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Abstract. In this note we provide new non-uniqueness examples for the continuity equation by
constructing infinitely many weak solutions with prescribed energy.
1. Introduction
In this paper we consider the continuity equation for a bounded scalar function u : R × Rd → R
with a bounded divergence-free vector field b : R × Rd → Rd :
∂t u + div(ub) = 0,
div b = 0.
This equation appears in various problems of mathematical physics, in particular fluid mechanics
and kinetic theory. In the smooth setting (and assuming suitable integrability) the energy
E (t) :=
u2 (t, x) dx
of the solution u is conserved:
E (t) = 0.
Indeed, since b is divergence-free, by multiplying (1) with u, using the chain rule and integrating
over Rd one immediately obtains (3).
In many applications one has to study (1) in a nonsmooth setting. Roughly speaking, since (1)
is linear, the conservation of energy (3) implies uniqueness of weak solutions to the corresponding
initial-value problem for (1). In fact, conservation of energy is a consequence of the so-called
renormalization property, which was proved by [DL89] for any vector field b with Sobolev regularity and later extended by Ambrosio [Amb] to the case when b has bounded variation. We refer
to [DL08, AC14] for a detailed review of recent results in this direction.
On the other hand, when the regularity of the vector field b is too low, the conservation of
energy (3) fails in general. In a nonsmooth setting several counterexamples to the uniqueness, and
therefore to the conservation of energy, are known, see [Aiz78, CLR03, Dep03, ABC14, ABC13].
A similar phenomenon occurs in the context of the Euler equations. For example, in the papers
[Sch93, Shn97, DLS09] weak solutions of the Euler equations were constructed with compact
support in space time.
In particular the example in [Dep03] gives a bounded vector field b and a bounded scalar field
u which satisfy (1) and (2) such that
E (t) =
for t ≤ 0
for t > 0.
2010 Mathematics Subject Classification. Primary: 35F10, Secondary: 35A02.
Key words and phrases. Transport and continuity equations, Non-uniqueness, Non-conservation of energy, Renormalization, Convex integration.
In this paper, for any given nonnegative bounded function E : R → R which is continuous on
an open interval and zero outside we construct infinitely many pairs (b, u) satisfying (1) and (2),
such that E (t) = E(t) for a.e. t. Thus, in contrast with (4), we provide more general profiles for
the energy. Our results are also connected to the chain rule problem for the divergence operator,
see [ADLM07, BG14, CGSW].
We construct such pairs (b, u) using the method of convex integration, and our techniques are
similar to the ones used in [DLS09, Sz´e12]. The latter reference contains an appendix giving a
general framework for convex integration, but for the problem at hand we need to consider a
nonlinear constraint that depends on the points in the domain (as was the case e.g. in [DLS10],
albeit in a different functional setting). For this reason we adapt the framework from [Sz´e12]
to this more general situation (see §2). We then apply this abstract framework to the specific
situation of the continuity equation (see §3).
Finally, let us mention [CFG11, Shv11, BLFNL], where results were obtained by convex integration, respectively, that yield as a byproduct counterexamples to the energy conservation for
continuity equations. However, in these works the energy profile is always piecewise constant.
Acknowledgements. This research has been partially supported by the SNSF grants 140232
and 156112. This work was started while the third author was a PostDoc at the Departement
Mathematik und Informatik of the Universit¨at Basel. He would like to thank the department
for the hospitality and the support. The second author was partially supported by the Russian
Foundation for Basic Research, project no. 13-01-12460. The authors are grateful to S. Bianchini,
C. De Lellis, and L. Sz´ekelyhidi for the fruitful discussions about the topic of the paper.
2. Differential inclusions with non-constant nonlinear constraint
We start with the so-called Tartar framework (cf. e.g. [DLS09]). Consider a system of m linear
partial differential equations
Ai ∂ i z = 0
in an open set D ⊂ RD , where Ai are constant m × n matrices and z : D → Rn . Consider a
nonlinear constraint
z(y) ∈ Ky
for a.e. y in D, where Ky ⊂ Rn is a compact set for any y ∈ D.
For any y ∈ D let Uy := int conv Ky , where with conv we denote the convex hull of the set Ky
and with int we denote its interior. Let U ⊂ D be a bounded open set.
Definition 1 (Subsolutions). We say that z ∈ L2 (D) is a subsolution of (5), (6) if z is a weak
solution of (5) in D, z is continuous on U , (6) holds for a.e. y ∈ D \ U and
z(y) ∈ Uy
for any y ∈ U .
Definition 2 (Localized plane waves/wave cone). A set Λ ⊂ Rn is called wave cone if there exists
a constant C > 0 such that for any z¯ ∈ Λ there exists a sequence wk ∈ C0∞ (B1 (0); Rn ) solving (5)
in RD such that
• dist(wk (x), [−¯
z , z¯]) → 0 for all x ∈ B1 (0) uniformly as k → ∞,
• wk
0 in L2 as k → ∞,
• |wk |2 dy > C|¯
z |2 for all k ∈ N.
In the above definition we denoted the segment with endpoints x ∈ Rn and y ∈ Rn with
[x, y] := conv{x, y}. The functions wk are called localized plane waves. We make the following
Assumption 1. There exists a wave cone Λ dense in Rn .
Let K denote the set of all compact subsets of Rn , endowed with the Hausdorff metric dH . It
is well-known that K is a complete metric space.
Assumption 2 (Continuity of the nonlinear constraint). The map f : U
continuous and bounded in the Hausdorff metric.
y → Ky ∈ K is
Our main abstract result is the following:
Theorem 1. Suppose that Assumptions 1 and 2 hold. Suppose that z0 is a subsolution of (5),
(6). Then there exist infinitely many weak solutions z ∈ L2 (D) of (5) which agree with z0 a.e.
on D \ U and satisfy (6) for a.e. y ∈ D.
2.1. Geometric preliminaries. The next lemma shows that compact subsets of the interior of
the convex hull of a compact set K are stable with respect to sufficiently small perturbations of
K in the Hausdorff metric.
Lemma 1. Let K ⊂ Rn be a compact set. Then for any compact set C ⊂ int conv K there exists
ε > 0 such that for any compact set K ⊂ Rn with dH (K, K ) < ε we have
C ⊂ int conv K .
Figure 1. An illustration of Lemma 1 in the case when K = {L, M, N } and K = {L , M , N }.
Proof. Since int conv K is open, for any point x ∈ C there exists a simplex Sx with vertices
{vi }i=1..n+1 ⊂ conv K such that x belongs to the inner open simplex
Ix :=
λ i vi
λi ∈
2(n + 1) n + 1
λi = 1, i = 1..n + 1 .
Since C is a compact set and the inner simplices {Ix }x∈C cover C we can extract a finite subcover
{Ixk }k=1..m of C. Let us fix k ∈ 1..m and consider the simplex S := Sxk with vertices {vi }i=1..n+1 ⊂
conv K. Let I := Ixk denote the corresponding inner simplex.
If ε < dist(∂I, ∂S) then for any points vi ∈ Bε (vi ), i = 1..n + 1 one has
I ⊂ conv{v1 , v2 , . . . , vn+1 }.
Observe that for any ε > 0 and i = 1..n + 1 the ball Bε (vi ) contains a point vi ∈ conv K . Indeed,
by Caratheodory’s theorem vi = n+1
j=1 µj = 1.
j=1 µj zj for some zj ∈ K and µj ∈ [0, 1] with
Since dH (K, K ) < ε there exist points zj ∈ K such that zj ∈ Bε (zj ), where j = 1..n + 1. Let
vi :=
µj zj ,
then |vi − vi | ≤ n+1
j=1 µj |zj − zj | < ε. Hence by (8) we have I ⊂ conv{v1 , v2 , . . . , vn+1 } provided
that ε is small enough. But vi ∈ conv K , hence I ⊂ conv K . Since I is open we can also write
I ⊂ int conv K .
Since we have finitely many simplices, we can choose ε > 0 in such a way that the inclusion
Ixk ⊂ int conv K holds for any k = 1..m (provided that dH (K, K ) < ε). Then
C ⊂ ∪k=1..m Ixk ⊂ int conv K .
We will also need the following elementary lemma:
Lemma 2. Suppose that z ∈ C(U ; Rn ) where U ⊂ RD is an open set. Suppose that for any
y ∈ U we have a compact set Ky ⊂ Rn and the function y → Ky is continuous in the Hausdorff
metric. Then the function F : y → dist(z(y), Ky ) is continuous on U .
Proof. One can prove directly that the function (z, K) → dist(z, K) is continuous on Rn × K .
The function y → (z(y), Ky ) is continuous in view of the assumptions. Hence the function F is
continuous as a composition of continuous functions.
In general the distance from a point z to a compact set K does not control from below the
distance from z to the boundary of conv K. However the following lemma shows that there exists
a segment inside int conv K with midpoint z and length controlled from below by dist(z, K):
Lemma 3 (Geometric lemma). Let K ⊂ Rn be a compact set. For any z ∈ int conv K there exists
z¯ ∈ Rn such that
• [z − z¯, z + z¯] ⊂ int conv K
dist(z, K)
• |¯
z | ≥ 2n
(This is exactly Lemma 5.3 from [DLS12].)
2.2. Convex integration. The following lemma is the main building block of the convex integration scheme:
Lemma 4 (Perturbation lemma). Suppose that Assumptions 1 and 2 hold and z is a subsolution
of (5) and (6) such that
dist2 (z(y), Ky ) dy = ε > 0.
Then there exists δ = δ(ε) > 0 and a sequence {zk }k∈N of subsolutions of (5) and (6) such that
• zk = z on D \ U for any k ∈ N
• U |z − zk |2 dy ≥ δ for any k ∈ N
• zk
z in L2 (U ) as k → ∞.
Proof. Step 1. Let y ∈ U . Since z(y) ∈ Uy we can apply Lemma 3 to obtain z¯∗ (y) such that
[z(y) − z¯∗ (y), z(y) + z¯∗ (y)] ⊂ Uy ,
z∗ (y)| ≥
dist(z(y), Ky ),
Since Λ is dense in Rn and Uy is open we can find z¯(y) ∈ Λ such that
[z(y) − z¯(y), z(y) + z¯(y)] ⊂ Uy ,
z (y)| ≥
dist(z(y), Ky ).
Due to (9) there exists ρ(y) > 0
[z(y) − z¯(y), z(y) + z¯(y)] + B2ρ(y) (0) ⊂ Uy .
Hence using Assumption 2, Lemma 1 and the continuity of z we can find R(y) > 0 such that
[z(x) − z¯(y), z(x) + z¯(y)] + Bρ(y) (0) ⊂ Ux
for all x ∈ BR(y) (y) ⊂ U . Moreover, in view of Lemma 2 we can choose R(y) in such a way that
dist(z(x), Kx ) ≤ 2 dist(z(y), Ky )
for all x ∈ BR(y) (y).
Using Assumption 1 for any fixed y ∈ U we can construct a sequence {wy,k }k∈N ⊂ C0∞ (B1 (0))
such that
• wy,k (x) ∈ [−¯
z (y), z¯(y)] + Bρ(y) (0) for all x ∈ B1 (0) and k ∈ N,
• wy,k
0 in L2 as k → ∞,
• |wy,k |2 dx > C|¯
z (y)|2 for all k ∈ N.
Step 2. Let ε := U dist2 (z(y), Ky ) dy. The balls { Br (y) | y ∈ U , r ∈ (0, R(y))} cover U , so
using Vitali’s covering theorem (see e.g. [Bog07], Theorem 5.5.2) and the absolute continuity of
the Lebesgue integral we can find finitely many points {yi }i=1..N ⊂ U and radii ri ∈ (0, R(yi ))
such that
dist2 (z(y), Ky ) dy > ε,
where the balls Bi := Bri (yi ) are pairwise disjoint.
For each i = 1..N let us introduce the scaled and translated perturbations wi,k (x) :=
wyi ,k ( x−y
ri ). These functions belong to C0 (Bi ) and satisfy
(i) wi,k (x) ∈ [−¯
z (yi ), z¯(yi )] + Bρ(yi ) (0) for all x ∈ Bi , k ∈ N, i = 1..N ;
(ii) wi,k
0 in L2 as k → ∞ (for each fixed i = 1..N );
(iii) |wi,k |2 dx > C|¯
z (yi )|2 L D (Bi ) for all k ∈ N.
In view of (i) and (11) we have z(x) + wi,k (x) ∈ Ux for all x ∈ U and i = 1..N , hence
z + wi,k ∈ X0 . Since the balls Bi are pairwise disjoint the function
zk := z +
also belongs to X0 .
Using successively (iii), (10), (12) and (13) we obtain:
|z − zk | dy =
z (yi )|2 L D (Bi )
|wi,k (y)| dy > C
dist2 (z(yi ), Kyi )L D (Bi ) =
dist2 (z(yi ), Kyi ) dx
dist2 (z(x), Kx ) dx >
It remains to observe that since N is finite and the points yi are fixed we have zk
k → ∞.
z in L2 as
2.3. Proof of Theorem 1. We are now ready to prove our main abstract theorem.
Proof of Theorem 1. Let X0 denote a set of all subsolutions of (5) and (6) which agree with z0 on
D \U . Let X be the closure of X0 in the weak topology of L2 (U ), endowed with the corresponding
induced weak topology. Clearly any z ∈ X solves (5) and satisfies (2) a.e. on D \ U .
For any z ∈ X let us define
|z(y)|2 dy.
I(z) :=
This functional is a Baire-1 function on X. Indeed, for any j ∈ N let
|(ω1/j ∗ z)(y)|2 dy
Ij (z) :=
{ y∈U | dist(y,∂U )>1/j}
where for any ε > 0 we denote by ωε (·) = ε−D ω(·/ε) the standard convolution kernel. Then for
any j ∈ N the functional Ij is continuous on X, and for any z ∈ X we have Ij (z) → I(z) as
j → ∞.
In view of Assumption 2 X is a bounded subset of L2 (U ). Since the weak topology is metrizable
on the norm-bounded subsets of L2 (U ), we can consider X as a complete metric space with some
metric dX .
Then by Baire category theorem (see also Theorem 7.3 from [Oxt80]) the set
Y := { z ∈ X | I is continuous at z}
is residual in X (and hence is infinite). We claim that z ∈ Y implies J(z) = 0, where
J(z) :=
dist2 (z(y), Ky ) dy.
Indeed, suppose that J(z) = ε > 0 for some z ∈ Y . Let zj ∈ X0 be a sequence such that zj
in L2 (U ) as j → ∞. Since I is continuous at z this implies that I(zj ) → I(z) and consequently
zj → z in L2 (U ) as j → ∞.
Then in view of Assumption 2 we have J(zj ) → J(z) as j → ∞ and hence without loss of
generality we can assume that J(zj ) > ε/2 for all j ∈ N.
Applying Lemma 4 to zj for each j ∈ N we can find zj ∈ X0 such that dX (zj , zj ) < 2−j and
U |zj − zj | dy ≥ δ > 0, where δ = δ(ε) is independent of j.
Since dX (zj , z) ≤ dX (zj , zj ) + dX (zj , z) → 0 as j → ∞ we also have zj
z in L2 . Since z is a
point of continuity of I we also have zj → z in L (U ) as j → ∞. But then zj − zj → 0 in L2 (U ),
which contradicts the construction of zj .
3. Application to the continuity equation
In this section we apply our abstract framework to the case of the continuity equation.
Theorem 2. Suppose that d ≥ 2. Let E : R → R be a non-negative bounded function which is
continuous on some bounded open interval I ⊂ R and vanishes on R\I. Then there exist infinitely
many bounded, compactly supported u : R × Rd → R and b : R × Rd → Rd which satisfy (1) and
(2) in sense of distributions and such that
u2 (t, x) dx = E(t)
for a.e. t ∈ I.
Remark 1. It is well-known that a representative of u can be chosen such that the map t → u(t, ·) is
continuous with values in L2 equipped with the weak topology. Then the question arises whether
the assertion in the theorem holds even for every, and not just almost every, time t. We expect
this to be true: indeed this should follow by methods similar to those of [DLS10]. We will however
not pursue this question further in this article.
Remark 2. When d = 2 and f is a characteristic function of an interval, the statement of Theorem 2, essentially, follows from the example constructed in [Dep03]. This particular case of
Theorem 2 was also proved in [Gus11] using the convex integration method.
Remark 3. A similar problem can be addressed for more general equation of the form div(uB) = 0
instead of (1). For this equation the problem is stated as follows: given a distribution g is it
possible to construct compactly supported bounded functions u : Rn → R, B : Rn → Rn such that
div(uB) = 0,
div B = 0,
div(u2 B) = g ?
This is related to the so-called chain rule problem for the divergence operator [ADLM07]. When
n = 2 such a construction is not possible for g = 0 in view of [BG14], while for n ≥ 3 it is possible
and is obtained using convex integration and rank-2 laminates in [CGSW].
Let us put the continuity equation in the framework of the previous section. Fix a bounded
open set Ω ⊂ Rd . Let U := I × Ω and
F (t, x) := d
1Ω (x),
L (Ω)
where 1Ω denotes the characteristic function of Ω.
We consider equations (1) and (2) as a linear system
∂t u + divx m = 0,
divx b = 0
in D := R × Rd with u : D → R, m : D → Rd and b : D → Rd such that z := (u, m, b) satisfies
the constraint
z(y) ∈ Ky :=
(u, m, b)
m = ub,
|b| = 1,
u2 = F (y)
if y ∈ U
if y ∈ D \ U
for a.e. y = (x, t) ∈ D.
Suppose that z = (u, m, b) ∈ L∞ (D) satisfies (14) and (15) in sense of distributions and
moreover (16) holds a.e. in D. Then the couple (u, b) satisfies the assertion of Theorem 2.
Let us check the assumption of Theorem 2.
Lemma 5. Suppose that A, B ⊂ Rn are compact sets and r > 0 is such that
• for any z ∈ A there exists z ∈ B ∩ Br (z)
• for any z ∈ B there exists z ∈ A ∩ Br (z)
Then dH (A, B) < r.
Proof. Suppose that dH (A, B) ≥ r. Then without loss of generality we can assume that there
exists z ∈ A such that for any z ∈ B we have z ∈
/ Br (z ). But then the ball Br (z) cannot contain
any point of B, which leads to a contradiction.
Lemma 6. If F : U → R is continuous, bounded and non-negative then the map y → Ky is
continuous and bounded (w.r.t. dH ) on U .
Proof. Let f (y) := F (y). Let us fix y ∈ U . For any ε > 0 let δ > 0 be such that |f (y)−f (y )| < ε
for any y ∈ Bδ (y) ⊂ U . Let us prove that dH (Ky , Ky ) < 2ε for all y ∈ Bδ (y).
For any z ∈ Ky there exist σ ∈ {±1} and b ∈ Rd with |b| = 1 such that z = (σf (y), σf (y)b, b).
Then z := (σf (y ), σf (y )b, b) belongs to Ky and |z − z | ≤ 2|f (y) − f (y )|. Hence there exists
z ∈ Ky ∩ B2ε (z).
Similarly, for any z ∈ Ky there exist σ ∈ {±1} and b ∈ Rd with |b| = 1 such that z =
(σf (y ), σf (y )b, b). Then z := (σf (y), σf (y)b, b) belongs to Ky and |z − z | ≤ 2|f (y) − f (y )|.
Hence there exists z ∈ Ky ∩ B2ε (z ).
Therefore by Lemma 5 we have dH (Ky , Ky ) < 2ε.
Lemma 7. Assumption 1 holds for the system (14)–(16).
Proof. Let φ : D → R be a non-negative smooth function such that 0 ≤ φ ≤ 1 on D, φ = 0 on
D \ B1 (0) and φ = 1 on B1/2 (0).
Part 1. Suppose that d > 2. Let us show that Assumption 1 holds with Λ = R2d+1 . Fix u
¯ ∈ R,
¯ ∈ Rd and let z¯ = (¯
¯ Since d > 2 there exists a unit vector n ∈ Rd such
¯ ∈ Rd and b
¯ b).
u, m,
¯ = 0. Denote n
¯ =n·b
ˆ = (0, n), a
¯ = (¯
¯ For any k ∈ N define a
¯k : D → Rd+1 by
that n · m
u, m).
¯k (y) := a
ˆ (¯
n · ∇y (φΠk )) − n
a · ∇y (φΠk ))
where y = (t, x) and
Πk (y) :=
ˆ · y)
sin(k n
Observe that
¯k = (¯
divy a
a · ∇y )(ˆ
n · ∇y )(φΠk ) − (ˆ
n · ∇y )(¯
a · ∇y )(φΠk ) = 0.
¯k , then by the equation above we have ∂t uk +divx mk = 0.
Let (uk , mk ) denote the components of a
Similarly let
¯ · ∇x (φΠk )) − n(b
¯ · ∇x (φΠk ))
bk (t, x) := b(n
Then arguing as above div bk = 0.
Now we introduce wk := (uk , mk , bk ). Then
ˆ · y) + f Πk
wk (y) = z¯φ cos(k n
where f does not depend on k and vanishes on B1/2 (0).
On the other hand,
|wk |2 dy ≥
|wk |2 dy =
B1/2 (0)
ˆ · y) dy =
z |2 cos2 (k n
B1/2 (0)
z |2
B1/2 (0)
provided that k is sufficiently large.
ˆ · y)
1 + cos(2k n
z |2
dy ≥
|B1/2 (0)| (17)
¯ ∈ R × R2 × R2 with u
¯ b)
Part 2. Suppose that d = 2 and fix z¯ = (¯
u, m,
¯ = 0. Let us look for a
localized plane wave in the following form:
wk = (ak , bk )
ak (y) = ∇y × φA
bk (t, x) = ∇⊥
sin(kn · y)
sin kn · (t, x)
where n = (nt , nx ) ∈ R × R2 and A ∈ R3 are to be chosen and k ∈ N. Then, by construction
divy ak = 0,
divx bk = 0.
Then, we get
sin(kn · y)
where zˆ = (A × n, nx ) and f does not depend on k and vanishes on B1/2 (0).
In order to have zˆ = z¯ the vectors A and n must satisfy
wk = zˆφ cos(kn · y) + f
A × n = (¯
u, m),
n⊥ = b.
¯ ⊥ . Since u
From the second equation we immediately obtain that nx = −b
¯ = 0 there exists nt
such that n ⊥ (¯
u, m).
Then, we can always find A such that the first equation is satisfied. It
remains to observe that the estimate (17) holds also in the considered case. We thus have verified
Assumption 1 for Λ = R5 \ {¯
u = 0}.
Proof of Theorem 2. By symmetry of Ky for any y ∈ U we have 0 ∈ int conv Ky . On the other
hand Ky = {0} for any y ∈ D \ U . Therefore u ≡ 0, m ≡ 0 and b ≡ 0 is a subsolution of
(14)–(16). Then the result follows from Lemma 2, Lemma 7 and Theorem 1.
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