常小軍,路京京

(1. 吉林大學(xué) 數(shù)學(xué)學(xué)院,長(zhǎng)春 130012;2. 吉林大學(xué) 金融學(xué)院,長(zhǎng)春 130012)

0 引 言

考慮如下非線性波方程:

(1)

其中f∈C([0,π]×R2,R)且f關(guān)于t是2π-周期的.

記‖·‖r為空間Lr(Q)(r∈[1,∞))中范數(shù),(·,·)r為相應(yīng)的內(nèi)積. 定義Banach空間

其上的范數(shù)為

顯然,E具有分解E=E-⊕E0⊕E+. 定義E上的對(duì)偶

其中

定義作用泛函Φ:E→ R,

(2)

由(2)可得Φ∈C1(E,R). 顯然,Φ是強(qiáng)不定的泛函,并且Φ的臨界點(diǎn)對(duì)應(yīng)于式(1)的弱解,即

(3)

令X={u∈E:u(t,x)=u(π+t,π-x)},則X是E的閉子空間且X∩Ker□={0}. 顯然,X緊嵌入到Lr(R)(?r≥1),且X在□及f的作用下是不變的. 因此Φ|X的臨界點(diǎn)即為Φ的臨界點(diǎn). 不妨記Φ|X為Φ,X+=X∩E+,X-=X∩E-,則有X=X-⊕X+. 在X上定義范數(shù)‖u‖X?‖u‖E. 顯然,X是Banach空間,將其記為 ‖·‖. 重排□的特征值: …≤λ-,3≤λ-,2≤λ-,1<0<λ+,1≤λ+,2≤λ+,3≤…,相應(yīng)的特征函數(shù)分別記為ψ-,i,ψ+,k,i,k∈Z+.

1 主要結(jié)果

定理1假設(shè)下面條件滿(mǎn)足,則問(wèn)題(1)存在非平凡弱解:

1) 存在a1,a2>0,使得對(duì)于一致的(t,x)∈Q,有|f(t,x,s)|≤a1|s|+a2,?s∈R;

5)f(t,x,s)=f(π+t,π-x,s),?(t,x,s)∈Q×R.

注1由于波算子□的核空間是無(wú)窮維的,通常要求f(t,x,s)關(guān)于s是單調(diào)函數(shù). Coron[11]通過(guò)引入取不變子空間的技巧在不要求f單調(diào)的情形下研究了問(wèn)題(1). 本文結(jié)合Coron的技巧研究問(wèn)題(1),因此不要求f滿(mǎn)足單調(diào)性條件. 此外,這里不要求比率f(t,x,s)/s在s充分大時(shí)有極限,并且不要求f在無(wú)窮遠(yuǎn)處滿(mǎn)足共振或非共振條件,從而f(t,x,s)/s在s充分大時(shí)可能跨越波算子□的多個(gè)特征值.

注2Costa等[12]在如下非二次條件下,結(jié)合定理1中的條件2),3)研究了問(wèn)題(1)非平凡弱解的存在性:

關(guān)于 a.e. (t,x)∈Q一致成立,

且有

關(guān)于 a.e. (t,x)∈Q一致成立,

或

關(guān)于a.e. (t,x)∈Q一致成立.

顯然定理1的條件4)弱于非二次條件.

下面應(yīng)用強(qiáng)不定泛函的臨界點(diǎn)理論證明定理1. 類(lèi)似于文獻(xiàn)[12]中定理2.10的證明,只需證明泛函Φ有鞍點(diǎn)結(jié)構(gòu)并且滿(mǎn)足(C)c條件. 證明分兩個(gè)步驟: 泛函Φ有鞍點(diǎn)結(jié)構(gòu)和泛函Φ滿(mǎn)足緊性條件.

引理1在定理1的假設(shè)下,泛函Φ有鞍點(diǎn)結(jié)構(gòu).

證明: 由條件1)～3)知,存在ρ1,ρ2∈R,使得μ1<ρ1<λ+,m<ρ2<μ2,并且存在p>2和Cp>0,使得

?(t,x,u)∈Q×R.

對(duì)任意的j∈Z+,記Ej=X-⊕span{φ+,1,…,φ+,j}. 對(duì)任意的u∈(Em-1)⊥,

Φ(u)≥δm, ?u∈(Em-1)⊥, ‖u‖=rm.

易見(jiàn),存在充分大的Rm>0,使得當(dāng)u∈Em且滿(mǎn)足‖u‖≥Rm時(shí),有Φ(u)≤0. 證畢.

引理2在定理1的假設(shè)下,對(duì)任意的c∈R,泛函Φ滿(mǎn)足(C)c條件.

證明: 令{un}?X滿(mǎn)足

Φ(un)→c, (1+‖un‖)Φ′(un)→0.

(4)

矛盾. 因此,w0≠0. 從而|un(t,x)| → +∞,?(t,x)∈Q*,其中Q*∶={(t,x)∈Q:w0(t,x)≠0}. 結(jié)合假設(shè)條件1)和4),存在M>0,使得

矛盾. 因此{(lán)un}在X中有界. 進(jìn)而利用X緊嵌入到Lr(Q)(r≥1)可知,存在u0∈X,使得在X中un→u0. 證畢.

下面證明定理1. 由引理1知,存在泛函Φ的(C)c序列{un}使得c>0. 利用引理2,序列{un}在X中是一致有界的,且有u0∈X,使得在X中un→u0. 再利用標(biāo)準(zhǔn)的討論[12],可得Φ(u0)=c,并且有

?φ∈X.

從而u0是問(wèn)題(1)的非平凡弱解.

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