Proof. If $f_1(u,v)=0$ , then

\begin{equation*}u=0 \quad \mbox {or} \quad r\left (1-\dfrac {u}{K}\right )-\dfrac {cv}{m+bu}=0.\end{equation*}

It is easy to obtain that $v=r\left (1-\dfrac {u}{K}\right )(m+bu)/c$ has a maximum point $(u_M,v_M)$ , where

(2.1) \begin{equation} u_M=\dfrac {Kb-m}{2b},\quad v_M=\dfrac {r\left (1-\dfrac {u_M}{K}\right )(m+bu_M)}{c}. \end{equation}

When $Kb\gt m$ , then $u_M\gt 0$ , which shows that

(2.2) \begin{equation} v=p(u)=r\left (1-\dfrac {u}{K}\right )(m+bu)/c \end{equation}

is monotone increasing in $(\!-\infty, u_M)$ , while monotone decreasing in $(u_M,+\infty )$ . As a result, for $v\in (p(0),v_M)=(rm/c,v_M)$ , $u = h_{1}(v)$ is monotone increasing with respect to $v$ , and for $v\in (\!-\infty, v_M)$ , $u=h_2(v)$ is monotone decreasing with respect to $v$ . Direct calculations provide $u_M\lt K$ . Based on the expression of $v_M$ in (2.1), we easily deduce that $v_M\gt 0$ .

Proof. We omit the details because the proof is elementary.

Proof. (i) Obviously,

\begin{equation*}f_2(h_0(v),v)=-av\lt 0 \quad \mbox {for}\quad v\in (0,+\infty ).\end{equation*}

Then, we find that $u=h_2(v)\gt u^*_2$ for $v\in (\!-\infty, v^*_2)$ and $\beta cu/(m+bu)$ is monotone increasing with respect to $u$ . So

\begin{equation*}\dfrac {\beta ch_2(v)}{m+bh_2(v)}-a\gt \dfrac {\beta cu^*_2}{m+bu^*_2}-a=0.\end{equation*}

This shows that $f_2(h_2(v),v)\gt 0$ for $v\in (0,v^*_2)$ .

(ii) By direct calculations, we get

\begin{equation*}\frac {d}{dv}f_2(h_0(v),v)=-a\lt 0 \quad \mbox {for}\quad v\in (\!-\infty, +\infty ),\end{equation*}

\begin{equation*}\frac {d}{dv}f_2(h_2(v),v)=\left (\dfrac {\beta ch_2(v)}{m+bh_2(v)}-a\right )+v\dfrac {\beta cmh^{'}_R(v)}{(m+bh_2(v))^2}.\end{equation*}

Recall that $h^{'}_2(v)\lt 0$ for $v\in (\!-\infty, v_M)$ from the Proposition2.1 proof. Combining this with $-a+\beta ch_2(v^*_2)/(m+bh_2(v^*_2))=0,$ then we get $\frac {d}{dv}f_2(h_2(v),v)|_{v=v^*_2}\lt 0.$ By continuity, there exists a constant $\widetilde {d}\in (0,v^*_2)$ such that $\frac {d}{dv}f_2(h_2(v),v)\lt 0$ for $(\widetilde {d},v^*_2]$ .

Remark 1. It is easy to find that every regular solution of problem (3.1) satisfies

\begin{equation*}f(u(x),v(x))=f(\theta (v(x)),v(x))=0\quad \mbox {for all}\quad x\in \Omega, \end{equation*}

where $v=v(x)$ is a solution of the elliptic Neumann problem

(3.2) \begin{equation} d_2\Delta _\tau v+P(v)=0 \quad \mbox {for}\quad x\in \Omega \end{equation}

with $P(v)=g(\theta (v),v).$

Proof. We prove this lemma using the Rabinowitz bifurcation theorem of the variational equation [Reference Rabinowitz23]. Then, assume that

(i) $M$ is a real Hilbert space,

(ii) $X\in C^2(M,R)$ with $X'(u)=Lu+Z(u)$ ,

(iii) $L$ is linear and $Z(u)=o(||u||)$ at $u=0,$

(iv) $\lambda$ is an isolated eigenvalue of $L$ of a finite multiplicity. If these assumptions hold, by [Reference Rabinowitz23], we know that $(\lambda, 0)\in R\times M$ is a bifurcation point of

(3.4) \begin{equation} A(\mu, v)\equiv Lv+Z(v)-\mu v=0. \end{equation}

Thus, for $||v||\neq 0$ , the solution $(\mu, v)$ of the equation (3.4) is present in each neighbourhood of $(\lambda, 0)$ . And we apply the usual Sobolev space $M=W^{1,2}(\Omega )$ with the equivalent scalar product

\begin{equation*}\left \langle u,v\right \rangle _{W^{1,2}(\Omega )}=\int _\Omega \nabla u\cdot \nabla v dx+\int _\Omega uv dx\end{equation*}

and the functional

\begin{equation*}X(v)=\dfrac {\lambda _k+d_2}{2}\int _\Omega v^2dx+\int _\Omega S(v)dx\end{equation*}

with $S(v)=\int _{0}^{v}y(s)ds$ . It is easy to find that $X\in C(W^{1,2}(\Omega ),R)$ and $X(v)$ is differentiable in the $Fr\acute {e}chet$ sense for each $v\in W^{1,2}(\Omega )$ . By simple calculation, we have

\begin{equation*}DX(v)\zeta =(\lambda _k+d_2)\int _\Omega v\zeta dx+\int _\Omega y(v)\zeta dx\end{equation*}

with $DX\in C(W^{1,2}(\Omega )$ , $\mbox {Lin}(W^{1,2}(\Omega ),R)).$ The second $Fr\acute {e}chet$ derivative at the point $v\in W^{1,2}(\Omega )$ is represented by the bilinear form

\begin{equation*}\left \langle D^2X(v)\zeta, \kappa \right \rangle =(\lambda _k+d_2)\int _\Omega \zeta \kappa dx+\int _\Omega y'(v)\zeta \kappa dx.\end{equation*}

Next, we prove that $D^2X(v)\in C\left (W^{1,2}(\Omega ),\mbox {Lin}(W^{1,2}(\Omega ),\mbox {Lin}(W^{1,2}(\Omega ),R))\right )$ . For $v_n\rightarrow v$ in $W^{1,2}(\Omega )$ and $\zeta, \kappa \in W^{1,2}(\Omega )$ , we estimate

\begin{align*}|\left \langle (D^2X(v_n)-D^2X(v))\zeta, \kappa \right \rangle |&\le \int _\Omega |y'(v_n)-y'(v)||\zeta ||\kappa |dx \\&\le ||y''||_\infty \int _\Omega |v_n-v||\zeta ||\kappa |dx\\&\le ||y''||_\infty ||v_n-v||_3||\zeta ||_3||\kappa ||_3\\&\le ||y''||_\infty ||v_n-v||_{W^{1,2}}||\zeta ||_{W^{1,2}}||\kappa ||_{W^{1,2}}.\end{align*}

The last inequality comes from the Sobolev embedding assuming $N\le 6$ .

Particularly, we have for each test function $\zeta \in W^{1,2}(\Omega )$

\begin{equation*}X'(v)(\zeta )=(\lambda _k+d_2)\int _\Omega v\zeta dx+\int _\Omega y(v)\zeta dx\equiv Lv(\zeta )+H(v)(\zeta ).\end{equation*}

Therefore, we can find that $H(v)=o(||v||_{W^{1,2}})$ as $||v||_{W^{1,2}}\rightarrow 0$ by assuming $y=y(v)$ . It is easy to find that $\mu =1$ is an isolated eigenvalue of the operator $L$ with finite multiplicity. So, we obtain

\begin{equation*}Lv(\zeta )=\left \langle v,\zeta \right \rangle _{W^{1,2}(\Omega )}\quad \mbox {for all}\quad \zeta \in W^{1,2}(\Omega ),\end{equation*}

that is

\begin{equation*}(\lambda _k+d_2)\int _\Omega v\zeta dx=\int _\Omega \nabla v\cdot \nabla \zeta dx+\int _\Omega v\zeta dx\quad \mbox {for all}\quad \zeta \in W^{1,2}(\Omega ).\end{equation*}

Obviously, we can reduce to the eigenvalue problem for $\Delta _\tau$ . Now, the property that $\lambda _k$ is an isolated eigenvalue with finite multiplicity is applied. So, by the Rabinowitz Theorem [Reference Rabinowitz23], we can find that $(1,0)$ is a bifurcation point of (3.4) which means that there is a sequence of numbers $d_s\rightarrow d_2$ and nonzero $\left \{ {v_s}\right \}\subset W^{1,2}(\Omega )$ such that $||v_s||_{W^{1,2}}\rightarrow 0$ , satisfying

\begin{equation*}Lv_s(\zeta )+Z(v_s)(\zeta )-d_s\left \langle v_s,\zeta \right \rangle _{1,2}=0\quad \mbox {for all}\quad \zeta \in W^{1,2}(\Omega ),\end{equation*}

which is equivalent to the equation satisfied by the weak solutions $v_s\in W^{1,2}(\Omega )$ to problem (3.3)

\begin{equation*}-d_s\int _\Omega \nabla v_s\cdot \nabla \zeta dx+(\lambda _k+d_2-d_s)\int _\Omega v_s\zeta dx+\int _\Omega y(v_s)\zeta dx=0\end{equation*}

for all $\zeta \in W^{1,2}(\Omega )$ .

Proof. We construct non-constant solutions to the reaction–diffusion–ODE system by Proposition3.1. In the following, we define an open ball with a radius of $\rho \gt 0$ that is centred at $\overline v$ as $B_\rho (\overline v)$ . First, we show that only a finite number of $v_s$ can be constant in Proposition3.1. If there is a constant subsequence $\left \{ {v_{l_n}}\right \}$ that satisfies the equation (3.3) such that $v_{l_n}\rightarrow 0$ , then we know that $y'(0)=-\lambda _k$ , which is obviously a contradiction.

Next, since det $f_{u}(\overline u,\overline v)\neq 0$ , for all $V\in (B_\rho (\overline v))$ , we obtain that there is a $\rho \gt 0$ and a function $\theta \in C^2(B_\rho (\overline v))$ such that $\theta (\overline v)=\overline u$ and $f(\theta (V),V)=0$ . Then, for all $V\in (B_\rho (\overline v))$ , we prove that $P(V)\equiv g(\theta (V),V)$ satisfies $P(\overline v)=0$ and $P'(\overline v)=d_2\lambda _k\gt 0.$ It is easy to find that $P(\overline v)=g(\theta (\overline v),\overline v)=g(\overline u,\overline v)=0$ . In addition, differentiating the function $P(V)=g(\theta (V),V)$ gets

(3.8) \begin{equation} P'(V)=g_{u}(\theta (V),V)\theta '(V)+g_{v}(\theta (V),V). \end{equation}

On the other hand, we differentiate the equation $f(\theta (V),V)=0$ to obtain $f_{u}(\theta (V),V)\theta '(V)+f_{v}(\theta (V),V)=0$ , or, equivalently,

(3.9) \begin{equation} \theta '(V)=-f_{u}^{-1}(\theta (V),V)f_{v}(\theta (V),V). \end{equation}

In the end, choosing $V=\overline v$ , substituting equation (3.9) into equation (3.8) and by (3.5) we have

\begin{equation*}P'(\overline v)=-g_{u}(\theta (\overline v),\overline v)f_{u}^{-1}(\theta (\overline v),\overline v)f_{v}(\theta (\overline v),\overline v)+g_{u}(\theta (\overline v),\overline v) =-c_0a_0^{-1}b_0+d_0.\end{equation*}

Notice that

(3.10) \begin{equation} \begin{array}{l} -c_0a_0^{-1}b_0+d_0=\mbox {det}{\left (\begin{array}{cc} 1 & 0\\ 0 & -c_0a_0^{-1}b_0+d_0 \end{array}\right )}=\dfrac {1}{a_0}\mbox {det}{\left (\begin{array}{cc} a_0 & 0\\ c_0 & -c_0a_0^{-1}b_0+d_0 \end{array}\right )}\\\\ \qquad \qquad \qquad \;\;=\dfrac {1}{a_0}\mbox {det}\left ({\left (\begin{array}{cc} a_0 & b_0\\ c_0 & d_0 \end{array}\right )}{\left (\begin{array}{cc} 1 & -a^{-1}_0b_0\\ 0 & 1 \end{array}\right )}\right )=\dfrac {1}{a_0}\mbox {det}{\left (\begin{array}{cc} a_0 & b_0\\ c_0 & d_0 \end{array}\right )}.\end{array} \end{equation}

So, $P'(\overline v)=d_2\lambda _k\gt 0.$

Next, $\overline P$ represents an arbitrary extension of the function $P$ to the whole line $R$ that satisfies

(3.11) \begin{equation} \overline P\in C^2_b(R)\quad \mbox {and}\quad \overline P(V)=P(V)\quad \mbox {for all}\quad V\in (B_\rho (\overline v)). \end{equation}

By Proposition3.1, we have a sequence $d_s\rightarrow d_2$ such that

(3.12) \begin{equation} d_s\Delta _\tau v_s+(d_2-d_s)(v_s-\overline v)+\overline P(v_s)=0\quad \mbox {for}\quad x\in \Omega \end{equation}

has a non-constant solution $v_s\in W^{1,2}(\Omega )$ . Indeed, It is sufficient to search for these solutions in the form $v_s=\overline v+m_s$ , where $m_s$ satisfies

(3.13) \begin{equation} d_s\Delta _\tau m_s+(\overline P'(\overline v)+d_2-d_s)m_s+ y(m_s)=0\quad \mbox {in}\quad \Omega \end{equation}

with $\overline h'(\overline v)=d_2\lambda _k$ and $y(m_s)=\overline h(\overline v+m_s)-\overline h'(\overline v)m_s$ satisfies $y\in C^2_b(R)$ , $y(0)=0$ , and $y'(0)=0$ .

Next, we can find the solutions $m_s$ of problem (3.13) by Proposition3.1. Therefore, according to the standard elliptic theory, we have $||m_s||_{W^{2,2}(\Omega )}\rightarrow 0$ . By the bootstrap arguments using the elliptic $L_p$ estimates and the Sobolev embedding theorem, we know that $||m_s||_{W^{2,q}(\Omega )}\rightarrow 0$ for $q\gt \frac {N}{2}$ and hence $||m_s||_{L^\infty (\Omega )}\rightarrow 0$ . Particularly, by (3.11), if $||m_s||_\infty \le \rho$ , we get $\overline h(v_s)=\overline h(\overline v+m_s)=h(\overline v+m_s)=h(v_s)$ . So, the nontrivial solution of problem (3.12) is $v_s=\overline v+m_s$ , where $\overline h$ is changed to $\rho$ . In the end, we define $u_s=\theta (v_s)$ to get a nontrivial regular solution of problem (3.7).

Proof. We consider a solution $(\overline u_3,\overline v_3)$ of problem (3.14) and use Proposition3.2 with the constant stationary solution $(\overline u,\overline v)=(\overline u_3,\overline v_3)$ . Since $2\overline u_3\gt K$ , $1-\dfrac {2\overline u_3}{K}\lt 0$ , that is, $a_0\lt 0$ . By simple calculation, we can find that

\begin{equation*}\mbox {det}{\left (\begin{array}{cc} a_0 & b_0\\ c_0 & d_0 \end{array}\right )}= \left |\begin{array}{cc} r-\dfrac {2r\overline u_3}{K}-\dfrac {cm\overline v_3}{(m+b\overline u_3)^2} & -\dfrac {c\overline u_3}{m+b\overline u_3}\\[8pt] \dfrac {\beta cm\overline v_3}{(m+b\overline u_3)^2} & 0 \end{array}\right |=\dfrac {\beta cm\overline v_3}{(m+b\overline u_3)^2}\cdot \dfrac {c\overline u_3}{m+b\overline u_3}\gt 0. \end{equation*}

As a result, for some eigenvalue $\lambda _k\gt 0$ , we may select $d_2\gt 0$ to satisfy equation (3.6).

Remark 2. By classical nontrivial solution, we mean a solution $v(x)$ of (4.1) such that $v(x)\not \equiv 0$ , $v(x)\not \equiv v^*_2$ , $v(x)\in C^1({\bar {\Omega }})$ and $\Delta v(x)$ on the $\bar {\Omega }$ has jump discontinuity.

Remark 3. If $0\in \partial Q_{d_2}(s)$ , then the critical point of $Q_{d_2}$ is $s\in W^{1,2}(\Omega )$ , as stated in [7]. Once we obtain a critical point $s$ of $Q_{d_2}(s)$ , then $v=s+v^*_2$ is a critical point of $J_{d_2}(v)$ .

Proof. According to the definition of $\widetilde f_2^\gamma (v)$ in (4.3), we rewrite equation (4.6) as

(4.7) \begin{equation} Q_{d_2}(s)=R^*-\int _\Omega \int _{0}^{s} (h^\gamma (w+v^*_2)-h^\gamma (v^*_2))dwdx, \end{equation}

where $R^*=\dfrac {d_2}{2}\int _\Omega |\nabla s|^2dx+\dfrac {a}{2}\int _\Omega s^2dx$ . Clearly, $R^*$ is $C^1$ on $W^{1,2}(\Omega )$ and hence locally Lipschitz continuous. On the other hand, we know that $h^\gamma (v)=0$ if $v\lt \gamma$ , $h^\gamma (v)={\beta ch_2(v)}v/({m+bh_2(v)})$ if $\gamma \lt v\leq v_M$ and $h^\gamma (v)={\beta ch_2(v_M)}v_M/({m+bh_2(v_M)})$ if $v\gt v_M$ . Thus, we find that there exist a constant $a_1$ such that $|h^\gamma (v)-h^\gamma (v^*_2)|\lt a_1$ for all $v\in R$ , which means that $|h^\gamma (s+v^*_2)-h^\gamma (v^*_2)|\lt a_1$ for all $s\in R$ . Let $H(s)=\int _{0}^{s} (h^\gamma (w+v^*_2)-h^\gamma (v^*_2))dw$ and $B(s)=\int _\Omega H(s)dx$ . Then

\begin{equation*}|H(s_1)-H(s_2)|\le |\int _{s_2}^{s_1}\left |h^\gamma (w+v^*_2)-h^\gamma (v^*_2)|dw\right |\lt a_1|s_1-s_2|;\end{equation*}

so that

(4.8) \begin{align} |B({s_1})-B({s_2})|&\lt {a_1}{\int }_\Omega |{s_1}-{s_2}|dx\le {a_1}mes{(\Omega )^{1/2}}||{s_1} - {s_2}|{|_{{L^2}(\Omega )}} \nonumber\\ &\quad \le {a_1}mes{(\Omega )^{1/2}}||{s_1}-{s_2}|{|_{W^{1,2}(\Omega )}}. \end{align}

Therefore, $B(s)$ is a locally Lipschitz continuous function on $L^2(\Omega )$ and $W^{1,2}(\Omega )$ . From this, it can be concluded that $Q_{d_2}(s)$ is locally Lipschitz continuous on $W^{1,2}(\Omega )$ .

Proof. By Proposition4.3, we know

\begin{equation*} Q_{d_2}(s_n)=\dfrac {d_2}{2}\int _\Omega |\nabla s_n|^2dx+\dfrac {a}{2}\int _\Omega s_n^2dx-\int _\Omega H(s_n)dx \end{equation*}

\begin{equation*}\qquad \qquad \qquad \quad \;\;\;\geq \dfrac {1}{2}\mbox {min}\left \{ d_2,a\right \}||s_n||_{W^{1,2}(\Omega )}^2-a_1\mbox {mes}{(\Omega )^{1/2}}||s_n||_{W^{1,2}(\Omega )}.\end{equation*}

Thus, $\left \{ {s_n}\right \}$ is bounded in $W^{1,2}(\Omega )$ and there is a weakly convergent subsequence $\left \{ {s_{n_i}}\right \}$ with limit $s_0$ in $W^{1,2}(\Omega ).$ Since $W^{1,2}(\Omega )\rightarrow L^2(\Omega )$ is compact, $\left \{ {s_{n_i}}\right \}$ is strongly convergent in $L^2(\Omega )$ . Recalling Proposition4.3, both $B(s)$ and $Q_{d_2}(s)$ are locally Lipschitz continuous. So, according to Definition3, the generalized gradients of $B(s)$ and $Q_{d_2}(s)$ with respect to $s$ do exist and they are denoted by $\partial B(s)$ and $\partial Q_{d_2}(s)$ , respectively. Note that

(4.9) \begin{equation} \partial Q_{d_2}(s_n)\subset \left \{ Ls_n\right \}-\partial B(s_n), \end{equation}

where $L$ is an elliptic differential operator such that $Ls=-d_2\Delta s+as.$ We have applied Propositions (4) and (3) of [Reference Chang7] to prove (4.9). Since $\lambda (s_n)\rightarrow 0$ as $n\rightarrow \infty, $ there is a sequence $\rho _{n_i}\in \partial B(s_n)$ such that

\begin{equation*}Ls_{n_i}-\rho _{n_i}\rightarrow 0 \quad \mbox {in}\quad {(W^{1,2}(\Omega ))^*}.\end{equation*}

Because of $\rho _{n_i}\in \partial B(s_n)$ , $\left \{ {\rho _{n_i}}\right \}$ is bounded in $L^2(\Omega )$ . This can be demonstrated by noting that $B(s_{n_i})$ is locally Lipschitz continuous on $L^2(\Omega )$ according to (4.8). Therefore, there is a subsequence $\left \{ {n'_i}\right \}$ of $\left \{ {{n_i}}\right \}$ , which satisfies that $\left \{ {{\rho _{n'_i}}}\right \}$ is weakly convergent to $\rho _0$ in $L^2(\Omega )$ . Therefore, it is strongly convergent in $(W^{1,2}(\Omega ))^*$ . So,

\begin{equation*}Ls_{n'_i}\rightarrow \rho _0 \quad \mbox {in}\quad {(W^{1,2}(\Omega ))^*},\end{equation*}

which shows that $s_{n'_i}\rightarrow (L)^{-1}\rho _0$ in $W^{1,2}(\Omega ).$

Proof. (i) By (4.7), we know $Q_{d_2}(0)=0.$ Moreover, we can rewrite (4.7) as follows

(4.10) \begin{align} Q_{d_2}(s)&=\dfrac {d_2}{2}{\int }_\Omega |\nabla s|^2dx+\dfrac {1}{2}{\int }_\Omega (a-(h^\gamma )_v(v^*_2))s^2dx\nonumber \\ &\quad-{\int }_\Omega {\int }_{0}^{s} (h^\gamma (w+v^*_2)-h^\gamma (v^*_2)-(h^\gamma )_v(v^*_2)w)dwdx \end{align}

From the definitions of $f_2(h_2(v),v)$ in (4.2) and $h^\gamma (v)$ after (4.7), we know $f_2(h_2(v),v)=-av+h^\gamma (v),$ which implies $\partial _v f_{2}(h_2(v),v)=-a+(h^\gamma )_v(v).$ By the proof of Proposition2.3 (ii), we see $\partial _v f_{2}(h_2(v^*_2),v^*_2)\lt 0,$ so there is a constant $a_2\gt 0$ such that $a-(h^\gamma )_v(v^*_2)\gt a_2$ for all $x\in \overline {\Omega }$ .

Let $\Phi (s)=h^\gamma (s+v^*_2)-h^\gamma (v^*_2)-(h^\gamma )_v(v^*_2)s$ and $\Psi (s)=\int _{0}^{s}\Phi (w)dw.$ It is easy to see that $\Phi (s)=o(|s|)$ at $s=0$ uniformly in $x\in \overline {\Omega }$ . Thus, for any $\iota \gt 0$ , there exists a $\delta \gt 0$ such that $|\Psi (s)|\le \iota s^2$ if $|s|\le \delta$ . In addition, in the proof of Proposition4.3, we can recall the following that $|h^\gamma (w+v^*_2)-h^\gamma (v^*_2)|\lt a_1$ , which leads to the conclusion that for every $\varepsilon \in (1,(N+2)/(N-2))$ , there exists a constant $a_3\gt 0$ such that $|h^\gamma (w+v^*_2)-h^\gamma (v^*_2)|\lt a_1+a_3|w|^\varepsilon$ for all $w\in R$ . This means that there exists a constant $a_4\gt 0$ such that $|H(s)|=|\int _{0}^{s} (h^\gamma (w+v^*_2)-h^\gamma (v^*_2))dw|\le a_4|s|^{\varepsilon +1}$ for $|s|\gt \delta$ . Thanks to the Sobolev embedding theorem, we get

\begin{equation*}Q_{d_2}(s)=\dfrac {d_2}{2}\int _{|s|\gt \delta } |\nabla s|^2dx+\dfrac {a}{2}\int _{|s|\gt \delta } s^2dx-\int _{|s|\gt \delta } H(s)dx\end{equation*}

\begin{equation*}\qquad \qquad \qquad \qquad \;\;\;\;+\dfrac {d_2}{2}\int _{|s|\le \delta } |\nabla s|^2dx+\dfrac {1}{2}\int _{|s|\le \delta } (a-(h^\gamma )_v(v^*_2))s^2dx-\int _{|s|\le \delta }\Psi (s)dx\end{equation*}

\begin{equation*}\qquad \quad \geq \dfrac {d_2}{2}\int _{|s|\gt \delta } |\nabla s|^2dx+\dfrac {a}{2}\int _{|s|\gt \delta } s^2dx-c_5||s||^{\varepsilon +1}_{W^{1,2}(\Omega )}\end{equation*}

\begin{equation*}+\dfrac {d_2}{2}\int _{|s|\le \delta } |\nabla s|^2dx+(\dfrac {a_2}{2}-\iota )\int _{|s|\le \delta } s^2dx\end{equation*}

\begin{equation*}\qquad \qquad \qquad \qquad \;\;\;\geq a_6||s||^2_{W^{1,2}(\Omega )}-a_5||s||^{\varepsilon +1}_{W^{1,2}(\Omega )}=(a_6-a_5||s||^{\varepsilon -1}_{W^{1,2}(\Omega )})||s||^2_{W^{1,2} (\Omega )}\end{equation*}

with some positive constants $a_5$ and $a_6$ . Therefore, let $\rho =(a_6/2a_5)^{1/(\varepsilon -1)}$ , we can see that $Q_{d_2}(s)=0$ for $||s||_{W^{1,2}(\Omega )}\le \rho$ if and only if $s=0$ and that $Q_{d_2}(s)\geq (a_6/2)\rho ^2=\alpha$ for $||s||_{W^{1,2}(\Omega )}=\rho$ .

(ii) By selecting $e=-v^*_2$ , we derive $Q_{d_2}(e)=J_{d_2}(0)-J_{d_2}(v^*_2)\le 0$ from Case 1.

Proof. (i) Obviously, we know $Q_{d_2}(0)=0$ by (4.7). Similarly to (4.10), (4.5) can be rewritten as follows

(4.11) \begin{equation} Q_{d_2}(v)=\dfrac {d_2}{2}\int _\Omega |\nabla v|^2dx+\dfrac {a}{2}\int _\Omega v^2dx-\int _\Omega \int _{0}^{v} h^\gamma (w)dw dx. \end{equation}

Since $h^\gamma (v)=0$ if $v\lt \gamma$ , we know that there exists a $\delta _1\in (0,\gamma )$ such that $\int _{0}^{v} h^\gamma (w)dw=0$ for $|v|\le \delta _1$ . Then, by the definition of $h^\gamma (v)$ , we obtain $| h^\gamma (w)|\lt a_7$ with a constant $a_7\gt 0$ , which implies that for every $\varepsilon \in (1,(N+2)/(N-2))$ , there exists a constant $a_8\gt 0$ such that $|h^\gamma (w)|\lt a_7+a_8|w|^\varepsilon$ for all $w\in R$ . This shows that there is a constant $a_9\gt 0$ such that $|\int _{0}^{v} h^\gamma (w)dw|\le a_9|v|^{\varepsilon +1}$ for $|v|\gt \delta _1$ . Then repeating the proof of Proposition4.5, we can show that there exist positive constants $\rho _1$ , $\alpha _1$ such that $Q_{d_2}\geq \alpha _1$ if $||v||_{W^{1,2}(\Omega )}=\rho _1$ . (ii) Let $e_1=v^*_2$ . Then it is easy to know that $Q_{d_2}(e_1)=J_{d_2}(v^*_2)-J_{d_2}(0)\lt 0$ according to Case 2.

Proof of Theorem 4.1 . A critical point $v(x)$ of $Q_{d_2}$ has been found. In fact, $v(x)$ is a weak solution of (4.4). We can obtain that $v(x)$ is a classical solution of (4.4) by the elliptic regularity theorem [Reference Uhlenbeck29].

Now, we demonstrate that $0\le v(x)\le v^*_2$ . Because the proof for $0\le v(x)$ and $v(x)\le v^*_2$ are similar, we only show that $v(x)\le v^*_2$ . Let $v(x_1)=\mathop {\max }\limits _{x\in \overline {\Omega }}v(x)\gt v^*_2$ . If $x_1\in \Omega$ , then $d_2\Delta v|_{x=x_1}\le 0$ , which means that $\widetilde f_2^\gamma (v(x_1))\geq 0$ . But by the definition of $\widetilde f_2^\gamma (v)$ , we find $\widetilde f_2^\gamma (v(x_1))\lt 0$ , which is a contradiction. If $x_1\in {\Omega }$ , there is a ball $B_R(r_0)\subset \Omega$ centred at $r_0\in \Omega$ of radius $R$ , which satisfies $\partial \Omega \cap \overline B_R(r_0)=\left \{ {x_1}\right \}$ and $v(x)\lt v(x_1)$ in $B_R(r_0)$ . Since $v(x_1)\gt v^*_2$ , it follows from continuity that $v(x)\gt v^*_2$ in $B_R(r_0)$ , provided $r_0$ is sufficiently close to $x_1$ and $R$ is small enough. So, $\widetilde f_2^\gamma (v(x))\lt 0$ for $x\in B_R(r_0)$ , which shows that $d_2\Delta v\gt 0$ in $ B_R(r_0)$ . In addition, $v(x)\lt v(x_1)$ in $B_R(r_0)$ . By employing the Hopf boundary point Proposition (see, e.g., Chapters 8 and 9 of [Reference Gilbarg and Trudinger13]), we observe that $\partial _\tau v\gt 0$ . This is in contradiction with the boundary condition $\partial _\tau v=0$ . Thus, $v(x)\le v^*_2$ on $\overline {\Omega }$ . Since $f_2^\gamma (v)=\widetilde f_2^\gamma (v)$ for all $v\leq v_M$ , any classical solution $v(x)$ of (4.4) is also a classical solution of (4.1).

Finally, we demonstrate that $v(x)$ crosses $\gamma$ . Otherwise, we can suppose that $\gamma \lt v(x)\le v^*_2$ for all $x\in \overline {\Omega }$ . According to Proposition2.3 (i), we find that $f_2^\gamma (v)=f_2(h_2(v),v)\geq 0$ . So, $\int _\Omega f_2(h_2(v),v)dx\geq 0$ . In addition, integrating the first equation of (4.1) over $\Omega$ , we get $\int _\Omega f_2^\gamma (v)=\int _\Omega f_2(h_2(v),v)dx=0$ . This only holds if $v(x)\equiv v^*_2$ . So we have a contradiction. Similarly, we can demonstrate that $0\le v(x)\lt \gamma$ for all $x\in \overline {\Omega }$ is also invalid.

Remark 4. Suppose $v(x)$ is a solution of (4.1) and define

\begin{equation*}u(x)= \left \{\begin{array}{lll} 0 & v(x)\lt \gamma, \\ h_2(v(x)) & v(x)\gt \gamma .\\ \end{array} \right .\end{equation*}

Proof. We divide the proof into two steps.

Step 1. We consider the following initial value problems

(5.2) \begin{equation} \left \{\begin{array}{lll} d_2 v''-av=0,& x\in (0,n_0),\\ v'(0)=0,\qquad v(n_0)=\gamma, \\ \end{array} \right . \end{equation}

(5.3) \begin{align} \left \{\begin{array}{lll} d_2 v''+f_2(h_2(v),v)=0,& x\in (n_0,1),\\ v(n_0)=\gamma, \qquad v'(1)=0,\\ \end{array} \right . \\[8pt]\nonumber\end{align}

where $n_0\in (0,1)$ . It is easy to find that $W_0(x;\,n_0,d_2)$ is a unique monotone increasing solution to problem (5.2) for every $d_2\gt 0$ , where

\begin{equation*}W_0(x;\,n_0,d_2)=\dfrac {\gamma }{\mbox {cosh}\sqrt {a/d_2}n_0}\mbox {cosh}\sqrt {a/d_2}x.\end{equation*}

Next, we prove that problem (5.3) has a unique monotone increasing solution $W_1(x;\,n_0,d_2)$ for every $d_2\gt 0$ . Since $f_2(h_2(v^*_2),v^*_2)=0$ , $f_2(h_2(v),v)\gt 0$ for all $v\in (0,v^*_2)$ and $\partial _v f_{2}(h_2(v^*_2),v^*_2)\lt 0$ by Proposition2.3, there exist $0\lt n_1\lt n_2$ such that $-n_1(v-v^*_2)\lt f_2(h_2(v),v)\lt -n_2(v-v^*_2)$ for $v\lt v^*_2$ . In order to find a solution of (5.3), we study the following initial value problems for $i=1,2$

(5.4) \begin{equation} \left \{\begin{array}{lll} d_2 v''-n_i(v-v^*_2)=0 & x\in (n_0,1),\\ v(n_0)=\gamma, \quad v'(1)=0.\\ \end{array} \right . \end{equation}

Let $Y_i(x;\,n_0,d_2)$ be respective solutions of problems (5.4) for $i=1,2.$ Then it is easy to find that $Y_1(x;\,n_0,d_2)$ is a lower solution of (5.3) and $Y_2(x;\,n_0,d_2)$ is an upper solution of (5.3). For simplicity, we let $G_{n_0}(x;\,a)=\mbox {cosh}(a(1-x))/\mbox {cosh}(a(1-n_0)).$ Simple calculations yield that for $i=1,2$ , $Y_i(x;\,n_0,d_2)=(\gamma -v^*_2)G_{n_0}(x;\,\sqrt {n_i/d_2})+v^*_2.$ We claim that $Y_1(x;\,n_0,d_2)\lt Y_2(x;\,n_0,d_2)$ , let $Y(x;\,n_0,d_2)=Y_1(x;\,n_0,d_2)-Y_2(x;\,n_0,d_2)$ . Then $Y(x;\,n_0,d_2)$ satisfies

\begin{equation*}\left \{\begin{array}{lll} d_2 Y''=n_1(Y_1-v^*_2)-n_2(Y_2-v^*_2)\gt n_1Y& x\in (n_0,1),\\ Y(k;\,n_0,d_2)=0,\quad Y'(1;\,n_0,d_2)=0.\\ \end{array} \right .\end{equation*}

If $Y(x_M;\,n_0,d_2)=\mathop {\max }\limits _{x\in [n_0,1]}Y(x;\,n_0,d_2)\gt 0$ at some $x_M\in (n_0,1),$ then

\begin{equation*}0\geq d_2 Y''(x_M;\,n_0,d_2)\gt n_1Y(x_M;\,n_0,d_2)\gt 0.\end{equation*}

This is a contradiction. So, $Y_1(x;\,n_0,d_2)\leq Y_2(x;\,n_0,d_2).$ We verify that (5.3) has a solution $W_1(x;\,n_0,d_2)$ by using the upper and lower solution approach.

Next, we show the uniqueness of a solution for (5.3). By the comparison method, we can guarantee the existence of a maximal solution $W_M(x)$ and a minimal solution $W_m(x)$ such that

\begin{equation*}Y_1(x)\lt W_m(x)\lt W_M(x)\lt Y_2(x).\end{equation*}

Let $\widetilde f_2(v)=f_2(h_2(v),v)/v$ , then $\widetilde f_2(v)$ is strictly decreasing in $v$ . Due to $W_M(x)$ and $W_m(x)$ satisfying (5.3), we see

(5.5) \begin{equation} d_2 W_M''+\widetilde f_2(W_M)W_M=0, \qquad x\in (n_0,1), \end{equation}

and

(5.6) \begin{equation} d_2 W_m''+\widetilde f_2(W_m)W_m=0, \qquad x\in (n_0,1). \end{equation}

Multiply (5.5) by $W_m$ and multiply (5.6) by $W_M$ . Then we obtain

(5.7) \begin{align} 0&={\int }_{n_0}^{1}((d_2 W_M''+\widetilde f_2(W_M)W_M)W_m-(d_2 W_m''+\widetilde f_2(W_m)W_m)W_M)dx\nonumber\\&=d_2( W_M'(x)W_m(x)-W_m'(x)W_M(x))\Bigm |^1_{n_0}+{\int }_{n_0}^{1}(\widetilde f_2(W_M)-\widetilde f_2(W_m))W_MW_m dx\nonumber\\&=d_2\gamma (W_m'(n_0)-W_M'(n_0))+{\int }_{n_0}^{1}(\widetilde f_2(W_M)-\widetilde f_2(W_m))W_MW_m dx. \end{align}

Since $W_M(x)\gt W_m(x)$ in $(n_0,1]$ and $\widetilde f_2(v)$ is strictly decreasing in $v$ , then

\begin{equation*}W_m'(n_0)-W_M'(n_0)\le 0,\quad \widetilde f_2(W_M)-\widetilde f_2(W_m) \le 0, \quad n_0 \le x \le 1.\end{equation*}

Due to $W_MW_m\gt 0$ , $W_M\equiv W_m$ can be seen from equation (5.7). This shows that we have completed the proof of the uniqueness of a solution for (5.3). Moreover, combining $d_2 W_1''+f_2(h_2(W_1),W_1)=0$ with $f_2(h_2(W_1),W_1)\gt 0$ and $W_1'(1;\,n_0,d_2)=0$ , we get that $W_1(x;\,n_0,d_2)$ is monotone increasing in $x$ .

Step 2. By simple calculation, we get

(5.8) \begin{equation} W_0'(n_0;\,n_0,d_2)=\gamma \sqrt {a/d_2}\mbox {tanh}(\sqrt {a/d_2}n_0),\, W_1'(n_0;\,n_0,d_2)=\dfrac {1}{d_2}{\int }_{n_0}^{1}f_2(h_2(W_1),W_1)dx. \end{equation}

Define $\rho _0$ to be a sufficiently small positive number. If $n_0=1-\rho _0$ , then we find $W_0'(1-\rho _0;\,1-\rho _0,d_2)\gt W_1'(1-\rho _0;\,1-\rho _0,d_2)$ by (5.8), since $f_2(h_2(v),v)$ is bounded for all $x\in [0,v^*_2]$ . In addition, if $n_0=\rho _0$ , then $W_0'(\rho _0;\,\rho _0,d_2)\lt W_1'(\rho _0;\,\rho _0,d_2)$ for sufficiently small $\rho _0\gt 0$ . Assume $\Theta (n_0,d_2)=W_0'(n_0;\,n_0,d_2)-W_1'(n_0;\,n_0,d_2)$ . Thus, we obtain

(5.9) \begin{equation} \Theta (1-\rho _0,d_2)\gt 0,\quad \Theta (\rho _0,d_2)\lt 0. \end{equation}

It is easy to see that $\Theta (n_0,d_2)$ is continuous with respect to $n_0$ for all $d_2\gt 0$ . The combination of this and (5.9), there exists a $n_0^*$ such that $\Theta (n_0^*,d_2)=0$ and

\begin{equation*}V_{1,+}(x,d_2)= \left \{\begin{array}{lll} W_0(x;\,n_0^*,d_2) & \mbox {for} \quad x\in [0,n_0^*],\\ W_1(x;\,n_0^*,d_2) & \mbox {for} \quad x\in [n_0^*,1],\\ \end{array} \right .\end{equation*}

is a monotone increasing solution of (5.1) for all $d_2\gt 0$ .

Following, using $V_{1,+}(x,d_2)$ and its reflection, we create symmetric solutions to (5.1) starting from the monotone increasing solution $V_{1,+}(x,d_2)$ . For all $n\geq 2$ , define a function $V_{n,+}(x,\overline d_2)$ on $0\le x\le 1$ by

\begin{equation*}V_{n,+}(x,\overline d_2)= \left \{\begin{array}{lll} V_{1,+}(nx-2j;\,\overline d_2) & \mbox {for} \quad x\in [2j/n,(2j+1)/n],\\ V_{1,+}(2(j+1)-nx;\,\overline d_2) & \mbox {for} \quad x\in [(2j+1)/n,2(j+1)/n],\\ \end{array} \right .\end{equation*}

where $j=0,1,2,\cdots, [n/2]$ and $\overline d_2=n^2d_2$ . Then $V_{n,+}(x,\overline d_2)$ is a symmetric solution of (5.1). This completes the proof.

Proof. By $f_2(h_L(V_0),V_0)=f_2(0,V_0)=-aV_0,$ we know $d_2 V_0''=aV_0.$ By simple calculation

(5.13) \begin{equation} V_0(x;\,b^*_0)=b^*_0\mbox {cosh}\left (\sqrt {a/d_2}x\right )\quad \mbox {and}\quad V_0'(x;\,b^*_0)=b^*_0\sqrt {a/d_2}\mbox {sinh}\left (\sqrt {a/d_2}x\right ). \end{equation}

So, for all $x\gt 0,$ we can get $V_0(x;\,b^*_0)$ is positive and strictly increasing. By Proposition2.3 (i), we have $d_2 V_1''=-f_2(h_2(V_1),V_1)\lt 0.$ Thus, $V_1(x;\,b^*_1)$ is decreasing with respect to $x$ . Since $V_1'(0;\,b^*_1)=0,$ so $V_1'(x;\,b^*_1)\lt 0$ for $0\le x\lt x_{b^*_1}.$ These results show that $V_1(x;\,b^*_1)\gt 0$ is strictly monotone decreasing on $[0,x_{b^*_1})$ and $V_1(x_{b^*_1};b^*_1)=0.$

Proof. Since the assertions for $l_0$ and $\psi _0(l_0)$ may be treated similarly, it is sufficient to verify the statements for $l_1$ and $\psi _1(l_1)$ .

(i) Recall that $V_1(l_1(b^*_1);\,b^*_1)=\gamma$ . We differentiate it with respect to both sides with respect to $b^*_1$ and get

\begin{equation*}\dfrac {\partial {V_1}}{\partial x}\dfrac {dl_1}{db^*_1}+\dfrac {\partial {V_1}}{\partial b^*_1}=0,\end{equation*}

so

\begin{equation*}\dfrac {dl_1}{db^*_1}=-{\dfrac {\partial {V_1}}{\partial b^*_1}}/{\dfrac {\partial {V_1}}{\partial x}}.\end{equation*}

Moreover, let $\xi _1(x;\,b^*_1)=\dfrac {\partial {V_1}}{\partial x}(x;\,b^*_1)$ and $\eta _1(x;\,b^*_1)=\dfrac {\partial {V_1}}{\partial b^*_1}(x;\,b^*_1)$ . We get $\xi _1(x;\,b^*_1)\lt 0$ for all $x\in (0,l_1)$ by the proof of Proposition5.4. It is easy to find that $\eta _1(x;\,b^*_1)$ is a solution of

\begin{equation*}\left \{\begin{array}{lll} d_2\eta _1''+\dfrac {d}{dv}f_2(h_2(V_1),V_1)\eta _1=0 & \mbox {for} \quad x\in [0,1],\\ \eta _1'(0;\,b^*_1)=0, \quad \eta _1(0;\,b^*_1)=1.\\ \end{array} \right .\end{equation*}

We find that if $0\le x\le l_1$ , then $V_1(x;\,b^*_1)\in [\gamma, b^*_1]\subset (d^*,v^*_2].$ By Proposition2.3 (ii), we have $\dfrac {d}{dv}f_2(h_2(V_1),V_1)\lt 0$ for $V_1(x;\,b^*_1)\in [\gamma, b^*_1]\subset (d^*,v^*_2].$ So,

\begin{equation*}\eta _1''(0;\,b^*_1)=-\dfrac {1}{d_2}\dfrac {d}{dv}f_2(h_2(b^*_1),b^*_1)\eta _1(0;\,b^*_1)\gt 0.\end{equation*}

Expanding $\eta _1(x;\,b^*_1)$ near $x=0$ , we get

\begin{equation*}\eta _1(x;\,b^*_1)=\eta _1(0;\,b^*_1)+\eta _1'(0;\,b^*_1)x+\dfrac {1}{2}\eta _1''(0;\,b^*_1)x^2+\cdots \gt 0.\end{equation*}

Thus, we have $d_2\eta _1''=-\dfrac {d}{dV_1}f_2(h_2(V_1),V_1)\eta _1\gt 0$ for $\eta _1(x;\,b^*_1)\gt 0$ . Then, $\eta _1'(x;\,b^*_1)$ is strictly increasing with respect to $x$ . Since $\eta _1'(0;\,b^*_1)=0,$ we know $\eta _1'(x;\,b^*_1)\gt 0$ for $x\in (0,l_1]$ . Combining this with $\eta _1(0;\,b^*_1)=1$ , we get $\eta _1(x;\,b^*_1)\geq 1$ for all $x\in [0,l_1]$ . So,

\begin{equation*}\dfrac {dl_1}{db^*_1}=-{\dfrac {\partial {V_1}}{\partial b^*_1}}/{\dfrac {\partial {V_1}}{\partial x}}=-\dfrac {\eta _1(l_1(b^*_1);\,b^*_1)}{\xi _1(l_1(b^*_1);\,b^*_1)}\gt 0,\end{equation*}

and $l_1$ is strictly increasing in $b^*_1$ . Next, we show that $\mathop {\lim }\limits _{b^*_1\rightarrow \gamma }l_1(b^*_1)=0$ . Since $e^*\,:\!=f_2(h_2(\gamma ),\gamma )/d_2\gt 0$ and $V_1(x;\,b^*_1)$ satisfies

(5.14) \begin{equation} V_1(x;\,b^*_1)=b^*_1-\dfrac {e^*}{2}x^2+O(x^3) \quad \mbox {as}\quad x\rightarrow 0. \end{equation}

Therefore, by simple calculation, we have $l_1(b^*_1)=\sqrt {2(b^*_1-\gamma )/e^*}(1+o(1))$ as $b^*_1\rightarrow \gamma$ , which shows that $\mathop {\lim }\limits _{b^*_1\rightarrow \gamma }l_1(b^*_1)=0$ . As a result, the inverse of $l_1(b^*_1)$ exists and is represented as $b^*_1(l_1)$ .

(ii) Since $\psi _1(l_1)=\dfrac {\partial V_1}{\partial x}(l_1;\,b^*_1(l_1))=\xi _1(l_1;\,b^*_1(l_1)),$ from Proposition 4.2, we get

\begin{equation*}\dfrac {d\psi _1(l_1)}{dl_1}=\dfrac {\partial \xi _1}{\partial x}+\dfrac {\partial \xi _1}{\partial b^*_1}\dfrac {db^*_1}{dl_1}=\dfrac {\partial \xi _1}{\partial x}-\dfrac {\partial \xi _1}{\partial b^*_1}\dfrac {\xi _1(l_1;\,b^*_1(l_1))}{\eta _1(l_1;\,b^*_1(l_1))}\end{equation*}

\begin{equation*}\quad \quad \qquad =\dfrac {1}{\eta _1(l_1;\,b^*_1(l_1))}\left (\dfrac {\partial \xi _1}{\partial x}\eta _1-\dfrac {\partial \eta _1}{\partial x}\xi _1\right )(l_1;\,b^*_1(l_1))\end{equation*}

\begin{equation*}\quad \quad \qquad =\dfrac {1}{\eta _1(l_1;\,b^*_1(l_1))}\left (\dfrac {\partial \xi _1}{\partial x}\eta _1-\dfrac {\partial \eta _1}{\partial x}\xi _1\right )(0;\,b^*_1(l_1))\end{equation*}

\begin{equation*}=-\dfrac {f_2(h_2(b^*_1),b^*_1)}{d_2\eta _1(l_1;\,b^*_1(l_1))}\lt 0.\end{equation*}

Since $\left (\dfrac {\partial \xi _1}{\partial x}\eta _1-\dfrac {\partial \eta _1}{\partial x}\xi _1\right )'=\xi _1''\eta _1-\eta _1''\xi _1=0$ , we obtain $\dfrac {\partial \xi _1}{\partial x}\eta _1-\dfrac {\partial \eta _1}{\partial x}\xi _1$ is a constant. And by (5.14), we have $V_1^{'}(x;\,b^*_1)=-e^*x+O(x^2)$ as $x\rightarrow 0$ , so

\begin{equation*}\psi _1(l_1)=-\sqrt {2e^*(b^*_1-\gamma )}(1+o(1))\quad \mbox {as}\quad l_1\rightarrow 0.\end{equation*}

Hence, $\mathop {\lim }\limits _{l_1\rightarrow 0}\psi _1(l_1)=-\mathop {\lim }\limits _{b^*_1\rightarrow \gamma }\sqrt {2e^*(b^*_1-\gamma )}=0.$ As a result, $\psi _1(l_1)$ is strictly decreasing with respect to $l_1$ and the inverse $\psi ^{-1}_1(\alpha _1)$ of $\alpha _1$ does exist.

Proof. By the implicit function theorem and Proposition5.5, we know that $z_0(\alpha )$ and $z_1(\!-\alpha )$ are a strictly increasing functions of class $C^1$ . Then, it follows from Proposition5.5 (ii) that $\mathop {\lim }\limits _{l_0\rightarrow 0}\psi _0(l_0)=0$ and $\mathop {\lim }\limits _{l_1\rightarrow 0}\psi _1(l_1)=0$ . Thus, $\mathop {\lim }\limits _{\alpha \rightarrow 0}z_0(\alpha )=0$ and $\mathop {\lim }\limits _{\alpha \rightarrow 0}z_1(\!-\alpha )=0$ , which show $z_0(0)+z_1(0)=0$ . Since $\overline z_0=\mathop {\lim }\limits _{\alpha _0\rightarrow \overline \alpha }z_0(\alpha ),\overline z_1=\mathop {\lim }\limits _{\alpha _1\rightarrow -\overline \alpha }z_1(\alpha )$ , we have $z_0(\overline \alpha )+z_1(\!-\overline \alpha )=\overline z_0+\overline z_1.$

Proof of Theorem 5.2. Since $1\le \overline z_0+\overline z_1$ , $z_0(\alpha )+z_1(\!-\alpha )$ increases in $\alpha$ according to Proposition5.6, so there is a unique $\alpha ^*\in (0,\overline \alpha ]$ that satisfies $z_0(\alpha ^*)+z_1(\!-\alpha ^*)=1.$ Let $z_0(\alpha ^*)=\Upsilon ^*$ for such $\alpha ^*$ . Then the definitions of $z_0$ and $z_1$ yield $\psi _0(\Upsilon ^*)=\alpha ^*$ and $\psi _1(1-\Upsilon ^*)=-\alpha ^*$ . So, $\psi _0(\Upsilon ^*)+\psi _1(1-\Upsilon ^*)=0$ . Define

\begin{equation*}T_{1,+}(x)=\left \{\begin{array}{lll} V_0(x;\,a^*_0(\Upsilon ^*)) & \mbox {for} \ x\in [0,\Upsilon ^*],\\ V_1(1-x;\,a^*_1(1-\Upsilon ^*)) & \mbox {for} \ x\in [\Upsilon ^*,1],\\ \end{array} \right .\end{equation*}

where $a^*_0(\Upsilon ^*)=a^*_0(z_0(\alpha ^*))$ and $a^*_0(1-\Upsilon ^*)=a^*_1(z_1(\!-\alpha ^*))$ .

Then it becomes a unique increasing solution of (5.1). Then, for each integer $n\geq 2,$ we demonstrate the existence of $T_{n,+}(x)$ . Let $\overline \alpha _n=\alpha _0=-\alpha _1$ such that $z_0(\overline \alpha _n)+z_1(\!-\overline \alpha _n)=\dfrac {1}{n}$ . Since $ n\geq 2$ , we know $\dfrac {1}{n}\lt 1\le \overline z_0+\overline z_1$ . As a result, we can construct a unique monotone increasing solution $\overline Z_{1,+}(x)$ on $[0,{1}/{n}]$ , where $\overline T_{1,+}(x)=V_0(x;\,\overline b^*_0)$ with $\overline b^*_0=b^*_0(z_0(\overline \alpha _n))$ for $x\in [0,z_0(\overline \alpha _n)]$ and $\overline T_{1,+}(x)=V_1(({1}/{n})-x;\,\overline b^*_1)$ with $\overline b^*_1=b^*_1(z_1(\!-\overline \alpha _n))$ for $x\in [z_0(\overline \alpha _n),{1}/{n}]$ . Now define

\begin{equation*}T_{n,+}(x)=\left \{\begin{array}{lll} \overline T_{1,+}((2j/n)+x) & \mbox {for} \ x\in [2j/n,(2j+1)/n],\\ \overline T_{1,+}([2(j+1)/n]-x) & \mbox {for} \ x\in [(2j+1)/n,2(j+1)/n],\\ \end{array} \right .\end{equation*}

where $j=0,1,2,\cdots, [n/2]$ . Then $T_{n,+}(x)$ is an $n$ -mode symmetric solution of (5.1). This completes the proof.

Proof of Theorem 5.3 Since $z_0(\alpha )+z_1(\!-\alpha )$ is a strictly increasing function for $\alpha \in [0,\overline \alpha ]$ from Proposition5.6, whose range is the same as $[0,\overline z_0+\overline z_1]\subset [0,1]$ . If $N_0$ is the smallest positive integer greater than $1/(\overline z_0+\overline z_1)$ , it follows that there is a unique $\alpha ^*_n$ such that $z_0(\alpha ^*_n)+z_1(\!-\alpha ^*_n)={1}/{n}(n\geq N_0)$ . Similar to the proof of Theorem5.5, we can create a unique monotone increasing solution $\widetilde B_{1,+}(x)$ on $[0,z_0(\alpha ^*_n)+z_1(\!-\alpha ^*_n)]$ , where

\begin{equation*}\widetilde B_{1,+}(x)=\left \{\begin{array}{lll} V_0(x;\,b^*_0) & \mbox {for} \ x\in [0,z_0(\alpha ^*_n)],\\ V_1(z_0(\alpha )+z_1(\!-\alpha )-x;\,b^*_1) & \mbox {for} \ x\in [z_0(\alpha ^*_n),E]\\ \end{array} \right .\end{equation*}

with $b^*_0=b^*_0(z_0(\alpha ^*_n))$ , $b^*_1=b^*_1(z_1(\!-\alpha ^*_n))$ and $E=z_0(\alpha ^*_n)+z_1(\!-\alpha ^*_n)$ . Then we extend $\widetilde B_{1,+}(x)$ to the interval $[0,2E]$ by

\begin{equation*}\widetilde B_{2,+}(x)=\left \{\begin{array}{lll} \widetilde B_{1,+}(x) & \mbox {for} \ x\in [0,E],\\ \widetilde B_{1,+}(2E-x) & \mbox {for} \ x\in [E,2E].\\ \end{array} \right .\end{equation*}

We continue this process until $x$ reaches $x=1$ . Then $\widetilde B_{n,+}(x)$ is an $n$ -mode symmetric solution of (5.1).

Proof. The proof is simple. As a result, we omit the detail.

Proof. Assume that $\widetilde {\Phi }=(\widetilde {\phi },\widetilde {\psi })\in \mbox {ker}\widetilde {L}$ and let $\widetilde {\phi }=\sum _{ j}\widetilde {c_j}{\phi _j}$ , $\widetilde {\psi }=\sum _{ j}\widetilde {d_j}{\phi _j}$ , then $\sum _{ j}^\infty D_j{\left (\begin{array}{cc} \widetilde {c_j} \\ \widetilde {d_j} \end{array}\right )}\phi _j=0$ , where

\begin{equation*}D_j={\left (\begin{array}{cc} \widetilde {f_{11}} & \widetilde {f_{12}} \\ \widetilde {f_{21}} & -d_2L_j \end{array}\right )}.\end{equation*}

Obviously, $\mbox {det}D_j=0\Leftrightarrow d_2=\widetilde {d_2}$ . Let $d_2=\widetilde {d_2}$ , we get

\begin{equation*}\mbox {ker}\widetilde {L}=\mbox {span}\{ {\Phi _0}\},\quad \Phi _0= {\left (\begin{array}{cc} \phi _0 \\ \psi _0 \end{array}\right )}={\left (\begin{array}{cc} \mbox {cos} (\pi jx) \\ -\frac {\widetilde {f_{11}}}{\widetilde {f_{12}}}\mbox {cos} (\pi jx) \end{array}\right )}.\end{equation*}

Similar to this, it is simple to calculate an eigenvector $\Phi _0^*$ of $\widetilde {L}^*$ associated with $0$ having the following form

\begin{equation*}\mbox {ker}\widetilde {L}^*=\mbox {span}\{ {\Phi _0^*}\},\quad \Phi _0^*= {\left (\begin{array}{cc} \phi _0 \\ \psi _0 \end{array}\right )}={\left (\begin{array}{cc} \mbox {cos} (\pi jx) \\ -\frac {\widetilde {f_{11}}}{\widetilde {f_{21}}}\mbox {cos} (\pi jx) \end{array}\right )},\end{equation*}

where $\widetilde {L}^*$ is the adjoint operator of $\widetilde {L}$ , which is obtained by

\begin{equation*}\widetilde {L}^*={\left (\begin{array}{cc} \widetilde {f_{11}} & \widetilde {f_{21}} \\ \widetilde {f_{12}} & d_2\frac {d^2}{dx^2} \end{array}\right )}.\end{equation*}

Because $\mbox {rang}\widetilde {L}=(\mbox {ker}\widetilde {L}^*)^\bot$ , we have $\mbox {dim}\mbox {ker}\widetilde {L}^*=\mbox {codim}\,\mbox {rang}\,\widetilde {L}=1.$ Finally, since

\begin{equation*}\hat {L}=\dfrac {\partial \hat {L}}{\partial d_2}{\left (\begin{array}{cc} 0 & 0 \\ 0 & \frac {d^2}{dx^2} \end{array}\right )}\end{equation*}

and $\hat {L}\Phi _0\notin \mbox {rang}\,\widetilde {L}$ , we get that the conditions required for the standard bifurcation theorem to apply, based on the presence of a simple eigenvalue [Reference Crandall and Rabinowitz8], are satisfied.

Proof. According to the proof procedure of theorem6.2, we can get $\mbox {dim}\mbox {ker}\widetilde {L}=\mbox {codim}\,\mbox {rang}\,\widetilde {L}=1$ . Then $i\Phi \notin \mbox {rang}\,\widetilde {L}$ , where $\Phi _0$ satisfies $\mbox {ker}\,\widetilde {L}=\{ {\Phi _0}\}$ . Thus, it is clear that $\widetilde {L}$ possesses $0$ as an $i$ -simple eigenvalue according to the definition of a $K$ -simple eigenvalue presented in [Reference Wu30].

Proof. From (6.3) we get $(\lambda _j(d_2))^2-(\widetilde {f_{11}}-d_2L_j)\lambda _j(d_2)-\widetilde {f_{12}}\widetilde {f_{21}}-d_2\widetilde {f_{11}}L_j=0$ . Taking the derivative of both sides with respect to $d_2$ , we get $2\lambda _j(d_2)\lambda _j'(d_2)+L_j\lambda _j(d_2)-(\widetilde {f_{11}}-d_2L_j)\lambda _j'(d_2)-\widetilde {f_{11}}L_j=0.$ So, $\lambda _j(\widetilde {d_2})=0$ shows

\begin{equation*}\lambda _j'(d_2)=\dfrac {-L_j\widetilde {f_{11}}}{\widetilde {f_{11}}-L_j\widetilde {d_2}}=\dfrac {-\widetilde {f_{11}}^2L_j}{\widetilde {f_{11}}^2+\widetilde {f_{12}}\widetilde {f_{21}}}.\end{equation*}

Thus, if $\widetilde {f_{11}}^2+\widetilde {f_{12}}\widetilde {f_{21}}\lt 0$ , then $\lambda _j'(d_2)\gt 0$ , which shows that both $\lambda (l)$ and $-ld_2'(l)$ possess the same sign. But if $\widetilde {f_{11}}^2+\widetilde {f_{12}}\widetilde {f_{21}}\gt 0$ , then $\lambda _j'(d_2)\lt 0$ , which implies that both $\lambda (l)$ and $ld_2'(l)$ possess the same sign.

Proof. We only demonstrate the assertion (i) here, as we can approach the proof of the other assertions in a similar manner. By Proposition6.4, it is sufficient to compute $d_2'(l)$ to reveal the sign of $\lambda (l)$ . We expand $v(x,l)$ and $d_2(l)$ in $l$ to obtain

\begin{align*}v(x,l)&=v^*_3+l\widetilde n_1(x)+l^2\widetilde n_2(x)+l^3\widetilde n_3(x)+\cdots, \\ d_2(l)&=\widetilde {d_2}+l\widetilde k_1+l^2\widetilde k_2+l^3\widetilde k_3+\cdots .\end{align*}

Due to $d(v^*_3)=f_2(h_1(v^*_3),v^*_3)=0$ , we obtain

\begin{equation*}d(v)=C(v-v^*_3)+\dfrac {1}{2}D(v-v^*_3)^2+\dfrac {1}{6}E(v-v^*_3)^3+\cdots, \end{equation*}

where

\begin{equation*}C=\dfrac {\beta cmv^*_3}{(m+bh_1(v^*_3))^2}h_1'(v^*_3),\end{equation*}

\begin{equation*}D=\dfrac {-2\beta cmbv^*_3}{(m+bh_1(v^*_3))^3}(h'_1(v^*_3))^2+\dfrac {2\beta cm}{(m+bh_1(v^*_3))^2}h'_1(v^*_3)+\dfrac {\beta cmv^*_3}{(m+bh_1(v^*_3))^2}h''_1(v^*_3),\end{equation*}

\begin{equation*}E=\dfrac {6\beta cmb^2v^*_3}{(m+bh_1(v^*_3))^4}(h'_1(v^*_3))^3- \dfrac {6\beta cmb}{(m+bh_1(v^*_3))^3}(h'_1(v^*_3))^2-\end{equation*}

\begin{equation*}\dfrac {6\beta cmbv^*_3}{(m+bh_1(v^*_3))^3}h'_1(v^*_3)h''_1(v^*_3)+\dfrac {3\beta cm}{(m+bh_1(v^*_3))^2}h''_1(v^*_3)+\dfrac {\beta cmv^*_3}{(m+bh_1(v^*_3))^2}h'''_1(v^*_3).\end{equation*}

In addition, it is easy to find that $f_1(h_1(v^*_3),v^*_3)=0.$ So, we get $h_1'(v^*_3)=-\widetilde {f_{12}}/\widetilde {f_{11}}.$ Consequently, based on the previous definition of $\widetilde {f_{21}}$ and $\widetilde {d_2}$ , we know

\begin{equation*}C=\dfrac {\beta cmv^*_3}{(m+bh_1(v^*_3))^2}h_1'(v^*_3)=\dfrac {\beta cmv^*_3}{(m+bh_1(v^*_3))^2}(\dfrac {-\widetilde {f_{12}}}{\widetilde {f_{11}}})=-\dfrac {\widetilde {d_2}\widetilde {f_{11}}L_j}{\widetilde {f_{12}}}(\dfrac {-\widetilde {f_{12}}}{\widetilde {f_{11}}})=\widetilde {d_2}\pi ^2j^2.\end{equation*}

By substituting these expressions into (6.4), we can obtain a sequence of equations by assigning a value of zero to the coefficient of each power of $l$ .

(6.5) \begin{align} \widetilde {d_2}\widetilde n''_1&+C\widetilde n_1=0, \end{align}

(6.6) \begin{align} \widetilde {d_2}\widetilde n''_2+\widetilde k_1\widetilde n''_1&+\dfrac {1}{2}D\widetilde n_1^2+C\widetilde n_2=0, \end{align}

(6.7) \begin{align} \widetilde {d_2}\widetilde n''_3+\widetilde k_1\widetilde n''_2+\widetilde k_2\widetilde n''_1&+C\widetilde n_3+D\widetilde n_1\widetilde n_2+\dfrac {1}{6}E\widetilde n_1^3=0. \\[8pt]\nonumber\end{align}

By (6.5), we know that $\widetilde {d_2}\frac {d^2}{dx^2}+C$ has an eigenvalue of 0. So, equation (6.6) can be solved only if

(6.8) \begin{equation} \widetilde k_1{\int }_{0}^{1}\widetilde n''_1\widetilde n_1\mbox {d}x+\dfrac {1}{2}D{\int }_{0}^{1}\widetilde n^3_1\mbox {d}x=0. \end{equation}

When $\widetilde n_1(x)=\mbox {cos}(\pi jx)$ , equation (6.5) is satisfied. By simple calculations, it can be deduced that $\widetilde k_1=0$ when it is substituted into (6.8). Therefore, (6.6) is rewritten as

\begin{equation*}\widetilde {d_2}\widetilde n''_2+\dfrac {1}{2}D\widetilde n_1^2+C\widetilde n_2=0.\end{equation*}

Since $\widetilde n_1^2=(1+\mbox {cos}(2\pi jx))/2$ , then we know

\begin{equation*}\widetilde {d_2}\widetilde n''_2+C\widetilde n_2=-\dfrac {1}{2}D\dfrac {1+\mbox {cos}(2\pi jx)}{2}.\end{equation*}

By simple calculations, it can be deduced that

\begin{equation*}\widetilde n_2(x)=-\dfrac {D}{4C}-\dfrac {D}{4(C-4\widetilde {d_2}\pi ^2j^2)}\mbox {cos}(2\pi jx)=-\dfrac {D}{4C}+\dfrac {D}{12C}\mbox {cos}(2\pi jx).\end{equation*}

Now we consider equation (6.7). Since $\widetilde k_1=0$ , it follows that (6.7) has a solution if and only if

(6.9) \begin{equation} \widetilde k_2{\int }_{0}^{1}\widetilde n''_1\widetilde n_1\mbox {d}x+D{\int }_{0}^{1}\widetilde n_1^2\widetilde n_2\mbox {d}x+\dfrac {1}{6}E{\int }_{0}^{1}\widetilde n^4_1\mbox {d}x=0. \end{equation}

It can be directly calculated that

\begin{equation*}{\int }_{0}^{1}\widetilde n''_1\widetilde n_1\mbox {d}x=-\dfrac {1}{2}L_j,\quad {\int }_{0}^{1}\widetilde n^2_1\widetilde n_2\mbox {d}x=-\dfrac {5D}{48C},\quad {\int }_{0}^{1}\widetilde n^4_1\mbox {d}x=\dfrac {3}{8}.\end{equation*}

So, (6.9) becomes

\begin{equation*}-\dfrac {1}{2}\widetilde k_2L_j-\dfrac {5D^2}{48C}+\dfrac {E}{16}=0;\quad \mbox {so that}\quad \widetilde k_2=\dfrac {1}{24CL_j}(3CE-5D^2).\end{equation*}

Since $\widetilde k_1=0$ and $l{d_2}'(l)=l(\widetilde k_1+2l\widetilde k_2+O(l^2)),$ we have $l{d_2}'(l)=2l^2\widetilde k_2+O(l^3))$ . Thus, for $|l|$ sufficiently small, the sign of $l{d_2}'(l)$ is the same as that of $\widetilde k_2$ . This implies that the sign of $-\lambda (l)$ is the same as that of $\widetilde k_2$ if $\widetilde {f_{11}}^2+\widetilde {f_{12}}\widetilde {f_{21}}\lt 0$ , according to Proposition6.4. So, if $\widetilde {f_{11}}^2+\widetilde {f_{12}}\widetilde {f_{21}}\lt 0$ and $N\gt 0$ , then $\widetilde k_2\gt 0$ ; so that $\lambda (l)\lt 0.$