(4)
有唯一解
其中,格林函数G(t,s)定义为
其中,p(s)=aα+(α2-α)b-ληα(1-s).
证明由引理1.2可得,边值问题(4)等价于积分方程
c1tα-1+c2tα-2+c3tα-3+c4tα-4,
其中,ci∈R,i=1,2,3,4.
由u(0)=u′(0)=u″(0)=0可得c2=c3=c4=0.因此,
则
(5)
(α-1)c1,
(6)
(7)
故
所以,当t≤η时,
当t≥η时,
证毕.
引理1.4Green函数有如下性质:
3)p(s)>0且p(s)在[0,1]上为增函数;
4)G(t,s)>0,∀t,s∈(0,1).
证明当s≤t,s≤η时,
(α2-α)btα-1(1-s)α-2-p(0)(t-s)α-1-
λ(η-s)αtα-1}=
(α2-α)btα-1(1-s)α-1-p(0)(t-s)α-1-
λη
(α2-α)btα-1(1-s)α-1}≥
(α2-α)btα-1(1-s)α-1-ληα(1-s)αtα-1-
p(0)(t-s)α-1+(α2-α)btα-1(1-s)α-2s}=
p(0)(t-s)α-1+(α2-α)btα-1(1-s)α-2s}=
(α2-α)btα-1(1-s)α-2-
p(0)(t-s)α-1-λ(η-s)αtα-1}=
(α2-α)btα-1(1-s)α-1-p(0)(t-s)α-1-
λ(η-s)αtα-1+(α2-α)btα-1(1-s)α-2s}≤
(α2-α)btα-1(1-s)α-2s+ληα[tα-1(1-s)α-1-
t
(α2-α)btα-1(1-s)α-2s+
ληαtα-1(1-s)
(α2-α)btα-1(1-s)α-2s+ληα(1-s)
当0≤η≤s≤t≤1时,
(α2-α)btα-1(1-s)α-2-p(0)(t-s)α-1}=
(α2-α)btα-1(1-s)α-1-p(0)(t-s)α-1+
(α2-α)btα-1(1-s)α-2s}≥
(α2-α)btα-1(1-s)α-2-p(0)(t-s)α-1}=
(α2-α)btα-1(1-s)α-1-p(0)(t-s)α-1+
(α2-α)btα-1(1-s)α-2s}≤
(α2-α)btα-1(1-s)α-2s+ληαtα-1(1-s)α-1}=
(α2-α)btα-1(1-s)α-2s+ληαtα-1(1-s)α-1}≤
(α2-α)btα-1(1-s)α-2s+ληαtα-1(1-s)α-1}≤
当0≤t≤s≤η≤1时,
(α2-α)btα-1(1-s)α-2-λ(η-s)αtα-1}=
(α2-α)btα-1(1-s)α-1-λtα-1η
(α2-α)btα-1(1-s)α-2s}≥
(α2-α)btα-1(1-s)α-1-ληαtα-1(1-s)α+
(α2-α)btα-1(1-s)α-2s}=
(α2-α)btα-1(1-s)α-2s}=
(α2-α)btα-1(1-s)α-2-λ(η-s)αtα-1}=
(α2-α)btα-1(1-s)α-1-λ(η-s)αtα-1+
(α2-α)btα-1(1-s)α-2s}≤
当t≤s,η≤s时,
(α2-α)btα-1(1-s)α-2}=
(α2-α)btα-1(1-s)α-2s}≥
(α2-α)btα-1(1-s)α-2}=
(α2-α)btα-1(1-s)α-2s}≤
若结论1)和2)成立,由p(s)的表示式易知结论3)成立.而结论4)可由结论1)获得.证毕.
(i) ‖Sω‖≤‖ω‖,ω∈P∩∂Ω1,‖Sω‖≥‖ω‖,ω∈P∩∂Ω2;
(ii) ‖Sω‖≥‖ω‖,ω∈P∩∂Ω1,‖Sω‖≤‖ω‖,ω∈P∩∂Ω2;
2 正解的存在性
记
f0=
f0=
f∞=
f∞=
δ=
M=max
m=min
本文将使用如下条件:
(P)f(t,u):[0,1]×[0,∞)→[0,∞)是连续函数;
令E=C[0,1],令范数‖u‖=则E为Banach空间.令
则P为E中的锥.
证明类似于文献[6]的证明,本结论易证.
定理2.1假设条件(P)、(A1)和(A3)成立,则BVP(3)至少有2个正解u1和u2且0≤‖u1‖
证明由条件(A1)知,选取>0,满足f0->,∃0f(t,u)>(f0-)u>u,
∀t∈[0,1], 0≤u≤r0.
设r∈(0,r0),Ωr={u∈P|‖u‖因此
r=‖u‖,
因此,‖Au‖>‖u‖,∀u∈∂Ωr.
f(t,u)>(f∞-)u>u,
∀t∈[0,1],u≥H.
取R>R0max设ΩR={u∈P|‖u‖R=‖u‖,
即‖Au‖>‖u‖,∀u∈∂ΩR.
设Ωp={u∈P|‖u‖
因此
s(1-s)α-2)f(s,u(s))ds<
p=‖u‖.
即‖Au‖<‖u‖,∀u∈∂Ωp.
综上,由引理1.5知,结论成立.证毕.
定理2.2假设条件(P)、(A2)和(A4)成立,则BVP(3)至少有2个正解u1和u2且0≤‖u1‖
证明由条件(A2)知,选取>0,满足f0+<,∃0f(t,u)≤(f0+)u
∀t∈[0,1], 0≤u≤r0.
设r∈(0,r0),Ωr={u∈P|‖u‖因此
f(t,u)≤(f∞+)u
∀t∈[0,1],u≥H.
f(t,u)≤
∀u∈(0,R),t∈[0,1].
则对于∀u∈P且‖u‖=R,有
s(1-s)α-2)f(s,u(s))ds≤
(f∞+)RMs(1-s)α-2(2-s)ds<
f(t,u)≤L, ∀u≥0,t∈[0,1],
因此,设ΩR={u∈P|‖u‖设Ωp={u∈P|‖u‖
进而
故
p=‖u‖,
即‖Au‖>‖u‖,∀u∈∂Ωp.
综上,由引理1.5知结论成立.证毕.
推论2.4假设条件(P)成立,且f0=∞,f∞=0(次线性),则BVP(3)至少有一个正解.
推论2.6假设条件(P)成立,且f0=0,f∞=∞(超线性),则BVP(3)至少有一个正解.
例2.1边值问题
(8)
至少有2个正解u1和u2且0≤‖u1‖<1<‖u2‖.
事实上,设f(t,u)=uλ(t)+uμ(t),则
M=max
取p=1,则对0≤u≤p有
f(t,u(t))≤pλ+pμ=2<=p,
由定理2.1及注1知,边值问题(8)至少存在2个正解u1和u2且满足0≤‖u1‖<1<‖u2‖.
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