AN INITIAL BOUNDARY-VALUE PROBLEM FOR THE ZAKHAROV

Advances in Differential Equations Volume 15, Numbers 11–12 (2010), 1001–1031

AN INITIAL BOUNDARY-VALUE PROBLEM FOR THEZAKHAROV-KUZNETSOV EQUATION

Jean-Claude SautLaboratoire de Mathematiques, UMR 8628, Universite Paris-Sud et CNRS

Bat. 425, 91405 Orsay, France

Roger TemamThe Institute for Scientific Computing and Applied Mathematics

Indiana University, 831 East Third Street, Bloomington, Indiana 47405

(Submitted by: Reza Aftabizadeh)

Abstract. We introduce and study an initial and boundary-value prob-lem for the Zakharov-Kuznetsov equation posed on an infinite strip ofRd+1, d = 1, 2. After establishing a suitable trace theorem, we firstconsider the linearized case and define the corresponding semigroup onL2 and prove that it has a global smoothing effect. Then we proceedto the nonlinear case and use the smoothing effect to prove in both di-mensions the existence of (unique when d = 1) global weak solutions ofthe initial and boundary problem with null boundary conditions and L2

initial data.

1. Introduction

The Zakharov-Kuznetsov equation∂u

∂t+ c

∂u

∂x+ ∆

∂u

∂x+ u

∂u

∂x= f, (1.1)

where u = u(x, x⊥, t) with x⊥ = y or x⊥ = (y, z) has been introduced in[25] to describe the propagation of nonlinear ionic-sonic waves in a plasmasubmitted to a magnetic field directed along the x-axis; in (1.1), c ≥ 0 isthe sound velocity. More precisely (1.1) is the long wave limit of the Euler-Poisson system for ionized plasmas (see [14] for a rigorous derivation).Thetransport term c ∂u/∂x in (1.1) can of course be eliminated when x ∈ Rby changing the system of coordinates to one which moves with the soundvelocity. However, since we will be concerned with the initial and boundaryvalue problem, x will be restricted to an interval [0, L], and this change

Accepted for publication: May 2010.AMS Subject Classifications: 35Q53, 35A01, 35A02.

1001

1002 Jean-Claude Saut and Roger Temam

of coordinates is not possible. We will therefore keep the transport termhereafter. Note that, in the physical model the right-hand side of (1.1)vanishes, but f is added here for mathematical convenience and to be ableto treat possibly non-homogeneous boundary values (which is not done here).

It is worth noticing that, when u depends only on x and t, (1.1) reducesto the classical Korteweg-de Vries (KdV) equation.

The Cauchy problem for (1.1) in the 2-dimensional case ((x, y) ∈ R2)has been solved by Faminskii [6]. Actually, using the dispersive propertiesof the (unitary on L2(R2)) group et∆∂x , and the conservation of the twocanonical invariants of (1.1), Faminskii proved that the Cauchy problemfor (1.1) is globally well-posed for data in the Sobolev space H1(R2). TheCauchy problem for the 3-dimensional case was proven in [15] to be locallywell-posed in Hs(R3) for s > 9

8 by adapting the method used by Kenig in[13] for the KP I equation.

Concerning the boundary-value problem associated with (1.1) Faminskiiproved in [8] that the initial-boundary-value problem for (1.1) in the half-space (x, y) : x > 0 is globally well-posed for a Dirichlet data at x = 0and an initial value in some weighted L2-spaces.

We will consider here (1.1) posed on Ω × [0, T ], where Ω is the strip(x, x⊥) : 0 < x < 1, x⊥ ∈ Rd, d = 1, 2. A similar problem on the region(0, 1) × [0, T ] for the KdV equation has been investigated in particular byT. Colin and J.-M. Ghidaglia [4], T. Colin and M. Gisclon [5] and by J.Bona, S-M. Sun and B-Y. Zhang [2], using boundary conditions at x = 0and x = 1.

After the completion of this work, we were informed of the paper [9]where in particular the well-posedness of a non-homogeneous initial andboundary-value problem for the Zakharov-Kuznetsov equation in a finite orsemi-infinite strip of R2 is established. Faminskii uses boundary potentialsand thus the treatment of the traces and the strategy of proofs are different.

A key idea in [4] and [2] was to derive and use smoothing estimates forthe linearized problem (with ad hoc boundary conditions) which allow oneto compensate for the loss of an x derivative. Recall that starting with thepioneering paper of T. Kato [12], local smoothing effects of the Airy groupet∂

3x on R have been extensively used to deal with the pure Cauchy problem

for the KdV equation on the real line. It was also noticed in [1] that similarsmoothing effects are valid for the Airy equation posed on the quarter plane(x, t) : x ≥ 0, t ≥ 0 with a Dirichlet boundary condition at x = 0.

The Zakharov-Kuznetsov equation 1003

In the case of a finite domain (0, 1)× [0, T ] the smoothing effect is global.In particular, Colin and Ghidaglia [4] have shown that the linear problem

ut + uxxx = 0 on (0, 1)× [0, T ]u(0, t) = ux(1, t) = uxx(1, t) = 0,u(x, 0) = u0(x),

(1.2)

has smoothing properties very similar to that of the heat equation.A similar result was established in [2] for the problem

ut + ux + uxxx = 0 on (0, 1)× [0, T ]u(0, t) = u(1, t) = ux(1, t) = 0,u(x, 0) = u0(x).

(1.3)

Those smoothing effects are related to the well-known fact that the Airyfunction F−1(eiξ

3/3)(x) behaves, for x > 0, like the heat kernel (exponentialdecay).

The present work is in some sense an extension of [4] and [2] to amulti-dimensional situation which induces new interesting facts. For in-stance, the traces at x = 0, 1 are distributions in Rd and the smoothingeffect is global in the transverse variables.

The present paper is organized as follows. Two sections consider the linearproblem and have an independent interest as far as the traces and smoothingproperties are concerned. In Section 2 we justify the traces at x = 0, 1 bothfor strong solutions (Theorem 2.1) and weak solutions (Theorem 2.2). Weintroduce and study the semigroup generated by the unbounded operatorc ∂∂x + ∆ ∂

∂x in L2(Ω) together with suitable boundary conditions at x = 0, 1.Then we use the semigroup theory to establish the well-posedness of an

initial and boundary value problem for the linear Zakharov-Kuznetsov equa-tion on a strip (strong solutions). The existence of weak solutions (L2 initialdata) is obtained by a standard application of the semigroup theory. Fur-thermore we prove a global (with respect to all variables) smoothing effectfor weak solutions. In this context, it is similar to the corresponding one forthe heat equation, displaying the dissipative effect of boundary conditionson dispersive equations..

Section 3 is devoted to various regularity properties for strong solutions.We then consider in Section 4 the full nonlinear problem and derive by acompactness method a result of existence of global weak solutions in spacedimension 2 and 3 (d = 1, 2 respectively) and a result of uniqueness of suchsolutions in the two-dimensional case.


Finally we state and prove in the Appendix some trace theorems whichare used in this article.

In a work in preparation [21], we intend to establish other results of ex-istence of solutions in the nonlinear case and discuss some problems relatedto the controllability of the ZK equations as in e.g. [19], [18], [10] for theAiry and KdV equations.

2. The linear Zakharov-Kuznetsov equation

Our aim in this section is to study the linearized Zakharov-Kuznetsovequation, namely

∂u

∂t+ c

∂u

∂x+ ∆

∂u

∂x= f, (2.1)

considered in Ω = (0, 1)×Rd = (0, 1)x ×Rd, d = 1, 2. At times we will writeut, ux, etc., instead of ∂u/∂t, ∂u/∂x, etc., so that e.g. ∆u = uxx + ∆⊥u.,where ∆⊥ = ∂2

y or ∂2y + ∂2

z . This equation will be supplemented with thefollowing boundary and initial conditions:

u(0, x⊥, t) = u(1, x⊥, t) = ux(1, x⊥, t) = 0, (2.2)

u(x, x⊥, 0) = u0(x, x⊥), (x, x⊥) ∈ Ω. (2.3)

Results of existence, uniqueness, and regularity of solutions for (2.1)-(2.3)will be derived in Section 2.3, where we recast (2.1)-(2.3) in the form of anabstract evolution equation

du

du+Au = f, u(0) = u0, (2.4)

in which - A is the infinitesimal generator of a contraction semigroup. Beforewe reach this point, a number of technical results will be developed: a tracetheorem in Section 2.1, definition of the operator A in Section 2.2 with aregularity result for the functions in its domain. We also prove a resultof density of smooth functions in the space X defined below. We couldhave considered non-homogeneous boundary conditions in (2.2) (with ad hoccompatibility conditions) but we refrained to do so for the sake of simplicity.We intend to go back to this problem in [21].

2.1. A trace theorem for the steady problem. Equation (2.1) will beconsidered with c fixed arbitrarily in R, possibly c = 0. All the results inSection 2 are valid for such a c, and could possibly be simpler at times forc = 0. We consider the space

X = X(Ω) = u ∈ L2(Ω) : ∆ux + cux ∈ L2(Ω), (2.5)


which we endow with its natural Hilbert norm: |u|X(Ω) = (|u|2L2(Ω) + |∆ux +

cux|2L2(Ω))1/2. We will prove the following result.

Theorem 2.1. For u ∈ X = X(Ω), we can define the traces, on the linesx = 0 and x = 1, of the functions u, ux, uxx, and ∆u, and of their y, zderivatives:

γiu ∈ H−1/2(Rd), γiux ∈ H−3/2(Rd), γi∆u ∈ H−2(Rd),

γiuxx, γi∆⊥u, γiuxyy, γiuxzz ∈ H−5/2(Rd),

where γi is the trace on x = i, i = 0, 1. Furthermore, the trace operators arecontinuous from X(Ω) into the corresponding spaces.

Remark 2.1. i) The spaces in which these traces are defined may not beoptimal (they are very “large”). However what is important for the sequelis that these traces are defined.

ii) Of course, since the derivatives ∂y = ∂/∂y and ∂z = ∂/∂z are continu-ous from H−k(Rd) into H−k−1(Rd), we can also define the traces on x = 0and x = 1 of the y, z derivatives of arbitrary orders of these functions, withcontinuity in the natural spaces. The statement above on γiuxyy, γiuxzz isjust one of the many possible results.

Proof. For u ∈ X we set g = (∆u+ cu)x ∈ L2(Ω) and we define

g1 = g1(x, x⊥) =∫ x

0g(x′, x⊥)dx′;

this function belongs to H1x(0, 1;L2(Rd)) and therefore to L2(Ω), and g1−cu

belongs to L2(Ω) as well. Consequently the Dirichlet problem∆u1 = g1 − cu in Ω,u1 = 0 on ∂Ω (that is at x = 0 and x = 1),

(2.6)

possesses a unique solution u1 ∈ H10 (Ω). 1 We then observe that

(∆(u− u1))x = 0, (2.7)

so that ∆(u − u1) is a distribution ψ which does not depend on x ([22],page 54).

We set v = u− u1. Clearly, v ∈ L2x(0, 1;L2(Rd)).

Let us admit for the moment the following.

1We recall that the Poincare inequality holds in H10 (Ω) since Ω is bounded in the

direction 0x, which is sufficient. Hence |∇u|L2(Ω) is a norm on H10 (Ω) equivalent to the

H1 - norm.


Lemma 2.1. The distribution ψ belongs to H−2(Rd).

By the previous lemma, ψ belongs (in particular) to L2x(0, 1;H−2(Rd))

andvxx = −∆⊥v + ψ ∈ L2

x(0, 1;H−2(Rd)). (2.8)

By the intermediate derivatives theorem, vx ∈ L2x(0, 1; H−1(Rd)) (see [16],

Theorem 2.3).On the other hand one has ux = vx + u1x ∈ L2

x(0, 1;H−1(Rd)) and

uxxx = −∆⊥ux − cux + g ∈ L2x(0, 1;H−3(Rd)),

so that uxx ∈ L2x(0, 1;H−2(Rd)).

By the trace theorem (Theorem 3.1 of [16]), one deduces that

∂ju

∂xj∈ Cx([0, 1];H−(2j+1)/2(Rd)), j = 0, 1, 2.

From the above computations, ∆u ∈ L2x(0, 1;H−2(Rd)) and since ∆ux ∈

L2x(0, 1;H−1(Rd)), we obtain that ∆u ∈ Cx([0, 1];H−2(Rd)).The existence of the traces as stated in Theorem 2.1 follows; the continuity

of the trace operators is easy. Theorem 2.1 is proved once we have provenLemma 2.1.

Proof of Lemma 2.1. By the Fourier transform in x⊥, we find that v =v(x, η, ζ) (or v(x, η)) satisfies

vxx − k2v = ψ, (2.9)

where k = (η2+ζ2)12 , from which we infer that v is C∞ in x with values in the

space of distributions on Rd (with variables (η, ζ) or η). Two integrations inx of (2.8) give

v(x, .) = k2A+B + Cx+ ψx2

2, (2.10)

where A =∫ x

0 (x− x1)v(x1, .) dx1 and B = v(0, .), C = vx(0, .).By assumption, u and u1 and thus v belong to L2(Ω) and, by the Parseval

theorem, v belongs to L2x(0, 1;L2

η,ζRd)). The same is true of A and thus k2A

belongs to the space L2x(0, 1;L2

η,ζ(k−4Rd)), where L2

η,ζ(k−4Rd) is the L2 space

on Rd with weight k−4 = (η2 +ζ2)−2. Hence, thanks to (2.10), B+Cx+ x2

2 ψ

belongs to the same space and, by the Fubini theorem, B+Cx+ x2

2 ψ belongsto L2

η,ζ(k−4Rd) for almost every x. Since B, C ψ are independent of x and


the determinant1 x1 x2

1

1 x2 x22

1 x3 x23

= (x1 − x2)(x2 − x3)(x3 − x1),

one can find three different values of x for which B + Cx + x2

2 ψ is definedand belongs to L2

η,ζ(k−4Rd). Thus each of B, C and ψ belongs to this space,

which means that the Fourier inverse ψ of ψ belongs to H−2(Rd).

2.2. A trace theorem for the unsteady problem. We will need in thesequel to justify the existence of traces at x = 0, 1 for weak solutions of (2.1).This is the aim of the next theorem.

Theorem 2.2. Let f ∈ L2x(0, 1;L2(0, T ;H−2(Rd)) and u ∈ L2(0, T ;H1(Ω))

be a solution of (2.1) in the sense of distributions. Then, we can define thetraces, on the lines x = 0 and x = 1, of the functions u, ux, uxx,:

γiu ∈ H−1/3((0, T )× Rd), γiux ∈ H−1((0, T )× Rd),

γiuxx ∈ H−5/3((0, T )× Rd),

where γi is the trace on x = i, i = 0, 1. Furthermore the trace operators arecontinuous from L∞(0, T ;L2(Ω)) ∩ L2(0, T ;H1(Ω)) into the correspondingspaces.

Proof. We write equation (2.1) in the form

uxxx = f − cux − uxyy − uxzz − ut (2.11)

and we observe that, since u, ux ∈ L2(0, T ;L2(Ω)) = L2x(0, 1;L2((0, T )×Rd)),

then

ut ∈ L2x(0, 1;H−1(0, T ;L2(Rd))), uxyy, uxzz ∈ L2

x(0, 1;L2(0, T ;H−2(Rd))).

Thus, uxxx belongs to L2x(0, 1;H−1(0, T ;H−2(Rd)). One deduces again by

the trace theorem ([16]) that, for j = 0, 1, 2, 2

∂ju

∂xj∈ Cx([0, 1]; [L2((0, T )× Rd)), H−1(0, T ;H−2(Rd))](j+1/2)/3),

which implies that

∂ju

∂xj∈ Cx([0, 1];H−2(j+1/2)/3(0, T )× Rd)).

2We denote [X0, X1]θ the interpolate of order θ of the two Hilbert spaces X0 and X1

(see [16]).


Furthermore if a sequence of functions um satisfies (2.1) with f = fm, andif um → u in L2(0, T ;H1(Ω)) and fm → f in L2

x(0, 1;L2(0, T ;H−2(Rd))strongly (respectively weakly), then γium, γiumx, γiumxx converge to γiu,γiux, γiuxx in the corresponding trace spaces, strongly (respectively weakly).

2.3. The operator A and its domain. Thanks to Theorem 2.1 we arenow in position to define the operator A: its domain D(A) is

D(A) = u : u and ∆ux + cux ∈ L2(Ω), u(0, x⊥) = u(1, x⊥) (2.12)

= ux(1, x⊥) = 0, ∀x⊥ ∈ Rd

and, for u ∈ D(A),

Au = ∆ux + cux. (2.13)

We will now prove the following regularity result (see Theorem 3.2 below fora more precise result).

Theorem 2.3. D(A) is included in H10 (Ω) with a continuous imbedding:

D(A) ⊂ H10 (Ω). (2.14)

Proof of Theorem 2.3. This result is proven by partial regularization inx⊥. We first present and analyze this partial regularization in x⊥ and thenshow (2.14).

i) Partial regularization in x⊥. We let ρε = ρε(x⊥) be a family of regular-izing functions; that is, ρε(x⊥) = ε−dρ(x⊥/ε), ρ is C∞ with compact support,ρ ≥ 0, and

∫Rd ρ(x⊥) dx⊥ = 1. We denote by ρε∗⊥ the partial convolution

in x⊥; that is,

(ρε ∗⊥ u)(x, x⊥) =∫

Rdu(x, x⊥ − x′⊥)ρε(x′⊥)dx′⊥.

We will show that, if u ∈ X, then uε = ρε ∗⊥ u ∈ C2x([0, 1];H∞(Rd)), where

H∞(Rd) =⋂k∈NH

k(Rd).We set, as in the proof of Theorem 2.1,

gε = ∆uεx + cuεx = ρε ∗⊥ (∆u+ cu)x = ρε ∗⊥ g;

we see that gε ∈ L2x(0, 1;H∞(Rd)) since gε ∈ L2

x(0, 1;L2(Rd)). Similarlygε1 = gε1(x, x⊥) =

∫ x0 gε(x

′, x⊥) dx′ ∈ L2x(0, 1;H∞(Rd)).


By convolution of u1 with ρε, we see (compare to (2.6)) that uε1 = ρε ∗⊥ u1

is a solution of ∆uε1 = gε1 − cuε in Ω,uε1 = 0 on ∂Ω.

(2.15)

Note, for (2.15), that uε1 = ρε ∗⊥ u1 vanishes at x = 0 and 1 like u1, andthat

ρε ∗⊥ (u1xx + ∆⊥u1) = (ρε ∗⊥ u1)xx + ∆⊥(ρε ∗⊥ u1) = ∆uε1.

We then conclude as in (2.7) that (∆(uε − uε1))x = 0 and that ψε =∆(uε − uε1) is a distribution independent of x; in fact ψε = ρε ∗⊥ ψ. Henceψε ∈ H∞(Rd) and we conclude that ∆uε = gε1 − cuε + ψε belongs toL2x(0, 1;H∞(Rd)), and so does ∆⊥uε. Thus we successively obtain

uεxx ∈ L2x(0, 1;H∞(Rd)), uε ∈ C1

x([0, 1];H∞(Rd)), (2.16)

uεx and ∆⊥uεx ∈ Cx([0, 1];H∞(Rd)),

uεxxx = gε − cuεx −∆⊥uεx ∈ L2x(0, 1;H∞(Rd)), (2.17)

uε ∈ C2x([0, 1];H∞(Rd)). (2.18)

Furthermore, as ε→ 0, u being in X(Ω),

uε → u in X(Ω), (2.19)

and if u (or ux) vanishes at x = 0 or 1, so does uε:

uεx(1, x⊥) = (ρε ∗⊥ u)x(1, x⊥) = (ρε ∗⊥ ux)(1, x⊥) = 0.

In particular,If u ∈ D(A), then uε ∈ D(A). (2.20)

Note also that, as a consequence of (2.16) -(2.20), the intersection of D(A)with the following set is dense in D(A) :

u ∈ C2x([0, 1];H∞(Rd)), uxxx ∈ L2

x(0, 1;H∞(Rd)). (2.21)

ii) Proof of (2.14). As observed in (2.20), if u ∈ D(A), then so is uε;and, due to the regularity property (2.18) of uε, the following calculations,integrations by parts and cancellations of boundary terms are legitimate:∫

Ω(∆uεx + cuεx)xuεdxdx⊥ (2.22)

=∫

Rd[uεxxxuε +

c

2u2εx+ xu∆⊥u] |10 dx⊥


−∫

Ω[uεxxuε + uεxxxuεx +∇⊥(uεxx).∇⊥uε +

c

2u2ε]dxdx

⊥

= −∫

Rd[xu2

εx + uεuεx +12x|∇⊥u|2] |10 dx⊥

+∫

Ω[32|uεx|2 +

12|∇⊥uε|2 −

c

2u2ε]dxdx

⊥

=∫

Ω[32|uεx|2 +

12|∇⊥uε|2 −

c

2u2ε]dxdx

⊥.

As ε→ 0, the left-hand side of (2.22) remains bounded, as well as the term

(c/2)∫

Ωu2εdxdx

⊥.

We conclude that grad uε remains bounded in L2(Ω)d+1 and hence gradu ∈ L2(Ω)d+1, and u ∈ H1(Ω). Theorem 2.2 is proved.

Density of smooth functions in X(Ω). Before we proceed, we also provethe following density result which will be useful, using the same type ofmethods as before.

Lemma 2.2.X(Ω) ∩ C∞(Ω) is dense in X(Ω). (2.23)

Proof. We observe that X = X(Ω) is not a local space; that is, if u ∈ X andϕ ∈ D(Ω), then ϕu does not necessarily belong to X. To prove (2.23) we thenstart by extending the functions of X(Ω) to Rd+1: to a function u in X(Ω) wewill (linearly and continuously) associate a function u in Xloc(Rd+1), whichis the space of functions belonging to X((−n, n)x×Rd), for every n > 0. Wedefine u on (−1, 0)x×Rd using the Babitch procedure in the x variable [16];we set, for −1 < x < 0,

u(x, x⊥) = a1u(− x

16, x⊥) + a2u(−x

8, x⊥) + a3u(−x

4, x⊥);

a1, a2, a3 are such that u will be C2 at x = 0 if u is C2 at x = 0 (x > 0); thatis,

a1 + a2 + a3 = 1, −a1

16− a2

8− a3

4= 1,

a1

162+a2

82+a3

42= 1;

these equations define the a′is uniquely. We proceed in the same way on theother intervals k < x < k+ 1, k ∈ Z and thus u is defined on Rx×Rd. Usingthis C2 Babitch extension procedure, it is clear that if u and (∆u + cu)x ∈L2(Ω), then u and (∆u+ cu)x are in L2

x(−n, n;L2(Rd)) for every n > 0.


We then regularize this function u using a family of regularizing functionsσε which now operate in both variables x and x⊥. The functions uε =σε ∗ u are in C∞(Rd+1), and, as ε→ 0, uε converges to u and (∆uε + cuε)xconverges to (∆u + cu)x in L2

x(−n, n;L2(Rd)), for every n > 0, since u and(∆u + cu)x are in L2

x(−n, n;L2(Rd)) for every n > 0. By restriction of theuε to (0, 1) × Rd, we obtain C∞ functions of X(Ω) which converge to u inX(Ω); (2.23) is proved.

2.4. Existence and uniqueness of solutions for the linear equation.Let H = L2(Ω). We now consider the linear equation (2.1) with the bound-ary and initial conditions (2.2), (2.3). We recast this initial- and boundary-value problem as an abstract equation in H:

du

dt+Au = f, u(0) = u0, (2.24)

with A as in (2.13). We will establish the existence and uniqueness of so-lutions of (2.24) using the Phillips version of the Hille-Yosida theorem; forthat purpose we now need to define and characterize the adjoint A∗ of Aand its domain D(A∗).

The adjoint A∗. Assume that u ∈ D(A) and u ∈ H are smooth functions;then we have

(Au, u)H =∫

Ω(∆u+ cux)udxdx⊥ = I0 + I1, (2.25)

where

I0 = −∫

Ωu(∆u+ cu)xdxdx⊥,

and

I1 =∫

(uxx(1, x⊥)u(1, x⊥)− uxx(0, x⊥)u(0, x⊥) + ux(0, x⊥)ux(0, x⊥))dx⊥.

According to [20], D(A∗) consists of the u in H such that u → (Au, u)H iscontinuous on D(A) for the norm of H. Fix u ∈ H and consider first the u’sin D(A) which are C∞ with compact support in Ω. For such u’s

(Au, u)H = − < u, (∆u+ cu)x >,

and if u→ (Au, u)H is continuous on H for such u’s, then necessarily (∆u+cu)x ∈ L2(Ω). Now, if u and (∆u + cu)x belong to L2(Ω), the traces of uare defined at x = 0 and 1 as in Theorem 2.1 and the calculations above are


valid for all u in D(A) which are, say, in C∞(Ω). For such u’s, the mappingu→ (Au, u)H can only be continuous for the norm of H if

u(0, y) = u(1, x⊥) = ux(0, x⊥) = 0. (2.26)

Finally, if u ∈ X(Ω) and satisfies (2.26), then it can be approximated inX(Ω) by functions of the space (2.21) satisfying (2.26) (same reasoning asfor D(A)). Also any function u ∈ D(A) can be approximated in D(A) byfunctions belonging to D(A) and to the space (2.21). The integrations byparts in (2.25) are then justified by approximation and I1 = 0. This showsthat u ∈ D(A∗) and, finally,

D(A∗) = u ∈ X(Ω) : u satisfies (2.26) . (2.27)

Of course we can prove exactly as in Theorem 2.3 that

D(A∗) ⊂ H1(Ω). (2.28)

For u ∈ D(A∗), we setA∗u = −(∆u+ cu)x. (2.29)

Application of the Hille-Yosida Theorem. As indicated before, we want tosolve (2.24) by application of the Hille-Phillips-Yosida theorem. We need toshow that −A is the infinitesimal generator of a semigroup of contractionsin H; according to [24], [11], [17], it suffices to show that

i) A and A∗ are closed operators and their domains D(A) and D(A∗) aredense in H; and

ii) A and A∗ are positive:(Au, u)H ≥ 0, ∀u ∈ D(A),(A∗u, u)H ≥ 0, ∀u ∈ D(A∗).

(2.30)

The proof of i) is elementary. To show that (Au, u)H ≥ 0, for all u ∈D(A), we can assume that u belongs to the dense space (2.21), and then thefollowing calculations are legitimate:

(Au, u)H =∫

Ω(∆u+ cu)xudx (2.31)

=∫

Rd[uxxu−

12u2x −

12|∇⊥u|2 +

c

2u2] |10 dx⊥

= ( by (2.12) ) =12

∫Rdu2x(0, x⊥)dx⊥ ≥ 0.

The proof is similar for A∗.


Hence −A is the infinitesimal generator of a semigroup of contractions inH denoted S(t)t≥0. Then, by application of the Hille-Phillips-Yosida the-orem, we obtain the existence of strong solutions of the initial-value problem(2.30).

Theorem 2.4. Assume that u0 is given in D(A) and f given in L1loc(R+;H),

with f ′ = df/dt in L1loc(R+;H). Then the initial-value problem (2.24) pos-

sesses a unique solution u such that

u ∈ C([0, T ];D(A)) ∩ C1(0, T ;H) for any T > 0. (2.32)

Remark 2.2. i) Due to Theorem 2.1, (2.32) e.g. implies that

u ∈ C([0, T ];H1(Ω)). (2.33)

ii) If u0 is given in H and f in L1loc(R+;H), then the initial-value problem

(2.24) possesses a unique weak solution u ∈ C(R+;H), given by

u(t) = S(t)u0 +∫ t

0S(t− s)f(s)ds. (2.34)

iii) By temporal differentiation of (2.24), if the data are suitable, we canobtain more regularity for u. For instance, if, besides the hypotheses ofTheorem 2.4, f ′′ ∈ L1(0, T ;H) and u′(0) = f(0)−Au0 ∈ D(A), then

du

dt∈ C([0, T ];H) ∩ L∞(0, T ;D(A)),

d2u

dt2∈ L∞(0, T ;H), (2.35)

so that, with i), ut and Au ∈ C([0, T ];H1(Ω)).

We now discuss the smoothing properties of the linear problem (2.1) withthe boundary conditions (2.2), (2.3) when the initial data u0 ∈ H = L2(Ω).

We first consider the homogeneous problem (2.1)-(2.3) when f ≡ 0. Ourmain result establishes a global smoothing effect for the mild solutions (seeRemark 2.2, ii)).

Theorem 2.5. Let u0 ∈ L2(Ω) and f ≡ 0. Then there exists a uniquesolution u ∈ Cb(R+;L2(Ω))∩L2(R+;H1(Ω)) of (2.1)-(2.3), where the tracesare taken in the sense of Theorem 2.2.

Furthermore, u satisfies, for any t ∈ R+,

||u(·, t)||2L2(Ω) +∫ t

0

∫Rd|ux(0, x⊥, s)|2dx⊥ds =

∫Ωu2

0(x, x⊥)dxdx⊥, (2.36)∫Ωxu2(x, x⊥, t)dxdx⊥ +

∫ t

0

∫Ω

[3u2x(x, x⊥, s) + |∇⊥u(x, x⊥, s)|2]dxdx⊥ds


≤c′∫

Ωu2(x, x⊥, t)dxdx⊥ + (1 + t)

∫Ωu2

0(x, x⊥)dxdx⊥

≤ c3(1 + t)∫

Ωu2

0(x, x⊥)dxdx⊥. (2.37)

Corollary 2.1. Under the hypotheses of Theorem 2.5 one has for any t > 0

||u||L2(0,t;H1(Ω)) ≤ c4(1 + t)12 ||u0||L2(Ω). (2.38)

Remark 2.3. (2.36) implies a smoothing effect on the trace at x = 0,namely ux(0, ·, ·) ∈ L2((0, T ) × Rd). Recall that by Theorem 2.2 this traceis meaningful a priori in H−1/3((0, T )× Rd)).

Remark 2.4. (2.37) is a smoothing effect a la Kato but global in space(of parabolic type). This is an extension to the multidimensional case of thesmoothing result Proposition 2.1 of [2] which was established for the KdVequation.

Proof. The existence and uniqueness is deduced from standard semigrouptheory (see [17]). In fact one has u(t) = S(t)u0 and the ZK equation issatisfied (at least) in the distribution sense.

To prove (2.36), (2.37), we assume first that u0 ∈ D(A). In this case, theseproperties follow after multiplication of (2.1) by u and xu and integrationover Ω. In the general case we use a standard density argument (see the proofof Proposition 2.1 in[2] for details in the context of the KdV equation).

More precisely, if u0 ∈ L2(Ω), we consider a sequence u0n ∈ D(A) suchthat u0n → u0 in L2(Ω) and define un(., t) = S(t)u0n so that (2.36) and(2.37) hold true for un. These estimates pass to the limit yielding the desiredresult.

We now turn to the non-homogeneous case u0 ≡ 0, f 6= 0, f ∈ L1loc(R+;

L2(Ω)); that is, we consider

u(t) =∫ t

0S(t− s)f(s)ds.

As in the homogeneous case one has a smoothing property.

Theorem 2.6. Let f ∈ L1loc(R+;L2(Ω). Then there exists a unique solution

u ∈ C(R+;L2(Ω)) ∩ L2loc(R+;H1(Ω)) of (2.1)-(2.3), where again the traces

are taken in the sense of Theorem 2.2.Moreover, for any t ∈ R+,

||u||L∞(0,t;L2(Ω)) + ||ux(0, ·, ·)||L2(0×Rd×(0,t)) ≤ c5 ||f ||L1(0,t);L2(Ω), (2.39)

The Zakharov-Kuznetsov equation 1015∫Ωxu2(x, x⊥, t)dxdx⊥ +

∫ t

0

∫Ω

[3u2x(x, x⊥, s) + |∇⊥u(x, x⊥, s)|2]dxdx⊥ds

≤c6(1 + t)||f ||2L1(0,t;L2(Ω)). (2.40)

Proof. As before, the following calculations are justified by considering firstthe case when f is “smooth” (that is f ∈ L1

loc(R+;D(A)) and then passing tothe limit by density. The estimate (2.39) is obtained by multiplying (2.24) byu and performing integrations by parts which use the boundary conditions.

To prove (2.40) we multiply (2.24) by 2xu to obtain∫Ωxu2(x, x⊥, t)dxdx⊥ +

∫ t

0

∫Ω

[3u2x(x, x⊥, t) + |∇⊥u(x, x⊥, t)|2]dxdx⊥ds

= 2∫ t

0

∫Ωxf(x, y, s)u(x, x⊥, s)dxdx⊥ds+

∫ t

0

∫Ωu2(x, x⊥, s)dxdx⊥ds

≤ 2∫ t

0||x

12 f(., s)||L2(Ω)||x

12u(., s)||L2(Ω)ds

+∫ t

0

∫Ωu2(x, x⊥, s)dxdx⊥ds

≤ 2 sup0≤s≤t

[||x12 f(., s)||L2(Ω)]

∫ t

0||x

12 f(., s)||L2(Ω)ds

+∫ t

0

∫Ωu2(x, y, s)dxdx⊥ds

≤ sup0≤s≤t

||x12 f(., s)||2L2(Ω) +

[ ∫ t

0||x

12 f(., s)||L2(Ω)ds

]2

+∫ t

0

∫Ωu2(x, y, s)dxdx⊥ds, (2.41)

which yields (2.40) in view of (2.39).

3. Regularity properties for the linearZakharov-Kuznetsov equation

In this section our aim is to prove regularity properties for strong solutionsof the linearized ZK equations (that is when u0 ∈ D(A)).

Concerning the regularity results we first recall the result in Remark2.2 i), namely that the function u given by Theorem 2.4 ii) belongs toL∞(0, T ;H1(Ω)). Furthermore, we observe that the boundary conditions in(2.2) are invariant by tangential differentiation in y (or z). From this we


conclude that if e.g. u0y and fy satisfy the same hypotheses as u0 and f inTheorem 2.4 ii), then uy satisfies also the properties (2.32). This is true foras many tangential derivations as we want in y (and in z in dimension 3)up to C∞ regularity in y or z. The same is true for the time differentiationprovided certain compatibility conditions are satisfied and this has alreadybeen mentioned in Remark 2.2 i) and ii) (up to C∞ regularity in time aswell). Returning then to equation (2.1), we can obtain some regularity inthe x variable as well, up to C∞ regularity. To be more specific we now statean example of such a regularity result.

Theorem 3.1. Under the hypotheses of Theorem 2.4 (ii) (that is u0 ∈D(A), f, f ′ ∈ L∞(0, T ;H)), the solution u of (2.24), (2.32), satisfies

u ∈ C([0, T ];H2(Ω)) (3.1)

and∆ux ∈ L∞(0, T ;L2(Ω)). (3.2)

The proof of Theorem 3.1 is based on the following additional regularityresult for the functions in D(A).

Theorem 3.2. D(A) is included in H2(Ω)∩H10 (Ω) and there exist constants

c1, c′1 such that

|u|H2(Ω) ≤ c1|∆u|L2(Ω) ≤ c′1|Au|L2(Ω), (3.3)

for every u in D(A).

Proof of Theorem 3.2. If u ∈ D(A), then u ∈ H1(Ω) by Theorem 2.4,and ∆ux = Au − cux = f ∈ L2(Ω). Our first step will be to show that∆u ∈ L2(Ω) and that there exists a constant c′′1 such that

|∆u|L2(Ω) ≤ c′′1|f |L2(Ω). (3.4)

Note that ∆u ∈ L2(Ω) does not follow from ∆ux ∈ L2(Ω) because we do notknow that ∆u(x0, ·) belongs to L2(Rd) at any point x0 ∈ [0, 1]. This is partof what we now prove.

We denote by u(x, η, ζ) the Fourier transform of u with respect to thetransverse variables (y, z) and set k2 = η2 + ζ2. Integrating on the segment[x, 1] the equation

uxxx − k2ux = f

and using the boundary condition at x = 1, we readily obtain that (in whatfollows we will only denote the dependence of the functions on the first


variable x)

uxx(x)− k2u(x) = uxx(1)−∫ 1

xf(x1)dx1. (3.5)

We solve this ODE by the method of variation of constants, setting u = Kekx

and noticing that the boundary conditions u(0) = u(1) = ux(1) = 0 implythat K(0) = K(1) = Kx(1) = 0. After elementary computations, one readilychecks that (3.5) may be written as

d

dx(Kxe

2kx) = uxx(1)ekx −∫ 1

xf(x1)ekxdx1,

which by integration and using the fact that Ax(1) = 0 yields

Kx(x)e2kx = −1kuxx(1)(ek − ekx) +

∫ 1

x

∫ 1

x2

f(x1)ekx2dx1dx2.

Observing that∫ 1

x

∫ 1

x2

f(x1)ekx2dx2dx1 =∫ 1

x

∫ x1

xf(x1)ekx2dx2dx1

=1k

∫ 1

xf(x1)(ekx1 − ekx)dx1,

we deduce that

Kx(x) = −1kuxx(1)(ek(1−2x) − e−kx) +

1k

∫ 1

xf(x1)(ek(x1−2x) − e−kx)dx1.

Integrating this last equation between x and 1, and recalling that K(1) = 0,yields

K(x) =12kuxx(1)(

e−k

k+ek(1−2x)

k− 2e−kx

k)

− 1k

∫ 1

x

∫ 1

x2

f(x1)(ek(x1−2x2) − e−kx2)dx1dx2

=1

2k2uxx(1)(e−k + ek(1−2x) − 2e−kx)

− 12k2

∫ 1

xf(x1)[e−kx1 + ek(x1−2x) − 2e−kx]dx1.


In order to compute uxx(1) we use the fact that K(0) = 0 in the previousidentity and compute the integral to infer

uxx(1) =∫ 1

0f(x1)

ekx1 + e−kx1 − 2ek + e−k − 2

dx1,

so that

K(x) =1

2k2e−k + ek(1−2x) − 2e−kx

∫ 1

0f(x1)

cosh kx1 − 1cosh k − 1

dx1 (3.6)

− 12k2

∫ 1

xf(x1)[e−kx1 + ek(x1−2x) − 2e−kx]dx1.

With (2.40) we now have the following expression for u(x) = K(x)ekx:

u(x) =1k2

cosh k(x− 1)− 1cosh k − 1

∫ 1

0f(x1)[cosh kx1 − 1]dx1 (3.7)

− 1k2

∫ 1

xf(x1)[cosh k(x− x1)− 1]dx1.

We will use this formula for u(x) to prove that ∆u(1) ∈ L2(Rd). Indeed onehas ∆u(1) = uxx(1) since u(1) = 0 and by (3.7)

uxx(1) =∫ 1

0f(x1)

cosh kx1 − 1cosh k − 1

dx1.

Since the function multiplying f(x1) in the integrand is bounded by 1 inabsolute value, we conclude that

|∆u(1)|L2(Rd) = |uxx(1)|L2(Rd) ≤ |f |L2(Ω), (3.8)

which proves our claim.To conclude the proof of Theorem 3.2, we observe that

∆u(1)−∆u(x) =∫ 1

x∆ux(x1)dx1,

and the result follows since ∆ux = f ∈ L2(Ω) and

|∆u(1)|L2(Ω) ≤ |∆u(1)|L2(Rd).

Once (2.38) has been proved, we conclude that u ∈ H2(Ω) because of theDirichlet boundary condition satisfied by u, namely u = 0 at x = 0, 1. Theproof of (3.3) follows easily from the above analysis and Theorem 3.2 isproved.


Remark 3.1. In view of possible applications to e.g. time dependent prob-lems, we observe that with essentially the same proof as for Theorem 3.2,one can prove the following.

If Y is a Hilbert space, u ∈ L2(Ω;Y ) with ∆ux + cux ∈ L2(Ω;Y ), and ifu(0) = u(1) = ux(1) = 0, then u belongs to H2(Ω;Y ) ∩H1

0 (Ω;Y ).

Proof of Theorem 3.1. Because of Theorems 2.3 and 3.2 we know thatthe solution u of (2.24), (2.32), belongs to L∞(0, T ;D(A)) and hence toL∞(0, T ;H2(Ω)), which proves (3.1). Furthermore ∆ux = Au − cux ∈L∞(0, T ;L2(Ω)), which proves (3.2). Theorem 3.1 is proved.

To obtain (2.38), we multiply the equation for uε by xuε, integrate andintegrate by parts. The calculations are the same as in (2.22). We obtain(3.2) for uε and then (3.2) for u by letting ε→ 0.

Theorem 2.5 is proved.

4. The Nonlinear Zakharov-Kuznetsov equation

We now proceed and consider the initial-boundary-value problem for thefull (nonlinear) ZK equation in the strip 0 < x < 1, that is, (1.1), (2.2) and(2.3). Our main result in this case is as follows.

Theorem 4.1. We are given u0 in L2(Ω) and f in L2loc(R+;L2(Ω)),Ω =

(0, 1)× Rd, d = 1, 2. Then:(i) The initial- and boundary-value problem for the ZK equation, that is

(1.1), (2.2), (2.3), possesses a solution u,

u ∈ L∞(0, T ;L2(Ω)) ∩ L2(0, T ;H1(Ω)), (4.1)∫ T

0

∫Rd|∇u(x, x⊥, t)|2dxdx⊥dt <∞, (4.2)

for all T > 0.(ii) We assume here that d = 1. The solution u is then unique. Moreover,

u ∈ C(R;L2(Ω)) and the flow map u0 → u(., t) is continuous from L2(Ω) toL2(Ω).

Note that the boundary conditions u|x=0,1 make sense since u ∈ L2(0, T ;H1(Ω)). The boundary condition ux|x=1 makes sense in H−2((0, T )×Rd) asexplained in Lemma 4.1 and Lemma A.2.

Proof. i) We start with the proof of existence. The existence is proven byparabolic regularization; that is, for ε > 0 “small,” we consider the parabolic


equation

∂uε

∂t+ c

∂uε

∂x+ ∆

∂uε

∂x+ uε

∂uε

∂x+ ε

∂4uε

∂x4+∂4uε

∂y4+∂4uε

∂z4

= f, 3 (4.3)

supplemented with the initial and boundary conditions (2.2), (2.3) and theadditional boundary condition

∆uε(0, x⊥, t) = 0, x⊥ ∈ Rd, t ∈ (0, T ). (4.4)

Note that, since uεyy = uεzz = 0 at x = 0 (and x = 1), (4.4) is equivalent to

uεxx(0, x⊥, t) = 0, x⊥ ∈ Rd, t ∈ (0, T ). (4.5)

The following a priori estimates classically guarantee the existence of a so-lution uε for ε > 0 fixed and then allow us to pass to the limit ε → 0, thusproviding the existence of a solution for the ZK equation.

As for the linearized ZK equation these a priori estimates are obtained bymultiplying (4.3) by uε, then by xuε, integrating and integrating by partsin each case. The contributions of the linear terms cuεx and ∆uεx have beenalready studied, and we thus concentrate on the contributions of the otherterms. Note that the solutions uε to the parabolic problem are sufficientlyregular for the following calculations to be fully legitimate without any needof further regularization.

We drop the super index ε and start by multiplying equation (4.3) byu = uε. We find

•∫

Ω

∂u

∂tudxdx⊥ =

12d

dt|u|2L2(Ω),

•∫

Ωu∂u

∂xudxdx⊥ = 0,

•∫

Ωfudxdx⊥ ≤ 1

2|f |2L2(Ω) +

12|u|2L2(Ω)

• ε∫

Ω

(∂4u

∂x4+∂4u

∂y4+∂4u

∂z4

)udxdx⊥

= ε

∫Ω

(∣∣∣∂2u

∂x2

∣∣∣2 +∣∣∣∂2u

∂y2

∣∣∣+∣∣∣∂2u

∂z2

∣∣∣2)dxdx⊥ =: ε[u]22.

3Here and everywhere we consider the three-dimensional case; the terms involving zshould be dropped if d = 1.


Hence with (2.31) we find

d

dt|uε|2L2(Ω) +

∫Rd|uεx(0, x⊥, t)|2dx⊥ + 2ε [uε]22 ≤ |f |

2L2(Ω) + |uε|2L2(Ω). (4.6)

Using the Gronwall lemma we classically infer from (4.6) (and the fact thatu0 ∈ L2(Ω)) the following bounds independent of ε:

uε is bounded in L∞(0, T ;L2(Ω))√εuε is bounded in L2(0, T ;H2(Ω))

uεx(0, ·, ·) is bounded in L2(0, T ;L2(Rd)).(4.7)

For (4.7), we observe that [u]2 is a norm on

W =v ∈ H2(Ω), v = 0 at x = 0 and 1, vx = 0 at x = 1

equivalent to the H2-norm. Indeed it suffices to observe first that |w′′|L2(0,1)

is a norm on

Z =w ∈ H2(0, 1) : w(0) = w(1) = w′(1) = 0

,

and this follows promptly from the Poincare inequality applied to w′ and tow, so that, for v ∈W,

|v|L2(Ω) ≤ c′|vx|L2(Ω) ≤ c′′|vxx|L2(Ω).

Secondly, one shows classically by Fourier transform in y and z that

|vy|2L2(Ω) + |vz|2L2(Ω) ≤ c′(|v|2L2(Ω) + |vyy|2L2(Ω) + |vzz|2L2(Ω)

).

Finally for the derivatives vxy, vxy, one has classically, by integrations byparts taking into account the boundary conditions,∫

Ω|vxy|2dxdx⊥ =

∫Ωvxxvyydxdx

⊥ ≤ 12|vxx|2L2(Ω) +

12|vyy|2L2(Ω).

This last inequality is shown for v smooth and then by continuity for all vin W since the smooth functions are dense in W.

We now multiply (4.3) by xu(= xuε), integrate and integrate by parts:

•∫

Ω

∂u

∂txudxdx⊥ =

12d

dt|√xu|2L2(Ω),

•∣∣∣ ∫

Ωu∂u

∂xxudxdx⊥

∣∣∣ =∣∣∣ ∫

Ωx∂

∂x

(u3

3

)dxdx⊥

∣∣∣ = ( with (2.2))

=13|∫

Ωu3dxdx⊥| ≤ ( by interpolation, H1(Ω) ⊂ L3(Ω)4)


≤ c1|u|3/2L2(Ω)|∇u|3/2

L2(Ω)≤ 1

4|∇u|2L2(Ω) + c2|u|6L2(Ω),

•∫

Ωfxudxdx⊥ ≤ 1

2|f |2L2(Ω) +

12

∣∣√xu∣∣2L2(Ω)

,

• ε∫

Ω

(∂4u

∂y4+∂4u

∂z4

)xudxdx⊥ = ε

∣∣∣√x∂2u

∂y2

∣∣∣2L2(Ω)

+ ε∣∣∣√x∂2u

∂z2

∣∣∣2L2(Ω)

•∫ 1

0

∂2u

∂x4xudx = −

∫ 1

0

∂3u

∂x3

(u+ x

∂u

∂x

)dx

=∫ 1

0

(2uxxux + xu2

xx

)dx = − |ux (0, ·, ·)|2 +

∫ 1

0xu2

xxdx,

• ε∫

Ω

∂4u

∂x4xudxdx⊥ = ε

∣∣√xuxx∣∣2L2(Ω)− ε

∫Rd

∣∣∣ux(0, x⊥, t)∣∣∣2dx⊥.

Hence, together with (2.22), we arrive at

d

dt|√xuε|2L2(Ω) +

12|∇uε|2L2(Ω) + 2|uεx|2L2(Ω) (4.8)

+ 2ε|√xuεxx|2L2(Ω) + 2ε|

√xuεyy|2L2(Ω) + 2ε|

√xuεzz|2L2(Ω)

≤|f |2L2(Ω) + |√xuε|2L2(Ω) + c|uε|2L2(Ω)+

+ 2c2|uε|6L2(Ω) + 2ε∫

Rd|uεx(0, x⊥, t)|2dx⊥.

Taking into account the previous estimates (4.7) we obtain the followingestimates, also independent of ε :

∇uε,√εxuεxx,

√εxuεyy,

√εxuεzz, are bounded in L2(0, T ;L2(Ω)). (4.9)

From (4.7) and (4.9) we infer a bound on uεuεx on which we now elaboratebecause of our needs below. We write (dropping momentarily the ε):∫

Ω|uux|9/8dxdx⊥ =

∫Ω

(u2)38 (u6)

116 |ux|9/8dxdx⊥

≤ ( with H1(Ω) ⊂ L6(Ω) in dimension 3) ≤ |u|34

L2(Ω)|∇u|

2416

L2(Ω);

that is,

|uεuεx|L9/8(Ω) ≤ |uε|2/3L2(Ω)

|∇uε|4/3L2(Ω)

. (4.10)

4We use the interpolation inequalities corresponding to the worse case, d = 2.


The function in the right-hand side of (4.10) is bounded in L32 (0, T ); that is,

uεuεx is bounded in L3/2(0, T ;L9/8(Ω)),

and hence in L9/8(0, T ;L9/8(Ω)).(4.11)

Since L9 is the dual of L9/8 and H76 (Ω) ⊂ L9(Ω) in space dimension 3, we

also haveuεuεx is bounded in L9/8(0, T ;H−

76 (Ω)). (4.12)

Thanks to (4.7), (4.9), (4.12), equation (4.3) now gives∂uε

∂tis bounded (independently of ε) in L9/8(0, T ;H−3(Ω)). (4.13)

Although the estimate (4.13) is a very poor one, it allows us to show that thefamily uε is relatively compact in L2(0, T ;L2

loc(Ω)). As we said the estimatesabove are then sufficient to obtain the existence of uε for ε > 0 fixed, andthey permit also, in a second step, to pass to the limit ε → 0, using acompactness argument for the nonlinear term.

ii) Having shown that the limit u of (a subsequence extracted from) uε isa solution of (1.1) we want now to address the question of the boundary andinitial conditions. The initial condition u(x, x⊥, 0) = u0(x, x⊥) is satisfiedbecause, due in particular to (4.13), uε converges to u in C([0, T ];H−3

w (Ω)),where H−3

w (Ω) is H−3(Ω) equipped with the weak topology.Similarly the boundary conditions

u(0, x⊥, t) = u(1, x⊥, t) = 0

are satisfied since they are satisfied by uε and since uε converges to u weaklyin L2(0, T ;H1(Ω)).

For the existence, there remains to show that the boundary condition

ux(1, x⊥, t) = 0 (4.14)

is satisfied. This boundary condition is the object of Lemma 4.1 below wherewe show that ux(1, ·, ·) is defined when u ∈ L∞(0, T ;L2(Ω))∩L2(0, T ;H1(Ω))and u satisfies an equation like (1.1) (or (4.3)) and furthermore this tracedepends continuously on u in a suitable topology, so that uεx(1, ·, ·) = 0gives, at the limit, ux(1, ·, ·) = 0. Note that Lemma 4.1 is a slight variant ofTheorem 2.2).

Remark 4.1. Before we start the proof of (4.14), let us observe that (4.23)and the estimates leading to (4.1) allow us to state, in the nonlinear case,a result totally similar to Theorem 2.5; remarks similar to Remark 2.3 and2.4 are also valid in the nonlinear case.


We now pursue the proof of (4.14) and start with Lemma 4.1 which showsthat if u ∈ L∞(0, T ;L2(Ω)) ∩ L2(0, T ;H1(Ω)) satisfies equation (1.1) thenux(0, ·, ·) makes sense (as well as other traces).

Lemma 4.1. If u ∈ L∞(0, T ;L2(Ω)) ∩ L2(0, T ;H1(Ω)) satisfies equation(1.1), then

u, ux, uxx ∈ Cx([0, 1];Y ), Y = H−2((0, T )× Rd), (4.15)

and, in particular,ux|x=0,1, uxx|x=0,1 (4.16)

are well defined in Y. Furthermore these traces depend continuously on u ina sense made precise in the proof.

Proof. As in the proof of Theorem 2.2, we write equation (1.1) in the form

uxxx = f − cux − uxyy − uxzz − uux − ut (4.17)

and we observe that, since

u, ux ∈ L2(0, T ;L2(Ω)) = L2x(0, 1;L2((0, T )× Rd)),

it follows that

ut ∈ L2x(0, 1;H−1(0, T ;L2(Rd))),

uxyy, uxzz ∈ L2x(0, 1;L2(0, T ;H−2(Rd))).

Also according to (4.11),

uux ∈ L98x (0, 1;L

98 ((0, T )× Rd)) ⊂ L

98x (0, 1;H−2((0, T )× Rd)).

Thus uxxx belongs (at least) to the largest of these spaces; that is,

uxxx ∈ L98x (0, 1;H−2((0, T )× Rd)); (4.18)

(4.15) and (4.16) follow.Furthermore, if a sequence of functions um satisfies (1.1) with f = fm,

and um → u in L∞(0, T ;L2(Ω)) and L2(0, T ;H1(Ω)) and fm → f in L2(0, T ;L2(Ω)) (strongly), then um, umx, umxx converge to ux|x=0,1, uxx|x=0,1 in Y.If the convergence of um and fm are weak (weak-star for L∞), then theconvergences hold in Cx([0, 1];Yw) and in Yw. These convergences use a com-pactness argument based on the analog of (4.13), which is used to show thatumumx converges to uux.


iii) We now need to show that the boundary condition uεx(1, ., .) = 0,which is satisfied in a strong sense for ε > 0 (since uε ∈ L2(0, T ;H2(Ω)),“passes to the limit” to imply (4.14). It suffices here to use Lemma A.2 foruε a solution of (4.3), (2.2), (2.3) with p = 9

8 , Y = H−2((0, T ) × Rd), as inLemma 4.1 and

gε = f − uεt − cuεx −∆⊥uεx − uεuεx − εuεyyyy − εuεzzzz.

We observe that, since uε remains bounded in L∞(0, T ;L2(Ω)) ∩ L2(0, T ;H1

0 (Ω)) as ε → 0, the following functions remain bounded in the indicatedspaces:

uεx in L2x(0, 1;L2

t (0, T ;L2(Rd)), (4.19)

uεt in L2x(0, 1;H−1

t (0, T ;L2(Rd)), (4.20)

∆⊥uεx in L2x(0, 1;L2

t (0, T ;H−2(Rd)), (4.21)

uεyyyy + uεzzzz in L2x(0, 1;L2

t (0, T ;H−4(Rd)). (4.22)

We have thus shown that uεuεx remains bounded in L9/8(0, T ;L9/8(Ω)) =L

9/8x (0, 1;L9/8((0, T )× Rd).Finally gε remains bounded in the reflexive Banach space L

9/8x (0, 1;Y )

where

Y = H−1t (0, T ;L2(Rd)) + L2

t (0, T ;H−4(Rd)) + L9/8(0, T ;L9/8(Rd)).

We can apply Lemma A.2 from the Appendix with this space Y .

Remark 4.2. As in the stationary and linear cases, the traces in (4.16) aredefined in a weak sense (in a large space). We did not try to refine thesetrace results, as we only need to know that these traces exist and dependcontinuously on u.

iv) Finally, we conclude the proof of Theorem 4.1 by proving the unique-ness and the strong continuity properties.

Let u and v be two solutions of (1.1), (2.2), (2.3) and let w = u− v. Thenw satisfies

∂w

∂t+ c

∂w

∂x+ ∆

∂w

∂x= w

∂u

∂x+ v

∂w

∂x(4.23)

together with the boundary and initial conditions (2.2) and (2.3) (with w0 =0). Assuming that w is smooth we multiply (4.23) by (1+x)w and integrateover Ω. We find (compare also to (4.6) and (4.8)):d

dt(|w|2L2(Ω) + |

√xw|2L2(Ω)) + |∇w|2L2(Ω)


≤2∫

Ω(uxw2 + vwwx)dxdx⊥+ 2

∫Ω(xw2ux + xvwwx)dxdx⊥ + c

∫Ωw2dxdx⊥

=∫

Ω

[2(1 + x)w2ux − vw2 + (1 + x)vxw2

]dxdx⊥ + c

∫Ωw2dxdx⊥

≤ ( with σ(t) = |ux(t)|L2(Ω) + |vx(t)|L2(Ω) + |v(t)|L2(Ω)) 5

≤ c′σ(t)|w|2L4(Ω) + c

∫Ωw2dxdx⊥

≤ ( by interpolation H1/2 ⊂ L4 in dimension 2)

≤ c′σ(t)|w|L2(Ω)|∇w|L2(Ω) + c

∫Ωw2dxdx⊥

≤ 12|∇w|2L2(Ω) + c′(σ2(t) + 1)|w|2L2(Ω).

Hence,d

dt

(|w|2L2(Ω) + |

√xw|2L2(Ω)

)≤ c′(σ2(t) + 1)|w|2L2(Ω). (4.24)

Since w(0) = 0 and σ2 is an integrable function (∈ L1(0, T )), (4.24) implies,using the Gronwall lemma, that w(t) = 0 for all t ∈ (0, T ). The uniquenessfollows in dimension 2 once we have proven the first of the previous chain ofinequalities.

To do so, we regard w as a solution in the space L∞(0, T ;L2(Ω)) ∩L2(0, T ;H1(Ω)) of (4.23) written as

∂w

∂t+ c

∂w

∂x+ ∆

∂w

∂x= g, (4.25)

with g = wux + vwx.Furthermore, w satisfies the boundary conditions (2.2) and (2.3) and w =

0 at t = 0. As for (4.12), we can prove that

|g|L4/3(Ω) ≤ C′(|w|L4(Ω)|ux|L2(Ω) + |v|L4(Ω)|wx|L2(Ω))

≤ C ′(|w|1/2L2(Ω)

|∇w|1/2L2(Ω)

|ux|L2(Ω) + |v|1/2L2(Ω)

|∇v|1/2L2(Ω)

|wx|L2(Ω)),

so that g ∈ L4/3(0, T ;L4/3(Ω).By the Galerkin method, we can construct a solution of (4.25), (2.2), (2.3)

satisfying w = 0 at t = 0 and the energy type inequalityd

dt(|w|2L2(Ω) + |

√xw|2L2(Ω)) + |∇w|2L2(Ω) − c|w

2|L2(Ω) ≤∫

Ωgwdxdx⊥. (4.26)

5The c′, c′′, ci are constants which may be different in different places.


Note that the right-hand side of (4.26) makes sense since w ∈ L2(0, T ;H1(Ω))∩L∞(0, T ;L2(Ω)), and thus w ∈ L4(0, T ;L4(Ω) which is the dual of L4(0, T ;L4(Ω)).

Finally, we observe that, for such a g, the solution of the linear equation(4.25) with boundary conditions (2.2) and (2.3) and w(0) = 0 is necessarilyunique. Indeed, if w1, w2 are two such solutions, setting w = w1 − w2 andW (t) =

∫ t0 w(s)ds, then

∂W

∂t+ c

∂W

∂x+ ∆

∂W

∂x= 0. (4.27)

Since W ∈ L∞(0, T ;L2(Ω)) and since W and ∂W∂t = w satisfy the boundary

conditions (2.2) and (2.3), equation (4.27) shows that W ∈ L∞(0, T ;D(A)).Therefore, since (AW,W ) = 0,

d

dt|W |2L2(Ω) = 0,

which proves that W (t) = 0 for every t, and thus that w(t) = 0 for every t.To prove the strong continuity in time and the continuity of the flow map,

we can use for instance the classical “Bona-Smith approximation trick” (see[3]). Details are omitted.

This conclude the proof of Theorem 4.1.

Remark 4.3. The result in Theorem 4.1 displays the remarkable propertiesof the ZK equation posed on a strip of Rd+1, d = 1, 2, in contrast with theCauchy problem posed in the whole space, see ([6], [15]). In particular itis not known whether the Cauchy problem is well-posed for initial data inL2(Rd+1), d = 1, 2. A crucial point in the proof above is the fact that the ZKequation posed on a strip behaves essentially as a nonlinear heat equation“perturbed” by a dispersive term. This fact is illustrated when one performsthe change of unknown u(x, x⊥, t) = e−xv(x, x⊥, t). If u is a solution of theZK equation (1.1), then v solves

vt − 2v + 4vx − 3vxx −∆⊥v + ∆vx + e−x(vvx − v2) = e−xf. (4.28)

Remark 4.4. Theorem 4.1 implies obviously the global well-posedness ofthe initial-boundary-value problem for the ZK equation on a strip Ω =(0, L)×Rd, the estimates on the solution depending of course on L. It wouldbe interesting (for instance for numerical purposes) to obtain estimates in-dependent of L, allowing thus to pass to the limit as L → +∞ in order toobtain a solution of an initial-boundary-value problem on the infinite strip

5The c′, c′′, ci are constants which may be different in different places.


(0,+∞)×Rd. Such a result has been obtained in [5] for the KdV equation,under the boundary conditions u(0, t) = 0, ux(L, t) = uxx(L, t) = 0.

Remark 4.5. We could as well have treated the case of inhomogeneousboundary conditions on ∂Ω. We will consider this issue in [21] in the contextof controllability problems.

Appendix A

In this Appendix we give some trace theorems which are used in thearticle.

Lemma A.1. Let Y be a reflexive Banach space, c ∈ R, and assume thatu, g ∈ L1(0, 1;Y ) satisfy

uxxx + cux = g, (A.1)in the distribution sense on (0, 1) with values in Y . Then

u ∈ C2x([0, 1];Y ), uxxx ∈ L1

x(0, 1;Y ), (A.2)

and in particular the traces u(0), u(1), ux(0) and ux(1) are well defined inY and depend continuously on u, g, strongly from L1

x(0, 1;Y )2 into Y orweakly from Lpx(0; 1;Y )2 into Y if u, g ∈ Lpx(0, 1;Y ), for some p, 1< p < +∞.Proof. We observe that

uxx + cu = g1 + ψ, (A.3)

where g1(x) =∫ x

0 g(x′)dx′, and ψ ∈ Y (ψ a distribution independent ofx, see [22]). Hence, uxx ∈ L1

x(0, 1;Y ), u ∈ C1x([0, 1];Y ) and (A.1) gives

uxxx ∈ L1(0, 1;Y ) and u ∈ C2x([0, 1];Y ).

Therefore, the indicated traces (as well as e.g. uxx(0) and uxx(1)) are welldefined and strong and weak continuity results are easy.

The next lemma is a singularly perturbed version of Lemma A.1

Lemma A.2. Let Y be a reflexive Banach space and let p > 1. Assume thattwo sequences of functions uε, gε ∈ Lpx(0, 1;Y ) satisfy

uεxxx + εuεxxxx = gε,

uε(0) = uε(1) = uεx(1) = uεxx(0) = 0,(A.4)

with gε bounded in Lpx(0, 1;Y ) as ε → 0. Then uεxx (and hence uεx, uε) is

bounded in L∞(0, 1;Y ) as ε→ 0, and for any (strongly or weakly) convergentsubsequences uε → u, gε → g, uεx(1) converges to ux(1) in Y (weakly at least),and hence ux(1) = 0.


Remark A.1. i) It is likely that, in Lemma A.2, uεxxx is bounded in Lpx(0, 1;Y ) but this improvement is not needed here.

ii) As the proof shows, we can take p = 1, provided we drop the weakconvergence results in Lemma A.2 (or modify them in a suitable sense).

Proof. We drop the superscript ε for the moment and set v = uxx w = vx =uxxx; w satisfies

w + εwx = g. (A.5)Knowing w we can recover v using the boundary condition uxx = v(0) = 0,so that

v(x) =∫ x

0w(x1)dx1. (A.6)

Then we can recover u using the boundary conditions u(1) = ux(1) = 0, sothat

u(x) =∫ 1

x(x1 − x)v(x1)dx1. (A.7)

For u = uε to be a solution of the initial problem (A.4) we need w tosatisfy (A.5), and we need to enforce the last remaining boundary conditionu(0) = 0.

By integration of (A.5) we find

w(x) = w(0)e−x/ε +1ε

∫ x

0g(x1)e−(x−x1)/εdx1. (A.8)

We determine w(0) by enforcing the boundary condition u(0) = 0 whichgives

0 = u(0) =∫ 1

0x2v(x2)dx2 =

∫ 1

0

∫ x1

0w(x1)dx1dx2 =

12

∫ 1

0(1−x2

2)w(x1)dx2.

This last expression of u(0) is of the form

0 = u(0) = I1w(0) + I2 (A.9)

with

I1 =12

∫ 1

0(1− x2

2)e−x2/εdx2,

I2 =12ε

∫ 1

0(1− x2

2)∫ x2

0g(x1)e(x2−x1)/εdx1dx2.

An easy computation shows that I1 = ε2 +O(ε3), and that

|I2|Y ≤1ε

∫ 1

0

∫ x2

0|g(x1)|Y e−(x2−x1)/εdx1dx2 ≤ |g|L1(Y ).


Hence, (A.9) gives

|w(0)|Y ≤C

ε|g|L1(Y ), (A.10)

with C a constant independent of ε.Then we find

uxx(x) = v(x) =∫ x

0w(x1)dx1 = J1 + J2,

with

J1 = w(0)∫ x

0ex2/εdx2 = εw(0)(1− e−x/ε),

so that, with (A.10),|J1|Y ≤ C|g|L1(Y ),

and

J2 =1ε

∫ x

0

∫ x2

0g(x1)e−(x2−x2)/εdx1dx2.

The estimate for J2 is the same as that of I2, |J2|Y ≤ |g|L1(Y ).Finally, we find

|uεxx(x)|Y ≤ C|gε|L1(Y ),

which shows that uεxx remains bounded in L∞(0, 1;Y ) as ε → 0.. The ex-istence of the trace ux(1) and its linear continuous dependence on uε, gεfollow easily.

Acknowledgments. This work was partially supported by the NationalScience Foundation under the grant NSF-DMS-0906440, and by the ResearchFund of Indiana University. J.-C. S. acknowledges partial support fromproject ANR-07-BLAN-0250 of the Agence Nationale de la Recherche andfrom the Wolfgang Pauli Institute in Vienna.

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AN INITIAL BOUNDARY-VALUE PROBLEM FOR THE ZAKHAROV