Theorem 3.1
Suppose that conditions [\(H_{1}\)]-[\(H_{3}\)] hold and
$$\frac{M^{p}_{1}(m)T^{2p-1}(1+ \vert c \vert )(2m_{0}T+\alpha)}{(1- \vert c \vert )^{p}} < 1, $$
then there exist positive constants
\(A_{1}\), \(A_{2}\), \(A_{3}\)
and
ρ, which are independent of
λ
such that
$$A_{1}\leq u(t)\leq A_{2}, \quad\quad \vert v \vert _{0}\leq A_{3}, $$
where
\(u(t)\)
is any solution to the equation
\(Lx = \lambda Nx\), \(\lambda\in(0,1)\).
Proof
Consider the following operator equation:
$$Lx=\lambda Nx, \quad \lambda\in(0,1), $$
where L and N is defined by (2.2) and (2.3), respectively.
Define
$$\Omega_{1}= \Bigl\{ (u,v)^{\top}\in X: \min _{t\in[0,T]}u(t)>0, Lx=\lambda Nx ,\lambda\in(0,1) \Bigr\} . $$
If \(x=(u,v)^{\top}\in\Omega_{1}\), then \((u,v)\) satisfies
$$ \textstyle\begin{cases}u^{(m)}(t)=\lambda[A^{-1}\varphi_{q}(v)](t),\\ v^{(m)}(t)=-\lambda f(u(t))u'(t)-\lambda g(t,u(t-\tau))+ \lambda e(t). \end{cases} $$
(3.1)
From the first equation of (3.1), we get \(v(t)=\varphi _{p}(\lambda^{-1}(Au)^{(m)}(t))\), and combining with the second equation of (3.1) yields
$$ \bigl(\varphi_{p} \bigl((Au)^{(m)}(t) \bigr) \bigr)^{(m)}+\lambda^{p }f \bigl(u(t) \bigr)u'(t)+ \lambda^{p}g \bigl(t,u(t-\tau) \bigr)=\lambda^{p}e(t). $$
(3.2)
Integrating equation (3.2) on the interval \([0,T]\), we have
$$ \int_{0}^{T}g \bigl(t, u(t-\tau) \bigr)\,dt=0. $$
(3.3)
It follows from condition (1) in assumption \([H_{1}]\) that there exist positive constants \(D_{1}\), \(D_{2}\) and \(\sigma\in[0,T]\) such that
$$ D_{1}\leq u(\sigma)\leq D_{2}, $$
(3.4)
then we get
$$ \begin{aligned}[b] \vert u \vert _{0} &=\max _{t\in[0,T]} \bigl\vert u(t) \bigr\vert \leq\max _{t\in[0,T]} \biggl\vert u( \sigma)+ \int_{\sigma}^{t}u'(s)\,ds \biggr\vert \\ &\leq D_{2}+ \int_{0}^{T} \bigl\vert u'(s) \bigr\vert \,ds. \end{aligned} $$
(3.5)
Multiplying both sides of (3.2) by \((Au)(t)\) and integrating on the interval \([0,T]\), we get
$$\begin{aligned} \int_{0}^{T} \bigl\vert (Au)^{(m)}(t) \bigr\vert ^{p}\,dt &\leq \bigl(1+ \vert c \vert \bigr) \vert u \vert _{0} \int_{0}^{T} \bigl\vert f \bigl(u(t) \bigr) \bigr\vert \bigl\vert u' \bigr\vert \,dt \\ &\quad{} + \bigl(1+ \vert c \vert \bigr) \vert u \vert _{0} \int_{0}^{T} \bigl\vert g \bigl(t,u(t-\tau) \bigr) \bigr\vert \,dt \\ &\quad{} + \bigl(1+ \vert c \vert \bigr) \vert u \vert _{0} \int_{0}^{T} \bigl\vert e(t) \bigr\vert \,dt, \end{aligned} $$
which combining with [\(H_{3}\)] yields
$$ \begin{aligned}[b] \int_{0}^{T} \bigl\vert (Au)^{(m)}(t) \bigr\vert ^{p}\,dt &\leq \bigl(1+ \vert c \vert \bigr) \vert u \vert _{0} \int_{0}^{T} \bigl(\alpha \bigl\vert u(t) \bigr\vert ^{p-2}+\beta \bigr) \bigl\vert u'(t) \bigr\vert \,dt \\ &\quad{} + \bigl(1+ \vert c \vert \bigr) \vert u \vert _{0} \int_{0}^{T} \bigl\vert g \bigl(t,u(t-\tau) \bigr) \bigr\vert \,dt+ \bigl(1+ \vert c \vert \bigr) \vert u \vert _{0} \vert e \vert _{0}T \\ &\leq \bigl(1+ \vert c \vert \bigr)\alpha \vert u \vert ^{p-1}_{0} \int_{0}^{T} \bigl\vert u'(t) \bigr\vert \,dt+ \bigl(1+ \vert c \vert \bigr)\beta \vert u \vert _{0} \int_{0}^{T} \bigl\vert u'(t) \bigr\vert \,dt \\ &\quad{} + \bigl(1+ \vert c \vert \bigr) \vert u \vert _{0} \int_{0}^{T} \bigl\vert g \bigl(t,u(t-\tau) \bigr) \bigr\vert \,dt+ \bigl(1+ \vert c \vert \bigr) \vert u \vert _{0} \vert e \vert _{0}T. \end{aligned} $$
(3.6)
Write
$$\begin{aligned}& I_{+}= \bigl\{ t\in[0,T]:g \bigl(t, u(t-\tau) \bigr)\geq0 \bigr\} ; \\& I_{-}= \bigl\{ t\in[0,T]:g \bigl(t, u(t-\tau) \bigr)\leq0 \bigr\} . \end{aligned}$$
Then it follows from (3.3) and [\(H_{2}\)](1) that
$$ \begin{aligned}[b] \int_{0}^{T} \bigl\vert g \bigl(t,u(t-\tau) \bigr) \bigr\vert \,dt &= \int_{I_{+}}g \bigl(t,u(t-\tau) \bigr)\,dt- \int_{I_{-}}g \bigl(t,u(t-\tau) \bigr)\,dt \\ &=2 \int_{I_{+}}g \bigl(t,u(t-\tau) \bigr)\,dt \\ &\leq 2m_{0} \int_{0}^{T} u^{p-1}(t-\tau)\,dt+2 \int_{0}^{T}m_{1}\,dt \\ &\leq 2m_{0}T \vert u \vert ^{p-1}_{0}+2Tm_{1}. \end{aligned} $$
(3.7)
Substituting (3.7) into (3.6), combining with (3.5) and Lemma 2.2, we can have
$$\begin{aligned}& \int_{0}^{T} \bigl\vert (Au)^{(m)}(t) \bigr\vert ^{p}\,dt \\& \quad \leq 2 \bigl(1+ \vert c \vert \bigr)m_{0}T \vert u \vert ^{p}_{0}+ \bigl(1+ \vert c \vert \bigr)\alpha \vert u \vert ^{p-1}_{0} \int_{0}^{T} \bigl\vert u'(t) \bigr\vert \,dt \\& \quad\quad{} + \bigl(1+ \vert c \vert \bigr)\beta \vert u \vert _{0} \int_{0}^{T} \bigl\vert u'(t) \bigr\vert \,dt +2 \bigl(1+ \vert c \vert \bigr)m_{1}T \vert u \vert _{0}+ \bigl(1+ \vert c \vert \bigr) \vert e \vert _{0}T \vert u \vert _{0} \\& \quad \leq 2 \bigl(1+ \vert c \vert \bigr)m_{0}T^{p+1}M^{p}_{1}(m) \biggl(\frac{D_{2}}{TM_{1}(m)}+ \int_{0}^{T} \bigl\vert u^{(m)}(t) \bigr\vert \,dt \biggr)^{p} \\& \quad\quad{} + \bigl(1+ \vert c \vert \bigr)\alpha T^{p}M^{p}_{1}(m) \biggl(\frac{D_{2}}{TM_{1}(m)}+ \int_{0}^{T} \bigl\vert u^{(m)}(t) \bigr\vert \,dt \biggr)^{p-1} \cdot \int_{0}^{T} \bigl\vert u^{(m)}(t) \bigr\vert \,dt \\& \quad\quad{} + \bigl(1+ \vert c \vert \bigr)\beta T^{2}M^{2}_{1}(m) \biggl( \frac{D_{2}}{TM_{1}(m)}+ \int_{0}^{T} \bigl\vert u^{(m)}(t) \bigr\vert \,dt \biggr)\cdot \int_{0}^{T} \bigl\vert u^{(m)}(t) \bigr\vert \,dt \\& \quad\quad{} +2 \bigl(1+ \vert c \vert \bigr)m_{1}T^{2}M_{1}(m) \biggl( \frac{D_{2}}{TM_{1}(m)}+ \int_{0}^{T} \bigl\vert u^{(m)}(t) \bigr\vert \,dt \biggr) \\& \quad\quad{} + \bigl(1+ \vert c \vert \bigr) \vert e \vert _{0}T^{2}M_{1}(m) \biggl(\frac{D_{2}}{TM_{1}(m)}+ \int_{0}^{T} \bigl\vert u^{(m)}(t) \bigr\vert \,dt \biggr). \end{aligned}$$
(3.8)
It follows from conclusion (2) of Lemma 2.1 and by applying the Hölder inequality that
$$ \begin{aligned}[b] \int_{0}^{T} \bigl\vert u^{(m)}(t) \bigr\vert \,dt & = \int_{0}^{T} \bigl\vert \bigl(A^{-1}(Au)^{(m)} \bigr) (t) \bigr\vert \,dt \\ &\leq \frac{\int_{0}^{T} \vert (Au )^{(m)}(t) \vert \,dt}{ 1- \vert c \vert } . \end{aligned} $$
(3.9)
From the above inequality, we consider the following two cases:
Case 1
If \(\int_{0}^{T} \vert (Au )^{(m)}(t) \vert \,dt=0\), then \(\int_{0}^{T} \vert u^{(m)} (t) \vert \,dt=0\), it follows from (3.5) and Lemma 2.2 that
$$ \begin{aligned}[b] \vert u \vert _{0} &\leq D_{2}+ \int_{0}^{T} \bigl\vert u'(s) \bigr\vert \,ds\leq D_{2}+T \bigl\vert u' \bigr\vert _{0} \\ &\leq D_{2}+TM_{1}(m) \int_{0}^{T} \bigl\vert u^{(m)}(t) \bigr\vert \,dt \\ &=D_{2}. \end{aligned} $$
(3.10)
Case 2
If \(\int_{0}^{T} \vert (Au )^{(m)}(t) \vert \,dt>0\), then substituting (3.9) into (3.8), we can have
$$\begin{aligned}& \int_{0}^{T} \bigl\vert (Au)^{(m)}(t) \bigr\vert ^{p}\,dt \\& \quad \leq 2 \bigl(1+ \vert c \vert \bigr)m_{0}T^{p+1}M^{p}_{1}(m) \biggl(\frac{D_{2}}{TM_{1}(m)}+\frac{\int_{0}^{T} \vert (Au )^{(m)}(t) \vert \,dt}{1- \vert c \vert } \biggr)^{p} \\& \quad\quad{} + \bigl(1+ \vert c \vert \bigr)\alpha T^{p}M^{p}_{1}(m) \biggl(\frac{D_{2}}{TM_{1}(m)}+ \frac{\int_{0}^{T} \vert (Au )^{(m)}(t) \vert \,dt}{1- \vert c \vert } \biggr)^{p-1}\cdot \frac{\int_{0}^{T} \vert (Au )^{(m)}(t) \vert \,dt}{1- \vert c \vert } \\& \quad\quad{} + \bigl(1+ \vert c \vert \bigr)\beta T^{2}M^{2}_{1}(m) \biggl( \frac{D_{2}}{TM_{1}(m)}+\frac{\int_{0}^{T} \vert (Au )^{(m)}(t) \vert \,dt}{1- \vert c \vert } \biggr)\cdot\frac {\int_{0}^{T} \vert (Au )^{(m)}(t) \vert \,dt}{1- \vert c \vert } \\& \quad\quad{} +2 \bigl(1+ \vert c \vert \bigr)m_{1}T^{2}M_{1}(m) \biggl( \frac{D_{2}}{TM_{1}(m)}+\frac{\int_{0}^{T} \vert (Au )^{(m)}(t) \vert \,dt}{1- \vert c \vert } \biggr) \\& \quad\quad{} + \bigl(1+ \vert c \vert \bigr) \vert e \vert _{0}T^{2}M_{1}(m) \biggl(\frac{D_{2}}{TM_{1}(m)}+ \frac{\int_{0}^{T} \vert (Au )^{(m)}(t) \vert \,dt}{1- \vert c \vert } \biggr). \end{aligned}$$
(3.11)
Furthermore,
$$ \begin{aligned}[b] & \biggl(\frac{D_{2}}{TM_{1}(m)}+\frac{\int _{0}^{T} \vert (Au )^{(m)}(t) \vert \,dt}{1- \vert c \vert } \biggr)^{p} \\ &\quad =\frac{1}{(1- \vert c \vert )^{p}} \biggl( \int_{0}^{T} \bigl\vert (Au)^{(m)}(t) \bigr\vert \,dt \biggr)^{p} \biggl(1+\frac{\frac{D_{2}}{TM_{1}(m)}(1- \vert c \vert )}{\int_{0}^{T} \vert (Au)^{(m)}(t) \vert \,dt} \biggr)^{p}. \end{aligned} $$
(3.12)
By classical elementary inequalities, we see that there exists a \(l(p)>0\) which is dependent on p only, such that
$$ (1+x)^{p}< 1+(1+p)x, \quad x\in\bigl(0,l(p) \bigr]. $$
(3.13)
If \(\frac{\frac{D_{2}}{TM_{1}(m)}(1- \vert c \vert )}{\int _{0}^{T} \vert (Au)^{(m)}(t) \vert \,dt}>l(p)\), then \(\int _{0}^{T} \vert (Au)^{(m)}(t) \vert \,dt<\frac{\frac {D_{2}}{TM_{1}(m)}(1- \vert c \vert )}{l(p)}\). It follows from (3.5), (3.9) and Lemma 2.2 that
$$ \vert u \vert _{0} \leq D_{2}+TM_{1}(m) \int_{0}^{T} \bigl\vert u^{(m)}(t) \bigr\vert \leq D_{2}+\frac{ D_{2}}{l(p)}:=M_{1}. $$
(3.14)
If \(\frac{\frac{D_{2}}{TM_{1}(m)}(1- \vert c \vert )}{\int _{0}^{T} \vert (Au)^{(m)}(t) \vert \,dt}\leq l(p)\), then it follows from (3.12) and (3.13) that
$$ \begin{aligned}[b] & \biggl(\frac{D_{2}}{TM_{1}(m)}+\frac{\int _{0}^{T} \vert (Au)^{(m)}(t) \vert \,dt}{1- \vert c \vert } \biggr)^{p} \\ &\quad \leq\frac{1}{(1- \vert c \vert )^{p}} \biggl( \int_{0}^{T} \bigl\vert (Au)^{(m)} (t) \bigr\vert \,dt \biggr)^{p} \biggl(1+\frac{(p+1)\frac{D_{2}}{TM_{1}(m)}(1- \vert c \vert )}{\int_{0}^{T} \vert (Au)^{(m)}(t) \vert \,dt} \biggr) \\ &\quad = \frac{ ( \int_{0}^{T} \vert (Au)^{(m)}(t) \vert \,dt )^{p}}{(1- \vert c \vert )^{p}}+\frac{(p+1)\frac {D_{2}}{TM_{1}(m)}}{(1- \vert c \vert )^{p-1}}\cdot \biggl( \int_{0}^{T} \bigl\vert (Au)^{(m)}(t) \bigr\vert \,dt \biggr)^{p-1}. \end{aligned} $$
(3.15)
Substituting (3.15) into (3.11) and by applying the Hölder inequality, we can see that
$$\begin{aligned}& \int_{0}^{T} \bigl\vert (Au)^{(m)}(t) \bigr\vert ^{p}\,dt \\& \quad \leq \frac{M^{p}_{1}(m)T^{2p-1}(1+ \vert c \vert )(2m_{0}T+\alpha)}{(1- \vert c \vert )^{p}} \int_{0}^{T} \bigl\vert (Au)^{(m)}(t) \bigr\vert ^{p}\,dt \\& \quad\quad{} +\frac{ T^{\frac{2p^{2}-3p+1}{p}} M^{p-1}_{1}(m)D_{2}(1+ \vert c \vert ) [2m_{0}T(p+1)+\alpha p] }{(1- \vert c \vert )^{p-1}} \cdot \biggl( \int_{0}^{T} \bigl\vert (Au)^{(m)}(t) \bigr\vert ^{p} \,dt \biggr)^{\frac{p-1}{p}} \\& \quad\quad{} + \frac{(1+ \vert c \vert )\beta T^{4p-2} M^{2}_{1}(m)}{(1- \vert c \vert )^{2}} \biggl( \int_{0}^{T} \bigl\vert (Au)^{(m)}(t) \bigr\vert ^{p} \,dt \biggr)^{\frac{2}{p}} \\& \quad\quad{} + \frac{(1+ \vert c \vert )T^{\frac{2p-1}{p}}M_{1}(m) [\beta D_{2}+2m_{1}T+ \vert e \vert _{0} T] }{1- \vert c \vert }\cdot \biggl( \int_{0}^{T} \bigl\vert (Au)^{(m)}(t) \bigr\vert ^{p} \,dt \biggr)^{\frac{1}{p}} \\& \quad\quad{} + \bigl(1+ \vert c \vert \bigr)TD_{2} \bigl(2m_{1}+ \vert e \vert _{0} \bigr). \end{aligned}$$
(3.16)
It follows from \(\frac{M^{p}_{1}(m)T^{2p-1}(1+ \vert c \vert )(2m_{0}T+\alpha)}{(1- \vert c \vert )^{p}} <1\) and \(p\geq2\) that there exists a positive constant \(M_{2}>0\) such that
$$\int_{0}^{T} \bigl\vert (Au)^{(m)}(t) \bigr\vert ^{p}\,dt\leq M_{2}. $$
Then by (3.9), we can see that
$$ \begin{aligned}[b] \int_{0}^{T} \bigl\vert u^{(m)}(t) \bigr\vert \,dt &\leq \frac{\int_{0}^{T} \vert (Au )^{(m)}(t) \vert \, dt}{1- \vert c \vert } \\ &\leq \frac{T^{\frac{p-1}{p}} (\int_{0}^{T} \vert (Au )^{(m)}(t) \vert ^{p}\,dt )^{\frac{1}{p}}}{1- \vert c \vert } \\ & \leq\frac{T^{\frac{p-1}{p}} M_{2} ^{\frac{1}{p}}}{1- \vert c \vert } :=M_{3}, \end{aligned} $$
(3.17)
which together with (3.5) and Lemma 2.2 yields
$$\begin{aligned} \vert u \vert _{0} &\leq D_{2}+ \int_{0}^{T} \bigl\vert u'(s) \bigr\vert \,ds\leq D_{2}+T \bigl\vert u' \bigr\vert _{0} \\ &\leq D_{2}+TM_{1}(m) \int_{0}^{T} \bigl\vert u^{(m)}(t) \bigr\vert \,dt \\ &\leq D_{2}+ TM_{1}(m)M_{3}:=M_{4}. \end{aligned}$$
Therefore, in both Case 1 and Case 2, we obtain
$$ \vert u \vert _{0}\leq M_{4}. $$
(3.18)
From the second equation of (3.1), we can get
$$\begin{aligned} \int_{0}^{T} \bigl\vert v^{(m)}(t) \bigr\vert \,dt &\leq \lambda \biggl[ \int_{0}^{T} \bigl\vert f \bigl(u(t) \bigr) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt+ \int_{0}^{T} \bigl\vert g \bigl(t,u(t-\tau) \bigr) \bigr\vert \,dt \\ &\quad{} + \int_{0}^{T} \bigl\vert e(t) \bigr\vert \,dt \biggr], \end{aligned}$$
from which by applying [\(H_{3}\)] and (3.7), we have
$$ \begin{aligned}[b] \int_{0}^{T} \bigl\vert v^{(m)}(t) \bigr\vert \,dt &\leq \lambda \bigl[ \alpha T \vert u \vert ^{p-2}_{0} \bigl\vert u' \bigr\vert _{0}+ \beta T \bigl\vert u' \bigr\vert _{0}+ 2 m_{0}T \vert u \vert ^{p-1}_{0} \\ &\quad{} + 2 Tm_{1} + \vert e \vert _{0}T \bigr]. \end{aligned} $$
(3.19)
By Lemma 2.2, (3.17) and (3.18), we obtain
$$\begin{aligned} \int_{0}^{T} \bigl\vert v^{(m)}(t) \bigr\vert \,dt &\leq \lambda \bigl[ \alpha T \vert u \vert ^{p-2}_{0} \bigl\vert u' \bigr\vert _{0}+ \beta T \bigl\vert u' \bigr\vert _{0}+ 2 m_{0}T \vert u \vert ^{p-1}_{0} \\ &\quad{} + 2 Tm_{1} + \vert e \vert _{0}T \bigr] \\ \leq& \lambda \bigl[ \alpha TM^{p-2}_{4} M_{1}(m)M_{3}+ \beta T M_{1}(m)M_{3}+ 2m_{0}T M^{p-1}_{4} \\ &\quad{} + 2Tm_{1} + \vert e \vert _{0}T \bigr] . \end{aligned}$$
(3.20)
Moreover, integrating the first equation of (3.1) on the interval \([0,T]\), we have
$$\int_{0}^{T} \bigl\vert v(t) \bigr\vert ^{q-2}v(t)\,dt=0, $$
which implies that there exists \(\eta\in[0,T]\) such that \(v(\eta )=0\). Thus,
$$\begin{aligned} \bigl\vert v(t) \bigr\vert &= \biggl\vert \int_{\eta}^{t} v'(s)\,ds+v(\eta) \biggr\vert \leq \int_{0}^{T} \bigl\vert v'(s) \bigr\vert \,ds\leq T \bigl\vert v' \bigr\vert _{0}, \end{aligned}$$
by Lemma 2.2 and (3.20), we can obtain
$$ \begin{aligned}[b] \vert v \vert _{0}&=\max _{t\in[0,T]} \bigl\vert v(t) \bigr\vert \\ &\leq T \bigl\vert v' \bigr\vert _{0}\leq TM_{1}(m) \int_{0}^{T} \bigl\vert v^{(m)}(t) \bigr\vert \,dt \\ &\leq \lambda TM_{1}(m) \bigl[ \alpha TM^{p-2}_{4} M_{1}(m)M_{3}+ \beta T M_{1}(m)M_{3}+ 2m_{0}T M^{p-1}_{4} \\ &\quad{} + 2Tm_{1} + \vert e \vert _{0}T \bigr] \\ &\leq TM_{1}(m) \bigl[ \alpha TM^{p-2}_{4} M_{1}(m)M_{3}+ \beta T M_{1}(m)M_{3}+ 2m_{0}T M^{p-1}_{4} \\ &\quad{} + 2Tm_{1} + \vert e \vert _{0}T \bigr] \\ &:=A_{3} . \end{aligned} $$
(3.21)
On the other hand, from the second equation of (3.1) and [\(H_{2}\)], we can see that
$$ \begin{aligned}[b] v^{(m)}(t+\tau)&=- \lambda f \bigl(u(t+ \tau) \bigr)u'(t+\tau) \\ &\quad{} -\lambda \bigl[g_{1} \bigl(t+\tau, u(t) \bigr)+g_{0} \bigl(u(t) \bigr) \bigr] \\ &\quad{} +\lambda e(t+\tau). \end{aligned} $$
(3.22)
Furthermore, multiplying both sides of equation (3.22) by \(u'(t)\), we have
$$ \begin{aligned}[b] v^{(m)}(t+\tau)u'(t)&=- \lambda f \bigl(u(t+\tau) \bigr)u'(t+\tau)u'(t) \\ &\quad{} -\lambda \bigl[g_{1} \bigl(t+\tau, u(t) \bigr)+g_{0} \bigl(u(t) \bigr) \bigr] u'(t) \\ & \quad{} +\lambda e(t+\tau)u'(t). \end{aligned} $$
(3.23)
Let \(\sigma\in[0,T]\) be as in (3.4). For any \(t\in[\sigma ,T]\), integrating equation (3.23) on the interval \([\sigma,t]\), we get
$$\begin{aligned} \lambda \int_{u(\sigma)}^{u(t)}g_{0}(u)\,du&=\lambda \int_{\sigma}^{t}g_{0} \bigl(u(t) \bigr)u'(t)\,dt \\ &=- \int_{\sigma}^{t}v^{(m)}(t+ \tau)u'(t)\,dt - \lambda \int_{\sigma}^{t}f \bigl(u(t+\tau) \bigr)u'(t+ \tau)u'(t)\,dt \\ &\quad{} -\lambda \int_{\sigma}^{t}g_{1} \bigl(t+\tau,u(t) \bigr)u'(t)\,dt+\lambda \int_{\sigma}^{t} e(t +\tau)u'(t)\,dt, \end{aligned}$$
which together with (3.20) yields
$$\begin{aligned} &\lambda \biggl\vert \int_{u(\tau)}^{u(t)}g_{0}(u)\,du \biggr\vert \\ &\quad = \lambda \biggl\vert \int_{\tau}^{t}g_{0} \bigl(u(t) \bigr)u'(t)\,dt \biggr\vert \\ &\quad \leq \int_{0}^{T} \bigl\vert v^{(m)}(t+\tau) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt + \lambda \int_{0}^{T} \bigl\vert f \bigl(u(t+\tau) \bigr) \bigr\vert \bigl\vert u'(t+\tau) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt \\ &\quad \quad{}+\lambda \int_{0}^{T} \bigl\vert g_{1} \bigl(t+ \tau,u(t) \bigr) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt+ \lambda \int_{0}^{T} \bigl\vert e(t +\tau) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt \\ &\quad \leq\lambda \bigl\vert u' \bigr\vert _{0} \bigl[ \alpha TM^{p-2}_{4} M_{1}(m)M_{3}+ \beta T M_{1}(m)M_{3}+ 2m_{0}T M^{p-1}_{4}+ 2Tm_{1} + \vert e \vert _{0}T \bigr] \\ &\quad \quad{}+\lambda \bigl\vert u' \bigr\vert ^{2}_{0} \int_{0}^{T} \bigl\vert f \bigl(u(t+\tau) \bigr) \bigr\vert \,dt+ \lambda \bigl\vert u' \bigr\vert _{0} \int_{0}^{T} \bigl\vert g_{1} \bigl(t+ \tau,u(t) \bigr) \bigr\vert \,dt \\ &\quad \quad{}+\lambda \bigl\vert u' \bigr\vert _{0} \int_{0}^{T} \bigl\vert e(t +\tau) \bigr\vert \,dt. \end{aligned}$$
Furthermore, set
$$F_{M_{4}}=\max_{ \vert u \vert \leq M_{4}} \bigl\vert f(u) \bigr\vert \quad\text{and} \quad G_{M_{4}}=\max_{t\in[0,T], \vert u \vert \leq M_{4}} \bigl\vert g_{1}(t,u) \bigr\vert , $$
then we have
$$\begin{aligned} \lambda \biggl\vert \int_{u(\tau)}^{u(t)}g_{0}(u)\,du \biggr\vert & \leq \lambda \bigl\vert u' \bigr\vert _{0} \bigl[ \alpha TM^{p-2}_{4} M_{1}(m)M_{3}+ \beta T M_{1}(m)M_{3} \bigr] \\ &\quad{} + \lambda \bigl\vert u' \bigr\vert _{0} \bigl[ 2m_{0}T M^{p-1}_{4}+ 2Tm_{1} + \vert e \vert _{0}T \bigr] \\ & \quad{} +\lambda \bigl\vert u' \bigr\vert ^{2}_{0} TF_{M_{4}}+\lambda \bigl\vert u' \bigr\vert _{0}TG_{M_{4}}+ \lambda \bigl\vert u' \bigr\vert _{0}T \vert e \vert _{0}, \end{aligned}$$
by (3.17) and Lemma 2.2, we obtain
$$ \begin{aligned}[b] \biggl\vert \int_{u(\tau)}^{u(t)}g_{0}(u)\,du \biggr\vert & \leq M_{1}(m)M_{3} \bigl[ \alpha TM^{p-2}_{4} M_{1}(m)M_{3}+ \beta T M_{1}(m)M_{3} \bigr] \\ &\quad{} + M_{1}(m)M_{3} \bigl[2m_{0}T M^{p-1}_{4}+ 2Tm_{1} + \vert e \vert _{0}T \bigr] \\ & \quad{} + \bigl[M_{1}(m)M_{3} \bigr]^{2} TF_{M_{4}}+ M_{1}(m)M_{3}TG_{M_{4}} \\ &\quad{} + M_{1}(m)M_{3}T \vert e \vert _{0} \\ &< +\infty. \end{aligned} $$
(3.24)
According to condition (2) in [\(H_{2}\)], we see that there exists a constant \(M_{5}>0\) such that, for \(t\in[\sigma,T]\),
$$ u(t)\geq M_{5}. $$
(3.25)
In a similar way, we can handle the case of \(t\in[0,\sigma]\).
Let us define
$$0< A_{1}=\min\{D_{1},M_{5}\},\quad\text{and}\quad A_{2}=\max\{D_{2},M_{4}\}. $$
Then by (3.4), (3.10), (3.18) and (3.25), we can obtain
$$ A_{1}\leq u(t)\leq A_{2}. $$
(3.26)
Clearly, \(A_{1}\) and \(A_{2}\) are independent of λ. Therefore, the proof of Theorem 3.1 is complete. □
Theorem 3.2
Assume that all the conditions in Theorem
3.1
hold, then system (1.5) has at least one positive
T-periodic solution.
Proof
Set
$$\Omega= \biggl\{ x=(u,v)^{\top}\in X:\frac{A_{1}}{2}< u(t)< A_{2}+1, \vert v \vert _{0}< A_{3}+1 \biggr\} . $$
From (2.3) and (2.4), one can easily see that Ω is an open bounded subset of X and N is L-compact on Ω̅. Then the conditions (1) and (2) of Lemma 2.3 are satisfied.
In the following, we prove that condition (3) of Lemma 2.3 also holds.
Now, we let
$$\omega=Bx=B \begin{pmatrix}u \\v \end{pmatrix} = \begin{pmatrix}u-\frac{A_{1}+A_{2}}{2} \\v \end{pmatrix} . $$
Define a linear isomorphism
$$J:\quad \operatorname {Im}Q\rightarrow\ker L,\quad\quad J (u,v)= \begin{pmatrix}v \\ -u \end{pmatrix} , $$
and define
$$H(\mu,x)=\mu B x+(1-\mu) JQN x, \quad \forall(x,\mu)\in(\Omega\cap\ker L) \times[0,1]. $$
Then we can get
$$ \begin{aligned} K(\mu,x)= \begin{pmatrix}\mu u-\frac{\mu(A_{1}+A_{2})}{2} \\ \mu v \end{pmatrix} +\frac{1-\mu}{T} \begin{pmatrix}\int_{0}^{T}[f(u)u'+g(t, u)]\,dt \\ (1- \vert c \vert )^{-1}\int_{0}^{T}\varphi_{q}(v)\,dt \end{pmatrix} . \end{aligned} $$
(3.27)
In order to prove the condition (3) of Lemma 2.3 is also satisfied, firstly, we prove that \(K(\mu,x)\) is a homotopic mapping. By way of contradiction, i.e., suppose that there exist \(\mu_{0}\in [0,1]\) and \(x_{0}= \bigl({\scriptsize\begin{matrix}{}u_{0}\cr v_{0} \end{matrix}} \bigr) \in\partial(\Omega\cap\ker L)\) such that \(K(\mu_{0},x_{0})=0\). Then, substituting \(\mu_{0}\) and \(x_{0}\) into (3.27), we have
$$ K(\mu_{0},x_{0})= \begin{pmatrix}\mu_{0} u_{0}-\frac{\mu_{0}(A_{1}+A_{2})}{2} +(1-\mu_{0}) \overline{g}(u_{0}) \\ \mu_{0} v_{0}+\frac{1-\mu_{0}}{1- \vert c \vert }\varphi_{q}(v_{0}) \end{pmatrix} . $$
(3.28)
It follows from \(K(\mu_{0},x_{0})=0\) that
$$\mu_{0} v_{0}+\frac{1-\mu_{0}}{1- \vert c \vert }\varphi_{q}(v_{0})=0, $$
which together with \(\mu_{0}\in[0,1]\) gives \(v_{0}=0\). Thus, we can get \(u_{0}=\frac{A_{1}}{2}\) or \(A_{2}+1\). Furthermore, it follows from [\(H_{1}\)](2) that \(\overline{g} (\frac{A_{1}}{2})<0\) and \(\overline {g} (A_{2}+1)>0\), substituting \(u_{0}=\frac{A_{1}}{2}\) or \(A_{2}+1\) into (3.28), we can obtain
$$ \mu_{0} u_{0}- \frac{\mu_{0}(A_{1}+A_{2})}{2} +(1- \mu_{0}) \overline{g} (u_{0})< 0 $$
(3.29)
and
$$ \mu_{0} u_{0}- \frac{\mu_{0}(A_{1}+A_{2})}{2} +(1- \mu_{0}) \overline{g} (u_{0})>0. $$
(3.30)
From (3.29) and (3.30), we have \(K(\mu_{0},x_{0})\neq0\), which is a contradiction. Therefore \(K(\mu,x)\) is a homotopic mapping and \(x^{\top}K(\mu,x)\neq0\). For all \((x,\mu)\in(\partial\Omega \cap\ker L )\times[0,1]\), we get
$$\begin{aligned} \deg(JQN ,\Omega\cap\ker L ,0)&=\deg \bigl(K(0,x),\Omega\cap\ker L ,0 \bigr) \\ &=\deg \bigl(K(1,x),\Omega\cap\ker L ,0 \bigr) \\ &=\deg(Bx,\Omega\cap\ker L ,0) \\ &=\sum_{x\in B^{-1}(0)} \operatorname {sgn}\bigl(\det B'(x) \bigr) \\ &=1\neq0. \end{aligned}$$
Thus, the condition (3) of Lemma 2.3 is also satisfied. Therefore, we can conclude that equation (1.5) has at least one positive T-periodic solution. □