Navier–Stokes and Euler equations: Cauchy problem and ... · To prove this theorem, we shall need the following natural result, which is of independent interest. Its proof is given

Navier–Stokes and Eulerequations: Cauchy problem

and controllability

ARMEN SHIRIKYANCNRS UMR 8088, Department of MathematicsUniversity of Cergy–Pontoise, Site Saint-Martin

2 avenue Adolphe Chauvin95302 Cergy–Pontoise Cedex, France

E-mail: [email protected]

(version of May 14, 2008)

Abstract

The course is devoted to studying the Navier–Stokes and Euler sys-tems in a bounded domain. We begin with the investigation of the initial-boundary value problems for both equations. A complete proof of well-posedness is given in the 2D case and some results are announced in the3D case. We next turn to the problem of controllability. To avoid techni-cal difficulties, consideration is given to the 1D Burgers equation. We alsodiscuss the situation for Navier–Stokes and Euler equations and formulatesome open questions.

CONTENTS 2

Contents

1 Navier–Stokes Equations 31.1 Functional spaces and Leray projection . . . . . . . . . . . . . . 31.2 Reduction to an evolution equation . . . . . . . . . . . . . . . . . 71.3 Existence, uniqueness, and regularity in the 2D case . . . . . . . 101.4 Remarks on the 3D case . . . . . . . . . . . . . . . . . . . . . . . 17

2 Euler equations 172.1 Smooth vector fields and flows . . . . . . . . . . . . . . . . . . . 182.2 Reduction to an evolution equation in the 2D case . . . . . . . . 212.3 Existence and uniqueness in the 2D case . . . . . . . . . . . . . . 232.4 The 3D case: Beale-Kato-Majda theorem . . . . . . . . . . . . . . 25

3 Controllability 263.1 Cauchy problem for Burgers equation . . . . . . . . . . . . . . . 263.2 Formulation of the main result . . . . . . . . . . . . . . . . . . . 273.3 Agrachev–Sarychev approach . . . . . . . . . . . . . . . . . . . . 273.4 Details of proof of Theorem 3.3 . . . . . . . . . . . . . . . . . . . 30

3.4.1 Extension principle . . . . . . . . . . . . . . . . . . . . . . 313.4.2 Convexification principle . . . . . . . . . . . . . . . . . . 313.4.3 Saturating property . . . . . . . . . . . . . . . . . . . . . . 353.4.4 Case of a large control space . . . . . . . . . . . . . . . . 36

3.5 Remarks on the Navier–Stokes and Euler equations . . . . . . . 36

4 Appendix 38Gronwall inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38Comparison principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38Poincare and Friedrichs inequalities . . . . . . . . . . . . . . . . . . . 39Some boundary value problems for the Laplace operator . . . . . . . 39Friedrichs extension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40Brouwer and Leray–Schauder theorems . . . . . . . . . . . . . . . . . 41

5 Problems 42

Notation 44

Bibliography 46

1 NAVIER–STOKES EQUATIONS 3

1 Navier–Stokes Equations

Let us consider the Navier–Stokes system describing the motion of an incom-pressible fluid:

∂tu + 〈u,∇〉u− ∆u +∇p = f (t, x), div u = 0, (1.1)

u∣∣∂D = 0. (1.2)

Here D ⊂ Rd is a bounded domain with smooth boundary, u = (u1, . . . , ud)and p are unknown velocity field and pressure, f is a given external force, and

〈u,∇〉 =d

∑j=1

uj(t, x)∂

∂xj.

Our aim is to investigate the well-posedness of the Cauchy problem for (1.1),(1.2). This question was first studied by Leray [Ler34], whose results weredeveloped later by many others; see the books [Lad63, Lio69, Tem79, CF88,Soh01] and the references therein.

An essential ingredient of the modern theory of the Navier–Stokes systemis a detailed description of the functional spaces. They are introduced andstudied in the first subsection. We next show that the problem in question canbe reduced to a nonlocal evolution equation in a Hilbert space. The latter isstudied in detail for the 2D case. We conclude this section by formulating tworesults concerning the 3D system.

1.1 Functional spaces and Leray projection

Let D ⊂ Rd be a connected bounded domain with smooth boundary ∂D andlet L2(D, Rd) be the space of vector functions u = (u1, . . . , ud) on D whosecomponents are square-integrable. We denote by (·, ·) the natural scalar prod-uct in L2(D, Rd) and by ‖ · ‖ the corresponding norm. If u ∈ L2(D, Rd) is suchthat

div u =d

∑j=1

∂u∂xj

= 0 in D, (1.3)

where the equality is understood in the sense of distributions, then we saythat u is divergence free.

Definition 1.1. We shall say that a divergence-free vector field u ∈ L2(D, Rd)has zero normal component on ∂D if 1∫

D〈u,∇ϕ〉 dx = 0 for any ϕ ∈ H1(D). (1.4)

In this case, we shall write 〈u, n〉∣∣∂D = 0, where n stands for the outward unit

normal to ∂D.1Note that if u ∈ C1(D, Rd) is divergence free and satisfies (1.4), then the scalar product of u

with the normal vector to ∂D vanishes.


For any integer s ≥ 0, let Hs(D) be the Sobolev space of order s, let H10(D)

be the space of functions in H1(D) vanishing on ∂D, and let Hs(D) be the spaceof functions u ∈ Hs(D) with zero mean value:

〈u〉 :=1

vol(D)

∫D

u(x) dx = 0, (1.5)

where vol(D) stands for the volume of D. Similar spaces of Rd-valued func-tions will be denoted by Hs(D, Rd), etc.

Let us endow the space H10 = H1

0(D, Rd) with scalar product

(∇u,∇v) =d

∑i,j=1

∫D

∂iuj∂ivjdx, u, v ∈ H10(D, Rd).

The norm corresponding to this scalar product is given by ‖∇u‖. Denote byH−1 = H−1(D, Rd) the space of Rd-valued distributions on D such that

|(u, ϕ)| ≤ C ‖∇ϕ‖ for any ϕ ∈ C∞0 (D, Rd).

In other words, H−1 is the dual space of H10 . By the Riesz representation the-

orem, there is a natural isometry between H−1 and H10 , and it is easy to see

that it is given by the Laplacian. Namely, for any f ∈ H−1 there is a uniquefunction u f ∈ H1

0 such that

f (v) = (∇u f ,∇v) for any v ∈ H10 .

It follows from Proposition 4.4 that u f is the unique solution of (4.9). In whatfollows, the space H−1 is endowed with the natural scalar product

( f , g)−1 = g(u f ) = f (ug) = (∇u f ,∇ug), f , g ∈ H−1(D, Rd), (1.6)

and the corresponding norm is denoted by ‖ · ‖−1.Let us introduce the functional spaces

H(D) ={

u ∈ L2(D, Rd) : div u = 0 in D, 〈u, n〉∣∣∂D = 0

},

Z(D) = {u ∈ H1(D) : ∆u = 0 in D}.

We denote by ∇H10(D) the space of vector fields u ∈ L2(D, Rd) that are rep-

resentable in the form u = ∇p with some p ∈ H10(D) and define ∇Z(D) in

a similar way. In what follows, we shall often omit the domain D from thenotation and write L2, H, Z, etc. It follows from the Friedrichs and Poincareinequalities (see the Appendix) that ∇H1

0 and ∇Z are closed subspaces of L2.

Theorem 1.2 (Hodge-Kodaira decomposition). Let D ⊂ Rd be a bounded domainwith C1-smooth boundary ∂D. Then the space of square-integrable vector fields in Dis representable as the direct sum

L2(D, Rd) = H ⊕∇H10 ⊕∇Z. (1.7)


Proof. Let us take an arbitrary function u ∈ L2(D, Rd) and show that is repre-sentable in the form

u = v +∇w +∇z, (1.8)

where v ∈ H, w ∈ H10 , and z ∈ Z. Consider the elliptic boundary value

problems

∆w = div u in D, w∣∣∂D = 0, (1.9)

∆z = 0 in D, 〈∇z− u +∇w, n〉∣∣∂D = 0. (1.10)

By Proposition 4.4 and 4.5, problems (1.9) and (1.10) are uniquely solvable inthe spaces H1

0(D) and H1(D), respectively. Let us set v = u−∇w−∇z. Weneed to prove that v ∈ H. For any ϕ ∈ C∞

0 (D), we have

(v,∇ϕ) = (u−∇w−∇z,∇ϕ) = (div u− ∆w− ∆z, ϕ) = 0,

and therefore div v = 0 in D. Furthermore, the second relation in (1.10) impliesthat 〈v, n〉 = 0 on ∂D. We have thus proved (1.8).

To complete the proof of the theorem, it suffices to show that the spaces H,∇H1

0 , and ∇Z are pairwise orthogonal. This fact is a straightforward conse-quence of the definition and the density of C∞

0 (D) in the space H10(D).

Regularity of solutions for elliptic equations (see Prositions 4.4 and 4.5) andthe explicit description of the functions on the right-hand side of (1.8) implythe following result:

Corollary 1.3. Let u ∈ Hs(D, Rd) for some integer s ≥ 0. Then the functions v, w,and z entering decomposition (1.8) satisfy the inclusions v ∈ Hs and w, z ∈ Hs+1.Moreover, there is a constant Cs > 0 such that

‖v‖s + ‖w‖s+1 + ‖z‖s+1 ≤ Cs‖u‖s. (1.11)

In what follows, we denote by Π : L2(D, Rd) → H the orthogonal projec-tion in L2 associated with the Hodge–Kodaira decomposition (1.7). It is calledthe Leray projection. Corollary 1.3 implies that Π is continuous in any Sobolevspace Hs(D, Rd). Another useful consequence of Theorem 1.2 is the followingnecessary and sufficient condition for the representation of an L2 vector fieldas the gradient of a function.

Corollary 1.4. A vector function u ∈ L2(D, Rd) can be represented in the formu = ∇p for some p ∈ H1 if and only if

(u, ϕ) = 0 for any ϕ ∈ H. (1.12)

In this case, the function p is unique, and if u ∈ Hs(D, Rd) for some integer s ≥ 0,then p ∈ Hs+1(D).

Let V = V(D) be the space of divergence-free functions u ∈ H10(D, Rd) and

let V = V(D) = C∞0 (D, Rd) ∩ V. The latter space is not empty; for instance,

for any ψ ∈ C∞0 (D), the function (−∂2ψ, ∂1ψ, . . . ) belongs to V . The following

result of fundamental importance implies, in particular, that V is much bigger.


Theorem 1.5. A function u ∈ H−1(D, Rd) is representable in the form u = ∇p forsome p ∈ L2(D) if and only if

(u, ϕ) = 0 for any ϕ ∈ V(D). (1.13)

To prove this theorem, we shall need the following natural result, which isof independent interest. Its proof is given at the end of this subsection.

Proposition 1.6. Let p be a distribution in D such that ∇p ∈ H−1(D, Rd). Thenp ∈ L2(D). Moreover, there is a constant C > 0 depending only on D such that

‖p− 〈p〉‖ ≤ C‖∇p‖H−1 . (1.14)

Proof of Theorem 1.5. Step 1. Consider the operator ∇ : L2(D) → H−1(D, Rd)taking a function p to its gradient ∇p. By Problem 1, the image F(D) of ∇ is aclosed subspace in H−1(D, Rd). Furthermore, since the adjoint of ∇ coincideswith the operator div : H1

0 → L2 taking u to div u, in view of Problem 2, wehave

F(D) =(

Ker(div))⊥= (V(D))⊥

= { f ∈ H−1 : f (u) = 0 for any u ∈ V(D)}. (1.15)

Let us represent D as the union of an increasing sequence of connecteddomains Dk with smooth boundaries such that Dk ⊂ Dk+1. Suppose we haveproved that for any integer k ≥ 1 the restriction of u to Dk (which we denoteby uk) belongs F(Dk). In this case, we can find a function pk ∈ L2(Dk) suchthat uk = ∇pk. Arguing by induction, we can construct p ∈ L2

loc(D) such thatu = ∇p in D. Applying Proposition 1.6, we conclude that p ∈ L2(D).

Step 2. We now prove that uk ∈ F(Dk). In view of relation (1.15) appliedto Dk, it suffices to show that

uk(ϕ) = 0 for any ϕ ∈ V(Dk). (1.16)

Let {ωε, ε > 0} be a family of mollifying kernels. Since Dk ⊂ D, we see thatωε ∗ ϕ ∈ V(D) for ε� 1, and hence

u(ωε ∗ ϕ) = 0. (1.17)

On the other hand, the family ωε ∗ ϕ converges to ϕ in the space H1 as ε→ 0+.Passing to the limit in (1.17) as ε→ 0+, we arrive at (1.16).

Corollary 1.7. (i) The closure of V(D) in L2(D, Rd) coincides with H(D).

(ii) The closure of V(D) in H1(D, Rd) coincides with V(D).

Proof. (i) We need to show that if u ∈ H satisfies (1.13), then u = 0. Indeed,by Theorem 1.5, the function u can be written as u = ∇p for some p ∈ L2(D).Since u ∈ L2, we see that p ∈ H1(D). In view of Corollary 1.4, the function umust be orthogonal to H, and we conclude that u = 0.


(ii) Let u ∈ H−1(D, Rd) be such that u(ϕ) = 0 for any ϕ ∈ V(D). Therequired assertion will be proved if we show that u = 0 on V(D). By Theo-rem 1.5, we can write u = ∇p for some p ∈ L2(D). It follows that

∇p(ϕ) =∫

Dp(x) div ϕ(x) dx = 0 for any ϕ ∈ V(D).

This completes the proof of the corollary.

Proof of Proposition 1.6. We shall only outline the proof, leaving the details tothe reader. Let us remark ∆p = div u ∈ H−2(D), and therefore, by ellipticregularity, we conclude that p ∈ L2

loc(D). Let us denote by Td = Rd/2πZd thed-dimensional torus. It is straightforward to verify, using the Fourier series,that for any p ∈ L2(Td) with zero mean value we have

‖p‖ ≤ C ‖∇p‖−1. (1.18)

Approximating any function p ∈ L2(D) by smooth functions with compactsupport, we can prove easily that (1.14) holds for any p ∈ L2(D). Thus, itsuffices to show that any distribution p such that ∇p ∈ H−1 belongs to L2.This is a local result, and by rectifying the boundary, we can assume withoutloss of generality that D = (−1, 1)d. In this case, for any α ∈ (0, 1) the functionpα(x) = p(αx) belongs to L2(D) and therefore

‖pα − 〈pα〉‖ ≤ C‖∇pα‖ ≤ C1.

Since the unit ball in L2 is weakly compact, we can find a sequence αn → 1and a function p ∈ L2 such that pαn − 〈pαn〉 → p weakly in L2. On the otherhand, pα → p in the sense of distributions. It follows that 〈pαn〉 converges to aconstant c ∈ R, and p = p + c ∈ L2.

In what follows, we shall need also the dual space of V; it is denoted by V∗.Since V is a closed subspace of H1

0 , we can identify V∗ with the quotient spaceof H−1 over the closed subspace

{u ∈ H−1 : u(ϕ) = 0 for any ϕ ∈ V} = {u ∈ H−1 : u = ∇p for some p ∈ L2}.

It is straightforward to verify that the Leray projection admits a continuousextension from H−1 to V∗; see Problem 5 for a hint.

1.2 Reduction to an evolution equation

In what follows, we assume that d = 2 or 3. Let us fix T > 0 and denote by JTthe time interval [0, T]. If there is no ambiguity, we shall omit the subscript T.

Definition 1.8. Let f ∈ L2(J, H−1). A pair of distributions (u, p) in the domain(0, T)× D is called a weak solution of (1.1), (1.2) if

u ∈ L∞(J, L2) ∩ L2(J, H10), p = ∂tq for some q ∈ L∞(J, L2),

and Eqs. (1.1) are satisfied in the sense of distributions.


The regularity imposed on (u, p) by the above definition is dictated by astandard energy estimate. Namely, let (u, p) be a smooth solution of (1.1), (1.2).Multiplying the first equation of (1.1) by 2u and integrating over D, we derive

∂t‖u‖2 + 2‖∇u‖2 + 2(〈u,∇〉u +∇p, u) = 2( f , u). (1.19)

The third term on the left-hand side of (1.19) vanishes, and the right-hand sidecan be estimated by C‖ f ‖2

−1 + ‖∇u‖2. Substituting this into (1.19), integratingin time, and taking the supremum over t ∈ J, we obtain

supt∈J

(‖u(t)‖2 +

∫ t

0‖∇u‖2ds

)≤ ‖u(0)‖2 + C‖ f ‖2

L2(J,H−1).

This justifies the choice of the functional space for u. Furthermore, integratingthe first equation of (1.1) in time, we derive

∇(∫ t

0p(s) ds

)= u(0)− u(t) +

∫ t

0

(f + ∆u− 〈u,∇〉u

)ds.

The right-hand side of this relation belongs to L∞(J, H−1), and therefore∫ t

0 pdsmust be an element of L∞(J, L2).

The following proposition gives some additional information on weak solu-tions. It also suggests how to reduce the Navier–Stokes system to an evolutionequation.

Proposition 1.9. Let (u, p) be a weak solution of (1.1), (1.2). Then

u ∈ XT := L∞(J, H) ∩ L2(J, V), (1.20)

〈u,∇〉u ∈ L1(J, H−1), ∆u ∈ L2(J, H−1), (1.21)

∂tu ∈ L1(J, V∗), u ∈ C(J, V∗), (1.22)

and we have the relation

∂tu + 〈u,∇〉u− ∆u = f in V∗ for almost every t ∈ J. (1.23)

Proof. Inclusions (1.20) and (1.21) are easy to establish, and we confine our-selves to the proof of the first relation in (1.21). For any u ∈ V, we have

〈u,∇〉u =d

∑i=1

ui∂iu =d

∑i=1

∂i(uiu). (1.24)

Using the continuous embedding H1 ⊂ L4, which is valid for d ≤ 4, we derive

‖uiu‖ ≤ C1‖u‖2L4 ≤ C2‖∇u‖2. (1.25)

Since the derivation is continuous from L2 to H−1, relations (1.24) and (1.25)imply that

‖〈u,∇〉u‖−1 ≤ C3‖∇u‖2. (1.26)


This estimate implies the required inclusion.We now prove (1.22) and (1.23). It follows from the first equation of (1.1)

that

u(t) +∇q(t) = u +∫ t

0( f + ∆u− 〈u,∇〉u) ds for almost every t ∈ J, (1.27)

where u is an element of H−1, and the equality holds in H−1. Applying bothsides of (1.27) to a test function ϕ ∈ V , we obtain

(u(t), ϕ) = (u, ϕ) +∫ t

0( f + ∆u− 〈u,∇〉u, ϕ) ds for almost every t ∈ J.

Since V is dense in V, we conclude that

u(t) = u +∫ t

0( f + ∆u− 〈u,∇〉u) ds, (1.28)

where the equality holds in V∗ for almost every t ∈ J. Relation (1.28) impliesinclusions (1.22), as well as equality (1.23).

Proposition 1.9 suggests that projecting the first equation of (1.1) to thespace V∗, we can reduce the Navier–Stokes system to an evolution equation.Guided by this idea, let us consider the equation

∂tu + Lu + B(u, u) = g(t), (1.29)

where g ∈ L2(J, V∗), and we set

Lu = −Π∆, B(u, v) = Π (〈u,∇〉u). (1.30)

Using Problem 5 and (1.24) – (1.26), it is straightforward to see that the linearoperator L : V → V∗ and the bilinear form B(u, v) : V ×V → V∗ are continu-ous.

Definition 1.10. A distribution u(t, x) in the domain (0, T)×D is called a weaksolution of (1.29) if u ∈ XT , ∂tu ∈ L1(J, V∗), and relation (1.29) holds in V∗ foralmost every t ∈ J.

Proposition 1.9 shows that if (u, p) is a weak solution of (1.1), (1.2), then uis a weak solution for (1.29) with g = Π f . The converse assertion is also true.

Proposition 1.11. Let f ∈ L2(J, H−1) and let u(t, x) be a weak solution of Eq. (1.29)with g = Π f . Then there is q ∈ L∞(J, L2) such that the pair (u, p) with p = ∂tq is aweak solution of (1.1), (1.2).

Proof. Relation (1.29) implies that u ∈ C(J, V∗), and

u(t)− u +∫ t

0

(Lu + B(u, u)−Π f (t)

)ds = 0 for any t ∈ J, (1.31)


where u ∈ V∗, and the equality holds in V∗. Let us set

h(t) = u(t)− u +∫ t

0

(〈u,∇〉u− ∆u− f (t)

)ds.

Then h ∈ L∞(J, H−1), and in view of (1.31), there is a subset J0 ⊂ J of fullmeasure such that

(h(t), ϕ) = 0 for any ϕ ∈ V and t ∈ J0.

By Theorem 1.5 and Proposition 1.6, there is a function q ∈ L∞(J, L2) such thath(t) = ∇q(t). This relation is equivalent to the first equation of (1.1). Thesecond equation follows from the inclusion u ∈ L2(J, V).

From now on, we study the evolution equation (1.29).

1.3 Existence, uniqueness, and regularity in the 2D case

We begin with the investigation of the Cauchy problem for the Stokes equation.Namely, we consider the problem

∂tu + Lu = g(t), (1.32)u(0) = u0, (1.33)

where u0 ∈ H and g ∈ L2(J, V∗) are given functions. We shall show that L isa positive (self-adjoint) operator in H with discrete spectrum and then applythe Hille-Yosida theorem to construct a solution of (1.32), (1.33). We shall needthe concept of Friedrichs extension for semi-bounded symmetric operators; itis briefly described in the Appendix.

We define an operator L0 in H by D(L0) = V and L0u = −Π∆u for u ∈ V .It is straightforward to show that L0 satisfies inequalites (4.15) and (4.16) withM = 0 and therefore, by Theorem 4.6, admits a self-adjoint extension L calledthe Friedrichs extension. The following proposition describes the operator Land some of its properties.

Proposition 1.12. The domain D(L) of L coincides with the space H2(D, Rd) ∩ V,and Lu = −Π∆u for u ∈ D(L). Moreover, L has a compact resolvent, and there is anorthonormal basis of H consisting of the eigenfunctions of L.

Proof. Step 1. We first show thatD(L) = H2(D, Rd)∩V. Indeed, the norm ‖ · ‖qdefined in the Appendix coincides with ‖ · ‖1, and therefore, by Corollary 1.7,we have

D(Q) = V, Q(u, v) = (∇u,∇v).

Recall that

D(L) = {u ∈ V : there is f ∈ H such that (∇u,∇v) = ( f , v) for any v ∈ V}.(1.34)

This implies immediately that D(L) ⊃ H2(D, Rd) ∩ V. We now prove theconverse inclusion. To this end, we need the following lemma.


Lemma 1.13. For any f ∈ H−1(D, Rd) the problem

−∆u +∇p = f , div u = 0, (1.35)

has a unique solution (u, p) ∈ V × L2. Moreover, if f ∈ L2, then u ∈ H2 andp ∈ H1.

The existence and uniqueness of solution is a straightforward consequenceof the Riesz representation theorem. The regularity of solution in the generalcase follows from the elliptic theory; see [ADN64]. However, in the 2D case, asimpler proof can be given by reducing (1.35) to a biharmonic equation. Thecorresponding argument is outlined in Problem 8.

Now let u ∈ D(L). Then, by (1.34), there is f ∈ H such that

(∆u + f , ϕ) = 0 for any ϕ ∈ V . (1.36)

By Theorem 1.5, there is p ∈ L2 such that ∆u + f = ∇p. Thus, (u, p) is theunique solution of problem (1.35). Since f ∈ L2, in view of Lemma 1.13, thefunction u ∈ V must belong to H2. Furthermore, it follows from (1.36) thatf = Lu = −Π∆u.

Step 2. The remaining assertions will be proved if we show that for anyf ∈ H the equation Lu = f has a unique solution u ∈ D(L). Indeed, if thisproved, then λ = 0 is in the resolvent set of L, and the inverse L−1 is compact.Thus, L is a self-adjoint operator with compact resolvent, and therefore it has adiscrete spectrum; see Chapters VI–VIII of [RS80] for details.

To prove the unique solvability inD(L) of the equation Lu = f , it suffices tonote that it is equivalent to problem (1.35), and therefore all the claims followfrom Lemma 1.13.

Proposition 1.12 and The Hille–Yosida theorem enable one to prove the ex-istence and uniqueness of solution for the Stokes equation. Namely, we havethe following result whose proof is left to the reader.

Proposition 1.14. For any u0 ∈ H and g ∈ L2(J, V∗), problem (1.32), (1.33) has aunique solution u ∈ YT := C(J, H) ∩ L2(J, V), which satisfies the relation

‖u(t)‖2 + 2∫ t

0‖∇u(s)‖2ds = ‖u(0)‖2 + 2

∫ t

0(g(s), u(s)) ds, 0 ≤ t ≤ T.

(1.37)Moreover, if u0 ∈ V and g ∈ L2(J, H), then u ∈ C(J, V) ∩ L2(J,D(L)).

From now on until the end of this subsection, we assume that d = 2. Thenext step in the construction of solution of the Navier–Stokes system is theinvestigation of the quasilinear equation

∂tu + Lu + B(v1, u) + B(u, v2) = g(t), (1.38)

where v1, v2, and g are given functions.


Proposition 1.15. For any v1, v2 ∈ YT , g ∈ L2(J, V∗), and u0 ∈ H, problem (1.38),(1.33) has a unique solution u ∈ YT , which satisfies the inequality

‖u‖YT ≤ C(‖u0‖+ ‖g‖L2(J,V∗)

), (1.39)

where C > 0 does not depend on v1, g, and u0.

Proof. Step 1: Uniqueness and a priori estimate. It suffices to establish inequal-ity (1.39). If u, v1, v2 ∈ YT , then the function B(v1, u) + B(u, v2) belongs toL2(J, V∗); see Problem 9. Therefore, by (1.37), we can write

‖u(t)‖2 + 2∫ t

0‖∇u(s)‖2ds = ‖u0‖2 + 2

∫ t

0(g− B(v1, u)− B(u, v2), u) ds.

(1.40)Now note that

(B(v1, u), u) = 0,

|(B(u, v2), u)| ≤ C1‖v2‖1 ‖u‖2L4 ≤ C2‖v2‖1 ‖u‖ ‖∇u‖.

Substituting this into (1.40) and using the Cauchy inequality, we derive

Eu(t) ≤ ‖u0‖2 + C3

∫ t

0‖g‖2

V∗ds + C3

∫ t

0‖v2‖2

1Eu(s) ds, (1.41)

where we set

Eu(t) = ‖u(t)‖2 +∫ t

0‖∇u‖2ds. (1.42)

Applying the Gronwall inequality (see (4.2)), we obtain

Eu(t) ≤ C4 exp(∫ T

0‖v2‖2

1ds) (‖u0‖2 + ‖g‖2

L2(J,V∗)), t ∈ J.

This implies the required estimate (1.39).

Step 2: Existence. We now use the contraction mapping principle to con-struct a solution. We fix τ > 0 and define the ball

BR = {u ∈ Yτ : ‖u‖Yτ≤ R}.

Let F : BR → Yτ be a mapping that takes u to the solution u ∈ Yτ of theproblem

∂tu + Lu = g− B(v1, u)− B(u, v2), u(0) = u0.

We claim that there are constants R and τ depending only on the norms of thefunctions v1, v2, g and u0 such that F is a contraction of the ball BR into itself.Indeed, it follows from (1.37) that

‖u(t)‖2 + 2∫ t

0‖∇u‖2ds ≤ ‖u0‖2 + 2

∫ t

0(g− B(v1, u)− B(u, v2), u) ds

≤ ‖u0‖2 +∫ t

0

(C5‖g‖2

V∗ + ‖∇u‖2 + C6(‖v1‖2

L4 + ‖v2‖2L4

)‖u‖2

L4

)ds.


By interpolation inequality, it follows that

supt∈JEu(t) ≤ ‖u0‖2 + C5‖g‖2

L2(J,V∗) +18

∫ t

0‖∇u‖2ds

+ C7

(‖v1‖4

L4(Qτ) + ‖v2‖4L4(Qτ)

)sup

s∈[0,t]‖u(s)‖2. (1.43)

Let us choose

R = 3(‖u0‖2 + C5‖g‖2

L2(J,V∗)

)1/2.

Then, for sufficiently small τ > 0, the right-hand side of (1.43) is smallerthan R2/4, whence it follows that ‖u‖Yτ

≤ R. This proves that F maps theball BR into itself. A similar argument shows that F is a contraction.

Theorem 1.16. For any u0 ∈ H and g ∈ L2(J, V∗), problem (1.29), (1.33) has aunique solution u ∈ YT , which satisfies relation (1.37).

We shall need the following lemma; we refer the reader to Problem 11 forsome hints how to establish it.

Lemma 1.17. Let u ∈ L2(J, V) be such that u ∈ L2(J, V∗). Then u ∈ C(J, H) and

‖u‖2C(J,H) ≤ C ‖u‖L2(J,V)‖u‖L2(J,V∗). (1.44)

In particular, if u ∈ XT is a weak solution of (1.29), then u ∈ YT .

Proof of Theorem 1.16. Step 1: Uniqueness and a priori estimate. Let u ∈ YT be asolution. Then, by Problem 9, we have B(u, u) ∈ L2(J, V∗). Hence, by (1.37),we derive

‖u(t)‖2 + 2∫ t

0‖∇u(s)‖2ds = ‖u(0)‖2 + 2

∫ t

0(g− B(u, u), u)ds.

Since (B(u, u), u) = 0, we conclude that (1.37) remains valid for solutionsof (1.29). Furthermore, using the inequality

|(g, u)| ≤ ‖g‖V∗‖∇u‖ ≤ 12(‖g‖2

V∗ + ‖∇u‖2),we see that

Eu(t) ≤ ‖u(0)‖2 +∫ t

0‖g‖2

V∗ds.

It follows that

‖u‖Yτ≤ 2

(‖u0‖+ ‖g‖L2(Jτ ,V∗)

)for any τ ∈ JT . (1.45)

Suppose now that u1, u2 ∈ YT are two solutions for (1.29), (1.33). Then theirdifference u = u1 − u2 vanishes at t = 0 and satisfies the equation

∂tu + Lu = B(u2, u) + B(u, u1).


Using (1.37), we derive

‖u(t)‖2 + 2∫ t

0‖∇u(s)‖2ds = 2

∫ t

0

(B(u2, u) + B(u, u1), u

)ds. (1.46)

The first term under the integral on the right-hand side of (1.46) vanishes, whilethe second can be estimated by

|B(u, u1), u)| ≤ C1‖∇u1‖ ‖u‖ ‖∇u‖ ≤ 12(‖∇u‖2 + C2

1‖∇u1‖2 ‖u‖2).Substituting this into (1.46), we obtain

Eu(t) ≤ C21

∫ t

0‖∇u1‖ Eu(t) ds. (1.47)

Applying the Gronwall inequality, we conclude that u ≡ 0.

Step 2: Reduction to local existence. We now show that it suffices to establishthe existence of solution on a small time interval [0, τ] depending on the initialfunction and the right-hand side. Indeed, let us set

T∗ = sup{τ > 0 : there is a solution u ∈ Yτ for (1.29), (1.33)}.

Suppose that T∗ < T. It follows from (1.45) that u ∈ XT∗ and therefore, byLemma 1.17, we have u ∈ YT∗ . Let us set u∗ = u(T∗). By assumption, wecan solve Eq. (1.29) with the initial condition u(T∗) = u∗ on a small interval[T∗, T∗ + τ]. We thus obtained a solution u ∈ YT∗+τ ; this contradicts the defini-tion of T∗. Thus, T∗ must coincide with T, and repeating the above argument,we obtain a solution u ∈ YT .

Step 3: Construction of a local solution. The solution is sought in the formu = z + v, where z = exp(−tL)u0, and v is an unknown function. Substitutingthis into (1.29), we obtain the following problem for v:

∂tv + Lv + B(v, v) + B(v, z) + B(z, v) = g, v(0) = 0, (1.48)

where g = g− B(z, z) ∈ L2(Jτ , V∗). Let us fix positive constants τ ≤ 1 and Mand denote by K(τ, M) the set of functions v ∈ Yτ such that

Ev(t) ≤ M∫ t

0‖g(s)‖2

V∗ds for 0 ≤ t ≤ τ. (1.49)

It is clear that K(τ, M) is a closed subset in the space Yτ . Let us define a map-ping F : K(τ, M) → Yτ that takes v to the solution v ∈ Yτ of the problem(cf. (1.48))

∂tv + Lv + B(v, v) + B(v, z) + B(z, v) = g, v(0) = 0. (1.50)

It is clear that a function v ∈ Yτ is the solution of (1.48) if and only if it is afixed point for F . We claim that for an appropriate choice of τ and M, the


mapping F is a contraction of the set K(τ, M) into itself. Indeed, it followsfrom (1.37) and the relation (B(w, v), v) = 0 that

‖v(t)‖2 + 2∫ t

0‖∇v(s)‖2ds = 2

∫ t

0

(g− B(v, z), v

)ds. (1.51)

Now note that

|(g, v)| ≤ 14‖∇v‖2 + ‖g‖2

V∗ ,

|(B(v, z), v)| ≤ C2‖v‖2L4‖∇z‖ ≤ 1

4‖∇v‖2 + C2

2‖v‖2‖∇z‖2.

Substituting these inequalities into (1.51) and carrying out some simple trans-formations, we derive

Ev(t) ≤ 2∫ t

0‖g‖2

V∗ds + C3

∫ t

0‖∇z‖2Ev(s) ds. (1.52)

Application of the Gronwall inequality shows that (1.49) holds with

M = 2 exp(

C3

∫ 1

0‖∇z‖2ds

). (1.53)

Thus, the mapping F takes the set K(τ, M) into itself for any τ ∈ (0, 1]. Let usprove that it is a contraction for sufficiently small τ.

Take any vi ∈ K(τ, M), i = 1, 2, and set vi = F (vi). Then the differencev = v1 − v2 vanishes at t = 0 and satisfies the equation

∂tv + Lv + B(v, z) + B(z, v) = −B(v, v2)− B(v1, v),

where v = v1 − v2. Using again (1.37) and repeating the above arguments, wederive

Ev(t) ≤ C4

∫ t

0

(‖∇z‖2‖v‖2 + ‖v2‖2‖∇v‖2 + ‖∇v2‖2‖v‖2) ds

≤ C4

∫ t

0‖∇z‖2Ev(s) ds + C5‖v‖2

Yτsups∈Jt

Ev2(s).

Applying the Gronwall inequality and using (1.49) with v = v2, we derive

Ev(t) ≤ C6‖v‖2Yτ

∫ τ

0‖g‖2

V∗ds for 0 ≤ t ≤ τ,

where C6 > 0 does not depend on v1 and v2. The above inequality impliesthat F is a contraction for τ � 1 and, hence, has a fixed point v ∈ Yτ . Thiscompletes the proof of the theorem.

We now turn to the problem of smoothness of solution. The following resultshows that the weak solution corresponding to smooth data is smooth.


Theorem 1.18. Let u0 ∈ V and g ∈ L2(J, H). Then the solution u ∈ YT of (1.29),(1.33) belongs to the space C(J, V) ∩ L2(J, H2).

Proof. Using the methods applied for proving Theorem 1.16, it is not difficultto show that for any R > 0 there is τ > 0 such that problem (1.29), (1.33) hasa unique solution u ∈ C(Jτ , V) ∩ L2(Jτ , H2) for any u0 ∈ V and g ∈ L2(Jτ , H)satisfying the inequality

‖u0‖+ ‖g‖L2(Jτ ,H) ≤ R.

Therefore, it suffices to establish that the solution remains bounded in H1. Tothis end, we shall need the lemma below, whose proof is a straightforwardconsequence of Holder and interpolation inequalities.

Lemma 1.19. There is a constant C > 0 such that

|(B(u, u), Lu)| ≤ C ‖u‖1/2‖∇u‖ ‖Lu‖3/2 for any u ∈ H2 ∩ H10 . (1.54)

We now establish an a priori estimate for smooth solutions. We shall confineourselves to a formal derivation, leaving it to the reader to justify the calcula-tions. Taking the scalar product in L2 of (1.29) with 2tLu, we obtain

∂t(t‖∇u‖2)− ‖∇u‖2 + 2t‖Lu‖2 + 2t(B(u, u), Lu) = 2t(g, Lu). (1.55)

In view of the Cauchy inequality and Lemma 1.19, we have

|2t(g, Lu)| ≤ t2‖Lu‖2 + 2t‖g‖2,

|2t(B(u, u), Lu)| ≤ C1t‖u‖1/2‖∇u‖ ‖Lu‖3/2 ≤ t2‖Lu‖2 + C2t‖∇u‖4‖u‖2.

Substituting these inequalities into (1.55), we derive

∂t(t‖∇u‖2)+ t‖Lu‖2 ≤ ‖∇u‖2 + 2t‖g‖2 + C2t‖∇u‖4‖u‖2.

Integration in time results in

ϕ(t) ≤∫ t

0h(s) ds + C2

∫ t

0‖∇u‖2‖u‖2 ϕ(s) ds, (1.56)

where we set

ϕ(t) = t‖∇u(t)‖2 +∫ t

0s‖Lu(s)‖2ds, h(t) = ‖∇u(t)‖2 + 2t‖g(t)‖2.

Applying the Gronwall inequality to (1.56), we obtain

t‖∇u(t)‖2 +∫ t

0s‖Lu(s)‖2ds ≤ C3

(T, ‖u0‖, ‖g‖L2(J,H)

).

This estimate shows, in particular, that the solution remains bounded in H1 onany finite time inteval.

2 EULER EQUATIONS 17

1.4 Remarks on the 3D case

In the 3D case, the situation is much more complicated. Roughly speaking, onecan prove existence of a global weak solution and existence and uniqueness ofa local strong solution. However, it is an open question if there is a functionalclass in which (global) existence and uniqueness hold simultaneously.

In this section, we confine ourselves to the formulation of two results: localexistence and uniqueness of smooth solutions and weak-strong uniqueness.As before, we consider Eq. (1.29).

Definition 1.20. Let T > 0. A function u(t, x) defined in the domain (0, T)×Dis called a strong solution of (1.29) if

u ∈ C(J, V) ∩ L2(J, H2), ∂tu ∈ L2(J, H),

and Eq. (1.29) holds for almost every t ∈ J.

The proofs of the following two results can be found in [Tay97, Soh01].

Theorem 1.21 (local well-posedness). For any R > 0 there is T = T(R) > 0 suchthat if u0 ∈ V and g ∈ L2(JT , H) satisfy the inequality ‖u0‖ + ‖g‖L2(JT ,H) ≤ R,then Eq. (1.29) has a unique strong solution u(t, x) defined on JT and satisfying theinitial condition u(0) = u0.

Theorem 1.22 (weak-strong uniqueness). Let u be a strong solution of (1.29) de-fined on JT and let v be a weak solution of (1.29) that is defined on the same intervaland satisfies the energy inequality

‖v(t)‖2 + 2∫ t

0‖∇v(s)‖2ds ≤ ‖v(0)‖2 + 2

∫ t

0(g(s), v(s)) ds

for almost every 0 ≤ t ≤ T. In this case, if u(0) = v(0), then u ≡ v.

2 Euler equations

In this section, we study the initial-boundary value problem for the incom-pressible Euler system

∂tu + 〈u,∇〉u +∇p = f (t, x), div u = 0, x ∈ D, (2.1)

where D ⊂ Rd is a bounded domain with smooth boundary, and d = 2 or 3. Tosimplify the presentation, we assume that D is simply-connected, i.e., any twocontinuous curves with the same endpoints are homotopic. Let us emphasise,however, that the main results remain valid in the general case. Equations (2.1)are supplemented with the boundary and initial conditions

〈u, n〉∣∣∂D = 0, (2.2)

u(0, x) = u0(x), (2.3)


where n stands for the outward unit normal to ∂D. As for the Navier–Stokessystem, the problem in question is well posed in the 2D case and locally wellposed in the 3D case. Our presentation follows essentially the classical works[Wol33, Kat67].

We begin with a study of some properties of smooth vector fields and cor-responding flows. These results are used in the next two subsections to proveexistence and uniqueness of a global smooth solution in the 2D case. In con-clusion, we discuss briefly some results on the 3D system.

2.1 Smooth vector fields and flows

For an integer, k ≥ 0, we denote by Ck(D) the space of k time continuouslydifferentiable functions v : D → R. If s > 0 is a non-integer, then we write [s]for the integer part of s and denote by Cs(D) the space of functions v ∈ C[s](D)whose derivatives of order [s] are Holder continuous with exponent γ = s− [s].This space is endowed with the natural norm

|v|s = max|α|≤[s]

supx∈D|∂αv(x)|+ max

|α|=[s]sup

0<|x−y|≤1

|∂αv(x)− ∂αv(y)||x− y|γ .

We write Ck = Ck(D, Rd) and Cs = Cs(D, Rd) for similar spaces of Rd-valuedfunctions on D.

Let us fix a time-dependent vector field u ∈ L∞(J, Cs), where J = [0, T] ands > 1 is a non-integer, and consider the ordinary differential equation

x = u(t, x). (2.4)

Proposition 2.1. Suppose that u ∈ L∞(J, Cs) satisfies (2.2) for almost every t ∈ J.Then, for any y ∈ D, Eq. (2.4) has a unique solution x ∈ W1,∞(J, Rd) that satisfiesthe initial condition

x(0) = y. (2.5)

Moreover, the function ϕ : J × D → Rd taking (t, y) to x(t) belongs to W1,∞(J, Cs).

Proof. The local existence, uniqueness, and regularity are standard; for instance,see [CL55, Har82]. The fact that solutions are global is implied by the followingsimple observation. Since the vector field is tangent to ∂D, the solution startingfrom a point y ∈ ∂D remains on the boundary. It follows that if y ∈ D, thenϕ(t, y) ∈ D for any t ∈ J, an therefore a blow-up cannot occur.

Let us denote by LL(D) the space of log-Lipschitz functions, that is, thespace of v ∈ C(D) such that

‖v‖LL := supx∈D|v(x)|+ sup

0<|x−y|≤1

|v(x)− v(y)|λ(|x− y|) < ∞,

where λ(r) = r(| ln r| + 1). We write LL = LL(D, Rd) for the correspondingspace of vector functions.


Proposition 2.2. Let u ∈ L∞(J, LL) be such that (2.2) holds for almost every t ∈ J.Then for any y ∈ D problem (2.4), (2.5) has a unique solution x ∈ W1,∞(J, Rd).Moreover, there is γ > 0 depending only on M := ‖u‖L∞(J,LL) such that the func-tion ϕ(t, y) defined in Proposition 2.1 belongs to the Holder space C1,γ(Q, Rd), whereQ = J×D. Finally, the norm of ϕ in C1,γ(Q, Rd) is bounded by a constant dependingonly on M.

Proof. The local existence is a version of the Peano theorem (see [CL55]), and ifwe have uniqueness, then the same argument as in the proof of Proposition 2.1shows that all solution remain confined in D. Let us sketch the proof of theclaims about uniqueness and Holder continuity.

Let y1, y2 ∈ D be two initial points and let x1, x2 ∈ W1,∞(J, Rd) be thecorresponding solutions of (2.4). Let z(t) = |x1(t)− x2(t)|2. Differentiating zand using the log-Lipschitz property of u, we derive

z ≤ −Mz ln( z

C

),

where C > 1 is a constant not depending on y1 and y2, and the inequalityholds on any interval I ⊂ J such that x1(t) 6= x2(t) for t ∈ I. Applying thecomparison principle (see the Appendix), we obtain

z(t) ≤ C z(s)exp(−M(t−s)) for t, s ∈ I, s ≤ t. (2.6)

This inequality implies the uniqueness and the Holder continuity of the func-tion ϕ(t, y) in y with exponent γ = exp(−MT).

Let us show that ϕ is Lipschitz continuous in t. Since ∂t ϕ(t, y) = u(t, ϕ(t, y)),integrating in time, we see that

|ϕ(t, y1)− ϕ(t, y2)| ≤ ‖u‖L∞ |t1 − t2| for t1, t2 ∈ I, y ∈ D.

This completes the proof of the proposition.

From now on we assume that d = 2 and D is simply-connected, whichmeans that D is homeomorphic to the closed unit disc in R2. For a vector fieldu = (u1, u2) and a scalar function v, we set

curl u = ∂1u2 − ∂2u1, curl v = (∂2v,−∂1v).

The following proposition gives a necessary and sufficient condition for a vec-tor function to be representable as a gradient; cf. Theorem 1.5.

Proposition 2.3. Let D ⊂ R2 be a simply-connected bounded domain with smoothboundary and let s ≥ 0. Then a vector field u ∈ Cs(D) can be written as u = ∇pfor some p ∈ Cs+1(D) if and only if curl u = 0 in D. In this case, there is a constantCs > 0 not depending on u(x) such that

|p− 〈p〉|s+1 ≤ Cs|u|s. (2.7)


Proof. It is clear that if u = ∇p for some p ∈ Cs+1(D), then curl u = 0. Let usprove the converse implication.

By Theorem 1.5, we know that if u ∈ L2 satisfies (1.13), then u = ∇p forsome p ∈ H1(D). In view of Problem 4, in this case we have p ∈ C1(D). Thus,it suffices to show that if curl u = 0, then (1.13) holds. To this end, note that

0 = (curl u, ψ) = (u, curl ψ) for any ψ ∈ C∞0 (D).

It follows that the required assertion will be proved if we show that any ϕ ∈ Vcan be written as ϕ = curl ψ for some ψ ∈ C∞

0 (D).Let us fix any x0 ∈ ∂D and define

ψ(x) =∫

Γ(x0,x)(−ϕ2dy1 + ϕ1dy2), x ∈ D,

where Γ(x0, x) ⊂ D is an arbitrary smooth curve with the endpoints x0 and x.Since D is simply-connected, the Stokes formula (see [Tay97]) and the relationdiv ϕ = 0 imply that ψ is well defined. It is straightforward to verify thatψ ∈ C∞

0 (D) and curl ψ = ϕ. The proof of (2.7) is left to the reader.

For any ω ∈ Cs(D), we denote by ∆−1D ω the unique solution of the equation

∆u = ω satisfying the Dirichlet boundary condition. It is well known that,for any non-integer s > 0, the operator ∆−1

D is continuous from Cs to Cs+2,see [GT01].

Proposition 2.4. Let D ⊂ R2 be a simply-connected bounded domain with smoothboundary and let s > 0 be a non-integer. Then for any ω ∈ Cs(D) the problem

curl u = ω, div u = 0, 〈u, n〉∣∣∂D = 0, (2.8)

has a unique solution u ∈ Cs+1(D, R2). This solution is given by

u = − curl(∆−1D ω). (2.9)

Moreover, if ω ∈ C(D), then the function u(x) defined by (2.9) belongs to LL(D, R2)and satisfies the inequality

‖u‖LL ≤ C‖ω‖L∞ . (2.10)

Proof. It is easy to see that the function u defined by (2.9) belongs to Cs+1(D, R2)and satisfies (2.8). Moreover, if v ∈ Cs+1 is another solution, then w = u− vbelongs to H ∩ Cs+1 and satisfies the relation curl w = 0. By Proposition 2.3,there is p ∈ Cs+1(D) such that w = ∇p. Since ∇H1 is orthogonal to H, weconclude that w = 0.

We now assume that ω ∈ C(D) and prove that u ∈ LL(D, R2). Let usdenote by G(x, y) the Green function of the Dirichlet problem for the Laplaceoperator in the domain D. In other words, G is the kernel of the operator ∆−1

D :(∆−1

D ω)(x) =

∫D

G(x, y)ω(y) dy. (2.11)


It is well known that for any bounded domain with smooth boundary there isa constant C1 > 0 such that

|∂αxG(x, y)| ≤ C1 |x− y|−|α| for x, y ∈ D, (2.12)

where α = (α1, α2) is an arbitrary non-zero multi-index such that |α| ≤ 2. Itfollows from (2.9) and (2.11) that

u(x) = −∫

D(curlx G)(x, y)ω(y) dy. (2.13)

Combining this with (2.12), we obtain

|u(x)| ≤ C2‖ω‖L∞ for x ∈ D. (2.14)

Furthermore, for z ∈ D and r > 0, let us define the set Dr(z) = {y ∈ D :|y− z| < r} and denote by Dc

r(z) its complement in D. Let us fix any pointsx1, x2 ∈ D and set d = |x1 − x2|. It follows from (2.13) and (2.12) that

|u(x1)− u(x2)| ≤∫

D

∣∣((curlx G)(x1, y)− (curlx G)(x2, y))ω(y)

∣∣ dy

=∫

Dc2d(x1)

+∫

D2d(x1)

≤ C3‖ω‖L∞

(d sup

z∈D

∫Dc

d(z)|y− z|−2dy

+∫

D2d(x1)

(|y− x1|−1 + |y− x2|−1) dy

)A simple computation shows that the integrals on the right-hand sides can beestimated by C4d(| ln d|+ 1). Combining this with inequality (2.14), we see thatu ∈ LL(D, R2) and (2.10) holds.

2.2 Reduction to an evolution equation in the 2D case

Let us fix a constant T > 0 and a non-integer s > 0. We denote y Cs the spaceof scalar functions p ∈ Cs(D) with zero mean value. The following definitionconcerns both 2D and 3D cases.

Definition 2.5. Let s > 1 be a non-ineteger and let f ∈ L1(J, Cs). A pair offunctions (u, p) is called a classical solution of the Euler system (2.1), (2.2) if

u ∈ L∞(JT , Cs) ∩W1,1(JT , Cs−1), p ∈ L1(JT , Cs), (2.15)

and Equations (2.1), (2.2) are satisfied in the sense of distributions.

We now show how to reduce formally the 2D Euler system to an evolutionequation. Let us note that if div u = 0, then

curl(〈u,∇〉u

)= 〈u,∇〉 curl u. (2.16)


Therefore, setting ω = curl u and g = curl f and applying the operator curl tothe first relation in (2.1), we obtain the transport equation

∂tω + 〈u,∇〉ω = g(t, x). (2.17)

Taking into account Proposition 2.4, we see that u andω must be connected byrelation

u(t, x) = −(curl ∆−1

D ω)(t, x) for t ∈ J. (2.18)

The following proposition shows that the Euler system is essentially equivalentto the evolution problem (2.17), (2.18).

Proposition 2.6. Let s > 1 be a non-integer, let f ∈ L1(J, Cs), and let (u, p)be a classical solution of the Euler system. Then the function ω = curl u belongsto L∞(J, Cs−1) and satisfies Eq. (2.17), in which g = curl f , and u(t, x) is givenby (2.18). Conversely, if ω ∈ L∞(J, Cs−1) is a solution of Eqs. (2.17), (2.18) withg = curl f , then there is a unique function p ∈ L1(J, Cs) such that the pair (u, p) isa classical solution of (2.1), (2.2).

Proof. The fist part of the proposition is a simple consequence of the abovecalculations. Let us show that to any solution ω ∈ L∞(J, Cs−1) of (2.17), (2.18)there corresponds a unique classical solution of the Euler system.

Relations (2.17), (2.18) imply that

u ∈ L∞(J, Cs), ∂tu = −(curl ∆−1

D)(g− 〈u,∇〉ω) ∈ L1(J, Cs−1). (2.19)

Let us consider the function h = f − ∂tu− 〈u,∇〉u. It follows from (2.19) thath ∈ L1(J, Cs−1). Moreover, in view of (2.17), we have

curl h = g− ∂tω− 〈u,∇〉ω = 0 in the sense of distributions.

Hence, using Proposition 2.3, we conclude that h = ∇p for some p ∈ L1(J, Cs),and the pair (u, p) is a classical solution of the Euler system. The uniquenessof p is a straightforward consequence of Proposition 2.3.

Let us define the space

Csσ = {u ∈ Cs : div u = 0 in D, 〈u, n〉|∂D = 0}.

Proposition 2.6 implies that if u0 ∈ Csσ for some s > 1 and ω ∈ L∞(J, Cs−1) is a

solution of (2.17), (2.18) such that

ω(0, x) = ω0(x), (2.20)

where ω0 = curl u0, then one can construct a unique function p ∈ L1(J, Cs)such that (u, p) is a classical solution of (2.1) – (2.3). In what follows, whentalking about solutions of the Euler system, we shall often omit the function p,because it is uniquely determined by the vector field u.


2.3 Existence and uniqueness in the 2D case

Theorem 2.7. Let D ⊂ R2 be a simply-connected bounded domain with a smoothboundary and let s > 1 be a non-integer. Then for any u0 ∈ Cs

σ problem (2.1) – (2.3)has a unique classical solution (u, p).

Proof. We first outline the scheme of the proof. To prove the uniqueness, we as-sume that there are two classical solutions u1 and u2. In this case, the differenceu = u1 − u2 satisfies the equations

∂tu + 〈u1,∇〉u + 〈u,∇〉u2 +∇p = 0, div u = 0. (2.21)

Multiplying the first equation by 2u, integrating over D, and carrying out somesimple transformations, we arrive at the differential inequality

∂t‖u(t, ·)‖2 ≤ C ‖u(t, ·)‖2.

Application of the Gronwall inequality shows that u ≡ 0.To prove the existence of a solution, we construct a solution of the transport

equation (2.17). It is based on the Leray–Schauder fixed point theorem; see theAppendix. We define a mapping F that takes a continuous scalar functionω(t, x) defined on the cylinder Q = J × D to the solution of the problem

∂tω + 〈u,∇〉ω = g(t, x), ω(0, x) = ω0(x), (2.22)

where g = curl f , ω0 = curl u0, and u = − curl ∆−1D ω. It turns out that F is a

continuous mapping of an appropriately chosen compact subset of C(Q) intoitself. By the Leray–Schauder theorem, it must have a fixed point ω ∈ C(Q).Some further argument shows that ω has the needed regularity.

We now turn to the accurate proof. It is divided into three steps.

Step 1: Uniqueness. Suppose that (ui, pi), i = 1, 2, are two classical solutionsof the Euler system. Setting u = u1 − u2 and p = p1 − p2, we easily verifythat (2.21) holds. Let us set r(t) = ‖u(t, ·)‖2 and calculate ϕ(t). Using the firstrelation in (2.21), we derive

r(t) = 2(∂tu, u) = −2(〈u1,∇〉u + 〈u,∇〉u2 +∇p, u), (2.23)

where the equality holds for almost every t ∈ J. Now note that

(∇p, u) = 0, (〈u1,∇〉u, u) = 0,

|(〈u,∇〉u2, u)| ≤ C1‖∇u2‖L∞‖u‖2.


r(t) ≤ C2r(t) for almost every t ∈ J,

where C2 > 0 is a constant depending only on u2. Integrating in time, applyingthe Gronwall inequality, and recalling that r(0) = 0, we conclude that r ≡ 0.


Step 2: Fixed point argument. We now assume that 1 < s < 2 and construct asolution ω ∈ L∞(J, Cs−1) of problem (2.17), (2.18). We fix positive constants M,C, δ and a non-decreasing function ρ(r) ≥ 0 defined for r ≥ 0 and vanishingat r = 0. Denote by K = K(M, C, δ, ρ) the set of functions ω ∈ C(Q) such that

|ω(t, x)| ≤ M, |ω(t1, x1)−ω(t2, x2)| ≤ ρ(|t1 − t2|) + C |x1 − x2|δ

for all (t1, x1), (t2, x2) ∈ Q. By the Arzela–Ascoli theorem, K is a compact con-vex set in C(Q). Let us define an operator F : K → C(Q) taking a functionω ∈ K to the solution ω(t, x) of problem (2.22). We claim that, for an appropri-ate choice of M, C, δ, and ρ, the operator F is well defined, maps the set K intoitself, and is continuous.

Indeed, let us denote by ϕ(t, y) the flow associated with the time-dependentvector field u(t, x) and by ψ(t, x) the inverse of ϕ regarded as a function of y.Since ω ∈ C(Q) and ‖ω‖L∞ ≤ M, Proposition 2.4 implies that

‖u‖L∞(J,LL) ≤ M1,

where M1 > 0 depends only on M. Proposition 2.2 now implies that

|ψ(t, x)| ≤ C1, |ψ(t1, x1)− ψ(t2, x2)| ≤ C2(|t1 − t2|+ |x1 − x2|γ), (2.24)

and similar estimates holds for ϕ. Here C1, C2, and γ are some constants de-pending only on M.

Now recall that the solution of problem (2.22) is uniquely determined bythe relation

ω(t, ϕ(t, y)) = ω0(y) +∫ t

0g(τ, ϕ(τ, y)

)dτ, (t, y) ∈ Q. (2.25)

Setting x = ϕ(t, y), we can rewrite (2.25) in the form

ω(t, x) = ω0(ψ(t, x)) +∫ t

0g(τ, ϕ(τ, ψ(t, x))

)dτ, (t, x) ∈ Q. (2.26)

It follows from (2.24) and (2.26) that

|ω(t, x)| ≤ ‖ω0‖L∞ +∫ t

0‖g(τ, ·)‖L∞ dτ,

|ω(t, x1)−ω(t, x2)| ≤ C3|x1 − x2|γ(s−1) + C4|x1 − x2|γ2(s−1),

|ω(t1, x)−ω(t2, x)| ≤ C3|t1 − t2|s−1 + C4|t1 − t2|γ(s−1) +∣∣∣∣∫ t2

t1

‖g(τ, ·)‖L∞ dτ

∣∣∣∣,where C3 = C3(ω0, M) and C4 = C4(g, M) are some positive constants. Choos-ing

M = ‖ω0‖L∞ +∫ T

0‖g(τ, ·)‖L∞ dτ, C = C3 + C4,

ρ(r) = C3rs−1 + C4rγ(s−1) + sup0≤t1−t2≤r

∫ t2

t1

‖g(τ, ·)‖L∞ dτ, δ = γ2(s− 1),


we see that F (K) ⊂ K.We now show that F is continuous. Let {ωn} ⊂ K be a sequence that

converges to ω ∈ K. We set ωn = F (ωn) and ω = F (ω). Since K is compactin C(Q), there is nj → ∞ such that {ωnj} converges to a function ω ∈ K. Onthe other hand, the continuous dependence of solutions to ODE’s on the vectorfield implies that ωn(t, x) must converge to ω(t, x) for any (t, x) ∈ Q. It followsthat ω ≡ ω, and the entire sequence {ωn} converges to ω.

We have thus shown that F : K → K is a continuous mapping. By theLeray–Schauder theorem, there is ω ∈ K ⊂ L∞(J, Cδ) such that F (ω) = ω. Itis clear that ω is a solution of (2.17), (2.18).

Step 3: Regularity. To complete the proof of the theorem, it remains toshow that ω ∈ L∞(J, Cs−1). Suppose that 1 < s < 2. Since ω ∈ L∞(J, Cδ),Proposition 2.4 implies that u ∈ L∞(J, C1+δ). By Proposition 2.1, we see thatϕ, ψ ∈ W1,∞(J, C1+δ). Combining this with the explicit formula (2.26), we con-clude that ω ∈ L∞(J, Cs−1).

We now assume that s ∈ (k, k + 1) for some integer k ≥ 2 and it is alreadyproved that ω ∈ L∞(J, Cr−1) for any r < k. Repeating the above argument,we see that u ∈ L∞(J, Cr) and ϕ, ψ ∈ W1,∞(J, Cr). Using again representa-tion (2.26) for ω, we obtain the required result. The proof is complete.

2.4 The 3D case: Beale-Kato-Majda theorem

We now discuss some results concerning the 3D Euler system. For a smoothvector field u = (u1, u2, u3), we define the vorticity as

curl u = (∂2u3 − ∂3u2, ∂3u1 − ∂1u3, ∂1u2 − ∂2u1).

Let us set ω = curl u and g = curl f . Applying the operator curl to the firstequation in (2.1) and taking into account the relations

curl(∇p) = 0, curl(〈u,∇〉u

)= 〈u,∇〉ω− 〈ω,∇〉u,

we obtain the following system for ω:

∂tω + 〈u,∇〉ω− 〈ω,∇〉u = g(t, x). (2.27)

Compared to the 2D case, system (2.27) contains the additional nonlinear term〈ω,∇〉u. If we try to construct a solution with the help of a fixed point argu-ment similar to that applied in the previous subsection, we shall encounter thefollowing difficulty. The estimates that can be obtained for solutions of the as-sociated transport equations are not sufficient to construct an invariant subset.This can by done only if T is sufficiently small. More precisely, we have thefollowing result (see [EM70, Tem76]).

Theorem 2.8. Let D ⊂ R3 be a bounded domain with smooth boundary and let s > 1be a non-integer. For any R > 0 there is T = T(R) > 0 such that if u0 ∈ Cs

σ(D, R3),f ∈ L1(JT , Cs), and

|u0|s + ‖ f ‖L1(JT ,Cs) ≤ R,

3 CONTROLLABILITY 26

then problem (2.1) – (2.3) has a unique classical solution (u, p) on the interval JT .

It is still an open question to decide if a blow-up can happen for smoothinitial data. It turns out, however, that if a classical solution cannot be contin-ued beyond some point, then the L∞ norm of the vorticity must blow up. Thefollowing result is established in [BKM84] for the case of the entire space or thetorus and in [Fer93] for a general bounded domain.

Theorem 2.9. Let T > 0 be a constant and let s > 1 be a non-integer. For anyu0 ∈ Cσ(D, R3) and f ∈ L1(JT , Cs), let us set

T∗ = sup{τ ∈ JT : problem (2.1) – (2.3) has a classical solution on [0, τ]}.

In this case, if T∗ < T, then ∫ T∗

0‖ω(t)‖L∞ dt = ∞.

3 Controllability

In this section, we discuss the problem of controllability for some nonlinearPDE’s. Our presentation is based on the approach introduced by Agrachevand Sarychev in [AS05]. For simplicity, we study in detail only the case of the1D Burgers equation

∂tu− ν∂2xu + u∂xu = f (t, x), (3.1)

where x ∈ (0, π), t > 0, ν > 0 is a parameter, and f is a given function.Equation (3.1) is supplemented with the boundary and initial conditions

u(0, t) = u(π, t) = 0, (3.2)u(0, x) = u0(x). (3.3)

We first recall a theorem on well-posedness of problem (3.1) – (3.3). We next de-scribe the concept of controllability we are interested in and formulate the mainresult. The principal ideas of the Agrachev–Sarychev approach are presentedin Subsection 3.3, and the details are given in Subsection 3.4. We conclude thissection by discussing some results on the Navier–Stokes and Euler systems andformulating two open questions.

3.1 Cauchy problem for Burgers equation

Let us fix a constant T > 0 and set Q = JT × [0, π]. To simplify the notation,we shall write

H = L2(0, π), H10 = H1

0(0, π), ZT = C(J, H) ∩ L2(J, H10).

The following result can be established by using the same methods as in thecase of Navier–Stokes system. However, the corresponding arguments aremuch simpler.


Proposition 3.1. For any f ∈ L2(Q) and u0 ∈ H, problem (3.1) – (3.3) has a uniquesolution u ∈ ZT . Moreover, the operator R : H × L2(J, H) → ZT taking a pair(u0, f ) to the solution u ∈ ZT is uniformly Lipschitz continuous on bounded subsets.

In what follows, we denote by Rt : H × L2(Q) → H the restriction of R attime t ∈ J. Proposition 3.1 implies thatRt satisfies the inequality

‖Rt(u0, f )−Rt(v0, g)‖ ≤ C(R)(‖u0 − v0‖+ ‖ f − g‖L2(Q)

), (3.4)

where u0, v0 and f , g belong to the balls of radius R centred at origin in thespaces H and L2(Q), respectively, and C(R) > 0 is a constant depending onlyon R.

3.2 Formulation of the main result

We now fix a finite-dimensional space E ⊂ H and assume that

f (t, x) = h(t, x) + η(t, x), (3.5)

where h ∈ L2(Q) is a given function and η is a control with range in E.

Definition 3.2. We shall say that problem (3.1), (3.2), (3.5) is controllable at time Tby an E-valued control if for any constant ε > 0, any functions u0, u ∈ H, andany finite-dimensional subspace F ⊂ H there is a control η ∈ C∞(J, E) suchthat

‖RT(u0, h + η)− u‖ < ε, (3.6)PFRT(u0, h + η) = PFu, (3.7)

where PF : H → H stands for the orthogonal projection in H onto F.

The following theorem is the main result of this section. It was first estab-lished in [AS05] for a more complicated case of the Navier–Stokes equation.

Theorem 3.3. Let E be the vector space spanned by the function sin x and sin(2x).Then, for any ν > 0, T > 0, and h ∈ L2(Q), problem (3.1), (3.2), (3.5) is controllableat time T by an E-valued control.

We now present the scheme of the proof of the above theorem; the detailsare given in Subsection 3.4.

3.3 Agrachev–Sarychev approach

We shall show that the required result is a consequence of the so-called uniformapproximate controllability. We then outline the proof of the latter property.


Reduction to uniform approximate controllability

Definition 3.4. Let us fix a constant ε > 0, a function u0 ∈ H, and a compactset K ⊂ H. Problem (3.1), (3.2), (3.5) is said to be (ε, u0,K)-controllable at time Tby an E-valued control if there is a continuous mapping Ψ : K → L2(J, E) suchthat

supu∈K‖RT(u0, h + Ψ(u))− u‖ < ε. (3.8)

In what follows, the time T and the control space E are fixed, and we shallsimply say that Eq. (3.1) is (ε, u0,K)-controllable.

Definition 3.5. Problem (3.1), (3.2), (3.5) is said to be uniformly approximatelycontrollable if it is (ε, u0,K)-controllable for any ε > 0, u0 ∈ H, and K b H.

Theorem 3.3 is a consequence of the following result. Its proof is sketchedbelow.

Theorem 3.6. Let E be the vector span of the functions sin x and sin(2x). Then forany ν > 0 and h ∈ L2(Q) problem (3.1), (3.2), (3.5) is uniformly approximatelycontrollable by an E-valued control.

Proof of Theorem 3.3. Let us fix a constant ε > 0, functions u0, u ∈ H, and afinite-dimensional space F ⊂ H. Without loss of generality, we can assumethat u ∈ F; otherwise, we can replace F by the larger space spanned by Fand u.

Let us denote by BF(R) the ball in F of radius R centred at origin and de-fine K = BF(‖u‖ + ε). Since K is a compact subset of H, in view of Theo-rem 3.6, we can construct a continuous mapping Ψ : K → L2(J, E) satisfyinginequality (3.8). Furthermore, since K ⊂ H is compact and C∞(J, E) is densein L2(J, E), we can assume that the range of Ψ is contained in C∞(J, E); other-wise, we can replace the function Ψ by its convolution with a mollifying kernel.Let us consider the mapping

Φ : K → F, Φ(u) = PFRT(u0, h + Ψ(u)).

It follows from (3.8) that Φ is a continuous mapping satisfying the inequality

supu∈K‖Φ(u)− u‖ < ε.

Proposition 4.8 (see the Appendix) implies that the image of Φ contains the ballBF(‖u‖). In particular, there is u ∈ K such that Φ(u) = u. Setting η = Ψ(u),we see that

PFRT(u0, h + η) = u. (3.9)

Furthermore, it follows from (3.8) and (3.9) that

‖RT(u0, h + η)− u‖ = ‖RT(u0, h + η)− PFRT(u0, h + η)‖≤ ‖RT(u0, h + Ψ(u))− u‖ < ε,

where we used the facts that u ∈ F and that PF is an orthogonal projection.This completes the proof of Theorem 3.3.


Scheme of the proof of Theorem 3.6

Let us fix a constant ε > 0, a function u0 ∈ H, and a compact set K b H. Weneed to show that Eq. (3.1) is (ε, u0,K)-controllable by an E-valued control.

Step 1: Extension principle. Let G ⊂ H2 ∩ H10 be an arbitrary finite-di-

mensional subspace. Along with (3.1), consider the equation

∂tu− ν∂2x(u + ζ(t, x)) + (u + ζ(t, x))∂x(u + ζ(t, x)) = h(t, x) + η(t, x), (3.10)

where η and ζ are G-valued control functions. This is a Burgers-type equa-tion, and using the same methods as for the Navier–Stokes system, one canprove that the Cauchy problem for (3.10) is well posed. We shall say that prob-lem (3.10), (3.2) is (ε, u0,K)-controllable by G-valued controls (η, ζ) if there is acontinuous mapping Ψ : K → L2(J, G× G) such that

supu∈K‖RT(u0, Ψ(u))− u‖ < ε, (3.11)

where Rt : H × L2(J, G× G) → H stands for the operator that takes the triple(u0, η, ζ) to the solution u(t, ·) of problem (3.10), (3.2), (3.3).

Even though Eq. (3.10) is “more controlled” than Eq. (3.1), it turns out thatthe property of uniform approximate controllability is equivalent for them.Namely, we have the following result.

Proposition 3.7. Equation (3.1) is (ε, u0,K)-controllable if and only if so is prob-lem (3.10), (3.2).

Step 2: Convexification principle. Now let N ⊂ H2 ∩ H10 be another finite-

dimensional subspace such that

N ⊂ G, B(N) ⊂ G, (3.12)

where B(u) = u∂xu. Denote by F (N, G) the intersection of H2 ∩ H10 with the

vector space spanned by the functions of the form2

η + ξ∂x ξ + ξ∂xξ, (3.13)

where η, ξ ∈ G and ξ ∈ N. It is easy to see thatF (N, G) is a well-defined finite-dimensional space. The following proposition, which is an infinite-dimensionalanalogue of the well-known convexification principle for controlled ODE’s (e.g.,see [AS04, Theorem 8.7]), is a key point of the proof of Theorem 3.6.

Proposition 3.8. Let N and G be finite-dimensional subspaces in H2 ∩ H10 that sat-

isfy (3.12). Then (3.10), (3.2) is (ε, u0,K)-controllable by a G × G-valued control ifand only if (3.1), (3.2), (3.5) is (ε, u0,K)-controllable by an F (N, G)-valued control.

2Note that a function of the form (3.13) does not necessarily belong to H2 ∩ H10 , and therefore

the space F (N, G) may be not larger than G.


Step 3: Saturating property. Propositions 3.7 and 3.8 imply the followingresult, which is a kind of “relaxation property” for the controlled Navier–Stokessystem.

Proposition 3.9. Let N and G be finite-dimensional subspaces in H2 ∩ H10 that sat-

isfy (3.12). Then Eq. (3.1) is (ε, u0,K)-controllable by a G-valued control if and onlyif it is (ε, u0,K)-controllable by an F (N, G)-valued control.

We now introduce the subspaces Ek = {sin(jx), 1 ≤ j ≤ k}, so that thespace E defined in Theorem 3.3 coincides with E2. We wish to apply Proposi-tion 3.9 to the subspaces N = E1 and G = Ek.

Lemma 3.10. For any integer k ≥ 2, we have F (E1, Ek) = Ek+1.

Proposition 3.9 and Lemma 3.10 imply that, for any integer k ≥ 2, prob-lem (3.1), (3.2), (3.5) is (ε, u0,K)-controllable by an Ek-valued control if andonly if it is (ε, u0,K)-controllable by an Ek+1-valued control. Thus, Theorem 3.6will be established if we find an integer N ≥ 2 such that Eq. (3.1) is (ε, u0,K)-controllable by an EN-valued control. We shall be able to do that due to thesaturating property

∞⋃k=2

Ek is dense in H, (3.14)

which is a straightforward consequence of the definition of Ek.

Step 4: Case of a large control space. It is easy to construct a continuousmapping Ψ0 : K → L2(J, H) such that

supu∈K‖RT(u0, h + Ψ0(u))− u‖ < ε. (3.15)

Since K ⊂ H is a compact set, the image Ψ0(K) is compact in L2(J, H). Us-ing (3.14), it is not difficult to approximate Ψ0, within any accuracy δ > 0, by acontinuous function Ψ : K → L2(J, H) with range in L2(J, EN):

supu∈K‖Ψ0(u)−Ψ(u)‖ < δ. (3.16)

Since the function Rt(u0, h + η) is Lipschitz continuous on bounded subsets,inequalities (3.15) and (3.16) with δ � 1 imply (3.8). This completes the proofof Theorem 3.6.

3.4 Details of proof of Theorem 3.3

To simplify the presentation, we shall assume that K consists of a single pointu ∈ H. The proof in the general case can be carried out by similar arguments,following carefully the dependence of all the objects on the final point u; cf.the papers [AS05, Shi07] and Problems 12, 13. In what follows, the constant ε,the function u0, and the subset K = {u} are fixed, and we shall say simplyε-controllable rather than (ε, u0,K)-controllable.


3.4.1 Extension principle

In this subsection, we prove Proposition 3.7. It is clear that if problem (3.1), (3.2)is ε-controllable, then so is problem (3.10), (3.2), because it suffices to take ζ ≡ 0.Let us establish the converse assertion.

Let (η, ζ) ∈ L2(J, G) be an arbitrary control such that

‖RT(u0, η, ζ)− u‖ < ε. (3.17)

In view of continuity of RT(u0, η, ζ) with respect to ζ ∈ L2(J, H), there is noloss of generality in assuming that

ζ ∈ C∞(J, G), ζ(0) = ζ(T) = 0. (3.18)

Consider the function u(t, x) = Rt(u0, η, ζ) + ζ(t, x). It is straightforwardto see that it belongs to ZT and satisfies Eqs. (3.1), (3.2) with η = η + ∂t ζ ∈L2(J, G). Moreover, it follows from (3.17) and (3.18) that

u(0) = u0, ‖u(T)− u‖ = ‖RT(u0, η, ζ)− u‖ < ε.

Thus, problem (3.1), (3.2), (3.5) is ε-controllable.

3.4.2 Convexification principle

Let us prove Proposition 3.8. It follows from the extension principle that ifproblem (3.10), (3.2) is ε-controllable by a G× G-valued control, then Eq. (3.1)is ε-controllable by a G-valued control and all the more by an F (N, G)-valuedcontrol. The proof of the converse assertion is divided into several steps. Weneed to show that if η1 ∈ L2(J, H) is an F (N, G)-valued control such that

‖RT(u0, h + η1)− u‖ < ε, (3.19)

then there are η, ζ ∈ L2(J, G) such that

‖RT(u0, η, ζ)− u‖ < ε. (3.20)

Step 1. We first show that it suffices to consider the case in which η1 is apiecewise constant function. Indeed, suppose Proposition 3.8 is proved in thatcase and denote G1 = F (N, G). For a given η1 ∈ L2(J, G1), we can find asequence {ηm} of piecewise constant G1-valued functions such that

‖η1 − ηm‖L2(J,G1) → 0 as m→ ∞.

By continuity ofRt, there is an integer n ≥ 1 such that

‖RT(u0, h + ηn)− u‖ < ε. (3.21)


Since the result is true for piecewise constant controls, for any δ > 0 there areη, ζ ∈ L2(J, G) such that

‖RT(u0, h + ηn)− RT(u0, η, ζ)‖ < δ. (3.22)

Comparing (3.21) and (3.22), for a sufficiently small δ > 0 we arrive at (3.20).

Step 2. We now consider the case of piecewise constant G1-valued controls.A simple iteration argument combined with the continuity ofRt and Rt showsthat it suffices to consider the case of one interval of constancy. Thus, we shallassume that η1(t) ≡ η1 ∈ G1.

We shall need the lemma below, whose proof is given at the end of thissubsection. Recall that B(u) = u∂xu.

Lemma 3.11. For any η1 ∈ F (N, G) and any δ > 0 there is an integer k ≥ 1,constants αj > 0, and vectors η, ζ j ∈ G, j = 1, . . . , k, such that

k

∑j=1

αj = 1, (3.23)

∥∥∥η1 − B(u)−(

η −k

∑j=1

αj(

B(u + ζ j)− ν∂2xζ j))∥∥∥ ≤ δ for any u ∈ H1. (3.24)

We fix a small δ > 0 and choose constants αj > 0 and vectors η, ζ j ∈ Gsatisfying (3.23), (3.24). Let us consider the equation

∂tu− ν∂2xu +

k

∑j=1

αj(

B(u + ζ j(x))− ν∂2xζ j(x)

)= h(t, x) + η(x). (3.25)

This is a Burgers-type equation, and using the same arguments as in the caseof the Navier–Stokes system, it can be proved that problem (3.25), (3.2), (3.3)has a unique solution u ∈ ZT . On the other hand, we can rewrite (3.25) in theform

∂tu− ν∂2xu + u∂xu = h(t, x) + η1(x)− rδ(t, x), (3.26)

where rδ(t, x) stands for the function under sign of norm on the left-hand sideof (3.24) in which u = u(t, x). Since Rt is Lipschitz continuous on boundedsubsets, there is a constant C > 0 depending only on the L2 norm of η1 suchthat

‖RT(u0, h + η1)− u(T)‖ = ‖RT(u0, h + η1)−RT(u0, h + η1 − rδ)‖

≤ C‖rδ‖L2(J,H) ≤ C√

Tδ,

where we used inequality (3.24). Combining this with (3.19), we see that ifδ > 0 is sufficiently small, then

‖u(T)− u‖ < ε. (3.27)


We shall show that there is a sequence ζm ∈ L2(J, G) such that

‖RT(u0, η, ζm)− u(T)‖ → 0 as m→ ∞. (3.28)

In this case, inequalities (3.27) and (3.28) with m � 1 will imply the requiredestimate (3.20) in which ζ = ζm.

Step 3. Following a classical idea in the control theory, we define a sequenceζm ∈ L2(J, G) by the relation ζm(t) = ζ(mt/T), where ζ(t) is a 1-periodic G-valued function such that

ζ(t) = ζ j for 0 ≤ t− (α1 + · · ·+ αj−1) ≤ αj, j = 1, . . . , k.

Let us rewrite (3.25) in the form

∂tu− ν∂2x(u + ζm(t, x)) + B(u + ζm(t, x)) = h(t, x) + η(x) + fm(t, x),

where we set fm = fm1 + fm2,

fm1 = −ν∂2xζm + ν

k

∑j=1

αj∂2xζ j, (3.29)

fm2 = B(u + ζm)−k

∑j=1

αjB(u + ζ j). (3.30)

We now define an operator K : L2(J, H) → ZT that takes a function f to thesolution u(t, x) of the equation

∂tu− ν∂2xu = f (t, x),

supplemented with initial and boundary conditions (3.2), (3.3) with u0 = 0. Inother words,

(K f )(t) =∫ t

0eν(t−s)A f (s) ds,

where A stands for the operator d2

dx2 with the domainD(A) = H2 ∩H10 . Setting

vm = u− K fm, we see that vm ∈ ZT satisfies the equation

∂tv− ν∂2x(v + ζm) + B(v + ζm + K fm) = h + η. (3.31)

Suppose we have shown that

‖K fm‖ZT → 0 as m→ ∞. (3.32)

Then, by the Lipschitz continuity of the resolving operator for (3.31) on boundedsubsets, we have

‖RT(u0, η, ζm)− u(T)‖ ≤ ‖RT(u0, η, ζm)− vm(T)‖+ ‖K fm(T)‖ → 0


as m→ ∞. Thus, it remains to prove (3.32).

Step 4. We first note that { fm} is a bounded sequence in L2(J, H). It followsthat

‖K fm‖C(J,H1) + ‖K fm‖L2(J,H2) ≤ C1, (3.33)

where we denote by Ci, i = 1, 2, . . . , positive constants not depending on m.Furthermore, we have the interpolation inequalities

‖v‖ ≤ C2‖v‖1/21 ‖v‖

1/2−1 , ‖v‖1 ≤ C3‖v‖2/3

2 ‖v‖1/3−1 for v ∈ H2 ∩ H1

0 .


‖K fm‖ZT ≤ ‖K fm‖C(J,H) + ‖K fm‖L2(J,H1)

≤ C4

(‖K fm‖1/2

C(J,H−1) + ‖K fm‖1/3L2(J,H−1)

).

Thus, convergence (3.32) will be established if we show that

‖K fm‖C(J,H−1) → 0 as m→ ∞. (3.34)

Step 5. To prove (3.34), we write

(K fm)(t) = Fm(t) + Gm(t), (3.35)

where

Fm(t) =∫ t

0fm(s) ds, Gm(t) = ν

∫ t

0Aeν(t−s)AFm(s) ds.

Since ‖AeτA‖L(H,H−1) ≤ C5τ−1/2 for τ > 0, where ‖ · ‖L(H,H−1) stands for theusual norm of operators from H to H−1, we have

‖Gm‖C(J,H−1) ≤ ν supt∈[0,T]

∫ t

0‖Aeν(t−s)A‖L(H,H−1)‖Fm(s)‖ ds

≤ C6 ‖Fm‖C(J,H).

Comparing this with (3.35), we see that (3.34) will be established if we showthat

‖Fm‖C(J,H) → 0 as m→ ∞. (3.36)

This convergence is a straightforward consequence of relations (3.29) and (3.30);cf. [Shi06, Section 3.3]. The proof of Proposition 3.8 is complete.

Proof of Lemma 3.11. It suffices to find functions η, ζ j ∈ G, j = 1, . . . , m, suchthat ∥∥∥η1 − η +

k

∑j=1

B(ζ j)∥∥∥ ≤ δ. (3.37)


If such vectors are constructed, then we can set k = 2m,

αj = αj+m =12

, ζ j = −ζ j+m = ζ j for j = 1, . . . , m.

To construct η, ζ j ∈ G satisfying (3.37), note that if η1 ∈ F (N, G), then thereare functions ηj, ξ j ∈ G and ξ j ∈ N such that

η1 =k

∑j=1

(ηj − ξ j∂x ξ j − ξ j∂xξ j

). (3.38)

Now note that, for any ε > 0,

ξ j∂x ξ j + ξ j∂xξ j = B(εξ j + ε−1ξ j)− ε2B(ξ j)− ε−2B(ξ j).


η1 −k

∑j=1

(ηj + ε−2B(ξ j)

)+

k

∑j=1

B(εξ j + ε−1ξ j) = ε2k

∑j=1

B(ξ j).

Choosing ε > 0 sufficiently small and setting

η =k

∑j=1

(ηj + ε−2B(ξ j)

), ζ j = εξ j + ε−1ξ j,

we arrive at (3.37).

3.4.3 Saturating property

Let us prove Lemma 3.10 and the inclusion B(E1) ⊂ E2. For ξ = sin(jx) andξ = sin x, we have

ξ∂x ξ + ξ∂xξ = sin(jx) cos x + j sin x cos(jx)

=12((j + 1) sin(j + 1)x− (j− 1) sin(j− 1)x

). (3.39)

It follows that B(E1) ⊂ E2 and F (E1, Ek) ⊂ Ek+1. Furthermore, taking j = kin (3.39), we write

sin(k + 1)x =k− 1k + 1

sin(k− 1)x +2

k + 1(sin(kx) ∂x sin x + sin x ∂x sin(kx)

).

This relation implies that the function sin(k + 1)x belongs to F (E1, Ek) andtherefore Ek+1 ⊂ F (E1, Ek).


3.4.4 Case of a large control space

We wish to construct a control η ∈ L2(J, EN) with a large integer N ≥ 2 suchthat

‖RT(u0, h + η)− u‖ < ε. (3.40)

To this end, consider the function uµ(t, x) defined as

uµ(t, x) = T−1(teµAu + (T − t)eνtAu0), (3.41)

where A denotes the operator d2

dx2 with the domain D(A) = H2 ∩ H10 , and

µ > 0 is a small constant that will be chosen later. The function uµ belongs tothe space ZT and satisfies Eqs. (3.1) – (3.3) and (3.5), in which

η = ηµ := ∂tuµ − ν∂2xuµ + uµ∂xuµ − h. (3.42)

This function belongs to L2(J, H). Furthermore,

‖uµ(T)− u‖ = ‖eµAu− u‖ → 0 as µ→ 0. (3.43)

Choosing µ > 0 sufficiently small in (3.43) and approaching ηµ ∈ L2(J, H) bycontinuous H-valued functions, we can find η ∈ C(J, H) such that

‖RT(u0, h + η)− u‖ < ε. (3.44)

Let us denote by Pk : H → H the orthogonal projection in H onto thesubspace Ek. In view of the saturating property (3.14), we have

supt∈[0,T]

‖Pkη(t)− η(t)‖ → 0 as k→ ∞.

By continuity ofRt, we obtain

‖RT(u0, h + Pkη)−RT(u0, h + η)‖ → 0 as k→ ∞.

Combining this with (3.44), we see that for a sufficiently large N ≥ 1 the func-tion η = PN η satisfies (3.40). This completes the proof of Theorem 3.6 in thecase K = {u}.

3.5 Remarks on the Navier–Stokes and Euler equations

Let us turn to the problem of controllability for the Navier–Stokes and Eulerequations. It was established by Agrachev and Sarychev [AS05, AS06] that, forboth problems considered on a 2D torus, the velocity field can be controlledin the sense of Definition 3.2 by a finite-dimensonal external force. Their re-sults were extended to other boundary conditions by Rodrigues [Rod06] andto 3D Navier–Stokes equations on a torus by the author [Shi06, Shi07]. Thecase of a general Riemannian manifold satisfying some topological constraints


was studied in [AS07, Rod07]. More recently, some progress has been towardscontrolling the finite-dimensional projections of the pressure [Ner08].

Let us formulate without proof a result on controllability of the Navier–Stokes system on the 2D torus T2 = R2/2πZ2. Denote by E ⊂ L2(T2, R2) thesix-dimensional vector space spanned by the functions(

10

),(

01

),(

sin x2

0

),(

cos x2

0

),(

sin(x1 + x2)− sin(x1 + x2)

),(

cos(x1 + x2)− cos(x1 + x2)

).

Consider the control problem

∂tu + 〈u,∇〉u− ∆u +∇p = h(t, x) + η(t, x), div u = 0, (3.45)u(0, x) = u0(x), (3.46)

where x ∈ T2, t ∈ JT , h ∈ L2(JT ×T2) is a given function, and η is a controlwith range in E. Let H be the space of divergence-free functions that belongto L2(T2, R2). The following result is established in [AS06].

Theorem 3.12. For any ε > 0, any functions u0, u ∈ H, and any finite-dimensionalsubspace F ⊂ H, there is a control η ∈ C∞(JT , E) such that the weak solution (u, p)of problem (3.45), (3.46) satisfies the relations

‖u(T, ·)− u‖ < ε, PFu(T, ·) = PFu,

where PF : H → H denotes the orthogonal projection onto F.

The above-mentioned results on controllability of the Navier–Stokes andEuler systems have one common point: for all of them, the boundary condi-tions ensure that the Leray projection commutes with the Laplacian. It is achallenging problem to construct an example for which that property does nothold and still the controllability is true. For instance, the following question iscompletely open.

Problem 1. Let us define the strip D = R × (0, 1) and consider the controlledNavier–Stokes system (3.45) in D supplemented with the periodicity condition in thehorizontal direction and the Dirichlet condition in the vertical one:

u(x1, 0) = u(x1, 1) = 0, u(x1 + 1, x2) = u(x1, x2), p(x1 + 1, x2) = p(x1, x2).

Is it possible to find a finite-dimensional vector space E ⊂ L2 such that Eq. (3.45) iscontrollable in the sense of Definition 3.2 by an E-valued control η(t, x)?

Another question of interest concerns the problem of exact controllability. Itis well known that Eq. (3.45) considered on the torus T2 is exactly controllableby an external force η supported by a given open set ω ⊂ T2; see the referencesin [Fur00, Cor07]. Nothing is known for a similar problem in the case of a finite-dimensional control (see [Shi08] for the case of the Euler system).

Problem 2. Suppose that x ∈ T2 and h ∈ C∞(R+ × T2). Let (u, p) be a weaksolution of Eq. (3.45) with η ≡ 0. Is it possible to find a finite-dimensional spaceE ⊂ L2 such that for any u0 ∈ H there is a time T > 0 and a control η ∈ L2(JT , E)for which u(T) = u(T), where (u, p) stands for the weak solution of (3.45), (3.46)?

4 APPENDIX 38

4 Appendix

Gronwall inequality

Proposition 4.1. Let J ⊂ R be a closed interval, let τ ∈ J, let b ≥ 0 be a constant,and let ϕ ∈ C(J) and a ∈ L1(J) be non-negative functions such that

ϕ(t) ≤∣∣∣∣∫ t

τa(s)ϕ(s) ds

∣∣∣∣+ b for t ∈ J. (4.1)

Then

ϕ(t) ≤ b exp∣∣∣∣∫ t

τa(s) ds

∣∣∣∣ for t ∈ J. (4.2)

Proof. It suffices to establish (4.1) for t ≥ τ; the general case can be reduced tothe former by the change of variable t = τ − s. Consider the function

ψ(t) = exp(−A(t))(∫ t

τa(s)ϕ(s) ds + b

),

where A(t) =∫ t

τ a(s) ds. Inequality (4.1) implies that

ψ(t) = a(t) exp(−A(t))(

ϕ(t)−∫ t

τa(s)ϕ(s) ds− b

)≤ 0,

whence it follows thatψ(t) ≤ ψ(τ) = b.

Recalling the definition of ψ, we obtain the inequality

ϕ(t) ≤ ψ(t) exp(A(t)) ≤ b exp(A(t)),

which coincides with (4.2) for t ≥ τ.

Comparison principle

Proposition 4.2. Let Λ ⊂ R be an open interval, let V ∈ C1(Λ), and let x and y betwo absolutely continuous functions on J = [0, T] such that

x(t), y(t) ∈ Λ for t ∈ J, (4.3)x(t) ≤ V(x(t)), y(t) = V(y(t)) for almost every t ∈ J. (4.4)

In this case, if x(0) ≤ y(0), then x(t) ≤ y(t) for all t ∈ J.

Proof. For any ε > 0 we denote by yε(t) the solution of the problem

z = V(z) + ε, z(0) = y(0).

4 APPENDIX 39

It follows from (4.3) that yε is defined on an interval [0, Tε], where Tε ≤ T andTε → T as ε→ 0. Moreover,

yε(t)→ y(t) as ε→ 0 for any t ∈ [0, T). (4.5)

Let us setτε = max{t ∈ [0, Tε] : x(s) ≤ yε(s) for 0 ≤ s ≤ t}.

If τε < Tε, then x(τε) = yε(τε). Furthermore, it follows from (4.4) and thedefinition of yε that x(τε) < yε(τε). We conclude that x(t) < yε(t) on someinterval [τε, τε + δ]. This contradicts the definition of τε, and therefore τε = Tε.

We have thus shown that x(t) ≤ yε(t) for 0 ≤ t ≤ Tε. Passing to the limitas ε→ 0 and taking into account (4.5), we arrive at the required result.

Poincare and Friedrichs inequalities

The following elegant result is taken from Konkov’s lecture notes [Kon02].

Theorem 4.3. Let D ⊂ Rd be a bounded domain with C1 boundary and let p be acontinuous functional on H1(D) that does not vanish on non-zero constants. Thenthere is C > 0 such that

‖u‖1 ≤ C(‖∇u‖+ |p(u)|

)for any u ∈ H1(D). (4.6)

Applying the above theorem to the functionals

p1(u) =∫

Du dx, p2(u) =

∫∂D

u dσ,

we obtain the Poincare and Friedrich inequalities:

‖u− 〈u〉‖ ≤ C ‖∇u‖ for any u ∈ H1(D) (4.7)

‖u‖ ≤ C ‖∇u‖ for any u ∈ H10(D). (4.8)

Some boundary value problems for the Laplace operator

Let us consider the Dirichlet problem for the Laplacian:

−∆u = f , u∣∣∂D = 0, (4.9)

where f ∈ H−1(D) is a given function. Recall that a function u ∈ H10(D) is

called a solution for (4.9) if∫D〈∇u,∇ϕ〉 dx = f (ϕ) for any ϕ ∈ C∞

0 (D).

The following result is well known; its proof can be found in [GT01, Chapter 8].

4 APPENDIX 40

Proposition 4.4. For any function f ∈ H−1(D), problem (4.9) has a unique solutionu ∈ H1

0(D), which satisfies the inequality

‖u‖1 ≤ C ‖ f ‖−1, (4.10)

where C > 0 does not depend on f . Moreover, if f ∈ Hs(D) for an integer s ≥ 0,then u ∈ Hs+2(D), and there is a constant Cs > 0 such that

‖u‖s+2 ≤ Cs ‖ f ‖s, (4.11)

A similar result is true for the Neumann problem with inhomogeneousboundary condition. Namely, let us fix a divergence-free function g ∈ L2(D, Rd)and consider the problem

∆u = 0, 〈∇u− g, n〉∣∣∂D = 0. (4.12)

A function u ∈ H1(D) is called a solution for (4.12) if∫D〈∇u− g,∇ϕ〉 dx = 0 for any ϕ ∈ H1(D).

Recall that H1(D) stands for the space of function in H1(D) with zero meanvalue. We have the following existence, uniqueness, and regularity theorem.

Proposition 4.5. For any divergence-free function g ∈ L2(D, Rd), problem (4.12)has a unique solution u ∈ H1(D), which satisfies the inequality

‖u‖1 ≤ C ‖g‖. (4.13)

Moreover, if g ∈ Hs(D) for an integer s ≥ 1, then u ∈ Hs+1(D), and

‖u‖s+1 ≤ Cs ‖g‖s. (4.14)

Proof. We confine ourselves to the proof of existence of solution, because theregularity and inequality (4.14) are standard; see [GT01] or [Tay97, Section 5.7].Let us endow H1(D) with the scalar product (∇u,∇v) and consider a contin-uous functional ` : H1(D)→ R defined by the relation

`(ϕ) =∫

Dg∇ϕ dx, ϕ ∈ H1(D).

By the Riesz representation theorem (see [Yos95, Section III.6]), there is a uniqueu ∈ H1(D) such that `(ϕ) = (∇u,∇ϕ) for any ϕ ∈ H1(D). This relation coin-cides with (4.12).

Friedrichs extension

Let H be a Hilbert space with scalat product (·, ·) and the norm ‖ · ‖ and let L0be a semi-bounded symmetric operator in H with domain D(L0):

(L0u, v) = (u, L0v) for u, v ∈ D(L0), (4.15)

(L0u, u) ≥ −M‖u‖2 for u ∈ D(L0), (4.16)

where M ≥ 0 does not depend on u. Let us define an operator L by the follow-ing rule:

4 APPENDIX 41

• denote by D(Q) ⊂ H the closure of D(L0) with respect to the norm‖u‖q =

((L0u, u)+ (M + 1)‖u‖2)1/2 and define a quadratic form onD(Q)

by the relationQ(u, v) = lim

n→∞(L0un, vn), (4.17)

where {un}, {vn} ⊂ D(L0) are arbitrary sequences converging to u and v,respectively, in the norm ‖ · ‖q.

• denote by D(L) ⊂ H the space of vectors u ∈ D(Q) for which there isfu ∈ H such that

Q(u, v) = ( fu, v) for any v ∈ D(Q);

denote by L : D(L)→ H the operator taking u to fu.

A proof of the following important result can be found in [Yos95, RS80].

Theorem 4.6. The operator L is well defined. Moreover, it is self-adjoint and satisfiesthe inequality (Lu, u) ≥ −M‖u‖2 for u ∈ D(L).

Brouwer and Leray–Schauder theorems

The well-known Brouwer theorem says that any continuous application of theunit ball in Rn into itself has a fixed point. The following result is a generalisa-tion of the Brouwer theorem to the infinite-dimensional case.

Theorem 4.7 (Leray–Schauder). Let K be a compact convex subset of a Banachspace X and let F : K → K be a continuous mapping. Then F has at least onefixed point.

A proof of the above theorem can be found in [Tay97]. The following propo-sition gives a sufficient condition for the image of a continuous mapping tocontain a ball.

Proposition 4.8. Let F be a finite-dimensional vector space, let R > ε > 0 be someconstants, and let Φ : BF(R)→ F be a continuous operator such that

‖Φ(u)− u‖F ≤ ε for any u ∈ BF(R). (4.18)

Then Φ(BF(R)) ⊃ BF(R− ε).

Proof. Let us fix any u ∈ BF(R− ε) and consider the continuous function

Ψ : BF(R)→ F, Ψ(u) = u−Φ(u) + u.

It follows from (4.18) that

‖Ψ(u)‖F ≤ ‖u‖F + ‖Φ(u)− u‖F ≤ R.

Thus, Ψ is a continuous function from the ball BF(R) into itself. By the Brouwertheorem, Ψ has a fixed point v ∈ BF(R). It is easy to see that Φ(v) = u.

5 PROBLEMS 42

5 Problems1. Let D ⊂ Rd be a bounded domain with smooth boundary.

(a) Prove that the image of the operator ∇ : Hk+1(D) → Hk(D, Rd) is closedfor any integer k ≥ 0. Hint: Use the Poincare inequality.

(b) Prove that the image of the operator ∇ : L2(D) → H−1(D, Rd) is closed.Hint: Use Proposition 1.6.

2. Let X and Y be Banach spaces, let A ∈ L(X, Y), and let A∗ ∈ L(Y∗, X∗) be theadjoint operator of A. Show that

(Im A)⊥ = Ker A∗, (5.1)

where for a vector space F ⊂ Y we denote by F⊥ the space of linear functionals` ∈ Y∗ such that `( f ) = 0 for any f ∈ F. Furthermore, if Y is a reflexive spaceand F is a closed subspace of Y, then

Im A = (Ker A∗)⊥, (5.2)

where for a vector space L ⊂ Y∗ we denote by L⊥ the space of vectors f ∈ Y suchthat `( f ) = 0 for any ` ∈ L.

3. Give the details of the proof of Proposition 1.6.

4. Let p ∈ H1(D) be such that ∇p ∈ C(D). Show that p ∈ C1(D).

5. Show that the Leray projection can be extended by continuity to an operatorfrom H−1 to V∗. Hint: It suffices to show that

|(Πu, ϕ)| ≤ ‖u‖−1 ‖ϕ‖1 for any ϕ ∈ V.

6. Show that any function f ∈ H−1(D, Rd) can be written as f = v +∇p, wherev ∈ H−1(D, Rd), div v = 0, and p ∈ L2(D). Hint: Use the Hodge–Kodairadecomposition for the function u f = −∆−1 f ∈ H1

0 .

7. Show that the operators L : V → V∗ and B(u, v) : V ×V → V∗ defined by (1.30)are continuous.

8. The aim of this problem is to construct a solution of problem (1.35) in any dimen-sion and to prove the regularity of solution for d = 2.

(a) Show that (1.35) is equivalent to (1.36). Use then the Riesz representationtheorem and Theorem 1.5 to construct a solution (u, p) ∈ V × L2.

(b) Suppose that d = 2. Show that for any u ∈ V there is ψ ∈ H2 ∩ H10 such that

u = curl ψ = (−∂2ψ, ∂1ψ).

(c) Show that if (u, p) ∈ V× L2 is a solution of problem (1.35) with f ∈ L2, then−∆2ψ = ∂1 f2 − ∂2 f1. Use this and elliptic regularity for the biharmonicoperator to prove that ψ ∈ H3, u ∈ H2, and p ∈ H1.

9. Show that ‖B(u, v)‖−1 ≤ C ‖u‖L4‖v‖L4 for u, v ∈ L4(D) with div u = 0. Use thisand an interpolation inequality to show that, in the 2D case, we have

‖B(u, v)‖L2(J,V∗) ≤ C ‖u‖XT‖v‖XT for u, v ∈ XT . (5.3)

5 PROBLEMS 43

10. Show that for any divergence-free vector field u = (u1, u2, u3) we have

curl(〈u,∇〉u

)= 〈u,∇〉ω− 〈ω,∇〉u, (5.4)

Show also that if u = (u1, u2, 0), where u1 and u2 do not depend on x3, then

ω = (0, 0, ω3), curl(〈u,∇〉u

)= (0, 0, 〈u,∇〉ω3).

11. The aim of this problem is to prove Lemma 1.17.

(a) Let u ∈ C1(R, V). Show that

‖u(t)‖2 = ‖u(s)‖2 + 2∫ t

s(u(r), u(r)) dr for any s ≤ t. (5.5)

(b) Let u ∈ L2(R, V) be such that u ∈ L2(R, V) and u(t) = 0 for |t| ≥ T. Showthat u satisfies (5.5).

(c) Prove Lemma 1.17.

12. The aim of this problem is to establish the extension and convexification prin-ciples for an arbitrary compact set K ⊂ H; see Subsection 3.4. Let T > 0 bea constant and let N and G be finite-dimensional subspaces in H2 ∩ H1

0 satisfy-ing (3.12).

(a) Let Ψ : K → L2(JT , G×G) be a continuous mapping for which (3.11) holds,let ωn ∈ C∞

0 (R) be a family of mollifying kernels, and let χn ∈ C∞0 (JT) be a

sequence such that 0 ≤ χn ≤ 1 and χn(t) = 1 for 1n ≤ t ≤ T − 1

n . We writeΨ(u) = (η(t; u), ζ(t; u)) and set

ζn(t; u) =(ωn ∗ ζ(·; u)

)(t), ηn(t; u) = η(t; u) + ∂t

(χn(t)ζn(t; u)

),

where we extended ζ(·; u) by zero outside JT . Show that

supu∈K‖RT(u, h + ηn(·; u))− u‖ < ε for n� 1.

(b) Let Ψ : K → L2(JT ,F (N, G)) be a continuous function satisfying (3.8). Forintegers s ≥ 1 and r ≥ 0, denote by Ir,s(t) the indicator function of theinterval s−1[rT, (r + 1)T). Construct a finite set A = {ηl

1, l = 1, . . . , m} ⊂F (N, G) and continuous functions clr : K → R such that the mapping

Ψ1(t; u) =m

∑l=1

s−1

∑r=0

clr(u)Ir,s(t)ηl1

satisfies the inequality

supu∈K‖RT(u, h + Ψ1(·; u))− u‖ < ε.

Combine this with Lemma 3.11 to show that there is a continuous mappingΨ : K → L2(JT , G× G) for which (3.11) holds.

13. Let Ek ⊂ H be an increasing sequence of subspaces for which (3.14) holds. Showthat for any constant ε > 0 there is an integer N ≥ 1 and a continuous function Ψ :K → L2(JT , EN) satisfying (3.8). Hint: Use relations (3.41) and (3.42) to constructfirst a continuous mapping Ψ0 : K → L2(JT , H) such that (3.15) holds. Show thenthat the mapping Ψ(u) = PNΨ0(u), where Pk is the orthogonal projection to Ek,possesses the required property for N � 1.

14. Use the compactness of the embedding H1 ⊂ L2 to prove Theorem 4.3.

5 PROBLEMS 44

NotationWe denote by Rd the d-dimensional Euclidean space, by 〈·, ·〉 the standard scalar prod-uct in Rd, and by | · | the corresponding norm.

If f is a distribution and ϕ is a test function, then we write f (ϕ) or ( f , ϕ) for the valueof f on ϕ.

For a bounded domain D ⊂ Rd and constants p ∈ [1, ∞], s ∈ Z, and k ∈ Z+, wedenote by Lp(D) the standard Lebesgue space, by Hs(D) the Sobolev space of order s,and by Ck(D) the space of k times continuously differentiable functions on D. TheHolder space Cs(D) for a non-integer s > 0 and the corresponding norm are defined inSection 2.1.

If X is a subspace in L1(D), then X stands for the space of functions u ∈ X whose meanvalue is zero (see (1.5)).

For ν, γ ∈ (0, 1], T > 0, and a bounded domain D ⊂ Rd, we write Q = (0, T)× D anddenote by Cν,γ(Q) the space of continuous functions u on Q such that

∥∥u∥∥

Cν,γ := sup(t,x)∈Q

|u(t, x)|+ sup(t1,x1) 6=(t2,x2)

|u(t1, x1)− u(t2, x2)||t1 − t2|ν + |x1 − x2|γ

< ∞.

If T > 0 is a constant and X is a Banach space, then we write J = [0, T] and denoteby C(J, X) the space of continuous functions f : J → X with the natural norm. For1 ≤ p < ∞, let Lp(J, X) be the completion of C(J, X) with respect to the norm

‖ f ‖Lp(J,X) =(∫ T

0‖ f (t)‖p

Xdt)1/p

.

L∞(J, X) is defined as the space of functions f : J → X for which there is a boundedsequence { fn} ⊂ C(J, X) such that fn(t) → f (t) for almost every t ∈ J. This space isendowed with the norm

‖ f ‖L∞(J,X) = ess supt∈J

‖ f (t)‖X .

For 1 ≤ p ≤ ∞, we define W1,p(J, X) as the space of functions f ∈ Lp(J, X) that can bewritten in the form

f (t) = f0 +∫ t

0g(s) ds, 0 ≤ s ≤ T,

where g ∈ Lp(J, X) and f0 ∈ X. This space is endowed with the norm

‖ f ‖W1,p(J,X) = ‖ f ‖Lp(J,X) + ‖g‖Lp(J,X).

The spaces XT and YT are defined in Propositions 1.9 and 1.14, respectively.

We denote by C, C1, C2, . . . unessential positive constants.

REFERENCES 45

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Navier–Stokes and Euler equations: Cauchy problem and ... · To prove this theorem, we shall need the following natural result, which is of independent interest. Its proof is given