Multiplying both sides of Eq. (
2.1) by
\((Au)(t)\) and integrating on the interval
\([0,T]\), it is clear that
$$ \begin{aligned}[b] & \int ^{T}_{0}\bigl(\phi _{p} \bigl((Au)'(t)\bigr)\bigr)'(Au) (t)\,dt+\lambda P \int ^{T}_{0}u'(t) (Au) (t)\,dt+ \lambda \int ^{T}_{0}g\bigl(u(t)\bigr) (Au) (t)\,dt \\ &\quad =\lambda \int ^{T}_{0}e(t) (Au) (t)\,dt. \end{aligned} $$
(2.18)
Substituting
\(\int ^{T}_{0}(\phi _{p}(Au)'(t))'(Au)(t)\,dt=-\int ^{T}_{0}|(Au)'(t)|^{p}\,dt\) and
\(P\int ^{T}_{0}u'(t)u(t)\,dt=0\) into Eq. (
2.18), we have
$$\begin{aligned} \int ^{T}_{0} \bigl\vert (Au)'(t) \bigr\vert ^{p}\,dt&=-\lambda Pc \int ^{T}_{0}u'(t)u(t-\tau )\,dt+ \lambda \int ^{T}_{0}g\bigl(u(t)\bigr) \bigl(u(t)-cu(t-\tau )\bigr)\,dt \\ &\quad {}-\lambda \int ^{T}_{0}e(t) \bigl(u(t)-cu(t-\tau )\bigr)\,dt. \end{aligned}$$
(2.19)
Furthermore, we deduce
$$ \int ^{T}_{0}u'(t)u(t-\tau )\,dt= \int ^{T}_{0}u(t-\tau )\,du(t)= \int ^{T} _{0}u(t-\tau )\,du(t-\tau )=0. $$
(2.20)
From condition (
\(\mathrm{H}_{2}\)) and
\(u(t)>0\), we see that
$$ \int ^{T}_{0}g\bigl(u(t)\bigr)u(t)\,dt\leq \alpha \int ^{T}_{0}\bigl(u(t)\bigr)^{p}\,dt+ \beta \int ^{T}_{0}u(t)\,dt. $$
(2.21)
Substituting Eqs. (
2.20) and (
2.21) into (
2.19), applying the Hölder inequality, we obtain
$$ \begin{aligned}[b] \int ^{T}_{0} \bigl\vert (Au)'(t) \bigr\vert ^{p}\,dt &\leq \alpha \int ^{T}_{0}\bigl(u(t)\bigr)^{p}\,dt+ \beta \int ^{T}_{0}u(t)\,dt+|c| \int ^{T}_{0} \bigl\vert g\bigl(u(t)\bigr) \bigr\vert \bigl\vert u(t-\tau ) \bigr\vert \,dt \\ &\quad {}+\bigl(1+ \vert c \vert \bigr)\|u\| \int ^{T}_{0}\bigl|e(t)\bigr|\,dt \\ &\leq \alpha T\|u\|^{p}+\beta T\|u\|+|c|\|u\| \int ^{T}_{0} \bigl\vert g\bigl(u(t)\bigr) \bigr\vert \,dt \\ &\quad {}+\bigl(1+ \vert c \vert \bigr) \|u\|T^{\frac{1}{2}}\|e \|_{2}. \end{aligned} $$
(2.22)
From Eq. (
2.10) and condition (
\(\mathrm{H}_{2}\)), we get
$$ \begin{aligned}[b] \int ^{T}_{0} \bigl\vert g\bigl(u(t)\bigr) \bigr\vert \,dt&=2 \int _{g(u(t))\geq 0}g\bigl(u(t)\bigr)\,dt- \int ^{T}_{0}e(t)\,dt \\ &\leq 2\alpha \|u\|^{p-1}T+2\beta T+\|e\|_{2}T^{\frac{1}{2}}. \end{aligned} $$
(2.23)
Substituting Eqs. (
2.17) and (
2.23) into (
2.22), we see that
$$ \begin{aligned}[b] &\int ^{T}_{0} \bigl\vert (Au)'(t) \bigr\vert ^{p}\,dt \\ &\quad \leq \alpha \bigl(1+2 \vert c \vert \bigr)T \|u\|^{p}+\bigl(1+2 \vert c \vert \bigr) \bigl( \beta T+T^{\frac{1}{2}}\|e\|_{2}\bigr)\|u\| \\ &\quad \leq \alpha \bigl(1+2 \vert c \vert \bigr)T \biggl(d_{2}+ \int ^{T}_{0} \bigl\vert u'(t) \bigr\vert \,dt \biggr)^{p}+N _{1} \biggl(d_{2}+ \int ^{T}_{0} \bigl\vert u'(t) \bigr\vert \,dt \biggr) \\ &\quad = \alpha \bigl(1+2 \vert c \vert \bigr)T \biggl( \biggl( \int ^{T}_{0} \bigl\vert u'(t) \bigr\vert \,dt \biggr)^{p}+pd _{2} \biggl( \int ^{T}_{0} \bigl\vert u'(t) \bigr\vert \,dt \biggr)^{p-1} +\cdots +d_{2}^{p} \biggr) \\ &\quad \quad {}+N_{1} \biggl(d_{2}+ \int ^{T}_{0} \bigl\vert u'(t) \bigr\vert \,dt \biggr), \end{aligned} $$
(2.24)
where
\(N_{1}:=(1+2|c|)(\beta T+T^{\frac{1}{2}}\|e\|_{2})\). Applying Lemma
2.1, we have
$$ \begin{aligned}[b] \int ^{T}_{0} \bigl\vert u'(t) \bigr\vert \,dt&= \int ^{T}_{0} \bigl\vert \bigl(A^{-1}Au' \bigr) (t) \bigr\vert \,dt \\ &\leq \frac{\int ^{T}_{0}|(Au)'(t)|\,dt}{|c|-1} \\ &\leq \frac{T^{\frac{1}{q}} (\int ^{T}_{0}|(Au)'(t)|^{p}\,dt ) ^{\frac{1}{p}}}{|c|-1}, \end{aligned} $$
(2.25)
since
\(\frac{1}{p}+\frac{1}{q}=1\) and
\(|c|>1\). We apply the inequality
$$ (x+y)^{k}\leq x^{k}+ y^{k}, \quad \mbox{for } x, y>0, 0< k< 1. $$
Substituting Eqs. (
2.24) into (
2.25), we have
$$\begin{aligned} \int ^{T}_{0} \bigl\vert u'(t) \bigr\vert \,dt \leq &\frac{\alpha ^{\frac{1}{p}}(1+2|c|)^{ \frac{1}{p}}T\int ^{T}_{0}|u'(t)|\,dt}{|c|-1} \\ &{}+\frac{\alpha ^{\frac{1}{p}}(1+2|c|)^{\frac{1}{p}}(pd_{2})^{ \frac{1}{p}}T (\int ^{T}_{0}|u'(t)|\,dt )^{\frac{p-1}{p}}}{|c|-1} \\ &{}+\cdots +\frac{T^{\frac{1}{q}} (N_{1}^{\frac{1}{p}}+(pd_{2} ^{p-1})^{\frac{1}{p}} ) (\int ^{T}_{0}|u'(t)|\,dt )^{ \frac{1}{p}}+ T^{\frac{1}{p}}(d_{2}+ (N_{1}d_{2})^{\frac{1}{p}})}{|c|-1}. \end{aligned}$$
Since
\(\alpha ^{\frac{1}{p}}(1+2|c|)^{\frac{1}{p}}T<|c|-1\), we know that there exists a positive constant
\(M_{1}'\) such that
$$ \int ^{T}_{0} \bigl\vert u'(t) \bigr\vert \,dt\leq M_{1}'. $$
(2.26)
From Eqs. (
2.17) and (
2.26), we have
$$ \|u\|\leq d_{2}+\frac{1}{2} \int ^{T}_{0} \bigl\vert u'(t) \bigr\vert \,dt\leq d_{2}+\frac{M _{1}'}{2}:=M_{1}. $$
The proof is left for the reader, being the same as that of Theorem
1.1. □