! i;j , with i;j as in (57), de…ne continuous linear maps on W , with values ~1;0 2 B 1 3 , and ~i;j 2 B 5 3 , i = 2; 3, j = 0; 1. 3 , ~1;1 2 B ;1; 2
;2;2
;2;2
Proof. The bounds (119)-(124) are identical to the bounds (113)-(118), and the proof is therefore the same as for Proposition 15.
A
Derivation of the integral equations
In order to derive the integral equations (56), (57) we note that the equations (41)-(44) are of the form ^1; Q ^0; Q ^1; Q ^ 0 ) and with z_ = Lz + q, with z = (^ ! ; ^; ^; ^), q = (Q L(k) =
L1 (k) 0 0 L2 (k)
;
(125)
;
(126)
where 0 ik + 1
L1 (k) =
ik 0
and where 0 ik
L2 (k) =
ik 0
:
(127)
Then, we have that L1 = S1 D1 S1 1 , where S1 = and where D1 is the diagonal matrix with entries S2 (k) =
1
1
ik
ik
;
, and furthermore that L2 = S2 D2 S2 1 , where
and 1
1
ik jkj
ik jkj
and where D2 is the diagonal matrix with entries jkj and
and that 1
S2 (k) =
;
1 2 1 2
ik 2jkj ik 2jkj
(129)
jkj. We have that
ik 2 ik 2
1 2 1 2
S1 1 =
(128)
; !
(130)
:
(131)
Let S(k) = and z = Sr. Then r_ = Dr + S
1
S1 (k) 0 0 S2 (k)
;
(132)
q with S
1
(k) =
S1 1 (k) 0 0 S2 1 (k)
18
:
(133)
Let r = (^ !+ ; ! ^ ; ^+ ; ^ ). Using the de…nitions we …nd that from (41)-(44), 1^ ik ^ @y ! ^+ = ! ^+ + Q Q0 ; 1 2 2 1^ ik ^ @y ! ^ = ! ^ + Q Q0 ; 1+ 2 2 1^ ik ^ @y ^+ = jkj ^+ Q1 + Q0 ; 2 2jkj 1^ ik ^ @y ^ = jkj ^ Q1 Q0 : 2 2jkj
(134) (135) (136) (137)
Note that, in component form, we have for z = Sr : ! ^=! ^+ + ! ^
;
(138)
( ! ^+ + ! ^ ); ik ^ = ^+ + ^ ; ^ = ik ( ^+ + ^ ) : jkj
(139)
^=
(140) (141)
^0; Q ^ 1 ), a classical representation of solutions to (134)–(137) is (we use from now on t instead For given (Q of the y for the “time variable”): Z ik ^ 1 1 (t s) ^ Q0 (k; s) ds ; (142) e Q1 (k; s) ! ^ + (k; t) = 2 t Z 1 t ^ 1 (k; s) + ik Q ^ 0 (k; s) ds ; ! ^ (k; t) = ! (k)e (t 1) + e (t s) Q (143) 2 1 Z 1 ^+ (k; t) = 1 ^ 1 (k; s) ik Q ^ 0 (k; s) ds ; (144) ejkj(t s) Q 2 t jkj Z 1 t jkj(t s) ^ ik ^ jkj(t 1) ^ (k; t) = (k)e Q1 (k; s) + Q0 (k; s) ds : (145) e 2 1 jkj The functions !
and
are determined by the boundary condition (3). At t = 1 we have Z 1 1 (1 s) ^ ik ^ ! ^ + (k; 1) = Q0 (k; s) ds ; e Q1 (k; s) 2 1
! ^ (k; 1) = ! (k) ; Z 1 ^+ (k; 1) = 1 ejkj(1 2 1 ^ (k; 1) = (k) :
s)
ik ^ Q0 (k; s) jkj
^ 1 (k; s) Q
ds ;
(146) (147) (148) (149)
Substituting (142)-(145) into (138)-(141) and the result into (38) and (21) we get, when evaluating at t = 1, 0 = ! (k) + (k) Qv (k) ; where Z 1 1 e Qv (k) = 2 1 Z 1 1 e 2 1
(1 s) (1 s) ik
and 0=
ik
ejkj(1
! (k)
s)
ejkj(1
jkj ik 19
(k)
^ 1 (k; s) ds Q s)
ik jkj
^ 0 (k; s) ds ; Q
Qu (k) ;
(150)
where Qu (k) =
Z 1 1 e (1 2 1 Z 1 1 e (1 2 1
Since 1
ik
jkj ik
jkj ik
= (jkj + )
;
(152)
Qv (k) Qu (k)
1 1
ik
(151)
1 1
ik
we …nd that ! (k) (k)
^ 0 (k; s) ds : Q
s)
= (jkj + )
jkj ik
^ 1 (k; s) ds Q
ik
1
1
ik
ejkj(1
s)
s) jkj
ejkj(1
s)
;
(153)
from which we get that Z 1 1 (jkj + )2 e 2 1 ik Z 1 1 (jkj + )2 e 2 1
! (k) =
jkj(jkj + ) jkj(1 e ik
(1 s)
2
(1 s)
2(jkj + )ejkj(1
s)
^ 1 (k; s) ds Q
s)
^ 0 (k; s) ds ; Q
(154)
and that Z 1 1 (jkj + )2 jkj(1 e (k) = 2 1 ik Z 1 1 (jkj + )2 jkj(1 e 2 1 jkj
(jkj + ) e ik
s)
2
s)
2(jkj + )e
^ 1 (k; s) ds Q
(1 s)
(1 s)
^ 0 (k; s) ds : Q
(155)
Substituting ! ~ given by (154) into (139) gives, after splitting the integral over [1; 1] into an integral over [1; t] and over [t; 1], the representation in (56) for ^. Similarly, substituting ! ~ given by (154) into (138) gives the representation in (56) for ! ^ , substituting ~ given by (155) into (141) gives the representation in (57) for ^, and substituting ~ given by (155) into (140) gives the representation in (57) for ^.
B
Basic bounds
B.1
Continuity of semi-groups
We have: 0
Proposition 19 Let
0
,
1 1 + jkj
0
uniformly in k 2 R and t Proof. For 1
t 1
1 + jkj
j
j
0
t
0
+ 1
0
0
e
jkj(t 1)
jkj
0, and let
0
t
0
1
0
,
0
0
,
0
const:
t
> 0. Then, we have the bound
1 1 t 0 1 + (jkj t2 )
const:
t
1. Similarly, for positive
with
0
0
0
+
1 1 t 0 1 + (jkj t)
0
;
0+ 0
0
0 and
0+ 0
> 0 we have
;
1.
2 and jkj 0 e
0
0 with (t 1)
e
uniformly in k 2 R and t the bound 1 1 + jkj
0
,
(t 1)
1 we have that j
j
0
t
1 t
0
const:
20
const:
1 1 t 0 1 + (jkj t2 )
0
0+ 0
;
and that 1 1 + jkj Next, for 1
t
0
e
jkj(t 1)
t
0
jkj
0
1
const:
t
1 1 t 0 1 + (jkj t)
const:
0
0+ 0
:
2 and jkj > 1 we have that 1 1 + jkj
0
(t 1)
e
1
const:
0
1 + jkj 1
const:
j
j
0
1 t
(t 1)
e
0
jkj
0
t
0
(j
j (t
0
1 + jkj 1 1 const: 0 2 t 1 + (jkj t )
j
j
0
0
1
const:
0
0
1))
0
1 + jkj
0+ 0
:
0+ 0
and similarly that 1 1 + jkj
jkj(t 1)
e
0
const: const:
1 0
1 + jkj 1
0
t
0
jkj
t jkj(t 1)
e
0
1
0
jkj
1 + jkj 1 1 const: 0 t 1 + (jkj t)
(jkj (t
0
0
0
1))
jkj
0
0
1
const:
1 + jkj
0
0+ 0
:
0+ 0
Finally, for t > 2 and k 2 R we have 1 + jkj t
2
0
0
0
+
const: 1 + jkj t
e
2
const: 1 + jkj t2 const: 1 + const: 1 +
j j
jkj
jkj
0
0
+
0
0
0
+
0
0
0
0
0+ 0
)
+
0
2(
j
j
e
0+
)
0
tj
1 2
1
t
e2
2(
j
tj
!
t
0
1 t
t
+
0
2(
j
(t 1)
j
j
tj tj
0
0
0
0
+ )
0
j
tj e
1 2
t
!
const: ;
and similarly that 1 + (jkj t)
0
0
+
0
const: 1 + (jkj t)
B.2
e
jkj(t 1)
0
0
+
0
e
Convolution with the semi-group e
(jkj t) 1 2
jkjt
0
t
1
0
t (jkj t)
0
const: :
t
In order to bound the integrals over the interval [1; t] we systematically split them into integrals over 1+t [1; 1+t 2 ] and integrals over [ 2 ; t] and bound the resulting terms separately. We have: 21
Proposition 20 Let
0, r
0 and (t 1)
e
Z
0 and
t+1 2
ej
+1
j(s 1)
1
uniformly in t
j
j
0. Then, (s
1)
;r (k; s)
s
ds
8 1 > const: ~ (k; t); if > + 1 > > > t > > > > < log(1 + t) const: ~ (k; t); if = + 1 > t > > > > > +1 > > : const: t ~ (k; t); if < + 1 t
(156)
1 and k 2 R.
Proof. We have that (t 1)
e
Z
t+1 2
ej
j(s 1)
1 j t 21
(t 1) j
e
e
t
const:
j
j
j
t
e
t
1)
1 2
;r (k; s)
s
;r (k; 1)
+1
1
(s
j
j
Z
t+1 2
ds
1) s 1 8 1, if > + 1 > > > > < log(1 + t), if = + 1 ;1 (k; 1) > > > > : +1 t , if < + 1
j
(s
From the last expression (156) follows using Proposition 19 Proposition 21 Let e uniformly in t Proof. If
0, r (t 1)
Z
0,
2 R, and
ej
j(s 1)
2 f0; 1g. Then,
t t+1 2
j
1 s
j
;r (k; s)
ds
const: t 1+
;r (k; t)
;
(157)
1 and k 2 R. = 0 we have that Z t e (t 1) ej t+1 2
and (157) follows, and if e
(t 1)
j(s 1)
1 s
const: t
;r (k; s) ds
;r (k; t)
Z
t
ds ; t+1 2
= 1 we have that Z
const: t
t t+1 2
ej
j(s 1)
;r (k; t)e
j
(t 1)
j
Z
1 s
;r (k; s)
ds
t t+1 2
ej
j(s 1)
j
j ds
const: t
;r (k; t)
;
and (157) follows. Using Hölder’s inequality the proposition can also be proved for intermediate values of , but this is not needed here. Next we have:
22
Proposition 22 Let
ik uniformly in t
ej
0, r
> 1, and Z 1 e ej j(t 1) t Z 1 e e (t 1)
j(t 1)
0,
2 f0; 1g. Then, (s 1)
j
(s 1)
j
t
1 s 1 j s j
;r (k; s)
ds
;r (k; s)
ds
const: t 1+ const: t 2+
;r (k; t)
;
(158)
;r (k; t)
;
(159)
1 and k 2 R.
Proof. We …rst prove (158). If = 0 we have that Z 1 1 j j(t 1) e (s 1) e ;r (k; s) ds s t and (158) follows, and if Z 1 e ej j(t 1)
1
1 ds ; s
t
= 1 we have that (s 1)
t
j
j
1 s
;r (k; s)
ej
j(t 1)
e
1 t 1 t
ds
and (158) follows. We now prove (159). For jkj ik
;r (k; t)
Z
j ;r (k; t)e
j(t 1)
;r (k; t)
e
j
j ds
1 we have that
(t 1)
j(t 1)
(s 1)
e
;
const:
j
jkj
j
ej
j(t 1)
j(t 1)
j
j(t
const:ej
(t 1)
1)
t;
and the bound on (159) now follows as in the the proof of (158). For jkj ej
1
t
const:ej
ik
Z
1 we have that
j(t 1)
;
and the bound on (159) now again follows as in the the proof of (158). The proposition can also be proved for intermediate values of , but this is not needed here.
B.3
Convolution with the semi-group e
jkjt
In order to bound the integrals over the interval [1; t] we systematically split them into integrals over 1+t [1; 1+t 2 ] and integrals over [ 2 ; t] and bound the resulting terms separately. We have: Proposition 23 Let
0, r e
0 and jkj(t 1)
0 and
Z
t+1 2
ejkj(s
1
uniformly in t
1 and k 2 R.
+1 1)
jkj
0. Then, (s
1) s
;r (k; s)
8 1 > (k; t); if > + 1 const: > > > t > > > > < log(1 + t) const: (k; t); if = + 1 > t > > > > > > > t +1 : (k; t); if < + 1 const: t
Proof. The proof is as for Proposition 20. Next we have: 23
ds
(160)
Proposition 24 Let
0, r e
uniformly in t
jkj(t 1)
0, Z
2 R, and
t t+1 2
ejkj(s
1)
jkj
2 f0; 1g. Then, 1 s
;r (k; s)
ds
const: t 1+
;r (k; t)
;
1 and k 2 R.
Proof. The proof is as for Proposition 21. Using Hölder’s inequality the proposition can also be proved for intermediate values of , but this is not needed here. Next we have: Proposition 25 Let
0, r
jkj jkj(t e ik uniformly in t
1)
0,
> 1, 2 [0; 1] Then, Z 1 1 jkj(t 1) e jkj(s 1) jkj e s Zt 1 1 e jkj(s 1) jkj e jkj(t 1) s t
;r (k; s)
ds
;r (k; s)
ds
const: t 1+ const: t 1+
;r (k; t)
;
(161)
;r (k; t)
;
(162)
1 and k 2 R.
Proof. For jkj < 1=t and 0
Peter Wittwer
Université de Toulouse, IMT
Département de Physique Théorique
Toulouse, France
Université de Genève, Switzerland
[email protected]
[email protected]
Abstract We consider the problem of a body moving within an incompressible ‡uid at constant speed parallel to a wall, in an otherwise unbounded domain. This situation is modeled by the incompressible NavierStokes equations in an exterior domain in a half space, with appropriate boundary conditions on the wall, the body, and at in…nity. Here we prove existence of stationary solutions for this problem for the simpli…ed situation where the body is replaced by a source term of compact support.
Contents 1 Introduction
1
2 Reduction to an evolution equation
3
3 Functional framework
7
4 Proof of main lemmas 4.1 Proof of Lemma 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Proof of Lemma 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10 11 12
A Derivation of the integral equations
18
B Basic bounds B.1 Continuity of semi-groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B.2 Convolution with the semi-group e t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B.3 Convolution with the semi-group e jkjt . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20 20 21 23
1
Introduction
The present paper is the main step in an e¤ort to develop the mathematical framework which is necessary for the precise computation of the hydrodynamic forces that act on a body that moves at small constant speed parallel to a wall in an otherwise unbounded space …lled with a ‡uid. A very important practical application of such a situation is the description of the motion of bubbles rising in a liquid parallel to a nearby wall. Interesting recent experimental work is described in [14] and in [6]. Numerical studies can be found in [2], [5], [13], and [15]. The computation of hydrodynamic forces is reviewed in [10]. In what follows we consider the situation of a single bubble of …xed shape which rises with constant velocity in a regime of Reynolds numbers less than about …fty. The resulting ‡uid ‡ow is then laminar. The Stokes equations provide a good quantitative description (forces determined within an error of one Supported in part by the Swiss National Science Foundation.
1
percent) only for Reynolds numbers less than one. For the larger Reynolds numbers under consideration the Navier-Stokes equations need to be solved in order to obtain precise results. The vertical speed of the bubble depends on the drag, and the distance from the wall at which the bubble rises requires one to …nd the position relative to the wall where the transverse force is zero. Since at low Reynolds numbers the transverse forces are orders of magnitude smaller than the forces along the ‡ow, this turns out to be a very delicate problem which needs to be solved numerically with the help of high precision computations. But, if done by brute force, such computations are excessively costly even with today’s computers. In [3] we have developed techniques that lead for similar problems to an overall gain of computational e¢ ciency of typically several orders of magnitude. See also [4] and [10]. These techniques use as an input a precise asymptotic description of the ‡ow. The present work is an important step towards the extension of this technique to the case of motions close to a wall. In what follows we consider the two dimensional case. For convenience later on we place the position of the wall at y = 1. Namely, let x = (x; y), let + = f(x; y) 2 R2 j y > 1g, let B + be a compact set with smooth boundary @B, and let e1 = (1; 0). Then, in a frame comoving with the body, the Navier-Stokes equations are u ru which have to be solved in the domain
=
@x u+ u
+
rp = 0 ; r u=0;
(1) (2)
n B, subject to the boundary conditions
u(x; 1) = 0 ; lim u(x) = 0 ;
x2R;
x!1
uj@B =
e1 :
(3) (4) (5)
The standard technique to solve this problem is to prove the existence of weak solutions. Such solutions are constructed by considering a nested sequence of …nite domains that converge to + . Existence follows by a compactness argument. See for example [17], [8], [9], for the case of = R3 n B, and [12], [11] for the case of = + n B. Weak solutions constructed this way are smooth, the only shortcoming of the method is that only very little information is obtained about the behavior of the solutions at in…nity. In order to obtain such information a classic way is to consider the problem in an appropriately chosen weighted Sobolev space. Such methods are well developed for the case of isotropic weights, but become very technical if, as in the present case, anisotropic weights are needed. See for example [7], [1]. In this paper, we propose a new strategy which takes advantage of this anisotropy. In order to obtain detailed information at in…nity, we construct a classical solution in a function space which is motivated by the theory of dynamical systems. Namely, we chose the coordinate y to play the role of time and rewrite (1)-(5) as a system of evolution equations with respect to this variable. Information on the large time behavior of the dynamical system then naturally provides detailed information at in…nity. In order to get a system of ordinary di¤erential equations we use the Fourier transform in the x coordinate. We then choose the function spaces which are well adapted to the problem. These spaces come up naturally once the problem is formulated in this form. However, to use our techniques based on the Fourier transform we need that the problem be formulated on all of + . This is achieved as follows. Let u ~ be a smooth solution of the above problem and let ~ be the corresponding stream function, i.e., u ~ = ( @y ~; @x ~). One can then always use a smooth cut o¤ ~ function such that the function = is equal to ~ outside a su¢ ciently large disk D + containing B and zero inside some smaller disk D1 (but su¢ ciently large to still contain B). Inside B, is also de…ned equal zero. Let u = ( @y ; @x ). Then, since u = u ~ in the complement of D, we …nd that u satis…es (1), (2) for a certain smooth force term F of compact support. Motivated by these remarks we consider in what follows the equation @x u+ u = F + u ru + rp ;
(6)
in the domain + , subject to the incompressibility conditions (2) and the boundary conditions (3) and (4), and with F a smooth vector …eld with compact support in + , i.e., F 2 Cc1 ( + ). 2
The following theorem is our main result (see Section 3, Theorem 7 and Theorem 8 for a precise formulation): Theorem 1 For all F 2 Cc1 ( + ) with F su¢ ciently small in a sense to be de…ned below, there exist a unique vector …eld u = (u; v) 2 H 1 ( + ) and a function p satisfying the Navier-Stokes equations (6), (2) in + subject to the boundary conditions (3) and (4). Moreover, there exist constants C1 , C2 such that, uniformly in (x; y); ju(x; y)j C1 =y 3=2 and jv(x; y)j C2 =y 3=2 . To our knowledge this is the …rst proof of existence which provides detailed information on the behavior of solutions at in…nity. The proof that the functions F, obtained from the original exterior problem by the truncation procedure, are su¢ ciently small to apply the present theorem will be given in a subsequent paper. See [16]. The rest of this paper is organized as follows. In Section 2 we reduce the equation (6) and (2) to a set of integral equations for an evolution equation for which the coordinate y plays the role of time. In Section 3 we formulate the problem as a functional equation and prove the uniqueness of solutions. Existence of solutions is proved in Section 4.
2
Reduction to an evolution equation
Let u = (u; v) and F = (F1 ; F2 ). Then, the Navier-Stokes equations (6) are equivalent to ! = @y u + @x v ; @x !+ ! = q + ; @x u + @y v = 0 ;
(7) (8) (9)
q = @x (u!) + @y (v!) ; = @y F1 + @x F2 :
(10) (11)
where
The function ! is the vorticity of the ‡uid. Once the equations (7)-(9) are solved, the pressure p can be obtained by solving the equation p = r (F + u ru) in
+,
subject to the Neumann boundary condition
@y p(x; 1) = @y2 v(x; 1) : We now rewrite (7)-(9) as an evolution equation with y playing the role of time. Let q0 = u! ; q1 = v! ;
(12) (13)
Q0 = q0 + F2 ; Q1 = q1 F1 ;
(14) (15)
Q = @x Q0 + @y Q1 :
(16)
and let furthermore
and Then, instead of (7)-(8), we get the system of equations @y ! =
;
(17)
@y = @y u = @y v =
@x2 !
(18) (19) (20)
+ @x ! + Q ; ! + @x v ; @x u : 3
As we will see later on, the equations (17)-(20) have a special algebraic structure which permits to rewrite the solution as a sum of functions with di¤erent asymptotic behavior at in…nity, and this splitting will make the analysis simpler. With this in mind, we set v=!+ u = @x From (20) we then get that we get that
+ @y
= @y v =
@y u =
;
(21)
1
+
:
@x u =
! + @x v =
(22)
, and therefore that @y ! + @x ! + @x
=
:
, and from (19) (23)
Taking the derivative of (23) with respect to x gives, using (22), that @y + @y = @y @x u = @x ! + @x2 ! + @x2 , which, using (18), gives that ( @x2 ! + @x ! + Q) + @y = @x ! + @x2 ! + @x2 , and therefore we get that @y = @x2 + Q. We conclude that, instead of the system of equations (17)-(20), we can solve the system of equations @y ! =
;
@y = @y =
@x2 !
@y =
(24) ;
(25) (26)
+ @x Q0 + @y Q1 ;
(27)
@x2
+ @x ! + @x Q0 + @y Q1 ;
with v given by (21) and u given by (22). We now make a second change of variables which allows to express u in a more direct way. This will lead to additional signi…cant simpli…cations. Namely, we set = @x + Q1 ; = @x + Q1 :
(28) (29)
Substituting (28)-(29) into (24)-(27) we get that @y ! = @x + Q1 ; @x @y = @y = @x @y =
@x2 ! @x @x2
(30)
+ @x ! + @x Q0 ; Q1 ;
(31) (32)
+ @x Q0 :
(33)
All the terms on the right hand side containing only y-derivatives have disappeared and we can therefore instead of (30)-(33) solve the equations @y ! @y @y @y
= @x + Q1 ; = @x ! + ! + Q0 ; = @x Q1 ; = @x + Q0 ;
(34) (35) (36) (37)
with v given by (21) and with u given by u=
+
:
(38)
We now convert (34)-(37) into a system of ordinary di¤erential equations by taking the Fourier transform in the x-direction. De…nition 2 Let f^, g^ be complex valued functions de…ned almost everywhere on the inverse Fourier transform f = F 1 [f^] by Z 1 f (x; y) = F 1 [f^](x; y) = e ikx f^(k; y) dk ; 2 R 4
+.
Then, we de…ne
(39)
and ^ = f^ g^ by ^ (k; y) = (f^ g^)(k; y) =
Z
f^(k
k 0 ; y)^ g (k 0 ; y) dk 0 ;
R
whenever the integrals make sense.
We note that for functions f , g which are smooth and of compact support in f = F 1 [f^], and that f g = F 1 [f^ g^], where Z ^ f (k; y) = F[f ](k; y) = eikx f (x; y) dx ;
+
we have that
(40)
R
and similarly g^ = F[g]. With these de…nitions we formally have in Fourier space, instead of (34)-(37), the equations @y ! ^=
^1 ; ik ^ + Q
(41)
^0 ; @y ^ = (ik + 1)^ !+Q ^1 ; @y ^ = ik ^ Q
(43)
@y ^ =
^0 : ik ^ + Q
(44)
^ 0 = q^0 + F^2 ; Q ^ 1 = q^1 F^1 ; Q
(45)
(42)
From (14), (15) we get
(46)
from (12), (13) we get 1 (^ u ! ^) ; 2 1 (^ v ! ^) ; q^1 = 2
q^0 =
(47) (48)
and instead of (38) and (21) we have the equations ^+ ^ ; v^ = ! ^+ ^:
u ^=
(49) (50)
It is (41)-(50) that we solve in Section 3 in appropriate function spaces. We also show that the constructed solutions correspond via inverse Fourier transform to strong solutions of (2), (3), (6) with …nite Dirichlet integral. We now rewrite (41)-(50) as a system of integral equations (see Appendix A for a detailed derivation). From now we will use s, t 1 instead of y for the time variable, and , 0 for time di¤erences. We set p = k 2 ik ; (51) and de…ne, for k 2 R n f0g and
0, the functions Kn by,
1 e ; for n = 1; 2 ; 2 1 K3 (k; ) = e e ; 2 ik
Kn (k; ) =
(52) (53)
and the functions Gn by, 1 jkj e ; for n 2 1; 2 ; 2 1 jkj jkj G3 (k; ) = e e jkj : 2 ik
Gn (k; ) =
5
(54) (55)
We furthermore de…ne, for t 1, and n = 1, 2, 3, the intervals In by, I1 = [1; t], and In = [t; 1), otherwise. Using this notation, a representation of a classical solution to (41)-(44), which satis…es the boundary condition (3), in the sense that u ^(k; 1) = ^(k; 1) + ^(k; 1) = 0 and v^(k; 1) = ! ^ (k; 1) + ^(k; 1) = 0, is: X X X X ^= ^n;m ; ! ^= ! ^ n;m ; (56) m=0;1 n=1;2;3
^=
X
X
m=0;1 n=1;2;3
^n;m ;
^=
m=0;1 n=1;2;3
X
X
^n;m ;
(57)
m=0;1 n=1;2;3
with ^n;m (k; t) = Kn (k; t
1)
Z
gn;m (k; s
^ m (k; s) ds ; 1) Q
(58)
fn;m (k; s
^ m (k; s) ds ; 1) Q
(59)
kn;m (k; s
^ m (k; s) ds ; 1) Q
(60)
hn;m (k; s
^ m (k; s) ds ; 1) Q
(61)
In
! ^ n;m (k; t) = Kn (k; t
1)
^n;m (k; t) = Gn (k; t
1)
Z
Z
In
In
^n;m (k; t) = Gn (k; t
1)
Z
In
with Gn , Kn and In as de…ned above, with g1;0 (k; ) = g2;0 (k; ) =
ik ik ik ik
g3;0 (k; ) =
2
(jkj + )
e
e
+ 2 (jkj + ) e
(jkj + )
jkj
!
jkj (jkj + ) e ik
jkj
!
jkj (jkj + ) 2 e ik
jkj
)e
+ 2 (jkj + ) e
ik ik
g3;1 (k; ) = e
e
+
(jkj + ) e ik
2
(jkj + ) (1 + )e ik
ik
g3;m (k; ), f2;0 (k; ) =
jkj ik jkj (jkj + )2 jkj e + e ik jkj jkj jkj ik (jkj + )2 k2;0 (k; ) = ( + )e jkj ik jkj jkj ik jkj k3;0 (k; ) = e ; jkj jkj k1;1 (k; ) = ik jkj k2;1 (k; ) = ik
and h2;1 (k; ) =
(63)
!
;
(65)
;
(66) (67)
k1;0 (k; ) =
jkj
2
jkj
e
(jkj + ) e ik
jkj
2
(1
(jkj + ) )e ik
jkj
ik
(g2;0 (k; ) + 2e
), and
2(jkj + )e
;
(68)
2 (jkj + ) e
;
(69) (70)
(jkj + ) +2 e ik (jkj + ) +2 e ik
!
!
;
(71)
;
(72)
;
ik jkj k1;m (k; ), h3;m (k; ik jkj . jkj k2;1 (k; ) + 2e
and with h1;m (k; ) =
;
;
with f1;m (k; ) = ik g1;m (k; ), f3;m (k; ) = f2;1 (k; ) = ik g2;1 (k; ) 2e , with
k3;1 (k; ) = e
(62)
(64)
2
g2;1 (k; ) =
;
; 2
g1;1 (k; ) =
!
2
ik
(
e
jkj
(73) )=
ik jkj k3;m (k;
6
), h2;0 (k; ) =
ik jkj
k2;0 (k; ) + 2e
jkj
,
3
Functional framework
We start by de…ning adequate function spaces. Let , r ;r (k; t)
0 and k 2 R, and let
1 1 + (jkj tr )
=
:
(74)
Let furthermore (k; t) = ~ (k; t) =
;1 (k; t)
; ;2 (k; t) :
De…nition 3 Let R0 = R n f0g. We de…ne, for …xed of functions f 2 C(R0 [1; 1); C), for which the norm kf ; B
;p;q k
is …nite. Furthermore, we set W = B
= sup sup t 1 k2R0
; 27 ; 25
The following properties of the spaces B without mention: - if , - if
0
0, and p, p0 , q, q 0
> 1, p > 0 and q
0, then B
B
;p;q
;p;q
0, B
;p;q
to be the Banach space
jf (k; t)j (k; t) + t1q ~ (k; t)
1 tp
; 72 ; 52 ,
0, and p, q
and V = B
; 25 ;1
B
; 12 ;0
B
; 12 ;1 .
will be important below and will be routinely used \B
0 ;p0 ;q 0
Bminf
0;
;g;minfp0 ;pg;minfq 0 qg .
0, then
(k; t) 7!
1 tp
(k; t) 2 L2 (
+ ),
(k; t) 7!
1 ~ (k; t) 2 L2 ( tq
+)
:
Therefore, and because the Fourier transform is an isometry of L2 (R), we have that f = F L2 ( + ), whenever f^ 2 B ;p;q for some > 1, p > 0, q 0. - if > 1, p 0 and q 0, then f^ 2 B ;p;q is bounded by kf^; B Therefore, the function k 7! sup jf^( : ; t)j is in L1 (R).
;p;q k(1
1
[f^] 2
+ jkj)
, uniformly in t.
W ; (~ u1 ! ~ 2 ) ; 21 (~ v1 ! ~2) ;
(75)
t 1
Next, we rewrite the problem of solving (41)-(50) as a functional equation: Lemma 4 Let
> 1. Then, C :
V V ! ((~ !1 ; u ~1 ; v~1 ); (~ !2 ; u ~2 ; v~2 )) 7 !
1 2
de…nes a continuous bilinear map. Lemma 5 Let
> 1. Then, L :
W ! ~0; Q ~1) 7 ! (Q
V (~ !; u ~; v~) ;
(76)
de…nes a continuous linear map. Here, (~ !; u ~; v~) = (^ ! ; ^ + ^; ! ^ + ^), with (^ ! ; ^; ^; ^) given in terms of ^ ^ ~ ~ the integral equations (56)-(73), with (Q0 ; Q1 ) = (Q0 ; Q1 ). The maps C and L are studied in Section 4.1 and Section 4.2, respectively. Now let F = (F1 ; F2 ) 2 ~ = (F~1 ; F~2 ) = (F[F1 ]; F[F2 ]) be the Fourier transform of F. Note that (F~2 ; F~1 ) 2 W Cc1 ( + ), and let F for all > 1. 7
De…nition 6 Let
> 1. A triple (~ !; u ~; v~) is called an
-solution if:
(i) (~ !; u ~; v~) 2 V ; (ii) (~ !; u ~; v~) = L[C[(~ !; u ~; v~); (~ !; u ~; v~)] + (F~2 ; F~1 )] : With this de…nition at hand we can now give a precise formulation of Theorem 1: ~ = (F~1 ; F~2 ) be the Fourier Theorem 7 (Existence) Let > 1, F = (F1 ; F2 ) 2 Cc1 ( + ), and let F ~ ~ transform of F. If k(F2 ; F1 ); W k is su¢ ciently small, then there exists a unique -solution (~ !; u ~; v~) in V , with k(~ !; u ~; v~); V k C k(F~2 ; F~1 ); W k, for some constant C depending only on the choice of . Proof. Let " := k(F~2 ; F~1 ); W k. Since > 1, we have by Lemma 4 and Lemma 5 that the map N : V ! V , N [x] = L[C[x; x] + (F~2 ; F~1 )] is continuous. We now show that for " small enough there is a constant such that N is a contraction on the ball U = fx 2 V j kx; V k < g. Namely, let x 2 U . Then, by Lemma 4 there exists a constant C1 such that kC[x; x]; W k C1 ( )2 , and therefore kC[x; x] + (F~2 ; F~1 ); W k C1 ( )2 + " . Using now Lemma 5 it follows that there exists a constant C2 such that kN [x]; V k C2 (C1 ( )2 + " ). We set C = 2C2 . Now, we assume that "
1, and r, s 0 and let a, b be continuous functions from R0 satisfying the bounds (see (74) for the de…nition of ;r and ;s , respectively), ja(k; t)j jb(k; t)j
;r (k; t)
; (k; t) : ;s
Then, the convolution a b is a continuous function from R j(a b) (k; t)j uniformly in t
1 tr
const:
[1; 1) to C
[1; 1) to C and we have the bound
;s (k; t)
+
1 ts
;r (k; t)
;
(90)
1, k 2 R.
Proof. We only prove (90). Since the functions ;r and case k 0. Cutting the integral into two parts we have, j(a b) (k; t)j ;s (k=2; t)
const:
1 tr
Z
;s
k=2 ;r (k 1
0
0
; t) dk +
;r (k=2; t)
are even in k, it su¢ ces to consider the
Z
1
k 0 ; t) dk 0
;s (k
k=2
;s (k; t)
+
1 ts
;r (k; t)
;
and (90) follows. Corollary 10 Let, for i = 1; 2,
i
> 1, and pi , qi
0. Let fi 2 B
i ;pi ;qi
, and let
= minf 1 ; 2 g ; p = minfp1 + p2 + 1; p1 + q2 + 2; p2 + q1 + 2g ; q = minfq1 + q2 + 2; p1 + q2 + 1; p2 + q1 + 1g : Then f1 f2 2 B
;p;q
and there exists a constant C, depending only on kf1 f2 ; B
;p;q k
C kf1 ; B
1 ;p1 ;q1
k kf2 ; B
i,
such that
2 ;p2 ;q2
k
Proof. Using Proposition 9 we …nd that 1 1 1 1 (k; t) + q1 ~ 1 (k; t) (k; t) + q2 ~ 2 (k; t) tp1 1 t tp2 2 t const: const: minf 1 ; 2 g (k; t) + q1 +q2 +2 ~ minf 1 ; 2 g (k; t) p +p +1 1 2 t t const: const: const: const: + p1 +q2 +2 1 (k; t) + p1 +q2 +1 ~ 2 (k; t) + q1 +p2 +1 ~ 1 (k; t) + q1 +p2 +2 t t t t and the claim follows after regrouping of the terms involving and ~, respectively. Now let (~ !1 ; u ~1 ; v~1 ), (~ !2 ; u ~2 ; v~2 ) 2 V = B j 2 f1; 2g that u ~i ! ~ j 2 B ; 27 ; 25 , and that k~ ui ! ~j ; B
; 27 ; 52 k
B
; 52 ;1
const:k~ ui ; B
B
; 12 ;0
; 21 ;0 k
; 12 ;1 .
k~ !j ; B
2
(k; t) ;
Using Corollary 10 we …nd for i,
; 52 ;1 k
const:k(~ !i ; u ~i ; v~i ); V k k(~ !j ; u ~j ; v~j ); V k Similarly we …nd that v~i ! ~j 2 B k~ vi ! ~j ; B We conclude that
( 21
u ~1
! ~ 2 ; 21
; 27 ; 52 , ; 27 ; 25 k
and that const:k~ vi ; B
k~ !j ; B
; 52 ;1 k
const:k(~ !i ; u ~i ; v~i ); V k k(~ !j ; u ~j ; v~j ); V k v~1 ! ~2) 2 W = B
1 1 u ~1 ! ~2; v~1 ! ~ 2 ); W k 2 2 This completes the proof of Lemma 4. k(
; 12 ;1 k
; 72 ; 25
B
; 72 ; 25 ,
and that
const:k(~ !1 ; u ~1 ; v~1 ); V k k(~ !2 ; u ~2 ; v~2 ); V k
11
4.2 Let
Proof of Lemma 5 be as de…ned in (51), and de…ne =
by 1 2
Re( ) =
We have that j j = (k 2 + k 4 )1=4
q p 2 k 2 + k 4 + 2k 2 :
jkj1=2 + jkj
23=4 j j
and that jkj
j
Therefore, we have in particular that for
j
p
j j
2j
(91)
23=4 (1 + jkj) ; j:
(92) (93)
0 e
jkj
e
:
(94)
In what follows we prove Lemma 5 by providing bounds for the norms of ~, ! ~ , ~, and ~ in terms of ~ ~ the norms of Q0 and Q1 . We systematically use the notation introduced above, but, for simplicity, we set 1 1 (95) (k; s) = 7 (k; s) + 5 ~ (k; s) ; s2 s2 ~0; Q ~ 1 ); W k with C a constant independent of k and t. This constant may be di¤erent and kQk = C k(Q from instance to instance changing even within the same line. For ~ we have: Proposition 11 Let gi;j be as given in Section 2. Then we have the bounds j)ej
jg1;0 (k; )j
const:(1 + j
jg2;0 (k; )j
const:(1 + jkj)e
jg3;0 (k; )j
const:e
j
const:e const:ej
jg1;1 (k; )j jg2;1 (k; )j jg3;1 (k; )j
jkj
j
j
j
j2
minf1; j
2
g;
;
minf1; j
const:(1 + jkj)e e
j
(97) jg ;
(98)
for jkj 1 j for jkj > 1
jkj
(96)
;
;
(99) (100)
;
(101)
uniformly in 0 and k 2 R0 (and uniformly in k 2 R0 , jkj the case of (99)).
1 and k 2 R0 , jkj > 1, respectively, for
Proof. From (62) we get that jg1;0 (k; )j
const:(1 + j
j)ej
j
:
Expanding the exponential functions in (62) the …rst two terms cancel, so that 2
g1;0 (k; ) = (e +2
1
(jkj + ) e ik
and therefore, since for all z 2 C with Re(z) ez
(jkj + ) ik
)
jkj
e
1 + jkj
1+ ;
(102)
0 and N 2 N0 ,
PN
1 n n=0 n! z z N +1
12
const: ;
(103)
and for all z 2 C with Re(z) > 0
PN
1 n n=0 n! z z N +1
ez
const:eRe(z) ;
(104)
we …nd from (102) using (92) that jg1;0 (k; )j
j2
const:j
2
(1 + j
j)ej
j
:
This completes the proof of (96). The bounds (97) and (98) follow using (92), and (94). We now prove (99). Using (92) we …nd from (65) for k 1 that jg1;1 (k; )j For jkj that
const:(1 + j
j)ej
j
:
1 we use the fact that if we expand the exponential functions in (65) the …rst term cancels, so ! 2 (jkj + ) jkj (jkj + ) e jkj g1;1 (k; ) = (e 1) + e 1 2 1 : (105) ik ik ik
From (105) we …nd, using (92), (103), and (104) that jg1;1 (k; )j
j j2 j e jkj
const:
j
+ j j2
+
const:ej
j
:
To prove (100) we use that 2
(jkj + ) 2k 2 + 2 jkj j j = ; ik ik and the result again follows using (92), and (94). The bound (101) is trivial. 1+
As a consequence of Proposition 11 we have: ~0; Q ~ 1 ) 7! ~ de…nes a continuous linear map from W to B ; 3 ;0 . Proposition 12 Let > 1. Then, (Q 2 ~0; Q ~ 1 ) 7! ~i;j , with ~i;j as given in (56), de…nes continuous linear maps on W , with More precisely, (Q values ~i;0 2 B ; 25 ; 32 , i = 1; 2; 3, ~1;1 2 B ; 32 ;0 , ~2;1 2 B ; 52 ; 23 , and ~3;1 2 B ; 23 ; 12 . Proof. Using (96), Proposition 20, and Proposition 21 we …nd, with the notation (95), that Z t j~1;0 (k; t)j kQk e (t 1) (1 + j j)ej j(s 1) minf1; j j2 (s 1)2 g (k; s) ds 1 Z t 1 1 (t 1) (k; s) + 5 ~ (k; s) = kQk e (1 + j j)ej j(s 1) minf1; j j2 (s 1)2 g 7 2 2 s s 1 1 1 1 1 kQk 2 t 2 ~ (k; t) + 5 (k; s) + 3 ~ (k; t) ; t t2 t2 and therefore ~1;0 2 B
; 25 ; 32 .
Using (97) and Proposition 25 we …nd that Z 1 (t 1) j~2;0 (k; t)j kQke (1 + jkj)e jkj(s 1) (k; s) ds t
(t 1)
kQke and therefore ~2;0 2 B
; 52 ; 32 .
j~3;0 (k; t)j
e
jkj(t 1)
Using (98) with minf1; j kQk kQk
ik 1 5
t2
e
(t 1)
e
1
t
(k; t) +
5 2
jg
(t 1)
(k; t) +
3
t2
~ (k; t)
13
3
t2
~ (k; t)
;
j j and Proposition 22 we …nd that Z 1 e (s 1) j j (k; s) ds t
1
1
;
ds
and therefore ~3;0 2 B
; 52 ; 32 .
Using (99), Proposition 20 and Proposition 21, we …nd for jkj > 1 that
j~1;1 (k; t)j
kQk e kQk
and for jkj
(t 1)
Z
t
ej
j(s 1)
1
j
j (k; s) ds
1 1 ~ (k; t) + 7 t t2
(k; t) +
Z
1) (k; s) ds
1 5
t2
~ (k; t)
;
1 that kQk e
(t 1)
kQk e
(t 1)
+ kQk e
(t 1)
j~1;1 (k; t)j
t
ej
Z
t+1 2
1 Z t
t+1 2
kQk ~ (k; t) + and therefore ~1;1 2 B
; 32 ;0 .
ej
j(s 1)
ej
(s
(k; t) +
3
t2
1) (k; s) ds 1
j(s 1)
1
(t 1)
kQk e
; 52 ; 32 .
Z
1
s 1
(k; s) +
5 2
~ (k; t)
1
t2
1 3
s2
~ (k; s)
ds
;
(t 1)
e
jkj(s 1)
(1 + jkj) e
t
kQke
1
jkj(t 1)
(k; s) ds
(k; t) +
5
t2
1 3
t2
~ (k; t)
;
Finally, using (101) and Proposition 22 we …nd that
j~3;1 (k; t)j
kQk
e
ik 1
kQk and therefore ~3;1 2 B
(s
Using (100) and Proposition 25 we …nd that
j~2;1 (k; t)j
and therefore ~2;1 2 B
j(s 1)
1
t
(t 1)
Z
(t 1)
e
1
(s 1)
e
(k; s) ds
t
3 2
1
(k; t) +
1
t2
~ (k; t)
;
; 23 ; 12 .
For ! ~ we have: Proposition 13 Let fi;j be as given in Section 2. Then we have the bounds jf1;0 (k; )j
const:ej
jf2;0 (k; )j
)e
jf3;0 (k; )j
const:(jkj + jkj
const:e
jf1;1 (k; )j
const:(1 + j
j
jf2;1 (k; )j
const: (1 + jkj) e
jf3;1 (k; )j uniformly in
j
minfj 1=2
const:e
minf1; j j)e
jkj
minf1; j
j3
j; j
jkj
2
g;
(106)
;
(107)
j g;
(108)
2
j
minf1; j
j g;
;
(110)
jg ;
(111)
0 and k 2 R0 .
Proof. Since ik
const:
j j 1+j j
14
(109)
const: minf1; j
jg ;
(106) follows immediately from (96), (107) from (97), (108) from (98), (110) from (100), and (111) from (101). Finally, in order to prove (109), we note that 2
f1;1 (k; ) = e
(jkj + ) e ik
+
and therefore we …nd using (92) that jf1;1 (k; )j functions in (112) we see that
2
jkj (jkj + ) e ik
2
f1;1 (k; ) = (e
(jkj + ) ik
1) +
j)ej
const:(1 + j e
1
2
jkj
j
(112)
. Expanding the exponential
jkj (jkj + ) e ik
and therefore we …nd using (92), (103), and (104) that jf1;1 (k; )j
;
jkj
const:(1 + j
1
;
j)ej
j
j
j .
As a consequence of Proposition 11 we have: ~0; Q ~ 1 ) 7! ! Proposition 14 Let > 1. Then, (Q ~ de…nes a continuous linear map from W to B ; 52 ;1 . ~ ~ More precisely, (Q0 ; Q1 ) 7! ! ~ i;j , with ! ~ i;j as in (56), de…nes continuous linear maps on W , with values ! ~ 1;0 2 B ; 27 ; 52 , ! ~ 2;0 2 B ;3;2 , ! ~ 3;0 2 B ; 52 ; 23 , ! ~ 1;1 2 B ; 52 ;1 , and ! ~ i;1 2 B ; 52 ; 32 , i = 2; 3. Proof. Using (106), Proposition 20, and Proposition 21 we …nd that j~ !1;0 (k; t)j
(t 1)
kQk e
;3;2 .
j~ !3;0 (k; t)j
kQk e
(t 1)
kQk e
(t 1)
j~ !1;1 (k; t)j
; 52 ; 32 .
kQk
kQk e
; 25 ;1 .
j3 (s 1)2 g (k; s) ds !
j; j 1 5
t2
~ (k; s)
;
Z
1
t
1 t3
jkj(t 1)
e
j2 g
(t 1)
e
ik 1
1=2
(jkj + jkj
j
jkj(s 1)
(k; s) ds
1 ~ (k; t) t2
(k; t) +
;
j we get from (108) and Proposition 22 that Z 1 e (s 1) j j (k; s) ds
(t 1)
e
)e
t
(k; t) +
5
t2
1 3
t2
~ (k; t)
;
Using (109) and Proposition 20 we …nd that
kQk and therefore ! ~ 1;1 2 B
minfj
(k; s) +
Using that minf1; j
kQk and therefore ! ~ 3;0 2 B
j(s 1)
Using (107) and Proposition 25 we …nd that
j~ !2;0 (k; t)j
and therefore ! ~ 2;0 2 B
ej
1
1
; 27 ; 52 .
t
t2 1 ~ (k; t) + 7 t3 t2
kQk and therefore ! ~ 1;0 2 B
Z
(t 1)
Z
1
t
j)ej
(1 + j
1 1 ~ (k; t) + 5 t t2
(k; t) +
j(s 1)
1 3
t2
minf1; j
~ (k; t)
j(s
1)g (k; s) ds
;
Using (110) and Proposition 25 we …nd that
j~ !2;1 (k; t)j
kQk
1 e 2
kQk e
(t 1)
Z
t
(t 1)
e
1
(1 + jkj) e
jkj(t 1)
15
1 t
5 2
jkj(s 1)
(k; t) +
(k; s) ds 1 3
t2
~ (k; t)
;
and therefore ! ~ 2;1 2 B j~ !3;1 (k; t)j
; 52 ; 32 .
Finally, using (111) and Proposition 22 we …nd that
kQk kQk
and therefore ! ~ 3;1 2 B
e
ik 1
(t 1)
Z
(t 1)
e
1
(s 1)
e
t
(k; t) +
5
t2
1 3
t2
~ (k; t)
minf1; j
jg (k; s) ds
;
; 25 ; 32 .
For ~ we have: Proposition 15 Let ki;j be as given in Section 2. Then we have the bounds
uniformly in
3=2
jk1;0 (k; )j
const:(1 + jkj)ejkj minf1; jkj
jk2;0 (k; )j
const:(1 + jkj)e
jk3;0 (k; )j
e
jk1;1 (k; )j
const:(1 + jkj)e
jk2;1 (k; )j
const:(1 + jkj)e
jk3;1 (k; )j
e
jkj
jkj
g;
(113)
;
(114)
;
(115) jkj
jkj
2
1=2
jkj
minf1; (1 + jkj
1=2
) jkj
g;
(116)
;
(117)
;
(118)
0 and k 2 R0 . const:(1 + jkj)ejkj . Expanding the exponential functions
Proof. From (68) we get that jk1;0 (k; )j in (68) we …nd that 2
k1;0 (k; ) = ejkj
1
jkj
+
(jkj + ) ik
e
jkj
1 + jkj
2
jkj (jkj + ) e ik
1+
;
and therefore using (104) and (103) that jk1;0 (k; )j
2
const: jkj
2 jkj
e
3=2
const:(1 + jkj) jkj
2
+ jkj
2
2 jkj
e
(1 + jkj) + j
3
j
2
:
This completes the proof of (113). Using (92) and (94) we …nd (114), (115), (117) and (118). From (71) we get that jk1;1 (k; )j const:(1 + jkj)ejkj . Expanding the exponential functions in (71) we …nd that ! 2 jkj (jkj + ) (jkj + ) jkj jkj k1;1 (k; ) = e 1 e 1 +2 e 1 ; ik ik ik and therefore using (104) and (103) that jk1;1 (k; )j
const: jkj ejkj + jkj (1 + jkj) + (1 + jkj) j 1=2
const:(1 + jkj)(1 + jkj1=2 ) jkj
j
ejkj :
As a consequence of Proposition 15 we have: ~0; Q ~ 1 ) 7! ~ de…nes a continuous linear map from W to B ; 1 ; 3 . Proposition 16 Let > 1. Then, (Q 2 2 ~0; Q ~ 1 ) 7! ~i;j , with ~i;j as in (57), de…ne continuous linear maps on W , with values More precisely, (Q ~1;0 2 B 3 , ~1;1 2 B 1 3 , and ~i;j 2 B 5 3 , i = 2; 3, j = 0; 1. ;1; 2
;2;2
;2;2
16
Proof. Using (113), Proposition 23, and Proposition 24 we …nd that Z t 3=2 ~1;0 (k; t) kQk e jkj(t 1) (1 + jkj)ejkj(s 1) minf1; jkj (s
1)2 g (k; s) ds
1
kQk and therefore ~1;0 2 B
1
t3=2
t1=2
(k; t) +
1
t
(k; t) +
5 2
1
3
t2
~ (k; t)
;
;1; 23 .
Using (114) and Proposition 25 we …nd that Z 1 jkj(t 1) ~2;0 (k; t) (1 + jkj)e jkj(s 1) (k; s) ds kQk e t
kQke and therefore ~2;0 2 B
; 52 ; 32 .
~3;0 (k; t)
and therefore ~3;0 2 B
1
2jkj(t 1)
t
(k; t) +
5 2
1
3
t2
~ (k; t)
Using (115) and Proposition 25 we …nd that Z 1 jkj jkj(t 1) jkj(t 1) e jkj(s e kQk e ik t 1 1 kQk (k; t) + 3 ~ (k; t) ; 5 2 t t2
;
1)
(k; s) ds
Using (116), Proposition 23, and Proposition 24 we …nd that Z t 1=2 kQk e jkj(t 1) (1 + jkj)ejkj(s 1) minf1; (1 + jkj1=2 ) jkj (s 1)g (k; s) ds
~1;1 (k; t)
; 32 ; 12 .
1
kQk and therefore ~1;1 2 B
1
t1=2
(k; t) +
1
t
5 2
(k; t) +
1
3
t2
~ (k; t)
;
; 21 ; 32 .
Using (117) and Proposition 25 we …nd that Z 1 jkj(t 1) ~2;1 (k; t) kQk e (1 + jkj)e jkj(s 1) (k; s) ds t
kQke and therefore ~2;1 2 B
; 52 ; 32 .
~3;1 (k; t)
and therefore ~3;1 2 B
1
2jkj(t 1)
t
(k; t) +
5 2
1
3
t2
~ (k; t)
Using (118) and Proposition 25 we …nd that Z 1 jkj jkj(t 1) jkj(t 1) e e kQk e jkj(s ik t 1 1 (k; t) + 3 ~ (k; t) ; kQk 5 2 t t2
;
1)
(k; s) ds
; 52 ; 32 .
For ~ we have: Proposition 17 Let hi;j be as given in Section 2. Then we have the bounds
uniformly in
3=2
jh1;0 (k; )j
const:(1 + jkj)ejkj minf1; jkj
jh2;0 (k; )j
const:(1 + jkj)e
jh3;0 (k; )j
e
jh1;1 (k; )j
const:(1 + jkj)e
jh2;1 (k; )j
const:(1 + jkj)e
jh3;1 (k; )j
e
jkj
jkj
g;
(119)
;
(120)
;
(121) jkj
jkj
2
jkj
;
1=2
minf1; (1 + jkj ;
1=2
) jkj
g;
(122) (123) (124)
0 and k 2 R0 . 17
Proof. The bounds (119)-(124) immediately follow from (113)-(118) using the de…nitions. As a consequence of Proposition 17 we have: ~0; Q ~ 1 ) 7! ~ de…nes a continuous linear map from W to B ; 1 ; 3 . Proposition 18 Let > 1. Then, (Q 2 2 ~ ~ ~ ~ More precisely, (Q0 ; Q1 ) 7! i;j , with i;j as in (57), de…ne continuous linear maps on W , with values ~1;0 2 B 1 3 , and ~i;j 2 B 5 3 , i = 2; 3, j = 0; 1. 3 , ~1;1 2 B ;1; 2
;2;2
;2;2
Proof. The bounds (119)-(124) are identical to the bounds (113)-(118), and the proof is therefore the same as for Proposition 15.
A
Derivation of the integral equations
In order to derive the integral equations (56), (57) we note that the equations (41)-(44) are of the form ^1; Q ^0; Q ^1; Q ^ 0 ) and with z_ = Lz + q, with z = (^ ! ; ^; ^; ^), q = (Q L(k) =
L1 (k) 0 0 L2 (k)
;
(125)
;
(126)
where 0 ik + 1
L1 (k) =
ik 0
and where 0 ik
L2 (k) =
ik 0
:
(127)
Then, we have that L1 = S1 D1 S1 1 , where S1 = and where D1 is the diagonal matrix with entries S2 (k) =
1
1
ik
ik
;
, and furthermore that L2 = S2 D2 S2 1 , where
and 1
1
ik jkj
ik jkj
and where D2 is the diagonal matrix with entries jkj and
and that 1
S2 (k) =
;
1 2 1 2
ik 2jkj ik 2jkj
(129)
jkj. We have that
ik 2 ik 2
1 2 1 2
S1 1 =
(128)
; !
(130)
:
(131)
Let S(k) = and z = Sr. Then r_ = Dr + S
1
S1 (k) 0 0 S2 (k)
;
(132)
q with S
1
(k) =
S1 1 (k) 0 0 S2 1 (k)
18
:
(133)
Let r = (^ !+ ; ! ^ ; ^+ ; ^ ). Using the de…nitions we …nd that from (41)-(44), 1^ ik ^ @y ! ^+ = ! ^+ + Q Q0 ; 1 2 2 1^ ik ^ @y ! ^ = ! ^ + Q Q0 ; 1+ 2 2 1^ ik ^ @y ^+ = jkj ^+ Q1 + Q0 ; 2 2jkj 1^ ik ^ @y ^ = jkj ^ Q1 Q0 : 2 2jkj
(134) (135) (136) (137)
Note that, in component form, we have for z = Sr : ! ^=! ^+ + ! ^
;
(138)
( ! ^+ + ! ^ ); ik ^ = ^+ + ^ ; ^ = ik ( ^+ + ^ ) : jkj
(139)
^=
(140) (141)
^0; Q ^ 1 ), a classical representation of solutions to (134)–(137) is (we use from now on t instead For given (Q of the y for the “time variable”): Z ik ^ 1 1 (t s) ^ Q0 (k; s) ds ; (142) e Q1 (k; s) ! ^ + (k; t) = 2 t Z 1 t ^ 1 (k; s) + ik Q ^ 0 (k; s) ds ; ! ^ (k; t) = ! (k)e (t 1) + e (t s) Q (143) 2 1 Z 1 ^+ (k; t) = 1 ^ 1 (k; s) ik Q ^ 0 (k; s) ds ; (144) ejkj(t s) Q 2 t jkj Z 1 t jkj(t s) ^ ik ^ jkj(t 1) ^ (k; t) = (k)e Q1 (k; s) + Q0 (k; s) ds : (145) e 2 1 jkj The functions !
and
are determined by the boundary condition (3). At t = 1 we have Z 1 1 (1 s) ^ ik ^ ! ^ + (k; 1) = Q0 (k; s) ds ; e Q1 (k; s) 2 1
! ^ (k; 1) = ! (k) ; Z 1 ^+ (k; 1) = 1 ejkj(1 2 1 ^ (k; 1) = (k) :
s)
ik ^ Q0 (k; s) jkj
^ 1 (k; s) Q
ds ;
(146) (147) (148) (149)
Substituting (142)-(145) into (138)-(141) and the result into (38) and (21) we get, when evaluating at t = 1, 0 = ! (k) + (k) Qv (k) ; where Z 1 1 e Qv (k) = 2 1 Z 1 1 e 2 1
(1 s) (1 s) ik
and 0=
ik
ejkj(1
! (k)
s)
ejkj(1
jkj ik 19
(k)
^ 1 (k; s) ds Q s)
ik jkj
^ 0 (k; s) ds ; Q
Qu (k) ;
(150)
where Qu (k) =
Z 1 1 e (1 2 1 Z 1 1 e (1 2 1
Since 1
ik
jkj ik
jkj ik
= (jkj + )
;
(152)
Qv (k) Qu (k)
1 1
ik
(151)
1 1
ik
we …nd that ! (k) (k)
^ 0 (k; s) ds : Q
s)
= (jkj + )
jkj ik
^ 1 (k; s) ds Q
ik
1
1
ik
ejkj(1
s)
s) jkj
ejkj(1
s)
;
(153)
from which we get that Z 1 1 (jkj + )2 e 2 1 ik Z 1 1 (jkj + )2 e 2 1
! (k) =
jkj(jkj + ) jkj(1 e ik
(1 s)
2
(1 s)
2(jkj + )ejkj(1
s)
^ 1 (k; s) ds Q
s)
^ 0 (k; s) ds ; Q
(154)
and that Z 1 1 (jkj + )2 jkj(1 e (k) = 2 1 ik Z 1 1 (jkj + )2 jkj(1 e 2 1 jkj
(jkj + ) e ik
s)
2
s)
2(jkj + )e
^ 1 (k; s) ds Q
(1 s)
(1 s)
^ 0 (k; s) ds : Q
(155)
Substituting ! ~ given by (154) into (139) gives, after splitting the integral over [1; 1] into an integral over [1; t] and over [t; 1], the representation in (56) for ^. Similarly, substituting ! ~ given by (154) into (138) gives the representation in (56) for ! ^ , substituting ~ given by (155) into (141) gives the representation in (57) for ^, and substituting ~ given by (155) into (140) gives the representation in (57) for ^.
B
Basic bounds
B.1
Continuity of semi-groups
We have: 0
Proposition 19 Let
0
,
1 1 + jkj
0
uniformly in k 2 R and t Proof. For 1
t 1
1 + jkj
j
j
0
t
0
+ 1
0
0
e
jkj(t 1)
jkj
0, and let
0
t
0
1
0
,
0
0
,
0
const:
t
> 0. Then, we have the bound
1 1 t 0 1 + (jkj t2 )
const:
t
1. Similarly, for positive
with
0
0
0
+
1 1 t 0 1 + (jkj t)
0
;
0+ 0
0
0 and
0+ 0
> 0 we have
;
1.
2 and jkj 0 e
0
0 with (t 1)
e
uniformly in k 2 R and t the bound 1 1 + jkj
0
,
(t 1)
1 we have that j
j
0
t
1 t
0
const:
20
const:
1 1 t 0 1 + (jkj t2 )
0
0+ 0
;
and that 1 1 + jkj Next, for 1
t
0
e
jkj(t 1)
t
0
jkj
0
1
const:
t
1 1 t 0 1 + (jkj t)
const:
0
0+ 0
:
2 and jkj > 1 we have that 1 1 + jkj
0
(t 1)
e
1
const:
0
1 + jkj 1
const:
j
j
0
1 t
(t 1)
e
0
jkj
0
t
0
(j
j (t
0
1 + jkj 1 1 const: 0 2 t 1 + (jkj t )
j
j
0
0
1
const:
0
0
1))
0
1 + jkj
0+ 0
:
0+ 0
and similarly that 1 1 + jkj
jkj(t 1)
e
0
const: const:
1 0
1 + jkj 1
0
t
0
jkj
t jkj(t 1)
e
0
1
0
jkj
1 + jkj 1 1 const: 0 t 1 + (jkj t)
(jkj (t
0
0
0
1))
jkj
0
0
1
const:
1 + jkj
0
0+ 0
:
0+ 0
Finally, for t > 2 and k 2 R we have 1 + jkj t
2
0
0
0
+
const: 1 + jkj t
e
2
const: 1 + jkj t2 const: 1 + const: 1 +
j j
jkj
jkj
0
0
+
0
0
0
+
0
0
0
0
0+ 0
)
+
0
2(
j
j
e
0+
)
0
tj
1 2
1
t
e2
2(
j
tj
!
t
0
1 t
t
+
0
2(
j
(t 1)
j
j
tj tj
0
0
0
0
+ )
0
j
tj e
1 2
t
!
const: ;
and similarly that 1 + (jkj t)
0
0
+
0
const: 1 + (jkj t)
B.2
e
jkj(t 1)
0
0
+
0
e
Convolution with the semi-group e
(jkj t) 1 2
jkjt
0
t
1
0
t (jkj t)
0
const: :
t
In order to bound the integrals over the interval [1; t] we systematically split them into integrals over 1+t [1; 1+t 2 ] and integrals over [ 2 ; t] and bound the resulting terms separately. We have: 21
Proposition 20 Let
0, r
0 and (t 1)
e
Z
0 and
t+1 2
ej
+1
j(s 1)
1
uniformly in t
j
j
0. Then, (s
1)
;r (k; s)
s
ds
8 1 > const: ~ (k; t); if > + 1 > > > t > > > > < log(1 + t) const: ~ (k; t); if = + 1 > t > > > > > +1 > > : const: t ~ (k; t); if < + 1 t
(156)
1 and k 2 R.
Proof. We have that (t 1)
e
Z
t+1 2
ej
j(s 1)
1 j t 21
(t 1) j
e
e
t
const:
j
j
j
t
e
t
1)
1 2
;r (k; s)
s
;r (k; 1)
+1
1
(s
j
j
Z
t+1 2
ds
1) s 1 8 1, if > + 1 > > > > < log(1 + t), if = + 1 ;1 (k; 1) > > > > : +1 t , if < + 1
j
(s
From the last expression (156) follows using Proposition 19 Proposition 21 Let e uniformly in t Proof. If
0, r (t 1)
Z
0,
2 R, and
ej
j(s 1)
2 f0; 1g. Then,
t t+1 2
j
1 s
j
;r (k; s)
ds
const: t 1+
;r (k; t)
;
(157)
1 and k 2 R. = 0 we have that Z t e (t 1) ej t+1 2
and (157) follows, and if e
(t 1)
j(s 1)
1 s
const: t
;r (k; s) ds
;r (k; t)
Z
t
ds ; t+1 2
= 1 we have that Z
const: t
t t+1 2
ej
j(s 1)
;r (k; t)e
j
(t 1)
j
Z
1 s
;r (k; s)
ds
t t+1 2
ej
j(s 1)
j
j ds
const: t
;r (k; t)
;
and (157) follows. Using Hölder’s inequality the proposition can also be proved for intermediate values of , but this is not needed here. Next we have:
22
Proposition 22 Let
ik uniformly in t
ej
0, r
> 1, and Z 1 e ej j(t 1) t Z 1 e e (t 1)
j(t 1)
0,
2 f0; 1g. Then, (s 1)
j
(s 1)
j
t
1 s 1 j s j
;r (k; s)
ds
;r (k; s)
ds
const: t 1+ const: t 2+
;r (k; t)
;
(158)
;r (k; t)
;
(159)
1 and k 2 R.
Proof. We …rst prove (158). If = 0 we have that Z 1 1 j j(t 1) e (s 1) e ;r (k; s) ds s t and (158) follows, and if Z 1 e ej j(t 1)
1
1 ds ; s
t
= 1 we have that (s 1)
t
j
j
1 s
;r (k; s)
ej
j(t 1)
e
1 t 1 t
ds
and (158) follows. We now prove (159). For jkj ik
;r (k; t)
Z
j ;r (k; t)e
j(t 1)
;r (k; t)
e
j
j ds
1 we have that
(t 1)
j(t 1)
(s 1)
e
;
const:
j
jkj
j
ej
j(t 1)
j(t 1)
j
j(t
const:ej
(t 1)
1)
t;
and the bound on (159) now follows as in the the proof of (158). For jkj ej
1
t
const:ej
ik
Z
1 we have that
j(t 1)
;
and the bound on (159) now again follows as in the the proof of (158). The proposition can also be proved for intermediate values of , but this is not needed here.
B.3
Convolution with the semi-group e
jkjt
In order to bound the integrals over the interval [1; t] we systematically split them into integrals over 1+t [1; 1+t 2 ] and integrals over [ 2 ; t] and bound the resulting terms separately. We have: Proposition 23 Let
0, r e
0 and jkj(t 1)
0 and
Z
t+1 2
ejkj(s
1
uniformly in t
1 and k 2 R.
+1 1)
jkj
0. Then, (s
1) s
;r (k; s)
8 1 > (k; t); if > + 1 const: > > > t > > > > < log(1 + t) const: (k; t); if = + 1 > t > > > > > > > t +1 : (k; t); if < + 1 const: t
Proof. The proof is as for Proposition 20. Next we have: 23
ds
(160)
Proposition 24 Let
0, r e
uniformly in t
jkj(t 1)
0, Z
2 R, and
t t+1 2
ejkj(s
1)
jkj
2 f0; 1g. Then, 1 s
;r (k; s)
ds
const: t 1+
;r (k; t)
;
1 and k 2 R.
Proof. The proof is as for Proposition 21. Using Hölder’s inequality the proposition can also be proved for intermediate values of , but this is not needed here. Next we have: Proposition 25 Let
0, r
jkj jkj(t e ik uniformly in t
1)
0,
> 1, 2 [0; 1] Then, Z 1 1 jkj(t 1) e jkj(s 1) jkj e s Zt 1 1 e jkj(s 1) jkj e jkj(t 1) s t
;r (k; s)
ds
;r (k; s)
ds
const: t 1+ const: t 1+
;r (k; t)
;
(161)
;r (k; t)
;
(162)
1 and k 2 R.
Proof. For jkj < 1=t and 0
