Firstly, let \(A: C_{T}\rightarrow C_{T}\) be the operator on \(C_{T}:=\{u\in C(\mathbb{R}, \mathbb{R}): u(t+T)\equiv u(t)\ \forall t\in \mathbb{R}\}\) given by
$$ (Au) (t):=u(t)-cu(t-\tau ) \quad \forall u\in C_{T} , t\in \mathbb{R} . $$
Lemma 2.1
(see [12])
The operator
A
has a continuous inverse
\(A^{-1}\)
on
\(C_{T}\), satisfying
-
(1)
\(\vert [A^{-1}f ](t) \vert \leq \frac{\|f\|}{|1-|c||}\), \(\forall f\in C_{T}\), where
\(\|f\|:= \max_{t\in \mathbb{R}}|f(t)| \).
-
(2)
\(\int ^{T}_{0} \vert [A^{-1}f ](t) \vert \,dt \leq \frac{1}{|1-|c||}\int ^{T}_{0}|f(t)|\,dt\), \(\forall f\in C_{T}\).
Secondly, we embed Eq. (1.2) into the following equation family with a parameter \(\lambda \in (0,1]\):
$$ \bigl(\phi _{p}(Au)'(t) \bigr)'+\lambda Pu'(t)+\lambda g\bigl(u(t)\bigr)=\lambda e(t). $$
(2.1)
The following lemma is a consequence of Theorem 3.1 of [6].
Lemma 2.2
Assume that there exist positive constants
\(E_{1}\), \(E_{2}\), \(E_{3}\)
and
\(E_{1}< E_{2}\)
such that the following conditions hold:
-
(1)
Each possible periodic solution
u
to Eq. (2.1) such that
\(E_{1}< u(t)< E_{2}\), for all
\(t\in [0,T]\)
and
\(\|u'\|< E_{3}\).
-
(2)
Each possible solution
C
to the equation
$$ g(C)-\frac{1}{T} \int ^{T}_{0} e(t)\,dt=0 $$
satisfies
\(C\in (E_{1},E_{2})\).
-
(3)
We have
$$ \biggl(g(E_{1})-\frac{1}{T} \int ^{T}_{0} e(t)\,dt \biggr) \biggl(g(E_{2})- \frac{1}{T} \int ^{T}_{0} e(t)\,dt \biggr)< 0. $$
Then Eq. (1.2) has at least one
T-periodic solution.
Proof of Theorem 1.1
Proof of Theorem 1.1
Firstly, integrating both sides of Eq. (2.1) over \([0,T]\), we get
$$ \int ^{T}_{0}\bigl[g\bigl(u(t)\bigr)-e(t)\bigr] \,dt=0. $$
(2.2)
In view of the mean value theorem of integrals, there exists a point \(\xi \in (0,T)\) such that
$$ g\bigl(u(\xi )\bigr)-e(\xi )=0. $$
From condition (\(\mathrm{H}_{1}\)) and \(u(t)\) being continuous, we have
$$ d_{1}\leq u(\xi )\leq d_{2}. $$
(2.3)
Multiplying both sides of Eq. (2.1) by \((Au)'(t)\) and integrating from 0 to T, we deduce
$$ \begin{aligned}[b] & \int ^{T}_{0}\bigl(\phi _{p} \bigl((Au)'(t)\bigr)\bigr)'(Au)'(t)\,dt+ \lambda P \int ^{T}_{0}u'(t) (Au)'(t) \,dt + \lambda \int ^{T}_{0}g\bigl(u(t)\bigr) (Au)'(t) \,dt \\ &\quad =\lambda \int ^{T}_{0}e(t) (Au)'(t)\,dt. \end{aligned} $$
(2.4)
Moreover,
$$ \int ^{T}_{0}\bigl(\phi _{p}(Au)'(t) \bigr)'(Au)'(t)\,dt= \int ^{T}_{0}(Au)'(t)\,d\phi _{p}(Au)'(t)=0 $$
(2.5)
and
$$ \begin{aligned}[b] \int ^{T}_{0}g\bigl(u(t)\bigr) (Au)'(t) \,dt&= \int ^{T}_{0}g\bigl(u(t)\bigr) \bigl(Au' \bigr) (t)\,dt \\ &= \int ^{T}_{0}g\bigl(u(t)\bigr) \bigl(u'(t)-cu'(t- \tau )\bigr)\,dt \\ &= \int ^{T}_{0}g\bigl(u(t)\bigr)\,du(t)-c \int ^{T}_{0}g\bigl(u(t)\bigr)\,du(t-\tau ) \\ &=-c \int ^{T}_{0}g\bigl(u(t)\bigr)\,du(t) \\ &=0, \end{aligned} $$
(2.6)
since \((Au)'(t)=(Au')(t)\) and \(du(t)=\frac{du(t-\tau )}{d(t-\tau )}\,dt=du(t- \tau )\).
Substituting Eqs. (2.5) and (2.6) into (2.4), we obtain
$$ P \int ^{T}_{0}u'(t) \bigl(Au'\bigr) (t)\,dt= \int ^{T}_{0}e(t) \bigl(Au'\bigr) (t) \,dt. $$
(2.7)
From Eq. (2.7), we arrive at
$$ \biggl\vert P \int ^{T}_{0} \bigl\vert u'(t) \bigr\vert ^{2}\,dt \biggr\vert = \biggl\vert Pc \int ^{T}_{0}u'(t)u'(t- \tau )\,dt+ \int ^{T}_{0}e(t) \bigl(Au'\bigr) (t) \,dt \biggr\vert . $$
From the Hölder inequality and \(\int ^{T}_{0}|u'(t-\tau )|^{2}\,dt= \int ^{T}_{0}|u'(t)|^{2}\,dt\), we see that
$$\begin{aligned} \vert P \vert \int ^{T}_{0} \bigl\vert u'(t) \bigr\vert ^{2}\,dt \leq & \vert P \vert \vert c \vert \int ^{T}_{0} \bigl\vert u'(t) \bigr\vert \bigl\vert u'(t- \tau ) \bigr\vert \,dt+\bigl(1+ \vert c \vert \bigr) \int ^{T}_{0} \bigl\vert e(t) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt \\ \leq & \vert P \vert \vert c \vert \biggl( \int ^{T}_{0} \bigl\vert u'(t) \bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}} \biggl( \int ^{T}_{0} \bigl\vert u'(t-\tau ) \bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}} \\ &{}+\bigl(1+ \vert c \vert \bigr) \biggl( \int ^{T}_{0} \bigl\vert e(t) \bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}} \biggl( \int ^{T}_{0} \bigl\vert u'(t) \bigr\vert \,dt \biggr)^{\frac{1}{2}} \\ = & \vert P \vert \vert c \vert \int ^{T}_{0} \bigl\vert u'(t) \bigr\vert ^{2}\,dt+\bigl(1+ \vert c \vert \bigr)\|e\|_{2} \biggl( \int ^{T} _{0} \bigl\vert u'(t) \bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}}, \end{aligned}$$
where \(\|e\|_{2}:= (\int ^{T}_{0}|e(t)|^{2}\,dt )^{ \frac{1}{2}}\). Obviously, \(|P|-|P||c|>0\), since \(|c|<1\). Hence, we deduce
$$ \biggl( \int ^{T}_{0} \bigl\vert u'(t) \bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}}\leq \frac{(1+|c|) \|e\|_{2}}{|P|-|P||c|}:=M_{1}'. $$
(2.8)
From Eqs. (2.3), (2.8) and the Hölder inequality, we see that
$$ \begin{aligned}[b] u(t)&=u(\xi )+ \int ^{t}_{\xi }u'(s)\,ds\leq d_{2}+\sqrt{T} \biggl( \int ^{T}_{0} \bigl\vert u'(t) \bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}} \\ &\leq d_{2}+\sqrt{T} \frac{(1+|c|) \|e\|_{2}}{|P|-|P||c|}:=M_{1}. \end{aligned} $$
(2.9)
On the other hand, from Eq. (2.2), it is clear that
$$ \begin{aligned}[b] \int ^{T}_{0} \bigl\vert g\bigl(u(t)\bigr) \bigr\vert \,dt&= \int _{g(u(t))\geq 0}g\bigl(u(t)\bigr)\,dt- \int _{g(u(t))\leq 0}g\bigl(u(t)\bigr)\,dt \\ &=2 \int _{g(u(t))\geq 0}g\bigl(u(t)\bigr)\,dt- \int ^{T}_{0}e(t)\,dt. \end{aligned} $$
(2.10)
Case (I). If \(\overline{e}:=\frac{1}{T}\int ^{T}_{0}e(t)\,dt\leq 0\), from Eq. (2.10), we have
$$ \int ^{T}_{0} \bigl\vert g\bigl(u(t)\bigr) \bigr\vert \,dt \leq 2 \int ^{T}_{0}\bigl(g^{+}\bigl(u(t) \bigr)-e(t)\bigr)\,dt, $$
where \(g^{+}(u):=\max \{g(u),0\}\). Since \(g^{+}(u(t))-e(t)\geq 0\), from condition (\(\mathrm{H}_{1}\)), we know \(u(t)\geq d_{2}\). Then we deduce
$$ \begin{aligned}[b] \int ^{T}_{0} \bigl\vert g\bigl(u(t)\bigr) \bigr\vert \,dt&\leq 2 \int ^{T}_{0}g^{+}\bigl(u(t)\bigr)\,dt+ \int ^{T}_{0} \bigl\vert e(t) \bigr\vert \,dt \\ &\leq 2T \bigl\Vert g^{+}_{M_{1}} \bigr\Vert +T^{\frac{1}{2}}\|e\|_{2}, \end{aligned} $$
(2.11)
where \(\|g^{+}_{M_{1}}\|:=\max_{d_{2}\leq u\leq M_{1}}g^{+}(u)\). As \((Au)(0)=(Au)(T)\), there exists a point \(t_{1}\in [0,T]\) such that \((Au)'(t_{1})=0\), from Eqs. (2.2), (2.8) and (2.11), we have
$$ \begin{aligned}[b] \bigl\Vert \phi _{p}\bigl((Au)'\bigr) \bigr\Vert &=\max _{t\in [t_{1},t_{1}+T]} \biggl\{ \biggl\vert \int ^{t}_{t_{1}}\bigl(\phi _{p} \bigl((Au)'(s)\bigr)\bigr)'\,ds \biggr\vert \biggr\} \\ &\leq \vert P \vert \int ^{T}_{0} \bigl\vert u'(t) \bigr\vert \,dt+ \int ^{T}_{0} \bigl\vert g\bigl(u(t)\bigr) \bigr\vert \,dt+ \int ^{T}_{0} \bigl\vert e(t) \bigr\vert \,dt \\ &\leq \vert P \vert T^{\frac{1}{2}} \biggl( \int ^{T}_{0} \bigl\vert u'(t) \bigr\vert ^{2}\,dt \biggr)^{ \frac{1}{2}}+ \int ^{T}_{0} \bigl\vert g\bigl(u(t)\bigr) \bigr\vert \,dt+ \int ^{T}_{0} \bigl\vert e(t) \bigr\vert \,dt \\ &\leq |P|T^{\frac{1}{2}} M_{1}^{\prime \frac{1}{2}}+2T \bigl\Vert g^{+}_{M_{1}} \bigr\Vert +2T ^{\frac{1}{2}}\|e \|_{2}:=M_{2}'. \end{aligned} $$
(2.12)
We claim that there exists a positive constant \(M_{2}^{**}>M_{2}'+1\) such that, for all \(t\in \mathbb{R,}\)
$$ \bigl\Vert (Au)' \bigr\Vert \leq M_{2}^{**}. $$
(2.13)
In fact, if \(x'\) is not bounded, there exists a positive constant \(M_{2}''\) such that \(\|u'\|>M_{2}''\) for some \(u'\in \mathbb{R}\). Therefor, it is clear that
$$ \bigl\Vert \phi _{p}(Au)' \bigr\Vert = \bigl\Vert \phi _{p}\bigl(Au'\bigr) \bigr\Vert = \bigl\Vert Au' \bigr\Vert ^{p-1}=\bigl(1+|c|\bigr)^{p-1} \bigl\Vert u' \bigr\Vert ^{p-1}\geq \bigl(1+ \vert c \vert \bigr)^{p-1}M_{2}^{\prime \prime p-1}:=M_{2}^{*}. $$
Then we have a contradiction. So, Eq. (2.13) holds. By Lemma 2.1 and Eq. (2.13), we get
$$ \begin{aligned}[b] \bigl\Vert u' \bigr\Vert &= \bigl\Vert A^{-1}Au' \bigr\Vert = \bigl\Vert A^{-1}(Au)' \bigr\Vert \\ &\leq \frac{ \Vert (Au)' \Vert }{1-|c|} \\ &\leq \frac{M_{2}^{**}}{1-|c|}:=M_{2}, \end{aligned} $$
(2.14)
since \(|c|<1\).
Case (II). If \(\overline{e}>0\), from Eq. (2.10), we obtain
$$ \int ^{T}_{0} \bigl\vert g\bigl(u(t)\bigr) \bigr\vert \,dt\leq 2 \int ^{T}_{0}g^{+}\bigl(u(t)\bigr)\,dt. $$
Since \(g^{+}(u(t))\geq 0\), from condition (\(\mathrm{H}_{1}\)), we know that there exists a positive constant \(d_{2}^{*}\) such that \(u(t)\geq d_{2}^{*}\). Therefore, we see that
$$\begin{aligned} \int ^{T}_{0} \bigl\vert g\bigl(u(t)\bigr) \bigr\vert \,dt&\leq 2 \int ^{T}_{0}g^{+}\bigl(u(t)\bigr)\,dt \\ &\leq 2T \bigl\Vert g^{+}_{M} \bigr\Vert , \end{aligned}$$
where \(\|g^{+}_{M}\|:=\max_{d_{2}^{*}\leq u\leq M_{1}}g^{+}(x)\). Similarly, we deduce \(\|u'\|\leq M_{2}\).
Multiplying both sides of Eq. (2.1) by \(u'(t)\) and integrating on the interval \([\xi ,t]\), where \(\xi \in [0,T]\) is defined in Eq. (2.3), we get
$$ \begin{aligned}[b] \lambda \int ^{u(t)}_{u(\xi )}g(v)\,dv&=\lambda \int ^{t}_{\xi }g\bigl(u(s)\bigr)u'(s) \,ds \\ &=- \int ^{t}_{\xi }\bigl(\phi _{p} \bigl((Au)'(s)\bigr)\bigr)'u'(s)\,ds- \lambda P \int ^{t}_{ \xi }|u'(s)|^{2}\,ds \\ &\quad {}+\lambda \int ^{t}_{\xi }e(s)u'(s)\,ds. \end{aligned} $$
(2.15)
Furthermore, from Eqs. (2.12), (2.8) and (2.14), we get
$$\begin{aligned} \lambda \biggl\vert \int ^{u(t)}_{u(\xi )}g(v)\,dv \biggr\vert \leq & \int ^{t}_{ \xi } \bigl\vert \bigl(\phi _{p}\bigl((Au)'(s)\bigr)\bigr)' \bigr\vert \bigl\vert u'(s) \bigr\vert \,ds+\lambda P \int ^{t}_{\xi } \bigl\vert u'(s) \bigr\vert ^{2}\,ds \\ &{}+\lambda \int ^{t}_{\xi } \bigl\vert e(s) \bigr\vert \bigl\vert u'(s) \bigr\vert \,ds \\ \leq &\lambda M_{2}M_{2}'+\lambda \vert P \vert \bigl(M_{1}'\bigr)^{2}+\lambda M_{2}T^{ \frac{1}{2}}\|e\|_{2}:=\lambda M_{3}'. \end{aligned}$$
From the repulsive condition (\(g_{1}\)), we know that there exists a constant \(M_{3}>0\) such that
$$ u(t)\geq M_{3}, \quad \forall t\in [\xi ,T]. $$
(2.16)
Similarly, we can discuss \(t\in [0,\xi ]\).
From Eqs. (2.3), (2.9), (2.14) and (2.16), it is obvious that a periodic solution u to Eq. (2.1) satisfies
$$ E_{1}< u(t)< E_{2}, \qquad \bigl\Vert u' \bigr\Vert < E_{3}. $$
Then the condition (1) of Lemma 2.2 is satisfied. For a possible solution C to the equation
$$ g(C)-\frac{1}{T} \int ^{T}_{0}e(t)\,dt=0, $$
we have \(C\in (E_{1},E_{2})\). Hence, the condition (2) of Lemma 2.2 holds. Finally, it is clear that the condition (3) of Lemma 2.2 is also satisfied. In fact, from condition (\(\mathrm{H}_{1}\)), we can get
$$ g(E_{1})-\frac{1}{T} \int ^{T}_{0}e(t)\,dt< 0 $$
and
$$ g(E_{2})-\frac{1}{T} \int ^{T}_{0}e(t)\,dt>0. $$
Using Lemma 2.2, it is concluded that Eq. (1.2) has at least one positive periodic solution. □
Proof of Theorem 1.2
Proof of Theorem 1.2
The same strategy and notation are followed as in the proof of Theorem 1.1. Then we see that
$$ u(t)\leq d_{2}+ \int ^{T}_{0} \bigl\vert u'(t) \bigr\vert \,dt. $$
(2.17)
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. □