Tu(t)=10H(t,s)-1(10H1(s,)w()f(,u(),u())d)ds 10e(s)ds-12(101e()w()d)k2b=b。 故(Tu(t))=min0t1Tu(t)b。 (C2) 当‖u‖a时,由条件(III)及
Tu(t)=∫10H(t,s)-1(∫10H1(s,τ)w(τ)f(τ,u(τ),u′(τ))dτ)ds≥
ρ∫10e(s)dsψ-12(∫10ρ1e(τ)w(τ)dτ)k2b=b。
故α(Tu(t))=min0≤t≤1Tu(t)≥b。
(C2) 当‖u‖≤a时,由条件(III)及式(3)~式(4)可知 ‖Tu‖≤a。
(C3)当u∈P(α,b,c)且‖Tu‖>d时,由条件(II)可知α(Tu)>b。
综上所述, 引理1的三个条件都成立, 式(1)至少存在三个对称正解u1,u2和u3,且满足
‖u1‖ 由于该定理给出了式(1)至少存在三个对称正解的充分条件, 所以丰富了文献[5]的结论。
参考文献:
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