一類具非局部邊界條件的擬線性拋物方程解的整體存在與爆破

孟繁慧,高文杰

(1.長(zhǎng)春金融高等專科學(xué)校,長(zhǎng)春 130028;2.吉林大學(xué) 數(shù)學(xué)學(xué)院,長(zhǎng)春 130012)

0 引 言

考慮如下具非局部邊界條件和內(nèi)部吸收項(xiàng)的擬線性拋物型方程:

(1)

其中:Ω是N(N≥1)中具有光滑邊界?Ω的有界區(qū)域;a為正常數(shù);u0(x),f(s)和k(x,y)分別滿足如下假設(shè)條件:

(H1)f∈C([0,∞))∩C1((0,∞)),滿足f(0)≥0,f′(s)>0,s∈(0,∞);

許多物理、 化學(xué)或生物種群動(dòng)力學(xué)現(xiàn)象都可以用具非局部源的拋物型方程描述.例如,文獻(xiàn)[1]指出非局部反應(yīng)項(xiàng)能更精確地描述可壓縮活性氣體的燃燒過(guò)程.當(dāng)f(s)=sp(0

(2)

在某種特殊情形下解的整體存在和衰減性質(zhì);文獻(xiàn)[6]給出了問(wèn)題(2)在k(x,y)≥0,g(x,u)=cu(其中c是一個(gè)沒(méi)有任何限制的常數(shù))情形下解的一個(gè)整體上界和特殊情形時(shí)邊界值的衰減性質(zhì)； 文獻(xiàn)[7]建立了問(wèn)題(2)的比較原理,并對(duì)一般形式的g(x,u)證明了問(wèn)題(2)古典解的局部存在性,對(duì)g(x,u)=c(x)u的特殊情形,證明了解的整體存在性和指數(shù)增長(zhǎng)性質(zhì).當(dāng)g(x,u)關(guān)于u的增長(zhǎng)超線性時(shí),問(wèn)題(2)的解可能在有限時(shí)刻爆破.特別地,文獻(xiàn)[8]在g(x,u)=g(u)情形下討論了問(wèn)題(2),用上下解的方法給出了此時(shí)問(wèn)題正解在有限時(shí)刻爆破的充分條件,此外,還給出了當(dāng)g(u)=up及g(u)=eu時(shí)解的爆破速率估計(jì).

考慮如下具非局部邊界條件和局部反應(yīng)項(xiàng)的多孔介質(zhì)方程:

(3)

其中:m,p>1為常數(shù);初值u0(x)和權(quán)函數(shù)k(x,y)滿足與問(wèn)題(1)相同的假設(shè).文獻(xiàn)[9]給出了問(wèn)題(3)存在整體解的充分必要條件,并得到了解的爆破速率估計(jì).文獻(xiàn)[10-16]研究了具非局部邊界條件的拋物方程或方程組,得到了一些有意義的結(jié)果.

上述研究表明,問(wèn)題(2),(3)解的增長(zhǎng)或衰減性質(zhì)依賴于g(x,u)的增長(zhǎng),這與具齊次邊界條件的半線性方程解的性質(zhì)相似.另一方面,由于非局部邊值的存在性,解的漸近性質(zhì)也依賴于權(quán)函數(shù)k(x,y).基于此,本文研究問(wèn)題(1)正古典解的漸近性質(zhì),考慮非線性擴(kuò)散、 非局部項(xiàng)、 吸收項(xiàng)和非局部邊界條件對(duì)解漸近性質(zhì)的綜合影響.

1 比較原理

則稱v(x,t)是問(wèn)題(1)的一個(gè)下解.改變定義1中不等號(hào)的方向可以得到上解的定義.如果函數(shù)v(x,t)既是問(wèn)題(1)的一個(gè)上解又是一個(gè)下解,則稱其為問(wèn)題(1)的一個(gè)(古典)解.

問(wèn)題(1)古典解的局部存在性可以用標(biāo)準(zhǔn)壓縮映像不動(dòng)點(diǎn)定理得到[7,17].正古典解的唯一性可由下述引理推出.

(4)

2 解的整體存在和爆破

證明: 考慮如下常微分方程初值問(wèn)題:

(5)

證明: 先構(gòu)造問(wèn)題(1)的一個(gè)整體存在且下方有界的上解.記λ1和φ1(x)>0(x∈Ω)分別是如下特征值問(wèn)題的第一特征值和相應(yīng)的特征函數(shù):

-Δφ=λφ,x∈Ω;φ(x)=0,x∈?Ω.

(6)

(7)

Δv=div(v)

可得

(9)

另一方面,由式(7)可知,對(duì)任意的x∈?Ω都有

(10)

(11)

證明: 通過(guò)構(gòu)造一個(gè)下方有界且爆破的下解完成證明.考慮如下Cauchy問(wèn)題:

(12)

(14)

(15)

式(13)～(15)表明,v(x,t)是問(wèn)題(1)的一個(gè)下方有界的爆破下解.再次應(yīng)用命題1可知u(x,t)也在有限時(shí)刻爆破.證畢.

還可證明當(dāng)非局部項(xiàng)的系數(shù)a適當(dāng)大時(shí),對(duì)任意滿足假設(shè)(H3)的k(x,y),問(wèn)題(1)的解都在有限時(shí)刻爆破.為此,記ψ(x)為下述橢圓問(wèn)題的唯一正解:

-Δψ(x)=1,x∈Ω;ψ(x)=0,x∈?Ω.

(16)

證明: 考慮如下齊次Dirichlet邊值問(wèn)題:

(17)

(H4) 存在常數(shù)δ>0,使得對(duì)任意的x∈Ω,有

這里δ是滿足δ+1≥a|Ω|的正常數(shù).

引理2假設(shè)(H2)～(H4)成立,且u(x,t)在T時(shí)刻爆破,則對(duì)任意的(x,t)∈Ω×[0,T),有Δu≤0.

設(shè)w(x,t)=Δu(x,t),則由式(1)可知對(duì)任意的(x,t)∈Ω×(0,T),有

wt=upΔw+2pup-1u·w+pu-1utw+p(p-1)u-2ut-2pup-1-upw.

(18)

注意到ut≥0,0

wt-upΔw-2pup-1u·w≤pu-1utw-upw, (x,t)∈Ω×(0,T).

(19)

最后由假設(shè)條件(H4)可知

w(x,0)=Δu0(x)≤0,x∈Ω.

(21)

結(jié)合式(19)～(21)及引理1可得w(x,t)=Δu≤0,(x,t)∈Ω×(0,T).證畢.

引理3假設(shè)(H2)～(H4)成立,且u(x,t)在T時(shí)刻爆破.則存在正常數(shù)C1,使得U(t)≥C1(T-t)-1/p.

證明: 由方程(1)、U(t)的定義及引理2可得

U′(t)≤a|Ω|Up+1(t).

(22)

對(duì)式(22)在(t,T)上積分并注意到當(dāng)t→T時(shí)U(t)→∞,得U(t)≥C1(T-t)-1/p,這里C1=(ap|Ω|)-1/p.證畢.

引理4假設(shè)(H2)～(H4)成立,且u(x,t)在T時(shí)刻爆破.則有ut≥δup+1,(x,t)∈Ω×(0,T).

證明: 記J(x,t)=ut-δup+1.直接計(jì)算可得

由Young不等式和H?lder不等式可知,對(duì)任意的正常數(shù)θ,都有

對(duì)任意的(x,t)∈?Ω×(0,T),有

將式(27)代入式(26)得

(28)

此外,假設(shè)(H4)蘊(yùn)含

J(x,0)≥(δ-δ)u0(x)p+1=0,x∈Ω.

(29)

對(duì)ut≥δup+1在區(qū)間(t,T)上直接積分可得

u(x,t)≤C2(T-t)-1/p, (x,t)∈Ω×(0,T),

(30)

這里C2=(δp)-1/p.綜合U(t)≥C1(T-t)-1/p和式(30),得:

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