USA-Chile Workshop on Nonlinear Analysis, Electron. J. Diff. Eqns., Conf. 06, 2001, pp. 45–53. http://ejde.math.swt.edu or http://ejde.math.unt.edu ftp ejde.math.swt.edu or ejde.math.unt.edu (login: ftp)

First moments of energy and convergence to equilibrium ^{∗}

Jérôme Busca

Abstract

A basic question is to establish convergence to equilibrium for globally defined solutions to evolution problems. The purpose here is to emphasize the role of symmetry. In particular, it is proved that in some cases the first moments of energy are constant on the ω-limit set of the solution. This key property is used to prove convergence in two model evolution problems. This communication is based on two joint works with P. Felmer [3] and M.A. Jendoubi, P. Polacik [4].

1 Introduction and Main Results

A basic question in the study of evolution problems is the following: do globally defined in time solutions converge to an equilibrium? In case the problem is dissipative, one can typically prove that the ω−limit set (i.e. the set of all accumulation points of the solution u) ω(u) is included in the set of the solutions of some limiting stationary equation (steady states). This is usually done thanks to some appropriate Lyapunov energy functional.

If the set of steady states contains a continuum, then the convergence issue is whether the solution actually selects one of them as t → +∞, that is, whether ω(u) is a singleton.

In this generality, or even if one specializes to nonlinear parabolic evolution problems for instance, the question is still open, and appears to be surprinsingly difficult. Couterexamples in the non-autonomous case suggest that limitations do exist (see [21]). Partial results are available for instance in the analytic setting [12] [14] [22], in one dimension [18] [23], or under assumptions on the linearized operator, either explicitly stated as such, or resulting from the nature of the specific problem [10] [11] [17].

There is a large literature devoted to these questions, and I will not attempt to give any review of the results. For a very clear account on this, I refer to [5] and [13].

∗Mathematics Subject Classifications: 35B50, 35A05.

Key words: Parabolic Equations, Equilibrium, Convergence.

\begin{matrix}c \\ ◯ \end{matrix}2001 Southwest Texas State University.

Published January 8, 2001.

Partially supported by ECOS/CONICYT project C99E06

It is my purpose here to show that when the solutions enjoy some symmetry, the convergence question can be solved. The simplest nontrivial case is probably when the problem is posed in the whole space, is translation and rotationally invariant, and the set of positive steady states is made of all translates of a single radial solution. In this case, proving convergence of positive solutions of the evolution problem is tantamount to proving that ω(u) cannot contains more than one of such translates. To this purpose, I intend to introduce a method that makes use of first moments of the energy, a tool which appears to be new. These moments will be shown to assume constant values on the ω−limit set, just as energy does. However, unlike the latter, they are able to discriminate, meaning taking different values on, distinct translates.

For the interested reader I mention references [19] [20] where a thorough investigation of the links between symmetry and convergence is given, is the context of stable equilibria (different from the one I address here).

Rather than elaborating on this in full generality, let me select two simple instances where one can easily highlight the main underlying idea. These examples are taken from joint works with P. Felmer [3] and M.A. Jendoubi, P. Polacik [4]. These two evolutions problems will turn out to share the same stationary equation, namely the so-called scalar field equation

\begin{matrix}Δw − w + w^{p} = 0 in ℝ^{N} \\ w > 0, w(x) → 0 as |x| → ∞ \end{matrix}  (1.1)

with p subcritical, i.e. 1 < p < (N + 2)/(N − 2), N ≥ 3, an assumption that I make throughout this paper. By the well-known results of Berestycki-Lions [1], Gidas, Ni and Nirenberg [9] and Kwong [15] we know that the set of solutions to (1.1) is made of all translates of a unique positive radial solution. Note that the ground state condition

w(x) → 0 as |x| → ∞  (1.2)

is an essential piece of information here.

I now turn to the description of the two model problems.

1) A dissipative case: a nonlinear parabolic equation. Let us consider a globally defined in time weak positive solution u = u(x, t) ∈ C ([0, +∞), H^{1}(ℝ^{N})) to the following problem:

\begin{matrix}u_{t} = Δu − u + u^{p} in ℝ^{N} × (0, +∞) \\ u(·, 0) = u_{0}(·) ∈ C_{0}^{∞}(ℝ^{N}) \end{matrix}  (1.3)

We have the following convergence result:

Theorem 1.1 ([4, 7, 8]) Under the above assumptions, u(·, t) converges (in the H^{1}(ℝ^{N}) sense) either to zero or to a solution of (1.1).

2) A conservative case: a nonlinear elliptic equation. Let us now consider an entire positive solution u = u(x), x = (x_{1}, ⋯, x_{N+1}) to the scalar field equation

Δ_{ℝ^{N+1}}u − u + u^{p} = 0

in ℝ^{N+1}, and suppose we are interested in solutions which are not necessarily ground states in the sense of (1.2). The simplest case is to assume that u goes to zero in a cylindrical set of directions. This leaves out one variable (x_{N+1}, say) in the direction of which one wants to study the possible asymptotic behaviour of u. For that reason, it is convenient to think of x_{N+1} as time, and recast the problem as:

u_{tt} + Δu − u + u^{p} = 0 in ℝ^{N} × (−∞, +∞)
u(·, t) → 0 as |x| → ∞, uniformy in t ∈ ℝ,

(1.4)

with (x, t) = (x_{1}, ⋯, x_{N}, x_{N+1}) and Δ = ∑_{1≤i≤N} \frac{∂^{2}}{∂x_{i}^{2}} , the Laplace operator in

ℝ^{N}.

Since this problem is conservative, soliton-like solution may exists, so it is natural to assume that

u_{t}(x, t) → 0 as t → +∞, for all x ∈ ℝ^{N}.  (1.5)

Under these assumptions, one has the following convergence results:

Theorem 1.2 ([3]) Let u be a bounded weak solution to (1.4) satisfying (1.5). Then u(·, t) converges (in the H^{1}(ℝ^{N}) sense) either to zero or to a solution of (1.1) as t → +∞.

Since time is reversible in (1.4), it is straightforward to get the same result as t → −∞ if one assumes the equivalent of (1.5) as t → −∞. Moreover, it is an interesting fact that the right- and left-hand limits actually turn out to coincide. I refer to [3] for a proof of this.

Theorem 1.1 has been obtained in earlier independent works by Cortazar, Elgueta, del Pino [7] and Feireisl and Petzeltová [8] by different methods. Theorem 1.1 and Theorem 1.2 are quite particular cases of the results in [4] and [3] respectively. I simplified the setting here in order to draw a parallel between these two results.

In the sequel, I denote by case 1, or parabolic case (resp. 2 or elliptic case) the situation prevailing in Theorem 1.1 (resp. 1.2).

2 Sketch of the proofs

The key role is played by the energy functional

E(u(·, t)) = ∫ \{\frac{1}{2}|∇u|^{2} − F (u)\} dx,  (2.1)

where F (s) = ∫_{0}^{s} f(ζ) dζ = −\frac{1}{2}s^{2} + \frac{1}{p + 1}s^{p+1}, f(ζ) = −ζ + ζ^{p}, and its first

“moments” (for want of a better name):

E_{i}(u(·, t)) = ∫ x_{i} \{\frac{1}{2}|∇u|^{2} − F (u)\} dx,  (2.2)

i = 1, ⋯, N. As for the ω−limit set, it is defined as usual by:

ω(u) = ∩_{T >0} \overline{∪_{t≥T} \{u(·, t)\}}  (2.3)

in the parabolic case. In the elliptic case, some care is needed. One introduces the {v(t)}_{t∈ℝ}, defined as v(t)(x, τ) = u(x, t + τ), for all (x, τ) ∈ ℝ^{N} × [0, 1]. The correct notion of (right-hand) ω−limit set turns out to be in this case:

ω(u) = ∩_{T >0} \overline{∪_{t≥T} \{v(t)\}}.

Here the closures are taken for instance in the C^{2}(ℝ^{N}) (resp. C^{2}(ℝ^{N} × [0, 1])) topology.

The relevant properties of these functions are summarized in the following result.

Proposition 2.1 In both cases 1 and 2 we have:

a) ω(u) is either {0} or made of positive steady states, i.e. solutions to (1.1)

b) E is constant on ω(u)

c) Each function E_{1}, ⋯, E_{N} assumes a constant value on ω(u).

Note that in case 2, it is part of result a) that the functions in ω(u) do not depend on τ ∈ [0, 1].

Theorem 1.1 and 1.2 are simple consequences of this proposition. Indeed, suppose for contradiction that ω(u) were to contain two distinct translates of the radial solution of (1.1), say w_{1} and w_{2}. Up to a Euclidean change in co-ordinates, w_{1}(x) = w_{1}(|x|) and w_{2}(x) = w_{2}(|x − αe_{1}|) with α ≠ 0. Now E_{1}(w_{1}) = E_{1}(w_{2}) yields:

0 = E_{1}(w_{1}) = E_{1}(w_{2})
= ∫ x_{1} \{\frac{1}{2}|∇w_{2}|^{2} − F (w_{2})\} dx
= ∫ (x_{1} − α) \{\frac{1}{2}|∇w_{2}(|x − αe_{1}|)|^{2} − F (w_{2}(|x − αe_{1}|))\} dx + αE(w_{2})
= αE(w_{2}).

Here and in the sequel, unless otherwise specified, all integrals in space range over ℝ^{N}.

Multiplying the stationary equation in (1.1) by w_{2} and integrating by parts, it is straightforward to see that E(w_{2}) ≠ 0, a contradiction. Convergence then follows easily from the fact that ω(u) is a singleton, by compactness arguments that I do not reproduce here.

Lemma 2.1 ∃ε_{0} > 0 ∃C > 0 such that ∀α = (α_{1}, ⋯, α_{N}) ∈ ℕ^{N}, ∀k ∈ {0, 1} in case 1 ({0, 1, 2} in case 2), |α| + k ≤ 2,

|∂_{x}^{α}∂_{t}^{k}u(x, t)| ≤ Ce^{−ε_{0}|x|} for all (x, t) ∈ ℝ^{N} × ℝ_{+} (resp. ℝ).

Proof: It results from Corollary 3.1 in [7] and Lemma 2.1 in [3], to which I refer. The proof is based on comparison principles, together with Harnack inequality and blow-up arguments in the parabolic case.

We now turn to the proof of part a) and b) in Proposition 2.1. Case 1: It is well-known that t ↦ E(u(·, t)) is decreasing since:

\frac{d}{dt} \{∫ \{\frac{1}{2}|∇u|^{2} − F (u)\} dx\} = − ∫ u_{t}^{2}dx.  (2.4)

Hence ∫_{0}^{+∞}dt ∫ u_{t}^{2}dx < ∞. Making use of Lemma 2.1, it is standard to infer

part a) and b) in Proposition 2.1. Case 2: Let us test equation (1.4) with u_{t}:

∫_{t}^{t′} ds ∫ {u_{t}u_{tt} + Δuu_{t} + f(u)u_{t}} dx = 0.

Integrating by part (see Lemma 2.1) results in:

E(u(·, t′)) − E(u(·, t)) = \frac{1}{2} ∫ u_{t}^{2}(x, s)dx|_{s=t}^{t′} .

Hence by assumption (1.5) and Lemma 2.1, clearly ∃ lim_{t→+∞} E(u(·, t)), Now, test-

ing (1.4) with ϕ ∈ C_{0}^{∞}(ℝ^{N}) and integrating by parts results in:

∫_{t}^{t+1} ds ∫ {u_{tt} + Δu + f(u)} ϕ(x)dx = 0
∫ u_{t}(s, x)ϕ(x)dx|_{s=t}^{t+1} + ∫_{t}^{t+1} ds ∫ {uΔϕ + f(u)ϕ} dx = 0.

Since the first term in this last expression goes to zero as t → +∞ by assumption (1.5) and Lemma 2.1, it is clear from the definition of ω(u), Lemma 2.1, and (1.5) again that any v = v(x, τ) ∈ ω(u) is actually independent of τ and satisfies ∀ϕ ∈ C_{0}^{∞}(ℝ^{N}) ∫ {vΔϕ + f(v)ϕ} dx = 0. Since v ∈ C^{2}, it implies that v is a steady state. This completes the proof of part a) and b) in Proposition 2.1. I shall know sketch the proof of part c) in Proposition 2.1 in Case 1 and 2. Case 1: That the E_{i}’s are constant on ω(u) rely on the following lemma:

Lemma 2.2 i) We have \frac{d}{dt}E_{i} (u(·, t)) = − ∫_{ℝ^{N}} x_{i}u_{t}^{2} dx ii) There exists δ > 0 such that

∫_{0}^{+∞} dt ∫ e^{δ|x|}u_{t}^{2} dx < +∞.

Hence the limits lim_{t→+∞} E_{i} (u(·, t)), i = 1, ⋯, N, are well-defined.

Proof of i) Denoting by u_{i} the derivative of u with respect to x_{i} one has:

\frac{d}{dt}E_{i} (u(·, t)) = − ∫ u_{t} \{∇ · (x_{i}∇u) + x_{i}f(u)\} dx
= − ∫ x_{i}u_{t}^{2} dx − ∫ u_{t}u_{i} dx

∫ u_{t}u_{i} dx = ∫ {Δu + f(u)) u_{i} dx
= − ∫ ∇u · ∇u_{i} dx + ∫ f(u)u_{i} dx = 0 .

Proof of ii) Taking polar co-ordinates x = (r, θ), define for r > 0:

H^{T} (r) = \frac{1}{2} ∫_{0}^{T} dt ∫_{S^{N−1}} u_{t}^{2}(r, θ, t) dθ.

Denoting by Δ_{r} = \frac{∂^{2}}{∂r^{2}} + \frac{(N − 1)}{r} \frac{∂}{∂r} the radial Laplace operator, we have: We have:

Δ_{r}H^{T} = ∫_{0}^{T} dt ∫_{S^{N−1}} u_{t}Δ_{r}u_{t} dθ + ∫_{0}^{T} dt ∫_{S^{N−1}} |∇_{r}u_{t}|^{2} dθ
≥ −\frac{1}{2} ∫_{S^{N−1}} u_{t}^{2}(0, r, θ) dθ − ∫_{0}^{T} dt ∫_{S^{N−1}} f′(u)u_{t}^{2} dθ,

hence:

Δ_{r}H^{T} (r) ≥ −ψ(r) + αH^{T} (r) ∀r ≥ R_{0},

for some positive constants α, R_{0} and some ψ ∈ C_{0}^{∞}(ℝ^{N}), ψ ≥ 0.

Now a simple comparison argument with the solutions of Δ_{r}g_{0} − αg_{0} = −ψ, g_{0}(r_{0}) = 0, g_{0}(r) → 0 as r → ∞ and Δ_{r}g_{1} − αg_{1} = 0, g_{1}(r_{0}) = 1, g_{1}(r) → 0 as r → ∞ implies ∀r_{0} ≥ R_{0} ∃δ > 0 ∃C > 0 ∀T > 0 ∀r ≥ r_{0} 0 ≤ H^{T} (r) ≤ C (1 + H^{T} (r_{0})) e^{−2δr}. Since ∫_{0}^{+∞}dt ∫ u_{t}^{2} dx < +∞ by Fubini’s Theorem H^{T} (r_{0}) → H^{∞}(r_{0}) < ∞ as T → ∞ for a.e. r_{0}. Hence H^{∞}(r) ≤ Ce^{−2δr} ∀r ≥ r_{0}. Case 2: Let us define

Ẽ_{i} = E_{i} − \frac{1}{2} ∫ x_{i}u_{t}^{2}dx.

Differentating Ẽ_{i} results in:

\frac{d}{dt}Ẽ_{i}(u) = − ∫ u_{t} \{∇ · (x_{i}∇u) + x_{i}f(u)\} dx − ∫ x_{i}u_{t}u_{tt} dx

integrating by parts:

\frac{d}{dt}Ẽ_{i}(u) = ∫ u_{t}u_{i}dx  (2.5)

differentiating once more:

\frac{d^{2}}{dt^{2}} Ẽ_{i}(u) = ∫ \{Δu + f(u)\} u_{i}dx − ∫ u_{t}u_{it}dx
= − ∫ ∇u · ∇u_{i} − f(u)u_{i} + u_{t}u_{i}
= − ∫ \frac{∂}{∂x_{i}} \{\frac{1}{2}|∇u|^{2} − F (u) + \frac{u_{t}^{2}}{2} \} dx = 0 ,

thanks to Fubini’s Lemma. Note that I have repeatedly used Lemma 2.1 here. Hence Ẽ_{i}(u) = αt + β for constants α, β. Now (2.5) together with assumption

(1.5) imply lim_{t→∞} \frac{d}{dt}Ẽ_{i}(u(·, t)) = 0. Thus α = 0, hence t ↦ Ẽ_{i}(u(·, t)) is constant. Hence E_{i} is constant on ω(u) by (1.5). This completes the proof of part c) in Proposition 2.1, hence that of Theorem 1.2.

Acknowledgments It is my pleasure to thank the organizers of the USA-Chile Congress on Nonlinear Analysis for their invitation. This paper is based on a contribution to the Nonlinear Analysis 2000 meeting (Courant Institute of Mathematical Sciences, New York, May 2000).

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Jérôme Busca Laboratoire de Mathématiques et de Physique Théorique Université Francois Rabelais Parc de Grandmont 37200 Tours, France email: busca@univ-tours.fr