existence and continuation for euler equations · 2014. 3. 24. · 3.compressible euler =...

Existence and Continuation for Euler Equations

David DriverZhuo Min ‘Harold’ Lim

22 March, 2014

Contents

1 Introduction 11.1 Physical Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Notation and Conventions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2 Solutions of the Euler Equation 22.1 Equations of Compressible Fluid Flow . . . . . . . . . . . . . . . . . . . . . . . . . 22.2 Existence in Theorem 2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.3 Constructive Existence for the Incompressible Euler Equations . . . . . . . . . . . 10

3 The Cauchy Problem - A Direct Approach 153.1 Removal of the Pressure Term . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163.2 Regularisation by Mollifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.3 Global Existence and Uniqueness for the Regularised Pressureless Equation . . . . 193.4 Local Existence to the Euler Equation . . . . . . . . . . . . . . . . . . . . . . . . . 22

4 Vorticity and Continuation 28

1 INTRODUCTION Page 1 of 31

1 Introduction

1.1 Physical Background

Most equations arising from the study of continuum mechanics take the form of balance laws. Acommon (and reasonable) postulate is that of conservation of mass, which is expressed as

∂ρ

∂t+ divx (ρu) = 0

for the mass density ρ = ρ(t, x) : I×Ω→ [0,∞) and the velocity vector field u = u(t, x) : I×Ω→R3, where I is some time interval and Ω ⊆ R3 expresses the material body.

The Navier-Stokes equation is the standard equation expressing the balance of momentum. Inits most general form it reads

ρ∂u

∂t+ u · ∇u = divxT + f

where T : I × Ω → R3×3 is the stress tensor expressing surface forces within the fluid, andf : I × Ω → R3 expresses the body force. The tensor T depends on u and its derivatives, andits form is a characteristic of the fluid; this form should always be specified in the study of theNavier-Stokes equation.

For this essay, we restrict our attention to the case of an incompressible fluid, i.e. we assumethe fluid has constant density ρ ≡ const, so mass conservation becomes

divx u = 0 .

Moreover, we also assume that the fluid is at constant temperature, so that we do not have toconsider any heat generation or transport in an additional energy balance equation; and we assumethat there are no body forces, i.e. f ≡ 0.

More importantly, we make the strong assumption that the only surface acting on the fluid isthe pressure, i.e. T = −p1. Then the Navier-Stokes equation reduces to the Euler equation,

ρ∂u

∂t+ u · ∇u = −∇p .

This assumption is obtained by formally neglecting viscous stresses in the fluid. Such an approxi-mation is known to be valid for fluid flow at high Reynolds numbers (i.e. high speed, low viscosity),away from any boundaries of the flow; indeed, most known fluids will form viscous boundary layersnear boundaries at high-speed flow, where viscous stresses can no longer be neglected, even if thefluid is of very low viscosity. For simplicity we shall work in the whole space Ω = R3 to avoid anyboundary issues, and we shall work in non-dimensional variables with ρ ≡ 1.

In summary, we will concern ourselves with the Cauchy problem of the 3-dimensional incom-pressible Euler equation in whole space. These read

∂tu+ u · ∇u = −∇pdiv u = 0

u(t, ·) = u0

for the velocity field u and pressure field p, which, for t ∈ R sufficiently close to zero, should definefunctions u(t) : R3 → R3 and p(t) : R3 → R3.

1.2 Notation and Conventions

In this essay we shall always define the Fourier transform of a Schwartz function f ∈ S(Rd) by

f̂(ξ) :=

∫Rd

e−iξ·xf(x) dx ,

2 SOLUTIONS OF THE EULER EQUATION Page 2 of 31

so that the inverse transform is given by

f̂(ξ) :=

∫Rd

e−iξ·xf(x) dx .

We shall denote by Hs(Rd) the Sobolev space

Hs(Rd) :=

{f ∈ S ′(Rd)

∣∣∣∣∣ (1 + |ξ|2)s/2 f̂(ξ) ∈ L2(Rd)},

which is a Hilbert space equipped with the norm

‖f‖Hs :=(∫

Rd

∣∣∣f̂(ξ)∣∣∣2 (1 + |ξ|2)s dx) 12 .We remind the reader that when s ∈ Z≥0, this norm is equivalent to the usual norm for Sobolevspace of integer order, in terms of the sums of the L2 norms of the distributional derivatives upto order s.

2 Solutions of the Euler Equation

We define the incompressible Euler equation to be the PDE given by

∂u

∂t+ (u · ∇)u = −∇ρ (1)

divu = 0 (2)

u(x, 0) = u0(x). (3)

The main result we present here is the following.

Theorem 2.1. Given u0 in Hs(R3) with ‖u0‖H3 ≤ N0 0

and a unique classical solution

u ∈ C([0, T ];H3) ∩ C1([0, T ];H2) (4)

to equation (1). Furthermore, if T1 = inf{T : Equation (1) has no solution on [0, T ]}, then∫ T10

maxx∈R3

|∇ × u(x, t)|dt =∞ (5)

This existence result can also be proved from the abstract Kato theory but here we give amore classical approach. The continuation part of this theorem is special to 3 dimensions. Wefirst consider the case of u taking values in a bounded set and give results for dimensions 1,2 and3.

2.1 Equations of Compressible Fluid Flow

Throughout this section we work with an open set G ⊂ Rm such that u : RN → G for N ∈ {1, 2, 3}.Considering this set G is natural since many physical quantities often can not take values in allof RN for example density and total energy should be positive. We can also define Fj : G→ Rmto be a prescribed non-linear function and S : G × RN × [0,∞) → Rm to be a smooth functiondescribing a source term. We can then consider the general equation

∂u

∂t+

N∑j=1

∂

∂xjFj(u) = S(u, x, t) (6)


We consider the compressible Euler equations,

Dp

Dt+ γpdivv = 0

ρDv

Dt+∇p = 0

DS

Dt= 0 (7)

for ρ = p1/γe−S0/γ and

D

Dt=

∂

∂t+

N∑j=1

vj∂

∂xj.

Equation (6) is more clearly seen to be a generalisation when written in a more classical form.[6] shows that this is the correct framework. Equation (1) is in some sense (see Theorem 2.14)the limit of equation (7) and equation (7) takes the form in (6) with S = S0. Since for us S isconstant the last equation in (7). Thus we will require the following theorem. We can see that itis, in spirit, close to Theorem 2.1.

Theorem 2.2 (Majda, [6]). Assume u0 ∈ Hs , s > N/2+1 and u0 : RN → G1 for G1 ⊂⊂ G. Thenthere exists T > 0 such that equation (6) has a unique classical solution u ∈ C1(RN × [0, T ], G2)for G2 ⊂⊂ G.

Furthermore,u ∈ C([0, T ], Hs) ∩ C1([0, T ], Hs−1) (8)

and T only depends on ‖u0‖s and G1.

We see that when N = 3, taking s = 3 in the assumptions will give us a similar result toTheorem (2.1) for bounded u.

The plan is now to make rigorous, the following heuristic arguments:

1. Show uniform stability for the compressible solution to equation (7) independent of the“Mach number”.

2. Construct solution for equation (1) by taking the limit as the “Mach number” (defined inDefinition 2.12) approaches zero.

3. Compressible Euler = Incompressible Euler +M(Linear Acoustics)+O(M2) for a Mach num-ber M .

We can show uniqueness easily from the energy estimation methods discussed below or see[11].

2.2 Existence in Theorem 2.2

We split the proof into several parts. We first prove

Theorem 2.3. Under the hypotheses of Theorem 2.2, there exists a unique classical solutionu ∈ C1([0, T ]× RN ) to equation (6) with

u ∈ L∞([0, T ], Hs) ∩ Cw([0, T ], Hs) ∩ Lip([0, T ], Hs−1)

Here we have used the notation Cw to be continuous functions on [0, T ] with values in theweak topology of Hs. That is to say, for any φ ∈ Hs,

(φ, u(t)) =∑|α|≤s

∫RNDαφ ·Dαu(t) dx

is a continuous function on [0, T ].Lip([0, T ], Hs−1) is the space of Lipschitz functions on [0, T ] with values considered in the norm

topology of Hs−1.Once we have proved Theorem 2.2, we can deduce Theorem 2.2 from Theorem 2.4 below.


Theorem 2.4. Any classical solution of equation (6) u taking values in G2 ⊂⊂ G satisfyingthe regularity as in the conclusion of Theorem 2.2 on some interval [0, T ] also has the regularityproperty that

u ∈ C([0, T ], Hs) ∩ C1([0, T ], Hs−1).

As in [6], we consider S = 0 for ease of exposition although the same proof will work for moregeneral S. We are interested in constant S = S0 and so this doesn’t take away understandingfrom our situation. The proof of Theorem 2.2 splits up into several key ideas:

1. Consider the linearised problem with mollified initial data.

2. Use an iteration scheme to construct a solution to the linear equation.

3. Show boundedness in the “high” norm.

4. Show contraction in the “low” norm.

The so called high and low norms come about since it is easier to find a contraction with a lowernorm and combining this with the result for the high norm, we can get the required results in thenorm that we are interested in - the high norm. This will become more transparent in the proofbelow.

When S = 0, any constant u0 is a solution of (6). We can then consider solutions u(x, t) = u0+vand thus obtain the linearised equations,

∂v

∂t+

N∑j=1

Aj(u0)∂v

∂xj= 0, t > 0, x ∈ RN (9)

v(x, 0) = v0(x) (10)

where Aj(u) =∂Fj∂u , j = 1, . . . , N are m×m Jacobian matrices.

In many physical equations, for u taking values in G, we can find a positive definite symmetricmatrix A0(u) smoothly varying in u with

1.cI ≤ A0(u) ≤ c−1I, (11)

A0 = A∗0 with uniform constant c whenever u takes values in G1, G1 ⊂⊂ G

2. A0(u)Aj(u) = Ãj(u), where Ãj(u) = Ã∗j (u), for j = 1, . . . , N .

We add the assumption that this A0 exists for our proof. In the main case we are interested in,we can see that this holds with

A0(u) =

(γp)−1 0ρ(p, S)I30 1

. (12)One key result used in the proof Theorem 2.2 is the following stability lemma due to Friedrichs’

energy estimates [6].

Lemma 2.5. Suppose v is a solution to the equations

A0(u)∂v

∂t+

N∑j=1

Ãj(u)∂v

∂xj−B(u, x, t)v = f(t) (13)

v(x, 0) = v0(x) (14)

for u a C1 function taking values in G1 ⊂⊂ G and B a smoothly varying m×m matrix function.Then,

maxt∈[0,T ]‖v‖0 ≤ c−1 exp(

1

2c−1‖divA +B +B∗‖L∞T

)[‖v(0)‖0 +

∫ T0

‖f(s)‖0 ds

](15)

where A = (A0, Ã1, . . . , ÃN )


This result follows from considering the energy defined as E(t) = (A0(u)v, v) and using Gron-wall’s inequality on ∂∂tE(t). This will also be useful later.

Finally, before the proof of Theorem 2.2 we present a result due to [8] which gives sharpinequalities for products of functions.

Theorem 2.6 (Moser-type Inequalities). 1. For f, g ∈ Hs, s > N/2 we have f · g ∈ Hs andmoreover ‖f · g‖s ≤ Cs‖f‖s‖g‖s.

2. For f, g ∈ Hs ∩ L∞ and |α| ≤ s,

‖Dα(fg)‖0 ≤ Cs(‖f‖L∞‖Dsg‖0 + ‖g‖L∞‖Dsf‖0)

3. For f ∈ Hs, Df ∈ L∞ , g ∈ Hs−1 ∩ L∞ and |α| ≤ s,

‖Dα(fg)− fDα(g)‖0 ≤ Cs(‖Df‖L∞‖Ds−1g‖0 + ‖g‖L∞‖Dsf‖0)

4. Assume g(u) is a smooth vector-valued function on G and u is a continuous function withu(x) ∈ G1 for all x ∈ RN and G1 ⊂⊂ G and u ∈ Hs ∩ L∞. Then for s ≥ 1,

‖Dsg(u)‖0 ≤ Cs‖∂g

∂u‖Cs−1(G1)‖u‖

s−1L∞ ‖D

su‖0

The proof of these results can be found in [8] and partial proof is given in the section below(see Lemmas 3.9 and 3.14).

of Theorem 2.2. As mentioned above we will split the proof into various parts.Step 1: First of all we smooth the initial data using standard mollification tricks. We choose afunction j ∈ C∞0 (RN ) to be non-negative, supported in the unit ball and to have integral equal to1. Then we define jε(x) = 1/ε

N j(x/ε) and set Jεu(x) = jε ∗ u(x). We also set

εk = 2−kε0

u(k)0(x) = Jεku0

for ε0 > 0 to be chosen later and k ∈ N.Step 2: The key observation here is that in order to construct a smooth solution to equation(6), it suffices to differentiate the nonlinear term and apply the symmetrising matrix from above.Thus we consider the quasi-linear system (dropping the tilde notation)

A0(u)∂u

∂t+

N∑j=1

Aj(u)∂u

∂xj= 0 (16)

u(x, 0) = u0(x)

Now we can consider a classical iteration scheme.Our initial estimate is u(0)(x, t) = u

(0)0 (x) and we inductively define u

(k+1)(x, t) via the solutionto the linear equation

A0(u(k))

∂u(k+1)

∂t+

N∑j=1

Aj(u(k))

∂u(k+1)

∂xj= 0 (17)

u(k+1)(x, 0) = u(k+1)0 (x) (18)

First of all, we need to show that these iterates are actually well defined.Let G2 be an open set such that G1 ⊂ G2 and G2 ⊂⊂ G. Since by assumption, s > N/2, we

have‖v‖L∞(G2) ≤ C‖v‖s (19)

by a Sobolev embedding.


From the basic properties of mollification and the fact that u0 takes values in G1, it followsthat there exists an R and ε0 such that if

‖u− u(0)0 ‖s ≤ R, then u(x) ∈ G2 for all x. (20)

and for the same ε0 we have‖u0 − u(k)0 ‖s ≤ cR/4 (21)

for all k = 0, 1, . . . , and with c from the equation (11) i.e. cI ≤ A0(u) ≤ c−1I for all u ∈ G2.Combining these facts we can inductively solve for u(k+1) on some time interval [0, Tk] and in factu(k+1) ∈ C∞([0, Tk]× RN ) where Tk is the largest time such that

‖u(k) − u(0)0 ‖s,Tk ≤ R

holds where‖w‖s,T ≡ max

t∈[0,T ]‖w‖s

for w ∈ L∞([0, T ], Hs). In order for the problem (16) to have a solution on some interval ofpositive measure we need there to exist a time T∗ > 0 such that Tk > T∗ for all k ≥ 1. Thus weneed “boundedness in the high norm”. For now we will assume the following lemma.

Lemma 2.7. There exist constants L > 0 and T∗ > 0 such that the solutions u(k+1)(x, t) defined

above satisfy

‖u(k+1) − u(0)0 ‖s,T∗ ≤ R (22)

‖∂u(k+1)

∂t‖s−1,T∗ ≤ L (23)

We will defer the proof of Lemma 2.7 until later to keep the proof more transparent.Step 3: The next task is to find a norm, a so called “low norm” such that we have convergenceof u(k) to a function u. In fact we will look for a norm such that

∞∑k=1

‖u(k+1) − u(k)‖ ≤ ∞.

This is not necessarily true in the high norm. Note that we need the norm to be strong enoughso that the expressions in equation (16) converge to the desired limiting equation.

A naive choice of norm could be ‖ · ‖s,T∗∗ for T∗∗ ≤ T∗ but if we use the energy estimate in(15) it can be shown that

‖u(k+1) − u(k)‖s,T∗∗ ≤ αk‖u(k) − u(k−1)‖s,T∗∗ .

This is almost the contraction that we would like but unfortunately αk depends on ‖uk‖s+1 andthis is not bounded. It will turn out that it is sufficient to work in the norm ‖ · ‖0,T∗∗ . In fact wehave the following lemma for contraction in the low norm

Lemma 2.8. There exists a time T∗∗ ≤ T∗, α < 1 and {βk}∞k=1 with∑|βk|


Note that for any norm, if ‖u(k+1)−u(k)‖ ≤ α‖u(k)−u(k−1)‖+βk with α < 1 and∑∞k=1 |βk| <

∞, then∞∑k=1

‖u(k+1) − u(k)‖ N/2 + 1, we have that

u(k) → u, in C([0, T∗∗], C1(RN )). (29)

We see that we also get convergence of the time derivative of u(k). In particular, from equation(17) we have

∂u(k+1)

∂t= −A−10 (u(k))

N∑j=1

Aj(u(k))

∂u(k+1)

∂xj

and so∂u(k)

∂t→ ∂u

∂t, in C([0, T∗∗]× RN )

The proof thatu ∈ Cw([0, T ], Hs) ∩ Lip([0, T∗∗], Hs−1)

follows from strong convergence in C([0, T∗∗], Hs′) as shown above. For more details see [6] but

here we skip over this for brevity since the techniques are fairly standard. The proof is nowcomplete, up to the proof of Lemma 2.7.Step 5: For the proof of Lemma 2.7 we proceed by induction For k = 0, T0 = ∞ and so thebase case is easy. We set v(k+1) = u(k+1) − u(0)0 . Then we have

A0(u(k))

∂v(k+1)

∂t+

N∑j=1

Aj(u(k))

∂v(k+1)

∂xj= −

N∑j=1

Aj(u(k))

∂

∂xju(0)0 ≡ hk (30)

v(k+1)(x, 0) = u(k+1)0 (x)− u(0)(x)

By the induction hypothesis, u(k) ∈ G2 for all (x, t) ∈ RN × [0, T∗] for some T∗ to be chosenlater. We now ignore the superscripts above for improved readability. Consider u, v ∈ C∞ on


RN × [0, T∗] with u taking values in G2.

A0(u)∂v

∂t+

N∑j=1

Aj(u)∂v

∂xj= h (31)

v(x, 0) = v0(x)

Also, we use the shorthand notation vα = Dαv for |α| ≤ s. We can then obtain a similar equation

to the above plus some error terms and when α = 0 we get the same result as above. We alsoomit the u to improve presentation.

A0∂vα∂t

+

N∑j=1

Aj∂vα∂xj

= A0Dα(A−10 h) + Fα (32)

vα(x, 0) = Dαv0(x)

where Fα represents the commutator terms given by

Fα =

N∑j=1

A0

[(A−10 Aj

∂vα∂xj

−Dα(A−10 Aj∂v

∂xj)

](33)

We are now in a position to proceed by induction. We assume that the result of Lemma 2.7 holds

for u(k) and for some L and T∗ and estimate ‖u(k+1) − u(0)0 ‖s,T∗ . We will then be done, since Rdetermines a choice of L, independently of T∗, by simply studying the equation for

∂u(k+1)

∂t and

using Moser-type inequalities (see Theorem 2.6 above). In order to obtain a bound on ‖u(k+1) −u(0)0 ‖s,T for some T , we must control the right hand side of equation (32) we can use the energy

estimates in (15) and see that we will be done if we can show that for some C depending on G2,‖u‖s, R and s only, we have∑

1≤|α|≤s

‖Fα‖20 +∑|α|≤s

‖A0Dα(A−10 h)‖20 ≤ C(‖v‖2s + 1) (34)

If we temporarily assume this bound, we will be done. Indeed, applying the energy estimate to

v(k+1)α and summing over all α, we see that

‖u(k+1) − u(0)0 ‖s,T ≤ c−1e(LC+C)T (‖u(k+1)0 − u

(0)0 ‖s + TC) (35)

with c from equation (11) and

‖u(k+1)0 − u(0)0 ‖s ≤ cR/2.

Thus, we get for example

‖u(k+1) − u(0)0 ‖s,T ≤ e(L+1)CT (R/2 + TC)

and from here, it is easy to find a fixed T∗ such that

‖u(k+1) − u(0)0 ‖s,T ≤ R

as required for the inductive step.Step 6: Now all that remains, is to show the estimate in equation (34). We prove this boundusing Theorem 2.6.

Now we improve this result to show Theorem 2.2.

of Theorem 2.4. If u(x, t) is the local solution from Theorem 2.2 we see from the proof above thatA0(u(x, t)) ∈ C([0, T∗], C(RN )) and cI ≤ A0(u(x, t)) ≤ c−1I for (x, t) ∈ RN × [0, T∗∗]. We writeA0(t) ≡ A0(u(x, t)) in this case. For t ∈ [0, T∗∗], we define the norm for v ∈ Hs

‖v‖2s,A0(t) ≡∑|α|≤s

∫RN

(Dαv,A0(t)Dαv) dx.


From above, we have equivalence of norms, i.e.

c‖v‖2s ≤ ‖v‖2s,A0(t) ≤ c−1‖v‖2s. (36)

It is also straightforward to show that

lim supt→0

‖v(t)‖2s,A0(t) = lim supt→0

‖v(t)‖2s,A0(0) (37)

for all v(t) ∈ Cw([0, T∗∗], Hs).With these observations in mind, we can simplify our task by reducing the proof to showing

the result in a simpler setting.

1. It suffices to show that u ∈ C([0, T ], Hs) from the explicit form of equation (16) since thentrivially u ∈ C1([0, T ], Hs−1).

2. We will only show strong right continuity at t = 0 as the same argument gives us strongright continuity at every 0 ≤ t < T and equation (16) and the argument below are reversiblein time, and thus strong right continuity on [0, T ) is equivalent to strong left continuity on(0, T ].

Recall the general result that if {wn} is a weakly convergent sequence to w in a Hilbert space,then wn converges strongly to w if and only if ‖w‖ ≥ lim supn→∞ ‖wn‖. Applying this to Hs withnorm ‖ · ‖s,A0(0), we see that strong right continuity at t = 0 in Hs is given if we have

‖v‖2s,A0(t) ≥ lim supt→0

‖v‖2s,A0(t).

This will be proved using equation (37) and “taking limsup as t tends to 0 in following Lemma”.

Lemma 2.9. Let u be a local solution from Theorem 2.2 on some interval [0, T∗∗]. Then thereexists a function f ∈ L1([0, T∗∗]) such that

‖u(t)‖2s,A0(t) ≤ ‖u0‖2s,A0(0)

+

∫ t0

|f(s)|ds

for all t ∈ (0, T∗∗].

Proof of Lemma 2.9: Recall that uk ∈ C∞ ∩ Hs. By the energy techniques mentionedbefore, it is possible to show that for 0 ≤ t ≤ T∗∗ we have

∂

∂t

∑|α|≤s

∫RN

(Dαu(k+1), A0(u(k))Dαu(k+1)) =

∫RN

(divA(u(k))u(k+1), u(k+1))+2

∫RN

(F k+1s , u(k+1)).

(38)

Here we have defined F k+1s in the following way, with the notation u(k+1)α ≡ Dαu(k+1),

F k+1s =∑

1≤|α|≤s1≤j≤N

A0(u(k))

[A0(u

(k))−1Aj(u(k))

∂u(k+1)α

∂xj−Dα(A−10 Aj(u(k))

∂u(k+1)

∂xj)

](39)

By Theorem 2.6 and Lemma 2.7, we can bound the right hand side of equation (38) by someabsolutely integrable function f on [0, T∗∗]. We now integrate both sides of equation (38) between0 and t and rearrange to see that∑|α|≤s

∫RN

(Dαu(k+1)(t), A0(u(k)(t))Dαu(k+1)(t))

=∑|α|≤s

∫RN

(Dαu(k+1)0 , A0(u

(k)0 )D

αu(k+1)0 ) +

∫ t0

|f(s)|ds


Since {u(k)0 } is defined via a mollification procedure, it is simple to see from standard convergenceresults that

limk→∞

∑|α|≤s

∫RN

(Dαu(k+1)0 , A0(u

(k)0 )D

αu(k+1)0 ) = ‖u0‖2s,A0(0). (40)

From equation (28) we see that

limk→∞

maxt∈[0,T∗∗]

‖A0(u(k)(t))−A0(u(t))‖L∞ = 0. (41)

We also have weak convergence as a result of Theorem 2.2 in the sense that, for a fixed t > 0, wehave

lim supk→∞

(u(k+1)(t), u(k+1)(t))s,A0(t) ≥ (u(t), u(t))s,A0(t). (42)

We can now conclude the proof of Lemma 2.9 and thus the proof of Theorem 2.4 by noting thatby (36) and (37) we have that

‖u(t)‖2s,A(t) ≤ lim supk→∞

∑|α|≤s

∫(Dαu(k+1), A0(u

(k))Dαu(k+1)).

We now state useful corollaries of Theorem (2.2) and the sharp continuation principle in [6,p.46]. We do not give the proof here.

Proposition 2.10. Assume u0 ∈ Hs for some s > N/2 + 1 and that u is a classical solutionof equation (6) on some interval [0, T ] with u ∈ C1([0, T ] × RN ). Then u ∈ C([0, T ], Hs). Inparticular, if u0 ∈ ∩sHs on any interval [0, T ] where u ∈ C([0, T ], Hs0) for some s0 > N/2 + 1,then u ∈ C∞([0, T ]× RN ).

The next corollary is a continuation principle which characterises the behaviour of the solutionas we approach the endpoint of the maximum interval of the solution to our equation.

Proposition 2.11. Assume u0 ∈ Hs for some s > N/2 + 1. Then [0, T ) with T < ∞ is themaximal interval of Hs existence if and only if either

1. ‖ut‖L∞ + ‖Du‖L∞ as t↗ T , or

2. u(x, t) escapes every compact subset K ⊂⊂ G, as t↗ T .

For a special case of this, see Theorem 3.12.

2.3 Constructive Existence for the Incompressible Euler Equations

Notice that the above framework can not be directly applied to the incompressible Euler equations.For this reason we consider the incompressible equations as the limiting case of the compressibleequations that we just studied. More specifically, we are now ready to prove the following resultwhich states that the solutions for the compressible Euler equations converge as the Mach numbertends to zero and the resulting function satisfies the incompressible Euler equations. That is, wewant to show that the equations of isentropic compressible fluid flow

Dρ

Dt+ ρdivv = 0 (43)

Dv

Dt+

1

ρ∇p = 0 (44)

p = Aργ , γ > 1. (45)


converge as the “Mach number (defined below in Definition 2.12) tends to zero” to the incom-pressible Euler equations given by

ρ0Dv∞

Dt+∇p∞ = 0 (46)

divv∞ = 0. (47)

The advantage of this approach is that we obtain a constructive existence of a solution. First ofall we “non-dimensionalise” the compressible equations. We consider the initial data given by

ρ(x, 0) = ρ0(x), v(x, 0) = v0(x) (48)

and we set ρmax = max ρ0(x) and |vmax| = max |v0(x)|. We then normalise and change variablesaccording to

ṽ ≡ v|vmax|

, ρ̃ =ρ

ρmax(49)

x′ = x, t′ = |vmax|t. (50)

We then obtain the non-dimensional form of the compressible Euler equations

Dρ̃

Dt′+ ρ̃divṽ = 0 (51)

ρ̃Dṽ

Dt′+ λ2∇p(ρ̃) = 0 (52)

where we write

λ2 = (dp

dρ(ρm)

1

|vmax|2)(γA)−1

which has no dimension.

Definition 2.12. Using the notation above, the Mach number M is defined to be the ratio of

typical fluid speed (|vmax|) to the typical sound speed (c(ρmax) ≡(

dpdρ

)− 12). i.e

M =|vmax|c(ρmax)

(53)

and so λ = M−1(γA)−12 .

We now change notation again for convenience by removing the primes and tildes and writethe equation in terms of (pλ, vλ) satisfying

(γpλ)−1Dpλ

Dt+ divvλ = 0 (54)

ρ(pλ)Dvλ

Dt+ λ2∇pλ = 0 (55)

where ρ = (pλ)1/γe−a/γ and a is a constant source term. The initial conditions are given by

pλ(x, 0) = pλ0 (x), vλ(x, 0) = vλ0 (x). (56)

If we proceed in a formal manner by expanding, for example pλ = p0 + λ−1p1 + λ

−2p2 +O(λ−3)

and similarly for vλ and use the orthogonal decomposition of L2(RN ) into divergence free vectorfields and gradient fields (see Hodge decomposition in Lemma 3.2), we obtain hints to the therequired convergence but it turns out that we do not actually obtain the correct result. For amore detailed overview of the physical interpretation and formal derivation of the convergence,see [6, Section 2.4]. Here, for brevity of exposition, we go straight into important stability resultsfor convergence.

It turns out that the correct asymptotic expansion of (pλ, vλ) is given by

pλ = P0 + λ−2(p∞(x, t) + p1(x, t, λ)) +O(λ

−3) (57)

vλ = v∞ + λ−1(v1(x, t, λ)) +O(λ−2). (58)


Theorem 2.13. Assume the initial data (pλ0 (x), vλ0 (x)) satisfies the stability estimate

‖λ(pλ0 (x)− P0)‖s0 + ‖vλ0 (x)‖s0 ≤ R (59)

for some fixed R > 0, λ ≥ 1 and s0 = [N/2] + 2. Then there is a fixed time interval [0, T0] withT0 > 0 and λ0(R) such that for all λ ≥ λ0(R) the compressible Euler equations have a classicalsolution on [0, T0] and

(λ(pλ − P0), vλ) ∈ C([0, T0]), Hs0) ∩ C1([0, T0], Hs0−1) (60)

and‖λ(pλ − P0)‖s0,T0 + ‖vλ‖s0,T0 ≤ R′ (61)

for some fixed constant R′ > 0.

We are now in a position to prove the main result of this section.

Theorem 2.14. Consider the solutions of the system

γ−1(pλ)Dpλ

Dt+ divvλ = 0

ρ(pλ)Dvλ

Dt+ λ2∇pλ = 0 (62)

with initial data given by

vλ(x, 0) = v∞0 + λ−1v10(x), divv

∞0 = 0

pλ(x, 0) = P0 + λ−2p10(x) (63)

belonging to Hs0 and satisfying the stability estimate (59). From Theorem 2.13 we can find afixed time interval [0, T0] where a classical solution exists. Then as λ → ∞ there exists v∞ ∈L∞([0, T0], H

s0) such thatvλ → v∞ in C([0, T0], Hs0−εloc ) (64)

for any ε > 0.In fact,

v∞ ∈ C([0, T0], Hs0) ∩ C1([0, T0], Hs0−1) (65)

and is a classical solution of the incompressible Euler equations. In other words, there exists a p∞

such that

ρ0Dv∞

Dt= −∇p∞ (66)

divv∞ = 0 (67)

v∞(x, 0) = v∞0 (x). (68)

where ρ0 = ρ(P0). Moreover, the mean pressure p∞ is the limit of pλ in the weak-* sense. More

specifically,∇pλ → ∇p∞ weak-* in L∞([0, T0], Hs0−1).

We give a sketch of the proof here. The proof can be found in full detail in [6].

Proof. As before we wish to show results in a low norm and use estimates in a high norm torecover the regularity. Here we show compactness in a low norm. It is convenient to write this inthe abstract setting and so we use the notation below for the time being. We wish to look at

Aλ0∂uλ

∂t+

N∑j=1

Aλj∂uλ

∂xj= 0, uλ(x, 0) = uλ0 (x), (69)


where uλ = (λ(pλ − P0), vλ)> and

Aλj = Aj(λ−1uλ, uλ) + λA0j , 1 ≤ j ≤ N (70)

Aλ0 = A0(λ−1uλ) (71)

for A0j constant symmetric matrices and cI ≤ A0(v) ≤ c−1I for |v| bounded and for these v,Aj(v, y) are smoothly varying for arbitrary y ∈ Rm [6].

Differentiating with respect to time and writing wλ for (uλ)t we see that

Aλ0∂wλ

∂t+

N∑j=1

Aλj∂uλ

∂xj+Bλwλ = 0 (72)

wλ(x, 0) = −A−10 (λ−1uλ0 )(N∑j=1

Aλj (uλ0 )∂uλ0∂xj

) (73)

and

Bλ = λ−1uλt∂A0∂u

∣∣∣λ−1uλ

+

N∑j=1

(∂

∂uAλj (λ

−1uλ, uλ))uxj (74)

From the properties of the matrices in (70) and Theorem 2.13 we see that we can uniformly boundBλ. In particular,

‖Bλ(t)‖L∞ ≤ C, for 0 ≤ t ≤ T0. (75)

This means that we can use the energy estimate (15) again applied to wλ and see that

‖uλt ‖0,T0 ≤ C‖uλt (0)‖0, for λ ≥ λ0(R). (76)

Thus we can bound ‖uλt ‖0,T0 by R′ assuming that we can bound ‖uλt (0)‖0 independently of λ.By the smoothness of Aλj and since λ is bounded below, we can see that, from equation (72), asufficient condition for this is that ∥∥∥∥∥λ

N∑j=1

A0j∂

∂xjuλ0 (x)

∥∥∥∥∥0

≤ R̃ (77)

for some R̃ > 0.In our case, the initial conditions given in the theorem satisfy this. Thus going back to our

specific case we have the estimate,

λ‖pλt ‖0,T0 + ‖vλt ‖0,T0 ≤ R′. (78)

From Theorem 2.13 we also have the bound

λ‖pλ − P0‖s0,T0 + ‖vλ‖s0,T0 ≤ R′. (79)

By using the interpolation inequality in (26), with s′ = s0 − ε and the two bounds above give usequicontinuity in Hs0−εloc and uniform boundedness is clear. Therefore, by Arzela-Ascoli, we canchoose a convergent subsequence such that

vλ → v∞, in C([0, T0], Hs0−εloc ) (80)

for some v∞ ∈ L∞([0, T0], Hs0) from the uniform bound on the sequence above. In a similar waywe also have v∞ ∈ Lip([0, T0], L2) ∩ C([0, T ], C1(RN )).

Now that we have a candidate solution to the incompressible Euler equations we can show thatit is indeed a solution. By the equation (62) and the above estimates we see that

λ‖divvλ‖0,T0 ≤ λ∥∥∥γ−1(pλ)−1 Dpλ

Dt

∥∥∥0,T0≤ C


for C ≡ C(R′). Therefore from the above considerations we can take limits as λ→∞ to see that

divv∞ = 0 (81)

We now show that v∞ satisfies a weak form of equation (66). To this end we take w to bean arbitrary vector valued function with w ∈ Hs0 ∩ S for S the Schwartz space on RN . Supposealso that divw = 0. We also define φ ∈ C∞0 ((0, T0)) to be a smooth real valued test function. Wewrite (·, ·)0 for the L2 inner product.

By the orthogonal decomposition of L2 into divergence free vector fields and gradient fields wesee that

(w, λ2∇pλ)0 = 0

Therefore when we take the inner product of w with the second equation in (62), multiply by thetest function φ and then integrate over t and use integration by parts to obtain

0 =

∫ T00

−∂φ∂t

(w, ρλvλ)0 + φ(t)(w, ρλvλ · ∇vλ)0 − λ−1φ(t)(w,

∂ρ

∂p(pλ)

∂p̃λ

∂tvλ)0 dt (82)

where ρλ = ρ(pλ) and as before p̃λ = λ(pλ − P0).Therefore by (78) and (79) we have

ρλ → ρ(P0) = ρ0, uniformly on [0, T0]× RN

and ∥∥∥∂p̃λ∂t

vλ∥∥∥0,T0≤ C(R′), (for λ sufficiently large)

respectivelyFrom the result in (80) and dominated convergence theorem, since w ∈ S we also have con-

vergence in the first two integrands above. More specifically,

(w, ρλvλ · ∇vλ)0 → (w, ρ0v∞ · v∞)0 (83)(w, ρλ)0 → (w, ρ0v∞)0 (84)

uniformly on [0, T0]× RN . So we can pass the limit in equation (82) as λ→∞ to see that

0 =

∫ T00

(−∂φ∂t

(w, ρ0v∞)0 + φ(w, ρ0v

∞ · ∇v∞)0)

dt (85)

for all φ ∈ C∞0 ((0, T0)) and w ∈ Hs ∩ S with divw = 0. These w are dense in the space of L2functions with these properties. Thus we have shown v∞ to be a weak solution to equation (66)with v∞(x, 0) = v∞0 (x) and

v∞ ∈ Lip([0, T0], L2) ∩ C([0, T0], C1) ∩ L∞([0, T0], Hs0).

s0 = [N/2] + 2 and from above,

1

ρ0P(v∞ · ∇v∞) ∈ L∞([0, T0], Hs0−1)

where P is the orthogonal projection from L2 onto the space of divergence-free vector fields. Theweak equation above tells us that

∂v∞

∂t= −P(v∞ · ∇v∞)

in the sense of distributions. Therefore, v∞ ∈ Lip([0, T0], Hs0−1) ∩ L∞([0, T0], Hs0). Com-bining everything together we note that also v∞ ∈ C([0, T0], C1(RN )) ∩ Lip([0, T0], Hs0−1) ∩L∞([0, T0], H

s0) ∩ C([0, T0], Hs′) for s′ < s. Finally, we can show that we also have ∂v

∞

∂t ∈

3 THE CAUCHY PROBLEM - A DIRECT APPROACH Page 15 of 31

C([0, T0] × RN ) and so v∞ ∈ C1([0, T0] × RN ) and is therefore a classical solution of (66). Thefact that v∞ ∈ C([0, T0], Hs0) ∩ C1([0, T0], Hs0−1) follows by a similar argument to the proof ofTheorem 2.4. The pressure term can also be shown to converge (see [9]). The idea is that weknow that

P(∂v∞

∂t+ v∞ · ∇v∞) = 0

and so∂v∞

∂t+ v∞ · ∇v∞ = 1

ρ0∇p∞

for some p∞ ∈ L∞([0, T0], Hs+1) and by the convergence in the divergence-free part of L2 we canshow the required convergence.

Notice that when N = 3 we have an analogous result to that of Theorem 2.1. We end thissection with a continuation result.

Theorem 2.15. Under the assumptions of Theorem 2.14, the interval [0, T∗) with T∗ < ∞ is amaximal interval of Hs existence if and only if

max1≤i,j≤N

∥∥∥∥∂v∞i∂xj∥∥∥∥L∞→∞, as t↗ T∗ (86)

For a proof see [6]. This result gives a hint to the vorticity result (Theorem 4.1).

3 The Cauchy Problem - A Direct Approach

In this section we concern ourselves with the Cauchy problem of the 3-dimensional incompressibleEuler equation in whole space. These read

∂tu+ u · ∇u = −∇pdiv u = 0

u(t, ·) = u0

for the velocity field u and pressure field p, which, for t ∈ R sufficiently close to zero, should definefunctions u(t) : R3 → R3 and p(t) : R3 → R3 that satisfy the Euler equations in some sense. Ourpresentation follows that of [7].

The goal is to prove a local existence result. To this end, a fruitful strategy is to view solutionsto the Euler equation as flows on some function, and to invoke a fixed point theorem to guaranteelocal existence and uniqueness of such flows. We shall use the following generalisation of thestandard Picard theorem of existence and uniqueness of solutions to ODE in finite-dimensionalnormed spaces.

Theorem 3.1. Let X be a Banach space and U ⊆ X be an open subset. Suppose F : U → X isa locally Lipschitz mapping.

Then for any x0 ∈ U there exists T > 0 such that the following ODE in X,

dx

dt= F (x) , x(0) = x0 ,

has a unique solution x ∈ C1([−T, T ], U).

The proof is an obvious adaptation of the standard proof using the contraction mapping theorem,so we shall omit it here.


3.1 Removal of the Pressure Term

An awkward feature of the Euler equation (and also of the Navier-Stokes equation) is the presenceof the pressure term, which is not governed by a dynamical equation. In fact, pressure is notreally an independent variable, but serves a role analogous to a Lagrange multiplier enforcing theincompressibility condition div u = 0. Because of this, it is more natural to work in the Sobolevspaces of divergence-free vector fields,

V s(Rd)

:={v ∈ Hm

(Rd,Rd

) ∣∣∣ div v = 0} , s ∈ R≥0 .More precisely, V s(Rd) is the closure in Hs(Rd,Rd) of the C∞ divergence-free vector fields.

Lemma 3.2 (Hodge decomposition). For s ≥ 0, every vector field v ∈ Hs(Rd,Rd) has a uniqueorthogonal decomposition

v = w +∇ϕwith w ∈ V s(Rd) and ϕ ∈ S ′(Rd), ∇ϕ ∈ Hs(Rd).

We therefore define the Leray projection operator P : Hs(Rd,Rd)→ V s(Rd).

Proof. We first consider uniqueness. Since u is divergence-free, the Fourier transform of w mustsatisfy ξ ⊥ F [w](ξ) ∀ ξ ∈ Rd. The orthogonality condition then forces

F [w](ξ) = F [v](ξ)− ξ · F [v](ξ)|ξ|2

ξ .

This proves uniqueness of w and hence of ϕ.For existence, we simply define w and ϕ by their Fourier transforms,

F [ϕ](ξ) := −iξ · F [v](ξ)|ξ|2

, F [w](ξ) := F [v](ξ)− ξ · F [v](ξ)|ξ|2

ξ ,

It is easy to check that this works.

Remark 3.3. The Leray projection operator P has the following properties:

(i) P commutes with distributional derivatives: For m ∈ Z≥0,

PDα = DαP on Hm(Rd), for |α| ≤ m .

(ii) P commutes with mollifiers Jε (see the next section for the definition),

PJε = JεP on Hm(Rd), for ε > 0 .

The verification of these properties is straightforward. We shall make use of them without explicitcitation.

Using Leray projection, we can eliminate the pressure term to obtain a dynamical equation forthe velocity field only. We will then concentrate on solving the resulting pressureless equation. Inother words, we look for solutions in the following sense.

Definition 3.4. Let m ∈ Z with m ≥ 3, let u0 ∈ V m(R3) and let T > 0. If

u ∈ C0((−T, T ), V m(R3)

)∩ C1

((−T, T ), V m−1(R3)

)satisfies the pressureless Euler equation,

∂u

∂t+ P (u · ∇u) = 0

u(0) = u0 ,(87)

we say u is a strong solution to the Euler equation in V m(R3) up to time T .


Remark 3.5.

(i) Given a strong solution u, we can recover the pressure by orthogonally projecting (u · ∇)uto the space of gradients via Hodge decomposition (note that ∂tu is a divergence-free vectorfield and so has no gradient component under Hodge decomposition).

(ii) We impose m ≥ 3 in the definition in order to ensure, using Lemma 3.10 below, (u · ∇)u ∈C0((−T, T ), Hm−1(R3)).

Moreover, we have the following uniqueness result.

Theorem 3.6. Let m ∈ Z with m ≥ 3, let T > 0, and let u0 ∈ V m(R3). Then there can be atmost one strong solution u with initial data u0 up to time T .

Proof. Suppose u1, u2 are any two strong solutions up to time T , with the same initial data u0.Denote w := u1 − u2. Then

d

dt

(1

2‖w‖2H0

)= −

∫R3

[u1 · ∇u1 − u2 · ∇u2] · w dx

= −∫R3

[u1 · ∇w + w · ∇u2] · w dx

Since ∫R3

[u1 · ∇w] · w dx =3∑i=1

∫R3

(u1 · ∇wi)wi dx =3∑i=1

∫R3u1 · ∇

(1

2w2i

)dx

= −∫R3

1

2|w|2div u1 dx = 0 ,

we have

d

dt

(1

2‖w‖2H0

)= −

∫R3

[w · ∇u2] · w dx ≤∫R3|w|2|∇u| dx

≤ ‖∇u‖L∞ ‖w‖2H0 .

Using Gronwall’s inequality, we find

‖w(t)‖H0 ≤ ‖w(0)‖H0 exp(

2

∣∣∣∣∫ t0

‖∇u(s)‖L∞ ds∣∣∣∣) = 0

as required.

Therefore the notion of a strong solution seems to be a reasonably good one to work in.

3.2 Regularisation by Mollifiers

The Euler equation contains spatial derivative terms, and thus the vector fields generating the floware unbounded on Sobolev spaces (including the Lebesgue spaces). As such we cannot directlyapply Theorem 3.1 to the Euler equation. To overcome this problem, we shall initially work witha regularisation of the pressureless Euler equation.

Specifically, define the non-negative smooth function η : Rd → R by

η(x) :=

{const · exp

(− 11−|x|2

), if |x| ≤ 1

0 , otherwise

where the positive constant is chosen so that∫Rdη(x) dx = 1 .


For ε > 0 we define the smooth functionηε : Rd → R by

ηε(x) := ε−dη

(xε

),

and we define the mollifier Jε : L1loc(Rd)→ C∞(Rd) to be the convolution

Jεf := ηε ∗ f .

We will use standard properties of these mollifiers without citation; these and their proofs maybe found, for example, in [4], Chapter 5 and Appendix C. We also remind the reader (see Remark3.3) that Leray projection commutes with mollifiers.

We consider the following regularisation of the pressureless Euler equation.

∂tuε + PJε [(Jεuε) · ∇ (Jεuε)] = 0

uε(0, ·) = u0(88)

This equation arises from applying Leray projection to the pressureless Euler equation (87). Wewill prove global existence and uniqueness for the Cauchy problem (88); see Corollary 3.13 below.The hope is that one can then take the limit ε → 0 in the regularisation parameter, to recover astrong solution to the Euler equation.

To end off this section, we prove the following estimates which we will need in the sequel.

Lemma 3.7. For m, k ∈ Z≥0 and 0 < ε < 1, we have the estimates

‖Jεf‖Hm+k ≤C(d,m, k)

εk‖f‖Hm ,∥∥JεDkf∥∥L∞ ≤ C(d, k)εd/2+k ‖f‖H0 .

Proof. For f ∈ Hm(Rd) and α, β ∈ Zd≥0 with |α| ≤ k and |β| ≤ m, we have

Dα+βJεf = (Dαηε) ∗ (Dβf) .

Using the Schwarz inequality, we find

∥∥Dα+βJεf∥∥2L2 = ∫Rd

∣∣∣∣∫Rd

Dαηε(y) Dβf(x− y) dy

∣∣∣∣2 dx≤∫Rd

(∫Rd|Dαηε(y)|

∣∣Dβf(x− y)∣∣2 dy)(∫Rd|Dαηε(y)|dy

)dx

= ‖Dαηε‖2L1∥∥Dβf∥∥2

L2

so that ∥∥Dα+βJεf∥∥L2 ≤ ‖Dαηε‖L1 ∥∥Dβf∥∥L2 = ‖Dαη‖L1ε|α| ∥∥Dβf∥∥L2≤‖Dαη‖L1

εk∥∥Dβf∥∥

L2.

Summing over α and β proves the first assertion.The second assertion is even easier: For |α| ≤ k, the Schwarz inequality gives immediately

‖JεDαf‖L∞ = ‖(Dαηε) ∗ f‖L∞ ≤ ‖D

αηε‖L2 ‖f‖L2 =‖Dαη‖L2εd/2+|α|

‖f‖L2

≤‖Dαη‖L2εd/2+k

‖f‖L2

As before, sum over α to obtain the result.


Lemma 3.8. For m ∈ Z≥1, there is a constant C = C(d,m) > 0 such that

‖Jεf − f‖Hm−1 ≤ Cε ‖f‖Hm

holds for all f ∈ Hm(Rd).

Proof. We first consider the case m = 1. For θ > 0 we have

JεJθf(x)− Jθf(x) =∫B(0,ε)

ηε(z) (Jεf(x− z)− Jεf(x)) dz

=

∫B(0,ε)

ηε(z)

∫ 10

z · Jε∇f(x− sz) ds dz ,

so that, using Hölder’s inequality, we get

|JεJθf(x)− Jθf(x)| ≤ ε∫B(0,ε)

∫ 10

ηε(z) |Jε∇f(x− sz)| ds dz

= ε

[∫B(0,ε)

∫ 10

ηε(z) |Jε∇f(x− sz)|2 ds dz

] 12

.

Squaring both sides and integrating over Rd, we obtain

‖JεJθf − Jθf‖L2 ≤ ε ‖Jθ∇f‖L2 ≤ Cε ‖Jθf‖H1 .

Since θ > 0 was arbitrary, taking θ → 0 gives the assertion for m = 1.The case m > 1 is proved by applying the above argument to each of the weak derivatives of

f up to order m− 1.

3.3 Global Existence and Uniqueness for the Regularised PressurelessEquation

The goal of this subsection is to prove a global existence and uniqueness result for the regularisedpressureless equation (88). Here the term “global” refers to the solution existing for all time t ∈ R.Due to mollification, the vector fields in (88) are now locally Lipschitz, albeit with the Lipschitzconstant depending as negative powers of the regularisation parameter ε. Despite this, the Picardtheorem (Theorem 3.1) will still apply to give local existence and uniqueness. After this we canpatch these local solutions together to form a maximal solution, and we then prove that thismaximal solution is necessarily a global solution.

First we need a preliminary estimate.

Lemma 3.9. For m ∈ Z≥0 there is a constant C = C(d,m) > 0 such that for all u, v ∈ L∞ ∩Hm(Rd),

‖uv‖Hm ≤ C (‖u‖L∞ ‖v‖Hm + ‖u‖Hm ‖v‖L∞) .

Proof. The claim is trivial if m = 0, so we focus on the case m ≥ 1. Let α ∈ Zd≥0 with |α| ≤ m.If |α| = 0, then clearly

‖Dα(uv)‖L2 = ‖uv‖L2 ≤ ‖u‖L∞ ‖v‖L2 + ‖u‖L2 ‖v‖L∞ .

Otherwise, for 1 ≤ |α| ≤ m, we have, using Leibniz’s differentiation formula and Hölder’s inequal-ity,

‖Dα(uv)‖L2 ≤ C(d,m)∑β≤α

∥∥Dβu Dα−βv∥∥L2

≤ C(d,m)∑β≤α

∥∥Dβu∥∥L2|α|/|β|

∥∥Dα−βv∥∥L2|α|/|α−β|


By the Gagliardo-Nirenberg interpolation inequality,∥∥Dβu∥∥L2|α|/|β|

≤ C(d,m) ‖u‖1−|β|/|α|L∞∥∥∥D|α|u∥∥∥|β|/|α|

L2,∥∥Dα−βv∥∥

L2|α|/|α−β|≤ C(d,m) ‖v‖|β|/|α|L∞

∥∥∥D|α|v∥∥∥1−|β|/|α|L2

.

Plugging these into our estimates above and using Young’s inequality, we find, for 1 ≤ |α| ≤ m,

‖Dα(uv)‖L2 ≤ C(d,m)∑β≤α

(‖u‖L∞

∥∥∥D|α|v∥∥∥L2

)1− |β||α| (‖v‖L∞ ∥∥∥D|α|u∥∥∥L2

) |β||α|

≤ C(d,m)(‖u‖L∞

∥∥∥D|α|v∥∥∥L2

+ ‖v‖L∞∥∥∥D|α|u∥∥∥

L2

).

The result now follows by summing over all α ∈ Z≥0 with 0 ≤ |α| ≤ m.

From the preceding result, it is easy to see that we have a Banach algebra structure on Hm(Rd)whenever m > d/2. We state it as a lemma here, along with an immediate corollary regardinguniqueness of strong solutions to the Euler equation. We will make heavy use of this estimate inthe next subsection.

Lemma 3.10. For m ∈ Z,m > d/2, there exists a constant C = C(d,m) > 0 such that

‖uv‖Hm ≤ C ‖u‖Hm ‖v‖Hm for all u, v ∈ Hm(Rd).

In other words, Hm(Rd) is a Banach algebra.

Proof. This is immediate from the previous lemma and the Sobolev inequality ‖·‖L∞ ≤ C(d,m)‖·‖Hm for Hm(Rd) when m > d/2.

Theorem 3.11. Let m ∈ Z≥0 and u0 ∈ V m(R3), and let ε > 0. Then

(i) There exists a unique solution uε ∈ C1([−T, T ], V m(R3)) to (88), for some T = T (‖u0‖Hm , ε).

(ii) For T > 0 and uε ∈ C1([−T, T ], V 0(R3)) a solution to (88), we have

‖uε(t)‖H0 = ‖u0‖H0 for all t ∈ [−T, T ] .

Proof. We write (88) in the form ∂tuε = Fε(u

ε), where Fε : Vm(R3)→ V m(R3) is given by

Fε(v) := −PJε [(Jεv) · ∇ (Jεv)] .

For v1, v2 ∈ V m(R3), we have, from Lemma 3.9, the estimate

‖Fε(v1)− Fε(v2)‖Hm ≤ ‖PJε [(Jεv1) · ∇ (Jε (v1 − v2))]‖Hm+ ‖PJε [(Jε (v1 − v2)) · ∇ (Jεv2)]‖Hm

≤ C [‖Jεv1‖L∞ ‖Jε∇ (v1 − v2)‖Hm+ ‖Jεv1‖Hm ‖Jε∇ (v1 − v2)‖L∞+ ‖Jε (v1 − v2)‖L∞ ‖Jε∇v2‖Hm+ ‖Jε (v1 − v2)‖Hm ‖Jε∇v2‖L∞ ]

By Lemma 3.7 we have the bound

‖Fε(v1)− Fε(v2)‖Hm ≤C

ε5/2+m(‖v1‖H0 + ‖v2‖H0) ‖v1 − v2‖Hm (89)

where C = C(d,m) > 0.Thus Fε is locally Lipschitz, so Theorem 3.1 guarantees the existence and uniqueness of a flow

uε ∈ C1([−T, T ], V m), for some T = T (‖u0‖Hm , ε) > 0. The proof of (i) is complete.


Next, take the H0 inner product of (88) with the solution uε to obtain

1

2

d

dt

(‖uε‖2H0

)= −

∫R3uε · PJε [(Jεuε) · ∇ (Jεuε)] dx

= −∫R3

(Jεuε) · [(Jεuε) · ∇ (Jεuε)] dx

= −12

∫R3

(Jεuε) · ∇(

(Jεuε)2)

dx

=1

2

∫R3

[div (Jεuε)] (Jεuε)2 dx

= 0 .

This proves (ii).

As with ODE in normed spaces, the Picard theorem (Theorem 3.1) allows us to patch local-in-time solutions to give, for each initial data u0 ∈ Hm(R3), a unique maximal solution uε ∈C1((T−(u0), T+(u0)), V

m(R3)) to the regularised pressureless equation (88). We now show that,in fact, the maximal solution is global, i.e. (T−(u0), T+(u0)) = (−∞,∞).

For this, we need the following general result for ODE on Banach spaces.

Theorem 3.12. Let X be a Banach space and U ⊆ X be an open subset, and let F : U → X belocally Lipschitz. Suppose x ∈ C1((a, b), X) is the maximal solution to the ODE

dx

dt= F (x) , x(0) = x0 ∈ U .

If b −∞.

Proof. Assume b < ∞ and (i) does not hold; we have to prove (ii). Since (i) does not hold, wehave

M := 1 + lim supt↗b

‖F (x(t))‖X 0 such that

‖x(t)− x(s)‖X ≤∫ ts

‖F (x(τ))‖X dτ

≤M |t− s|

whenever b− δ < s ≤ t < b.This shows the existence of x(b−) := limt↗b x(t). If x(b−) ∈ U , then we could apply Theorem

3.1 to find a local-in-time solution starting from x(b−); patching this solution to x gives a solutionwhich exists beyond time b, which contradicts the maximality of x.

Corollary 3.13. The unique maximal solution u ∈ C1((a, b), V m(R3)) to the regularised pres-sureless equation (88) is global, i.e. a = −∞ and b = +∞.

Proof. Using Equation (89) with v1 = uε(t), v2 = 0, we find

d

dt

(‖uε(t)‖2Hm

)≤ 2 ‖uε(t)‖Hm ‖Fε (u

ε(t))‖Hm

≤ C(d,m)ε5/2+m

‖u0‖H0 ‖uε(t)‖2Hm .


By Gronwall’s inequality,

‖uε(t)‖2Hm ≤ ‖u0‖2Hm exp

(C(d,m)

ε5/2+m‖u0‖H0 |t|

)for all t ≥ 0.

By solving (88) we obtain a similar estimate for negative times. Hence, in fact,

‖uε(t)‖2Hm ≤ ‖u0‖2Hm exp

(C(d,m)

ε5/2+m‖u0‖H0 |t|

)for all t ∈ R.

Now we apply Theorem 3.12 with X = V m(R3) and U = X; since neither of the conclusions holdfor any finite b > 0, we must have b = +∞. A similar argument gives a = −∞.

3.4 Local Existence to the Euler Equation

Thus far we have proved global existence and uniqueness to the regularised pressureless equation(88) for each ε > 0. Much of this subsection will be devoted to showing that we can take the limitε→ 0.

Lemma 3.14. For m ∈ Z≥1 there is a constant C = C(d,m) > 0 such that for all u, v ∈L∞ ∩Hm(Rd),∑

|α|≤m

‖Dα(uv)− uDαv‖L2 ≤ C (‖∇u‖L∞ ‖v‖Hm−1 + ‖u‖Hm ‖v‖L∞) .

Proof. Observe that∑|α|≤m

‖Dα(uv)− uDαv‖L2 ≤ C(d,m)∑

|β|≤m−1

∥∥Dβ ((∇u)v)∥∥L2

≤ C(d,m) ‖(∇u)v‖Hm−1 .

By Lemma 3.9,

‖(∇u)v‖Hm−1 ≤ C(d,m) (‖∇u‖L∞ ‖v‖Hm−1 + ‖∇u‖Hm−1 ‖v‖L∞)≤ C(d,m) (‖∇u‖L∞ ‖v‖Hm−1 + ‖u‖Hm ‖v‖L∞) .

This completes the proof.

Lemma 3.15. Let m ∈ Z≥1 and u0 ∈ V m(R3). Then the unique solution uε ∈ C1(R, V m(R3)) to(88) satisfies the estimate

1

2

d

dt

(‖uε‖2Hm

)≤ C ‖Jε∇uε‖L∞ ‖u

ε‖2Hm

for some constant C = C(m) > 0.

Proof. For α ∈ Z3≥0, |α| ≤ m, we take the α-th derivative of (88), and then take the inner productwith Dαuε. This gives

1

2

d

dt

(‖Dαuε‖2L2

)= − (Dαuε,DαPJε [(Jεuε) · ∇ (Jεuε)])L2

=− (DαJεuε,Dα [(Jεuε) · ∇ (Jεuε)])L2=− (DαJεuε,Dα [(Jεuε) · ∇ (Jεuε)]− (Jεuε) · ∇ (DαJεuε))L2− (DαJεuε, (Jεuε) · ∇ (DαJεuε))L2

The last term on the right vanishes due to the divergence theorem:

(DαJεuε, (Jεuε) · ∇ (DαJεuε))L2 =∫R3

1

2(Jεuε) · ∇

[(Jεuε)2

]dx

= −∫R3

1

2[div (Jεuε)] (Jεuε)2 dx = 0 .


Now, summing our estimate over α and applying Lemma 3.14 gives

1

2

d

dt

(‖uε‖2Hm

)≤ ‖uε‖Hm

∑|α|≤m

‖Dα [(Jεuε) · ∇ (Jεuε)]− (Jεuε) · ∇ (DαJεuε)‖L2

≤ C(m) ‖uε‖Hm (‖∇Jεuε‖L∞ ‖∇Jεu

ε‖Hm−1 + ‖Jεuε‖Hm ‖∇Jεu

ε‖L∞)

≤ C(m) ‖Jε∇uε‖L∞ ‖uε‖2Hm

as required.

Proposition 3.16. Let m ∈ Z with m ≥ 3. Then the solution uε to (88) with initial datau0 ∈ V m(R3) satisfies

‖uε(t)‖Hm ≤‖u0‖Hm

1− C1|t| ‖u0‖Hmfor |t| < 1

C1 ‖u0‖Hm

where C1 = C1(m) > 0 is any constant satisfying Lemma 3.15.

Proof. We first consider the case t ≥ 0. Since m ≥ 3, Sobolev embedding gives

‖Jε∇uε‖L∞ ≤ ‖∇uε‖L∞ ≤ ‖∇u

ε‖Hm−1 ≤ ‖uε‖Hm .

Thus Lemma 3.15 gives, for any θ > 0,

1

2

d

dt

(‖uε‖2Hm

)≤ C1 ‖uε‖3Hm ≤ C1

(θ + ‖uε‖2Hm

) 32

.

Therefore, for 0 ≤ t < 1/(C1√θ + ‖u0‖2Hm), we have, by direct integration,

1√θ + ‖u0‖2Hm

− 1√θ + ‖uε(t)‖2Hm

≤ C1t ,

so that √θ + ‖uε(t)‖2Hm ≤

√θ + ‖uε(t)‖2Hm

1− C1t√θ + ‖uε(t)‖2Hm

.

Since θ > 0 was arbitrary, the limit θ → 0 gives the assertion for t ≥ 0. The case t < 0 is provedsimilarly by applying the above argument to the backwards solution of (88).

Proposition 3.17. Fix initial data u0 ∈ V 3(R3). Then the family {uε}0 2 is some constant depending onour choice of C1, but not on ε, θ or t.

Proof. First suppose t ≥ 0. From (88) it is immediate that

d

dt

(∥∥uε − uθ∥∥2H0

)= 2

(Jε [(Jεuε) · ∇ (Jεuε)]− Jθ

[(Jθuθ

)· ∇(Jθuθ

)], uε − uθ

)L2

.


Now,

Jε [ (Jεuε) · ∇ (Jεuε)]− Jθ[(Jθuθ

)· ∇(Jθuθ

)]= (Jε − Jθ) [(Jεuε) · ∇ (Jεuε)] + Jθ [((Jε − Jθ)uε) · ∇ (Jεuε)]

+ Jθ[(Jθ(uε − uθ

))· ∇ (Jεuε)

]+ Jθ

[(Jθuθ

)· ∇ ((Jε − Jθ)uε)

]+ Jθ

[(Jθuθ

)· ∇(J θ(uε − uθ

))]=: A1(t) +A2(t) +A3(t) +A4(t) +A5(t) .

Using Lemma 3.8, Lemma 3.9, Sobolev embedding and Proposition 3.16 in this order, we find

‖A1(t)‖L2 ≤ C(ε+ θ) ‖(Jεuε) · ∇ (Jεuε)‖H1

≤ C(ε+ θ) (‖uε‖L∞ ‖∇uε‖H1 + ‖u

ε‖H1 ‖∇uε‖L∞)

≤ C(ε+ θ) ‖uε‖2H3

≤ C(ε+ θ)(

‖u0‖H31− C1t ‖u0‖H3

)2.

Similarly we can prove the following estimates for A2 and A4,

‖A2(t)‖L2 ≤ C(ε+ θ)(

‖u0‖H31− C1t ‖u0‖H3

)2,

‖A4(t)‖L2 ≤ C(ε+ θ)(

‖u0‖H31− C1t ‖u0‖H3

)2.

For A3, we have, by Sobolev embedding and Proposition 3.16,

‖A3(t)‖L2 ≤ C ‖∇Jεuε‖L∞

∥∥uε − uθ∥∥L2≤ C ‖uε‖H3

∥∥uε − uθ∥∥L2

≤ C‖u0‖H3

1− C1t ‖u0‖H3∥∥uε − uθ∥∥

L2.

Finally, (A5(t), u

ε − uθ)L2

=1

2

∫R3

(Jθuθ

)· ∇[(Jθuθ

)2]dx

= −12

∫R3

[div(Jθuθ

)] (Jθuθ

)2dx = 0 .

Putting the above together, there is a constant C2 > 2, independent of ε, θ or t, such that

d

dt

(∥∥uε(t)− uθ(t)∥∥2H0

)≤ C1C2(ε+ θ)

(‖u0‖H3

1− C1t ‖u0‖H3

)2 ∥∥uε(t)− uθ(t)∥∥H0

+ C1C2‖u0‖H3

1− C1t ‖u0‖H3∥∥uε(t)− uθ(t)∥∥2

H0.

Thus, for any κ > 0, we have

d

dt

([∥∥uε(t)− uθ(t)∥∥2H0

+ κ] 1

2

(1− C1t ‖u0‖H3)C2

)≤ C1C2(ε+ θ) ‖u0‖2H3 (1− C1t ‖u0‖H3)

C2−2 ≤ C1C2(ε+ θ) ‖u0‖2H3 .

Directly integrating, we find[∥∥uε(t)− uθ(t)∥∥2H0

+ κ] 1

2

(1− C1t ‖u0‖H3)C2 ≤

√κ+ C1C2(ε+ θ) ‖u0‖2H3 t .

The desired result for t ≥ 0 is obtained by taking the limit κ → 0. For the case t < 0, apply theabove argument to the backwards solution of (88).


In fact, the bound in Proposition 3.17 will provide bounds in higher Sobolev spaces as well,thanks to the following interpolation inequality for Sobolev spaces. With these bounds we willobtain a local existence result for strong solutions to the Euler equation.

Proposition 3.18. Given s > 0, there exists a constant C = C(d, s) > 0 such that

‖f‖Hr ≤ C ‖f‖1−r/sH0 ‖f‖

r/sHs for all f ∈ H

s(Rd)

whenever 0 < r < s.

This inequality is a special case of [1], Theorem 5.2. We refer the reader there for details of theproof.

Theorem 3.19. Let m ∈ Z with m ≥ 4, and suppose u0 ∈ V m(R3). Let T1 be as in Proposition3.17.

Then there exists a strong solution u in V m−1(R3) to the Euler equation up to time T1, suchthat uε(t)→ u(t) in V m−1(R3), locally uniformly in time t, where uε is the unique global solutionto the regularised equation (88).

Proof. Let T ∈ (0, T1). By Propositions 3.18 and 3.16 respectively, we have, for any t ∈ [−T, T ],∥∥uε(t)− uθ(t)∥∥Hm−1

≤ C(m)∥∥uε(t)− uθ(t)∥∥ 1m

H0

∥∥uε(t)− uθ(t)∥∥1− 1mHm

≤ C (m, ‖u0‖Hm , T )∥∥uε(t)− uθ(t)∥∥1/m

H0

Using Proposition 3.17,

supt∈[−T,T ]

∥∥uε(t)− uθ(t)∥∥Hm−1

≤ C (m, ‖u0‖Hm , T ) (ε+ θ)1m .

Therefore, for any sequence εk ↘ 0 of regularisation parameters, {uεk}k∈N is a Cauchy sequence inthe Banach space C0([−T, T ], V m−1(R3)), and thus has a limit u ∈ C0([−T, T ], V m−1(R3)). It iseasy to see that this limit u does not depend on the choice of the sequence {εk}k∈N of regularisationparameters, i.e. uε → u in C0([−T, T ], V m−1(R3)) as ε→ 0.

Since T ∈ (0, T1) was arbitrary, we see that actually u ∈ C0((−T1, T1), V m−1(R3)) and uε → uin the topology of local uniform convergence of C0((−T1, T1), V m−1(R3)).

Next, using Lemma 3.10,

‖uε · ∇uε − u · ∇u‖Hm−2 ≤ ‖(uε − u) · ∇uε‖Hm−2 + ‖u · ∇ (u

ε − u)‖Hm−2≤ C (‖uε − u‖Hm−2 ‖∇u

ε‖Hm−2+ ‖u‖Hm−2 ‖∇ (u

ε − u)‖Hm−2)≤ C (‖uε‖Hm + ‖u‖Hm) ‖u

ε − u‖Hm−1

so that Proposition 3.16 gives

‖uε · ∇uε − u · ∇u‖Hm−2 ≤C (m, ‖u0‖Hm)

T1 − |t|‖uε − u‖Hm−1 .

Similarly, using Lemma 3.10 (this is where we need m ≥ 4) and Lemmas 3.7 and 3.8, we have

‖ (Jεuε) · ∇ (Jεuε)− uε · ∇uε‖Hm−2≤ C (‖(Jε − uε) · ∇ (Jεuε)‖Hm−2 + ‖u

ε · ∇ (Jεuε − uε)‖Hm−2)≤ C (‖Jεuε − uε‖Hm−2 ‖∇Jεu

ε‖Hm−2 + ‖uε‖Hm−2 ‖∇ (Jεu

ε − uε)‖Hm−2)≤ C ‖uε‖Hm ‖Jεu

ε − uε‖Hm−1≤ Cε ‖uε‖2Hm

so that Proposition 3.16 gives

‖(Jεuε) · ∇ (Jεuε)− uε · ∇uε‖Hm−2 ≤C (m, ‖u0‖Hm)

(T1 − |t|)2ε .


Finally, bearing in mind Lemma 3.8 and using Lemma 3.10 and Proposition 3.16, we find

‖Jε [(Jεuε) · ∇ (Jεuε)]− (Jεuε) · ∇ (Jεuε)‖Hm−2≤ Cε ‖(Jεuε) · ∇ (Jεuε)‖Hm−1≤ Cε (‖Jεuε‖Hm−1 ‖∇ (Jεu

ε)‖Hm−1)

≤ Cε ‖Jεuε‖2Hm

≤C (m, ‖u0‖Hm)

(T1 − |t|)2ε .

Therefore we have shown, with C = C(m, ‖u0‖Hm) > 0,

‖Jε [(Jεuε) · ∇ (Jεuε)]− u · ∇u‖Hm−2 ≤ C

(ε

(T1 − |t|)2+‖uε − u‖Hm−1

T1 − |t|

)

By the same argument as above, this shows that (Jεuε) · ∇(Jεuε) → u · ∇u in the topology oflocal uniform convergence of C0((−T1, T1), V m−2(R3)).

Since uε solves the regularised pressureless Euler equation (88), we have

uε(t)− u0 = −∫ t0

PJε [(Jεuε(s)) · ∇ (Jεuε(s))] ds , for t ∈ (−T1, T1)

as a Bochner integral (see [10], Section V.5 for details) in V m−2(R3). In the limit ε → 0, localuniform convergence gives

u(t)− u0 = −∫ t0

P [u(s) · ∇u(s)] ds , for t ∈ (−T1, T1)

in V m−2(R3). This shows that u ∈ C1((−T1, T1), V m−2(R3)), and

∂tu = −P [u · ∇u] , u(0) = u0 , in V m−2(R3) .

Hence u is a strong solution to the Euler equation.

Although the previous result was by no means trivial, there is one aspect in which it is ratherunsatisfactory: we have required the initial data to be in V m(R3), to prove existence of a weaksolution in V m−1(R3). It would have been desirable to prove an existence result such that u(t) ∈V m(R3).

The loss of spatial derivatives has its roots back in our definition of a strong solution: we haverequired that a strong solution should be continuous in time. It turns out that, for initial datain V m(R3), the solution constructed in the preceding theorem does indeed describe a possiblydiscontinuous path in V m(R3). We have the following result.

Theorem 3.20. Let m ∈ Z with m ≥ 4, and suppose u0 ∈ V m(R3). Let u be the strong solutionin V m−1(R3) to the Euler equation up to time T1 > 0 constructed in Theorem 3.19. Then

u ∈ L∞loc((−T1, T1), Hm(R3)) ∩ Liploc((−T1, T1), Hm−1(R3)) .

Moreover,C0((−T1, T1), Hmw (R3)) .

Proof. Fix T ∈ (0, T1) and let t ∈ [−T, T ]. Since {uε(t)}0


This shows that vt = u(t), so u(t) ∈ Hm(R3) and uε′k(t) ⇀ u(t). Since both t and the sequence

{εk}k∈N are arbitrary, we have uε(t) ⇀ u(t) in Hm(R3) as ε → 0, for each t ∈ [−T, T ]. ByProposition 3.16,

‖u(t)‖Hm ≤ lim infε→0 ‖uε(t)‖Hm ≤

‖u0‖Hm1− C1T ‖u0‖Hm

.

Hence u ∈ L∞([−T, T ], Hm(R3)).Next from ∂tu = −P(u · ∇u), we have, by Lemma 3.10 and Proposition 3.16,∥∥∥∥∂u∂t

∥∥∥∥Hm−1

≤ C ‖u‖Hm−1 ‖∇u‖Hm−1 ≤ C ‖u‖2Hm ≤ C (‖u0‖Hm , T ) .

Integrating, we find, for 0 ≤ s < t ≤ T , that

‖u(t)− u(s)‖Hm−1 ≤∫ ts

∥∥∥∥∂u∂t (τ)∥∥∥∥Hm−1

dτ ≤ C (‖u0‖Hm , T ) (t− s) .

This shows that u ∈ Lip([−T, T ], Hm−1(R3)).Finally, note that for any ψ ∈ H−(m−1)(R3), we have

|〈ψ, u(t)− u(s)〉| ≤ ‖ψ‖H−(m−1) ‖u(t)− u(s)‖Hm−1≤ C (‖u0‖Hm , T ) ‖ψ‖H−(m−1) |t− s| .

Therefore 〈ψ, u(s)〉 → 〈ψ, u(t)〉 as s → t. Since H−(m−1)(R3) is dense in H−m(R3), and u ∈L∞([−T, T ], Hm(R3)), we deduce that 〈ψ, u(s)〉 → 〈ψ, u(t)〉 as s → t, for each ψ ∈ Hm(R3), i.e.u(s) ⇀ u(t) as s→ t. Hence we conclude that u ∈ C0([−T, T ], Hmw (R3)).

We finally arrive at the main result of this section.

Theorem 3.21. Let m ∈ Z with m ≥ 4, and suppose u0 ∈ V m(R3). Let u be the strong solutionin V m−1(R3) to the Euler equation up to time T1 > 0 constructed in Theorem 3.19. Then

u ∈ C0((−T1, T1), V m(R3)) ∩ C1((−T1, T1), V m−1(R3)) .

In other words, u is a strong solution up to time T1.

Proof. Since V m(R3) is a Hilbert space, it is uniformly convex. In Theorem 3.20 we have shownthat u ∈ C0((−T1, T1), Hmw (R3)). If we could show that t 7→ ‖u(t)‖Hm is upper-semicontinuous,i.e.

lim sups→0

‖u(t+ s)‖Hm ≤ ‖u(t)‖Hm for t ∈ (−T1, T1) ,

then from uniform convexity (see [3], Proposition 3.32) it will follow that u is continuous as a mapu : (−T1, T1)→ V m(R3).

For this, fix t ∈ (−T1, T1), and let vε solve the regularised pressureless Euler equation with ini-tial data vε(0) = u(t). As before, we have vε(t) ⇀ v(t) in Hm(R3), with v ∈ L∞((−δ, δ), V m(R3))being a strong solution of the Euler equation in V m−1(R3), up to some time δ > 0. Local unique-ness of the strong solution (Theorem 3.6) guarantees that v(s) = u(t+ s). By Proposition 3.16,

‖u(t+ s)‖Hm ≤ lim infε→0 ‖vε(s)‖Hm ≤

‖u(t)‖Hm1− C1 |s| ‖u(t)‖Hm

for |s| < δ .

Hence we indeed havelim sups→0

‖u(t+ s)‖Hm ≤ ‖u(t)‖Hm

as required.

4 VORTICITY AND CONTINUATION Page 28 of 31

4 Vorticity and Continuation

In this short section our goal is to study the way(s) in which solutions to the Cauchy problem forthe Euler equation can fail to be global, i.e. fail to exist for all time. It turns out that, for theEuler equation, the failure of global existence is characterised by the rapid growth of the vorticity.Specifically, we have the following result.

Theorem 4.1. Let m ∈ Z with m ≥ 4, and suppose u is a strong solution to the Euler equationin V m(R3) up to time T∗ > 0, and that u cannot be continued beyond time T∗, i.e. there does notexist

ũ ∈ C0([0, T ), V m(R3) ∩ C1([0, T ), V m−1(R3))

with T > T∗, which extends u.Then necessarily ∫ T∗

0

‖ω(t)‖L∞ dt =∞ .

where ω(t) := rot u(t) is the vorticity of the velocity field u.

This section is devoted to the proof of Theorem 4.1. The presentation follows that of [2]. A keytool here is the Biot-Savart law, which allows us to recover the (divergence-free) velocity field fromits vorticity. We need the following facts from potential theory.

Lemma 4.2. Let f : R3 → R be bounded and locally Hölder continuous. Then its Newtonianpotential

Φ[f ](x) := − 14π

∫R3

f(y)

|x− y|dy

belongs to C2(R3) and satisfies

∇Φ[f ](x) = 14π

∫R3

x− y|x− y|3

f(y) dy

and 4Φ[f ] = f .

This result is fairly standard and is a special case of well-known results, so we omit the proof andrefer the interested reader to, for example, [5], Lemmas 4.1 and 4.2.

Proposition 4.3 (Biot-Savart Law). Let ω ∈ V 2(R3). Then there exists a unique u ∈ V 3(R3)with rot u = ω. Moreover we have the exact formula

u(x) = − 14π

∫R3

x− y|x− y|3

∧ ω(y) dy .

Proof. We deal with uniqueness first. Consider the Fourier transforms û and ω̂. The conditiondiv u = 0 forces ξ · û(ξ) = 0, and the requirement rot u = ω forces iξ ∧ û(ξ) = ω̂(ξ). From this itfollows that

û(ξ) = iξ ∧ ω̂(ξ)|ξ|2

.

This proves uniqueness. For existence we can simply define û by the preceding formula, and checkthat it works.

It remains to prove the representation formula. Let Ψ : R3 → R3 be the Newtonian potentialof ω. Since the Sobolev embedding H2(R3) ↪→ C0,1/2b (R3) holds, Lemma 4.2 applies, and we havediv Ψ = 0, 4Ψ = ω, and

−rot Ψ(x) = − 14π

∫R3

x− y|x− y|3

∧ ω(y) dy .

Now 4Ψ = ∇(div Ψ)− rot rot Ψ, so we identify the solution u = −rot Ψ.


Corollary 4.4. There exists a constant C > 0 independent of u, such that

‖∇u‖L2 ≤ C ‖rot u‖L2 for every u ∈ V3(R3).

Proof. From the previous proof, the Fourier transforms of Diuj and ω := rot u are related by

F [Diuj ] (ξ) = −ξi(ξ ∧ ω̂(ξ))j|ξ|2

.

That is, F [Du] (ξ) = S(ξ)ω̂(ξ) for a matrix-valued function S : R3 → R3×3, which is boundedindependently of ξ ∈ R3. The result now follows.

Using the formula given by the Biot-Savart law, we have the following estimate, which will beessential to the proof of Theorem 4.1.

Lemma 4.5. There is a constant C > 0 such that any u ∈ V 3(R3) satisfies

‖∇u‖L∞ ≤ C[1 + ‖rot u‖L2 +

(1 + log+ ‖u‖H3

)‖rot u‖L∞

]where log+ := max(0, log).

Proof. For convenience denote ω := rot u. We write the Biot-Savart law as

u(x) = − 14π

∫R3

x− y|x− y|3

∧ ω(y) dy =:∫R3K(x− y)ω(y) dy

with K a matrix valued function R3 → R3×3.For ρ ∈ (0, 1], pick a smooth cut-off function ζρ such that

ζρ ≡ 1 on B(0, ρ) , ζρ ≡ 0 on R3 \ B(0, 2ρ) , ‖∇ζρ‖L∞ ≤C

ρ

where C is a constant independent of ρ. We write

∇u(x) =∫R3K(x− y)∇ω(y) dy

=

∫{|x−y|≤2ρ}

ζρ(x− y)K(x− y)∇ω(y) dy

+

∫{ρ


To estimate I3, note |∇K| ≤ C| · |−3 implies that |∇K| ∈ L2(R3 \ B(0, 1)), so

|I3| ≤ C ‖ω‖L2 .

Putting the above estimates together, we find

‖∇u‖L∞ ≤ C[ρ

14 ‖u‖H3 + ‖ω‖L2 + (1− log ρ) ‖ω‖L∞

].

Now we choose ρ as follows: If ‖u‖H3 ≤ 1, set ρ = 1; otherwise set ρ = ‖u‖−4H3 . In either case weobtain the asserted bound.

Proof of Theorem 4.1. Assume for a contradiction that T∗

REFERENCES Page 31 of 31

References

[1] R. A. Adams, J. J. F. Fournier. Sobolev Spaces (2nd Edition). Academic Press: Pure andApplied Mathematics (2003).

[2] J. T. Beale, T. Kato, A. Majda. Remarks on the Breakdown of Smooth Solutions for the 3-DEuler Equations. Comm. Math. Phys. 94, 61-66 (1984).

[3] H. Brezis. Functional Analysis, Sobolev Spaces and Partial Differential Equations. Springer:Universitext (2011).

[4] L. C. Evans. Partial Differential Equations (2nd Edition). American Mathematical Society:Graduate Studies in Mathematics (2010).

[5] D. Gilbarg, N. S. Trudinger: Elliptic Partial Differential Equations of Second Order (2ndedition, revised 3rd printing). Springer: Classics in Mathematics (1998).

[6] A. Majda. Compressible Fluid Flow and Systems of Conservation Laws in Several SpaceVariables. Springer-Verlag New York Inc. (1984)

[7] A. J. Majda, A. L. Bertozzi. Vorticity and Incompressible Flow. Cambridge University Press:Cambridge Texts in Applied Mathematics (2002).

[8] J. Moser. A rapidly convergent iteration method and nonlinear differential equations. Ann.Scuola Norm. Sup. Pisa 20, 499-535 (1966).

[9] S. Klainerman and A. Majda. Singular Limits of Quasilinear Hyperbolic Systems with LargeParameters and the Incompressible Limit of Compressible Fluids. Comm. Pure Appl. Math.34, 481-524 (1981).

[10] K. Yosida. Functional Analysis (6th Edition). Springer: Grundlehren der mathematischenWissenschaften (1980).

[11] R. Courant and D. Hilbert Methods of Mathematical Physics Volume II Interscience Publish-ers (1966).

IntroductionPhysical BackgroundNotation and Conventions

Solutions of the Euler EquationEquations of Compressible Fluid FlowExistence in Theorem 2.2Constructive Existence for the Incompressible Euler Equations

The Cauchy Problem - A Direct ApproachRemoval of the Pressure TermRegularisation by MollifiersGlobal Existence and Uniqueness for the Regularised Pressureless EquationLocal Existence to the Euler Equation

Vorticity and Continuation

existence and continuation for euler equations · 2014. 3. 24. · 3.compressible euler =...

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