1 Introduction

The classical curve shortening flow equation in a plane is

$$\begin{aligned} \frac{\partial X(u,t)}{\partial t}=\kappa N, \end{aligned}$$
(1.1)

where \(X(u,t)=(x(u,t),y(u,t)):S^{1}\,\times \,[0,T)\rightarrow \,{\mathbb {R}}^{2}\) is a family of closed planar curves, \(\kappa \) is the curvature and N is the unit normal vector. Gage and Hamilton [1,2,3] have studied this curve shortening flow and have proved it shrinks to a round point in finite time. Then another natural question arises for expanding evolution flow for curves. Chow–Tsai [4] have studied the expanding flow such as

$$\begin{aligned} \frac{\partial X(u,t)}{\partial t}=-G\left( \frac{1}{\kappa }\right) N, \end{aligned}$$
(1.2)

where G is a positive smooth function with \(G^{'}>0\) everywhere. Andrews [5] and Tsai [6] have studied more general expanding flows, especially flows with anisotropic speeds. They have obtained deep results too. Later, people began to study curve flow problems preserving some geometric quantities. Gage [7] has considered an area-preserving flow

$$\begin{aligned} \frac{\partial X(u,t)}{\partial t}=\left( \kappa -\frac{2\pi }{L}\right) N, \end{aligned}$$
(1.3)

where L is the length of the evolving curve, and he proved that the length of the curve is non-increasing and finally converge to a circle. In [8], Pan-Yang considered a length-preserving curve flow

$$\begin{aligned} \frac{\partial X(u,t)}{\partial t}=\left( \frac{L}{2\pi }-\frac{1}{\kappa }\right) N. \end{aligned}$$
(1.4)

They have proved that the enclosed area of curve is non-decreasing and finally converge to a circle.

Motivated by their works, we now investigate a non-local area-preserving curvature flow for closed convex planar curves. Let \(X(u,t)=(x(u,t),y(u,t)):S^{1}\,\times \,[0,T)\rightarrow \,{\mathbb {R}}^{2}\,\) be a family of closed planar curves with \(X(u,0)=X_0(u)\) being a closed convex initial curve, then the evolution problem considered is defined as:

$$\begin{aligned} \frac{\partial X(u, t)}{\partial t} =\left( P-\frac{2A}{L}\right) N, \end{aligned}$$
(1.5)

where \(P=-<X,N>\) is the Minkowski support function, N is the unit inward pointing normal vector, A(t) and L(t) are the enclosed area and the length of the curve, respectively. The main result of this paper is the following theorem.

Theorem 1.1

A closed convex plane curve which evolves according to (1.5) remains convex, decreases its length and preserves the enclosed area, becomes more and more circular during the evolution process, and finally converges to a finite circle in Hausdorff sense with radius \(\sqrt{\frac{A}{\pi }}\) as \(t\rightarrow \infty \).

2 Limits of evolving curves

In this section we first derive a set of evolution equations along (1.5). Since monotonicity formulas are immediate consequences, we prove them in this section while the existence and convexity will be proved in the next section. Let s(u, t) be the arc-length parameter of X(u, t), then \(\frac{\partial s}{\partial u}=\vert X_{u}\vert \frac{\partial }{\partial s}\). Let \(T=X_{s}\) be the unit tangent vector.

Lemma 2.1

Suppose \(\frac{\partial X}{\partial t}=\phi N\), then the evolution equations for ds, Tand N are

$$\begin{aligned} (ds)_{t}=-\phi \kappa ds,\quad T_{t}=\phi _{s}N,\quad N_{t}=-\phi _{s}T. \end{aligned}$$

Proof

Since \(ds=\vert X_{u}\vert du\) one has

$$\begin{aligned} (ds)_{t}=\frac{\partial \vert X_{u}\vert }{\partial t} du=\frac{<X_{u},X_{ut}>}{\vert X_{u}\vert }du=<T,(\phi N)_{u}>du=<T,\phi N_{u}>du=-\phi \kappa ds. \end{aligned}$$

By direct computations one has

$$\begin{aligned} \frac{\partial T}{\partial t}&=\frac{X_{ut}}{\vert X_{u}\vert }-\frac{<X_{u},X_{ut}>X_{u}}{\vert X_{u}\vert ^{3}}\\&=\frac{1}{\vert X_{u}\vert }\left( X_{tu}-<X_{tu},T>T\right) \\&=\frac{\partial }{\partial s}(\phi N)-<\frac{\partial }{\partial s}(\phi N),T>T\\&=\phi _{s}N. \end{aligned}$$

Taking time derivative of \(<T,N>=0\) we have \(<T_{t},N>+<T,N_{t}>=0\). Note that \(N_{t}\perp N\) and we get \( N_{t}=-\phi _{s}T\). \(\square \)

Lemma 2.2

Suppose \(\frac{\partial X}{\partial t}=\phi N\), then the evolution equations for L(t), A(t) are

$$\begin{aligned} \frac{dL(t)}{dt}=-\int ^{L}_{0}\phi \kappa ds,\quad \frac{dA(t)}{dt}=-\int ^{L}_{0}\phi ds. \end{aligned}$$
(2.1)

Proof

The derivative of L(t) is a direct consequence of \((ds)_{t}=-\phi \kappa ds\). By the Green’s formula we have \(A=-\frac{1}{2}\int ^{L}_{0}<X,N>ds\), thus

$$\begin{aligned} \frac{dA(t)}{dt}&=-\frac{1}{2}\int ^{L}_{0}<X_{t},N>+<X,N_{t}>-<X,N>\phi \kappa ds\\&=-\frac{1}{2}\int ^{L}_{0}\phi +<X,-\phi _{s}T>-<X,\phi T_{s}>ds\\&=-\frac{1}{2}\int ^{L}_{0}\phi +\frac{\partial }{\partial s}<X,-\phi T>+<X_{s},\phi T>ds. \end{aligned}$$

Since X is a closed curve we have \(\int ^{L}_{0}\frac{\partial }{\partial s}<X,-\phi T>ds=0\). Plugging in \(X_{s}=T\) we get the formula for the derivative of A(t). \(\square \)

Lemma 2.3

If X(u, t) evolves under the equations defined by (1.5), then, during the evolution process, the length of the evolving curve is decreasing and the enclosed area keeps constant.

Proof

In (2.1) plug in \(\phi =P-\frac{2A}{L}\), one has

$$\begin{aligned} L^{'}(t)=-\int ^{L}_{0}\left( P-\frac{2A}{L}\right) \kappa ds=-L+\frac{4\pi A}{L}=\frac{4\pi A-L^{2}}{L}\le 0, \end{aligned}$$
(2.2)

and

$$\begin{aligned} A^{'}(t)=-\int ^{L}_{0}\left( P-\frac{2A}{L}\right) ds=-2A+2A=0, \end{aligned}$$
(2.3)

where we have used the facts \(\int ^{L}_{0}P\kappa ds=L, \int ^{L}_{0}Pds=2A\) and the classical isoperimetric inequality \(L^{2}-4\pi A\ge 0\). \(\square \)

Corollary 2.4

The length of the evolving curve X(u, t) is given by

$$\begin{aligned} L=\sqrt{4\pi A+\left( L^{2}_{0}-4\pi A\right) e^{-2t}}, \end{aligned}$$
(2.4)

where \(L_{0}\) is the length of initial curve \(X_{0}(u)\), and therefore L(t) goes to a constant \(\sqrt{4\pi A}\) as the time t goes to infinity.

Proof

From (2.2), we find that \(L^{'}(t)=-L+\frac{4\pi A}{L}\) is actually an ordinary differential equation. Since the area enclosed by the evolving curve is constant \(A(t)=A_{0}\) \(\big (A_{0}\) is enclosed area of initial curve \(X_{0}(u)\big )\), then solving this equation yields (2.4) and therefore L(t) goes to a constant \(\sqrt{4\pi A}\) as the time t goes to infinity. \(\square \)

Lemma 2.5

The isoperimetric deficit \(L^{2}-4\pi A\) of the evolving curve is decreasing during the evolution process and converges to zero as the time t goes to infinity.

Proof

$$\begin{aligned} \frac{d\left( L^{2}-4\pi A\right) }{dt}=2LL_{t}-4\pi A_{t}=2L\left( -L+\frac{4\pi A}{L}\right) -0=-2\left( L^{2}-4\pi A\right) \le 0. \end{aligned}$$
(2.5)

Integrating this yields

$$\begin{aligned} L^{2}-4\pi A=\left( L^{2}_{0}-4\pi A\right) e^{-2t}. \end{aligned}$$
(2.6)

Therefore, when \(t\rightarrow \infty \), there holds \(L^{2}-4\pi A\rightarrow 0\). \(\square \)

By the Bonnesen inequality (see [9])

$$\begin{aligned} L^{2}-4\pi A\ge \pi ^{2}\left( r_{out}-r_{in}\right) ^{2}, \end{aligned}$$

one gets that the difference between the inner and outer radii decreases to zero. Thus the evolving curve converges to a circle in the Hausdorff metric.

3 Existence and convexity

In this section we use Gage-Hamilton’s trick in [3]. Let \(\theta \) be the angle between T and the positive direction of the x axis and \(\tau =t\). With parameters \(\theta \) and \(\tau \) the evolution equation of curvature \(\kappa \) is

$$\begin{aligned} \frac{\partial \kappa }{\partial \tau } =\kappa ^{2}\left( \frac{\partial ^{2}}{\partial \theta ^{2}} \left( P-\frac{2A}{L}\right) +P-\frac{2A}{L}\right) . \end{aligned}$$
(3.1)

Lemma 3.1

A solution \(\kappa (\theta ,\tau )\) to (3.1) with initial value \(\kappa _{0}(\theta )\) exists for all time. Moreover, if \(\kappa _{0}(\theta )>0\) then \(\kappa (\theta ,\tau )>0\).

Proof

By (3.1) and \(P_{\theta \theta }+P=\frac{1}{\kappa }\) one has

$$\begin{aligned} \frac{\partial \kappa }{\partial \tau }&=\kappa ^{2}\left( \frac{\partial ^{2}}{\partial \theta ^{2}}\left( P-\frac{2A}{L}\right) +P-\frac{2A}{L}\right) \\&=\kappa ^{2}\left( \frac{1}{\kappa }-\frac{2A}{L}\right) . \end{aligned}$$

Set \(W=(\frac{1}{\kappa }-\frac{L}{2\pi })e^{\tau }\), then we have

$$\begin{aligned} \frac{\partial W}{\partial \tau }&=\left( -\frac{1}{\kappa ^{2}}\frac{\partial \kappa }{\partial \tau }-\frac{L_{\tau }}{2\pi }\right) e^{\tau }+\left( \frac{1}{\kappa }-\frac{L}{2\pi }\right) e^{\tau }\\&=\left( \frac{2A}{L}-\frac{1}{\kappa }+\frac{L}{2\pi }-\frac{2A}{L}\right) +\left( \frac{1}{\kappa }-\frac{L}{2\pi }\right) e^{\tau }=0. \end{aligned}$$

Since \(W(\theta ,0)=\frac{1}{\kappa _{0}}-\frac{L_{0}}{2\pi }\), then we get

$$\begin{aligned} W(\theta ,\tau )=\frac{1}{\kappa _{0}}-\frac{L_{0}}{2\pi }. \end{aligned}$$
(3.2)

Furthermore, we can write explicitly that

$$\begin{aligned} \kappa (\theta ,\tau )=\left[ \left( \frac{1}{\kappa _{0}}-\frac{L_{0}}{2\pi }\right) e^{-\tau }+\frac{L}{2\pi }\right] ^{-1}, \end{aligned}$$
(3.3)

combined with (2.4), It can also be expressed as

$$\begin{aligned} \kappa (\theta ,\tau )=\big [(\frac{1}{\kappa _{0}}-\frac{L_{0}}{2\pi })e^{-\tau }+\frac{\sqrt{4\pi A+(L^{2}_{0}-4\pi A)e^{-2\tau }}}{2\pi }\big ]^{-1}. \end{aligned}$$
(3.4)

Now we begin to prove the convexity of evolving curve.

When \(\frac{1}{\kappa _{0}}\ge \frac{L_{0}}{2\pi }\). Obviously, in this case, \(\kappa >0\) for all time.

When \(\frac{1}{\kappa _{0}}<\frac{L_{0}}{2\pi }\). Let \(\alpha =(\frac{1}{\kappa _{0}}-\frac{L_{0}}{2\pi })e^{-\tau }, \beta =\frac{\sqrt{4\pi A+(L^{2}_{0}-4\pi A)e^{-2\tau }}}{2\pi }\), respectively. (Clearly, \(\beta >0\)) Then we calculate:

$$\begin{aligned} \beta ^{2}-\alpha ^{2}&=\frac{4\pi A+\left( L^{2}_{0}-4\pi A\right) e^{-2\tau }}{4\pi ^{2}}- \left( \frac{1}{\kappa _{0}}-\frac{L_{0}}{2\pi }\right) ^{2}e^{-2\tau }\\&=\frac{A}{\pi }+\frac{L^{2}_{0}e^{-2\tau }}{4\pi ^{2}}-\frac{Ae^{-2\tau }}{\pi } -\frac{e^{-2\tau }}{\kappa ^{2}_{0}} -\frac{L^{2}_{0}e^{-2\tau }}{4\pi ^{2}}+\frac{L_{0}e^{-2\tau }}{\pi \kappa _{0}}\\&=\frac{1}{\kappa _{0}}\left( \frac{L_{0}}{\pi }-\frac{1}{\kappa _{0}}\right) e^{-2\tau } +\frac{A}{\pi }\left( 1-e^{-2\tau }\right) . \end{aligned}$$

Since \(\kappa _{0}>0\), then \(\beta ^{2}-\alpha ^{2}>0\). Combined with \(\beta >0\), we can easily get \(\beta >\alpha \). By (3.4), then we have \(\kappa >0\). This completes the proof. \(\square \)

Furthermore, when \(\tau \rightarrow \infty \), from (3.4), we obtain

$$\begin{aligned} \lim \limits _{\tau \rightarrow \infty }\kappa (\theta ,\tau )=\sqrt{\frac{\pi }{A}}. \end{aligned}$$

Since the radius of curvature \(\rho =\frac{1}{\kappa }\), then we have \(\lim _{\tau \rightarrow \infty }\rho (\theta ,\tau )=\sqrt{\frac{A}{\pi }}\).

Next, we study the following curve flow problem that is equivalent to (1.5) and prove its existence. To this end, we will deal equivalently with the evolution equation for support function of the evolving curve.(see [10, 11])

$$\begin{aligned} \frac{\partial X(\theta ,\tau )}{\partial \tau }=\left( P-\frac{2A}{L}\right) N(\theta ) -\frac{\partial \left( P-\frac{2A}{L}\right) }{\partial \theta }T(\theta ). \end{aligned}$$
(3.5)

Lemma 3.2

For a solution \(X(\theta ,\tau )\) to (3.5), the evolution equation of the support function satisfies

$$\begin{aligned} \frac{\partial P}{\partial \tau }=\frac{2A}{L}-P. \end{aligned}$$
(3.6)

Proof

By the definition of the support function \(P(\theta ,\tau )=-<X,-N>\) and \(\frac{\partial N}{\partial \tau }=0\), we get

$$\begin{aligned} \frac{\partial P}{\partial \tau }&=-<\frac{\partial X}{\partial \tau },N>+<F,\frac{\partial N}{\partial \tau }>\\&=-<\left( P-\frac{2A}{L}\right) N-\frac{\partial \left( P-\frac{2A}{L}\right) }{\partial \theta }T,N>\\&=\frac{2A}{L}-P. \end{aligned}$$

\(\square \)

Lemma 3.3

The support function P satisfies

$$\begin{aligned} P(\theta ,\tau )=\left( P(\theta ,0)+\frac{L_{0}}{2\pi }\right) e^{-\tau }+\frac{\sqrt{4\pi A+\left( L^{2}_{0}-4\pi A\right) e^{-2\tau }}}{2\pi }. \end{aligned}$$
(3.7)

Proof

Set \(U=(P-\frac{L}{2\pi })e^{\tau }\), then we have

$$\begin{aligned} \frac{\partial U}{\partial \tau }&=\left( P_{\tau }-\frac{L_{\tau }}{2\pi }\right) e^{\tau } +\left( P-\frac{L}{2\pi }\right) e^{\tau }\\&=\left( \frac{2A}{L}-P+\frac{L}{2\pi }-\frac{2A}{L}\right) +\left( P-\frac{L}{2\pi }\right) e^{\tau }=0. \end{aligned}$$

Since \(U(\theta ,0)=P(\theta ,0)-\frac{L_{0}}{2\pi }\), then we get

$$\begin{aligned} U(\theta ,\tau )=P(\theta ,0)-\frac{L_{0}}{2\pi }, \end{aligned}$$

combined with (2.4), we can complete the proof. \(\square \)

What’s more, when \(\tau \rightarrow \infty \), from (3.7), we obtain

$$\begin{aligned} \lim \limits _{\tau \rightarrow \infty }P(\theta ,\tau )=\sqrt{\frac{A}{\pi }}. \end{aligned}$$

Lemma 3.4

Suppose \(P(\theta ,\tau ):[0,2\pi ]\,\times \,[0,\infty )\rightarrow \,R\) is the smooth solution of the Eq. (3.7), the radius of curvature \(P_{\theta \theta }+P>0\) and the initial curve \(X_{0}=-P(\theta ,0)N(\theta )+\frac{\partial }{\partial \theta }P(\theta ,0)T(\theta )\). Then there exist a unique solution X(u, t) satisfying the Eq. (1.5) and the support function of curve is \(P(\theta ,\tau )\).

Proof

We know that any convex curve on the plane can be uniquely represented by the support function, so \({\widetilde{X}}:[0,2\pi ]\,\times \,[0,\infty )\) can be expressed as:

$$\begin{aligned} {\widetilde{X}}=-PN+\frac{\partial P}{\partial \theta }T. \end{aligned}$$

Because the unit tangent T and the unit normal vector N are independent of time \(\tau \), then we can get:

$$\begin{aligned} \frac{\partial {\widetilde{X}}}{\partial \tau }&=-\frac{\partial P}{\partial \tau }N-\frac{\partial N}{\partial \tau }P+\frac{\partial ^{2} P}{\partial \tau \partial \theta }T+\frac{\partial P}{\partial \theta }\frac{\partial T}{\partial \tau }\\&=-\frac{\partial P}{\partial \tau }N+\frac{\partial ^{2} P}{\partial \tau \partial \theta }T\\&=\left( P-\frac{2A}{L}\right) N+\frac{\partial }{\partial \theta }\left( \frac{2A}{L}-P\right) T. \end{aligned}$$

We make a parameter transformation, let \(\theta =\theta (u,t),\,\tau =t\), then \(\theta \) satisfies the following equation:

$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} \frac{\partial \theta }{\partial t}=-\kappa \frac{\partial }{\partial \theta }\left( \frac{2A}{L}-P\right) \\ \theta (u,0)=u \end{array}\right. } \end{aligned} \end{aligned}$$
(3.8)

We see that \(\theta (u,t)\) is the only solution of the above equation, then we have

$$\begin{aligned} \frac{\partial X(u,t)}{\partial t}&=\frac{\partial {\widetilde{X}}(\theta ,\tau )}{\partial t}\\&=\frac{\partial {\widetilde{X}}}{\partial \theta }\frac{\partial \theta }{\partial t}+ \frac{\partial {\widetilde{X}}}{\partial \tau }\frac{\partial \tau }{\partial t}\\&=\left( -P_{\theta }N-PN_{\theta }+P_{\theta \theta }T+P_{\theta }T_{\theta }\right) \frac{\partial \theta }{\partial t}+\frac{\partial {\widetilde{X}}}{\partial \tau }\\&=-\left( -P_{\theta }N+PT+P_{\theta \theta }T+P_{\theta }N\right) \kappa \frac{\partial }{\partial \theta }\left( \frac{2A}{L}-P\right) \\&\quad +\left( P-\frac{2A}{L}\right) N+\frac{\partial }{\partial \theta }\left( \frac{2A}{L}-P\right) T\\&=\left( P-\frac{2A}{L}\right) N. \end{aligned}$$

This completes the proof. \(\square \)