劉 令,王國銘,謝嘉寧
(1.吉林建筑大學 基礎(chǔ)科學部,長春130118;2.吉林大學 數(shù)學學院,長春130012;3.東北財經(jīng)大學 數(shù)學與數(shù)量經(jīng)濟學院,遼寧 大連116025)
考慮如下初邊值問題:
為方便,引進如下符號:
這里
設(shè)Lr(·)(Ω)表示如下可測函數(shù)空間:
其范數(shù)為
易驗證Lr(·)(Ω)是Banach空間[4,6].由范數(shù)定義有
根據(jù)文獻[4,6]可知
令B1=max{B,m1/p}.定義
則
引理1 函數(shù)E(t)關(guān)于時間變量t是非增的.
證明:由問題(1)及式(8),對函數(shù)E(t)求導得
證明:由式(5),(8)有
其中α=‖▽um‖.易驗證h(α)在區(qū)間(0,α1)上單調(diào)遞增,在區(qū)間(α1,+∞)上單調(diào)遞減且h(α)→-∞,當α→+∞時,h(α1)=E1,其中E1,α1的定義如式(6),(7).由于E(0)<E1,故存在α2>α1,使得h(α2)=E(0).令α0=‖▽‖,由式(11)有h(α0)≤E(0)=h(α2),進一步還可證明α0≥α2.
假設(shè)式(9)不成立,即存在t0>0,使得‖▽um(·,t0)‖<α2.由于‖▽um(·,t)‖關(guān)于時間t連續(xù),故可選取t1>0,使得‖▽um(·,t1)‖>α1.進一步,根據(jù)函數(shù)h(α)的單調(diào)性,有
這與引理1矛盾.
定義 H(t)=E1-E(t).
引理3 對所有t>0,均有
證明:由引理1,有H′(t)≥0,即 H(t)≥H(0)>0,?t≥0.根據(jù)式(8),(11)有
定理1 假設(shè)r(x)滿足式(2),(3),且如下條件成立:
則問題(1)的解在有限時刻爆破.
由式(8),(12),有
其中
由式(4),有
由Lr(·)+m(Ω)?Lm+1(Ω),有
又根據(jù)式(15)~(17),有
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