Proof. First, we transform ( P ) into a fixed boundary problem as in [Reference Chen and Friedman3, Reference Du and Lin6]. Let $\xi _1(y)$ and $\xi _2(y)$ in $C^{3}(\mathbb {R})$ be two nonnegative functions with $\xi _2(y)=\xi _1(-y)$ and

$$ \begin{align*} \xi_1(y)=1 \hspace{7.5pt}\text{if }|y-h_0|<{h_0}/{4},\hspace{7.5pt} \xi_1(y)=0 \hspace{7.5pt}\text{if } |y-h_0|>{h_0}/{2},\hspace{7.5pt} |\xi_1'(y)|<{6}/{h_0} \hspace{7.5pt}\text{for } y\in{\mathbb{R}}. \end{align*} $$

If we set

$$ \begin{align*} \begin{cases} v_U(t,y):=w_U(t,y+\xi_{2}(y)(g(t)+h_{0}))=w_U(t,x), & y<0,\\ v_L(t,y):=w_L(t,y+\xi_{1}(y)(h(t)-h_{0}))=w_L(t,x), & y>0, \end{cases} \end{align*} $$

then the free boundary problem becomes

(2.1) $$ \begin{align} \begin{cases} \partial_t v_U-A_U \partial_{yy}{v}_U-(B_U+C_U-\beta_U\sqrt{A_U})\partial_y{v}_U=f(v_U), & t>0,\ y\in(-h_0,0),\\ \partial_t v_L-A_L\partial_{yy}{v}_L-(B_L+C_L-\beta_L\sqrt{A_L})\partial_y{v}_L=f(v_L), & t>0,\ y\in(0,h_0),\\ v_U(t,0)=v_L(t,0), a_U\partial_{y}v_{U}(t,0)=a_L\partial_{y}v_{L}(t,0), & t>0,\\ v_U(0,z)=w_0(z), v_L(0,y)=w_0(y), & z\in[-h_0,0],\ y\in[0,h_0],\\ \end{cases} \end{align} $$

where, for $y<0$ , $t>0$ ,

$$ \begin{align*} \sqrt{A_U(t,y)}&:=\frac{1}{1+\xi_2'(y)(g(t)+h_0)},\\ B_U(t,y)&:=-\frac{\xi_2"(y)(g(t)+h_0)}{[1+\xi_2'(y)(g(t)+h_0)]^3}, \\ C_U(t,y)&:=\frac{\xi_2(y)g'(t)}{1+\xi_2'(y)(g(t)+h_0)} \end{align*} $$

and, for $y>0$ , $t>0$ ,

$$ \begin{align*} \sqrt{A_L(t,y)}&:=\frac{1}{1+\xi_1'(y)(h(t)-h_0)},\\ B_L(t,y)&:=-\frac{\xi_1"(y)(h(t)-h_0)}{[1+\xi_1'(y)(h(t)-h_0)]^3}, \\ C_L(t,y)&:=\frac{\xi_1(y)h'(t)}{1+\xi_1'(y)(h(t)-h_0)}. \end{align*} $$

Denote $h_{1}=-\mu w^{\prime }_{0}(h_{0})$ , $g_1=\mu w^{\prime }_{0}(-h_{0})$ . For $ 0<T\leq {h_{0}}/{16(1+h_{1}+g_1)}+1:=T_1$ , define

$$ \begin{align*} D_{g, T} & :=\{g\in C^1([0,T]):g(0)=-h_0, g^{\prime}(0)=-g_1, \|g^{\prime}+g_1\|_{C([0,T])}\leq 1 \}, \\ D_{h, T} & :=\{h\in C^1([0,T]):h(0)=h_0, h^{\prime}(0)=h_1, \|h^{\prime}-h_1\|_{C([0,T])}\leq 1 \}. \end{align*} $$

Then, $D_T:=D_T^g \times D_T^h$ is a bounded and closed convex set of $C^1([0,T])\times C^1([0,T])$ . For $T\in (0, T_1)$ , we define a subspace of $D_{T_1}$ by $D^T_{T_1}=D^T_{g, T_1}\times D^T_{h, T_1}$ with

$$ \begin{align*} D^T_{g, T_1} & :=\{g\in D_{g, T_1}: g(t)=g(T) \text{ in } [T, T_1] \}, \\ D^T_{h, T_1} & :=\{h\in D_{h, T_1}: h(t)=h(T) \text{ in } [T, T_1]\}. \end{align*} $$

Clearly, we can extend each $(g, h)\in D_T$ to be a member of $D^T_{T_1}$ as indicated above. In what follows, we will always identify $D^T_{T_1}$ with $D_T$ . For each $(g, h)\in D_T=D^T_{T_1}\subseteq D_{T_1}$ , we apply the extension trick of [Reference Wang15, Reference Wang and Du16], the $L^p$ estimates and the Sobolev embedding theorem to show that problem (2.1) has a unique solution $(v_U, v_L)$ with

$$ \begin{align*} \|v_U\| _{C^{{(1+\alpha)}/{2},{1+\alpha}}(\Omega_{T_1}^U)}+ \|v_L\| _{C^{{(1+\alpha)}/{2},{1+\alpha}}(\Omega_{T_1}^L)}\leq C_{T_1} (\|v_U\| _{W_p^{1,2}(\Omega_{T_1}^U)}+\|v_L\| _{W_p^{1,2}(\Omega_{T_1}^L)})\leq C_{1}, \end{align*} $$

where $\Omega _{T_1}^U=[0,T_1]\times [-h_0,0]$ , $\Omega _{T_1}^L=[0,T_1]\times [0, h_0]$ , $C_{1}$ is a constant dependent on $\mu $ , $h_{0}$ , and $\| w_{0}\|_{C^{2}([-h_{0},h_{0}])}$ , $C_{T_1}$ and $C_{T_1}$ depend on $h_0$ , $T_1$ and $\alpha $ .

For $t\in [0,T_1]$ , define

$$ \begin{align*} \hat{h}(t)=h_0-\int_{0}^{t}\mu\partial_y{v_L}(\tau,h_0) \,{d}\tau,\quad \hat{g}(t)=-h_0-\int_{0}^{t}\mu\partial_y{v_U}(\tau,-h_0) \,{d}\tau. \end{align*} $$

Then, $\hat {h}'(t)=-\mu \partial _y{v_L}(t,h_0)$ and $\hat {h}'(0)=-\mu w^{\prime }_{0}(h_{0})=h_1$ , so that $\hat {h}'\in C^{{\alpha }/{2}}([0,T_1])$ with $\|\hat {h}'\| _{C^{{\alpha }/{2}}([0,T_1])}\leq \mu C_{1}=:C_{2}$ . Similarly, $\hat {g}'\in C^{{\alpha }/{2}}([0,T_1])$ with $\|\hat {g}'\| _{C^{{\alpha }/{2}}([0,T_1])}\leq C_{2}$ .

Next, define the mapping $\mathcal {F}:D_T=D^T_{T_1}\to C^1([0,T_1])\times C^1([0,T_1])$ by $\mathcal {F}(g,h)=(\hat {g}, \hat {h})$ . We show that if $T>0$ is small, then $\mathcal {F}$ has a fixed point by the Schauder fixed point theorem. For any $T\in (0, T_1)$ , by the above results,

$$ \begin{align*} \|\hat{h}^{\prime}-h_1\|_{C([0,T])}\leq C_2 T^{{\alpha}/{2}} \quad\text{and }\quad \|\hat{g}^{\prime}+g_1\|_{C([0,T])}\leq C_2 T^{{\alpha}/{2}}. \end{align*} $$

Taking $T< T_2:=\min \{T_1, C_2^{-{2}/{\alpha }}\}$ , we conclude that $\mathcal {F}$ maps $D_T$ into itself. Consequently, by the Schauder fixed point theorem, $\mathcal {F}$ has at least one fixed point which also means that problem ( P ) admits at least one solution $(w_U,w_L,g,h)$ .

Next, we show the uniqueness of the solution to problem ( P ). For this purpose, let $(w_{U}^i,w_{L}^i,g_i,h_i)$ (for $i=1,2$ ) be two solutions of problem ( P ) for $t\in [0,T]$ , and set

$$ \begin{align*} \begin{cases} v_U^i(t,y):=w_U^i(t,y+\xi_{2}(y)(g_i(t)+h_{0})), & y<0,\\ v_L^i(t,y):=w_L^i(t,y+\xi_{1}(y)(h_i(t)-h_{0})), & y>0. \end{cases} \end{align*} $$

From the above estimates,

$$ \begin{align*} \|{v}^i_U\| _{C^{{(1+\alpha)}/{2},{1+\alpha}}(\Omega_{T}^U)}+ \|{v}^i_L\| _{C^{{(1+\alpha)}/{2},{1+\alpha}}(\Omega_{T}^L)}\leq C_{1}, \quad \|{g_i}'\| _{C^{{\alpha}/{2}}([0,T])}\leq C_{2},\quad \|{h_i}'\| _{C^{{\alpha}/{2}}([0,T])}\leq C_{2}. \end{align*} $$

Define $\tilde {g}(t):=g_1(t)-g_2(t)$ , $\tilde {h}(t):=h_1(t)-h_2(t)$ , $V_U(t,y):={v}_U^1(t,y)-{v}_U^2(t,y)$ and $V_L(t,y):={v}_L^1(t,y)-{v}_L^2(t,y)$ . Since $T\leq T_1$ , we can check that for any $t\in [0,T]$ ,

$$ \begin{align*} \begin{cases} \partial_tV_U-A_U^2\partial_{yy}V_U-(B_U^2+C_U^2-\beta_U\sqrt{A_U^2})\partial_yV_U=\tilde{f}_U(t,y), & t\in(0,T],\ y\in(-h_0,0),\\ \partial_tV_L-A_L^2\partial_{yy}V_L-(B_L^2+C_L^2-\beta_L\sqrt{A_L^2})\partial_yV_L=\tilde{f}_L(t,y), & t\in(0,T],\ y\in(0,h_0),\\ V_U(t,-h_0)=0=V_L(t,h_0), & t\in(0,T],\\ V_U(t,0)=V_L(t,0), a_U\partial_yV_U(t,0)=a_L\partial_yV_L(t,0), & t\in(0,T],\\ \end{cases} \end{align*} $$

with $V_U(0,z)=V_L(0,y)=0$ for $z\in [-h_0,0]$ , $y\in [0,h_0]$ and

$$ \begin{align*} \tilde{f}_U(t,y) & =f(v_{U}^1)-f(v_{U}^2)+(A_U^1-A_U^2)\partial_{yy}{v}_{U}^1 \\&\quad +\bigg[B_{U}^1-B_{U}^2+C_{U}^1-C_{U}^2-\beta_U\bigg(\sqrt{A_{U}^1}-\sqrt{A_{U}^2}\bigg)\bigg]\partial_{y}{v}_{U}^1,\\\tilde{f}_L(t,y) & =f(v_{L}^1)-f(v_{L}^2)+(A_L^1-A_L^2)\partial_{yy}{v}_{L}^1 \\&\quad +\bigg[B_{L}^1-B_{L}^2+C_{L}^1-C_{L}^2-\beta_L\bigg(\sqrt{A_{L}^1}-\sqrt{A_{L}^2}\bigg)\bigg]\partial_{y}{v}_{L}^1, \end{align*} $$

where $A^i_j$ , $B^i_j$ and $C^i_j$ are the coefficients of problem (2.1) with $(g_i,h_i)$ in place of $(g, h)$ for $j\in \{U, L\}$ . We apply the extension trick of [Reference Wang15, Reference Wang and Du16], the $L^p$ estimates for parabolic equations and the Sobolev embedding theorem to see that

(2.2) $$ \begin{align} \|V_U\|_{C^{{(1+\alpha)}/{2},1+\alpha}(\Omega_{T}^U)}+\|V_L\|_{C^{{(1+\alpha)}/{2},1+\alpha}(\Omega_{T}^L)}\leq C_3(\|\tilde{g}\|_{C^1([0,T])}+\|\tilde{h}\|_{C^1([0,T])}), \end{align} $$

where $C_3$ depends only on $h_{0}$ , $\| w_{0}\|_{C^{2}}$ , $C_1,\ C_2$ and $\alpha $ , but is independent of $T\in (0,T_1]$ .

Recalling that $\tilde {g}'(0)=g_1'(0)-g_2'(0)$ , from (2.2),

$$ \begin{align*} \|\tilde{g}\|_{C^1([0,T])}\leq 2 T^{{\alpha}/{2}} \|\tilde{g}'\|_{C^{{\alpha}/{2}}([0,T])}\leq C_4 T^{{\alpha}/{2}} (\|\tilde{g}\|_{C^1([0,T])}+\|\tilde{h}\|_{C^1([0,T])}), \end{align*} $$

where $C_4:=2\mu C_3$ . Similarly, $ \|\tilde {h}\|_{C^1([0,T])}\leq C_4 T^{{\alpha }/{2}} (\|\tilde {g}\|_{C^1([0,T])}+\|\tilde {h}\|_{C^1([0,T])})$ . Consequently,

(2.3) $$ \begin{align} \|\tilde{g}\|_{C^1([0,T])}+\|\tilde{h}\|_{C^1([0,T])}\leq 2C_4 T^{{\alpha}/{2}} (\|\tilde{g}\|_{C^1([0,T])}+\|\tilde{h}\|_{C^1([0,T])}). \end{align} $$

If we take $T<\min \{T_1,\ C_2^{-{2}/{\alpha }},\ (4C_4)^{-{2}/{\alpha }}\}$ , it then follows from (2.3) that

$$ \begin{align*} \|\tilde{g}\|_{C^1([0,T])}+\|\tilde{h}\|_{C^1([0,T])}\leq \tfrac{1}{2}(\|\tilde{g}\|_{C^1([0,T])}+\|\tilde{h}\|_{C^1([0,T])}). \end{align*} $$

This implies that $\tilde {g}\equiv 0\equiv \tilde {h}$ , which implies that $V_U\equiv 0\equiv V_L$ , establishing uniqueness of local solutions of problem ( P ). By the Schauder estimates, $w_U(t,x)\in C^{1+\alpha /2,2+\alpha }((0,T]\times [g(t),0])$ and $w_L(t,x)\in C^{1+\alpha /2,2+\alpha }((0,T]\times [0, h(t)])$ .

Proof. Let $[0,T_{\mathrm {max}})$ be the maximal time interval of existence for the solution. We assume $T_{\mathrm {max}}<\infty $ . Let $K_0:=\|w_0\|_{L^{\infty }}+1$ . From Lemma 2.1,

$$ \begin{align*} 0\leq w_U(t,x), w_L(t,y)\leq K_0 \quad\text{for }t\in(0,T_{\mathrm{max}}),\ x\in[g(t),0],\ y\in[0,h(t)]. \end{align*} $$

Furthermore, $f(s)=s-s^2 \leq K_0$ for $s\in [0,K_0]$ . For $(t,x)\in (0,T_{\mathrm {max}})\times (h(t)-M^{-1}, h(t))$ , define $u_L(t,x):=K_0[2M(h(t)-x)-M^2(h(t)-x)^2]$ , where

$$ \begin{align*} M:=\max\bigg\{h_{0}^{-1}, \sqrt{\dfrac{1}{2} \beta_L}+\sqrt{\dfrac{1}{4}\beta_L^2+1}, \sqrt{\dfrac{1}{2}\beta_U}+\sqrt{\dfrac{1}{4}\beta_U^2+1}, \frac{4\|w_{0}\|_{C^{1}([0,h_{0}])}}{3K_{0}}\bigg\}. \end{align*} $$

From direct calculations and the comparison principle,

$$ \begin{align*} -g'(t), h'(t)\leq 2\mu K_0M:=K_1\quad \text{for } t\in(0, T_{\mathrm{max}}). \end{align*} $$

Similar to the proof of Theorem 2.3, by the $L^p$ theory,

$$ \begin{align*} (w_U,w_L)\in W^{1,2}_p([0,T]\times [g(t), 0])\times W^{1,2}_p([0,T]\times [0, h(t)]) \end{align*} $$

for any $p>1$ and $T\in (T_{\mathrm {max}}/2, T_{\mathrm {max}})$ . From the Sobolev embedding theorem, for any $\alpha \in (0,1)$ and some large enough $p>1$ depending on $\alpha $ ,

(2.4) $$ \begin{align} \|w_L\|_{C^{{(1+\alpha)}/{2},{1+\alpha}}([0, T]\times [0, h(t)])}+\|w_U\|_{C^{{(1+\alpha)}/{2},{1+\alpha}}([0, T]\times [g(t), 0])}\leq C_{p,\alpha}, \end{align} $$

where $C_{p,\alpha }$ is a positive constant independent of $T\in (T_{\mathrm {max}}/2, T_{\mathrm {max}})$ .

Choose $t_n\in (0,T_{\mathrm {max}})$ with $t_n\nearrow T_{\mathrm {max}}$ , and regard $(w_U(t_n, \cdot ), w_L(t_n, \cdot ), g(t_n), h(t_n))$ as the initial data. By (2.4) and the properties of g and h, we can repeat the proof of Theorem 2.3 to conclude that there is $s_0>0$ depending on $C_{p,\alpha }$ but independent of n such that problem ( P ) has a unique solution $(w_U, w_L, g, h)$ for $t\in [t_n, t_n+s_0]$ . It follows that problem ( P ) admits a solution $(w_U, w_L,g,h)$ , which is defined for $t\in [0,t_n+s_0]$ . As $t_n+s_0>T_{\mathrm {max}}$ for $n\gg 1$ , this contradicts the definition of $T_{\mathrm {max}}$ , and so we see that $T_{\mathrm {max}}=\infty $ , as desired. The proof is complete.

Proof. (i) We can check directly that if $l_2 \geq {2\pi }/{\sqrt {4-\beta _{L}^2}}$ , then the principal eigenvalue $\tilde {\lambda }_1$ of

$$ \begin{align*} -\varphi" + \beta_{L}\varphi'-\varphi= \lambda \varphi\ \text{ in }\ (0, l_2), \quad \varphi(0) =0=\varphi(l_2), \end{align*} $$

satisfies $\tilde {\lambda }_1\leq 0$ . By the comparison principle, we obtain the desired result.

(ii) Given $0<l_{1}^{1}<l_{1}^{2}< {2\pi }/{\sqrt {4-\beta _{L}^2}}$ , let $(\varphi _{L,i}, \varphi _{U,i})$ be the positive principal eigenfunction pair of problem (3.1) corresponding to $\lambda _{1}^{i}$ with $l_1$ replaced by $l_{1}^{i}$ for $i=1,\ 2$ . Then, $\varphi _{U,i}(-l_{1}^{i}) = 0=\varphi _{L,i}(l_2)$ and

$$ \begin{align*} \begin{cases} -\varphi_{U,i}" + \beta_{U}\varphi_{U,i}'-\varphi_{U,i}= \lambda_1^{i}\varphi_{U,i}, & -l_{1}^{i} < x < 0, \\ -\varphi_{L,i}" + \beta_{L}\varphi_{L,i}'-\varphi_{L,i}= \lambda_1^{i}\varphi_{L,i}, & 0 < x < l_2, \\ \varphi_{U,i}(0) = \varphi_{L,i}(0), \quad a_{U}\varphi_{U,i}'(0) = a_{L}\varphi_{L,i}'(0). \end{cases} \end{align*} $$

Motivated by [Reference Meng, Ge and Lin13, Reference Xu, Lin and Santos17], multiply the equation for $\varphi _{L,1}$ by $\varphi _{L,2}$ and the equation for $\varphi _{L,2}$ by $\varphi _{L,1}$ , subtract the resulting equations, and integrate over $(0,l_2)$ to obtain

(3.3) $$ \begin{align} (\lambda_1^{1}-\lambda_1^{2})\int_{0}^{l_2}\varphi_{L,1}\varphi_{L,2}\,{d}x= \varphi_{L,2}(0)\varphi_{L,1}^{\prime}(0)-\varphi_{L,1}(0)\varphi_{L,2}^{\prime}(0). \end{align} $$

Similarly,

(3.4) $$ \begin{align} (\lambda_1^{1}-\lambda_1^{2})\int_{-l_{1}^{1}}^{0}\varphi_{U,1}\varphi_{U,2}\, {d}x= \varphi_{U,1}(0)\varphi_{U,2}^{\prime}(0)-\varphi_{U,2}(0)\varphi_{U,1}^{\prime}(0)+\varphi_{U,2}(-l_{1}^{1})\varphi_{U,1}^{\prime}(-l_{1}^{1}). \end{align} $$

We claim that $\Phi = \varphi _{L,2}(0)\varphi _{L,1}^{\prime }(0)-\varphi _{L,1}(0)\varphi _{L,2}^{\prime }(0)> 0$ . Otherwise, $\Phi \leq 0$ . It follows from (3.3) that $\lambda _1^{1} \leq \lambda _1^{2}$ . Moreover,

$$ \begin{align*} \varphi_{U,2}(0)\varphi_{U,1}^{\prime}(0)-\varphi_{U,1}(0)\varphi_{U,2}^{\prime}(0) =a_L a_U^{-1}[\varphi_{L,2}(0)\varphi_{L,1}^{\prime}(0)-\varphi_{L,1}(0)\varphi_{L,2}^{\prime}(0)]\leq0. \end{align*} $$

This, together with (3.4), implies that $\lambda _1^{1}>\lambda _1^{2}$ , which is a contradiction. This proves our claim and that $\lambda _1^{1}>\lambda _1^{2}$ . The monotonicity of $\lambda _1(l_1,l_2)$ in $l_1>0$ is established.

Next, we prove that $\lambda _{0} := \lim _{l_{1} \to 0} \lambda_1 (l_{1}, l_{2})> 0$ . The monotonicity of $\lambda _1(l_1,l_2)$ in $l_1>0$ implies the existence of the limit $\lambda _0$ . Letting $l_1\to 0$ in problem (3.1) gives

$$ \begin{align*} -\varphi^{\prime\prime}_{L} + \beta_{L}\varphi^{\prime}_{L}-\varphi_{L}= \lambda_0 \varphi_{L} \text{ for } x\in(0, l_2),\quad \varphi_{L}(0) =0=\varphi_{L}(l_2). \end{align*} $$

Since $l_2<{2\pi }/{\sqrt {4-\beta _{L}^2}}$ , we see that $\lambda _0>0$ . Moreover, since $\beta _{U}\in (0,2)$ , it follows from a similar argument to that in item (i) that $\lambda (l_{1}, l_{2}) < 0$ when $l_1 \geq {2\pi }/{\sqrt {4-\beta _U^2}}$ . This, together with the above arguments, implies the existence and uniqueness of $\ell ^{*}(l_2)$ .

Finally, setting $\varphi _j(x):=e^{\beta _jx/2}\psi _{j}(x)$ for $j=U,\ L$ and $l_1=\ell ^{*}(l_2)$ in (3.1), we see that there are four constants $C_1>0$ , $C_2>0$ , $\theta _1$ and $\theta _2$ such that

$$ \begin{align*} \psi_L(x)=C_1\cos\sqrt{1- \tfrac{1}{4}\beta_L^2}(x-\theta_1),\quad \psi_U(x)=C_2\cos\sqrt{1-\tfrac{1}{4}\beta_U^2}(x-\theta_2), \end{align*} $$

on the respective intervals $(0,l_2)$ and $(-\ell^*(l_2),0)$ . Since $\psi _{U}(-\ell ^{*}(l_2)) = \psi _{L}(l_2) = 0$ , then

$$ \begin{align*}(l_{2}-\theta_{1})\sqrt{1- \tfrac{1}{4}\beta_{L}^{2}} = ({\pi}/{2}) = (\ell^{*}(l_2)+\theta_2)\sqrt{1- \tfrac{1}{4}\beta_{U}^{2}}. \end{align*} $$

Recalling that $a_{U}\psi _{U}^{\prime }(0)=a_{L}\psi _{L}^{\prime }(0)$ and $\psi _{L}(0)=\psi _{U}(0)$ , a direct calculation yields

$$ \begin{align*} a_{U}\sqrt{4- \beta_U^2} \tan \theta_2\sqrt{1- \tfrac{1}{4}\beta_U^2}= a_{L}\sqrt{4- \beta_L^2} \tan \theta_1\sqrt{1- \tfrac{1}{4}\beta_L^2}. \end{align*} $$

This, together with the above results, proves (3.2). Further, $\ell ^{*}(l_2)$ is decreasing in $l_2$ , $\ell ^{*}(l_2)\to 0$ as $l_2\to {2\pi }/{\sqrt {4-\beta _{L}^2}}$ and $\ell ^{*}(l_2)\to {2\pi }/{\sqrt {4-\beta _U^2}}$ as $l_2\to 0$ . The monotonicity of $\ell ^{*}(l_2)$ yields the existence and uniqueness of $\hbar ^*$ .

Proof. (i) $\Rightarrow $ (ii) Without loss of generality, assume that $h_{\infty }< \infty $ . Aiming for a contradiction, assume that $\lambda _1(-g_\infty , h_\infty )<0$ . Then, there is $t_1>0$ large enough such that $\lambda _1(-g(t_1), h(t_1))<0$ . Now, we consider the auxiliary problem:

$$ \begin{align*} \begin{cases} \partial_{t}\underline{w}_{U}-\partial_{xx}\underline{w}_{U}+\beta_{U}\partial_{x}\underline{w}_{U}=\underline{w}_{U}-\underline{w}_{U}^{2}, & g(t_1)<x<0,\ t>t_1,\\ \partial_{t}\underline{w}_{L}-\partial_{xx}\underline{w}_{L}+\beta_{L}\partial_{x}\underline{w}_{L}=\underline{w}_{L}-\underline{w}_{L}^{2}, & 0<x<\xi(t),\ t>t_1,\\ \underline{w}_{L}(t,\xi(t))=0=\underline{w}_{U}(t,g(t_1)),& t>t_1,\\ \xi(t_1)=h(t_1),\ \xi'(t)=-\mu\partial_{x}\underline{w}_{L}(t,\xi(t)), & t>t_1,\\ \underline{w}_{U}(t,0)=\underline{w}_{L}(t,0), \ a_U\partial_{x}\underline{w}_{U}(t,0)=a_L\partial_{x}\underline{w}_{L}(t,0), & t>t_1,\\ \underline{w}_{L}(t_1,x)=w_L(t_1, x),\ \underline{w}_{U}(t_1,y)=w_U(t_1,y), & x\in[0,h(t_1)],\ y\in[g(t_1),0]. \end{cases} \end{align*} $$

Clearly, $(\underline {w}_U, \underline {w}_{L})$ is a subsolution of problem ( P ). Thus, $\xi (t)\leq h(t)$ and $\xi (\infty )<\infty $ . It follows from a similar argument to that in [Reference Du and Guo5, Lemma 2.2] that

$$ \begin{align*} \|\underline{w}_{L}(t,\cdot)- \Phi_L(\cdot)\|_{C^2([h_0, \xi(t)])} \to 0,\quad \text{as } t\to\infty, \end{align*} $$

where $(\Phi _U, \Phi _{L})$ is the solution pair of

$$ \begin{align*} \begin{cases} -\Phi_{U}"+\beta_{U}\Phi^{\prime}_{U}=\Phi_{U}-\Phi_U^{2}, & g(t_1)<x<0,\\ -\Phi_L"+\beta_{L}\Phi^{\prime}_{L}=\Phi_{L}-\Phi_{L}^{2}, & 0<x<\xi(\infty), \\ \Phi_{L}(\xi(\infty))=0=\Phi_{U}(g(t_1)),\\ \Phi_U(0)=\Phi_{L}(0), \ a_U\Phi^{\prime}_{U}(0)=a_L\Phi^{\prime}_{L}(0). \end{cases} \end{align*} $$

It follows that $\xi '(t) \to -\mu \Phi _{L}'(\xi (\infty ))>0$ , which contradicts $\xi (\infty )<\infty $ .

(ii) $\Rightarrow $ (i) Since $\beta _{L},\ \beta _{U}\in (0,2)$ , if assertion (ii) holds, we can check that $h_\infty <{2\pi }/{\sqrt {4-\beta _{L}^2}}$ and $g_\infty>-{2\pi }/{\sqrt {4-\beta _U^2}}$ , which implies assertion (i).

(iii) $\Rightarrow $ (ii) We use a proof by contradiction and assume that, for some small $\varepsilon>0$ , there exists $t_2>0$ large such that $\lambda _\varepsilon :=\lambda _1(-g(t_2)-\varepsilon , h(t_2)-\varepsilon )<0$ . Let $(\varphi _U, \varphi _L)$ be the corresponding positive eigenfunction pair with $\max \{\|\varphi _U\|_{L^\infty },\ \|\varphi _L\|_{L^\infty }\}=1$ . Set

$$ \begin{align*} \begin{cases} u_U(t,x)=\delta\varphi_U(x) & \text{for } t\geq0,\ x\in[g(t_2)+\varepsilon,0], \\ u_L(t,x)=\delta\varphi_L(x) & \text{for } t\geq0,\ x\in[0,h(t_2)-\varepsilon], \end{cases} \end{align*} $$

where $0<\delta <\min \{-\lambda _\varepsilon , \min _{x\in [0,h(t_2)-\varepsilon ]}w_L(t_2,x), \min _{x\in [g(t_2)+\varepsilon ,0]}w_U(t_2,x)\}$ . It follows from Lemma 2.1 that $w_U(t+t_2,x) \geq u_U(t,x)$ for $(t,x)\in [0,\infty )\times [0, h(t_2)-\varepsilon ]$ , which contradicts assertion (iii).

(ii) $\Rightarrow $ (iii) As $\beta _{L},\ \beta _{U}\in (0,2)$ , it follows from the proof of Lemma 3.1 and $\lambda _1(-g_\infty , h_\infty )\geq 0$ that $h_\infty $ and $g_\infty $ are both bounded. This, together with [Reference Du and Lou7, Theorem 1.1], implies assertion (iii) and completes the proof.

Proof. Choosing $h_1\in (h_0, \hbar ^*)$ , it follows from Lemma 3.1 that (3.1) with $l_1=l_2=h_1$ admits a positive principal eigenvalue, denoted by $\lambda ^*$ , whose corresponding positive eigenfunction pair, denoted by $(\varphi ^*_U, \varphi ^*_L)$ , can be normalised by $\|\varphi ^*_U\|_{L^\infty }+\|\varphi ^*_L\|_{L^\infty }=1$ . There exists $x^*\in (h_1/2, h_1)$ such that

$$ \begin{align*} (\varphi^*_U)'(x)>0>(\varphi^*_L)'(y)\quad\text{for } x\in[-h_1, -x^*],\ y\in[x^*,h_1]. \end{align*} $$

Set $ \delta \kern1.5pt{:=}\kern1.5pt\min \{{h_1}/{h_0}-1, \lambda _*, 1\} , \eta \kern1.5pt{: =}\kern1.5pt\max \{h_1-({\delta }/{2}) h_0,h_0,x^*\}, \varepsilon _0\kern1.5pt{:=}\kern1.5pt\min \{\varphi ^*_U(-\eta )$ , $\varphi ^*_L(\eta ) \}. $ Then, $\varepsilon _0 \leq 1$ and there is $\varepsilon _1 = \varepsilon _1(\delta )>0$ such that $ 2 \mu \varepsilon _1 [(\varphi ^*_U)'(-h_1)-(\varphi ^*_L)'(h_1)] < \delta ^2 h_0. $ Define

$$ \begin{align*} \begin{cases} v_U(t,x)=\varepsilon_0 \varepsilon_1 e^{-\delta t} \varphi^*_U(x) & \text{for } t\geq0,\ x\in[-h_1,0], \\ v_L(t,x)=\varepsilon_0 \varepsilon_1 e^{-\delta t} \varphi^*_L(x) & \text{for } t\geq0,\ x\in[0,h_1]. \end{cases} \end{align*} $$

If $w_0$ is chosen to be sufficiently small such that $w_0\leq \varepsilon _0 \varepsilon _1 \varphi ^*_U$ in $[-h_0,0]$ , and $w_0\leq \varepsilon _0 \varepsilon _1 \varphi ^*_L$ in $[0,h_0]$ , it follows from the comparison principle that

$$ \begin{align*} \begin{cases} w_U(t,x)\leq v_U(t,x) & \text{for } t\in[0,\tau),\ x\in[g(t),0], \\ w_L(t,x) \leq v_L(t,x) & \text{for } t\in[0,\tau),\ x\in[0, h(t)], \end{cases} \end{align*} $$

where $\tau :=\sup \{t>0 : h(t)<h_1,\ g(t)>-h_1\}$ . We prove that $\tau =\infty $ . Once this is done, we see that $(g_\infty , h_\infty )$ is bounded, and that $\|w_U(t,\cdot )\|_{L^\infty ([g(t),0])}\to 0$ and $\|w_L(t,\cdot )\|_{L^\infty ([0, h(t)])}\to 0$ as $t\to \infty $ , so the vanishing conclusion is established.

In what follows, we prove $\tau =\infty $ by contradiction. Assume that $\tau <\infty $ . Without loss of generality, we may assume that $h(\tau )=h_1$ . Motivated by [Reference Sun, Lou and Zhou14], define

$$ \begin{align*} k(t):=h_0 (1+\delta-\tfrac{1}{2}\delta e^{-\delta t}), \quad \overline{w_L}(t,x):=\varepsilon_1 e^{-\delta t} \varphi^*_L(x-k(t)+h_1) \end{align*} $$

for $t \geq 0$ , $x\in I_1(t):=[\eta +k(t)-h_1,k(t)]$ . We check that $\eta +k(t)-h_1 \geq h_0$ and

$$ \begin{align*} \begin{cases} \partial_{t}\overline{w_L}-\partial_{xx}\overline{w_L}+\beta_{L}\partial_{x}\overline{w_L}-\overline{w_L}+\overline{w_L}^{2}\geq 0, & t \geq 0,\ x\in I_1(t),\\ k^{\prime}(t)=\dfrac{\delta^{2} h_{0}}{2} e^{-\delta t} \geq -\mu \varepsilon_{1} e^{-\delta t} (\varphi^*_L)'(h_1) =-\mu \partial_{x}\overline{w_L}(t, k(t)), & t>0. \end{cases} \end{align*} $$

We claim that $k(t)\geq h(t)$ for $t\in [0,\tau ]$ . As $\eta \leq h_1$ , the desired conclusion is true when $\eta +k(t)-h_1\geq h(t)$ . We consider the case where $\{ 0\leq t\leq \tau : h(t)>\eta + k(t)-h_1\}\not = \emptyset $ consists of some intervals and $[s_1, s_2]$ is one of them. By the choice of $\eta $ , we see that $s_1>0$ , $h(s_1)=\eta + k(s_1)-h_1$ . As $\eta + k(t)-h_1>h_0$ for $t\geq 0$ ,

$$ \begin{align*} w_L(t, \eta+ k(t)-h_1) \leq \varepsilon_0 \varepsilon_1 e^{-\delta t} \leq \varepsilon_1 e^{-\delta t} \varphi^*_L(t, \eta) \equiv \overline{w_L}(t, \eta+ k(t)-h_1),\quad t\in [s_1, s_2]. \end{align*} $$

Hence, by comparison, $h(t)< k(t)$ in $[s_1, s_2]$ . This proves our claim and shows that

$$ \begin{align*} h(\tau)\leq k(\tau)< k(\infty)\leq h_1, \end{align*} $$

which contradicts $h(\tau )=h_1$ . Thus, $\tau =\infty $ , which completes the proof.

Proof. We first check that $[-\hbar ^*,\hbar ^*]\subset [g(t),h(t)]$ for any $t>0$ . By Lemma 3.2, $-g_\infty =h_\infty =\infty $ and $(w_U,w_L)\nrightarrow (0, 0)$ as $t \to \infty $ locally uniformly in $(-\infty ,0]\times [0,\infty )$ .

In what follows, we prove that $(w_U,w_L)$ converges to $(1, 1)$ locally uniformly as $t \to \infty $ . Setting $\bar {w}(t)=1+\|w_0\|_{L^\infty ([-h_0,h_0])}e^{-t}$ , we can check that

(3.5) $$ \begin{align} \begin{cases} \limsup_{t\to\infty}\|w_U(t,\cdot)\|_{L^\infty([g(t),0])}\leq \lim_{t\to\infty}\bar{w}(t)=1,\\ \limsup_{t\to\infty}\|w_L(t,\cdot)\|_{L^\infty([0,h(t)])}\leq \lim_{t\to\infty}\bar{w}(t)=1. \end{cases} \end{align} $$

As $h_0\geq \hbar ^*$ , there is $\epsilon>0$ small such that $[-\hbar ^*-\epsilon , \hbar ^*+\epsilon ]\subseteq (g(1)+\epsilon , h(1)-\epsilon )$ . It follows from Lemma 3.1 that (3.1) with $l_1=l_2=\hbar ^*+\epsilon $ admits a negative principal eigenvalue, whose corresponding positive eigenfunction pair $(\varphi ^\epsilon _U, \varphi ^\epsilon _L)$ , can be normalised by $\|\varphi ^\epsilon _U\|_{L^\infty }+\|\varphi ^\epsilon _L\|_{L^\infty }=1$ . We can find $\rho>0$ small such that

(3.6) $$ \begin{align} w_U(1,x)\geq \rho\varphi^\epsilon_U(x),\quad w_L(1,y)\geq \rho\varphi^\epsilon_L(y)\quad\text{for } x\in[-\hbar^*-\epsilon, 0],\ y\in[0, \hbar^*+\epsilon]. \end{align} $$

Let $(\underline {w}_U, \underline {w}_L, \underline {g}, \underline {h})$ be the solution of ( P ) with initial function $(\rho \varphi ^\epsilon _U, \rho \varphi ^\epsilon _L)$ . We can check directly that $-\underline {g}_\infty =\underline {h}_\infty =\infty $ . By the standard parabolic comparison principle,

$$ \begin{align*} \underline{w}_U(t,x)\geq \rho\varphi^\epsilon_U(x),\quad \underline{w}_L(t,y)\geq \rho\varphi^\epsilon_L(y)\quad\text{for } t>0,\ x\in[-\hbar^*-\epsilon, 0],\ y\in[0, \hbar^*+\epsilon]. \end{align*} $$

It follows from Lemma 2.1 that $(\underline {w}_U, \underline {w}_L)$ is nondecreasing with respect to t. Define $(\underline {w}_{U,\infty }(x), \underline {w}_{L,\infty }(x)):=(\lim _{t\to \infty }\underline {w}_U(t,x), \lim _{t\to \infty }\underline {w}_L(t,x))$ . Note that $-\underline {g}_\infty =\underline {h}_\infty =\infty $ . By a similar argument to that in [Reference Du, Lou, Peng and Zhou8], $(\underline {w}_{U,\infty }(x), \underline {w}_{L,\infty }(x))$ is a positive solution of

$$ \begin{align*} \begin{cases} -\partial_{xx}W_{U}+\beta_{U}\partial_{x}W_{U}-W_{U}+W_{U}^{2}=0, & -\infty<x<0,\\ -\partial_{xx}W_{L}+\beta_{L}\partial_{x}W_{L}-W_{L}+W_{L}^{2}=0, & 0<x<\infty,\\ W_{U}(0)=W_{L}(0), \ a_U\partial_{x}W_{U}(0)=a_L\partial_{x}W_{L}(0), \end{cases} \end{align*} $$

and satisfies $0<\underline {w}_{U,\infty }\leq 1$ , $0<\underline {w}_{L,\infty }\leq 1$ . Thanks to [Reference Du, Lou, Peng and Zhou8, Theorem 1.1], the above problem admits a unique positive solution pair $(1,1)$ . Thus,

(3.7) $$ \begin{align} (\underline{w}_U(t,\cdot), \underline{w}_L(t,\cdot))\to (1,1) \quad \text{locally uniformly as } t\to\infty. \end{align} $$

Moreover, from (3.6) and Lemma 2.1,

$$ \begin{align*} \underline{w}_U(t,x)\leq w_U(t+1,x), \quad \underline{w}_L(t, y)\leq w_L(t+1,y)\quad\text{for } t>0,\ x\in[\underline{g}(t), 0],\ y\in[0, \underline{h}(t)]. \end{align*} $$

From these estimates and (3.7),

$$ \begin{align*} \begin{cases} \liminf_{t\to\infty}w_U(t,x)\geq 1 &\text{locally uniformly in } (-\infty,0],\\ \liminf_{t\to\infty}w_L(t,x)\geq 1 &\text{locally uniformly in } [0, +\infty). \end{cases} \end{align*} $$

From this, together with (3.5), $(w_U,w_L)\to (1,1) \text { locally uniformly as } t\to \infty $ , which completes the proof of this lemma.