Exponential stability of a traveling wave for an area preserving curvature motion with two endpoints moving freely on a line

Abstract

The asymptotic behavior of solutions to an area-preserving curvature flow of planar curves in the upper half plane is investigated. Two endpoints of the curve slide along the horizontal axis with prescribed fixed contact angles. First, by establishing an isoperimetric inequality, we prove the global existence of the solution. We then study the asymptotic behavior of solutions with concave initial data near a traveling wave.

Keywords

curvature flow free boundary problem motion of planar curves area-preserving isoperimetric ratio traveling wave

1. Introduction

In this paper, we deal with the motion by area preserving curvature flow of planar curves having two moving end points on the x-axis with fixed interior contact angles to this axis. The problem is formulated as follows. For any curve γ lying on the upper half-plane having two end points, we shall denote the left and right interior angles by ${Ang}_{-} (γ)$ and ${Ang}_{+} (γ)$ . For any given such curve $Γ^{0} = Γ (0)$ , our problem is to look for a family of curves ${Γ (t)}_{t ⩾ 0}$ having two end points on the x-axis with the same fixed contact angles as above evolve by the area preserving curvature flow equation: $\begin{matrix} (1.1) & V = κ - \frac{\int_{Γ (t)} κ d s}{\int_{Γ (t)} d s} = κ + \frac{ψ_{+} + ψ_{-}}{L (t)}, {Ang}_{\pm} (Γ (t)) = {Ang}_{\pm} (Γ (0)) . \end{matrix}$ Here V is the outward normal velocity of the curve, $L (t)$ is the length of $Γ (t)$ and κ is the (signed) curvature of $Γ (t)$ . If the solution curve is represented as a graph, the problem is reduced to the following free boundary problem (P): $\begin{array}{rcl} (1.2) & u_{t} = \frac{u_{x x}}{1 + u_{x}^{2}} + \frac{ψ_{+} + ψ_{-}}{L (t)} \sqrt{1 + u_{x}^{2}}, x \in (l_{-} (t), l_{+} (t)), t > 0, \\ (1.3) & u (l_{\pm} (t), t) = 0, t > 0, \\ (1.4) & u_{x} (l_{\pm} (t), t) = \mp tan ψ_{\pm}, t > 0, \\ (1.5) & u (x, 0) = u^{0} (x), x \in [l_{-}^{0}, l_{+}^{0}], l_{\pm} (0) = l_{\pm}^{0}, \end{array}$ where $L (t) = \int_{l_{-} (t)}^{l_{+} (t)} \sqrt{1 + u_{x}^{2} (x, t)} d x$ and we assume that $ψ_{\pm} \in (0, π / 2)$ , $- \infty < l_{-}^{0} < l_{+}^{0} < \infty$ . Note that the arclength parameter s is given by $s (x, t) : = \int_{l_{-} (t)}^{x} \sqrt{1 + u_{x}^{2} (x, t)} d x$ . The purpose of this paper is to prove that the curve remains uniformly bounded and to study the asymptotic behavior of the solution as $t \to \infty$ for the problem (P). Throughout this paper, we also assume $u^{0} \in C^{2} ([l_{-}^{0}, l_{+}^{0}])$ and $\begin{matrix} (1.6) & u^{0} (l_{\pm}^{0}) = 0, u_{x}^{0} (l_{\pm}^{0}) = \mp tan ψ_{\pm}, {(u^{0})}_{x x} < 0 . \end{matrix}$ Note that these conditions also yield the positivity $u^{0} (x) > 0$ on $(l_{-}^{0}, l_{+}^{0})$ .

As is easily seen, a direct calculation yields $\begin{matrix} (1.7) & \{\begin{matrix} A^{'} (t) = 0, \\ L^{'} (t) = l_{+}^{'} (t) cos ψ_{+} - l_{-}^{'} (t) cos ψ_{-} + \frac{{(ψ_{+} + ψ_{-})}^{2}}{L (t)} - \int_{0}^{L (t)} κ^{2} d s, \end{matrix} \end{matrix}$ where $A (t) = \int_{l_{-} (t)}^{l_{+} (t)} u (x, t) d x$ (the area enclosed by the graph $Γ (t)$ and x-axis). Here we used the identities $\begin{matrix} (1.8) & l_{\pm}^{'} (t) = \pm cot ψ_{\pm} u_{t} (l_{\pm} (t), t), \end{matrix}$ which follow from differentiating $u (l_{\pm} (t), t) = 0$ by t and using the boundary condition (1.4). For this reason, Eq. (1.1) is so-called area-preserving curvature flow, which arises in application fields [4,8,9,14,16] such as phase transitions, image processing and droplet model, etc. For the curve shortening flow $V = κ$ , there is extensive literature on this subject both for simple closed curves and non-simple ones see [2,10,11] for example. As for the above free boundary problem for $V = κ$ , we refer to the work [5] and the references cited therein. Recently, the authors of [12] consider the flow $V = κ + c_{0}$ with a constant $c_{0} > 0$ under the same boundary condition. They first concern the classification of solutions. There are three possibility for the behavior of the solution, that is, area expanding, area bounded and area shrinking types. The detail asymptotic behavior of solutions and eventual concavity property are considered. They show that solutions are asymptotically self-similar both in the area expanding and the area shrinking cases. For the area bounded case, solutions converge to either a stationary solution or a traveling wave. Our main result explained in the below is similar to the convergence result for the area bounded case of [12], but the proof in our paper is completely different. The reason is that the intersection argument of [12], which was exceedingly useful, cannot be applied to our problem (P).

Our first main result is related to global existence and boundedness of concave solutions:

Theorem 1.1 (Globally bounded).

Any concave solution of (P) exists globally in time. The solution $u (x, t)$ and all derivatives are uniformly bounded. Furthermore, $L (t)$ remains bounded from above and below by two positive constants as $t \to \infty$ .

To prove this theorem, we first need to show that the length of the curve is uniformly bounded. In order to obtain it, we show that the aspect ratio of the curve remains bounded by using an idea similar to Proposition 6.2 of [12]. Combining this and the local existence theorem, we can prove the solution exists globally in time.

The second main result of this paper is related to the asymptotic behavior of the solution. A traveling wave of the problem (P) is a solution that has the form $u^{*} (x, t) = U (x - c t - a)$ , where c denotes the wave speed, $U (ξ)$ is called the profile of the wave and a is an arbitrary constant that adjusts the phase. Substituting this form into (1.2)–(1.4), we obtain $\begin{matrix} (1.9) & \{\begin{matrix} \frac{U_{ξ ξ}}{1 + U_{ξ}^{2}} + c U_{ξ} + \frac{ψ_{+} + ψ_{-}}{L^{*}} \sqrt{1 + U_{ξ}^{2}} = 0 in (- b, b), \\ U (\pm b) = 0, U_{ξ} (\pm b) = \mp tan ψ_{\pm}, \end{matrix} \end{matrix}$ where $L^{*} = \int_{- b}^{b} \sqrt{1 + U_{ξ}^{2} (ξ)} d ξ$ . Here $b > 0$ is some constant that shift the center of the support of $U$ to the origin. We shall discuss the existence of traveling wave solutions in Section 4. Roughly speaking, for any given area parameter $A^{*}$ (the area enclosed by the traveling wave and x-axis), there exists concave traveling wave solution and that is unique up to translation. In order to find a traveling wave solution, we analyze the corresponding evolution equation for the curvature, whose stationary solution corresponds to a traveling wave by following the idea of [12].

The second main theorem of this paper is the following.

Theorem 1.2 (Asymptotic shape).

Let $(u, l_{\pm})$ be a solution of (P) with (1.6) and let $U (ξ)$ be a solution of (1.9) with $\int_{l_{-}^{0}}^{l_{+}^{0}} u_{0} (x) d x = \int_{- b}^{b} U (ξ) d ξ$ . Let $κ (θ, t) : = u_{x x} / {(1 + u_{x}^{2})}^{3 / 2}$ be the curvature of the solution curve, where $θ (x, t) = arctan u_{x} (x, t)$ . Let $κ^{*} (θ) : = U_{ξ ξ} / {(1 + U_{ξ}^{2})}^{3 / 2}$ with $θ (ξ) = arctan U_{ξ} (ξ)$ . Then there exists a positive constant $ε_{0} > 0$ such that for any solution u satisfying the condition $∥ κ (\cdot, t) - κ^{*} ∥_{L^{\infty} ([- ψ_{+}, ψ_{-}])} < ε_{0}$ , there exists a constant $a \in R$ such that $l_{\pm} (t) - c t \to a \pm b$ as $t \to + \infty$ and $u (x, t) \to U (x - c t - a)$ uniformly on $R$ . Here we regard that $u \equiv 0$ outside the interval $[l_{-} (t), l_{+} (t)]$ and $U \equiv 0$ outside the interval $[- b, b]$ .

The argument of the proof is based on the spectral analysis of the equation for the curvature, but the linearized operator of the equation for the curvature around the stationary solution is difficult to handle, because of the non-local term arising from the area preserving property. Then we recover the original variable $(u, l_{\pm})$ of (P) from κ to conclude the uniform convergence.

The rest of this paper is organized as follows. In Section 2, we review from [12] the local existence and uniqueness of the solution to the problem (P) and also the continuous dependence of solutions on the initial data. In Section 3, we prove Theorem 1.1, by showing the preservation of the concavity and the boundedness of the length. We also provide some derivative estimates. In Section 4, we analyze the stationary problem of the equation for the curvature. In Section 5, we study more detailed asymptotic and prove Theorem 1.2.

2. Short time existence

The authors of [12] consider the local existence of the solution for $V = κ + c_{0}$ , where $c_{0} > 0$ is a constant. They first rewrite the evolution equation to the angle variable $θ = arctan u_{x}$ and use the normalized arclength parameter as the space variable. The problem will be converted into a simple semilinear problem, which makes it possible to apply a well-documented general theory for semilinear problems. They assume $u_{0} \in h^{1 + α} ([l_{-}^{0}, l_{+}^{0}])$ with $α \in (1 / 2, 1)$ for technical reasons, where $h^{1 + α}$ is the little Hölder space. This makes the assumption slightly stronger than that in [3], but their approach is more elementary and more self-contained; furthermore, the continuous dependence result is considered. As before, we introduce $\begin{array}{l} h^{α} (I) : = {f \in C^{α} (I) | lim_{δ \to 0} sup_{x, y \in I, | x - y | ⩽ δ} \frac{| f (x) - f (y) |}{| x - y |^{α}} = 0}, \\ h^{k + α} (I) : = {f \in C^{k} (I) | f^{(k)} \in h^{α} (I)}, \end{array}$ where $k \in N$ , $0 < α < 1$ and I is a closed interval. We shall write the statement of the local existence theorem as [12].

Theorem 2.1.
Let the initial data satisfy $\begin{matrix} u^{0} (x) > 0 for x \in (l_{-}^{0}, l_{+}^{0}), u^{0} (l_{\pm}^{0}) = 0, u_{x}^{0} (l_{\pm}^{0}) = \mp tan ψ_{\pm}, u^{0} \in h^{1 + α} ([l_{-}^{0}, l_{+}^{0}]) \end{matrix}$ for some $l_{-}^{0} < l_{+}^{0}$ with $α \in (1 / 2, 1)$ . Assume also that $\begin{matrix} {∥ u^{0} ∥}_{C^{1 + α} ([l_{-}^{0}, l_{+}^{0}])} + | l_{+}^{0} | + | l_{-}^{0} | ⩽ N \end{matrix}$ for some constant $N > 0$ . Then there exist $T_{1}$ and $N_{1} > 0$ depending only on N such that the problem (P) has a unique classical solution $\begin{matrix} u \in C^{1 + α, (1 + α) / 2} ({\overline{D}}_{T_{1}}) \cap C^{2 + α, 1 + α / 2} (D_{T_{1}}), l_{\pm} \in C ([0, T_{1}]) \cap C^{1 + α / 2} ((0, T_{1}]), \end{matrix}$ where $D_{T_{1}} : = {(x, t) ∣ l_{-} (t) ⩽ x ⩽ l_{+} (t), 0 < t ⩽ T_{1}}$ , and that this solution satisfies $\begin{matrix} {∥ u (\cdot, t) ∥}_{C^{1 + α} ([l_{-} (t), l_{+} (t)])} + | l_{+} (t) | + | l_{-} (t) | ⩽ N_{1}, t \in [0, T_{1}] . \end{matrix}$ Furthermore, $u (x, t)$ is $C^{\infty}$ in $D_{T_{1}}$ and $l_{\pm} (t)$ is $C^{\infty}$ in $(0, T_{1}]$ , and for any $T_{0} \in (0, T_{1})$ and $k \in N$ , there exists a constant $N_{2}$ depending only on N, k and $T_{0}$ such that $\begin{matrix} (2.1) & ∥ u ∥_{C^{k + 2 + α, (k + 1 + α) / 2} (D_{T_{1}} ∖ D_{T_{0}})} + ∥ l_{+} ∥_{C^{(k + 1 + α) / 2} ([T_{0}, T_{1}])} + ∥ l_{-} ∥_{C^{(k + 1 + α) / 2} ([T_{0}, T_{1}])} ⩽ N_{2} . \end{matrix}$

Now we shall outline the strategy of the proof, which is similar to that of Section 7 in [12]. We define a function $Θ (s, t)$ by $Θ (s (x, t), t) = arctan u_{x} (x, t)$ . Then Θ satisfies $\begin{array}{rcl} Θ_{t} = Θ_{s s} + {\frac{ψ_{+} + ψ_{-}}{L (t)} (Θ - ψ_{-} - cot ψ_{-}) - cot ψ_{-} Θ_{s} (0, t) + \int_{0}^{s} Θ_{s}^{2} d \tilde{s}} Θ_{s}, s \in (0, L (t)), \\ Θ (0, t) = ψ_{-}, Θ (L (t), t) = - ψ_{+} \end{array}$ for $t > 0$ along with the following free boundary condition: $\begin{array}{l} L^{'} (t) = & cot ψ_{+} (Θ_{s} (L (t), t) + \frac{ψ_{+} + ψ_{-}}{L (t)}) + cot ψ_{-} (Θ_{s} (0, t) + \frac{ψ_{+} + ψ_{-}}{L (t)}) \\ + \frac{{(ψ_{+} + ψ_{-})}^{2}}{L (t)} - \int_{0}^{L (t)} Θ_{s}^{2} d s, t > 0 . \end{array}$ Next, we normalize the interval $[0, L (t)]$ to a fixed interval $[0, 1]$ by using the transformation $\begin{matrix} (2.2) & τ = \int_{0}^{t} \frac{d \tilde{t}}{L^{2} (\tilde{t})}, z = \frac{s}{L (t)} . \end{matrix}$ Then the above problem is reduced to the following problem for $v (z, τ) = Θ (s, t)$ , $η (τ) = log L (t)$ : $\begin{matrix} (2.3) & \{\begin{matrix} v_{τ} = v_{z z} + {P (z, τ) + Q (τ) z} v_{z}, & z \in (0, 1), τ > 0, \\ η^{'} = Q (τ), & τ > 0, \\ v (0, τ) = ψ_{-}, v (1, τ) = - ψ_{+}, & τ > 0, \end{matrix} \end{matrix}$ where $\begin{array}{l} P (z, τ) = (ψ_{+} + ψ_{-}) (v - ψ_{-} - cot ψ_{-}) - (cot ψ_{-}) v_{z} (0, τ) + \int_{0}^{z} v_{z}^{2} d z, \\ Q (τ) = cot ψ_{+} (v_{z} (1, τ) + ψ_{+} + ψ_{-}) + cot ψ_{-} (v_{z} (0, τ) + ψ_{+} + ψ_{-}) + {(ψ_{+} + ψ_{-})}^{2} - \int_{0}^{1} v_{z}^{2} d z . \end{array}$ The initial conditions for (2.3) are given by $\begin{matrix} v^{0} (z) : = v (z, 0) = arctan {u_{x}^{0} (ζ^{- 1} (z))}, η^{0} : = η (0) = log L (0), \end{matrix}$ where $ζ^{- 1}$ is the inverse function of $\begin{matrix} z = ζ (x) = \frac{1}{L (0)} \int_{l_{-}^{0}}^{x} \sqrt{1 + {(u_{x}^{0})}^{2}} d x . \end{matrix}$ By (2.2), we have $\begin{matrix} t = t (τ) : = \int_{0}^{τ} e^{2 η (τ)} d τ, s = s (z, τ) : = z e^{η (τ)}, Θ (s (z, t), t (τ)) = v (z, τ) . \end{matrix}$ This implies that, once the solution $(v, η)$ of (2.3) is obtained, one can recover the original solution $(u, l_{\pm})$ of (P) as follows: $\begin{matrix} (2.4) & \{\begin{matrix} u (x, t) = \int_{0}^{z (x, t)} e^{η (τ (t))} sin v (z, τ (t)) d z, \\ l_{-} (t) = l_{-}^{0} - \int_{0}^{τ (t)} \frac{e^{η (τ)} (v_{z} (0, τ) + ψ_{+} + ψ_{-})}{sin ψ_{-}} d τ, \\ l_{+} (t) = l_{-} (t) + \int_{0}^{1} e^{η (τ (t))} cos v (z, τ (t)) d z, \end{matrix} \end{matrix}$ where $z (x, t)$ and $τ (t)$ are defined through the relations: $\begin{matrix} t = \int_{0}^{τ (t)} e^{2 η (τ)} d τ, x = l_{-} (t) + \int_{0}^{z (x, t)} e^{η (τ (t))} cos v (z, τ (t)) d z . \end{matrix}$ Note that $(v, η)$ determine $(τ (t), l_{\pm} (t))$ first, and $(z (x, t), u (x, t))$ is determined later from these ones. We apply the standard analytic semigroup theory to (2.3) in the same setting as in [12] and, by using (2.4), we can prove Theorem 2.1. We can also check that the following theorem holds from the proof of Theorem 7.2 of [12] easily. Here $T (u^{0}, l_{\pm}^{0})$ is the maximal existence time of the solution $(u, l_{\pm})$ to the problem (P). Theorem 2.2.
Let the functions $u^{0} \in C^{2} ([l_{-}^{0}, l_{+}^{0}])$ and $u_{n}^{0} \in C^{2} ([l_{n, -}^{0}, l_{n, +}^{0}])$ satisfy $\begin{array}{l} u^{0} > 0 in (l_{-}^{0}, l_{+}^{0}), u^{0} (l_{\pm}^{0}) = 0, u_{x}^{0} (l_{\pm}^{0}) = \mp tan ψ_{\pm}, \\ u_{n}^{0} > 0 in (l_{n, -}^{0}, l_{n, +}^{0}), u_{n}^{0} (l_{n, \pm}^{0}) = 0, {(u_{n}^{0})}_{x} (l_{n, \pm}^{0}) = \mp tan ψ_{\pm} \end{array}$ for all $n \in N$ . Suppose that $\begin{matrix} lim_{n \to \infty} l_{n, \pm}^{0} = l_{\pm}^{0}, lim_{n \to \infty} {∥ u_{n}^{0} - u^{0} ∥}_{L^{\infty} (R)} = 0, \end{matrix}$ where we set $u^{0}, u_{n}^{0} \equiv 0$ outside of $[l_{-}^{0}, l_{+}^{0}]$ and $[l_{n, -}^{0}, l_{n, +}^{0}]$ , respectively. Furthermore, we assume that there exists a constant $N > 0$ such that $\begin{matrix} {∥ u^{0} ∥}_{C^{2} ([l_{-}^{0}, l_{+}^{0}])}, {∥ u_{n}^{0} ∥}_{C^{2} ([l_{n, -}^{0}, l_{n, +}^{0}])} ⩽ N for all n \in N . \end{matrix}$ Then for any $t \in (0, T (u^{0}, l_{\pm}^{0}))$ , it holds that $t < T (u_{n}^{0}, l_{n, \pm}^{0})$ for sufficiently large n and that $\begin{matrix} lim_{n \to \infty} l_{n, \pm} (t) = l_{\pm} (t), lim_{n \to \infty} {∥ u_{n} (t) - u (t) ∥}_{L^{\infty} (R)} = 0 . \end{matrix}$
Remark 2.1.
The problem (2.3) is well-posed in $C^{α} ([0, 1])$ for $α \in (0, 1 / 3)$ as in [12]. This implies that the problem (4.1) is well-posed $C ([ψ_{+}, ψ_{-}])$ . If two initial datum of the problem (4.1) in Section 4 are sufficiently close uniformly, then the corresponding solutions become sufficiently close in $C^{\infty} ([ψ_{+}, ψ_{-}])$ for sufficiently small $t > 0$ , by the smoothing effect. This also holds for the generalized problem (5.1) in Section 5.

3. Some a priori estimates

In this section, we provide some preliminaries for later purposes. By Theorem 2.1, the solution $(u (x, t), l_{\pm} (t))$ is $C^{\infty}$ in x, t for $t > 0$ . Hence we are able to calculate higher derivatives of $u (x, t)$ in the following proof. Let $[0, T)$ be the maximal time interval for the existence of a solution u to (P). In this subsection, we shall use the idea of [12] (or [5]) to derive some a priori bounds for $u_{x}$ , $u_{t}$ and $u_{x x}$ , respectively. In the following, we shall introduce the following notations: $\begin{array}{l} ψ_{min} : = min {ψ_{+}, ψ_{-}}, ψ_{max} : = max {ψ_{+}, ψ_{-}}, \\ h (t) : = max_{x \in [l_{-} (t), l_{+} (t)]} u (x, t), l (t) : = l_{+} (t) - l_{-} (t) . \end{array}$

Lemma 3.1.
The following estimate holds: $\begin{matrix} (3.1) & - tan ψ_{+} < u_{x} (x, t) < tan ψ_{-}, x \in (l_{-} (t), l_{+} (t)), t \in [0, T) . \end{matrix}$ Furthermore $\begin{matrix} (3.2) & κ (0, t) < 0, κ (L (t), t) < 0, 0 < t < T . \end{matrix}$
Proof.
Let v be a solution of (2.3). By the concavity of the initial data $\begin{matrix} min {min_{z \in [0, 1]} v^{0}, \mp ψ_{\pm}} = - ψ_{+}, max {max_{z \in [0, 1]} v^{0}, \mp ψ_{\pm}} = ψ_{-} . \end{matrix}$ It follows from the strong maximum principle that $- ψ_{-} < v < ψ_{+}$ in $(0, 1) \times [0, T)$ . Thus (3.1) follows. Note that $v (0, τ) = ψ_{-}$ and $v (1, τ) = - ψ_{+}$ . Thus, by applying the Hopf lemma, we get $\begin{matrix} L (t) κ (0, t) = v_{z} (0, t) < 0, L (t) κ (L (t), t) = v_{z} (1, t) < 0 . □ \end{matrix}$
Lemma 3.2 (Preservation of concavity).

Any solution of the problem (P) remains strictly concave for all $t \in [0, T)$ .

Proof.
Let $W (s, t) = κ (s, t) e^{μ t}$ , where μ is a constant to be chosen later. If we use the s variable, as in [9], the evolution equation for the curvature is written as $\begin{matrix} κ_{t} = κ_{s s} + κ^{2} (κ + \frac{ψ_{+} + ψ_{-}}{L (t)}) . \end{matrix}$ Note that we are able to use this equation even the curve is not necessarily concave. From this equation, we obtain the following differential equation for $W (s, t)$ : $\begin{matrix} (3.3) & W_{t} (s, t) = W_{s s} (s, t) + (κ^{2} (s, t) + \frac{ψ_{+} + ψ_{-}}{L (t)} κ (s, t) + μ) W (s, t) . \end{matrix}$ Let $\tilde{Γ} (t)$ denote the closed curve consisting of $Γ (t)$ and its mirror image below the x-axis. The length of $\tilde{Γ} (t)$ equals $2 L (t)$ and the area enclosed by $\tilde{Γ} (t)$ equals $2 A (t)$ . By the isoperimetric inequality (see, for example, [6, p. 33]), we have $\begin{matrix} (3.4) & {(2 L (t))}^{2} ⩾ 4 π (2 A (t)) = 8 π A (0) . \end{matrix}$ If we choose $μ > \frac{{(ψ_{+} + ψ_{-})}^{2}}{8 π A (0)}$ , then $\begin{array}{l} κ^{2} (s, t) + \frac{ψ_{+} + ψ_{-}}{L (t)} κ (s, t) + μ & = {(κ (s, t) + \frac{ψ_{+} + ψ_{-}}{2 L (t)})}^{2} + μ - \frac{{(ψ_{+} + ψ_{-})}^{2}}{4 L^{2} (t)} \\ ⩾ μ - \frac{{(ψ_{+} + ψ_{-})}^{2}}{8 π A (0)} > 0 . \end{array}$

Now fix a time $0 < t_{1} < T$ and define $\begin{array}{l} W_{max} (t) = sup {W (s, t) ∣ 0 ⩽ s ⩽ L (t)}, \\ β (t_{1}) = max {W_{max} (0), sup_{t \in [0, t_{1}]} W (0, t), sup_{t \in [0, t_{1}]} W (L (t), t)} . \end{array}$ Note that $β (t_{1}) < 0$ from (3.2). Suppose that at some time $0 < t < t_{1}$ , $W_{max} (t) = β$ where $β (t_{1}) < β < 0$ . We assume $t_{max} \in (0, t_{1})$ is the smallest time such that $W (s_{max}, t_{max}) = β$ . Since $β (t_{1}) < β < 0$ implies that $s_{max} \in (0, L (t_{max}))$ , we have $\begin{matrix} W_{t} (s_{max}, t_{max}) ⩾ 0, W_{s s} (s_{max}, t_{max}) ⩽ 0, W (s_{max}, t_{max}) = β < 0 . \end{matrix}$ This contradicts the fact that W satisfies (3.3) and proves that $\begin{matrix} κ (s, t) ⩽ W_{max} (t) e^{- μ t} ⩽ β (t_{1}) e^{- μ t} < 0 . □ \end{matrix}$

The following lower bound follows immediately from the isoperimetric inequality (3.4).
Lemma 3.3.
The inequality $L (t) ⩾ \sqrt{2 π A (0)}$ holds for all $t \in [0, T)$ .

Next we prove the upper bound of time derivatives.
Corollary 3.4.
Let $(u, l_{\pm})$ be the solution of the problem (P). For all $0 ⩽ t ⩽ T$ and $l_{-} (t) ⩽ x ⩽ l_{+} (t)$ , the following estimates hold: $\begin{array}{l} (3.5) & u_{t} (x, t) ⩽ \frac{(ψ_{+} + ψ_{-})}{\sqrt{2 π A (0)} cos ψ_{max}}, \\ (3.6) & l_{+}^{'} (t) ⩽ \frac{(ψ_{+} + ψ_{-}) cot ψ_{+}}{\sqrt{2 π A (0)} cos ψ_{max}}, - l_{-}^{'} (t) ⩽ \frac{(ψ_{+} + ψ_{-}) cot ψ_{-}}{\sqrt{2 π A (0)} cos ψ_{max}}, \\ (3.7) & L^{'} (t) ⩽ \frac{(ψ_{+} + ψ_{-}) cot ψ_{+} cos ψ_{+}}{\sqrt{2 π A (0)} cos ψ_{max}} + \frac{(ψ_{+} + ψ_{-}) cot ψ_{-} cos ψ_{-}}{\sqrt{2 π A (0)} cos ψ_{max}} . \end{array}$
Proof.
Lemmas 3.2–3.3 and (1.2) immediately give us an upper bound (3.5). Combining the identity (1.8) and (3.5), we obtain (3.6). By (1.7) and the Cauchy inequality, we obtain $L^{'} (t) ⩽ l_{+}^{'} (t) cos ψ_{+} - l_{-}^{'} (t) cos ψ_{-}$ . This and (3.6) yield (3.7). □

Next we shall give a lower bound for $u_{t}$ and $u_{x x}$ that is related to the height of the solution. Since the problem is non-local, we cannot use intersection number argument as in [12], we assume concavity for the initial data. The proof is similar to that of Lemma 2.2 in [12], except for this difference, but we write its proof for the reader’s convenience.
Lemma 3.5.
For any solution $(u, l_{\pm})$ of the problem (P), we have $\begin{array}{rcl} (3.8) & u_{t} (x, t) ⩾ M_{1} + \frac{ψ_{+} + ψ_{-}}{L (t)}, \\ (3.9) & u_{x x} (x, t) ⩾ M_{1} \end{array}$ for $0 ⩽ t < T$ , $l_{-} (t) ⩽ x ⩽ l_{+} (t)$ , where $\begin{matrix} M_{1} = min {- \frac{ψ_{max} tan ψ_{max}}{d (t)}, \frac{4 {min}_{x \in [l_{-}^{0}, l_{+}^{0}]} u_{x x}^{0} tan ψ_{max}}{ψ_{min}}, min_{x \in [l_{-}^{0}, l_{+}^{0}]} u_{x x}^{0} (x)} < 0 \end{matrix}$ and $d (t) : = {min}_{τ \in [0, t]} h (τ)$ .
Proof.
We introduce $\begin{matrix} J (x, t) : = K_{-} (t) u (x, t) + arctan u_{x} (x, t) - ψ_{-}, \end{matrix}$ where $K_{-} (t)$ is a positive function given by $\begin{matrix} (3.10) & K_{-} (t) : = max {\frac{ψ_{-}}{d (t)}, \frac{4 | {min}_{x \in [l_{-}^{0}, l_{+}^{0}]} u_{x x}^{0} |}{ψ_{-}}} . \end{matrix}$ Note that $K_{-} (t)$ is non-decreasing, since $d (t)$ is non-increasing. We define a differential operator $Q$ by $\begin{matrix} Q U : = \frac{1}{1 + u_{x}^{2}} U_{x x} + \frac{(ψ_{+} + ψ_{-}) u_{x}}{L (t) \sqrt{1 + u_{x}^{2}}} U_{x} . \end{matrix}$ By a simple calculation, we have $\begin{matrix} (3.11) & J_{t} - Q J = K_{-}^{'} (t) u + \frac{(ψ_{+} + ψ_{-}) K_{-} (t)}{L (t) \sqrt{1 + u_{x}^{2}}} > 0, \end{matrix}$ where the derivative $K_{-}^{'} (t)$ is to be understood in a generalized sense. Let $x_{max} (t)$ be the maximum point of the function $u (x, t)$ , then $\begin{matrix} (3.12) & J (x_{max} (t), t) ⩾ \frac{h (t) ψ_{-}}{d (t)} - ψ_{-} ⩾ 0, 0 ⩽ t < T . \end{matrix}$ By substituting the boundary values and initial value, we can check that $\begin{array}{rcl} (3.13) & J (l_{-} (t), t) = 0, 0 ⩽ t < T, \\ (3.14) & J (x, 0) ⩾ \frac{4 | {min}_{x \in [l_{-}^{0}, l_{+}^{0}]} u_{x x}^{0} |}{ψ_{-}} u^{0} (x) + arctan u_{x}^{0} (x) - ψ_{-} ⩾ 0, x \in [l_{-}^{0}, x_{max} (0)] . \end{array}$ This is a consequence of the following calculation. Let $r_{1} : = min {x \in (l_{-}^{0}, x_{max} (0)) ∣ arctan u_{x}^{0} (x) = ψ_{-} / 2}$ . Then, for $x \in [r_{1}, x_{max} (0)]$ , $\begin{matrix} \frac{ψ_{-} - arctan u_{x}^{0} (x)}{u^{0} (x)} ⩽ \frac{ψ_{-}}{u^{0} (r_{1})} = 2 \frac{ψ_{-} - arctan u_{x}^{0} (r_{1})}{u^{0} (r_{1})} . \end{matrix}$ On the other hand, for $x \in (l_{-}^{0}, r_{1}]$ , we have $\begin{matrix} \frac{ψ_{-} - arctan u_{x}^{0} (x)}{u^{0} (x)} ⩽ \frac{- \int_{l_{-}^{0}}^{x} \frac{u_{x x}^{0}}{1 + {(u_{x}^{0})}^{2}} d x}{(tan \frac{ψ_{-}}{2}) (x - l_{-}^{0})} ⩽ - \frac{2}{ψ_{-}} min_{z \in [l_{-}^{0}, x_{max} (0)]} u_{x x}^{0} (z) . \end{matrix}$

The function $x_{max} (t)$ is continuous, because of Lemma 3.2 and the regularity of the solution. Thus (3.12), (3.13) and (3.14) imply that $J ⩾ 0$ on the parabolic boundary of the domain ${x \in (l_{-} (t), x_{max} (t)), t \in (0, T)}$ . Hence by (3.11) and the maximum principle, $\begin{matrix} (3.15) & J (x, t) ⩾ 0 for 0 ⩽ t < T and x \in [l_{-} (t), x_{max} (t)] . \end{matrix}$ Hence, by (3.13) and (3.15), we have $J_{x} (l_{-} (t), t) ⩾ 0$ for $0 ⩽ t < T$ , which implies $\begin{matrix} (3.16) & \frac{u_{x x}}{1 + u_{x}^{2}} ⩾ - K_{-} (t) tan ψ_{-}, 0 ⩽ t < T, \end{matrix}$ where $K_{-} (t)$ is given in (3.10). Similarly, replacing J by the function $\begin{matrix} K_{+} (t) u - arctan u_{x} - ψ_{+}, where K_{+} (t) : = max {\frac{ψ_{+}}{d (t)}, \frac{4 | {min}_{x \in [l_{-}^{0}, l_{+}^{0}]} u_{x x}^{0} |}{ψ_{+}}}, \end{matrix}$ and arguing as above in the region $x_{max} (t) < x < l_{+} (t)$ , we obtain $\begin{matrix} (3.17) & \frac{u_{x x}}{1 + u_{x}^{2}} ⩾ - K_{+} (t) tan ψ_{+}, 0 ⩽ t < T . \end{matrix}$

Now we introduce a function $D : = u_{x x} / (1 + u_{x}^{2})$ . Then a direct calculation shows $\begin{matrix} \begin{matrix} D_{t} & = \frac{1}{1 + u_{x}^{2}} D_{x x} + [- \frac{2 u_{x} u_{x x}}{{(1 + u_{x}^{2})}^{2}} + \frac{(ψ_{+} + ψ_{-}) u_{x}}{L (t) \sqrt{1 + u_{x}^{2}}}] D_{x} + \frac{ψ_{+} + ψ_{-}}{L (t) \sqrt{1 + u_{x}^{2}}} D^{2} \\ ⩾ \frac{1}{1 + u_{x}^{2}} D_{x x} + [- \frac{2 u_{x} u_{x x}}{{(1 + u_{x}^{2})}^{2}} + \frac{(ψ_{+} + ψ_{-}) u_{x}}{L (t) \sqrt{1 + u_{x}^{2}}}] D_{x} . \end{matrix} \end{matrix}$ Note also that $D (x, 0) ⩾ {min}_{x} u_{x x} (x, 0)$ . Combining these with (3.16) and (3.17), and the comparison principle, we conclude $\begin{matrix} D (x, t) ⩾ min {- \frac{ψ_{max} tan ψ_{max}}{d (t)}, - \frac{4 | {min}_{x \in [l_{-}^{0}, l_{+}^{0}]} u_{x x}^{0} | tan ψ_{max}}{ψ_{min}}, min_{x \in [l_{-}^{0}, l_{+}^{0}]} u_{x x}^{0} (x)} . \end{matrix}$ Therefore, the desired inequalities (3.8) and (3.9) follow from (1.2) and $| u_{x} | ⩽ tan ψ_{max}$ . The proof of Lemma 3.5 is complete. □

The next geometric lemma is the most technical step to prove Theorem 1.1. Here we again assume the concavity for the solution, which is not assumed in Proposition 6.2 of [12] for the shrinking case.
Lemma 3.6.
There exists a constant $M > 0$ that depends on initial curve such that $L (t) ⩽ M$ for all $t \in [0, T)$ .
Proof.
We define the following positive constants: $\begin{array}{l} M_{2} = max {\frac{(ψ_{+} + ψ_{-}) cot ψ_{+} cos ψ_{+}}{\sqrt{2 π A (0)} cos ψ_{max}}, \frac{(ψ_{+} + ψ_{-}) cot ψ_{-} cos ψ_{-}}{\sqrt{2 π A (0)} cos ψ_{max}}}, \\ M_{3} = {(\sqrt{2 π A (0)} cos ψ_{max})}^{3 / 2} . \end{array}$ Let $r_{-} (t) = l_{-} (t) + M_{3} {(l (t))}^{- 1 / 2}$ . The inequality $r_{-} (t) ⩽ l_{+} (t)$ holds by the following argument. By the definition of $L (t)$ and Lemma 3.3, we get $\begin{matrix} l (t) ⩾ L (t) cos ψ_{max} ⩾ \sqrt{2 π A (0)} cos ψ_{max} . \end{matrix}$ This implies $\begin{array}{l} l_{+} (t) - r_{-} (t) & = l (t) - M_{3} {(l (t))}^{- 1 / 2} \\ ⩾ \sqrt{2 π A (0)} cos ψ_{max} - M_{3} {(\sqrt{2 π A (0)} cos ψ_{max})}^{- 1 / 2} = 0 . \end{array}$ Thus, we have $r_{-} (t) ⩽ l_{+} (t)$ . Define $\begin{matrix} s (r_{-} (t), t) = \int_{l_{-} (t)}^{r_{-} (t)} \sqrt{1 + u_{x}^{2}} d x . \end{matrix}$ By (1.7) and (3.6), we have $\begin{matrix} (3.18) & L^{'} (t) ⩽ 2 M_{2} + \frac{ψ_{+} + ψ_{-}}{L (t)} - \int_{0}^{L (t)} k^{2} d s ⩽ 2 M_{2} + \frac{ψ_{+} + ψ_{-}}{L (t)} - \int_{0}^{s (r_{-} (t), t)} k^{2} d s . \end{matrix}$ We estimate the last term of (3.18). Let $x_{max} (t)$ be the maximum point of $u (x, t)$ at time t. By the concavity of the curve, the area of the triangle whose apexes are $(l_{-} (t), 0)$ , $(l_{+} (t), 0)$ and $(x_{max} (t), u (x_{max} (t), t))$ is smaller than the area that is enclosed by $Γ (t)$ and x-axis. That implies $\begin{matrix} \frac{l (t) h (t)}{2} ⩽ A (t) = A (0) . \end{matrix}$ Furthermore, the concavity of the curve also gives us the inequality $u_{x} (x, t) ⩾ u_{x} (r_{-} (t), t)$ for $l_{-} (t) ⩽ x ⩽ r_{-} (t)$ . Combining those inequalities, (1.3) and $\begin{matrix} (3.19) & l (t) ⩾ L (t) cos ψ_{max}, \end{matrix}$ we obtain $\begin{array}{l} u_{x} (r_{-} (t), t) & ⩽ \frac{\sqrt{l (t)}}{M_{3}} \int_{l_{-} (t)}^{r_{-} (t)} u_{x} d x = u (r_{-} (t), t) \frac{\sqrt{l (t)}}{M_{3}} \\ (3.20) & ⩽ h (t) \frac{\sqrt{l (t)}}{M_{3}} ⩽ \frac{2 A (0)}{M_{3} \sqrt{l (t)}} ⩽ \frac{2 A (0)}{M_{3} \sqrt{L (t) cos ψ_{max}}} . \end{array}$ By the Hölder inequality and the inequality $\begin{matrix} s (r_{-} (t), t) ⩽ \frac{r_{-} (t) - l_{-} (t)}{cos ψ_{max}}, \end{matrix}$ we have $\begin{array}{l} {(\int_{0}^{s (r_{-} (t), t)} κ d s)}^{2} & ⩽ s (r_{-} (t), t) \int_{0}^{s (r_{-} (t), t)} κ^{2} d s \\ (3.21) & ⩽ \frac{M_{3}}{\sqrt{l (t)} cos ψ_{max}} \int_{0}^{s (r_{-} (t), t)} κ^{2} d s . \end{array}$ If $\begin{matrix} (3.22) & L (t) ⩾ \frac{1}{cos ψ_{max}} {(\frac{2 A (0)}{M_{3} tan ψ_{-}})}^{2}, \end{matrix}$ then $\begin{matrix} ψ_{-} ⩾ arctan (\frac{2 A (0) {(L (t) cos ψ_{max})}^{- 1 / 2}}{M_{3}}) . \end{matrix}$ This fact and (3.20) imply $\begin{array}{l} {(\int_{0}^{s (r_{-} (t), t)} κ d s)}^{2} & = {ψ_{-} - arctan (u_{x} (r_{-} (t), t))}^{2} \\ (3.23) & ⩾ {ψ_{-} - arctan (\frac{2 A (0) {(L (t) cos ψ_{max})}^{- 1 / 2}}{M_{3}})}^{2} . \end{array}$ Combining (3.19), (3.21) and (3.23), we have $\begin{array}{l} \int_{0}^{s (r_{-} (t), t)} κ^{2} d s & ⩾ \frac{\sqrt{l (t)} cos ψ_{max}}{M_{3}} {ψ_{-} - arctan (\frac{2 A (0) {(L (t) cos ψ_{max})}^{- 1 / 2}}{M_{3}})}^{2} \\ ⩾ \frac{\sqrt{L (t)} {cos}^{3 / 2} ψ_{max}}{M_{3}} {ψ_{-} - arctan (\frac{2 A (0) {(L (t) cos ψ_{max})}^{- 1 / 2}}{M_{3}})}^{2} . \end{array}$ Therefore, $\begin{matrix} L^{'} (t) ⩽ 2 M_{2} + \frac{ψ_{+} + ψ_{-}}{L (t)} - \frac{\sqrt{L (t)} {cos}^{3 / 2} ψ_{max}}{M_{3}} {ψ_{-} - arctan (\frac{2 A (0) {(L (t) cos ψ_{max})}^{- 1 / 2}}{M_{3}})}^{2} . \end{matrix}$ This implies that $L (t)$ is non-increasing if the right-hand side of this inequality is negative and (3.22) holds. Note that these two conditions are satisfied if $L (t)$ is sufficiently large. Hence Lemma 3.6 has proved. □

Now we are at the stage to prove Theorem 1.1. Proof of Theorem 1.1.
By (3.9) and Lemmas 3.2 and 3.6, $u_{x x}$ is uniformly bounded from below and negative. Thus $v_{z} = L (t) κ < 0$ and $η^{'} = Q (τ)$ is uniformly bounded. By Theorem 2.1, the solution $(u, l_{\pm})$ exists globally in time and its curvature and its length are uniformly bounded as $t \to \infty$ . The uniform boundedness of higher derivatives follows from (2.1). □

4. Equation for the curvature

The time evolution of the curvature for strictly concave solution for $V = κ + c_{0}$ is discussed in Section 2 of [12]. The argument here is quite similar to that of [12], except for the proof of the uniqueness of traveling wave for any given area parameter $A^{*}$ . From Lemma 3.2 in Section 3, the solution preserve concavity for $t > 0$ if initial curve is concave, and the function $κ (θ, t) = u_{x x} / {(1 + u_{x}^{2})}^{3 / 2}$ , where $θ (x, t) : = arctan u_{x} (x, t)$ , is well-defined. By repeating a similar calculation as [12], we have $\begin{matrix} (4.1) & \{\begin{matrix} κ_{t} = κ^{2} (κ_{θ θ} + κ + \frac{ψ_{+} + ψ_{-}}{L (t)}), & - ψ_{+} < θ < ψ_{+}, t > 0, \\ κ_{θ} = cot θ (κ + \frac{ψ_{+} + ψ_{-}}{L (t)}), & θ = \mp ψ_{\pm}, t > 0, \\ κ (θ, 0) = κ_{0} (θ), & - ψ_{+} < θ < ψ_{-}, \end{matrix} \end{matrix}$ where $L (t) = \int_{- ψ_{+}}^{ψ_{-}} - \frac{d θ}{κ (θ, t)}$ and the initial data $κ_{0}$ satisfies the following additional condition: $\begin{matrix} \int_{- ψ_{+}}^{ψ_{-}} \frac{sin θ}{κ_{0} (θ)} d θ = 0 . \end{matrix}$ The meaning of this condition is the following. In order that a solution of (4.1) to represents a curve satisfying $u (l_{\pm} (t), t) = 0$ , the above condition is needed. It is not difficult to show that $\begin{matrix} (4.2) & \int_{- ψ_{+}}^{ψ_{-}} \frac{sin θ}{κ (θ, t)} d θ = 0 for all t ⩾ 0 . \end{matrix}$

By considering a solution of (4.1) whose shape is independent of time, we obtain the following proposition for the existence of traveling wave solution.

Proposition 4.1.
For any given $ψ_{\pm} \in (0, π / 2)$ and $L^{} > 0$ , there exist unique constants* $b > 0$ , $c \in R$ and a unique concave function $U (ξ)$ that satisfy (1.9). Furthermore, c satisfies $\begin{matrix} c \{\begin{matrix} > \\ = \\ < \end{matrix}\} 0 if and only if ψ_{-} \{\begin{matrix} > \\ = \\ < \end{matrix}\} ψ_{+}, \end{matrix}$ and the correspondence between $A^{}$ and* $L^{}$ is one-to-one, where* $A^{}$ is the area enclosed by the graph of the solution* $U$ and x-axis.
Remark 4.1.
We shall consider only for concave curve in this paper. The eventual concavity property, which means that all solution of the problem (P) becomes concave for any large $t ⩾ 0$ , will be investigated in the forthcoming paper. If we use the eventual concavity, we can prove the uniqueness of traveling wave without concavity assumption for initial data. We also remark that we cannot apply the eventual concavity results of [12], because their zero number arguments do not work for our non-local problem.
Proof of Proposition 4.1.
Under the assumption (4.2), $κ_{t} = 0$ and $\begin{matrix} \frac{d}{d t} L (t) = \frac{d}{d t} \int_{- ψ_{+}}^{ψ_{-}} - \frac{d θ}{κ (θ, t)} = 0 . \end{matrix}$ Thus $L (t) = L^{}$ , where $L^{}$ is some given positive constant. Therefore, a pair of the stationary solution $κ^{}$ and the constant c in (4.1)–(4.2) satisfies $\begin{array}{l} (4.3) & κ^{} = - c sin θ - \frac{ψ_{+} + ψ_{-}}{L^{}}, \\ (4.4) & \int_{- ψ_{+}}^{ψ_{-}} \frac{sin θ}{c sin θ + \frac{ψ_{+} + ψ_{-}}{L^{}}} d θ = 0 . \end{array}$ Now we find $κ^{}$ and c that satisfy these two equations. Since $- κ^{} = c sin θ + \frac{ψ_{+} + ψ_{-}}{L^{}} > 0$ , we obtain $\begin{matrix} (4.5) & - \frac{ψ_{+} + ψ_{-}}{L^{} sin ψ_{-}} < c < \frac{ψ_{+} + ψ_{-}}{L^{} sin ψ_{+}} . \end{matrix}$ The restriction (4.2) again gives us the following necessary condition: $\begin{matrix} \int_{- ψ_{+}}^{ψ_{-}} \frac{sin θ}{\frac{ψ_{+} + ψ_{-}}{L^{}} + c sin θ} d θ = 0 ⟹ \int_{0}^{ψ_{-}} \frac{sin θ}{\frac{ψ_{+} + ψ_{-}}{L^{}} + c sin θ} d θ = \int_{0}^{ψ_{+}} \frac{sin θ}{\frac{ψ_{+} + ψ_{-}}{L^{}} - c sin θ} d θ . \end{matrix}$ This implies that $c > 0$ if and only if $ψ_{-} > ψ_{+}$ so does the other inequality and the equality. Let us define $\begin{matrix} F (c) : = \int_{0}^{ψ_{-}} \frac{sin θ}{\frac{ψ_{+} + ψ_{-}}{L^{}} + c sin θ} d θ, G (c) : = \int_{0}^{ψ_{+}} \frac{sin θ}{\frac{ψ_{+} + ψ_{-}}{L^{}} - c sin θ} d θ \end{matrix}$ for any fixed $ψ_{\pm}$ . Without loss of generality, we can assume $ψ_{-} > ψ_{+}$ , thus $c > 0$ . We can easily check that $F (c)$ is monotone decreasing and $G (c)$ is monotone increasing for $c > 0$ . Also $F (0) > G (0)$ by the assumption $ψ_{-} > ψ_{+}$ . By taking limit $c ↗ \frac{ψ_{+} + ψ_{-}}{L^{} sin ψ_{+}}$ and applying the intermediate theorem, we conclude that there exists a unique c for the pair $(ψ_{\pm}, L^{})$ . Consequently, for any given $L^{} \in (0, \infty)$ , there exists a unique constant c for which the solution of (4.3) satisfying the constraint (4.4) exists. Let us also mention that the constant $L^{}$ is related with the solution by $L^{} = \int_{- ψ_{+}}^{ψ_{-}} - \frac{d θ}{κ_{0} (θ)}$ . Therefore, for any given length $L^{}$ , there exists a unique stationary solution $κ^{} (θ) = κ^{} (θ; L^{})$ of (4.1) satisfying $\begin{matrix} \{\begin{matrix} 0 = κ_{θ θ}^{} + κ^{} + \frac{ψ_{+} + ψ_{-}}{\int_{- ψ_{+}}^{ψ_{-}} - \frac{d θ}{κ^{} (θ)}}, & - ψ_{+} < θ < ψ_{+}, t > 0, \\ κ_{θ}^{} = cot θ (κ^{} + \frac{ψ_{+} + ψ_{-}}{\int_{- ψ_{+}}^{ψ_{-}} - \frac{d θ}{κ^{} (θ)}}), & θ = \mp ψ_{\pm}, t > 0 . \end{matrix} \end{matrix}$ Now, since $κ^{} = u_{x x}^{} / {(1 + {(u^{})}_{x}^{2})}^{3 / 2}$ and $sin θ = u_{x}^{} / {(1 + {(u^{})}_{x}^{2})}^{1 / 2}$ , the solution $u^{}$ corresponding to the above stationary solution (4.3) satisfies $\begin{matrix} (4.6) & \frac{u_{x x}^{}}{1 + {(u_{x}^{})}^{2}} + c u_{x}^{} + \frac{ψ_{+} + ψ_{-}}{L^{}} \sqrt{1 + {(u_{x}^{})}^{2}} = 0 . \end{matrix}$ Consequently, $u_{t}^{} = - c u_{x}^{}$ , which means that $u^{}$ is a traveling wave solution of the form $u^{} (x, t) = U (x - c t + a)$ , where $U$ is a solution of (1.9) and a is some constant. Conversely, if $u^{}$ is a traveling wave solution of the form $u^{} (x, t) = U (x - c t + a)$ with a strictly concave function $U$ , then it is clear that the corresponding curvature function $κ^{}$ satisfies (4.3) along with the constraint (4.4). It is easy to check that the set of solutions for (4.6) is invariant under the transformation $\begin{matrix} (x, u^{}, c) \mapsto (λ x, λ u^{}, λ^{- 1} c) \end{matrix}$ for any $λ > 0$ . Note that $L^{} \mapsto λ L^{}$ and $A^{} \mapsto λ^{2} A^{}$ by this transformation, where $A^{}$ is the area enclosed by the graph of the solution $u^{}$ and the x-axis. Hence the correspondence between $A^{}$ and $L^{}$ is unique. In summary, for any given area $A^{}$ , there exists a unique traveling wave solution up to translation. □

5. Spectral analysis

Our aim of this section is to prove the exponential stability of a traveling wave solution for the problem (P). For this purpose, we consider the following generalized problem of (1.1): $\begin{matrix} (5.1) & V = κ + \frac{ψ_{+} + ψ_{-}}{L (t)} - \frac{l_{+}^{'} (t)}{L (t)} H, {Ang}_{\pm} (Γ (t)) = {Ang}_{\pm} (Γ (0)), \end{matrix}$ where H is defined by $\begin{matrix} (5.2) & H = - \int_{- ψ_{+}}^{ψ_{-}} \frac{sin θ}{κ_{0} (θ)} . \end{matrix}$ Note that the problem (5.1) include the problem (1.1) when $H = 0$ . If the solution curve is represented as a graph, the problem is reduced to the following free boundary problem: $\begin{array}{l} (5.3) & u_{t} = \frac{u_{x x}}{1 + u_{x}^{2}} + (\frac{ψ_{+} + ψ_{-}}{L (t)} - \frac{l_{+}^{'} (t)}{L (t)} H) \sqrt{1 + u_{x}^{2}}, x \in (l_{-} (t), l_{+} (t)), t > 0, \\ (5.4) & u (l_{-} (t), t) = 0, u (l_{+} (t), t) = H, t > 0, \\ (5.5) & u_{x} (l_{\pm} (t), t) = \mp tan ψ_{\pm}, t > 0, \\ (5.6) & u (x, 0) = u^{0} (x), x \in [l_{-}^{0}, l_{+}^{0}], l_{\pm} (0) = l_{\pm}^{0} . \end{array}$ For this equation we define $A (t) : = \int_{l_{-} (t)}^{l_{+} (t)} u (x, t) d x$ , then it is easy to check that $A^{'} (t) = 0$ . Therefore, this equation also satisfies the area preserving property.

Remark 5.1.

By using Eq. (5.8) in below and integrating by part, we can easily check that $\begin{array}{l} \frac{d}{d t} (- \int_{- ψ_{+}}^{ψ_{-}} \frac{sin θ}{κ (θ, t)} d θ) & = \int_{- ψ_{+}}^{ψ_{-}} (κ_{θ θ} + κ + (\frac{ψ_{+} + ψ_{-}}{L (t)} - \frac{l_{+}^{'} (t)}{L (t)} H)) sin θ d θ \\ = {[κ_{θ} sin θ - κ cos θ - (\frac{ψ_{+} + ψ_{-}}{L (t)} - \frac{l_{+}^{'} (t)}{L (t)} H) cos θ]}_{- ψ_{+}}^{ψ_{-}} = 0 . \end{array}$ Therefore the problem (5.1) has two Hamiltonians. One is the area function $A (t)$ and the other is the height difference of the two endpoints.

The problem (5.1) can be also written as $\begin{matrix} (5.7) & V = κ + \frac{(ψ_{+} + ψ_{-}) sin ψ_{+} - κ (- ψ_{+}, t) H}{H + L (t) sin ψ_{+}}, {Ang}_{\pm} (Γ (t)) = {Ang}_{\pm} (Γ (0)) . \end{matrix}$ Here we have used the equation $\begin{matrix} l_{+}^{'} (t) = \frac{κ (- ψ_{+}, t) + \frac{ψ_{+} + ψ_{-}}{L (t)} - \frac{l_{+}^{'} (t)}{L (t)} H}{sin ψ_{+}}, t > 0, \end{matrix}$ which is obtained by differentiating (5.4) with respect to t, and then using (5.3) and (5.5) along with the identity $Θ_{s} (s, t) = u_{x x} / {(1 + u_{x}^{2})}^{3 / 2}$ . The solution of the problem (5.7) exists locally in time. This can be checked by following the argument of Theorem 1.1. Furthermore, by the continuous dependence of semi-group to the parameter H, the solution of (5.7) exists globally in time provided that the parameter H is sufficiently close to zero [1,15].

We note that there is an important difference between (5.1) and (1.1). If we consider the problem (1.1) for $H \neq 0$ , there is no traveling wave, which exists when $H = 0$ . On the other hand, the problem (5.1) has a two parameter family of a traveling wave parametrized by H and L. Furthermore, the linearized operator associated to this problem has 0-eigenvalue that is semisimple with multiplicity two. Our strategy is embedding the problem (1.1) into (5.1) and prove the exponential convergence of the curvature. The problem (4.1) is then changed to $\begin{matrix} (5.8) & \{\begin{matrix} κ_{t} = κ^{2} (κ_{θ θ} + κ + \frac{ψ_{+} + ψ_{-}}{L (t)} - \frac{l_{+}^{'} (t)}{L (t)} H), & - ψ_{+} < θ < ψ_{+}, t > 0, \\ κ_{θ} = cot θ (κ + \frac{ψ_{+} + ψ_{-}}{L (t)} - \frac{l_{+}^{'} (t)}{L (t)} H), & θ = \mp ψ_{\pm}, t > 0, \\ κ (θ, 0) = κ_{0} (θ), & - ψ_{+} < θ < ψ_{-} . \end{matrix} \end{matrix}$ By a similar argument as Section 4, the curvature of a traveling wave is represented as $\begin{matrix} (5.9) & κ^{*} (θ, H, L) = - \frac{ψ_{+} + ψ_{-}}{L} + \frac{c H}{L} - c sin θ, \end{matrix}$ where c is a unique constant depends on H and L. For any initial data $u^{0}$ , the value $A (0) = A^{*}$ is determined. From the structure of a family of traveling waves of the problem (1.1) for $H = 0$ , the correspondence between $A^{*}$ and $L^{*}$ is one-to-one. In this section, for this given $L^{*}$ , we denote $κ^{*} (θ, 0, L^{*})$ by $κ^{*} (θ)$ to shorten the notation.

Remark 5.2.

Let us explain that the problem (1.1) have traveling wave solution only when $H = 0$ . The relations $L^{*} = \int_{- ψ_{+}}^{ψ_{-}} - \frac{d θ}{κ^{*}}$ and (5.2) yield $\begin{matrix} \int_{- ψ_{+}}^{ψ_{-}} \frac{sin θ}{ψ_{+} + ψ_{-} + L^{*} c sin θ} d θ = \frac{H}{L^{*}}, \int_{- ψ_{+}}^{ψ_{-}} \frac{d θ}{ψ_{+} + ψ_{-} + L^{*} c sin θ} = 1 . \end{matrix}$ Thus $\begin{matrix} c H + ψ_{+} + ψ_{-} = \int_{- ψ_{+}}^{ψ_{-}} \frac{L^{*} c sin θ}{ψ_{+} + ψ_{-} + L^{*} c sin θ} d θ + \int_{- ψ_{+}}^{ψ_{-}} \frac{(ψ_{+} + ψ_{-}) d θ}{ψ_{+} + ψ_{-} + L^{*} c sin θ} = ψ_{+} + ψ_{-} \end{matrix}$ and $H = 0$ is needed for the existence of a traveling wave. On the other hand, for the problem (5.1), the above relations change to $\begin{matrix} \int_{- ψ_{+}}^{ψ_{-}} \frac{sin θ}{ψ_{+} + ψ_{-} + c (- H + L^{*} sin θ)} d θ = \frac{H}{L^{*}}, \int_{- ψ_{+}}^{ψ_{-}} \frac{d θ}{ψ_{+} + ψ_{-} + c (- H + L^{*} sin θ)} = 1 . \end{matrix}$ By a similar calculation, the following condition is needed for the existence of a traveling wave $\begin{matrix} (5.10) & \int_{- ψ_{+}}^{ψ_{-}} \frac{- H + L^{*} sin θ}{ψ_{+} + ψ_{-} + c (- H + L^{*} sin θ)} d θ = 2 H or c = 0 . \end{matrix}$ Let us define $ν : = c L^{*}$ . Then this equation becomes $\begin{matrix} F (ν, h^{*}) : = \int_{- ψ_{+}}^{ψ_{-}} \frac{- h^{*} + sin θ}{ψ_{+} + ψ_{-} + ν (- h^{*} + sin θ)} d θ = 2 h^{*} or ν = 0, \end{matrix}$ where $h^{*} = H / L^{*}$ . Here $h^{*}$ is a parameter that is invariant under the scaling transformation centered at the origin. The function $F (ν, h^{*})$ is strictly monotone decreasing for ν and $\begin{matrix} - \frac{ψ_{+} + ψ_{-}}{- h^{*} + sin ψ_{-}} < ν < \frac{ψ_{+} + ψ_{-}}{h^{*} + sin ψ_{+}} . \end{matrix}$ By taking the limit, we can check that (5.10) has a solution for each $(H, L^{*})$ as in Section 4. Furthermore, the parameter ν depends only on $h^{*}$ , $ψ_{+}$ and $ψ_{-}$ .

Now let us start to consider the linearized operator and its spectrum. We introduce the weighted $L^{2}$ space: $\begin{matrix} L_{*}^{2} ([- ψ_{+}, ψ_{-}]) : = {ϰ \in L^{2} ([- ψ_{+}, ψ_{-}]) ∣ ∥ ϰ ∥_{*} < \infty} \end{matrix}$ equipped with the inner product $\begin{matrix} {(f, g)}_{*} = \int_{- ψ_{+}}^{ψ_{-}} f g \frac{d θ}{{(κ^{*})}^{2}} . \end{matrix}$ The linearized operator $L : D (L) \subset L_{*}^{2} ([- ψ_{+}, ψ_{-}]) \to L_{*}^{2} ([- ψ_{-}, ψ_{+}])$ of (5.8) at $κ = κ^{*} (θ)$ becomes the following: $\begin{matrix} L (ϰ) : = {(κ^{*})}^{2} (ϰ_{θ θ} + ϰ + \frac{1}{L^{*}} \int_{- ψ_{+}}^{ψ_{-}} \frac{ϰ}{κ^{*}} d θ) . \end{matrix}$ In fact, by substituting $κ^{*} + ε v$ to (5.8), and equating the coefficients of $ε^{1}$ , we can check that $\begin{array}{l} L (v) & = {(κ^{*})}^{2} {v_{θ θ} + v - \frac{ψ_{+} + ψ_{-}}{{(L^{*})}^{2}} \int_{- ψ_{+}}^{ψ_{-}} \frac{v}{{(κ^{*})}^{2}} - \frac{c}{L^{*}} \int_{- ψ_{+}}^{ψ_{-}} \frac{v sin θ}{{(κ^{*})}^{2}} d θ} \\ = {(κ^{*})}^{2} {v_{θ θ} + v + \frac{1}{L^{*}} \int_{- ψ_{+}}^{ψ_{-}} \frac{- (ψ_{+} + ψ_{-}) - c L^{*} sin θ}{L^{*}} \frac{v}{{(κ^{*})}^{2}} d θ} \\ = {(κ^{*})}^{2} {v_{θ θ} + v + \frac{1}{L^{*}} \int_{- ψ_{+}}^{ψ_{-}} \frac{v}{κ^{*}} d θ} . \end{array}$ Here we used $\begin{matrix} \begin{matrix} L (t) = - \int_{- ψ_{+}}^{ψ_{-}} \frac{1}{κ^{*} + ε v} d θ = L^{*} + ε \int_{- ψ_{+}}^{ψ_{-}} \frac{v}{{(κ^{*})}^{2}} d θ + O (ε^{2}), \\ l_{+}^{'} (t) = c + O (ε), \\ H = - \int_{- ψ_{+}}^{ψ_{-}} \frac{sin θ}{κ^{*} + ε v} d θ = ε \int_{- ψ_{+}}^{ψ_{-}} \frac{v sin θ}{{(κ^{*})}^{2}} d θ + O (ε^{2}) \end{matrix} \end{matrix}$ to get the above operator. The domain of the operator $L$ is $\begin{matrix} D (L) : = {ϰ \in H_{*}^{2} ([- ψ_{+}, ψ_{-}]) | ϰ_{θ} = cot θ (ϰ + \frac{1}{L^{*}} \int_{- ψ_{+}}^{ψ_{-}} \frac{ϰ}{κ^{*}} d θ) at θ = \mp ψ_{\pm}}, \end{matrix}$ where $H_{*}^{2} ([- ψ_{+}, ψ_{-}])$ is the associated Sobolev space of $L_{*}^{2} ([- ψ_{+}, ψ_{-}])$ . It is easy to check that the kernel of the operator $L$ is $\begin{matrix} Ker (L) = span {1, sin θ}, \end{matrix}$ by using the relations $\begin{matrix} L^{*} = \int_{- ψ_{+}}^{ψ_{-}} - \frac{d θ}{κ^{*}}, \frac{1}{L^{*}} \int_{- ψ_{+}}^{ψ_{-}} \frac{sin θ}{κ^{*}} d θ = 0 . \end{matrix}$

By a simple calculation we can see that the following operator $L^{*} : D (L^{*}) \subset L_{*}^{2} ([- ψ_{+}, ψ_{-}]) \to L_{*}^{2} ([- ψ_{+}, ψ_{-}])$ is the adjoint operator of $L : D (L) \subset L_{*}^{2} ([- ψ_{+}, ψ_{-}]) \to L_{*}^{2} ([- ψ_{+}, ψ_{-}])$ : $\begin{matrix} L^{*} (ϕ) (θ) = {(κ^{*} (θ))}^{2} (ϕ_{θ θ} (θ) + ϕ (θ)) + \frac{κ^{*}}{L^{*}} δ L (ϕ), \end{matrix}$ where $\begin{matrix} δ L (ϕ) = \int_{- ψ_{+}}^{ψ_{-}} ϕ (θ) d θ + cot ψ_{-} ϕ (ψ_{-}) + cot ψ_{+} ϕ (- ψ_{+}) . \end{matrix}$ The domain of this operator is $\begin{matrix} D (L^{*}) = {ϕ \in H_{*}^{2} ([- ψ_{+}, ψ_{-}]) ∣ ϕ_{θ} (θ) = cot θ ϕ (θ) at θ = \mp ψ_{\pm}} . \end{matrix}$ This operator seems to be easier to handle, since the boundary condition is not non-local. Moreover, the linearized operator $L^{*}$ is related to the linearization of distance function, and that is well analyzed in the field of differential geometry [7,16]. By using the similar argument as Lemma 2 of [8], we can show the operator $L^{*} : D (L^{*}) \subset L_{*}^{2} ([- ψ_{+}, ψ_{-}]) \to L_{*}^{2} ([- ψ_{+}, ψ_{-}])$ is sectorial and the spectrum consists of infinite countably many eigenvalues with finite multiplicities. Since $L^{*}$ is adjoint operator of $L$ , the operator $L : D (L) \subset L_{*}^{2} ([- ψ_{+}, ψ_{-}]) \to L_{*}^{2} ([- ψ_{-}, ψ_{+}])$ is also sectorial and it has infinite countable many eigenvalues. Thus, to consider the spectrum of the linear operator $L$ , we analyze the spectrums of the linear operator $L^{*}$ . Our approach is different from those of [7,16], since we do not need to introduce a technical curvelinear coordinate, which is invented to overcome the difficulty of the free boundary motion.

Lemma 5.1.

Let $S^{*} (θ)$ be the support function of the traveling wave whose curvature is $κ^{*}$ , and whose center is the left end point of the traveling wave. More precisely it is defined by $\begin{matrix} (5.11) & S^{*} (θ) : = - sin θ \int_{ψ_{-}}^{θ} \frac{cos θ}{κ^{*}} d θ + cos θ \int_{ψ_{-}}^{θ} \frac{sin θ}{κ^{*}} d θ . \end{matrix}$ Then, the kernel of the operator $L^{*}$ is represented as the following: $\begin{matrix} (5.12) & Ker (L^{*}) = span {S^{*}, sin θ} . \end{matrix}$ Furthermore, all non-zero eigenvalues are real and negative.

Proof.

The boundary condition of $D (L^{*})$ for $S^{*}$ can be checked from the above representation formula (5.11). By the fundamental property of the support function [17], we obtain $\begin{matrix} (5.13) & S_{θ θ}^{*} + S^{*} = \frac{1}{κ^{*}} . \end{matrix}$ By integrating this and using the boundary condition, we have $\begin{matrix} (5.14) & δ L (S^{*}) = - L^{*} . \end{matrix}$ Therefore, $L^{*} (S^{*}) = 0$ holds. We can also see that the function $sin θ$ belongs to $D (L^{*})$ , and by a simple calculation, we have $\begin{array}{l} L^{*} (sin θ) = & {(κ^{*})}^{2} ({(sin θ)}_{θ θ} + sin θ) \\ - \frac{ψ_{+} + ψ_{-}}{{(L^{*})}^{2}} (\int_{- ψ_{+}}^{ψ_{-}} sin θ d θ + cot ψ_{-} sin ψ_{-} - cot ψ_{+} sin ψ_{+}) = 0, \end{array}$ hence $sin θ \in Ker (L^{*})$ . To show $Ker (L^{*}) = span {S^{*}, sin θ}$ , we introduce the following linear operator $B : D (B) \subset L_{*}^{2} ([- ψ_{+}, ψ_{-}]) \to L_{*}^{2} ([- ψ_{+}, ψ_{-}])$ : $\begin{array}{l} B (ϕ) : = {(κ^{*})}^{2} (ϕ_{θ θ} + ϕ), \\ D (B) : = {ϕ \in L_{*}^{2} ([- ψ_{+}, ψ_{-}]) ∣ ϕ_{θ} (- ψ_{+}) = - cot ψ_{+} ϕ (- ψ_{+}), ϕ_{θ} (ψ_{-}) = cot ψ_{-} ϕ (ψ_{-})} . \end{array}$ It is not difficult to check that $B$ is self-adjoint with respect to the inner product ${(\cdot, \cdot)}_{*}$ in $L_{*}^{2} ([- ψ_{+}, ψ_{-}])$ and $Ker (B) = span {sin θ}$ . Any $ϕ \in Ker (L^{*})$ must satisfy $\begin{matrix} (5.15) & B (ϕ) = - \frac{κ^{*}}{L^{*}} δ L (ϕ) . \end{matrix}$ Hence $ϕ \in span {sin θ}$ if $δ L (ϕ) = 0$ . For the other case $δ L (ϕ) \neq 0$ , Eq. (5.15) becomes $\begin{matrix} B (- \frac{L^{*} ϕ}{δ L (ϕ)}) = κ^{*} . \end{matrix}$ By the Fredholm alternative, (5.13), (5.14) and ${(κ^{*}, sin θ)}_{*} = 0$ , the solution of $B (Φ) = κ^{*}$ is written as $Φ = S^{*} + r sin θ$ for any $r \in R$ , which implies $- \frac{L^{*} ϕ}{δ L (ϕ)} = S^{*} + r sin θ$ . Thus $Ker (L^{*}) \subset span {S^{*}, sin θ}$ in both cases. Therefore, $Ker (L^{*}) = span {S^{*}, sin θ}$ .

Now we prove the second statement. Let λ be a non-zero eigenvalue of $L^{*}$ , and $ϕ_{λ}$ be the corresponding eigenfunction. By a simple calculation, we can easily check that $\begin{matrix} {(L^{*} (ϕ), 1)}_{*} = {(L^{*} (ϕ), sin θ)}_{*} = 0 for any ϕ \in D (L^{*}) . \end{matrix}$ Combining this with $L^{*} (ϕ_{λ}) = λ ϕ_{λ}$ , we have $\begin{array}{l} 0 = {(L^{*} (ϕ_{λ}), 1)}_{*} = λ {(ϕ_{λ}, 1)}_{*} = λ \int_{- ψ_{+}}^{ψ_{-}} \frac{ϕ_{λ}}{{(κ^{*})}^{2}} d θ, \\ 0 = {(L^{*} (ϕ_{λ}), sin θ)}_{*} = λ {(ϕ_{λ}, sin θ)}_{*} = λ \int_{- ψ_{+}}^{ψ_{-}} \frac{ϕ_{λ} sin θ}{{(κ^{*})}^{2}} d θ . \end{array}$ Therefore $\begin{matrix} (5.16) & {(ϕ_{λ}, 1)}_{*} = {(ϕ_{λ}, sin θ)}_{*} = {(ϕ_{λ}, κ^{*})}_{*} = \int_{- ψ_{+}}^{ψ_{-}} \frac{ϕ_{λ}}{κ^{*}} d θ = 0 . \end{matrix}$ The operator $B$ is self-adjoint with respect to the inner product ${(\cdot, \cdot)}_{*}$ in $L_{*}^{2} ([- ψ_{+}, ψ_{-}])$ , hence we can apply the Sturm–Liouville theorem. Since the function $sin θ$ is the simple 0-eigenfunction of $B$ and it has only one zero point in $[- ψ_{+}, ψ_{-}]$ , the spectrum of $B$ consists of eigenvalues $\begin{matrix} \dots < μ_{k} < \dots < μ_{3} < μ_{2} = 0 < μ_{1} . \end{matrix}$ Furthermore, if we set $φ_{k}$ is the eigenfunction of the operator $B$ for the eigenvalue is $μ_{k}$ , then $φ_{1}$ is positive for every point on $[- ψ_{+}, ψ_{-}]$ and $φ_{2} = sin θ$ .

Our aim is to prove that $(L^{*} (ϕ_{λ}), ϕ_{λ}) < 0$ for any eigenfunction $ϕ_{λ}$ of the operator $L^{*}$ if $λ \neq 0$ . First we shall prove that any function $ϕ \in D (L^{*}) = D (B)$ can be represented by $ϕ = α S^{*} + β sin θ + w$ , where α, β are some constants and w is a linear combination of ${φ_{j}}_{j = 3}^{\infty}$ . All we need to prove is that ${(φ_{1}, S^{*})}_{*} \neq 0$ . Then $S^{*}$ and ${φ_{j}}_{j = 2}^{\infty}$ span the same space spanned by ${φ_{j}}_{j = 1}^{\infty}$ . This can be checked as the following: $\begin{matrix} {(φ_{1}, S^{*})}_{*} = λ_{1}^{- 1} {(λ_{1} φ_{1}, S^{*})}_{*} = λ_{1}^{- 1} {(B (φ_{1}), S^{*})}_{*} = λ_{1}^{- 1} {(φ_{1}, B (S^{*}))}_{*} = λ_{1}^{- 1} {(φ_{1}, κ^{*})}_{*} \neq 0 . \end{matrix}$ Therefore, there are constants $α_{λ}$ , $β_{λ}$ and a function $w_{λ}$ , that is a linear combination of ${φ_{j}}_{j = 3}^{\infty}$ such that $ϕ_{λ} = α_{λ} S^{*} + β_{λ} sin θ + w_{λ}$ . Moreover, since the function $sin θ$ is the 0-eigenfunction of $L^{*}$ , we can assume $ϕ_{λ} = α_{λ} S^{*} + w_{λ}$ without loss of generality. By (5.16), we get $\begin{matrix} {(L^{*} (ϕ_{λ}), ϕ_{λ})}_{*} - {(B (ϕ_{λ}), ϕ_{λ})}_{*} = \int_{- ψ_{+}}^{ψ_{-}} \frac{ϕ_{λ}}{L^{*} κ^{*}} δ L (ϕ_{λ}) d θ = \frac{δ L (ϕ_{λ})}{L^{*}} {(ϕ_{λ}, κ^{*})}_{*} = 0 . \end{matrix}$ By applying this, we can prove that $λ \in R$ , since $\begin{matrix} λ {(ϕ_{λ}, ϕ_{λ})}_{*} = {(L^{*} (ϕ_{λ}), ϕ_{λ})}_{*} = {(B (ϕ_{λ}), ϕ_{λ})}_{*} = {(ϕ_{λ}, B (ϕ_{λ}))}_{*} = {(ϕ_{λ}, L^{*} (ϕ_{λ}))}_{*} = \bar{λ} {(ϕ_{λ}, ϕ_{λ})}_{*} . \end{matrix}$ Furthermore, by substituting $ϕ_{λ} = α_{λ} S^{*} + w_{λ}$ into ${(L^{*} (ϕ_{λ}), ϕ_{λ})}_{*}$ , we get $\begin{array}{l} {(L^{*} (ϕ_{λ}), ϕ_{λ})}_{*} = {(B (ϕ_{λ}), ϕ_{λ})}_{*} & = α_{λ}^{2} {(B (S^{*}), S^{*})}_{*} + 2 α_{λ} {(B (S^{*}), w_{λ})}_{*} + {(B (w_{λ}), w_{λ})}_{*} \\ = α_{λ}^{2} {(κ^{*}, S^{*})}_{*} + 2 α_{λ} {(κ^{*}, w_{λ})}_{*} + {(B (w_{λ}), w_{λ})}_{*} . \end{array}$ Now by (5.16), the following holds: $\begin{matrix} {(ϕ_{λ}, κ^{*})}_{*} = α_{λ} {(κ^{*}, S^{*})}_{*} + {(κ^{*}, w_{λ})}_{*} = 0 . \end{matrix}$ Finally we conclude that $\begin{array}{l} λ {(ϕ_{λ}, ϕ_{λ})}_{*} & = {(L^{*} (ϕ_{λ}), ϕ_{λ})}_{*} = {(B (ϕ_{λ}), ϕ_{λ})}_{*} \\ ⩽ - α_{λ}^{2} {(κ^{*}, S^{*})}_{*} + λ_{3} {(w_{λ}, w_{λ})}_{*} = - 2 A^{*} + λ_{3} {(w_{λ}, w_{λ})}_{*} . \end{array}$ Here we use the standard property ${(κ^{*}, S^{*})}_{*} / 2 = A^{*}$ of the support function explained in [17]. Therefore, $λ < 0$ and the lemma has proved. □

Proposition 5.2.

The spectrum of $L$ consists of countable many negative eigenvalues and zero. Moreover, $\begin{matrix} (5.17) & Ker (L) = span {1, sin θ}, \end{matrix}$ and both geometric and algebraic multiplicity of the 0-eigenvalue are two.

Proof.

We already showed $L^{*}$ is adjoint operator of $L$ with respect to the inner product ${(\cdot, \cdot)}_{*}$ in $L_{*}^{2} ([- ψ_{+}, ψ_{-}])$ and $L^{*}$ has 0-eigenvalue and others are negative. Thus, $L$ also has 0-eigenvalue and negative ones. We have already checked that (5.17) holds and the geometric multiplicity of 0-eigenvalue is two. In order to prove the last statement, we assume that $Ker (L) \neq Ker (L^{2})$ . Then there exists $v \in Ker (L^{2})$ but $v \notin Ker (L)$ . This condition implies $L (v) \in Ker (L) ∖ {0}$ . Thus, by (5.17), there exists $(α, β) \in R^{2} ∖ {(0, 0)}$ such that $L (v) = α κ^{*} + β sin θ$ . By (5.12), we have $\begin{matrix} {(L (v), sin θ)}_{*} = {(L (v), S^{*})}_{*} = 0, L (v) \in Ker (L) ∖ {0} . \end{matrix}$ The function in $Ker (L) ∖ {0}$ that is perpendicular to $sin θ$ is proportional to $κ^{*}$ . This can be proved by the following calculation: $\begin{matrix} 0 = {(α κ^{*} + β sin θ, sin θ)}_{*} = α {(κ^{*}, sin θ)}_{*} + β {(sin θ, sin θ)}_{*} = β {(sin θ, sin θ)}_{*} . \end{matrix}$ Hence $L (v) = α κ^{*}$ for some $α \neq 0$ , but this is impossible since $\begin{matrix} {(L (v), S^{*})}_{*} = α {(κ^{*}, S^{*})}_{*} = 2 α A^{*} \neq 0 . \end{matrix}$ This contradiction shows that the algebraic multiplicity is two and the lemma has proved. □

Theorem 5.1.

Let $κ_{0} (θ)$ be the curvature of a curve whose endpoints are on x-axis and the area enclosed by the curve and x-axis is $A^{*}$ . Let $κ^{*} (θ)$ be a curvature of a traveling wave solution whose endpoints are on x-axis having the same area $A^{*}$ . Assume that $κ_{0} (θ)$ is sufficiently close to $κ^{*} (θ)$ uniformly on $[- ψ_{+}, ψ_{-}]$ , then the corresponding solution $κ (θ, t)$ of the problem (4.1) converges to $κ^{*} (θ)$ exponentially on $[- ψ_{+}, ψ_{-}]$ uniformly as $t \to \infty$ .

Proof.

We will apply the general theory Theorem 2.1 of [13] and Proposition 5.2. We set $X_{0}$ and $X_{1}$ of [13] by $X_{0} = L_{*}^{2} ([- ψ_{+}, ψ_{-}])$ and $X_{1} = H_{*}^{2} ([- ψ_{+}, ψ_{-}])$ . The set of the stationary solutions for the problem (5.1) is $\begin{matrix} E = {κ^{*} (θ, H, L) = - \frac{ψ_{+} + ψ_{-}}{L} - c sin θ + \frac{c H}{L} | H \in R, L > 0}, \end{matrix}$ where $c = c (H, L)$ is a function that depends on H and L. Define a function $F (μ, H, L)$ by $\begin{matrix} F (μ, H, L) = \int_{- ψ_{+}}^{ψ_{-}} \frac{- H + L sin θ}{ψ_{+} + ψ_{-} + μ (- H + L sin θ)} d θ - 2 H . \end{matrix}$ The function F is smooth with respect to μ, H and L, and from the proof in Remark 5.2, $\partial F / \partial μ \neq 0$ . By the implicit function theorem, the function $μ = c (H, L)$ that satisfies $\begin{matrix} (5.18) & F (c (H, L), H, L) = 0 \end{matrix}$ is a smooth function with respect to H and L. It is obvious that $κ^{*} (\cdot, H, L)$ is smooth in the domain ${(H, L) ∣ H \in R, L > 0}$ and dimension of the above manifold $E$ is two in the space $H_{*}^{2} ([- ψ_{+}, ψ_{-}])$ . As is easily seen, the constant L corresponds to the scaling of a traveling wave at some fixed point on x-axis and H corresponds to the parameter which changes the difference of the height of the two endpoints. To calculate the tangent space of the set of stationary solutions of (5.8) at $κ^{*} = κ^{*} (\cdot, 0, L^{*})$ in the space $H_{*}^{2} ([- ψ_{+}, ψ_{-}])$ , we differentiate $κ^{*} (\cdot, H, L)$ with respect to H and L at the point $(H, L) = (0, L^{*})$ . Then we have $\begin{array}{l} (5.19) & \frac{\partial}{\partial H} |_{(H, L) = (0, L^{*})} κ^{*} (θ, H, L) = \frac{c (0, L^{*})}{L^{*}} - \frac{\partial c (0, L^{*})}{\partial H} sin θ, \\ (5.20) & \frac{\partial c}{\partial L} |_{(H, L) = (0, L^{*})} κ^{*} (θ, H, L) = \frac{ψ_{+} + ψ_{-}}{{(L^{*})}^{2}} - \frac{\partial c (0, L^{*})}{\partial L} sin θ . \end{array}$ Now, we check that $\frac{c (0, L^{*})}{L^{*}} - \frac{\partial c (0, L^{*})}{\partial H} sin θ$ and $\frac{ψ_{+} + ψ_{-}}{{(L^{*})}^{2}} - \frac{\partial c (0, L^{*})}{\partial L} sin θ$ are linearly independent. If this does not hold, by comparing the first term of right-hand side of (5.19) and (5.20), we have $\begin{matrix} (5.21) & \frac{ψ_{+} + ψ_{-}}{c (0, L^{*}) L^{*}} \frac{\partial c (0, L^{*})}{\partial H} = \frac{\partial c (0, L^{*})}{\partial L}, c (0, L^{*}) \neq 0 . \end{matrix}$ By differentiating (5.18) with respect to H and L at $(H, L) = (0, L^{*})$ , we also have $\begin{matrix} \begin{matrix} F_{μ} (c (0, L^{*}), 0, L^{*}) \frac{\partial c (0, L^{*})}{\partial H} + F_{H} (c (0, L^{*}), 0, L^{*}) = 0, \\ F_{μ} (c (0, L^{*}), 0, L^{*}) \frac{\partial c (0, L^{*})}{\partial L} + F_{L} (c (0, L^{*}), 0, L^{*}) = 0 . \end{matrix} \end{matrix}$ It is easy to see $F_{μ} (c (0, L^{*}), 0, L^{*}) < 0$ , and hence combining this and (5.21), we have $\begin{matrix} (5.22) & \frac{ψ_{+} + ψ_{-}}{c (0, L^{*}) L^{*}} F_{H} (c, 0, L^{*}) = F_{L} (c, 0, L^{*}) . \end{matrix}$ By a simple calculation, we also have $\begin{array}{l} F_{H} (c (0, L^{*}), 0, L^{*}) = - \int_{- ψ_{+}}^{ψ_{-}} \frac{ψ_{+} + ψ_{-}}{{(κ^{*} (θ, 0, L^{*}))}^{2} {(L^{*})}^{2}} d θ - 2, \\ F_{L} (c (0, L^{*}), 0, L^{*}) = \int_{- ψ_{+}}^{ψ_{-}} \frac{(ψ_{+} + ψ_{-}) sin θ}{{(κ^{*} (θ, 0, L^{*}))}^{2} {(L^{*})}^{2}} d θ . \end{array}$ Substituting these equations into (5.22) and using $ψ_{+} + ψ_{-} \neq 0$ , we see that the following equality must hold: $\begin{matrix} 0 = \int_{- ψ_{+}}^{ψ_{-}} \frac{ψ_{+} + ψ_{-}}{{(κ^{*} (θ, 0, L^{*}))}^{2} {(L^{*})}^{2}} d θ + \int_{- ψ_{+}}^{ψ_{-}} \frac{c (0, L^{*}) sin θ}{{(κ^{*} (θ, 0, L^{*}))}^{2} L^{*}} d θ + 2 . \end{matrix}$ By substituting (5.9) into this, we get $\begin{matrix} \int_{- ψ_{+}}^{ψ_{-}} \frac{ψ_{+} + ψ_{-}}{{(κ^{*} (θ, 0, L^{*}))}^{2} {(L^{*})}^{2}} d θ + \int_{- ψ_{+}}^{ψ_{-}} \frac{c (0, L^{*}) sin θ}{{(κ^{*} (θ, 0, L^{*}))}^{2} L^{*}} d θ + 2 = \frac{1}{L^{*}} \int_{- ψ_{+}}^{ψ_{-}} \frac{- d θ}{κ^{*} (θ)} + 2 = 3 . \end{matrix}$ This is a contradiction. Thus, as we mentioned above, the tangent space of the set of stationary solutions of (5.8) is spanned by $\frac{c (0, L^{*})}{L^{*}} - \frac{\partial c (0, L^{*})}{\partial H} sin θ$ and $\frac{ψ_{+} + ψ_{-}}{{(L^{*})}^{2}} - \frac{\partial c (0, L^{*})}{\partial L} sin θ$ , which is the same space spanned by 1 and $sin θ$ . Combining this with Proposition 5.2, we have already shown that

the set of all equilibria $E$ of (5.8) is a smooth 2-dimensional manifold in $X_{1}$ ,

the tangent space of $E$ at $κ^{*}$ is $Ker (L)$ , and the eigenvalue 0 is semisimple,

$σ (L) ∖ {0} \subset {z \in C : Re z < 0}$ .

Combining these and Remark 2.1, we can apply Theorem 2.1 of [13] to conclude that

κ (\cdot, t)

converges to some element in

E

. We have already known that there exist

\tilde{H} \in R

and

\tilde{L} > 0

such that

κ (\cdot, t)

converges to

κ^{*} (\cdot, \tilde{H}, \tilde{L})

exponentially uniformly on

[- ψ_{+}, ψ_{-}]

t \to \infty

. Since (4.2) holds and

A (t)

is a preserved function, the following equations must be satisfied:

\begin{matrix} \tilde{H} = \int_{- ψ_{+}}^{ψ_{-}} \frac{sin θ}{κ^{*} (θ, \tilde{H}, \tilde{L})} d θ = 0, \int_{ψ_{-}}^{- ψ_{+}} \frac{cos θ}{κ^{*} (θ, \tilde{H}, \tilde{L})} \int_{ψ_{-}}^{θ} \frac{sin \tilde{θ}}{κ^{*} (\tilde{θ}, \tilde{H}, \tilde{L})} d \tilde{θ} d θ = A^{*}, \end{matrix}

and the correspondence between

A^{*}

and

L^{*}

is unique, hence

\tilde{H} = 0

and

\tilde{L} = L^{*}

. Therefore,

κ (θ, t)

converges to

κ^{*} = κ^{*} (\cdot, 0, L^{*})

exponentially on

[- ψ_{+}, ψ_{-}]

uniformly as

t \to \infty

, and the theorem has proven. □

Remark 5.3.

The reader might think that, we need to consider the spectrum of the linearized operator around $κ^{*}$ for Eq. (4.1). This is one strategy to prove Theorem 1.2, and it is possible to prove that all eigenvalues are negative or zero. Unfortunately, the zero eigenvalue is not semisimple and we cannot apply the stability theory to get the exponential convergence to $κ^{*}$ . In order to overcome this difficulty, we consider the problem (5.1).

Now we go back to the proof of our main theorem.

Proof of Theorem 1.2.

By differentiating (1.3) with respect to t, and then using (1.2) and (1.4), we get $\begin{array}{l} l_{\pm}^{'} (t) - c & = \frac{\pm 1}{sin ψ_{\pm}} {(κ (θ, t) + \frac{ψ_{+} + ψ_{-}}{L (t)}) - (κ^{*} (θ) + \frac{ψ_{+} + ψ_{-}}{L^{*}})} \\ = \frac{\pm 1}{sin ψ_{\pm}} {(κ (θ, t) - κ^{*} (θ)) + (\frac{ψ_{+} + ψ_{-}}{\int_{- ψ_{+}}^{ψ_{-}} \frac{d θ}{κ^{*} (θ)}} - \frac{ψ_{+} + ψ_{-}}{\int_{- ψ_{+}}^{ψ_{-}} \frac{d θ}{κ (θ, t)}})} . \end{array}$ Theorem 5.1 implies that κ converges to $κ^{*}$ exponentially fast. Thus $| l_{\pm}^{'} (t) - c | = O (e^{- λ t})$ and $| L (t) - L^{*} | = O (e^{- λ t})$ for some $λ > 0$ as $t \to \infty$ . In particular, $\begin{matrix} L (t) = \int_{- ψ_{+}}^{ψ_{-}} \frac{d θ}{κ (θ, t)} \to \int_{- ψ_{+}}^{ψ_{-}} \frac{d θ}{κ^{*} (θ)} = L^{*} as t \to \infty . \end{matrix}$ By integrating $| l_{\pm}^{'} (t) - c | = O (e^{- λ t})$ , we obtain the convergence of $l_{\pm} (t) - l_{\pm} (0) - c t$ as $t \to \infty$ . Hence, we can define $μ^{\pm} : = {lim}_{t \to \infty} (l_{\pm} (t) - c t) \in R$ . Let us reconstruct the solution curve $Γ (t)$ from the curvature $κ (θ, t)$ by the relation: $\begin{matrix} x (θ, t) = l_{-} (t) + \int_{θ}^{ψ_{-}} \frac{cos θ}{- κ (θ, t)} d θ, u (x (θ, t), t) = \int_{- ψ_{+}}^{θ} \frac{sin θ}{κ (θ, t)} d θ . \end{matrix}$ By a similar way, we again construct the curve for the graph of $u^{*}$ from the curvature $κ^{*}$ , whose shape represents a traveling wave solution: $\begin{matrix} x^{*} (θ, t) = μ^{-} + c t + \int_{θ}^{ψ_{-}} \frac{cos θ}{- κ^{*} (θ)} d θ, u^{*} (x^{*} (θ, t), t) = \int_{- ψ_{+}}^{θ} \frac{sin θ}{κ^{*} (θ)} d θ . \end{matrix}$ Therefore, as $t \to \infty$ , we conclude $\begin{array}{l} (5.23) & x (θ, t) - x^{*} (θ, t) = l_{-} (t) - (μ^{-} + c t) + \int_{θ}^{ψ_{-}} (\frac{cos θ}{- κ (θ, t)} - \frac{cos θ}{- κ^{*} (θ)}) d θ \to 0, \\ (5.24) & u (x (θ, t), t) - u^{*} (x^{*} (θ, t), t) = \int_{- ψ_{+}}^{θ} (\frac{sin θ}{κ (θ, t)} - \frac{sin θ}{κ^{*} (θ)}) d θ \to 0 \end{array}$ uniformly for $θ \in [- ψ_{+}, ψ_{-}]$ . Thus, boundedness of $u_{x}^{*}$ and the mean value theorem imply $\begin{matrix} \begin{matrix} | u (x (θ, t), t) - u^{*} (x (θ, t), t) | \\ ⩽ | u (x (θ, t), t) - u^{*} (x^{*} (θ, t), t) | + | u^{*} (x^{*} (θ, t), t) - u^{*} (x (θ, t), t) | \to 0 \end{matrix} \end{matrix}$ as $t \to \infty$ as long as $x (θ, t) \in [l_{-} (t), l_{+} (t)] \cap [μ^{-} + c t, μ^{+} + c t]$ . This and (5.23)–(5.24) give us pointwise convergence $u (x, t) \to u^{*} (x, t)$ on $R$ as $t \to \infty$ , where we consider zero extension as in the statement. Combining these with space derivative bounds, we conclude the uniform convergence. □

Footnotes

Acknowledgement

This work is partially supported by JSPS KAKENHI Grant no. 23740128.

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