In this section we conclude the proof of Theorem2.4by using the Baire category method.
5.1. The Baire Category Method
We introduce the notion of subsolution associated with (3.5), (3.6). The set of constraints K(x,t) is understood with respect to a from now on fixed bounded functione:×(0,T)→R+with(x,t)→e
x−12gt2en,t
being continuous on an open setU ⊂D. Furthermore, for simplicity of notation, in this subsection we will, as in the proof of Lemma4.10, denotey:=(x,t).
Definition 5.1.We say thatz:D→ Zis a subsolution of (3.5) associated with the set of constraintsKy, iff it is a weak solution of (3.5) in the sense of Definition3.1 inD,π(z)is continuous inU,z(y)∈ Kyholds for almost everyy∈D\U and
z(y)∈Uy =intKcoy for anyy∈U. (5.1) We have the following convex integration result:
Theorem 5.2.Suppose that there exists a subsolution z0 in the sense of Defini-tion5.1. Then there exist infinitely many weak solutions z:D→ Z of (3.5)which coincide with z0almost everywhere inD\U, satisfy z(y)∈Kyalmost everywhere inD, and for every open ball B⊂U the solutions satisfy the mixing property
Bμ+−μ(x,t)d(x,t)·
Bμ(x,t)−μ−d(x,t) >0. (5.2) Furthermore, among these weak solutions there exists a sequence{zk}k1such that π(zk)converges weakly toπ(z0)in L2(U;π(Z)).
The proof is similar to those in [12,33], the only main difference being that one has to carefully track the role of the projectionπ. However, since the existence of the pressure is implicit in Definition3.1due to the use of divergence-free test functions, this can be done without any serious difficulty.
The main building block of the proof is the following perturbation lemma.
Lemma 5.3.Suppose that there exists a subsolution z such that
U d(π(z(y)), π(Ky))2dy=ε >0.
Then there existδ=δ(ε) >0and a sequence of subsolutions{zk}k0such that
•zk =z inD\U, for any k0,
•
U |π(zk(y)−z(y))|2dyδ,for any k0,
•π(zk) π(z)in L2(U;π(Z))as k→ +∞.
To prove Lemma5.3, we will use the following result which can be found together with its proof as Lemma 2.1 in [12].
Lemma 5.4.Let K ⊂Rnbe a compact set. Then for any compact set C ⊂intKco there existsε >0such that for any compact set K⊂Rnwith dH(K,K) < εwe have C ⊂int(K)co.
Proof of Lemma5.3. Fixy∈U. From Lemma4.9it follows that there exists some C>0 independent ofyandz, and somez(y)¯ ∈such that
[z(y)− ¯z(y),z(y)+ ¯z(y)] ⊂Uy, |π(¯z(y))|Cd(π(z(y)), π(Ky)).
Now Lemma4.10, the continuity ofπ(z)and Lemma5.4applied to the projected sets imply that there existr(y),R(y) >0 such that
[z(y)− ¯z(y),z(y)+ ¯z(y)] +BR(y)(0)⊂Uy, d(π(z(y)), π(Ky))2d(π(z(y)), π(Ky)), for anyy∈Br(y)(y).
Using Lemma4.1, we find a sequence{zy,N}N0⊂Cc∞(B1(0))solving (3.5) such that
• zy,N(y)∈ [−¯z(y),z(y)] +¯ BR(y)(0)for ally∈B1(0),N 0,
• zy,N 0 inL2,
•
|π(zy,N)|2dy˜C|π(¯z(y))|2for allN 0.
From here on the proof is the same as Step 2 of the proof of Lemma 2.4 from [12], using a standard covering argument, therefore the details are left to the reader.
Proof of Theorem5.2. Let X0=
z∈L2(D;π(Z))such thatz=π(z)for some subsolutionzin the sense of Definition5.1satisfyingz=z0onD\U},
and X denote the closure of X0 with respect to the weak L2 topology. From Lemma4.10and the boundedness of the functioneit follows thatX0is bounded, thereforeXis metrizable, denote its metric bydX(·,·). Also since the existence of the pressure is implicit in Definition3.1due to the use of divergence-free test func-tions, it follows that for anyz ∈ X there exists a possibly distributional pressure qsuch that(z,q)is indeed a weak solution of (3.5).
We observe that the functionalI(z)=
U |z|2dyis a Baire-1 function onX. Indeed, setting
Ij(z)=
{y∈U:d(y,∂U)>1/j}|(z∗χj)(y)|2dy,
whereχj ∈Cc∞(B1/j(0))is the standard mollifying sequence, one obtains thatIj
is continuous onXand thatIj(z)→ I(z)as j → +∞.
It follows from the Baire category theorem that the set Y = {z∈ X: I is continuous atz} is residual inX. We claim that for anyz∈Y it follows that
J(z):=
U d(z(y), π(Ky))2dy=0.
Suppose the contrary. ThenJ(z)=ε >0 for somez∈Y, and letzj ∈ X0be a sequence which converges tozwith respect todX. Since I is continuous atz, it follows thatzj → zstrongly inL2(U;π(Z)). Note thatJ is continuous with respect to the strongL2-topology. Therefore we may assume thatJ(zj) > ε/2 for allzj.
Sincezj ∈ X0, there exists somezj :D → Z which is a subsolution in the sense of Definition 5.1and such that zj = π(zj).We may then apply Lemma 5.3 to deduce that there exists δ = δ(ε) > 0 and a subsolution z˜j such that
U |π(zj(y)− ˜zj(y))|2dy δandπ(zj− ˜zj) 0 weakly inL2. Sincezj = π(zj) → z and z ∈ Y, we conclude as before π(˜zj) → z strongly in L2 contradicting the fact thatπ(˜zj)andzj are uniformly bounded away from each other. We thus have showed that the set of solutions J−1(0)is residual inX.
The proof of the mixing property (5.2) follows by another application of the Baire category theorem and is exactly the same as in [6]. For convenience we briefly present it here as well. LetBbe an open ball contained inU. The set
XμB±=
z∈X :
Bμ±−μ(x,t)d(x,t)=0
is closed inXand has empty interior, sinceXμB±⊂X\X0. ThereforeJ−1(0)\XμB± is residual inX, as isJ−1(0)\
i
XμB+
i ∪XμB−
i
for any countable union of balls Bi ⊂U. By taking all balls(Bi)i∈N ⊂U with rational centers and radii we can
conclude the statement.
5.2. Conclusion
In order to prove our convex integration result for (1.1) we apply a transforma-tion similar to (3.4) to the differential inclusion (3.5), (3.6) and in particular also its relaxation. Recall from Section3that for a bounded domain⊂RnandT >0 we definedD=
(y,t)∈Rn×(0,T):y−12gt2en ∈ .
Now letz = (μ, w,m, σ,q)be a weak solution of (3.5) with some suitable initial data. Defining againy:=x+12gt2en, as well as
ρ(x,t)=μ (y,t) , v(x,t)=w (y,t)−gt en, u(x,t)=m(y,t)−μ (y,t)gt en, P(x,t)=q(y,t)+gt1
n(gtμ(y,t)−2mn(y,t)) , S(x,t)=σ (y,t)−gt(m(y,t)⊗en+en⊗m(y,t))
+g2t2μ(y,t)en⊗en−
gt1
n(gtμ(y,t)−2mn(y,t))
id, (5.3)
one obtains through lenghty but straightforward calculations that(ρ, v,u,S,P)is a weak solution of (2.1) with the same initial data. Also here the transformation can be inverted in an obvious way, mapping a solution of (2.1) to a solution of (3.5).
Furthermore, for a given functione:× [0,T)→R+the conditionz(y,t)∈ K(y,t) for y = x+ 12gt2,(x,t) ∈ ×(0,T)and with K(y,t) defined in (3.6) translates to(ρ, v,u,S,P)(x,t)∈K(x,t)withK(x,t)defined in (2.2). Similarly, if we defineU(y,t)to be the interior of the convex hull ofK(y,t)then by Proposition4.2 the conditionz(y,t)∈U(y,t)translates to(ρ, v,u,S,P)(x,t)∈U(x,t)withU(x,t)
defined in (2.3),(2.4). Since the transformation is an affine bijection, we also see thatU(x,t)is the interior of the convex hull ofK(x,t).
We have now all pieces together to prove our main result.
Proof of Theorem2.4. Letz=(ρ, v,u,S,P):×(0,T)→Z be a subsolution (in the sense of Definition2.3) of (1.1) associated withe : × [0,T) → R+ bounded and initial data(ρ0, v0)∈L∞()×L2(;Rn)satisfying (1.2). We also define the transformed mixing zone
U=
(y,t)∈Rn×(0,T):
y−1 2gt2en,t
∈U .
The inverse of the transformation (5.3) applied toz gives us a weak solution of (3.5) (in the sense of Definition3.1) which we callz=(μ, w,m, σ,q):D →Z. By the discussion of this section and Definition2.3,π(z)is continuous onU, z(y,t)∈ U(y,t) = intK(coy,t)for all(y,t)∈ U andz(y,t) ∈ K(y,t) for almost every(y,t)∈D\U.
In other wordszis a subsolution of the differential inclusion (3.5), (3.6) in the sense of Definition5.1(with mixing zoneU). Theorem5.2therefore provides us with infinitely many solutions of our differential inclusion (3.5), (3.6) which outside ofUagree withzand insideUsatisfy the mixing property (5.2), as well as with a sequence of solutions such thatπ(zk)convergesL2-weakly toπ(z).
One may then transfer these conclusions to the setting of Theorem2.4 via Lemma3.2.
Let us now briefly explain how to establish the admissibility of the obtained solutions, provided thatπ(z)is in addition of classC0([0,T];L2(;π(Z))). As before letzbe the corresponding transformed subsolution defined onD. Due to an improvement of the Tartar framework as in [7,17] one can show that the induced sequence{π(zk)}k∈Nnot only converges weakly inL2(D)toπ(z), but weakly on every time-sliceD(t)uniformly int ∈ [0,T]. It is in fact straightforward but quite lengthy to adapt the proof from [17] to our situation, therefore we omit the details, cf. also [7] and in particular Remark 2.3 therein. Transformingzkagain tozkwe conclude that the associated energies
Ek(x,t):= n
2e(x,t)−gt en·uk(x,t)−1
2ρk(x,t)g2t2+ρk(x,t)gxn
satisfy
Ek(x,t)dx→
Esub(x,t)dx
uniformly int ∈ [0,T]ask→ ∞, recall the definitions (2.5), (2.7).
However this does not yet allow us to conclude the admissibility of the induced solutions, since the difference
ε(t):=
Esub(x,0)−Esub(x,t)dx>0, t ∈(0,T)
goes to 0 as t 0. Nonetheless, similarly to [7, Definition 2.4] (but a lot less technical for our purposes) we can extend the definiton of the spaceX0, such that the sequence (or any solution obtained by the convex integration scheme) satisfies
gt en·(u(x,t)−uk(x,t))+ 1
2g2t2−gxn
(ρ(x,t)−ρk(x,t))dx ε(t),
for allt ∈ [0,T], k0. The statement follows.
6. Subsolutions
We now turn to the construction of Rayleigh–Taylor subsolutions. We start by observing that the relaxation inside the mixing zoneU ⊂ ×(0,T)given in Definition2.3can be equivalently rewritten (in the spirit of [6]) as the system
∂t(ρv+ f)+divS+ ∇P= −ρgen, divv=0,
∂tρ+div(ρv+ f)=0,
(6.1)
where
f := ρ+−ρ ρ+−ρ−
ne
ρ+(ρ−ρ−)ξ+ ρ−ρ−
ρ+−ρ− ne
ρ−(ρ+−ρ)η, for some functionsξ, η:U →Rnsatisfying
√ne
ρ√−ρρ+−ξ−ρ+−ρ
√ρ− η
=(ρ+−ρ−)(v+gt en), |ξ|<1, |η|<1 (6.2) inU. The condition onλmax(A(z))from (2.4) withureplaced byρv+ f is kept in accordance with Definition2.3.
Indeed, in order to see this, given a subsolutionz =(ρ, v,u,S,P)it suffices to set
ξ :=
ρ+
ne
u−ρ−v+(ρ−ρ−)gt en
ρ−ρ− , η:=
ρ−
ne
u−ρ+v+(ρ−ρ+)gt en
ρ+−ρ .
(6.3) Conversely, given f, it suffices to setu :=ρv+ f to obtain a subsolution in the sense of Definition2.3.
Proof of Theorem2.7. Now letn =2,T >0 and⊂R2be the rectangle stated in the Theorem. In view of the equivalent reformulation above our goal is to find a suitable combination of functionsξ, ηande, such that (6.1) has a solution satisfying the energy inequality (2.8) in a strict sense.
In fact we will look for one-dimensional solutions of (6.1), that is a subsolution zin the sense of Definition2.3, which is independent ofx1and satisfiesu =u2e2, ξ =ξ2e2,η=η2e2respectively. We further assumev≡0.
If we have chosenξ,η, then condition (6.2) implies thatein the mixing zone is determined by
√2e=
√ρ−ρ+(ρ+−ρ−)gt
√ρ−(ρ−ρ−)ξ2− √ρ+(ρ+−ρ)η2. (6.4) Note also that under condition (6.2) the denominator will always be positive for t >0. Outside the mixing zone we will havee= 12ρg2t2in accordance with (2.5).
The last equation in (6.1) then becomes
∂tρ+gt∂x2
(ρ+−ρ)(ρ−ρ−)(√ρ−ξ2+ √ρ+η2) (ρ−ρ−)√ρ−ξ2−(ρ+−ρ)√ρ+η2
=0. (6.5) We recall (1.3) and hence thatρ0only depends on the sign ofx2. Using the change of coordinatesρ(x,t)=y(x,gt2/2)and interpreting theξ2, η2as functions ofρ the unique entropy solution (cf. Section 3.4.4 in [19]) of (6.6) with Rayleigh–Taylor initial dataρ0to obtain that
ρ(x2,t)=
Observe that this already implies that the height of the mixing zone grows (up to a constant) like21gt2, more precisely we will have
U = G is indeed uniformly strictly convex and the above entropy solution exists, then defining
withu2 and S extended by 0 outside U, one truly obtains a subsolution in the sense of Definition2.3. Indeed the relaxed momentum equation holds by definition
of P, andS is chosen in a way, such that the trace free part ofA(z)vanishes. In which holds, since by our reformulation inequalities (2.3) are automatically satisfied forξ2, η2∈(−1,1)andedefined in (6.4).
Therefore, all that remains to do in order to finish the construction of RT-subsolutions is to find ξ2, η2 : (ρ−, ρ+) → (−1,1)such that G is uniformly strictly convex and to assure the admissibility (2.8) (in a strict sense fort >0) of the associated total energy (2.7).
By the transformationx2=12gt2G(ρ)the desired admissibility (2.8) in the strict sense is then equivalent to across∂U. Then partial integration shows that (6.9) is equivalent to
Iξ2,η2 :=
Inspired by the known families of subsolutions for the Muskat problem [33]
or the Kelvin–Helmholtz instability [34], it is of interest to investigate the limit case when one is in the boundary of the convex hull, instead of its interior, as this corresponds to the limiting mixing zone growth rates of these families. In our case this means to choose |ξ| = |η| = 1 throughout all of[ρ−, ρ+], that is ξ2≡ −η2≡1. Of course this will not lead to a strict subsolution inside the mixing zone, so we will later consider a slight perturbation in order to be into the interior of the convex hull.
Denote byQ0,G0,e˜0the functions associated with the choiceξ2≡ −η2≡1, that is
Q0(ρ)=(ρ−ρ−)√ρ−+(ρ+−ρ)√ρ+, e˜0(ρ)=1 2
ρ+ρ−(ρ+−ρ−)2 Q0(ρ)2 , G0(ρ)=(ρ+−ρ)(ρ−ρ−)(√ρ−− √ρ+)
Q0(ρ) = − (ρ+−ρ)(ρ−ρ−) ρ++ρ−+ √ρ+ρ−−ρ. Through further calculations, one obtains that
G0(ρ)=
√ρ−ρ+(ρ++ρ−)+2ρ−ρ+ (ρ++ρ−+ √ρ+ρ−−ρ)2 −1, G0(ρ)=2√ρ−ρ+(ρ++ρ−)+4ρ−ρ+
(ρ++ρ−+ √ρ+ρ−−ρ)3 ,
in particular one sees thatG0is uniformly strictly convex on[ρ−, ρ+]. Furthermore, using the transformation ρ = ρ++ρ−+ √ρ+ρ−−ρ−s and the abbreviation r:= √√ρρ+− we obtain
I1,−1
ρ− = r2+r
1+r
r2(1+r)2 s3 +3
4 −3r(1+r)2 4s2
r(1+r)2 s2 −1
ds=0. This means that with the choiceξ2≡ −η2≡1 there holds equality in (2.8) for any t >0.
We now turn to the perturbation. Letε >0 and consider
ξ2(ρ):=1+εξ(ρ), η¯ 2(ρ):= −1+εη(ρ),¯ (6.11) with functionsξ,¯ η¯ : [ρ−, ρ+] → Rsatisfyingξ <¯ 0,η >¯ 0 on(ρ−, ρ+)and ξ(ρ¯ ±)= ¯η(ρ±) =0. Again, the last condition allows the functionedefined via (6.4) to be continuous over the whole domain×(0,T).
We will look for asymptotic expansions of the associatedQ= Qε,G =Gε,
˜
e= ˜eεwith respect toε >0. It holds that Qε(ρ)=Q0(ρ)+ε
(ρ−ρ−)√ρ−ξ¯−(ρ+−ρ)√ρ+η¯
=:Q0(ρ)+εQ(ρ),¯
˜
eε(ρ)=1 2
ρ+ρ−(ρ+−ρ−)2
(Q0(ρ)+εQ(ρ))¯ 2 = ˜e0(ρ)−ερ+ρ−(ρ+−ρ−)2 Q(ρ)¯
Q0(ρ)3 +O(ε2)
=: ˜e0(ρ)+εe(ρ)¯ +O(ε2), Gε(ρ)=G0(ρ)+ε(ρ+−ρ)(ρ−ρ−)
Q20(ρ)
√ρ+ρ−(ρ+−ρ−)(ξ¯+ ¯η)+O(ε2)
=:G0(ρ)+εG¯(ρ)+O(ε2),
while the expansion ofIε:=I1+εξ,−¯ 1+εη¯reads as
Iε=ε uniformly convex for small enoughε >0. Moreover, in order to have admissibility forε >0 small enough, it suffices to haveI¯>0.
By integration by parts we rewrite I¯= − the endpoints and concentrates atρ¯sufficiently such thatρ+
ρ− ξ(ρ)¯ H1(ρ)dρ >0.
Then, regardless of the sign ofH2, one may clearly choose a functionρ → ¯η= ¯η(ρ) which is strictly positive on(ρ−, ρ+), vanishes at the endpoints, and is small enough such thatI¯>0. The caseH2(ρ) >¯ 0 can be treated similarly.
Finally, to conclude the proof of Theorem2.7, we will prove that in fact the first case H1(ρ) <¯ 0 is not possible, while H2(ρ) >¯ 0 is possible if and only if ρ
ρ+− > 4+23√10.
Let us first prove the second statement.H2(ρ) >¯ 0 is equivalent to Q0(ρ)(¯ ρ¯−ρ−)√ρ−
3
2G0(ρ)¯ − ˜e0(ρ)¯
−ρ+ρ−(ρ+−ρ−)G0(ρ) >¯ 0.
Plugging in the expressions for Q0,G0ande˜0, one obtains that this is equivalent to
¯
ρ2−(ρ++2ρ−)ρ¯+2
3ρ+3/2ρ1−/2+5
3ρ+ρ−+ρ−2 <0.
This is possible only if the discriminant with respect toρ¯is strictly positive, which
ρ+− >1. The statement then follows by taking for instanceρ¯= ρ++22ρ− ∈(ρ−, ρ+)due to
ρ
ρ+− > 4+23√10.
The caseH1(ρ) <¯ 0 being not possible is proven similarly, the same calcula-tions yield the conditionr12 −3r8 −83 >0, which is not possible forr>1.
It remains to compute the precise growth rates of the mixing zoneU given in (6.8). Observe that∂ξG(ρ±)=∂ηG(ρ±)=0, such thatξ(ρ¯ ±)= ¯η(ρ±)=0 implies
Gε(ρ±)=G0(ρ±)=
√ρ±− √ρ∓
√ρ∓ .
This concludes the proof of Theorem2.7.
We would like to point out that the conditionρ
ρ+− > 4+23√10only enters in the admissibility of the subsolutions, more precisely it comes from our construction above for assuring I¯ > 0. For an arbitrary ratio ρρ+
− > 1 the fact that in the un-perturbed caseI−1,1=0 shows that there exist infinitely many turbulently mixing solutions with the exact same growth ratesc±(t)violating the weak admissibility by an arbitrary small amount of energy.
Furthermore, we summarize the other ansatzes used during our construction and note that they can all be seen as not too restrictive for different reasons:
• The independence ofx1can be interpreted as an averaging in thex1direction.
• v≡0 for the subsolution is in harmony with the vanishing initial velocity and the fact that the subsolution corresponds to an averaging of solutions.
• ξ and η only depending on ρ allow us to find the density ρ as the unique entropy solution of a relatively simple conservation law, this generalizes the construction from [33,34], where the unique viscosity solution of a Burgers equation was considered. In fact a similar conservation law also appeared in the relaxation of the two-phase porous media flow with different mobilities by Otto [29]. Our intuition behind choosingξandηto be perturbations of±e2has been explained during the proof. Nonetheless, it would be interesting to see if other choices ofξ andηalso lead to admissible subsolutions.
• The continuity ofeacross∂U is not a huge jump from Definition2.6, which combined withv≡0 already implied thate= 12g2t2ρ+in{x2>0} ∩D\U, respectivelye=12g2t2ρ−in{x2<0} ∩D\U, and therefore the continuity of ein each of the three pieces{x2<0} ∩D\U,U and{x2>0} ∩D\U.
Finally, we would like to state further properties than those of the growth rates of the unperturbed “subsolution” associated withξ2 ≡1,η2 ≡ −1 in an explicit way. Inversion of the derivativeG0: [ρ−, ρ+] →"
−√ρ+√−√ρρ+ −,√ρ+√−√ρρ− −# shows that the density profile, defined in (6.7), inside the mixing zone is given by
ρ (x2,t)=ρ++√ρ+ρ−+ρ−−(√ρ++ √ρ−)√4ρ+ρ−
1+2xgt22
,
the relaxed momentumu2(x2,t)= gt G0(ρ(x2,t))andedefined in (6.4) inside U by
u2(x2,t)=gt(√ρ++√ρ−)
⎛
⎝√4ρ+ρ−
1+2xgt22
+√4ρ+ρ−
$ 1+2x2
gt2 −√ρ+−√ρ−
⎞
⎠ e(x2,t)= 1
2g2t2√ρ−ρ+
1+2x2
gt2
,
from which an interested reader can obtain a formula of the associated energy density Esub defined in (2.7). Here we would only like to state the conversion rate of total potential energy into total kinetic energy. Recall that the unperturbed
“subsolution” satisfies (2.8) with equality. Hence the total kinetic energy at time t 0 can be expressed as the difference in total potential energy, which is
(ρ0(x)−ρ(x,t))gx2dx= g3t4 8
ρ+
ρ−
(ρ0(G0(ρ))−ρ)
G0(ρ)2 dρ
= g3t4 8
ρ+
ρ−
G0(ρ)2dρ
= g3t4(√ρ++ √ρ−)(√ρ+− √ρ−)3
24√ρ+ρ− .
We conclude the paper by presenting a plot of the above density (blue) and mo-mentum (red) profiles for the choice ρ− = 1/4, ρ+ = 4, g = 1 at fixed time t =
2 g(√ρ+−√ρ−)
12
. At this specific time the mixing zone extends fromx2 =
−ρ+−1/2= −1/2 tox2=ρ−−1/2=2.
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Björn Gebhard,József J. Kolumbán&László Székelyhidi Jr.
Mathematisches Institut, Universität Leipzig,
Augustusplatz 10, 04109 Leipzig
Germany.
e-mail: bjoern.gebhard@math.uni-leipzig.de József J. Kolumbán
e-mail: jozsef.kolumban@math.uni-leipzig.de László Székelyhidi Jr.
e-mail: laszlo.szekelyhidi@math.uni-leipzig.de (Received February 22, 2020 / Accepted May 14, 2021)
Published online June 12, 2021
© The Author(s)(2021)