EXISTENCE OF THE GLOBAL ATTRACTOR FOR WEAKLY DAMPED, FORCED KDV EQUATIONS ON SOBOLEV SPACE OF NEGATIVE INDEX

Kotaro Tsugawa

Mathematical Institute, Tohoku University, Sendai 980-8578, Japan E-mail: k99d97@math.tohoku.ac.jp

1. Introduction

以下の様な弱いダンピング項と外力項が付いた KdV 方程式に対するグローバルアトラクターの存在について考える．

∂_{t}u + γu + ∂_{x}^{3}u + \frac{1}{2}∂_{x}u^{2} = f, (1.1)

u(x, 0) = u_{0}(x) ∈ Ḣ^{s}(𝕋) (1.2)

ここで， 𝕋 は 1 次元トーラス． u は 𝕋 × [0, ∞) から ℝ への関数とする．また， γ > 0 とし， f ∈ L̇^{2}(𝕋) は時間に依存しないものとする．ここで， L̇^{2}(𝕋) = {u ∈ L^{2}(𝕋); ∫_{𝕋} u(x) dx = 0}， Ḣ^{s}(𝕋) = {u ∈ H^{s}(𝕋); ∫_{𝕋} u(x) dx = 0} である. グローバルアトラクターの存在について，十分滑らかな初期値 (s = 2) に対しては，既に多くの研究が有る ([3],[4],[7])．ここではグローバルアトラクターの存在を保障する s の下限について考える．まず，時間局所解に関する結果を述べる．通常の KdV 方程式 (γ = 0, f = 0) に対して Bourgain ([1]) は s = 0 で時間局所適切性（解の一意存在と初期値に対する連続依存性）を示した．ここで用いられた手法は Fourier restriction norm method と呼ばれ，この手法を改良する事により s ≥ −1/2 で時間局所適切である事が示された ([6],[2])．逆に s < −1/2 の場合にはある条件下で適切でない事も示されている．つまり、 s ≥ −1/2 が時間局所解が存在するための optimal な条件である．時間局所解の存在時間は初期値の大きさのみよって決まり，通常の KdV 方程式は L^{2} 保存則が成立するので， s = 0 の場合には、時間局所解を繰り返し延長する事が出来て，自動的に時間大域解になる．これらの方法は(1.1)–(1.2) に対しても適用する事が出来る．そして(1.1) に対して u を掛けて 𝕋 × [0, t] 上で積分する事により，

∥u(t)∥_{L^{2}}^{2} < ∥u_{0}∥_{L^{2}}^{2} exp(−2γt) + γ^{−2}∥f∥_{L^{2}}^{2}

が得られる．よって、初期値に応じて十分大きな t をとる事により， ∥u(t)∥_{L^{2}} < 2γ^{−2}∥f∥_{L^{2}} が得られるので，吸収集合の存在がわかる．この評価式を利用して， Goubet は， s = 0 に対して，グローバルアトラクターの存在を示した ([5])．一方， KdV 方程式は無限個の保存量を持つが， s < 0 においては保存量は存在しない．よって，時間大域的評価が得られず， s < 0 で時間大域解を考えるのは難しいと思われる．しかし， Colliander, Keel, Staffilani, Takaoka and Tao は， s < 0 において定義される擬保存量を計算する事により，通常の KdV 方程式が s ≥ −1/2 において時間大域的適切である事を示した ([2])．この手法は I-method と呼ばれている．ここで，次のような問題が考えられる． (1.1)–(1.2) に対しても，この I-method は適用できるか？そして，それを利用してグローバルアトラクターの存在を示す事が出来るか？これに対し以下の結果を得た．

Theorem 1.1. We assume s ≥ −1/2. Then, there exist the semigroup S(t) and maps M_{1} and M_{2} such that S(t)u_{0} is the unique solution of (1.1)–(1.2) and

S(t)u_{0} = M_{1}(t)u_{0} + M_{2}(t)u_{0}, (1.3)

sup_{t>T_{1}} ∥M_{1}(t)u_{0}∥_{L^{2}} < K, (1.4)

and for t > T_{1}

∥M_{2}(t)u_{0}∥_{H^{s}} < K exp (−γ(t − T_{1})) (1.5)

where the constant K depending only on ∥f∥_{L^{2}} and γ and T_{1} depending only on ∥f∥_{L^{2}}, γ and ∥u_{0}∥_{H^{s}}.

Corollary 1.2. Let s ≥ −1/2. Then, equation (1.1)–(1.2) possess a global attractor in H^{s}, that is a bounded subset of L^{2}.

2. The proof of Theorem 1.1

We define m : ℝ ↦ ℝ be a smooth monotone ℝ-valued function such that

m(ξ) = \{ \begin{matrix}1, & |ξ| < N \\ N^{−s}|ξ|^{s}, & |ξ| > 2N \end{matrix} .

We define the operator I as following

{ \atop If(ξ)} = m(ξ)f̂(ξ).

Here, we summarize the properties of I. For any function f and s < 0, we have

∥f∥_{H^{s}} ≤ ∥If∥_{L^{2}} ≤ N^{−s}∥f∥_{H^{s}}

Let ĝ_{1} = f̂ |_{|ξ|<2N} and ĝ_{2} = f̂ |_{|ξ|>2N}. Then, we have

∥g_{1}∥_{L^{2}} ≤ 2^{−s}∥If∥_{L^{2}}, ∥g_{2}∥_{H^{s}} ≤ N^{s}∥If∥_{L^{2}}.

We apply “I-method” to (1.1)–(1.2). Then, we have the following a priori estimate.

Proposition 2.1. Let T > 0 be given and u be a solution of (1.1)–(1.2) on t ∈ [0, T]. We assume s ≥ −1/2, N^{3/5} > γ, N^{1/5} > C_{1}T and

∥Iu_{0}∥_{L^{2}}^{2} + \frac{1}{γ^{2}} ∥If∥_{L^{2}}^{2} exp(2γT) < N^{6/5}C_{2}, (2.6)

then we have

∥Iu(T)∥_{L^{2}}^{2} exp(2γT) < C_{3} (∥Iu_{0}∥_{L^{2}}^{2} + \frac{1}{γ^{2}} ∥Ig∥_{L^{2}}^{2} exp(2γT)) . (2.7)

We prove Theorem 1.1 by using Proposition 2.1.

Proof of Theorem 1.1. We choose T_{1} > 0 so that

(2.8) exp(2γT_{1}) > ∥u_{0}∥_{H^{s}}^{2}∥f∥_{L^{2}}^{−2} max{γ^{−10s/3}, (C_{1}T_{1})^{−10s},

(C_{2}/2)^{5s/(3+5s)}∥u_{0}∥_{H^{s}}^{−10s/(3+5s)}, (2γ^{−2}C_{2}^{−1}∥f∥_{L^{2}}^{2} exp(2γT_{1}))^{−5s/3}},

which may certainly be done because s ≥ −1/2 and T_{1} depends only on ∥f∥_{L^{2}},γ and ∥u_{0}∥_{L^{2}}. Put

N = max{γ^{5/3}, (C_{1}T_{1})^{5}, (C_{2}/2)^{−5/(6+10s)}∥u_{0}∥_{H^{s}}^{5/(3+5s)},

(2γ^{−2}C_{2}^{−1}∥f∥_{L^{2}}^{2} exp(2γT_{1}))^{5/6}}.

Then, we have

N^{3/5} > γ, N^{1/5} > C_{1}T_{1},
∥Iu_{0}∥_{L^{2}}^{2} ≤ N^{−2s}∥u_{0}∥_{H^{s}}^{2} = N^{6/5}N^{−(6+10s)/5}∥u_{0}∥_{H^{s}}^{2} < C_{2}N^{6/5}/2,
γ^{−2}∥If∥_{L^{2}} exp(2γT_{1}) < C_{2}N^{6/5}/2.

Therefore, from Proposition 2.1 we obtain

∥u(T_{1})∥_{H^{s}}^{2} ≤ ∥Iu(T_{1})∥_{L^{2}}^{2} < C_{3} (N^{−2s}∥u_{0}∥_{H^{s}}^{2} exp(−2γT_{1}) + γ^{−2}∥f∥^{2}) . (2.9)

¿From (2.8), we have

N^{−2s} exp(−2γT_{1}) < ∥u_{0}∥_{H^{s}}^{−2}∥f∥_{L^{2}}^{2}.

Therefore, we obtain

∥u(T_{1})∥_{H^{s}}^{2} < C_{3}(1 + γ^{−2})∥f∥_{L^{2}}^{2} < K_{1} (2.10)

where K_{1} depends only on ∥f∥_{L^{2}} and γ. We next fix T_{2} > 0 and solve (1.1)–(1.2) on [T_{1}, T_{1} + T_{2}] with initial data u(T_{1}). Let K_{2} > 0 be sufficiently large enough to satisfy

(2.11) K_{2} exp(2γt) > max{γ^{−10s/3}, (C_{1}t)^{−10s}

(C_{2}^{−1}K_{1})^{−5s/3}, (C_{2}^{−1}γ^{−2}∥f∥_{L^{2}}^{2} exp(2γt))^{−5s/3}}

for any t > 0, which may certainly be done because s ≥ −1/2 and K_{2} depends only on ∥f∥_{L^{2}} and γ. Put N^{−2s} = K_{2} exp(2γT_{2}). Then, from (2.11), the assumptions in Proposition 2.1 are satisfied. Therefore, we obtain

\begin{aligned}∥Iu(T_{1} + T_{2})∥_{L^{2}}^{2} & < C_{3} (N^{−2s}∥u(T_{1})∥_{H^{s}}^{2} exp(−2γT_{2}) + γ^{−2}∥f∥_{L^{2}}^{2}) \\ & < C_{3} (K_{1}K_{2} + γ^{−2}∥f∥_{L^{2}}^{2}) < K_{3} \end{aligned} (2.12)

where K_{3} depends only on ∥f∥_{L^{2}} and γ. For t > T_{1}, we define maps M_{1}(t) and M_{2}(t) such that

{ \atop M_{1}(t)u_{0}} = { \atop S(t)u_{0}} |_{|ξ|<2N}, { \atop M_{2}(t)u_{0}} = { \atop S(t)u_{0}} |_{|ξ|>2N} (2.13)

where S(t)u_{0} = u(t) and N = (K_{2} exp(2γ(t − T_{1})))^{−1/2s}. Then, for t > T_{1}, we have

∥M_{1}(t)u_{0}∥_{L^{2}}^{2} < ∥Iu(t)∥_{L^{2}}^{2} < K_{3}, (2.14)

∥M_{2}(t)u_{0}∥_{H^{s}}^{2} < N^{2s}∥Iu(t)∥_{L^{2}}^{2} < K_{2}^{−1}K_{3} exp(−2γ(t − T_{1})). (2.15)

Let K = max{K_{2}^{1/2}, K_{1}^{−1/2}K_{2}^{1/2}}. Then, we have (1.4) and (1.5).

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