近几年来，不动点理论已经成为研究带时滞积分微分方程的稳定性和周期解的主要方法之一^[1-15].例如，方程式

$ \begin{array}{l} \frac{{\rm{d}}}{{{\rm{d}}t}}x\left( t \right) = - a\left( t \right)x\left( t \right) + \frac{{\rm{d}}}{{{\rm{d}}t}}Q\left( {t,x\left( {t - g\left( t \right)} \right)} \right) + \\ G\left( {t,x\left( t \right),x\left( {t - g\left( t \right)} \right)} \right), \end{array} $

(1)

它的特别形式一直都受到许多研究人员的关注.本文关注的是以下非线性中立型微分方程

$ \begin{array}{l} \frac{{\rm{d}}}{{{\rm{d}}t}}x\left( t \right) = - a\left( t \right)h\left( {x\left( t \right)} \right) + \frac{{\rm{d}}}{{{\rm{d}}t}}Q\left( {t,x\left( {t - } \right.} \right.\\ \left. {\left. {g\left( t \right)} \right)} \right) + G\left( {t,x\left( t \right),x\left( {t - g\left( t \right)} \right)} \right), \end{array} $

(2)

其中a(t)，g(t)，h(t)和G(t, x, y)在各自定义的区间中是连续的，Q(t, x)是连续可导函数.由于方程式(2)出现了非线性项h(x(t))，不能直接使用不动点定理.因此本文需要加减一项非线性项.因此定义两个映射，其中一个是完全连续的，另外一个是大压缩的.

1 准备工作

设T>0，定义集合P_T={φ∈C( , ):φ(t+T)=φ(t)}，$ x\left( t \right)\left\| {\mathop {\max }\limits_{t \in \left[ {0, T} \right]} } \right.\left| {x\left( t \right)} \right| $，其中C是连续实值函数空间.则(P_T, .||)是Banach空间.对任意L>0，定义

$ {M_L} = \left\{ {\varphi \in {P_T}:\varphi \left\| { \le L} \right.,\left| {\varphi '} \right| \le L'} \right\}, $

(3)

其中L′是正实数.

本文做出以下假设：

$ a\left( t \right) = a\left( {t + T} \right),g\left( t \right) = g\left( {t + T} \right), $

(4)

其中g(t)是连续函数，和g(t)≥g^*>0.

$ \int_0^T {a\left( s \right){\rm{d}}s} > 0. $

(5)

同时假设Q(t, x)和G(t, x, y)是关于t的周期函数和关于x，y的Lipschitz函数，即

$ Q\left( {t + T,x} \right) = Q\left( {t,x} \right),G\left( {t + T,x,y} \right) = G\left( {t,x,y} \right) $

(6)

存在正数E₁，E₂，E₃使得

$ \left| {Q\left( {t,x} \right) - Q\left( {t,y} \right)} \right| \le {E_1}\left\| {x - y} \right\|, $

(7)

$ \left| {G\left( {t,x,y} \right) - G\left( {t,z,w} \right)} \right| \le {E_2}\left\| {x - z} \right\| + {E_3}\left\| {y - w} \right\|. $

(8)

引理1  假设式(4)和(6)成立.如果x(t)∈P_T，则x(t)是方程(1)的解当且仅当

$ \begin{array}{l} x\left( t \right) = Q\left( {t,x\left( {t - g\left( t \right)} \right)} \right) + {\left( {1 - {{\rm{e}}^{ - \int_{t - T}^t {a\left( u \right){\rm{d}}u} }}} \right)^{ - 1}} \times \\ \int_{t - T}^t {\left[ {a\left( u \right)H\left( {x\left( u \right)} \right) + G\left( {u,x\left( u \right),x\left( {u - g\left( u \right)} \right)} \right) - } \right.} \\ \left. {a\left( u \right)Q\left( {u,x\left( {u - g\left( u \right)} \right)} \right)} \right]{{\rm{e}}^{ - \int_u^t {a\left( s \right){\rm{d}}s} }}{\rm{d}}u, \end{array} $

(9)

其中H(x(u))=x(u)－h(x(u)).

证明  设x(t)∈P_T是式(1)的一个解.首先将式(1)写成以下形式

$ \begin{array}{*{20}{c}} {\frac{{\rm{d}}}{{{\rm{d}}t}}x\left( t \right) + a\left( t \right)x\left( t \right) = a\left( t \right)H\left( {x\left( t \right)} \right) + }\\ {\frac{{\rm{d}}}{{{\rm{d}}t}}Q\left( {t,x\left( {t - g\left( t \right)} \right)} \right) + G\left( {t,x\left( t \right),x\left( {t - g\left( t \right)} \right)} \right).} \end{array} $

上式两边同时乘以$ {\rm{e}}^{\int_0^t {a\left( s \right){\rm{d}}s}} $以及对t从t－T到t进行积分得$\smallint _{_{t - T}}^{^t}\left[ {x\left( u \right){{\rm{e}}^{\int_{_0}^{^t} a \left( s \right){\rm{d}}s}}} \right]'{\rm{d}}u = \smallint _{_{t - T}}^{^t}[a\left( u \right){\rm{ }}H\left( {x\left( u \right)} \right) + \frac{{\rm{d}}}{{{\rm{d}}u}}Q\left( {u, x\left( {u - g\left( u \right)} \right)} \right) + $$ G\left( {u, x{\rm{ }}\left( u \right), x\left( {u - g{\rm{ }}\left( u \right)} \right)} \right)]{{\rm{e}}^{\int_0^{^u} {a\left( s \right){\rm{d}}s} }}{\rm{d}}u.$

因此得到

$ \begin{array}{l} x\left( t \right){{\rm{e}}^{\int_0^t {a\left( s \right){\rm{d}}s} }} - x\left( {t - T} \right){{\rm{e}}^{\int_0^{t - T} {a\left( s \right){\rm{d}}s} }} = \int_{t - T}^t {\left[ {a\left( u \right)H\left( {x\left( u \right)} \right) + } \right.} \\ \left. {\frac{{\rm{d}}}{{{\rm{d}}u}}Q\left( {u,x\left( {u - g\left( u \right)} \right)} \right) + G\left( {u,x\left( u \right),x\left( {u - g\left( u \right)} \right)} \right)} \right]{{\rm{e}}^{\int_0^u {a\left( s \right){\rm{d}}s} }}{\rm{d}}u. \end{array} $

两边同时除以$ {{\rm{e}}^{\int_0^t {a\left( s \right){\rm{d}}s} }} $，以及由于x(t)=x(t+T), 得到

$ \begin{array}{l} x\left( t \right) = {\left( {1 - {{\rm{e}}^{ - \int_{t - T}^t {a\left( u \right){\rm{d}}u} }}} \right)^{ - 1}}\int_{t - T}^t {\left[ {a\left( u \right)H\left( {x\left( u \right)} \right) + } \right.} \\ \frac{{\rm{d}}}{{{\rm{d}}u}}Q\left( {u,x\left( {u - g\left( u \right)} \right)} \right) + G\left( {u,x\left( u \right),x\left( {u - } \right.} \right.\\ \left. {\left. {\left. {g\left( u \right)} \right)} \right)} \right]{{\rm{e}}^{ - \int_u^t {a\left( s \right){\rm{d}}s} }}{\rm{d}}u = Q\left( {t,x\left( {t - g\left( t \right)} \right)} \right) + \left( {1 - } \right.\\ {\left. {{{\rm{e}}^{ - \int_{t - T}^t {a\left( u \right){\rm{d}}u} }}} \right)^{ - 1}}\int\limits_{t - T}^t {\left[ {a\left( u \right)H\left( {x\left( u \right)} \right) + G\left( {u,x\left( u \right),x\left( {u - } \right.} \right.} \right.} \\ \left. {\left. {\left. {g\left( u \right)} \right)} \right) - a\left( u \right)Q\left( {u,x\left( {u - g\left( u \right)} \right)} \right)} \right]{{\rm{e}}^{ - \int_u^t {a\left( s \right){\rm{d}}s} }}{\rm{d}}u. \end{array} $

得证.

定义1 ^[15] 设(M, d)是一个完备度量空间和B：M→M.B是一个大压缩的.如果ϕ, φ∈M，当ϕ≠φ时，有d(Bϕ, Bφ) < d(ϕ, φ)和对任意ε>0，存在δ < 1，使得[ϕ, φ∈M, d(ϕ, φ)≥ε]⇒d(Bϕ, Bφ)≤δd(ϕ, φ).

下一个定理是Krasnoselskii’s不动点定理，它是构成我们重要结论的主要定理.

定理1 ^[15]  设M是Banach空间(S, ||.||)的一个有界凸非空子集.假设A和B将M投影到M，以及：

(1) 对所有x, y∈M, 有Ax+By∈M;

(2) A是连续的和AM包含在M的一个紧子集中;

(3) B是一个大压缩.

则存在一个z∈M，使得z=Az+Bz.

2 周期解的存在性

首先定义一个映射P：

$ \begin{array}{l} \left( {P\varphi } \right)\left( t \right) = Q\left( {t,\varphi \left( {t - g\left( t \right)} \right)} \right) + \left( {1 - } \right.\\ {\left. {{{\rm{e}}^{ - \int_{t - T}^t {a\left( u \right){\rm{d}}u} }}} \right)^{ - 1}}\int\limits_{t - T}^t {\left[ {a\left( u \right)H\left( {\varphi \left( u \right)} \right) + G\left( {u,\varphi \left( u \right),\varphi \left( {u - } \right.} \right.} \right.} \\ \left. {\left. {\left. {g\left( u \right)} \right)} \right) - a\left( u \right)Q\left( {u,\varphi \left( {u - g\left( u \right)} \right)} \right)} \right]{{\rm{e}}^{ - \int_u^t {a\left( s \right){\rm{d}}s} }}{\rm{d}}u \end{array} $

(10)

为了使用定理1，需要将映射P表示成两个映射之和，其中一个是完全连续映射, 另一个是压缩映射.

设(Pφ)(t)=Aφ(t)+Bφ(t), 其中A，B：P_T→P_T定义为

$ \begin{array}{l} \left( {A\varphi } \right)\left( t \right) = Q\left( {t,\varphi \left( {t - g\left( t \right)} \right)} \right) + \left( {1 - } \right.\\ {\left. {{{\rm{e}}^{ - \int_{t - T}^t {a\left( u \right){\rm{d}}u} }}} \right)^{ - 1}}\int_{t - T}^t {\left[ {G\left( {u,\varphi \left( u \right),\varphi \left( {u - g\left( u \right)} \right)} \right) - } \right.} \\ \left. {a\left( u \right)Q\left( {u,\varphi \left( {u - g\left( u \right)} \right)} \right)} \right]{{\rm{e}}^{ - \int_u^t {a\left( s \right){\rm{d}}s} }}{\rm{d}}u \end{array} $

(11)

和

$ \left( {B\varphi } \right)\left( t \right) = {\left( {1 - {{\rm{e}}^{ - \int_{t - T}^t {a\left( u \right){\rm{d}}u} }}} \right)^{ - 1}}\int_{t - T}^t {a\left( u \right)H\left( {\varphi \left( u \right)} \right){{\rm{e}}^{ - \int_u^t {a\left( s \right){\rm{d}}s} }}{\rm{d}}u} . $

(12)

还需要假设：

$ \left( {{E_2} + {E_3}} \right)L + \left| {G\left( {t,0,0} \right)} \right| \le \beta La\left( t \right), $

(13)

$ {E_1}L + \left| {Q\left( {t,0} \right)} \right| \le \alpha L, $

(14)

$ J\left( {2\alpha + \beta } \right) \le 1, $

(15)

其中α，β，L和J是常数，以及J≥3.

存在正数M，使得

$ \left| {\frac{{\rm{d}}}{{{\rm{d}}t}}Q\left( {t,x\left( {t - g\left( t \right)} \right)} \right)} \right| \le M,x \in {M_L}. $

(16)

引理2  假设条件(4)-(8)和(13)-(15)成立.L是式(3)中所定义的，A:M_L→M_L是连续的和将M投影到M的一个紧集中.

证明  在式(12)中变量的改变可以证明(Aφ)(T+t)=(Aφ)(t).注意到

$ \begin{array}{l} \left| {Q\left( {t,x} \right)} \right| \le \left| {Q\left( {t,x} \right) - Q\left( {t,0} \right)} \right| + \left| {Q\left( {t,0} \right)} \right| \le \\ {E_1}\left\| x \right\| + \left| {Q\left( {t,0} \right)} \right|, \end{array} $

以及

$ \begin{array}{l} \left| {G\left( {t,x,y} \right)} \right| \le \left| {G\left( {t,x,y} \right) - G\left( {t,0,0} \right)} \right| + \\ \left| {G\left( {t,0,0} \right)} \right| \le {E_2}\left\| x \right\| + {E_3}\left\| y \right\| + \left| {G\left( {t,0,0} \right)} \right|. \end{array} $

首先证明A将M_L投影到本身.因此，对任意φ∈M_L，有

$ \begin{array}{l} \left| {\left( {A\varphi } \right)\left( t \right)} \right| \le \left| {Q\left( {t,\varphi \left( {t - g\left( t \right)} \right)} \right)} \right| + \left( {1 - } \right.\\ {\left. {{{\rm{e}}^{ - \int_{t - T}^t {a\left( u \right){\rm{d}}u} }}} \right)^{ - 1}}\left| {\int_{t - T}^t {G\left( {u,\varphi \left( u \right),\varphi \left( {u - g\left( u \right)} \right)} \right){{\rm{e}}^{ - \int_u^t {a\left( s \right){\rm{d}}s} }}{\rm{d}}u} } \right| + \\ {\left( {1 - {{\rm{e}}^{ - \int_{t - T}^t {a\left( u \right){\rm{d}}u} }}} \right)^{ - 1}}\left| \begin{array}{l} \int_{t - T}^t {a\left( u \right)Q\left( {u,\varphi \left( {u - } \right.} \right.} \\ \left. {\left. {g\left( u \right)} \right)} \right){{\rm{e}}^{ - \int_u^t {a\left( s \right){\rm{d}}s} }}{\rm{d}}u \end{array} \right| \le \\ {E_1}\left\| \varphi \right\| + \left| {Q\left( {t,0} \right)} \right| + \\ {\left( {1 - {{\rm{e}}^{ - \int_{t - T}^t {a\left( u \right){\rm{d}}u} }}} \right)^{ - 1}}\left| \begin{array}{l} \int_{t - T}^t {\left[ {\left( {{E_2} + {E_3}} \right)\left\| \varphi \right\| + } \right.} \\ \left. {\left| {G\left( {u,0,0} \right)} \right|} \right]{{\rm{e}}^{ - \int_u^t {a\left( s \right){\rm{d}}s} }}{\rm{d}}u \end{array} \right| + \\ {\left( {1 - {{\rm{e}}^{ - \int_{t - T}^t {a\left( u \right){\rm{d}}u} }}} \right)^{ - 1}}\left| \begin{array}{l} \int_{t - T}^t {a\left( u \right)\left( {{E_1}\left\| \varphi \right\| + } \right.} \\ \left. {\left| {Q\left( {u,0} \right)} \right|} \right){{\rm{e}}^{ - \int_u^t {a\left( s \right){\rm{d}}s} }}{\rm{d}}u \end{array} \right| \le \\ {\left( {1 - {{\rm{e}}^{ - \int_{t - T}^t {a\left( u \right){\rm{d}}u} }}} \right)^{ - 1}}\left| {\int_{t - T}^t {\beta La\left( u \right){{\rm{e}}^{ - \int_u^t {a\left( s \right){\rm{d}}s} }}{\rm{d}}u} } \right| + 2\alpha L \le \\ \left( {2\alpha + \beta } \right)L \le \frac{L}{J} < L. \end{array} $

证毕.

接着证明A是连续的.设φ, ϕ∈M_L，令

$ \begin{array}{l} \eta = \mathop {\max }\limits_{t \in \left[ {0,T} \right]} {\left( {1 - {{\rm{e}}^{ - \int_{t - T}^t {a\left( s \right){\rm{d}}s} }}} \right)^{ - 1}},\theta = \mathop {\max }\limits_{t \in \left[ {0,T} \right]} \left| {a\left( t \right)} \right|,\\ \nu = \mathop {\max }\limits_{s \in \left[ {t - T,t} \right]} {{\rm{e}}^{ - \int_s^t {a\left( u \right){\rm{d}}u} }}. \end{array} $

(17)

对任意ε>0，取$ \delta = \frac{\varepsilon }{N} $，使得当||ϕ－φ||≤δ，有

$ \begin{array}{l} \left| {\left( {A\phi } \right)\left( t \right) - \left( {A\varphi } \right)\left( t \right)} \right| \le {E_1}\left\| {\phi - \varphi } \right\| + \\ \left| {{{\left( {1 - {{\rm{e}}^{ - \int_{t - T}^t {a\left( s \right){\rm{d}}s} }}} \right)}^{ - 1}}\int_{t - T}^t {\left( {{E_2} + {E_3}} \right)\left\| {\phi - \varphi } \right\|{{\rm{e}}^{ - \int_u^t {a\left( s \right){\rm{d}}s} }}{\rm{d}}u} } \right| + \\ \left| {{{\left( {1 - {{\rm{e}}^{ - \int_{t - T}^t {a\left( s \right){\rm{d}}s} }}} \right)}^{ - 1}}\int_{t - T}^t {a\left( u \right){E_1}\left\| {\phi - \varphi } \right\|{{\rm{e}}^{ - \int_u^t {a\left( s \right){\rm{d}}s} }}{\rm{d}}u} } \right| \le \\ \left[ {2{E_1} + \eta T\nu \left( {{E_2} + {E_3}} \right)} \right]\left\| {\phi - \varphi } \right\| < \varepsilon . \end{array} $

其中，N=2E₁+ηTν(E₂+E₃).证毕.

接着证明A是紧的.

设φ_n∈M_L，n是正整数.故有(Aφ_n)(t)≤L.直接计算，得$ \left( {A{\varphi _n}} \right)'\left( t \right) = \frac{{\rm{d}}}{{{\rm{d}}t}}Q\left( {t, {\varphi _n}\left( {t - g\left( t \right)} \right)} \right) - a\left( t \right)Q\left( {t, {\varphi _n}\left( {t - g\left( t \right)} \right)} \right) + G(u, {\varphi _n}(u), $$ {\varphi _n}\left( {u - g\left( u \right)} \right)) - a\left( t \right){\left( {1 - {e^{ - \smallint _{_{t - T}}^{^t}a\left( u \right){\rm{d}}u}}} \right)^{ - 1}}\smallint _{_{t - T}}^{^t}[G(u, {\varphi _n}\left( u \right), $$ {\varphi _n}\left( {u - g\left( u \right)} \right)) - a\left( u \right)Q\left( {u, {\varphi _n}\left( {u - g\left( u \right)} \right)} \right)]{{\rm{e}}^{ - \smallint _{_u}^{^t}a\left( s \right){\rm{d}}s}}{\rm{d}}u $.

由式(13)、(14)、(16)、(17)得

$ \begin{array}{l} \left| {{{\left( {A{\varphi _n}} \right)}^\prime }\left( t \right)} \right| \le M + \theta \alpha L + \theta \beta L + \theta \eta T\left[ {\left( {\theta \alpha L + } \right.} \right.\\ \left. {\left. {\theta \beta L} \right)\nu } \right] \le {\rm{d}}, \end{array} $

存在正实数d.因此数列(Aφ_n)是一致有界和等度连续的.由Ascoli-Arzela定理可知，A是紧的.

引理3 假设式(4)-(6)和或(13)-(15)成立.L是式(3)中所定义的，同时假设

$ \begin{array}{l} {\left( {1 - {{\rm{e}}^{ - \int_{t - T}^t {a\left( u \right){\rm{d}}u} }}} \right)^{ - 1}}\int_{t - T}^t {\left| {a\left( u \right)H\left( {\varphi \left( u \right)} \right)} \right|} \times \\ {{\rm{e}}^{ - \int_u^t {a\left( s \right){\rm{d}}s} }}{\rm{d}}u \le \frac{{J - 1}}{J}L, \end{array} $

(18)

对任意φ∈M_L成立，A，B分别是式(11)和(12)所定义的，对任意φ, ϕ∈M_L，有

$ A\varphi + B\phi :{M_L} \to {M_L}. $

证明  对任意φ, ϕ∈M_L，由B的定义和引理2的结论，得到

$ \begin{array}{l} \left| {\left( {A\phi } \right)\left( t \right) + \left( {B\varphi } \right)\left( t \right)} \right| \le \left| {Q\left( {t,\varphi \left( {t - g\left( t \right)} \right)} \right)} \right| + \\ {\left( {1 - {{\rm{e}}^{ - \int_{t - T}^t {a\left( u \right){\rm{d}}u} }}} \right)^{ - 1}}\left| {\int_{t - T}^t {G\left( {u,\varphi \left( u \right),\varphi \left( {u - g\left( u \right)} \right)} \right){{\rm{e}}^{ - \int_u^t {a\left( s \right){\rm{d}}s} }}} {\rm{d}}u} \right| + \\ {\left( {1 - {{\rm{e}}^{ - \int_{t - T}^t {a\left( u \right){\rm{d}}u} }}} \right)^{ - 1}}\left| {\int_{t - T}^t {a\left( u \right)Q\left( {u,\varphi \left( {u - g\left( u \right)} \right)} \right){{\rm{e}}^{ - \int_u^t {a\left( s \right){\rm{d}}s} }}} {\rm{d}}u} \right| + \\ {\left( {1 - {{\rm{e}}^{ - \int_{t - T}^t {a\left( u \right){\rm{d}}u} }}} \right)^{ - 1}}\left| {\int_{t - T}^t {\left| {a\left( u \right)H\left( {\phi \left( u \right)} \right)} \right|{{\rm{e}}^{ - \int_u^t {a\left( s \right){\rm{d}}s} }}} {\rm{d}}u} \right| \le \frac{L}{J} + \\ \frac{{\left( {J - 1} \right)L}}{J} = L. \end{array} $

因此Aφ+Bϕ∈M_L.

下一个引理将证明映射B在M_L上是大压缩的，这个引理的证明读者可以参考文献[15].最后，对函数h:R→R作出以下假设：

(1) h在区间U_L=[－L, L]上是连续可导的.

(2) h在区间U_L=[－L, L]上严格递增.

(3) $ \mathop {{\rm{sup}}}\limits_{s \in {U_L}} h'\left( s \right) \le 1 $.

(4) $ \left( {s - r} \right)\left\{ {\mathop {{\rm{sup}}}\limits_{t \in {U_L}} h'\left( t \right)} \right\} \ge h\left( s \right) - h\left( r \right) \ge \left( {s - r} \right)\left\{ {\mathop {{\rm{inf}}}\limits_{t \in {U_L}} h'\left( t \right)} \right\} \ge 0 $, 其中s, r∈U_L以及s≥r.

引理4 ^[15]  设函数h满足假设(1)~(4).L是式(3)所定义的，则映射B是一个大压缩.

定理2  假设条件式(4)~(8)、式(13)~(15)和式(18)成立，以及函数h满足假设(1)~(4)，则方程式(2)在M_L={ϕ∈P_T: ||ϕ|| ≤L, |ϕ′| ≤L′}存在一个周期解.

证明 由引理(1)可知，Aϕ+Bϕ∈M_L是方程式(2)的一个解，如果

$ \phi = A\phi + B\phi . $

根据引理(2)，A:M_L→M_L是连续的和M_L包含在M_L紧集中.Aϕ+Aφ∈M_L，对任意的ϕ, φ∈M_L.而且根据式(12)，B是一个大压缩.显然，满足Krasnoselskii’s不动点定理.因此，存在一个不动点ϕ∈M_L使得ϕ=Aϕ+Bϕ.因此方程(12)有一个周期为T的解.

3 实例

非线性微分方程:

$ \begin{array}{l} x'\left( t \right) = - \frac{1}{4}\left[ {\frac{{15}}{{16}}x\left( t \right) - \frac{L}{{16}}\sin \left( {\frac{{\left. {x\left( t \right)} \right)}}{L}} \right.} \right] + \\ \frac{{\rm{d}}}{{{\rm{d}}t}}\left[ {\frac{{{{10}^{ - 3}}}}{L}\sin \left( t \right){x^2}\left( {t - 1} \right)} \right] + \frac{{{{10}^{ - 3}}}}{L}\cos \left( t \right){x^2}\left( {t - 1} \right), \end{array} $

(19)

其中T=2π，$a\left( t \right) = \frac{1}{4} $，$ h\left( t \right) = \frac{{15}}{{16}}t - \frac{L}{{16}}\sin (\frac{t}{L}) $，$ H\left( {x\left( t \right)} \right) = \frac{1}{{16}}x\left( t \right) + \frac{L}{{16}}\sin (\frac{{x\left( t \right)}}{L}) $，$ Q\left( {t, x} \right) = \frac{{{{10}^{ - 3}}}}{L}{x^2}\sin t $，$ G\left( {t, x, y} \right) = \frac{{{{10}^{ - 3}}}}{L}{y^2}\cos t $，g(t)=1.

显然式(4)~(6)成立.通过直接计算，得

$ h'\left( t \right) = \frac{{15}}{{16}} - \frac{1}{{16}}\cos \left( {\frac{t}{L}} \right), $

故满足假设(1)~(4).

$ \begin{array}{l} \left| {\frac{{\rm{d}}}{{{\rm{d}}t}}Q\left( {t,x\left( {t - g\left( t \right)} \right)} \right)} \right| = \\ \left| {\frac{{{{10}^{ - 3}}}}{L}\left[ {{x^2}\left( {t - 1} \right)\cos \left( t \right) + 2x\left( {t - 1} \right)x'\left( {t - 1} \right)\sin \left( t \right)} \right]} \right| \le \\ {10^{ - 3}}\left( {L + 2L'} \right), \end{array} $

因此式(16)成立.

$ \begin{array}{l} {E_1} = {E_3} = 2 \times {10^{ - 3}},{E_2} = 0,\alpha = 2 \times {10^{ - 3}},\\ \beta = 8 \times {10^{ - 3}}. \end{array} $

由式(15)得

$ 3 \le J \le {\left( {1.2 \times {{10}^{ - 2}}} \right)^{ - 1}}, $

(20)

$ \eta = {\left( {1 - {{\rm{e}}^{ - \frac{{\rm{ \mathsf{ π} }}}{2}}}} \right)^{ - 1}},\theta = \frac{1}{4},\nu = 1. $

$ \begin{array}{l} {\left( {1 - {{\rm{e}}^{ - \int_{t - T}^t {a\left( u \right){\rm{d}}u} }}} \right)^{ - 1}}\int_{t - T}^t {\left| {a\left( u \right)H\left( {\varphi \left( u \right)} \right)} \right|{{\rm{e}}^{ - \int_u^t {a\left( s \right){\rm{d}}s} }}} {\rm{d}}u \le \\ {\left( {1 - {{\rm{e}}^{ - \frac{{\rm{ \mathsf{ π} }}}{2}}}} \right)^{ - 1}}\frac{{\rm{ \mathsf{ π} }}}{{16}}L \le 0.3L. \end{array} $

又由式(20)可知，$ \frac{2}{3}L \le \frac{{J - 1}}{J}L,L \in \left[ {3,{{\left( {1.2 \times {{10}^{ - 2}}} \right)}^{ - 1}}} \right] $.

故式(19)成立.

故式(19)在M_L={ϕ∈P_2π: ||ϕ|| ≤L, |ϕ′| ≤L′}上存在一个周期为2π的解.