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　　Tu(t)=∫?1?0H(t,s)?-1(∫?1?0H?1(s,τ)w(τ)f(τ,u(τ),u′(τ))?d?τ)?d?s≥
　　ρ∫?1?0e(s)?d?sψ-1?2(∫?1?0ρ?1e(τ)?w(τ)?d?τ)k?2b=b。??
　　故?α(Tu(t))=?min?0≤t≤1Tu(t)≥b。? ?
　　?(?C?2) 当‖u‖≤a时,由条件(?III?)及式(3)~式(4)可知 ‖Tu‖≤a。?
　　(?C?3)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?u?1,u?2和u?3?,且满足??
　　‖u?1‖　　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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　　[6] X ZHANG,M FENG,WGE. Symmetric positive solutions for p-Laplacian fourth-order differential equations with integral boundary conditions [J]. J. Comput. Appl. Math. 2008,222:561-573.?
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