daniel spirn and j. douglas wright- linear dispersive decay estimates for vortex sheets with surface...

8/3/2019 Daniel Spirn and J. Douglas Wright- Linear Dispersive Decay Estimates for Vortex Sheets with Surface Tension

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2 DANIEL SPIRN AND J. DOUGLAS WRIGHT

(43). We include a self-contained derivation of the governing equations for thissystem in Appendix 1.

Hou-Lowengrub-Shelley [15, 16] developed a highly accurate numerical schemeto study the evolution of z by using, as coordinates, the angle that the surfacemakes with respect to the horizontal, denoted (, t), and the jump in tangential

velocity across (t), denoted (, t). Their choice of coordinates was motivatedby a desire to reduce the stiffness of numerical simulations. While the curvature is a nonlinear function of z, it is proportional to . These coordinates arealso well-suited for rigorous analysis, see for example [2, 3, 5, 13].

Since the vortex sheet equations (43) are nonlinear at the highest order, well-posedness theory poses difficulties; however, local-in-time existence for waterwaves without surface tension was established by Wu [28]. When surface tensionis added, the system gains increased regularity and local well-posedness can beestablishedIguchi [17] does so via a graph method and Ambrose [2] uses amore Lagrangian approach. Both authors employ an energy method, and theirproofs take particular advantage of structure that is apparent at the linear

level. Building on the methods of Ambrose, Ambrose-Masmoudi [3] and Guo-Hallstrom-Spirn [13] improved the energy bounds to the form:

E CE(1 + E)k

for E suitably defined and k a fixed value. However, these bounds are farfrom implying small data global-in-time existence. Hence, establishing nonlinearstability of two dimensional vortex sheets remains elusive. In certain physicalsituations, such as when the heavier fluid is placed above the lighter fluid orwhen there is a large amount of shear, the flat equilibrium state is not stable.Linear instability in these cases has been studied by Hou-Lowengrub-Shelley[15, 16] and nonlinear instability was established by Guo-Hallstrom-Spirn in[13]. In higher dimensions there have been local-in-time results due to Ambrose-Masmoudi [3] on local well-posedness of irrotational vortex sheets and to Cheng-Coutand-Shkoller [8] on local well-posedness of vortex sheets with vorticity inthe interior of the fluids.

It is natural to ask whether there exist small-data global-in-time solutions.Typically global existence for nonlinear dispersive equations requires Strichartzestimates in a critical interpolation space and some form of local smoothing.This methodology is well understood for many equations which are closely re-lated to vortex sheets, such as the Korteweg-de Vries (KdV) and nonlinearSchrodinger (NLS) equationssee Bourgain [7], Constantin-Saut [10], Kato [18],Kenig-Ponce-Vega [19] and Tao [25]. Such estimates arise from a conserved en-ergy and a dispersive decay estimate on the linearized equation, so a startingpoint would be to rigorously establish dispersive estimates for the linearizedvortex sheet problem. In particular, one would like to show solutions of the


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DISPERSIVE ESTIMATES 3

linearized equations decay to zero at an algebraic rate as time evolves and thisis the main goal of this paper.

When the physical situation is one in which a heavy fluid, such as water,is below a more rarefied one, such as air, then there are heuristic argumentswhich indicate solutions disperse, see Whitham [27] Chapter 13 for example.

In more general situations, there is a complicated interplay between the effectsof surface tension and gravity, which tend to cause dispersion, and the shear,which tends to cause roll-up of the surface. Such interactions can be sortedout by means of techniques which arise in the study of oscillatory integrals.We consider dispersive estimates on the linearized equations describing vortexsheets with surface tension and gravity in two dimensions.

1.1. Linear dispersive estimates. The vortex sheet system is in equilibriumwhen the two fluids shear past one another along a perfectly flat interface, thatis, when (, t) = 0 and (, t) = . The equations can be found at (43). Thelinearization of the equations of motion about this equilibrium is given by

t =12H()

t =1

W +

2

2H() A

2 2Ag,

(1)

see also Siegel [23]. Here W is the Weber number and is inversely proportional tothe strength with which surface tension acts. The constant g is the acceleration

due to gravity and A :=2 11 + 2

where j is the density of fluid j. The Hilbert

transform H is given by the singular integral

Hf() :=1

P.V.R f()

d.The main result of this paper establishes linear dispersion for (1) in the case

when the background flow is quiescent, i.e. = 0. It is:

Theorem 1. Suppose that A > 0, g > 0, 1/W > 0 and = 0. If (, t) and(, t) solve (1) with initial data (, 0) = 0() and (, 0) = 0(), then forany 1 p :

(t)L + (t)L Ct1/3

0L1B2+

1p

p,1

+ 0L1B

12+1p

p,1

.

for t 1. (Here C > 0 is a nonessential constant.)Remark 1. Though Theorem 1 is specialized to = 0, we expect the same resultfor in a range determined by the strength of gravity and surface tension. Seethe discussion below.


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Remark 2. That the inital data is forced to lie in the intersection ofL1 and someBesov space may seem somewhat unnatural. We are free to choose p however.The most fitting choice is p = 1 since in this case Bs1,1 L1 and thus we onlyrequire 0 B31,1 and 0 B3/21,1 . Similarly if we take p = 2 and note thatHs

Bs2,1 then we need 0

L1

H5/2 and 0

L1

H1.

1.2. Methodology. The first step towards proving linear dispersive estimatesis to compute the explicit solution of equation (1) by means of the Fouriertransform. It is

(, t) = eic1t ic1 sin(()t)()

+ cos(()t)

0() + || sin(()t)()0()

(, t) = eic1t c21 2()||() sin(()t)0() +

ic1 sin(()t)()

+ cos(()t)

0()(2)

where c1 :=

A/4 and

(3) 2() :=1

2W||3

2

4

1 A

2

4

||2 + Ag || .

By examining (), one sees how the interaction of gravity, surface tensionand shear affects growth of solutions (see [1], [23]). For example, in the absenceof surface tension one has 1/W = 0. This implies that 2() < 0 for sufficientlylarge and thus ei()t grows extremely fast for large frequencies. It is this factthat is responsible for the Kelvin-Helmholtz instability and also the ill-posednessof the full nonlinear problem (see [12]). If solutions of (1) are to decay, it isnecessary that 2() > 0 for all . This occurs if and only if the inequality

(4) 18

W41 A242 < 4Ag

is met. Equivalently, this tells us that gravitation suppresses low frequencymodes in much the same way that surface tension suppresses those at highfrequencies. This is why both effects will be needed to induce dispersion.

In order to simplify the appearance of solution operator in (2) and the dis-persion relation () we make the following change of variables:

= (2W Ag)1/2, t = 21/4W1/4(Ag)3/4t, =

W

8Ag

1/4.

Doing so, setting = 0 and dropping the prime notation results in the equa-tion

t = Ht =

2 .

(5)


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The solution of this equation is

(, t) = cos(()t)0() + || sin(()t)()0()

(, t) = ()sin(()t)

|| 0() + cos(()t)0().

(6)

with

() =

|| + ||3.If we define the operators

S1(t)f :=F1

ei()tf()S2(t)f :=F

1

ei()t||

()f()

S3(t)f :=F1

ei()t

()

|

|f()

.

(7)

then Theorem 1 is an immediate consequence of

Theorem 2. There exists t0 0 such that for all t t0S1(t)fL

C

t1/3f

L1B12+1p

p,1

S2(t)fL C

t1/3fL1

and

S3(t)fL C

t1/3f

L1B2+1p

p,1

.

(Here p is any number in [1, ] and C is a nonessential constant.)Remark 3. The rate of decay of t1/3 is the (formally) optimal rate of decay forour system. Notice that for any situation in which (4) is met, () as 0+ and also as +. This implies the existence of a point at which() = 0. Thus this problem is more akin to the Airy equation (where thedispersion relation is 3 and has a decay rate of t1/3) than the Schrodingerequation (where the dispersion relation is 2 and has a decay rate of t1/2).

We divide our proof into three separate sections that handle S1 to S3 inorder. Our primary method is to use oscillatory integral estimates, such asthose of Constantin [9] or Kenig-Ponce-Vega [19]. Such tools are called Vander Corput estimates, see for example Stein [24], and they control integrals ofhighly oscillatory functions by the size of high derivatives of the phase function.In our case the phase function has a more unusual structure than those foundin [6, 19], therefore we need to develop slightly more general methods.


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The oscillatory integrals that we need to control are generally of the form

(8)

ba

eih(,)

g()d.

Here h(, ) is the phase and g is an additional factor. This is an example of

an oscillatory integral of the second kind and the goal will be to estimateit independently of the parameter . The theory for controlling such integralsis well-developed, but the methods typically require 1/g() to have compactsupport (see [14], [24], [26]). This is not the case here and in our applicationthere is the further complication that the function 1/g() may diverge to nearthe origin or alternately as || . As such, it is extremely important thatwe keep track of the interplay between the growth of 1/g() and the delicatecancellations due to the complex exponential. We prove a generalization of theVan der Corput estimate (Lemma 2 below) which allows us to control (8) by thesize of gh (the modified phase) and its derivatives. The proof of this lemmacan be found in Appendix 2.

Unlike more conventional phase functions, the Van der Corput lemmas are notsufficient to finish the oscillatory integral estimates. In general good dispersiveestimates are related to large curvature in the phase function, see [19, 24] etc.Unfortunately the curvature of the modified phase for S1 and S3 goes to zerofor large || and so we truncate the domain of integration so that we excludethe high frequencies where our decay estimates start to fail. This is a strategysimilar to one employed in [22] to prove dispersive estimates for wave equationsand is the reason why the Besov spaces appear in Theorem 1.

We briefly summarize the rest of the paper. Section 2 describes the Vander Corput lemmas that are used throughout the proof of the theorem alonga lemma for controlling the high frequency pieces. Sections 4-6 handle the L

estimates on S1 through S3. Appendix 1 has a self-contained derivation of theirrotational vortex sheet equation with surface tension. Finally, Appendix 2contains the proofs of the lemmas in Section 2.

1.3. Notations and Definitions. In this document we define the Fouriertransform of a function f() by

f() := F[f]() := 12

R

eif()d.

The Sobolev space of order s is denoted by Hs and we use as its norm f2Hs :=R(1 + 2)s |f()|2 d. We denote the inverse Fourier tranform of a function g()by g(). The convolution of f() and g() is

f g() :=R

f( )g()d.


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b. Additionally, if min[a,b] {|gh|} > 0 and (gh) = 0 at a finite number ofpoints in [a, b] we have

(11)

ba

eih()

g()d

C1

min[a,b]

|gh|1

.

The proof of this lemma is in Appendix 2.Notice that we estimate the oscillatory integrals above on truncated frequencydomains. The following lemma will be used to control the operators outside ofthese regions:

Lemma 3. Fix s 0, s 0, > 0 and 1 p . Suppose that () Cand |()| C||s. Then, for all t 1,

||t

()f() CtsfBs+s+1pp,1

.

The proof of this lemma is also in Appendix 2.

3. Formal stationary phase estimates

In this section we compute at a formal level the expected dispersive decayrate for the operator S1(t) when (4) is met in different physical circumstances.

3.1. Purely gravity waves: g = 1, 1/W = 0, = 0, A = 1. In this case,

notice that that () = ||1/2 and thus:

S1(t)f() =12

R

ei(+||1/2t)f()d.

In particular we will have to estimate the oscillatory integrals of the form:

ba

ei(+||1/2t)d = b

a

eit(+||1/2)d

where = /t. Notice that

2

2

+ ||1/2

= 14

||3/2

and is independent of . Importantly, this function is bounded away from zeroon any bounded interval [a, b]. Thus we have from Lemma 1 (with k = 2) that

b

a

ei(+||1/2t)d

C

min[a,b]

t ||3/2

1/2

C(b)t1/2.

Thus we expect that in the absence of surface tension, that the amplitude ofthe solution should decay like t1/2. Note that C(b) = O(b3/4) and thus thesize of the constant is large for high wave numbersthis can be handled via thetruncation methods discussed earlier.


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3.2. Purely capillary waves: g = 0, 1/W = 1, = 0, A = 1. Similarly, ifwe consider waves in which only surface tension acts, we must estimateb

a

eit(+||3/2)d

We have2

2

+ ||3/2 = 34

||1/2

and is independent of . As before, this function is bounded away from zero onany bounded interval [a, b] with a > 0. And so we have by the same reasoning:b

a

ei(+||1/2t)d

C(b)t1/2.(The constant C(b) diverges as b , but once again this is a complicationcan be handled by truncation.)

3.3. General case: g > 0, 1/W > 0,

0 and with (4) met. In this situa-

tion, the function () is more complicated than in the previous two situations.In particular, notice that when is close to 0 that

() c1 ||1/2and that as ||

() c2 ||1/2where both constants c1 and c2 are positive. Since

() blows up as 0and as , there is clearly a point 0 < slow < at which () achievesa minimum value and at which (slow) = 0. Thus it is impossible to applyLemma (1) with k = 2. It also happens that (slow) = 0 (at least for 0)and so Lemma (1) can used with k = 3 to achieve an overall time decay of t1/3.This is a surprising result: competition between surface tension and gravitationresults in dispersive decay which is slower than in the cases in which solely oneof these forces is present. This is related to the fact that in experimental studiesof small amplitude water waves, the presence of surface tension causes there tobe a non-zero lower bound on wave speedthis slowest speed is precisely(slow). The remainder of this paper is dedicated to rigorously justifying thedispersive decay in the situation with = 0.

4. Estimates for S1(t)

In this section we will control the operator S1. We have

S1(t)f() = 12R

eiei()tf()d=

12

R

eit(,)f()d.


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where (, ) := + () and = /t. At times we will suppress the depen-dence of on and simply write (). Similarly, (k) means k . Our goal isto estimate S1f independently of .

We break the integral into a low frequency and a high frequency piece:

|S1(t)f()| =1

2 t2/3

t2/3 +||t2/3 eit(,)f()d Ct2/3t2/3

eit(,)d

L

fL1 + C||t2/3

f() dWe apply Lemma 3 with s = 1/2, s = 0 and = 2/3 to control the highfrequency piece:

||t2/3

f() d Ct1/3 fB12+1p

p,1

.

This estimate along with the following proposition prove the first estimate in

Theorem 2.Proposition 1. Let(, ) = + () with () as above. Then

(12)

t2/3t2/3

eit(,)d

Ct1/3where C is independent of .

Proof. Notice that (k) is independent of if k 2. We compute

() = () =34 + 62 1

43()

and

() = () = sgn()36 154 + 152 + 3

85().

Notice that () = 0 at

0 :=

4

3 1

and that () = 0 only at = 1. Let us divide the interval over which theintegral in (12) is evaluated into the union of the following sets:

I0 := [

, ]

I1 := [0 , 0 + ] [0 , 0 + ] andI2 := [t2/3, 0 ] [0 + , ] [, 0 ] [0 + , t2/3].

Here and are small, positive constants, as yet undetermined.


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We will controlIj

eit()d for each j = 0, 1, 2. Trivially we have

I0

eit()d

2.

The zeros of lie in I1 and by chosing sufficiently small we can guaranteethat = 0 within this set. Therefore there is a constant C > 0 (whichdepends only on ) such that

minI1

|()| C.

Thus Lemma 2 implies I1

eit()d

C3C1/3 t1/3.In I2 the second derivative of is nonzero. So, by Lemma 2 we have

I2

eit()d C2t1/2minI2 ||1/2 .The minimum of || in I2 occurs at an endpoint of one of the intervals in I2or at 1, the zeros of . Let

m1 := min {|(1)|, |(0 )|, |(0 )|} .Note that m1 > 0 and depends only on .

From the expression for , we see that there are positive constants c1 andc0 so that |()| c13/2 and |(t2/3)| c0(t2/3)1/2 = c0t1/3. For sufficiently small and t sufficiently large we can conclude that c1

3/2

m1

c0t1/3 and sominI2

|| c0t1/3.

Therefore I2

eit()d

C2t1/2 c0t1/31/2 = C2c1/20 t1/3.All together, we have

t2/3

t2/3eit()d 2 + Cdt

1/3

where Cd := max

C2c1/20 , C3C

1/3

and is an arbitrary positive number.

This immediately implies (24) and we are done.


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5. Estimates for S2(t)

We have

S2(t)f() =12

R

eit(,)||

()

f()d

with (, ) as in the previous section. The convolution estimate shows that

|S2(t)f()| C

R

eit(,)||

()d

L

fL1 .

The primary result of this section (which implies the second estimate in Theorem2) is

Proposition 2. We have R

eit(,)||

()d

Ct1/3for all R and t t0.Proof. Notice that (, ) = (, ) and so

R

eit(,)||

()d

= R

eit(,)||

()d

.Thus we need only concern ourselves with 0.

In order to establish Proposition 2, we need to control the integral over R bymeans of Lemma 2. To do so we need estimates on the modified phase function

(, ) :=()

|| ( + ())

and its derivatives with respect . These are

= || = || + || =

||

=

||

+

||

=

||

=

||

+

||

Rewriting the modified phase function from above, we can say that

() =

()

||

( + ())

() = || + (32 1)(2 1) () =

||

+4(1 + 2)

(3 + 62 4)

.

(13)


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These imply

If K+() := () + ||

2then () 1/2.

If K() := () ||

2then () 1/2.

= K0() := ()

then () = 0.

Likewise, (, ) = 0 if and only if

= L() :=() (32 1)

(2 1)and (, ) = 0 if and only if

= M() :=4() (2 + 1) (3 + 62 4) .

In Figure 1 we plot the functions K+, K, K0, L and M when 0 and 0.Observe that and are zero simultaneously at one (and only one) point whichwe denote (, ). Note that 1/2 1/5.

Now we define the following sets which cover the upper half of the plane:

0 := {(, )| 0 and 0}1 := {(, )| 0 and 0 and / (K(), K+())}

2 := {(, )| 1 and [K(), K+()]}3 := {(, )| 1 1/2 and [K(), K+()]}

4 := {(, )| 1/2 1/5 and [K(), K+()]}and

5 := {(, )| 1/5 0 and [K(), K+()]} .We now show that in each of these sets, at least one of ||, || or | | is boundedaway from zero.

In 0,

(, )

()()

12

+3

2

3.

(14)


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!2 !1.8 !1.6 !1.4 !1.2 !1 !0.8 !0.6 !0.4 !0.2 00

0.5

1

1.5

2

2.5

3

3.5

4

!

"

#5

#4

#3

#2

#1

"=L(!)$!

= 0

"=K0(!)

$ = 0

"=M(!)$!!

= 0

"=K!

(!)

$ = !1/2

"=K+(!)

$ = 1/2

#1

(!*,"

*)

Figure 1. The regions j for S2.

By the definition of K

() we have

(15) |(, )| 1/2 if (, ) 1.There are no zeros of in the set 3 (since the graph of L does not pass

through this set). This set is compact and is continuous for all = 0. Thuswe can conclude that there is a constant c3 > 0 such that

(16) |(, )| c3 if (, ) 3.In exactly the same fashion, we see that there are no zeros of in 4 and thusthere is a constant c4 > 0 such that

(17) |(, )| c4 if (, ) 4.First we note that in the set 2 and 5, = 0. However, these sets are not

compact and thus we cannot immediately conclude that || is bounded awayfrom zero in these sets. Nevertheless we claim that

(18) |(, )| 1/8 if (, ) 2


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16/30


Since Ij() j, we can use the lower bounds in (14) to (19) and Lemma 2 toconclude:

5j=0

Ij()

eit(+()) ||()

d

Ct1 + t1/2 + t1/3 Ct1/3

if t is sufficiently large.

6. Estimates for S3(t).

In this section we will control the operator S3. We have

S3(t)f() =12

R

eit(,)()

||f()d

where (, ) is as in the previous sections.We divide the integral into two pieces as was done for S1(t):

|S3(t)f()| = 12

t2/9t2/9 +||t2/9 eit(,) ()||f()d Ct2/9t2/9

eit(,)()

|| dL

fL1 + C||t2/9

()||f() d.

For || 1, ()/|| C||1/2, so we apply Lemma 3 with s = 3/2, s = 1/2and = 2/9 to control the high frequency piece:

||t2/9 ()

|

|f()

d Ct1/3 f

B2+1pp,1

.

This estimate along with the following proposition proves the final estimate inTheorem 2.

Proposition 3. Let(, ) = + () with () as above. Then

(24)

t2/9t2/9

eit(,)()

|| d Ct1/3

where C is independent of .

Proof. We need only concern ourselves with 0 for the same reasons as inthe previous section. We need to control the integral over [t

2/9

, t2/9

] by meansof Lemma 2 and thus we need estimates on the modified phase function

(, ) :=||

()( + ())


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and its derivatives with respect . Explicitly, and its first two derivatives are

=

||

=||

+||

= ||

= ||

+

||

= ||

=

||

+

||

.

We can rewrite these as

() =

( + )

() = ||

+

4

1 4

() = ||

+ 8sgn()2(32 1)

(1 + 102 34) .(25)

This allows us the make the following observations about (, ).

If K+() := () + ()

4||then () 1/4.

If K() := () ()

4||then ()

1/4.

= K0() := ()

then () = 0.

Similarly (, ) = 0 if and only if

= L() := 4()1 4

and (, ) = 0 if and only if

= M() :=

8sgn()2 (32 1)()(1 + 102 34)

.

In Figure 2 we plot the functions K+, K, K0, L and M when 0 and 0.Observe that and are zero simultaneously at one (and only one) point whichwe denote (, ). Note that 1/2 1/5.


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!2 !1.8 !1.6 !1.4 !1.2 !1 !0.8 !0.6 !0.4 !0.2 00

0.5

1

1.5

2

2.5

3

3.5

4

!

"

#3

#4

#1

#5

#1

"=K+(!)

$ = 1/4

"=K!

(!)

$ = !1/4 "=K0(!)

$ = 0

"=M(!)$!!

= 0

"=L(!)$!

= 0

(!*,"

*)

#2

Figure 2. The regions j for S3.

Now we define the following sets which cover the region of the plane where 0 and || t2/9:

0 :=

(, )|0 t2/9 and 01 :=

(, )| t2/9 0 and 0 and / (K(), K+())

2 :=

(, )| t2/9 1 and [K(), K+()]

3 := {(, )| 1 1/2 and [K(), K+()]}4 := {(, )| 1/2 1/5 and [K(), K+()]}

and5 :=

{(, )

| 1/5

0 and

[K

(), K+()]

}.

In the sets 0, 1, 2, 3, 4 at least one of ||, || or | | is boundedaway from zero. In the set 5, it is true that || is bounded away from zero.Unfortunately, for large values of this set becomes very close to the = 0axis, where ||/() is very small. This is problematic when applying Lemma


19/30


2. We discuss how to deal with 5 in a moment. First we prove lower boundsin the other sets.

In 0,

(, ) ()

()

=32 + 1

2 (2 + 1)

1/2.

(26)

By the definition of K() we have

(27) |(, )| 1/4 if (, ) 1.There are no zeros of in the set 3 (since the graph of L does not pass

through this set). This set is compact and is continuous for all = 0. Thuswe can conclude that there is a constant c3 > 0 such that

(28) |(, )| c3 if (, ) 3.In exactly the same fashion, we see that there are no zeros of in 4 and thusthere is a constant c4 > 0 such that

(29) |(, )| c4 if (, ) 4.Note that in the set 2, = 0. This set has a boundary which depends on t.

Thus we cannot conclude that || is bounded away from zero in a way whichis independent of t. Still, we claim that

(30) |(, )| Ct2/9 if (, ) 2

First we cover the case in 2. In this situation, notice that

=

||()

=

(1 2)23()

0

for all (, ) 2. Thus , for fixed a , decreases with . This in turn impliesthat the minimum of must occur on the bottom boundary of 2that is tosay, on = K(). A direct calculation shows

(, K()) = 54 + 122 1

8(1 + 2)2(31)

The minimum value of this function occurs at the endpoint =

t2/9, and thus

there exists a constant C such that

(, K()) Ct2/9(32)which concludes the validation of (30).


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Now we start estimating the oscillatory integral in Proposition 3. Fix 0.If we let

Ij() := {|(, ) j}for j = 0,..., 5, then

R

eit(+())

||() d

5j=0Ij() e

it(+())

||() d .

Noting that ||/() is bounded away from zero in the sets 3 and 4, we applyLemma 2 and conclude using the bounds (26), (27), (28) and (29) above that

I0()I1()I3()I4()

eit(+()) ||()

d

Ct1 + t1/2 + t1/3 Ct1/3.In the set 2, the minimum of||/() occurs at when = t2/9. Specifically,

we have||

() Ct1/9.

Therefore Lemma 2 impliesI2()

eit(+()) ||()

d

Ct1/2t1/9t1/18 = Ct1/3.Finally we estimate the integral in I5(). First of all, there exists 1 such

that if < 1 then I5() is empty, and there is nothing to estimate. Second ofall, we claim that if 1 then(33) min

I5()

||()

C

and

(34) minI5()

|(, )| C2

.

If so, then Lemma 2 impliesI5()

eit(+()) ||()

d

Ct1/21/2 Ct1/2and we would be finished with Proposition 3. So we now check the claim.

Denote the left and right hand endpoints of I5() by 1() and 2() respec-tively. An examination of the functional forms ofK() shows that there existsC > 1 such that

1

C2 |j()| C

2

for j = 1, 2. It is clear that the minimum value of ||/() will occur on theright hand endpoint 2(), and this together with the estimates on 2() confirm(33). On the other hand, an examination of the shows that for 0 1/5


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this function is negative and decreasing for any . Thus the minimum of || inI5() will occur at 1(). The explicit formula for implies (34).

7. Strichartz estimates

Given the decay estimates and some natural L2 bounds, we can get the fol-lowing class of Strichartz estimates

Theorem 3. For any R we haveSjgL8t(B8,2) CgH+j

where j = { 916 , 916 , 2716}, respectively.We sketch the proof of this result.

Proof. Let {

n} be a partition of unity subordinate to the regions

U0 =

{|

| [0, 1/2]

}Un =

{|

| [2n2, 2n1]

}for n 1. Therefore, the support of n is in Un1 Un Un+1 if n 1 and inU0 U1 if n = 0. Let n() be the characteristic function of Un1 Un Un+1if n 1 and the characteristic function of U0 U1 otherwise. Let I0 := [0, 1/2]and In = [2

n2, 2n] if n 1.In order to prove the Strichartz estimates for the three operators simultane-

ously, let j(s) = {1,s3+ss

, ss3+s

} and the solution operators are defined asSjf = F1

ei(||)t 1j(||) f

. Let fn = F1 [nnf] then by a Paley-Littlewooddecomposition

Sj(t)fnL2 njnfL2 Cj(2n) fnL2and likewise from Theorem 2 (choosing p = 1)

Sj(t)fnL1 Ct1/32sjn fnL1where sj = {32 , 0, 3}. From Riesz-Thorin Interpolation Theorem we find thatfor 1 r 2 and 1r + 1q = 1 then

(35) SjfnLq Ct1

3(2

r1)2jn fnLr ,

where j = {3r 32 , 1r 1, 5r 2}.We use duality to complete the Strichartz estimates. In particular for a testfunction Sj(t)g, L2(R2+1) 1/2j g

L2

t

eit1/2j dt

L2


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We estimate the second term on the right hand side. In particulart

eit1/2j dt

L2 C

n=0

t

eit1/2j nndt2

L2

and each wavelet is bounded via our dispersive estimate (35).t

eit1/2j nndt2

L2

=

t

s

n (t), F1

ei(st)1j nn(s)L2

dsdt

t

s

2jn

|t s| 13 ( 2r1)n (t)Lr n (s)Lr dsdt

Therefore,

t eit

1/2j dt

2

L2

Ct

s

|t s| 13( 2r1)n=0

2jn

2 n (t) 2jn

2 n (s) dsdt

Ct

s

|t s| 13 ( 2r1) (t)Bj/2

r,2

(s)Bj/2

r,2

C2Lp(B

j/2

r,2 )

for p = 22 1

3( 2r1) =

6r7r2 by the classical Hardy-Sobolev-Littlewood inequality.

Choosing p r then r = 87

and

t

eit1/2j dt

L2

CL8/7(B

j8/7,2

)

where j = { 916 , 116 , 1916}, respectively. In particularSj(t)g, L2(R2+1) CgHj L8/7(Bj8/7,2

)

with j = {0, 12 , 12}, so by dualitySjgL8(Bj

8,2 ) CgHj .

And if we convolve our initial data g with (1

)/2 for some real number = j + then

SjgL8(B8,2)

CgH+jwhere j = { 916 , 916 , 2716}.


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8. Appendix 1: Derivation & Linearization of Equations

We start with a two-layer inviscid, incompressible, irrotational fluid in with + = R2, = and R2\+ = , where is a smooth,open curve parametrized by z(, t). Inside each domain we have

tu + (u ) u = 1p geydiv u = 0

(36)

where g is the gravitational constant. Since the fluids are irrotational, we findpotential functions and the Bernoulli equations

(37) t +1

2||2 + p

+ g y = 0.

inside . Since = 0 in then there is a double-layer potential and thePlemej Formulae for the velocity of the fluid on either side of the interface

(38) u[z()] = P V K2[z() z()] |z| [z()]d 12[z()]t[z()].where K2[x] =

x

|x|2 , see for example [20]. Here z() parametrizes . Set

V = 12 (u+ + u) to be the averaged velocity; hence12 (+ + ) = V.

We now have two equations

+ + = 2V and + = thence

+

= V +1

2t and

= V

1

2t.

This implies

(39)1

2

|+|2 + ||2 = |V|2 + 14

2.

We now consider the equation of motion for both + + and + . Wehave by the Bernoulli equations:

t (+ + ) +1

2

|+| + ||2+ 2g y = G = p++

p

t (+

) +1

2 |+| ||2

= H = p+

++

p

.

We solve for p+ and p and get

p+ = +2

(H+ G) and p =2

(H G) .


24/30


25/30


tz = V+Rt+Rt. But since tz = (ts)ei+itse

i = (ts)t+(t)sn,then t t = 0 and t n = imply

st = n tz = n V + Rts = t

tz = t

V + R.

Finally, we note that curvature is simple in the SSD coordinate system: n =t = ie

i = n, and so = . We get our system

st = n V + Rst =

1

W n V R [V t]

A

2t [st V] + 2V V + 2

+ 2g sin()

ts = t V + R.

(43)

There are steady states when = 0, = , R = R. Choosing a special Rsuch that R = t V then ts = 0, which simplifies the calculations. Inparticular we set R() = (t V) () + n Vd. In [15, 2], R is chosenso that the arclength remains constant in time, i.e. R = T (V t) t withT a prescribed function, see for example [2]. Note there is no gravity in theequations of motion considered in [2, 3]. It is straightforward to linearize (43)about the steady solution.

We note that the equations of motion for irrotational vortex sheets with nosurface tension was written down by Baker-Meiron-Orszag [4].

9. Appendix 2: Assorted Proofs

Proof. (Lemma 2)First we prove b. We haveb

a

eih()

g()d

= ba

1

igh()

eih()

d

1

igh()eih()|ba

+

ba

1

igh()

eih()d

2min[a,b] |gh|1 + ba 1gh()

d(44)

By assumption, (gh) is zero at a finite number of points in (a, b). Denote thesepoints by 1 < 2 < ... < n. Also let 0 =: a and n+1 = b. In any interval


26/30


(j, j+1) notice that gh is monotonic. Then

ba

1

gh()

d =n

j=0

j+1j

1

gh()

d

=n

j=0

j+1j 1gh() dCn

min[a,b]

|gh|1

.

Now we prove a.Special Casemin

[a,b]|g| 1. The proof is by induction on k. Let k = 2. The

assumptions tell us that gh is monotonic. We assume without loss of generalitythat (gh) is positive. Since gh is increasing, it crosses the -axis either one

time or not at all. We assume that it crosses at the point c (a, b). (If thecrossing occurs at and endpoint or not at all then things proceed in much thesame fashion.) Let 0 <


27/30


We now must control the integral in (45). In what follows, we make use ofthe fact that (gh) has a definite sign on [a, b]. To witR

1

igh()

eih()d

R

1

gh()

d

R 1

gh() d 1gh() |bc+

2 min[a,b] |(gh)| .

To control the integral from (a, c ) follows exactly the same pattern. Alltold, we have

b

a

eih()

g()d

2 + 8

min[a,b]|(gh)

|.

Choosing =2

min[a,b] |(gh)|yields

ba

eih()d

8min[a,b] |(gh)| .This conclude the base case k = 2.

Now assume the result for all 2 n < k. Without loss of generality, assumethat (gh)(k1)() is positive on [a, b]. This implies that (gh)(k2)() crosses the-axis in at most one point. We assume that this happens at c (a, b). (If thecrossing occurs at and endpoint or not at all then things proceed in much the

same fashion.) Let 0 <


28/30


Thus we have R

eih()

g()d

Ck1 min[a,b] (gh)(k1)1/(k1)

.

To control the integral from (a, c

) follows exactly the same pattern. All

told, we haveba

eih()

g()d

2 + 2Ck1 min[a,b] (gh)(k1)1/(k1)

.

Choosing =

min[a,b]

(gh)(k1)1/k yields

ba

eih()

g()d

(2 + 2Ck1)

min[a,b]

(gh)(k1)

1/k

and this completes the proof of Lemma 2 under the special case assumption.General Casemin[a,b]

|g| > 0. Let g := gmin[a,b] |g| . Then min[a,b] |g| = 1 and

we can use the result from the special case to concludeba

eih()

g()d

Ckmin[a,b] (gh)(k1)1/k

.

Rewriting the above in terms of g instead of g results in estimate (10) and thislemma is proven.

Proof. (Lemma 3) Fix p and let q be its Holder dual. First we compute2j+1||=2j

()f() C2j+1||=2j

||sf()j()d

= C2js

2j+1||=2j

fj() d C2jsfjLq 2j+1 2j 1p C2j( 1p+s)

fjLq .

(Here fj

:= j

f.) The Hausdorff-Young inequality statesfLq fLp fordual exponents p, q. Thus2j+1

||=2j

()f() C2j( 1p+s)fjLp.


29/30


Now let N(t) := log2(t)the smallest integer less than or equal to log2(t).Then we have

||t

()f()

d

j=N(t)

2j+1||=2j

()f()

d

C j=N(t)

2j(1p+s)fjLp

C

j=N(t)

2sj2j(1

p+s+s)fjLp

C2sN(t)

j=N(t)

2j(1

p+s+s)fjLp

CtsfBs+s+1pp,1

.

References

[1] Z. Alterman, Effect of Surface Tension on the Kelvin-Helmholtz Instability of Two Ro-tating Fluids, PNAS 47 (1961), no. 1, 224-227.

[2] Ambrose, D. Well-Posedness of Vortex Sheets with Surface Tension, SIAM J. Math.Anal., 35 (2003), 221244.

[3] Ambrose, D. and Masmoudi, N. The zero surface tension limit of two-dimensional waterwaves. Comm. Pure Appl. Math., 58 (2005), 12871315.

[4] G. Baker, D. Meiron, and S. Orszag. Generalized vortex methods for free-surface flowproblems, J. Fluid Mech., 123 (1982).

[5] J. T. Beale, T. Hou, and J. Lowengrub, Growth rates for the linearized motion of fluid

interfaces away from equilibrium, Comm. Pure Appl. Math., 46 (1993), 12691301.[6] M. Ben-Artzi and J.-C. Saut, Uniform decay estimates for a class of oscillatory integrals

and applications Differential Integral Equations 12 (1999), 137145.[7] J. Bourgain, Global solutions of nonlinear Schrdinger equations, American Mathematical

Society Colloquium Publications, 46. American Mathematical Society, Providence, RI,(1999).

[8] A. Cheng, D. Coutand, and S. Shkoller, On the Motion of Vortex Sheets with SurfaceTension in the 3D Euler Equations with Vorticity, (2007), preprint.

[9] P. Constantin, Decay Estimates for Schrodinger Equations, Commun. Math. Phys., 127(1990), 101-108.

[10] P. Constantin and J.-C. Saut Local smoothing properties of dispersive equations. J. Amer.Math. Soc. 1 (1988), 413439.

[11] De Godefroy, A. Nonlinear decay and scattering of solutions to a Bretherton equation in

several space dimensions. Electron. J. Differential Equations 2005, 141, 17.[12] D. Ebin, Ill-posedness of the Rayleigh-Taylor and Helmholtz problems for incompressible

fluids, Comm. Partial Differential Equations 13 (1988), no. 10, 1265-1295.[13] Guo, Y., Hallstrom, C., and Spirn, D. Dynamics near unstable, interfacial fluids, Comm.

Math. Phys., 270 (2007), 635689.


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[14] L. Hormander, The Analysis of Linear Partial Differential Operators I, Springer-Verlag,New York (1990).

[15] Hou, T., Lowengrub, J., and Shelley, M. Removing the stiffness from interfacial flowswith surface tension, J. Comp. Phys. 114 (1994).

[16] Hou, T., Lowengrub, J. and Shelley, M. The long-time motion of vortex sheets withsurface tension, Phys. Fluids, 9 (1997), 19331954.

[17] T. Iguchi, Well-posedness of the initial value problem for capillary-gravity waves, Funk-cial. Ekvac., 44 (2001), 219241.

[18] T. Kato, On the Korteweg-de Vries equation, Manuscripta Math. 29 (1979), 89-99.[19] Kenig, C., Ponce, G., and Vega, L. Oscillatory integrals and regularity of dispersive

equations. Indiana Univ. Math. J. 40 (1991), 3369.[20] A. Majda and A. Bertozzi Vorticity and incompressible flow. Cambridge Texts in Applied

Mathematics, 27. Cambridge University Press, Cambridge, (2002).[21] Ogawa, M. and Tani, A. Free boundary problem for an incompressible ideal fluid with

surface tension, Math. Models Methods Appl. Sci. 12 (2002), 17251740.[22] J. Shatah and M. Struwe, Geometric Wave Equations, American Mathematical Society,

Providence (1998).[23] Siegel, M. A study of singularity formation in the Kelvin-Helmholtz instability with sur-

face tension. SIAM J. Appl. Math. 55 (1995), 865891.

[24] Stein, E. M. Harmonic analysis: real-variable methods, orthogonality, and oscillatoryintegrals. Princeton Mathematical Series, 43. Monographs in Harmonic Analysis, III.Princeton University Press, Princeton, NJ, (1993).

[25] T. Tao, Low-regularity global solutions to nonlinear dispersive equations Surveys in anal-ysis and operator theory (Canberra, 2001), 1948, Proc. Centre Math. Appl. Austral.Nat. Univ., 40, Austral. Nat. Univ., Canberra, (2002).

[26] M. Taylor, Partial Differential Equations. Springer, New York, (1996).[27] Whitham, G. B. Linear and nonlinear waves. Pure and Applied Mathematics. Wiley-

Interscience, New York, (1974).[28] Wu, S. Well-posedness in Sobolev spaces of the full water wave problem in 2-D. Invent.

Math. 130 (1997), 3972.

School of Mathematics, University of Minnesota, Minneapolis, MN 55455

E-mail address: [email protected]

School of Mathematics, University of Minnesota, Minneapolis, MN 55455,

and, Department of Mathematics, Drexel University, Philadelphia, PA 19104

E-mail address: [email protected]

daniel spirn and j. douglas wright- linear dispersive decay estimates for vortex sheets with surface...

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