一類微分邊值問題的正解

一類微分邊值問題的正解

1微分方程邊值問題由于它廣泛應(yīng)用于各種應(yīng)用學(xué)科，并得到了廣泛的研究。假設(shè)如下：。{y″+λ(m∑i=1ai(t)[y(t)]αi-Μ)=0,0<t<1,ay(0)-by′(0)=0,cy(1)+dy′(1)=0.(1.1)其中ai∈C((0,1),(0,+∞)),1≤i≤m;k≥0,αi<0,1≤i≤k;αi≥0,k+1≤i≤m,αm>1,M>0為常數(shù),λ>0為參數(shù),a,b,c,d非負(fù),ρ=ac+ad+bc>0,bd=0.由于在(1.1)中,ai(t)可在t=0,1奇異,非線性項(xiàng)中的指數(shù)αi(1≤i≤k)小于零,因此這里的多項(xiàng)式型微分方程邊值問題為奇異的.注意到M>0,因此(1.1)又為半正微分方程邊值問題.對于非奇異半正微分方程邊值問題,目前已有一些研究(參看文獻(xiàn)).而對于奇異的半正微分方程邊值問題,目前所得結(jié)果不多,這方面研究結(jié)果可參看文獻(xiàn).微分邊值問題(1.1)中的非線性項(xiàng)不能為上述文獻(xiàn)條件所包含,因此這里的結(jié)論是新的.若y∈C且y(t)>0,t∈(0,1),y又滿足(1.1),則稱y為微分邊值問題(1.1)的正解.本文中我們將在一個(gè)特殊錐上使用不動(dòng)點(diǎn)指數(shù)和逼近方法,討論微分邊值問題(1.1)的正解.為方便起見,我們將給出如下假設(shè)(Η)0<he1=∫10e1(s)(m∑i=1ai(s))ds<+∞,其中e1(s)={s(1-s),b=d=0;(1-s),b>0,d=0;s,b=0,d>0.我們有如下主要結(jié)果定理1設(shè)(H)成立,則對于充分小的λ>0,方程(1.1)存在正解.2+c1-tn,bct0n[u]*={u,u≥0;0,u=0.對于任給0<λ<+∞,j∈N,x∈P,我們令A(yù)jλx(t)=λΜ∫10G(t,s)(m∑i=1ai(s)([x(s)-wλ(s)]*+1j)αi)ds,t∈.(2.1)其中wλ(t)=λΜ∫10G(t,s)ds,t∈?G(t,s)=ρ-1{(b+as)(d+c(1-t)),s≤t,(b+at)(d+c(1-s)),s≥t.引理1設(shè)un,u∈P,且un|→u(n|→+∞),則對于任給α>0,我們有uαn|→uα(n|→+∞).引理2對于任給0<λ<+∞,j∈N,Ajλ:P|→Q全連續(xù).證對于任給x∈P,令y(t)=Ajλx(t),t∈.又設(shè)t0∈,使得y(t0)=‖y‖.由于G(t,s)G(t0,s)={b+atb+at0,t,t0≤s,(b+at)(d+c(1-s))(b+as)(d+c(1-t0)),t≤s≤t0,d+c(1-t)d+c(1-t0),s≤t,t0,(b+as)(d+c(1-t))(b+at0)(d+c(1-s)),t0≤s≤t}≥e(t).因此,我們有y(t)=λ∫10G(t,s)G(t0,s)G(t0,s)(m∑i=1ai(s)([x(s)-wλ(s)]*+1j)αids≥∥y∥e(t),t∈.于是,我們知Ajλ:P|→Q.下面我們?nèi)プCAjλ:P|→Q為全連續(xù).事實(shí)上,對于任給zn,z∈P,使得zn|→z(n|→+∞)。令xn(s)=([zn(s)-wλ(s)]*+1j)-1,x(s)=([z(s)-wλ(s)]*+1j)-1,s∈.則我們有|Ajλzn(t)-Ajλz(t)|≤λm∑i=1∫10G(t,s)ai(s)|([zn(s)-wλ(s)]*+1j)αi-([z(s)-wλ(s)]*+1j)αi|ds≤λk∑i=1∥xn-αi-x-αi∥∫10G(s,s)ai(s)ds+λm∑i=k+1∥([zn-wλ]*+1j)αi-([z-wλ]*+1j)αi∥∫10G(s,s)ai(s)ds.(2.2)另一方面,我們易證∥xn-x∥≤j2∥zn-z∥.(2.3)由(2.3)和引理1,對于任給ε>0,存在N1使得∥xn-αi-x-αi∥≤ε2λkhii=1,2,\:,k,n≥Ν1?(2.4)∥([zn-wλ]*+1j)αi-([z-wλ]*+1j)αi∥≤ε2λ(m-k)hi,i=k+1,k+2,\:,m,n≥Ν1.(2.5)這里hi=∫10G(s,s)ai(s)ds.由(2.2),(2.4),(2.5),我們有|Ajλzn(t)-Ajλz(t)|≤k∑i=1ε2λkhiλhi+m∑i=k+1ε2λ(m-k)hiλhi≤ε,t∈.(2.6)由(2.6)有Ajλzn|→Ajλz(n|→+∞).因此,Ajλ:P|→Q為連續(xù).現(xiàn)在,設(shè)B?P為有界集.不失一般性,我們設(shè)存在L>0使得‖x‖≤L,x∈B.則有|Ajλz(t)|≤λk∑i=11jαi∫10G(s,s)ai(s)ds+λm∑i=k+1(L+1)αi∫10G(s,s)ai(s)ds?這表明AjλB有界.對于任給ε>0,由(H),我們知存在δ1>0使得k∑i=12jαi∫δ10G(s,s)ai(s)ds+m∑i=k+12(L+1)αi∫δ10G(s,s)ai(s)ds≤ε3λ?(2.7)k∑i=12jαi∫11-δ1G(s,s)ai(s)ds+m∑i=k+12(L+1)αi∫11-δ1G(s,s)ai(s)ds≤ε3λ.(2.8)令bi′=maxs∈[δ1,1-δ1]ai(s),b0′=max{b1′,\:,bm′}.因?yàn)镚(t,s)在×上一致連續(xù),對于上述ε>0,存在δ>0使得|G(t1,s)-G(t2,s)|≤ε3λb0′(k∑i=1j-αi+m∑i=k+1(L+1)αi),t1,t2∈,|t1-t2|≤δ,s∈.(2.9)由(2.7)-(2.9),對于任給x∈B,t1,t2∈,|t1-t2|≤δ,我們有|Ajλx(t1)-Ajλx(t2)|≤λ∫10|G(t1,s)-G(t2,s)|(m∑i=1ai(s)([x(s)-wλ(s)]*+1j)αi)ds≤λ∫δ10|G(t1,s)-G(t2,s)|(k∑i=1ai(s)1jαi+m∑i=k+1ai(s)(L+1)αi)ds+λ∫11-δ1|G(t1,s)-G(t2,s)|(k∑i=1ai(s)1jαi+m∑i=k+1ai(s)(L+1)αi)ds+λb0′∫1-δ1δ1|G(t1,s)-G(t2,s)|(k∑i=1j-αi+m∑i=k+1(L+1)αi)ds≤ε.這表明AjλB在上等度連續(xù).由Ascoli-Arezla定理,我們知AjλB在上為相對緊集.因此,Ajλ為全連續(xù)算子.證畢.?對于任給r≥1,令α0=min{α1,α2,\:,αk},再令G(r)=k∑i=12-αirαi-α0+m∑i=k+1(r+1)αir-α0?λ(r)=min{ρ(2Μ(b+a)(c+d))-1,r-α0+14he1(-α0+1)G(r)}.(2.10)則我們有如下引理3引理3對于任給0<λ<λ(r),j∈N,有i(Ajλ,Br,Q)=1.證對于任給0<λ<λ(r),令c(λ)=ρ-1(b+a)(d+c)Mλ,則c(λ)<c(λ(r))≤12.下面我們將去證明z≠μAjλz,μ∈,z∈?Br.(2.11)事實(shí)上,若(2.11)不成立,則存在z0∈?Br,μ0∈,使得z0=μ0Aλjz0.由于z0∈Q,故有z0(t)≥∥z0∥e(t)=re(t),t∈.(2.12)另一方面,我們有wλ(t)=λΜ∫10G(t,s)ds≤c(λ)e(t)≤c(λ)rz0(t).(2.13)由(2.12),(2.13),我們有z0(t)-wλ(t)≥(1-c(λ)r)z0(t)≥0,t∈.(2.14)于是,我們直接計(jì)算,有{z0″(t)+μ0λm∑i=1ai(t)(z0(t)-wλ(t)+1j)αi=0,0<t<1,az0(0)-bz′0(0)=0,cz0(1)+dz′0(1)=0.(2.15)由(2.15)知,z0″(t)≤0,t∈(0,1);z0(t)為上的凹函數(shù).我們對于(1)b=d=0;(2)b>0,d=0;(3)b=0,d>0.分別討論(1)b=d=0.此時(shí)由z0(t)的非平凡性及z0(t)為上的凹函數(shù)知,存在t0∈(0,1),使得z0′(t0)=0,z0′(t)≥0,t∈(0,t0);z0′(t)≤0,t∈(t0,1);z0(t0)=‖z0‖.由(2.14),(2.15),對于任給s∈(0,1),我們有-z0″(s)≤λ(m∑i=1ai(s))(k∑i=1(z0(s)-wλ(s)+1j)αi+m∑i=k+1(z0(s)-wλ(s)+1j)αi)≤λ(m∑i=1ai(s))(k∑i=1(1-c(λ)r)αi(z0(s))αi+m∑i=k+1(z0(s)+1)αi)≤λ(m∑i=1ai(s))(z0(s))α0G(r).(2.16)對(2.16)從t(t∈(0,t0))到t0積分,有z0′(t)≤λ[z0(t)]α0G(r)∫t0t(m∑i=1ai(s))ds.從而,我們有z0′(t)[z0(t)]-α0≤λG(r)∫t0t(m∑i=1ai(s))ds.再對上式從0到t0積分,有r-α0+1-α0+1=∫t00z0′(s)[z0(s)]-α0ds≤λG(r)∫t00ds∫t0s(m∑i=1ai(τ))dτ=λG(r)∫t00s(m∑i=1ai(s))ds.于是,我們有r-α0+1λ(-α0+1)G(r)≤11-t0∫t00s(1-s)(m∑i=1ai(s))ds.(2.17)同理,我們對(2.16)從t0到t(t∈(t0,1))積分,然后從t0到1積分,得r-α0+1λ(-α0+1)G(r)≤1t0∫1t0s(1-s)(m∑i=1ai(s))ds.(2.18)(2)b>0,d=0.由z0(0)≥0知,z0′(0)=abz0(0)≥0,因此,存在t0∈[0,1),使得z0′(t0)=0,‖z0‖=z0(t0).于是,我們有如下兩種情形(a)t0=0.由(2.15)與(2.16)類似,我們可知,對于任給s∈(0,1),有-z0″(s)≤λ(∑i=1mai(s))[z0(s)]α0G(r).對上式從0到t(t∈(0,1))積分,注意到z0(s)在(0,t)上為減的,有-z0′(t)[z0(t)]-α0≤λG(r)∫0t(∑i=1mai(s))ds?t∈[0,1).再對上式從0到1積分,有r-α0+1-α0+1≤λG(r)∫01(1-s)(∑i=1mai(s))ds.于是,我們有r-α0+1λ(-α0+1)G(r)≤∫01(1-s)(∑i=1mai(s))ds.(2.19)(b)t0>0.仿(1)的情形可知(2.17),(2.19)成立.(3)b=0,d>0.此時(shí)必存在t0∈(0,1],滿足z0′(t0)=0,z0(t0)=‖z0‖.我們有如下兩種情況(a)t0=1.仿(2)中(a)的情形,有r-α0+1λ(-α0+1)G(r)≤∫01s(∑i=1mai(s))ds.(2.20)(b)t0<1.此時(shí)(2.17),(2.18)成立.綜上,我們有r-α0+1λ(-α0+1)G(r)≤2∫01e1(s)(∑i=1mai(s))ds?此與(2.10)矛盾,故(2.11)成立.由不動(dòng)點(diǎn)指數(shù)的性質(zhì)我們知引理3成立.?引理4對于任給0<λ<λ(1),存在R(λ)>1,使得i(Aλj,BRλ,Q)=0.證取[α,β]?(0,1),令ε0=(b+aα)(d+c(1-β))(b+a)(d+c),R(λ)>max{1,[2αm(ε0αmλsupt∈∫αβG(t,s)am(s)ds)-1]1αm-1}.我們?nèi)プC,取ψ∈Q,對于任給0<λ<λ(1),有u≠Aλju+μψ,u∈?BR(λ),μ≥0.(2.21)事實(shí)上,若否,(2.21)不成立,則存在u0∈?BR(λ),使得u0≥Aλju0.此時(shí),對于任給0<λ<λ(1),我們知,c(λ)<c(λ(1))≤12.于是,有u0(t)-wλ(t)≥(1-c(λ))u0(t)≥(1-c(λ(1)))u0(t)≥12u0(t)≥0.(2.22)由于u0(t)∈Q,我們有u0(t)≥∥u0∥e(t)≥ε0∥u0∥,t∈[α,β].(2.23)由(2.22),(2.23),有u0(t)≥λ∫01G(t,s)(∑i=1mai(s)([u0(s)-wλ(s)]*+1j)αids=∫01G(t,s)(∑i=1mai(s)(u0(s)-wλ(s)+1j)αi)ds≥λ∫αβG(t,s)(am(s)(u0(s)-wλ(s))αm)ds≥∫αβG(t,s)am(s)2-αm[u0(s)]αmds≥2-αm(ε0∥u0∥)αmλ∫αβG(t,s)am(s)ds,t∈.于是,我們有R(λ)≥2-αm(ε0R(λ))αmλsupt∈∫αβG(t,s)am(s)ds.從而,有R(λ)≤[2αm(ε0αmλsupt∈∫αβG(t,s)am(s)ds)-1]1αm-1.此與R(λ)的選取矛盾,故(2.21)成立.由不動(dòng)點(diǎn)指數(shù)性質(zhì),我們知引理4成立.?3s,1.2.由引理2和引理3知,對于任給0<λ<λ(1),j∈N,有i(Aλj,BR(λ)\Bˉ(1),Q)=i(Aλj,BR(λ),Q)-i(Aλj,B(1),Q)=-1.從而,對于任給0<λ<λ(1),j∈N,Aλj在BR(λ)\Bˉ(1)中存在不動(dòng)點(diǎn)xj(t).即xj(t)=λ∫01G(t,s)(∑i=1mai(s)([xj(s)-wλ(s)]*+1j)αi)ds,t∈.(3.1)由于‖xj‖≥1,我們有wλ(t)≤c(λ)e(t)≤c(λ(1))∥xj∥xj(t)≤c(λ(1))xj(t)≤12xj(t),t∈.于是,有xj(t)-wλ(t)≥12xj(t)≥0,t∈.從而,由(3.1)我們有xj(t)=λ∫01G(t,s)(∑i=1mai(s)(xj(s)-wλ(s)+1j)αi)ds,t∈.(3.2)我們直接計(jì)算知{xj″+λ(∑i=1mai(t)(xj(t)-wλ(t)+1j)αi)=0,0<t<1,axj(0)-bxj′(0)=0,cxj(1)+dxj′(1)=0.(3.3)對于任給0<λ<λ(1),令Sλ={xj|j∈N,xj為Aλj的不動(dòng)點(diǎn)}.我們分以下三種情況討論(1)b=d=0.令a0=inf{tj|tj∈(0,1),x′(tj)=0,xj∈Sλ},b0=sup{tj|tj∈(0,1),x′(tj)=0,xj∈Sλ}.我們?nèi)菀鬃C明,0<a0,b0<1.對于任給η1,η2∈(0,1),由條件(H)知∫0η1s(∑i=1mai(s))ds<+∞,∫η21(1-s)(∑i=1mai(s))ds<+∞.于是,對于任給ε>0,存在δ1:12>δ1>0,使得∫0δ1s(∑i=1mai(s))ds<ε3,δ1∫121(1-s)(∑i=1mai(s))ds<ε3.(3.4)再取0<δ2<δ1,使得δ2∫δ112(∑i=1mai(s))ds<ε3.(3.5)由(3.4),(3.5),對于任給t∈(0,δ2),有0≤t∫t1(1-s)(∑i=1mai(s))ds≤∫tδ1s(∑i=1mai(s))ds+δ2∫δ112(∑i=1mai(s))ds+δ1∫121(1-s)(∑i=1mai(s))ds<ε.于是,我們有l(wèi)imt|→0+t∫t1(1-s)(∑i=1mai(s))ds=0.(3.6)同理可證limt|→1-(1-t)∫0ts(∑i=1mai(s))ds=0.(3.7)對于任給z∈R+,我們令I(lǐng):R+|→R+如下Ι(z)=∫0zu-α0du=1-α0+1z-α0+1.(3.8)易知,I:R+|→R+為單調(diào)增且為連續(xù)的.下面我們證明,對于任給0<λ<λ(1),I(Sλ(t))在上等度連續(xù).事實(shí)上,對于任給z∈Sλ,0<t1<t2<a0,我們有0≤Ι(z(t2))-Ι(z(t1))=∫z(t1)z(t2)u-α0du=∫t1t2[z(τ)]-α0z′(τ)dτ≤λG(R(λ))∫t1t2ds∫sb0(∑i=1mai(τ))dτ=λG(R(λ))(∫t1t2(s-t1)(∑i=1mai(s))ds+(t2-t1)∫t2b0(∑i=1mai(s))ds)≤λG(R(λ))(∫t1t2s(∑i=1mai(s))ds+(1-b0)-1(t2-t1)∫t2b0(1-s)(∑i=1mai(s)ds)≤(1-b0)-1λG(R(λ))(∫t1t2s(1-s)(∑i=1smai(s))ds+(t2-t1)∫t2-t11(1-s)(∑i=1mai(s))ds).(3.9)由(3.6),(3.9)易知,I(Sλ(t))在[0,a0]等度連續(xù).用I-1表示I在R+上的反函數(shù).易知,I-1在[0,I(R(λ))]上一致連續(xù).且0≤z(t2)-z(t1)=Ι-1(Ι(z(t2))-Ι-1(Ι(z(t1)),0<t1<t2≤a0,z∈Sλ.于是,我們知,Sλ在[0,a0]上等度連續(xù).同理,由(3.7)我們可證Sλ在[b0,1]上等度連續(xù).對于任給t∈[a0,b0],z∈Sλ,易知|z′(t)|≤λ[z(t)]α0G(R(λ))∫a0b0(∑i=1mai(s))ds.(3.10)再由z(t)∈Q,我們有z(t)≥∥z∥e(t)=∥z∥t(1-t)≥a0(1-b0)>0,t∈[a0,b0].(3.11)由(3.10),(3.11),對于任給z∈Sλ,t∈[a0,b0],我們有|z′(t)|≤λ[a0(1-b0)]α0G(R(λ))∫a0b0(∑i=1mai(s))ds.于是,我們知Sλ在[a0,b0]上等度連續(xù).從而Sλ在上為等度連續(xù)集合.(2)b>0,d=0.令b1=sup{tj|tj∈(0,1),x′(tj)=0,xj∈Sλ}.與(1)類似知,b1<1.由條件(H)知,limt|→0+t∫01-t(∑i=1mai(s))ds=limt|→0+t∫t1(∑i=1mai(1-τ))dτ=0.(3.12)再令I(lǐng):R+|→R+由(3.8)式定義.與(3.9)類似,對于任給0<λ<λ(1),b1<t1<t2≤1,我們有0≤Ι(z(t1))-Ι(z(t2))=∫z(t2)z(t1)u-α0du=∫t1t2-[z(τ)]-α0z′(τ)dτ≤λG(R(λ))∫t1t2ds∫0s(∑i=1mai(τ))dτ≤λG(R(λ))(∫t1t2(t2-s)(∑i=1mai(s))ds+(t2-t1)∫0t1(∑i=1mai(s))ds)≤λG(R(λ))(∫t1t2(1-s)(∑i=1mai(s))ds+(t2-t1)∫01-(t2-t1)(∑i=1mai(s))ds).(3.13)由(3.12),(3.13),我們易知,I(Sλ(t))在[b1,1]上為等度連續(xù)集合.從而Sλ(t)在[b1,1]上等度連續(xù).對于任給t∈[0,b1],z∈Sλ,我們有|z′(t)|≤λ[z(t)]α0G(R(λ))∫0b1(∑i=1mai(s))ds?(3.14)又有z(t)≥∥z∥e(t)=∥z∥(b+at)b+a(1-t)≥bb+a(1-t),t∈[0,b1].(3.15)由(3.14),(3.15),我們知|z′(t)|≤λ[bb+a]α0(1-b1)α0G(R(λ))∫0b1(∑i=1mai(s))ds,t∈[0,b1].于是,Sλ(t)在[0,b1]上為等度連續(xù)集合.從而Sλ(t)在上為等度連續(xù)的.(3)b=0,d>0.仿(2)可證Sλ(t)在上為等度連續(xù)集合.綜上,我們證得,對于任給0<λ<λ(1),Sλ(t)為上的等度連續(xù)集合.又顯然Sλ為有界集.由Ascoli-Arezla定理知,Sλ為C中的相對緊集.