Theorem 3.1
For problem (1.3), assume the following conditions hold:
-
\((g_{1})\)
:
-
(Balance condition) There exist positive constants
\(A_{1}\)
and
\(A_{2}\)
with
\(A_{1}< A_{2}\)
such that if
x
is a positive continuous
T-periodic function satisfying
\(\int_{0}^{T} g(t,x(t),x(t-\tau(t)))\,dt=0\), then
$$A_{1}\leq x(\varepsilon)\leq A_{2} $$
for some
\(\varepsilon\in[0,T]\).
-
\((g_{2})\)
:
-
(Degree condition) \(\overline{g}(x)<0\)
for all
\(x\in(0,A_{1})\)
and
\(\overline{g}(x)>0\)
for all
\(x>A_{2}\), where
\(\overline{g}(x)=\frac{1}{T} \int_{0}^{T} g(t,x(t),x(t-\tau(t)))\,dt\), \(x>0\).
-
\((g_{3})\)
:
-
(Decomposition condition) \(g(t,x(t),x(t-\tau (t)))=g_{1}(t,x(t-\tau(t)))+g_{0}(x(t))\), where
\(g_{0}\)
is a continuous function and there exist positive constants
\(a_{i}\), \(c_{i}\), \(i=1, 2\), and
b
such that
$$g \bigl(t,x(t),x \bigl(t-\tau(t) \bigr) \bigr)\leq a_{1}x(t)+a_{2}x \bigl(t-\tau(t) \bigr)+b, \quad (t,x)\in [0,T]\times(0,+\infty). $$
Meanwhile, \(\vert g_{1}(t,x)\vert \leq c_{1}x+c_{2}\).
-
\((g_{4})\)
:
-
(Strong force condition at
\(x=0\)) \(\int_{0}^{1} g_{0}(x)\,dx=-\infty\).
-
\((g_{5})\)
:
-
\(B:= ( \int_{0}^{T}\vert p(t)\vert ^{2}\,dt )^{\frac {1}{2}}+\sup_{t\in[0,T]}\vert p(t)\vert <+\infty\), \(\vert \beta \vert >c_{1}T\), and
$$\vert \beta \vert M_{1}+T \bigl[2(a_{1}+a_{2})M_{1}+2b+B \bigr]< 1, $$
where
\(M_{1}= A_{2}+ \frac{c_{1}A_{2}T+c_{2}T+B\sqrt{T}}{ \vert \beta \vert -c_{1}T}\).
Then equation (1.3) has at least one positive
T-periodic solution.
Proof
Let \(\Omega_{1}=\{x\in\bar{\Omega},Lx=\lambda Nx,\forall\lambda \in(0,1)\}\). If, \(\forall x\in\Omega_{1}\), we have
$$ \textstyle\begin{cases} u'(t)=\lambda\frac{\varphi_{q}(v(t))}{\sqrt{1-\varphi _{q}^{2}(v(t))}}=\lambda\phi(v(t)),\\ v'(t)=-\lambda\beta\phi(v(t))-\lambda g(t,u(t),u(t-\tau (t)))+\lambda p(t), \end{cases} $$
(3.1)
where \(v(t)=\phi^{-1} (\frac{u'(t)}{\lambda} )=\varphi _{p} (\frac{\frac{1}{\lambda}u'(t)}{\sqrt{1+\frac {(u'(t))^{2}}{\lambda^{2}}}} )\).
Integrating the second equation of (3.1) from 0 to T, we have
$$ \int_{0}^{T}g \bigl(t,u(t), u \bigl(t-\tau(t) \bigr) \bigr)\,dt=0. $$
(3.2)
Combining with \((g_{1})\), we can see that there exist positive constants \(A_{1}\), \(A_{2}\), and \(\varepsilon\in[0,T]\) such that
$$ A_{1}\leq u(\varepsilon)\leq A_{2}. $$
(3.3)
Therefore, we have
$$ \begin{aligned}[b] \vert u\vert _{0}&= \max _{t\in[0,T]} \bigl\vert u(t) \bigr\vert \leq\max _{t\in [0,T]} \biggl\vert u(\varepsilon)+ \int_{\varepsilon}^{t}u'(s)\,ds \biggr\vert \\ &\leq A_{2}+ \int_{0}^{T} \bigl\vert u'(s) \bigr\vert \,ds \\ &\leq A_{2}+\sqrt{T} \bigl\Vert u' \bigr\Vert _{2}. \end{aligned} $$
(3.4)
Multiplying the second equation of (3.1) by \(u'(t)\) and integrating on the interval \([0,T]\), we can get
$$\begin{aligned} 0&= \int_{0}^{T}v'(t)u'(t)\,dt \\ & = - \int _{0}^{T}\beta \bigl(u'(t) \bigr)^{2}\,dt-\lambda \int_{0}^{T}g \bigl(t,u(t), u \bigl(t-\tau(t) \bigr) \bigr)u'(t)\,dt+\lambda \int_{0}^{T} p(t)u'(t)\,dt. \end{aligned}$$
It follows from \((g_{3})\) that
$$\begin{aligned} \int_{0}^{T}\beta \bigl(u'(t) \bigr)^{2}\,dt&=- \int_{0}^{T}g \bigl(t,u(t), u \bigl(t-\tau(t) \bigr) \bigr)u'(t)\,dt+\lambda \int_{0}^{T} p(t)u'(t)\,dt \\ &=- \int_{0}^{T} \bigl[ g_{1} \bigl(t,u \bigl(t-\tau(t) \bigr) \bigr)+g_{0} \bigl(u(t) \bigr) \bigr]u'(t) \,dt+\lambda \int_{0}^{T} p(t)u'(t)\,dt \\ &=- \int_{0}^{T}g_{1} \bigl(t,u(t), u \bigl(t- \tau(t) \bigr) \bigr)u'(t)\,dt+\lambda \int _{0}^{T} p(t)u'(t)\,dt, \end{aligned}$$
i.e.,
$$\begin{aligned} \vert \beta \vert \int_{0}^{T} \bigl\vert u'(t) \bigr\vert ^{2}\,dt &\leq \int_{0}^{T} \bigl\vert g_{1} \bigl(t,u \bigl(t-\tau(t) \bigr)\bigr) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt+ \int_{0}^{T} \bigl\vert p(t) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt \\ &\leq \int_{0}^{T} \bigl(c_{1} \bigl\vert u \bigl(t-\tau(t) \bigr) \bigr\vert +c_{2} \bigr) \bigl\vert u'(t) \bigr\vert \,dt+ \int_{0}^{T} \bigl\vert p(t) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt \\ &\leq c_{1}\vert u\vert _{0}\sqrt{T} \bigl\Vert u' \bigr\Vert _{2}+c_{2}\sqrt{T} \bigl\Vert u' \bigr\Vert _{2}+B \bigl\Vert u' \bigr\Vert _{2}, \end{aligned}$$
which together with (3.4) gives
$$ \begin{aligned}[b]\vert \beta \vert \bigl\Vert u' \bigr\Vert ^{2}_{2} &\leq c_{1}\vert u\vert _{0}\sqrt{T} \bigl\Vert u' \bigr\Vert _{2}+c_{2}\sqrt{T} \bigl\Vert u' \bigr\Vert _{2}+B \bigl\Vert u' \bigr\Vert _{2} \\ &\leq c_{1} \bigl[ A_{2}+\sqrt{T} \bigl\Vert u' \bigr\Vert _{2} \bigr]\sqrt {T} \bigl\Vert u' \bigr\Vert _{2}+c_{2}\sqrt{T} \bigl\Vert u' \bigr\Vert _{2}+B \bigl\Vert u' \bigr\Vert _{2} \\ &= c_{1}T \bigl\Vert u' \bigr\Vert ^{2}_{2}+ (c_{1}A_{2} \sqrt{T}+c_{2}\sqrt {T}+B ) \bigl\Vert u' \bigr\Vert _{2}. \end{aligned} $$
(3.5)
It follows from \(\vert \beta \vert >aT\) that
$$ \bigl\Vert u' \bigr\Vert _{2}\leq\frac{c_{1}A_{2}\sqrt{T}+ c_{2}\sqrt{T}+B}{\vert \beta \vert -c_{1}T}. $$
(3.6)
Substituting (3.6) into (3.4), we obtain
$$ \vert u\vert _{0}\leq A_{2}+ \frac{c_{1}A_{2}T+c_{2}T+B\sqrt{T}}{ \vert \beta \vert -c_{1}T}:=M_{1}. $$
(3.7)
Furthermore, from the second equation of (3.1), we can get
$$ \begin{aligned}[b] \int_{0}^{T} \bigl\vert v'(t) \bigr\vert \,dt &\leq \int_{0}^{T}\vert \beta \vert \bigl\vert u'(t) \bigr\vert \,dt+ \lambda \int_{0}^{T} \bigl\vert g \bigl(t,u(t),u \bigl(t- \tau(t) \bigr) \bigr) \bigr\vert \,dt \\ &\quad{} +\lambda \int_{0}^{T} \bigl\vert p(t) \bigr\vert \,dt. \end{aligned} $$
(3.8)
Write
$$\begin{aligned}& I_{+}= \bigl\{ t\in[0,T]:g \bigl(t,u(t), u \bigl(t-\tau(t) \bigr) \bigr) \geq0 \bigr\} ; \\& I_{-}= \bigl\{ t\in[0,T]:g \bigl( t,u(t),u \bigl(t-\tau(t) \bigr) \bigr) \leq0 \bigr\} . \end{aligned}$$
Then we can get from (3.2) and \((g_{3})\)
$$ \begin{aligned}[b] & \int_{0}^{T} \bigl\vert g \bigl(t,u(t),u \bigl(t- \tau(t) \bigr) \bigr) \bigr\vert \,dt \\ &\quad = \int_{I_{+}}g \bigl(t,u(t),u \bigl(t-\tau(t) \bigr) \bigr)\,dt- \int _{I_{-}}g \bigl(t,u(t),u \bigl(t-\tau(t) \bigr) \bigr)\,dt \\ &\quad = 2 \int_{I_{+}}g \bigl(t,u(t),u \bigl(t-\tau(t) \bigr) \bigr)\,dt \\ &\quad \leq2 a_{1} \int_{0}^{T}u(t)\,dt+2a_{2} \int_{0}^{T} u \bigl(t-\tau(t) \bigr)\,dt+2 \int_{0}^{T}b\,dt \\ &\quad \leq2(a_{1}+a_{2})T \vert u\vert _{0}+2bT. \end{aligned} $$
(3.9)
Substituting (3.9) into (3.8) and combining with (3.6) and (3.7), we obtain
$$ \begin{aligned}[b] \int_{0}^{T} \bigl\vert v'(t) \bigr\vert \,dt&\leq \vert \beta \vert \sqrt{T} \bigl\Vert u' \bigr\Vert _{2} + \lambda \bigl[ 2(a_{1}+a_{2})T \vert u\vert _{0}+2bT \bigr] + \lambda BT \\ &\leq \vert \beta \vert M_{1}+T \bigl[ 2(a_{1}+a_{2})M_{1}+2b+B \bigr]. \end{aligned} $$
(3.10)
Integrating the first equation of (3.1) on the interval \([0,T]\), we have
$$\begin{aligned} \int_{0}^{T}\frac{\varphi_{q}(v(t))}{\sqrt{1-\varphi_{q}^{2}(v(t))}}\,dt=0. \end{aligned}$$
Then we can see that there exists \(\eta\in[0,T]\) such that \(v(\eta )=0\). It implies that
$$\begin{aligned} \bigl\vert v(t) \bigr\vert = \biggl\vert \int_{\eta}^{t} v'(s)\,ds+v(\eta) \biggr\vert \leq \int _{0}^{T} \bigl\vert v'(s) \bigr\vert \,ds, \end{aligned}$$
which combining with (3.10) gives
$$\begin{aligned} \bigl\vert v(t) \bigr\vert &\leq \int_{0}^{T} \bigl\vert v'(s) \bigr\vert \,ds \\ &\leq \vert \beta \vert M_{1}+T \bigl[2(a_{1}+a_{2})M_{1}+2b+B \bigr] \\ &:=\rho. \end{aligned}$$
Since \(\vert \beta \vert M_{1}+T(2(a_{1}+a_{2})M_{1}+2b+B)<1\), we have
$$ \vert v\vert _{0}=\max_{t\in[0,T]} \bigl\vert v(t) \bigr\vert \leq\rho< 1. $$
(3.11)
From (3.1), we can also have
$$ \bigl\vert u' \bigr\vert _{0} \leq\lambda\cdot\max _{t\in[0,T]}\frac {\vert v(t)\vert ^{q-1}}{\sqrt{1-v^{2(q-1)}}(t)}\leq\frac{\lambda\rho ^{q-1}}{1-\rho^{2(q-1)}}. $$
(3.12)
On the other hand, from the second equation of (3.1) and by \((g_{3})\), we can see that
$$\begin{aligned} v'(t)=- \beta u'(t)-\lambda g \bigl(t,u(t),u \bigl(t- \tau(t) \bigr) \bigr)+\lambda p(t). \end{aligned}$$
Take \(\omega\in[0,T]\), then
$$ \begin{aligned}[b] v'(\omega)&=- \beta u'( \omega)-\lambda g \bigl(\omega,u(\omega),u \bigl(\omega -\tau(\omega) \bigr) \bigr)+ \lambda p(\omega) \\ &=- \beta u'(\omega)-\lambda \bigl[g_{1} \bigl(\omega, u \bigl(\omega-\tau(\omega ) \bigr) \bigr)+g_{0} \bigl(u(\omega) \bigr) \bigr]+\lambda p(\omega). \end{aligned} $$
(3.13)
Multiplying both sides of equation (3.13) by \(u'(\omega)\), we have
$$ \begin{aligned}[b] v'(\omega)u'(\omega) &=- \beta u'(\omega)u'(\omega)-\lambda \bigl[g_{1} \bigl(\omega, u \bigl(\omega-\tau(\omega) \bigr) \bigr)+g_{0} \bigl(u( \omega) \bigr) \bigr] u'(\omega) \\ &\quad{} +\lambda p(\omega)u'(\omega). \end{aligned} $$
(3.14)
Let \(\varepsilon\in[0,T]\) be as in (3.3). For any \(\omega\in [\varepsilon,T]\), integrating equation (3.14) on the interval \([\varepsilon,T]\), we have
$$ \begin{aligned}[b] \lambda \int_{u(\varepsilon)}^{u(\omega)}g_{0}(u)\,du &= \lambda \int _{\varepsilon}^{\omega}g_{0} \bigl(u(\omega) \bigr)u'(\omega)\,d\omega \\ &=- \int_{\varepsilon}^{\omega}v'(\omega)u'( \omega)\,d\omega - \int_{\varepsilon}^{\omega}\beta \bigl(u'(\omega) \bigr)^{2} \,d\omega \\ &\quad{} -\lambda \int_{\varepsilon}^{\omega}g_{1} \bigl(\omega,u \bigl( \omega-\tau (\omega) \bigr) \bigr)u'(\omega)\,d\omega+\lambda \int_{\varepsilon}^{\omega} p(\omega)u'(\omega)\,d \omega. \end{aligned} $$
(3.15)
By (3.12) and (3.15), we get
$$\begin{aligned}& \lambda \biggl\vert \int_{u(\varepsilon)}^{u(\omega)}g_{0}(u)\,du \biggr\vert \\& \quad = \lambda \biggl\vert \int_{\varepsilon}^{\omega}g_{0} \bigl(u(\omega) \bigr)u'(\omega )\,d\omega \biggr\vert \\& \quad \leq \int_{\varepsilon}^{\omega} \bigl\vert v'(\omega )u'(\omega) \bigr\vert \,d\omega + \int_{\varepsilon}^{\omega} \bigl\vert \beta \bigl(u'( \omega) \bigr)^{2} \bigr\vert \,d\omega \\& \quad \quad{}+\lambda \int_{\varepsilon}^{\omega} \bigl\vert g_{1} \bigl( \omega,u \bigl(\omega -\tau(\omega) \bigr) \bigr)u'(\omega) \bigr\vert \,d\omega+\lambda \int_{\varepsilon }^{\omega} \bigl\vert p(\omega)u'( \omega) \bigr\vert \,d\omega \\& \quad \leq \bigl\vert u' \bigr\vert _{0} \int_{\varepsilon}^{\omega} \bigl\vert v'(t) \bigr\vert \,dt+\vert \beta \vert T \bigl\vert u' \bigr\vert ^{2}_{0}+\lambda G_{M_{1}}T \bigl\vert u' \bigr\vert _{0}+\lambda T\vert p\vert _{0} \bigl\vert u' \bigr\vert _{0} \\& \quad \leq \bigl\vert u' \bigr\vert _{0} \bigl[ \vert \beta \vert M_{1}+2T(a_{1}+a_{2})M_{1}+2bT+BT \bigr]+\vert \beta \vert T \bigl\vert u' \bigr\vert _{0} \\& \quad\quad{} +\lambda TG_{M_{1}}+\lambda T\vert p\vert _{0} ] \\& \quad \leq\frac{\lambda\rho^{q-1}}{1-\rho^{2(q-1)}} \biggl[\vert \beta \vert M_{1} +2T(a_{1}+a_{2})M_{1}+2bT+BT \\& \quad\quad{} +\vert \beta \vert T\frac{\lambda\rho^{q-1}}{1-\rho^{2(q-1)}}+\lambda TG_{M_{1}}+ \lambda T\vert p\vert _{0} \biggr], \end{aligned}$$
(3.16)
where \(G_{M_{1}}=\max_{\vert u\vert \leq M_{1}}g_{1}(t,u)\). It follows from (3.16) that
$$\begin{aligned} \biggl\vert \int_{u(\varepsilon)}^{u(\omega)}g_{0}(u)\,du \biggr\vert < + \infty. \end{aligned}$$
According to \((g_{4})\), we can see that there exists a constant \(M_{2}>0\) such that, for \(\omega\in[\varepsilon,T]\),
$$ u(\omega)\geq M_{2}. $$
(3.17)
For the case \(\omega\in[0,\varepsilon]\), we can handle it similarly. Thus, we have
$$\begin{aligned} u(t)\geq M_{2}, \quad \forall t\in[0,T]. \end{aligned}$$
(3.18)
Let us define
$$0< D_{1}=\min\{A_{1},M_{2}\} \quad\text{and}\quad D_{2}=\max\{A_{2},M_{1}\}. $$
Then by (3.3), (3.7), and (3.18) we can obtain
$$ D_{1}\leq u(t)\leq D_{2}. $$
(3.19)
Set
$$ \Omega= \biggl\{ x=(u,v)^{\top}\in X:\frac{D_{1}}{2}< u(t)< D_{2}+1,\vert v\vert _{0}< \rho_{1}< \frac{\rho+1}{2} \biggr\} . $$
Then the condition (1) of Lemma 2.1 is satisfied. Suppose that there exists \(x\in\partial\Omega\cap\ker L\) such that \(QN x=\frac{1}{T}\int_{0}^{T}{N}x(s)\,ds=(0,0)^{\top}\), i.e.,
$$ \textstyle\begin{cases} \frac{1}{T}\int_{0}^{T}\frac{\varphi_{q}(v(t))}{\sqrt {1-\varphi_{q}^{2}(v(t))}}\,dt=0,\\ \frac{1}{T}\int_{0}^{T} [-\beta\frac{\varphi _{q}(v(t))}{\sqrt{1-\varphi_{q}^{2}(v(t))}}-g(t, u(t),u(t-\tau(t) ))+ p(t) ]\,dt=0. \end{cases} $$
(3.20)
Since \(\ker L=\mathbb {R}^{2}\), and \(u\in{ \mathbb {R}}\), \(v\in{ \mathbb {R}}\) are constant, combining with the first equation of (3.20), we obtain
$$\begin{aligned} v=0< \rho_{1}. \end{aligned}$$
From the second equation of (3.20), we have
$$\begin{aligned} \frac{1}{T} \int_{0}^{T}g \bigl(t,u(t),u \bigl(t-\tau(t) \bigr) \bigr)\,dt=0. \end{aligned}$$
From \((g_{1})\) we can see that
$$\begin{aligned} \frac{D_{1}}{2}< D_{1}< A_{1} \leq u(t)\leq A_{2}< D_{2}< D_{2}+1, \end{aligned}$$
which contradicts the assumption \(x\in\partial\Omega\). So, for all \(x\in\ker L\cap\partial\Omega\), we have \(QN x\neq 0\). Then the condition (2) of Lemma 2.1 is satisfied.
In the following, we prove that the condition (3) of Lemma 2.1 is also satisfied.
Let
$$\begin{aligned} z=Kx=K \begin{pmatrix}u \\v \end{pmatrix} = \begin{pmatrix} u-\frac{A_{1}+A_{2}}{2} \\v \end{pmatrix} , \end{aligned}$$
then we have
$$\begin{aligned} x=z+ \begin{pmatrix} \frac{A_{1}+A_{2}}{2} \\ v \end{pmatrix} . \end{aligned}$$
Define \(J:\operatorname {Im}Q\rightarrow\ker L\) to be a linear isomorphism with
$$\begin{aligned} x=z+ \begin{pmatrix} v \\ -u \end{pmatrix} , \end{aligned}$$
and define
$$\begin{aligned} H(\mu,x)=\mu K x+(1-\mu)JQNx, \quad \forall(x,\mu)\in\Omega\times[0,1]. \end{aligned}$$
Then
$$ H(\mu,x)= \begin{pmatrix} \mu u-\frac{\mu(A_{1}+A_{2})}{2} \\ \mu v \end{pmatrix} + \frac{1-\mu}{T} \begin{pmatrix} \int_{0}^{T} [\frac{cv^{q-1} }{\sqrt{1-v^{2(q-1)}}}+g(t, u(t),u(t-\tau(t))) ]\,dt \\ \int_{0}^{T}\frac{v^{q-1}}{\sqrt{1-v^{2(q-1)}}}\,dt \end{pmatrix} . $$
(3.21)
Now we claim that \(H(\mu,x)\) is a homotopic mapping. Assume, by way of contradiction, that there exist \(\mu_{0}\in[0,1]\) and \(x_{0}=\bigl ({\scriptsize\begin{matrix}{}u_{0}\cr v_{0}\end{matrix}} \bigr ) \in\partial\Omega\) such that \(H(\mu_{0},x_{0})=0\).
Substituting \(\mu_{0}\) and \(x_{0}\) into (3.21), we have
$$ H(\mu,x)= \begin{pmatrix} \mu_{0} u_{0}-\frac{\mu_{0}(A_{1}+A_{2})}{2}+(1-\mu_{0}) \frac {cv^{q-1}_{0}}{\sqrt{1-v_{0}^{2(q-1)}}}+(1-\mu_{0}) \overline{g}(u_{0}) \\ \mu_{0} v_{0}+(1-\mu_{0})\frac{v^{q-1}_{0}}{\sqrt{1-v_{0}^{2(q-1)}}} \end{pmatrix} . $$
(3.22)
Since \(H(\mu_{0},x_{0})=0\), we can see that
$$\begin{aligned} \mu_{0} v_{0}+(1-\mu_{0})\frac{v^{q-1}_{0}}{\sqrt{1-v_{0}^{2(q-1)}}}=0. \end{aligned}$$
Combining with \(\mu_{0}\in[0,1]\), we obtain \(v_{0}=0\). Thus \(u_{0}=A_{1}\) or \(A_{2}\).
If \(u_{0}=A_{1}\), it follows from \((g_{2})\) that \(g (u_{0})<0\), then substituting \(v_{0}=0\) into (3.22), we have
$$ \begin{aligned}[b] &\mu_{0} u_{0}- \frac{\mu_{0}(A_{1}+A_{2})}{2}+(1-\mu_{0}) \frac{\beta v^{q-1}_{0}}{\sqrt{1-v_{0}^{2(q-1)}}}+(1- \mu_{0})\overline{g}(u_{0}) \\ &\quad =\mu_{0} u_{0}-\frac{\mu_{0}(A_{1}+A_{2})}{2} +(1- \mu_{0}) \overline{g}(u_{0}) \\ &\quad < \mu_{0} \biggl(u_{0}-\frac{A_{1}+A_{2}}{2} \biggr)< 0. \end{aligned} $$
(3.23)
If \(u_{0}=A_{2}\), it follows from \((g_{2})\) that \(g (u_{0})>0\), then substituting \(v_{0}=0\) into (3.22), we have
$$ \begin{aligned}[b] &\mu_{0} u_{0}- \frac{\mu_{0}(A_{1}+A_{2})}{2}+(1-\mu_{0}) \frac{\beta v^{q-1}_{0}}{\sqrt{1-v_{0}^{2(q-1)}}}+(1- \mu_{0})\overline{g}(u_{0}) \\ &\quad =\mu_{0} u_{0}-\frac{\mu_{0}(A_{1}+A_{2})}{2} +(1- \mu_{0}) \overline{g}(u_{0}) \\ &\quad >\mu_{0} \biggl(u_{0}-\frac{A_{1}+A_{2}}{2} \biggr)>0. \end{aligned} $$
(3.24)
Combining with (3.23) and (3.24), we can see that \(H(\mu _{0},x_{0})\neq0\), which contradicts the assumption. Therefore \(H(\mu,x)\) is a homotopic mapping and \(x^{\top}H(\mu,x)\neq0\), \(\forall(x,\mu )\in(\partial\Omega\cap\ker L)\times[0,1]\). Then
$$\begin{aligned} \deg( JQN,\Omega\cap\ker L,0)&=\deg \bigl(H(0,x),\Omega\cap\ker L,0 \bigr) \\ &=\deg \bigl(H(1,x),\Omega\cap\ker L,0 \bigr) \\ &=\deg(Kx,\Omega\cap\ker L,0) \\ &=\sum_{x\in K^{-1}(0)}\operatorname {sgn}\bigl\vert K'(x) \bigr\vert \\ &=1\neq0. \end{aligned}$$
Thus, the condition (3) of Lemma 2.1 is also satisfied. Therefore, by applying Lemma 2.1, we can conclude that equation (1.3) has at least one positive T-periodic solution. □