The steady state of the inclined problem

Xiaoding Yang

Abstract

The inclined problem is a problem that describes an open fluid flowing over an angled wall. It has broad applications in science and engineering. In this paper, we study the steady state of the inclined problem in two dimensions. The steady-state solution is depicted by the Euler-Lagrange equation of a given energy functional with a fixed contact angle as the boundary condition. By choosing a suitable maximal point to parameterize the surface of the fluid, we can construct a solution to this Euler-Lagrange equation via a shooting method in terms of the volume of the fluid. The construction works for any contact angle and any arbitrary inclined angle.

1 Introduction

Fluid mechanics is a fundamental branch of physics and engineering that deals with the behavior of fluids in motion and at rest. In recent years, an increasing number of math studies have focused on this field. One particular area of interest is fluid that flows over inclined surfaces, which is so-called fluid inclined problems. These problems arise in various natural and industrial applications such as the movement of water on sloped terrains, oil transport through pipelines, and the design of spillways and drainage systems[1].

When a fluid interacts with an inclined surface, several forces influence its motion, such as gravity, pressure, and surface tension. It is crucial to understand those effects if we want to predict flow behavior, optimize engineering designs, and improve efficiency in fluid-based systems. Even for the steady state (which means that the velocity of the fluid equals zero everywhere), discussing the balancing between different forces acting on fluids is quite interesting and important. It can give us some information about the moving fluid. Because of this, considerable effort has been devoted to analyze the shape of the equilibrium and steady state by mathematicians and engineers. But there is still a lot of work to do, including theoretical and computational problems.

Based on the practical problems we aim to study, the inclined problem typically falls into two distinct settings. The first involves the motion of a fluid flowing down an inclined surface—for instance, a thin film of fluid sliding down a tilted plane under the influence of gravity. The second setting concerns the motion of an open fluid flowing over an angled wall, as illustrated in Figure 1. This scenario can also be interpreted as a fluid mechanics problem within a triangular container, such as the cross-section of water in a channel. Both settings are common in everyday life and are important to discuss. In this paper, we focus exclusively on the second setting.

1.1 Formulation: Based on the discussion in the Background part, the purpose of this paper is to study the static viscous incompressible fluid occupied in an open-top vessel in two dimensions. The vessel is modeled as an open, connected, bounded triangular subset $\Gamma\subset\mathbb{R}^{2}$ that obeys the following pair of assumptions. See Figure 1. First, we posit that two inclined angles $\theta_{1}$ and $\theta_{2}$ that are defined as angles between the wall of the vessel and the horizontal plane are both angles between 0 and $\pi$ . Second, we assume that $\theta_{1}+\theta_{2}<\pi$ . Finally, it is assumed that the fluid occupies a subset $\Omega$ of the vessel $\Gamma$ , resulting in a free boundary where the fluid meets the air above the vessel.

In this article, we only discuss the steady state, which is equivalent to finding a static boundary surface. In the scenario depicted in Figure 1, the surface can be represented as a graph of x, which is denoted by v(x). However, this is not always the case; for instance, the boundary curve shown in Figure 3 cannot be represented as a graph of x. This observation forces us to describe the surface in polar coordinates. We choose the intersection point of two vessel walls (denoted point O in Figure 1) as the original point and then use $\rho(\theta)$ to represent the surface. Then we need to determine the equation that the boundary surface satisfies.

We first write down the energy functional with the conservation of the total volume in the polar coordinates.

{}E(\rho(\theta))=\frac{1}{3}g\int_{\theta_{2}}^{\pi-\theta_{1}}\rho^{3}\sin\theta d\theta+\int_{\theta_{2}}^{\pi-\theta_{1}}\sigma\sqrt{\rho^{2}+\rho^{\prime 2}}d\theta-[[\gamma]](\rho(\theta_{1})+\rho(\theta_{2}))

(1)

{}V=\mathcal{V}(\rho(\theta))=\frac{1}{2}\int_{\theta_{2}}^{\pi-\theta_{1}}\rho^{2}(\theta)d\theta

(2)

The first term of the functional E defined in (1) is the gravity energy and the second term is the surface tension energy. The third term is the free boundary energy caused by the interface between the fluid and the vessel. The second equation is the conservation of the total energy, where V is an arbitrary given fixed number. To make the model meaningful, we should assume that $-1\leq\frac{[[\gamma]]}{\sigma}\leq 1$ .

Using the variation method for this energy functional E, we can write down the Euler-Lagrange equation:

{}\begin{cases}g\rho^{2}\sin\theta+\sigma\frac{\rho}{\sqrt{\rho^{2}+\rho^{\prime 2}}}-\sigma\partial_{\theta}(\frac{\rho^{\prime}}{\sqrt{\rho^{2}+\rho^{\prime 2}}})-P_{0}\rho=0\\ \frac{\rho^{\prime}}{\sqrt{\rho^{2}+\rho^{\prime 2}}}(\theta_{2})=\frac{[[\gamma]]}{\sigma}\\ \frac{\rho^{\prime}}{\sqrt{\rho^{2}+\rho^{\prime 2}}}(\pi-\theta_{1})=-\frac{[[\gamma]]}{\sigma}\end{cases}

(3)

$P_{0}$ is the Lagrange multiplier that depends only on the given volume $V$ . The purpose of the paper is to construct a solution to the Euler-Lagrange equation (3). Our approach involves selecting a special point and utilizing equation (3) to construct a curve parametrized by the coordinate of this special point, which we designate as special parameters. We then represent the total volume $\mathcal{V}(\rho(\theta))$ and pressure $P_{0}$ in terms of these chosen parameters. We finally determine the value of special parameters via a shooting method by setting $\mathcal{V}=V$ , thereby obtaining the desired solution function.

We notice that the only unknown parameter in the equation system (3) is $P_{0}$ . We introduce the following shift to eliminate it:

{}(x,y)\rightarrow(x,y-\frac{P_{0}}{g})

(4)

We will see how this shift works in the following sessions.

To better understand the system of equations (3), we define a new variable $\psi$ as follows.

{}\begin{cases}\sin\psi(\theta)=\frac{\sin\theta\rho^{\prime}+\rho\cos\theta}{\sqrt{\rho^{2}+\rho^{\prime 2}}}\\ \cos\psi(\theta)=\frac{\rho\sin\theta-\rho^{\prime}\cos\theta}{\sqrt{\rho^{2}+\rho^{\prime 2}}}\end{cases}

(5)

which represents the slope angle in Cartesian coordinates. Due to the complexity of its formula, the definition may be difficult to grasp. But figure 2 can explicitly express the geometric meaning of $\psi$ .

1.3 The structure: We introduce the following definition.

	$\displaystyle\sin\psi_{1}$	$\displaystyle:=\sin\psi(\pi-\theta_{1})=\sin(\gamma+\frac{\pi}{2}-\theta_{1})=(-\frac{[[\gamma]]}{\sigma})\sin\theta_{1}+\sqrt{1-\frac{[[\gamma]]^{2}}{\sigma^{2}}}\cos\theta_{1}$		(6)
	$\displaystyle\sin\psi_{2}$	$\displaystyle:=\sin\psi(\theta_{2})=\sin(-\gamma-\frac{\pi}{2}+\theta_{2})=(\frac{[[\gamma]]}{\sigma})\sin\theta_{2}+\sqrt{1-\frac{[[\gamma]]^{2}}{\sigma^{2}}}\cos\theta_{2}$		(7)

where $\psi$ is defined in (5) and $\gamma\in[-\frac{\pi}{2},\frac{\pi}{2}]$ is defined as

{}\sin\gamma=-\frac{[[\gamma]]}{\sigma}

(8)

Then we divide our problem into two cases based on the sign of $\psi_{1}$ and $\psi_{2}$ .

1.3.1 $\psi_{1}$ and $\psi_{2}$ have different sign: In Section 2 we show that $\psi$ is a global variable for the boundary curve. See figure 3. Then we consider the equation system (10) which is equivalent to the original equation in (3).

{}\begin{cases}\frac{dx}{d\psi}=\sigma\frac{\cos\psi}{gy-P_{0}}\\ \frac{dy}{d\psi}=\sigma\frac{dy}{dx}\frac{dx}{d\psi}=\sigma\tan\psi\frac{\cos\psi}{gy-P_{0}}=\sigma\frac{\sin\psi}{gy-P_{0}}\\ \sin\psi(\pi-\theta_{1})=\sin\psi_{1}=(-\frac{[[\gamma]]}{\sigma})\sin\theta_{1}+\sqrt{1-\frac{[[\gamma]]^{2}}{\sigma^{2}}}\cos\theta_{1}\\ \sin\psi(\theta_{2})=\sin\psi_{2}=(\frac{[[\gamma]]}{\sigma})\sin\theta_{2}+\sqrt{1-\frac{[[\gamma]]^{2}}{\sigma^{2}}}\cos\theta_{2}\end{cases}

(9)

where:

\begin{cases}y(\psi)=(\rho\sin\theta)(\psi)\\ x(\psi)=(\rho\cos\theta)(\psi)\end{cases}

Then we apply the shift we introduced in (5) and the system (9) becomes:

{}\begin{cases}\frac{dx}{d\psi}=\sigma\frac{\cos\psi}{gy}\\ \frac{dy}{d\psi}=\sigma\frac{dy}{dx}\frac{dx}{d\psi}=\sigma\tan\psi\frac{\cos\psi}{gy}=\sigma\frac{\sin\psi}{gy}\\ \sin\psi(\pi-\theta_{1})=(-\frac{[[\gamma]]}{\sigma})\sin\theta_{1}+\sqrt{1-\frac{[[\gamma]]^{2}}{\sigma^{2}}}\cos\theta_{1}\\ \sin\psi(\theta_{2})=(\frac{[[\gamma]]}{\sigma})\sin\theta_{2}+\sqrt{1-\frac{[[\gamma]]^{2}}{\sigma^{2}}}\cos\theta_{2}\end{cases}

(10)

To derive a formula for the total volume of $\Omega$ , we first consider the equation in system (10) without the boundary condition, which is the following equation:

{}\begin{cases}\frac{dx}{d\psi}=\sigma\frac{\cos\psi}{gy}\\ \frac{dy}{d\psi}=\sigma\frac{dy}{dx}\frac{dx}{d\psi}=\sigma\tan\psi\frac{\cos\psi}{gy}=\sigma\frac{\sin\psi}{gy}\end{cases}

(11)

From the discussion in Chapter two, we can see that any solution satisfying the system (10) reaches the maximum at an interior point. We define them as follows.

	$\displaystyle u_{m}$	$\displaystyle=\sup_{\psi_{2}\leq\psi\leq\psi_{1}}y(\psi)=y(0)$		(12)
	$\displaystyle x_{m}$	$\displaystyle=x(\psi=0)=x(0)$		(13)

Using the existence and uniqueness theorem for ODE solution, we can obtain a solution curve by combining (11),(12) and (13) together. Then we have the following theorem for the total volume $\mathcal{V}(\rho(\theta))$ .

Theorem 1.1.

The functional $\mathcal{V}(\rho(\theta))$ and pressure $P_{0}$ can be represented as a function of $u_{m}$ . And we have the following two asymptotic properties of $\mathcal{V}$ :

	$\displaystyle\lim_{u_{m}\rightarrow 0}\mathcal{V}(u_{m})$	$\displaystyle=+\infty$		(14)
	$\displaystyle\lim_{u_{m}\rightarrow-\infty}\mathcal{V}(u_{m})$	$\displaystyle=0$		(15)

Consequently, for any V, there exists a $u_{m}$ and a solution to equation system (10) such that $\mathcal{V}(\rho(\theta))=\mathcal{V}(u_{m})=V$ . Furthermore, $\mathcal{V}(u_{m})$ is a monotone increasing function of $u_{m}$ when $[[\gamma]]<0$ which implies that for any given V, there exists a $u_{m}$ and a solution to (10) such that $\mathcal{V}(u_{m})=V$ . And when $[[\gamma]]<0$ , the choice of $u_{m}$ and the solution are unique.

Note: From Theorem 1.1, we can get a solution $(x(\psi),y(\psi))$ to (10) and $P_{0}$ for any given constant V. Therefore, $(x(\psi),y(\psi)+\frac{P_{0}}{g})$ is the solution to the original system.

1.3.2: Case two: $\psi_{1}$ and $\psi_{2}$ have the same sign

In this case we can show that the surface curve is a graph of x. See Figure 4. So we discuss the problem in Cartesian coordinates and use $u(x)$ to represent the surface curve. We have the equation system below which is equivalent to the system (3):

{}\begin{cases}\sigma\partial_{x}(\frac{\partial_{x}v}{\sqrt{1+|\partial_{x}u|^{2}}})=gv-P_{0}\\ \frac{\partial_{x}v}{\sqrt{1+|\partial_{x}v|^{2}}}(x_{1})=\sin\psi_{1}\\ \frac{\partial_{x}v}{\sqrt{1+|\partial_{x}v|^{2}}}(x_{2})=\sin\psi_{2}\end{cases}

(16)

Here, $\psi_{1}$ and $\psi_{2}$ are given by equations (6) and (7). Also:

\displaystyle(x,y)

\displaystyle=(\rho\cos\theta,\rho\sin\theta)

(17)

Then we apply the shift (5) again to change the system (18) to:

{}\begin{cases}\sigma\partial_{x}(\frac{\partial_{x}u}{\sqrt{1+|\partial_{x}u|^{2}}})=gu\\ \frac{\partial_{x}u}{\sqrt{1+|\partial_{x}u|^{2}}}(x_{1})=\sin\psi_{1}\\ \frac{\partial_{x}u}{\sqrt{1+|\partial_{x}u|^{2}}}(x_{2})=\sin\psi_{2}\end{cases}

(18)

We mainly discuss this system in Chapter 3.

Just like in the previous case, we examine the equation without boundary conditions first:

{}\sigma\partial_{x}(\sin\psi)=\sigma\partial_{x}(\frac{\partial_{x}u}{\sqrt{1+|\partial_{x}v|^{2}}})=gu

(19)

$\psi$ is defined by the equation (5). Then we choose another special point. We can show that the slope angle $\psi$ reaches its maximum value at some point. We have the following definitions:

{}\begin{cases}\psi_{m}:=\sup_{\theta_{2}\leq\theta\leq\theta_{1}}\psi(\theta)\\ u_{m}:=u(\theta_{m})\\ x_{m}:=x(\theta_{m})\end{cases}

(20)

Here $\theta_{m}$ is defined to be the angle such that $\psi(\theta)=\psi_{m}$ . Then we choose the point and angle defined by (20) as the initial conditions and get a solution by combining these conditions with (19). We will show the following theorem in Section 3:

Theorem 1.2.

We have two different circumstances.

•

When $\theta_{m}=\pi-\theta_{1}$ or $\theta_{m}=\theta_{2}$ (without loss of generality, we can assume that $\theta_{m}=\pi-\theta_{1}$ ), we have $\psi_{m}=\psi_{1}$ is a constant. We can choose $u_{m}$ as the independent variable. Then $\mathcal{V}(\rho(\theta))$ and $P_{0}$ can be represented as functions of $u_{m}$ . We have the following result:

$\lim_{u_{m}\rightarrow-\infty}\mathcal{V}(u_{m})=0$

and:

$V_{1}:=\max_{u_{m}\in(-\infty,0)}\mathcal{V}(u_{m})$

is a bounded constant. Therefore, there exist a $u_{m}$ such that $\mathcal{V}(u_{m})=V$ (defined in equation (2)) and a solution function of system (18) for any $V\leq V_{1}$ .
•

When $\theta_{m}\in(\theta_{2},\pi-\theta_{1})$ , we can show $u_{m}=0$ and we choose $\psi_{m}$ as the independent variable. Then $\mathcal{V}(\rho(\theta))$ and $P_{0}$ can be expressed as functions of $\psi_{m}$ and we can show the following result:

$\lim_{\psi_{m}\rightarrow 0}\mathcal{V}(\psi_{m})=+\infty$

and:

$V_{m}:=\min_{\psi_{m}\in(\max(\psi_{1},\psi_{2}),0)}\mathcal{V}(\psi_{m})$

is a positive constant. Consequently, there exist a $\psi_{m}$ such that $\mathcal{V}(\psi_{m})=V$ and a solution function of system (18) for any $V\geq V_{m}$

Finally, we have the following result:

Theorem 1.3.

$V_{1}\geq V_{m}$ , which means that there exists a solution function of the system (18) for any $V\geq 0$ and the solution is not unique when $V_{1}>V_{m}$ .

1.4 previous work: The study of the functional defined by (1) and (2) begins with analyzing the Euler-Lagrange equation, which is derived by Young[18], Laplace[13], and Gauss[3] along with the boundary condition of the contact angle using variational methods.

For fixed contact angle, Robert Finn investigated the existence of solutions in specific cases[2]. Finn proved the existence of an equilibrium of the sessile drop for any given volume V. In addition, he proved the nonexistence of the equilibrium of a liquid drop on an inclined wall. However, to our knowledge, general contact line problems involving different slopes remain largely unexplored.

In recent work, Guo-Tice[8, 7] studied the dynamic stability of the steady state when $\theta_{1}=\theta_{2}=\frac{\pi}{2}$ . The well-posedness theory of this problem is established by [9, 6, 19], while [4, 5] demonstrates certain decay properties of the water wave in a periodic domain and whole space, which indirectly imply the existence of the steady state. Beyond the contact line problem, Tice has analyzed the dynamics and stability of water waves in various settings related to the contact line problem in his papers [12, 10, 11, 16, 14, 15]. In particular, Tice-Wu[17] examined the dynamic stability of the steady state of the sessile drop which can be viewed as the flat version of the contact line problem where $\theta_{1}=\theta_{2}=0$ . Tice and Wu also showed the existence and stability of the steady state in their paper [17]. All of these problems involve dynamic contact points and contact angles, making them challenging to solve. Their work has inspired us to study a generalized version of the steady-state problem and has provided valuable insight. In this paper, our objective is to show the existence of a steady state for any given volume V and any incline angle $\theta_{1}$ and $\theta_{2}$ .

2 The first case: $\psi_{1}\cdot\psi_{2}<0$

In this chapter, we discuss the case that $\psi_{1}$ and $\psi_{2}$ have opposite signs. We begin by discussing the possible shape of the curve in Subsection 2.1 and then compute the total volume using the chosen parameter $u_{m}$ . Without loss of generality, we can assume that $\psi_{1}>0$ and $\psi_{2}<0$ are as shown in Figure 3 and $\theta_{1}>\theta_{2}$ .

2.1 The shape of the curve

In this subsection, we first assume that the equation system (10) has a solution function. We rewrite the system as follows.

\begin{cases}\frac{dx}{d\psi}=\sigma\frac{\cos\psi}{gy}\\ \frac{dy}{d\psi}=\sigma\frac{dy}{dx}\frac{dx}{d\psi}=\sigma\tan\psi\frac{\cos\psi}{gy}=\sigma\frac{\sin\psi}{gy}\\ \sin\psi(\pi-\theta_{1})=(-\frac{[[\gamma]]}{\sigma})\sin\theta_{1}+\sqrt{1-\frac{[[\gamma]]^{2}}{\sigma^{2}}}\cos\theta_{1}\\ \sin\psi(\theta_{2})=(\frac{[[\gamma]]}{\sigma})\sin\theta_{2}+\sqrt{1-\frac{[[\gamma]]^{2}}{\sigma^{2}}}\cos\theta_{2}\end{cases}

We first use it to study the shape of the solution curve. In fact, we want to show that the slope angle $\psi$ (defined by equation (5)) is a global legal variable for this solution curve. In addition, the y component of the curve reaches its maximum at an angle $\theta_{m}\in(\theta_{2},\pi-\theta_{1})$ since $\psi_{1}>0$ and $\psi_{2}<0$ . We let $u_{m}$ equal the maximum of the y component of the curve(the definition is given by (12)) and we have the following theorem for $\psi$ and $u_{m}$ .

Theorem 2.1.

(The shape of the surface curve) We define $\psi$ as in equation (5) and $u_{m}$ is the maximum of the y component of the solution curve. Then $u_{m}<0$ and $\psi$ are global legal parameters for the curve, meaning that the boundary curve can be represented as a function of $\psi$ .

Proof.

We first start looking at the shape of the curve. To begin with, we study the curve near the boundary point $(x_{1},y_{1})=(x(\pi-\theta_{1}),y(\pi-\theta_{1}))$ .

Case 1 $0\leq\psi(x_{1})<\frac{\pi}{2}$ :

In this setting, we know that $\sin\psi(x_{1})$ and $\cos\psi(x_{1})$ are both greater than 0. From the first and second equation in the system (10):

{}\begin{cases}\frac{dy}{d\psi}=\sigma\frac{\sin\psi}{gy}\\ \frac{dx}{d\psi}=\sigma\frac{\cos\psi}{gy}\end{cases}

(21)

We know that both $\frac{du}{d\psi}$ and $\frac{dx}{d\psi}$ are greater than 0 at the point $\psi(x_{1})$ due to our assumption. Therefore, the curve is locally a map of $\psi$ . We want to show that $y_{1}=y(\psi_{1})=y(\pi-\theta_{1})<0$ . We prove this by using a contradiction argument. We suppose that $y_{1}\geq 0$ and we separate the main proof into four steps.

Step 1: The local map of $\psi$ can be extended to a map of $\psi\in(\psi_{1},2\pi-\psi_{1})$ .

Using the assumption that $u_{1}\geq 0$ , we know from (21) that $\frac{du}{d\psi}$ and $\frac{dx}{d\psi}$ are equal to or greater than 0 near $\psi_{1}$ . So $u(\psi)$ and $x(\psi)$ are both an increasing function of $\psi$ near $\psi=\psi_{1}$ . Using a bootstrap argument, we can show that $u(\psi)\geq 0$ and $\frac{du}{d\psi}(\psi)\geq 0$ for any $\psi\in(\psi_{1},\frac{\pi}{2})$ . Then we want to extend the curve across the vertical point $\psi=\frac{\pi}{2}$ .

We can solve the first equation in the system (21) by integrating both sides of the equation from $\psi_{1}$ to any arbitrary angle $\psi$ to get the following equation:

{}\frac{1}{2}gy^{2}(\psi)-\frac{1}{2}gy_{1}^{2}=\sigma\cos\psi(x_{1})-\sigma\cos\psi

(22)

So when $\psi=\frac{\pi}{2}$ , $y(\frac{\pi}{2})$ is finite. Therefore, from the (21) we know that:

\frac{dy}{d\psi}(\frac{\pi}{2})=\frac{1}{gy(\frac{\pi}{2})}\neq 0

So, the solution function $(x(\psi),u(\psi))$ of the equation system (21) can be extended as a solution of the same system across the point $\psi=\frac{\pi}{2}$ .

When $\frac{\pi}{2}<\psi\leq\pi$ , we have $\cos\psi<0$ and $\sin\psi>0$ . Then $x(\psi)$ monotone decreasing and $y(\psi)$ monotone increasing. When $\psi=\pi$ , we define $x_{0}:=x(\pi)$ .

When $\pi<\psi<\frac{3}{2}\pi$ we have $x(\psi)$ monotone decreasing and $y(\psi)$ monotone decreasing if $y(\psi)>0$ . We will prove that $y(\psi)=y(2\pi-\psi)$ in step 2, which means that the curve is symmetric and $y(\psi)>0$ for any $\pi<\psi<\frac{3}{2}\pi$ . Therefore, $y(\psi)>0$ for any $\psi\in(\pi,\frac{3}{2}\pi)$ , which implies that $y(\psi)$ monotone decreasing. See Figure 3. Again, the curve can be extended across the point $\psi=\frac{3\pi}{2}$ due to the same reason as the point $\frac{\pi}{2}$ and finally it can be extended to the point $\psi=2\pi-\psi_{1}$ .

Step 2: The curve constructed in step 1 exhibits the property $y(\psi)=y(2\pi-\psi)$ , which implies the symmetry of the curve.

Applying the equation (22), We have the following equation:

{}\frac{1}{2}gy^{2}(\gamma)-\frac{1}{2}gy_{1}^{2}=\sigma\cos\psi_{1}-\sigma\cos\gamma

(23)

for any $\gamma\in(\psi_{1},2\pi-\psi_{1})$ . We set $\psi=2\pi-\gamma$ in equation (22) to obtain:

{}\frac{1}{2}gy^{2}(2\pi-\psi)-\frac{1}{2}gy_{1}^{2}=\sigma\cos\psi_{1}-\sigma\cos(2\pi-\gamma)=\sigma\cos\psi_{1}-\sigma\cos\gamma

(24)

We combine equations (23) and (24) together to show that:

{}y(\psi)^{2}=y^{2}(2\pi-\psi)

(25)

which means that $y(\psi)=y(2\pi-\psi)$ or $y(\psi)=-y(2\pi-\psi)$ . We assume that there exists a $\psi_{0}\in(\pi,2\pi-\psi_{1})$ such that $y(\psi_{0})=-y(2\pi-\psi_{0})$ . Then $2\pi-\psi_{0}\in(\psi_{1},\pi)$ . However, we know that $y(2\pi-\psi_{0})>0$ from the discussion in step 1. Then we should have $y(\psi_{0})<0$ . Therefore, there is a $\psi_{7}\in(2\pi-\psi_{0},\psi_{0})$ such that $y(\psi_{7})=y(2\pi-\psi_{7})=0$ . We know that either $\psi_{7}\in(\psi_{1},\pi)$ or $2\pi-\psi_{7}\in(\psi_{1},\pi)$ , which means that $y(2\pi-\psi_{7})>0$ or $y(\psi_{7})>0$ . This contradicts the definition of $\psi_{7}$ . So we always have:

y(\psi)=y(2\pi-\psi)

(26)

which is the result we want.

Step 3 The curve we constructed in Step 1 cannot be a solution curve to our system (10).

Using equation (10), we know that the curve should meet with the right wall with the contact angle $\psi_{2}$ . Because of this, we need to consider the point $\psi=\psi(x_{2})+2\pi=\psi_{2}+2\pi$ . We suppose that the curve meets the right wall at the point $\psi=2\pi+\psi_{2}$ . We want to show that this assumption is not true.

From equation (7), we know that $\psi_{3}:=2\pi+\psi(x_{2})=2\pi-\gamma-\frac{\pi}{2}+\theta_{2}$ is an angle between $\pi$ and $2\pi$ . More specifically, we have:

\psi_{3}=2\pi-\gamma-\frac{\pi}{2}+\theta_{2}<2\pi-(\gamma+\frac{\pi}{2}-\theta_{1})=2\pi-\psi_{1}

where we used the fact that $\theta_{1}>\theta_{2}$ and equation (6) and (7). This implies that $\psi_{3}\in(\pi,2\pi-\psi_{1})$ . We suppose that the corresponding point is $(x_{3},u_{3})=(x(\psi_{3}),y(\psi_{3}))$ . See Figure 5. We want to prove that the curve we constructed does not meet the right wall when $\psi\in(\psi_{1},\psi_{3})$ .

By definition:

\psi_{3}-\pi=2\pi+\psi(x_{2})-\pi=\frac{\pi}{2}-\gamma+\theta_{2}>\theta_{2}

We can show that for any angle $\psi\in(\pi+\theta_{2},\psi_{3})$ , $(x(\psi),u(\psi))$ is on the left hand side of the right wall. Then using the property $\theta_{1}>\theta_{2}$ , we can show that:

\psi_{4}:=2\pi-\psi_{3}=-\psi_{2}=\gamma+\frac{\pi}{2}-\theta_{2}>\gamma+\frac{\pi}{2}-\theta_{1}=\psi_{1}

which means that $\psi_{4}\in(\psi_{1},\pi)$ . In addition, we have $u(\psi_{4})=u(\psi_{3})$ by using step 2. See Figure 5. We should have $x(\psi_{4})>x(\psi_{3})$ to avoid self-intersection. So the point $(x(\psi_{4}),u(\psi_{4}))$ should be on the right hand side of the right wall. Therefore the curve intersects the wall at a point $(x(\psi_{5}),u(\psi_{5}))$ where $\psi_{4}<\psi_{5}<\psi_{3}$ which contradicts to the assumption.

Therefore, by applying the contact angle condition (7), the curve does not meet the right wall when $\psi_{1}<\psi\leq 2\pi-\psi_{3}$ . However, when $\psi_{6}=2\pi-\psi_{1}$ , we can use the Step 2 again to obtain $y(\psi_{6})=y(\psi_{1})$ . In addition, the curve needs to satisfy the condition $x(\psi_{6})<x(\psi_{1})$ to avoid self-intersection. This implies that the point lies left to the left wall. So the curve meets the left wall and even passes across it at some angle between $\psi_{1}$ and $2\pi-\psi_{1}$ , which is impossible. Hence, we conclude that $y_{1}<0$ .

Step 4: $u_{m}<0$ .

From Step 1 to Step 3, we have $y_{1}<0$ . Using the equation (19), we observe that $\psi(x)$ decreases as $y(x)<0$ . Suppose there is an angle $0<\psi_{8}<\psi_{1}$ such that $y(\psi_{8})=0$ . Applying the same discussion as for the $y_{1}\geq 0$ , we find that this assumption leads to a contradiction. So at the maximal point for u where $\sin\psi=0$ , we must have $u_{m}=y(0)<0$ . Consequently, for every $\psi\in(\psi_{2},\psi_{1})$ we obtain $y(\psi)<0$ ensuring that the equation (21) remains well defined for every $\psi\in(\psi_{2},\psi_{1})$ . Thus $\psi$ is a well-defined global parameter for the surface curve. This completes the proof of Case 1.

Case 2 $\frac{\pi}{2}\leq\psi(x_{1})<\pi$ . In this case, we also do the same discussion as above. Actually, it is even easier since we do not need to consider the extending problem when $\psi=\frac{\pi}{2}$ .

In conclusion, we have shown that $\psi$ is a legal parameter and $u_{m}<0$ .

∎

2.2 The existence of the solution function

From the previous subsection, we have the a priori discussion for the boundary curve, which gives us an idea about the shape of the steady state. In this subsection, we discuss the existence of the steady state.

Using Theorem 2.1, we now use $\psi$ to parameterize the curve ( $\psi$ is given by equation (5)). So, the original equation system (3) is equivalent to the equation system (10) after shifting $(x,y)\rightarrow(x,y-\frac{P_{0}}{g})$ . And we focus on analyzing the system (10) in this subsection.

Using the assumption that $\psi_{1}>0$ and $\psi_{2}<0$ , the curve reaches its maximum at a point $(x_{m},u_{m})=(x(\psi_{m}),y(\psi_{m}))$ . At this point, we must have $\psi(\theta_{m})=0$ . See figure 3. Using the existence and uniqueness theorem of ODEs, the solution of this system is uniquely determined once the initial conditions are specified. In other words, we construct the curve using the equation (11) with the initial conditions (12) and (13). Then we use these two new parameters $x_{m}$ and $u_{m}$ to derive a formula for the total volume instead of the unknown parameter $P_{0}$ .

Since the equation system (10) includes two boundary conditions, there is effectively only one independent variable. As a result, both the total volume $\mathcal{V}$ and $P_{0}$ can be expressed as functions of this variable, and then we set $\mathcal{V}=V$ to determine this variable.

A key challenge in applying the boundary condition in (10) is determining the position of the original point O (the junction of the two walls) after the coordinate change. To be specific, the y-component of its coordinate remains unknown. See the figure 3. In this figure, we know that the y component of O must be $-\frac{P_{0}}{g}$ to make the system compatible with the shift $(x,y)\rightarrow(x,y-\frac{P_{0}}{g})$ . However, since $P_{0}$ is still unknown at this point, we cannot use geometry relations to derive the coordinates of two contact points. Fortunately, we know that the curve hits the left wall and the right wall with contact angles $\psi_{1}$ and $\psi_{2}$ , which are fixed angles. So once the curve is constructed by the equation (10) with the chosen initial conditions, we can set $\psi=\psi_{1}$ or $\psi=\psi_{2}$ to derive their coordinates. This leads to the following theorem.

Theorem 2.2.

( Boundary Points) We suppose that the boundary curve contacts two walls at two contact points $(x_{1},u_{1})$ and $(x_{2},u_{2})$ . We have the following computation:

	$\displaystyle u_{1}=-\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1})}$		(27)
	$\displaystyle u_{2}=-\sqrt{u_{m}^{2}+\frac{2}{g}(1-\cos\psi_{2})}$		(28)
	$\displaystyle x_{1}=x_{m}-\int_{0}^{\psi_{1}}\frac{\cos\gamma}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\gamma)}}d\gamma$		(29)
	$\displaystyle x_{2}=x_{m}-\int_{0}^{\psi_{2}}\frac{\cos\gamma}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\gamma)}}d\gamma$		(30)
	$\displaystyle r(\psi):=\|x(\psi)-x_{m}\|=\|\int_{0}^{\psi}\frac{\cos\gamma}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\gamma)}}d\gamma\|$		(31)

Proof.

Step 1 Proof of (27):

Using the equation (22), we have:

{}\frac{1}{2}gy^{2}(\psi)-\frac{1}{2}gu_{1}^{2}=\sigma(\cos\psi_{1}-\cos\psi)

(32)

We know that $u_{1}=y(\psi_{1})$ and $\psi_{m}=0$ . So we set $\psi=\psi_{m}=0$ in the equation (32) to get:

\frac{1}{2}gu_{1}^{2}-\frac{1}{2}gu_{m}^{2}=\sigma(1-\cos\psi_{1})

which shows that:

{}\frac{1}{2}gu_{1}^{2}=\frac{1}{2}gu_{m}^{2}+\sigma(1-\cos\psi_{1})

(33)

Using Theorem 2.1, we know that $u_{m}<0$ . Since $(x_{m},u_{m})$ is the maximum point of the y component of the curve, it follows that $u_{1}<0$ as well. Taking square root on both sides of the equation (33), we obtain:

u_{1}=-\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1})}

which is exactly (27). We can also use the similar computation to show equation (28)

Step 2 Proof of (29):

Using the first equation in the system (10), we have:

{}\frac{dx}{d\psi}=\sigma\frac{\cos\psi}{gy(\psi)}

(34)

From equation (32) in the step 1 we can obtain:

{}y(\psi)=-\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}

(35)

We plug equation (35) back into the equation (34) to show that:

{}\frac{dx}{d\psi}=\frac{\cos\psi}{-g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}}

(36)

Then we integrate both sides of the equation from 0 to $\psi_{1}$ then we to derive the following formula:

x_{1}-x_{m}=-\int_{0}^{\psi_{1}}\frac{\cos\gamma}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\gamma)}}d\gamma

which is exactly what we want. In addition, we can use the same computation to derive (30).

Finally, the proof of (31) is just another simple application of the equation (36).

∎

We can then express the total volume in terms of $u_{m}$ and $x_{m}$ to. Before we derive this formula, we first establish a relationship between $x_{m}$ and $u_{m}$ using the geometry relation so that we only have one independent parameter.

Theorem 2.3.

There holds:

	$\displaystyle x_{m}=\frac{r_{1}\tan\theta_{1}-r_{2}\tan\theta_{2}-(u_{1}-u_{2})}{\tan\theta_{1}+\tan\theta_{2}}\leavevmode\nobreak\ \operatorname{when}\leavevmode\nobreak\ \theta_{1}\neq\frac{\pi}{2}$		(37)
	$\displaystyle x_{m}=r_{1}\leavevmode\nobreak\ \operatorname{when}\leavevmode\nobreak\ \theta_{1}=\frac{\pi}{2}$		(38)
	$\displaystyle P_{0}=g(x_{2}\tan\theta_{2}-u_{2})$		(39)

where:

	$\displaystyle r_{1}$	$\displaystyle:=r(\psi_{1})=x_{m}-x_{1}$		(40)
	$\displaystyle r_{2}$	$\displaystyle:=r(\psi_{2})=x_{2}-x_{m}$		(41)

Remark: Equations (37), (38), and (39) imply that $u_{1}$ , $u_{2}$ , $r_{1}$ , and $r_{2}$ are all functions of $u_{m}$ . Consequently, by applying this theorem, $x_{m}$ can also be expressed as a function of $u_{m}$ .

Proof.

We can show that $x_{m}=r_{1}$ when $\theta_{1}=\frac{\pi}{2}$ just by definition (31). So we only need to prove (37) and (39).

We suppose that the two walls intersect at the point $O$ . The x component of O is zero by our construction of the coordinate, while y component of the point, denoted as $y_{0}$ , is still unknown. Since the point $(x_{1},u_{1})$ lies on the left wall which is inclined at $\theta_{1}$ , we can derive the following result by using geometry relation:

{}y_{0}=u_{1}-(-x_{1})\tan\theta_{1}

(42)

Also, since $(x_{2},u_{2})$ lies on the right wall which inclined at angle $\theta_{2}$ , we have:

{}y_{0}=u_{2}-x_{2}\tan\theta_{2}

(43)

We then combine (32) and (33) together to get:

{}u_{1}-(-x_{1})\tan\theta_{1}=u_{2}-x_{2}\tan\theta_{2}

(44)

Then we plug the definition of $r_{1}$ and $r_{2}$ (40) and (41) into the equation(44) to get:

u_{1}-(r_{1}-x_{m})\tan\theta_{1}=u_{2}-(r_{2}+x_{m})\tan\theta_{2}

which implies the following relation:

x_{m}(\tan\theta_{1}+\tan\theta_{2})=u_{2}-u_{1}+r_{1}\tan\theta_{1}-r_{2}\tan\theta_{2}

which means that:

x_{m}=\frac{r_{1}\tan\theta_{1}-r_{2}\tan\theta_{2}-(u_{1}-u_{2})}{(\tan\theta_{1}+\tan\theta_{2})}

This is exactly the equation (37).

Also, using (42), we know that the coordinate of the connecting point O is $(0,y_{0})=(0,u_{2}-x_{2}\tan\theta_{2})$ . And we know that after the shift $(x,y)\rightarrow(x,y+\frac{P_{0}}{g})$ , coordinates of this point become $(0,0)$ . So we should have:

0=\frac{P_{0}}{g}+(u_{2}-x_{2}\tan\theta_{2})

which implies that:

P_{0}=g(x_{2}\tan\theta_{2}-u_{2})

This finishes the proof. ∎

Now we have the representation for the $P_{0}$ and $x_{m}$ . In order to continue to compute the area, we also care about the sign of $x_{m}$ . We have the following lemma:

lemma 2.4.

When $\theta_{1}=\theta_{2}$ , $x_{m}=0$ . When $\theta_{2}<\theta_{1}<\frac{\pi}{2}$ and $[[\gamma]]\leq 0$ , we fix the value of $\theta_{2}$ and $u_{m}$ . Then $x_{m}$ is an increasing function of $\theta_{1}$ .

Proof.

Using Theorem 2.3, we have the formula for $x_{m}$ :

{}x_{m}=\frac{r_{1}\tan\theta_{1}-r_{2}\tan\theta_{2}-(u_{1}-u_{2})}{\tan\theta_{1}+\tan\theta_{2}}

(45)

Case one $\theta_{1}=\theta_{2}$ :

We use $f(\theta_{1})$ to denote the numerator of the right hand side of equation (45):

{}f(\theta_{1})=r_{1}\tan\theta_{1}-r_{2}\tan\theta_{2}-(u_{1}-u_{2})

(46)

When $\theta_{1}=\theta_{2}$ , we know from (6) and (7) that $\psi_{1}=-\psi_{2}$ , which implies that $\cos\psi_{1}=\cos\psi_{2}$ . Substituting this into the four equations (The equation (27)-(30)) in theorem 2.2, we obtain:

u_{1}=u_{2}\leavevmode\nobreak\ \leavevmode\nobreak\ and\leavevmode\nobreak\ \leavevmode\nobreak\ r_{1}=r_{2}

So we should have $x_{m}=0$ by using equation (37).

Case two $\theta_{1}\neq\theta_{2}$ :

Then we want to compute the general case that $\theta_{1}>\theta_{2}$ . To prove this we need to compute $\frac{df}{d\theta_{1}}$ where f is defined as in equation (46).

First by using equations (27)-(30) in Theorem 2.2 we can show that:

{}r_{1}=x_{m}-x_{1}=\int_{0}^{\psi_{1}}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}}

(47)

{}r_{2}=x_{2}-x_{m}=\int_{\psi_{2}}^{0}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}}

(48)

and:

{}u_{1}-u_{2}=-\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1})}+\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{2})}

(49)

Then we plug (47), (48) and (49) into equation (46) to compute the first derivative of f with respect to $\theta_{1}$ :

$\displaystyle\frac{df(\theta_{1})}{d\theta_{1}}=$	$\displaystyle\frac{1}{\cos^{2}\theta_{1}}r_{1}+\frac{dr_{1}}{d\theta_{1}}\tan\theta_{1}-\frac{du_{1}}{d\theta_{1}}$
$\displaystyle=$	$\displaystyle\frac{1}{\cos^{2}\theta_{1}}\int_{0}^{\psi_{1}}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}}+\tan\theta_{1}\frac{\cos\psi_{1}}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1})}}\frac{d\psi_{1}}{d\theta_{1}}$
	$\displaystyle+\frac{\sin\psi_{1}}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1})}}\frac{d\psi_{1}}{d\theta_{1}}$	(50)

To continue to the next step, we need to introduce an inequality below :

{}\int_{0}^{\psi_{1}}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}}\geq\int_{0}^{\psi_{1}}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1})}}

(51)

We use Lemma 2.5 which we are going to prove after the main part of the theorem to show this inequality.

We apply inequality (51) to the $\eqref{equ:2.2.21}$ and obtain:

	$\displaystyle\frac{df}{d\theta_{1}}\geq$	$\displaystyle\frac{1}{g\cos^{2}\theta_{1}\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1})}}\int_{0}^{\psi_{1}}\cos\psi d\psi$
		$\displaystyle+\tan\theta_{1}\frac{\cos\psi_{1}}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1})}}\frac{d\psi_{1}}{d\theta_{1}}+\frac{\sin\psi_{1}}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1})}}\frac{d\psi_{1}}{d\theta_{1}}$

Also we know from (6) that $\psi_{1}=\gamma+\frac{\pi}{2}-\theta_{1}$ where $\gamma$ is the contact angle. So we have $\frac{d\psi(x_{1})}{d\theta_{1}}=-1$ . We plug it back into the inequality above and obtain:

$\displaystyle\frac{df(\theta_{1})}{d\theta_{1}}\geq$	$\displaystyle\frac{\sin\psi(x_{1})}{g\cos^{2}\theta_{1}\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi(x_{1}))}}-\tan\theta_{1}\frac{\cos\psi(x_{1})}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi(x_{1}))}}$
	$\displaystyle-\frac{\sin\psi(x_{1})}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi(x_{1}))}}$
$\displaystyle=$	$\displaystyle\frac{1}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi(x_{1}))}}(\frac{\sin\psi(x_{1})}{\cos^{2}\theta_{1}}-\tan\theta_{1}\cos\psi(x_{1})-\sin\psi(x_{1}))$
$\displaystyle=$	$\displaystyle\frac{1}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi(x_{1}))}}(\sin\psi(x_{1})\tan^{2}\theta_{1}-\tan\theta_{1}\cos\psi(x_{1}))$
$\displaystyle=$	$\displaystyle\frac{1}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi(x_{1}))}}\tan\theta_{1}(\sin\psi(x_{1})\tan\theta_{1}-\cos\psi(x_{1}))$
$\displaystyle=$	$\displaystyle\frac{1}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi(x_{1}))}}\tan\theta_{1}(\sin(\frac{\pi}{2}-\theta_{1}+\gamma)\tan\theta_{1}-\cos(\frac{\pi}{2}-\theta_{1}+\gamma))$
$\displaystyle=$	$\displaystyle\frac{1}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi(x_{1}))}}\tan\theta_{1}(\cos(\theta_{1}-\gamma)\tan\theta_{1}-\sin(\theta_{1}-\gamma))$
$\displaystyle=$	$\displaystyle\frac{1}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi(x_{1}))}}\tan\theta_{1}\cos(\theta_{1}-\gamma)(\tan\theta_{1}-\tan(\theta_{1}-\gamma))$	(52)

Since $[[\gamma]]\leq 0$ , we can use the definition of $\gamma$ (8) to show that:

\sin\gamma=-\frac{[[\gamma]]}{\sigma}\geq 0

which means that $\gamma\in[0,\frac{\pi}{2})$ . In addition, we have the condition $0<\theta_{1}<\frac{\pi}{2}$ . So we should have $\cos(\theta_{1}-\gamma)>0$ and $\tan\theta_{1}-\tan(\theta_{1}-\gamma)>0$ . We plug these results back into equation (52) and obtain:

\frac{df(\theta_{1})}{d\theta_{1}}\geq\frac{1}{g\sqrt{u_{m}^{2}+\frac{2}{g}(1-\cos\psi(x_{1}))}}\tan\theta_{1}\cos(\theta_{1}-\gamma)(\tan\theta_{1}-\tan(\theta_{1}-\gamma))\geq 0

which finishes the proof. ∎

Then it remains to prove the inequality (51). We have the following lemma:

lemma 2.5.

\int_{0}^{\psi_{1}}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}}\geq\int_{0}^{\psi_{1}}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi(x_{1}))}}

Proof.

First, if $\psi_{1}=\psi(\theta_{1})\in(0,\frac{\pi}{2})$ , the lemma is straightforward to prove. Since $\cos\psi>0$ for any $\psi\in(0,\psi_{1})$ and $\cos\psi\geq\cos\psi_{1}$ for any $\psi\in(0,\psi_{1})$ , we can show that:

{}\frac{\cos\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}}\geq\frac{\cos\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi(x_{1}))}}

(53)

Then we integrate on both sides of the inequality (53) from $0$ to $\pi$ and we have the inequality (51) proved.

If $\psi(x_{1})\in(\frac{\pi}{2},\pi)$ , we split the integral into three parts:

	$\displaystyle\int_{0}^{\psi_{1}}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}}=$	$\displaystyle\int_{0}^{\pi-\psi_{1}}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}}$
		$\displaystyle+\int_{\pi-\psi_{1}}^{\frac{\pi}{2}}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}}+\int_{\frac{\pi}{2}}^{\psi_{1}}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}}$

For the first term above, we use the fact that $\cos\psi\geq 0\geq\cos(\psi_{1})$ for any $\psi\in(0,\pi-\psi_{1})$ to obtain that:

{}\int_{0}^{\pi-\psi_{1}}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}}\geq\int_{0}^{\pi-\psi_{1}}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1})}}

(54)

For the second term and the third term we combine them together and using the fact $\cos(\pi-\psi)=-\cos(\psi)$ to obtain:

	$\displaystyle\int_{\pi-\psi_{1}}^{\frac{\pi}{2}}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}}+\int_{\frac{\pi}{2}}^{\psi_{1}}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}}$
$\displaystyle=$	$\displaystyle\int_{\pi-\psi_{1}}^{\frac{\pi}{2}}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1})}}+\int_{\frac{\pi}{2}}^{\psi_{1}}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1})}}$
	$\displaystyle+\int_{\pi-\psi_{1}}^{\frac{\pi}{2}}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}}-\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1})}}$
	$\displaystyle+\int_{\pi-\psi_{1}}^{\frac{\pi}{2}}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1})}}-\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1+\cos\psi)}}$
$\displaystyle=$	$\displaystyle\int_{\pi-\psi_{1}}^{\frac{\pi}{2}}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1})}}+\int_{\frac{\pi}{2}}^{\psi_{1}}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1})}}$	(55)
	$\displaystyle+\int_{\pi-\psi_{1}}^{\frac{\pi}{2}}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}}-\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1+\cos\psi)}}$	(56)

Here we use the fact that $\cos\psi=\cos(\pi-\psi)$ The terms in (55) are what we are going to keep and we want to show (56) is greater than zero. Using the fact that:

\cos\psi>0\leavevmode\nobreak\ \operatorname{when}\leavevmode\nobreak\ \psi\in(\pi-\psi_{1},\frac{\pi}{2})

we can show that:

g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}<g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1+\cos\psi)}

which implies that:

\int_{\pi-\psi_{1}}^{\frac{\pi}{2}}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}}-\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1+\cos\psi)}}>0

Therefore, we obtain the following result:

\int_{\pi-\psi_{1}}^{\frac{\pi}{2}}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}}+\int_{\frac{\pi}{2}}^{\psi_{1}}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}}

{}\geq\int_{\pi-\psi_{1}}^{\frac{\pi}{2}}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1})}}+\int_{\frac{\pi}{2}}^{\psi_{1}}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1})}}

(57)

We combine the inequality (57) with the inequality (54) and we can obtain the final result. ∎

Remark: Based on Theorem 2.4 we know that $x_{m}>0$ when $\theta_{2}<\theta_{1}<\frac{\pi}{2}$ and $[[\gamma]]<0$ . In addition, when $\theta_{1}>\frac{\pi}{2}$ , $x_{m}$ is obviously positive by definition (Equation (13)). So we always have $x_{m}>0$ when $\theta_{1}>\theta_{2}$ and $[[\gamma]]<0$

Also, Theorem 2.3 and Theorem 2.4 have already guaranteed that we can use $u_{m}$ to represent $P_{0}$ . Then we only need to compute the total volume $\mathcal{V}(u_{m})$ . We split the total volume into five parts. The first part is the region enclosed by $y=u_{1}$ , $x=x_{m}$ , and the surface curve $(x(\psi),y(\psi))$ where $\psi\in(0,\psi_{1})$ . And the second part is the region enclosed by $y=u_{2}$ , $x=x_{m}$ and the surface curve $(x(\psi),y(\psi))$ where $\psi\in(\psi_{2},0)$ . The third part is the triangular area enclosed by $y=u_{1}$ , $x=m$ and the left wall. The fourth part is the triangular area enclosed by $y=u_{2}$ , $x=0$ and the right wall. The fifth part is the square area enclosed by $y=u_{1}$ , $x=0$ , $y=u_{2}$ and $x=x_{m}$ . We use $\Omega_{1}$ to $\Omega_{5}$ to denote them and also use $\mathcal{V}_{1}$ to $\mathcal{V}_{5}$ to represent the volume. See Figure 6.

In addition, there might be some difference between the case $\theta_{1}=\frac{\pi}{2}$ and $\theta_{1}\neq\frac{\pi}{2}$ from Theorem 2.3. Therefore, we split our computation into two parts.

2.2.1 $\theta_{1}\neq\frac{\pi}{2}$

We assume that $\theta_{1}\neq\frac{\pi}{2}$ . Then we can compute the volume as follows:

	$\displaystyle\mathcal{V}_{1}$	$\displaystyle=\int_{u_{1}}^{u_{m}}r(v)dy=\int_{0}^{\psi_{1}}r(\psi)\frac{dy}{d\psi}d\psi=\int_{0}^{\psi_{1}}r(\psi)\frac{\sin\psi}{gy}d\psi$
		$\displaystyle=\int_{0}^{\psi_{1}}\int_{0}^{\psi}\frac{\cos\gamma d\gamma}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\gamma)}}\frac{\sin\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}}d\psi$		(58)

Here $r(\psi)$ is defined by equation (31).

Using the same computation, we have the following relations:

{}\mathcal{V}_{2}=\int_{\psi_{2}}^{0}\int_{0}^{\psi}\frac{\cos\gamma d\gamma}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\gamma)}}\frac{\sin\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}}d\psi

(59)

Also, we can know from the geometry that:

$\displaystyle\mathcal{V}_{3}$	$\displaystyle=\frac{1}{2}(r_{1}-x_{m})^{2}\tan\theta_{1}$	(60)
$\displaystyle\mathcal{V}_{4}$	$\displaystyle=\frac{1}{2}(r_{2}+x_{m})^{2}\tan\theta_{2}$	(61)
$\displaystyle\mathcal{V}_{5}$	$\displaystyle=x_{m}(u_{1}-u_{2})$	(62)

Then we want to find the relation between $\mathcal{V}=\mathcal{V}_{1}+\mathcal{V}_{2}+\mathcal{V}_{3}+\mathcal{V}_{4}+\mathcal{V}_{5}$ and $u_{m}$ .

Remark: Although Figure 6 only shows the case where $\theta_{1}\in(0,\frac{\pi}{2})$ , the representation of the total volume $\mathcal{V}$ has the same form when $\theta_{1}\in(\frac{\pi}{2},\pi)$ .

Theorem 2.6.

The two volume functional $\mathcal{V}_{1}$ and $\mathcal{V}_{2}$ are both increasing functions of $u_{m}$ . And when $u_{m}\rightarrow 0$ , $\mathcal{V}_{1}$ and $\mathcal{V}_{2}$ converge to $+\infty$ . When $u_{m}\rightarrow-\infty$ , $\mathcal{V}_{1}$ and $\mathcal{V}_{2}$ converge to $0$ .

Proof.

Since $\mathcal{V}_{1}$ and $\mathcal{V}_{2}$ have similar forms. We only need to prove this theorem for $\mathcal{V}_{1}$ .

Based on a simple observation of equation(58), $\lim_{u_{m\rightarrow-\infty}}\mathcal{V}_{1}=0$ since the denominator of the term inside the integral converges to $\infty$ .

We then show the monotonicity of the volume $\mathcal{V}_{1}$ . Taking derivative with respect to $u_{m}$ on both sides of the equation (58), we obtain the following expression:

	$\displaystyle\frac{d\mathcal{V}_{1}}{du_{m}}=$	$\displaystyle u_{m}\int_{0}^{\psi_{1}}\int_{0}^{\psi}\frac{\cos\tau d\tau}{g(u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\tau))^{\frac{3}{2}}}\frac{\sin\psi}{-g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}}d\psi$
		$\displaystyle+u_{m}\int_{0}^{\psi_{1}}\int_{0}^{\psi}\frac{\cos\tau d\tau}{g(u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\tau))^{\frac{1}{2}}}\frac{\sin\psi}{-g({u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)})^{\frac{3}{2}}}d\psi$		(63)

For the first term in the right hand side of equation (63) .we need to study the sign of the integral:

\int_{0}^{\psi}\frac{\cos\tau d\tau}{g(u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\tau))^{\frac{3}{2}}}

Then we define:

a(\psi)=\int_{0}^{\psi}\frac{\cos\tau d\tau}{g(u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\tau))^{\frac{3}{2}}}

We introduce lemma 2.9 which we will show the the proof after the main proof of this theorem to derive that $a(\psi)$ is positive. So we obtain that $u_{m}a(\psi)$ is negative. Also by using lemma 2.9 we can know that:

b(\psi):=\int_{0}^{\psi}\frac{\cos\tau d\tau}{g(u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\tau))^{\frac{1}{2}}}

is also positive. So $u_{m}b(\psi)$ is a negative function.

Therefore, we can rewrite the equation (63) to:

\frac{d\mathcal{V}_{1}}{du_{m}}=\int_{0}^{\psi_{1}}u_{m}a(\psi)\frac{\sin\psi}{-g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}}d\psi+\int_{0}^{\psi_{1}}u_{m}b(\psi)\frac{\sin\psi}{-g(u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi))^{\frac{3}{2}}}d\psi

We have $\sin\psi>0$ , $u_{m}a(\psi)<0$ and $u_{m}b(\psi)<0$ by using the lemma 2.9 and the fact that $\psi\in(0,\pi)$ . Thus, $\frac{d\mathcal{V}_{1}}{du_{m}}$ is non-negative. Therefore $\mathcal{V}_{1}$ is a monotone increasing function of $u_{m}$ .

Also, by using the formula for $\mathcal{V}_{1}$ we should have:

\mathcal{V}_{1}=\int_{0}^{\psi_{1}}\int_{0}^{\psi}\frac{\cos\tau d\tau}{-g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\tau)}}\frac{\sin\psi}{-g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}}d\psi

When $u_{m}=0$ , we then have:

\mathcal{V}_{1}=\int_{0}^{\psi_{1}}\int_{0}^{\psi}\frac{\cos\tau d\tau}{-g\sqrt{\frac{2\sigma}{g}(1-\cos\tau)}}\frac{\sin\psi}{-g\sqrt{\frac{2\sigma}{g}(1-\cos\psi)}}d\psi

We can use Taylor expansion of the function inside the integral near $\psi=0$ and $\tau=0$ to get:

\frac{\cos\tau}{-g\sqrt{\frac{2\sigma}{g}(1-\cos\tau)}}\frac{\sin\psi}{-g\sqrt{\frac{2\sigma}{g}(1-\cos\psi)}}\sim\frac{1}{4g}\frac{1}{\sin\gamma}\frac{\sin\psi}{\sin\psi}\sim\frac{1}{4g}\frac{1}{\gamma}

Since we know that:

\int_{0}^{\psi}\frac{1}{4g}\frac{1}{\gamma}d\gamma=+\infty

we can show that $\mathcal{V}_{1}(u_{m})=+\infty$ when $u_{m}=0$ . And then we conclude $\mathcal{V}_{1}\rightarrow+\infty$ when $u_{m}\rightarrow 0$ by using Fatou’s lemma and the fact that $\mathcal{V}_{1}$ is a monotone increasing function of $u_{m}$ .

We can use the same discussion to prove that $\mathcal{V}_{2}$ obeys the same rules. ∎

Remark: In reality, $-u_{m}a(\psi_{1})$ defined in the above theorem represents the derivative of $r_{1}$ , implying that $r_{1}$ is a monotone increasing function of $u_{m}$ . Similarly, for the same reason, $r_{2}$ is a monotone increasing function of $u_{m}$ . And the proof in Theorem 2.6 shows that:

{}\lim_{u_{m}\rightarrow 0}r_{i}=+\infty

(64)

and

{}\lim_{u_{m}\rightarrow-\infty}r_{i}=0

(65)

for $i=1,2$ .

Then we want to build some similar results for $\mathcal{V}_{3}$ , $\mathcal{V}_{4}$ , and $\mathcal{V}_{5}$ . We have the following theorem.

Theorem 2.7.

When $u_{m}\rightarrow-\infty$ , we have $\mathcal{V}_{i}\rightarrow 0$ for $i=3,4,5$ .

Proof.

We plug (37) into the equation (60) and get:

\mathcal{V}_{3}=\frac{1}{2}(\frac{r_{1}\tan\theta_{2}+r_{2}\tan\theta_{2}+(u_{1}-u_{2})}{\tan\theta_{1}+\tan\theta_{2}})^{2}\tan\theta_{1}

(66)

We know from equation (64) and equation (65) that $r_{1}\rightarrow 0$ and $r_{2}\rightarrow 0$ when $u_{m}\rightarrow-\infty$ . Also, from the fact that:

	$\displaystyle u_{1}-u_{2}$	$\displaystyle=-\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi(x_{1}))}+\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi(x_{2}))}$
		$\displaystyle=\frac{\cos\psi(x_{1})-\cos\psi(x_{2})}{\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi(x_{1}))}+\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi(x_{2}))}}$

From the equation above, we can observe that $u_{1}-u_{2}\rightarrow 0$ when $u_{m}\rightarrow-\infty$ . We then return to equation (66). When $u_{m}\rightarrow-\infty$ we have:

\mathcal{V}_{3}=\frac{1}{2}(\frac{r_{1}\tan\theta_{2}+r_{2}\tan\theta_{2}+(u_{1}-u_{2})}{\tan\theta_{1}+\tan\theta_{2}})^{2}\tan\theta_{1}\rightarrow 0

We can also plug (37) into (61) and obtain that:

\mathcal{V}_{4}=\frac{1}{2}(\frac{r_{1}\tan\theta_{1}+r_{2}\tan\theta_{2}-(u_{1}-u_{2})}{\tan\theta_{1}+\tan\theta_{2}})^{2}\tan\theta_{2}

Using the same reason, we have $\mathcal{V}_{4}\rightarrow 0$ when $u_{m}\rightarrow-\infty$ .

From above computation we can actually show that $u_{1}-u_{2}\rightarrow 0$ when $u_{m}\rightarrow-\infty$ . Combining this property with the relation (65), we can obtain:

x_{0}=\frac{r_{1}\tan\theta_{1}-r_{2}\tan\theta_{2}-(u_{1}-u_{2})}{\tan\theta_{1}+\tan\theta_{2}}\rightarrow 0\leavevmode\nobreak\ \operatorname{when}\leavevmode\nobreak\ u_{m}\rightarrow-\infty

Based on this result, we can show that:

\lim_{u_{m}\rightarrow-\infty}\mathcal{V}_{5}=\lim_{u_{m}\rightarrow-\infty}(u_{1}-u_{2})x_{0}=0

So we have the theorem proved. ∎

Theorem 2.8.

When $\theta_{1}\neq\frac{\pi}{2}$ For any fixed total volume $V\geq 0$ we can find at least one $u_{m}$ such that $\mathcal{V}(u_{m})=\mathcal{V}_{1}(u_{m})+\mathcal{V}_{2}(u_{m})+\mathcal{V}_{3}(u_{m})+\mathcal{V}_{4}(u_{m})+\mathcal{V}_{5}(u_{m})=V$ . Therefore, we have at least one solution function of the equation system (10) and thus a solution function of the equation system (3).

Proof.

When $u_{m}\rightarrow 0$ , we know that $\mathcal{V}(u_{m})\geq\mathcal{V}_{1}+\mathcal{V}_{2}\rightarrow+\infty$ using Theorem 2.6.

When $u_{m}\rightarrow-\infty$ , we know that $\mathcal{V}_{i}(u_{m})\rightarrow 0,1\leq i\leq 5$ from Theorem 2.6 and Theorem 2.7. So we have:

\lim_{u_{m}\rightarrow+\infty}\mathcal{V}(u_{m})=\lim_{u_{m}\rightarrow+\infty}(\mathcal{V}_{1}+\mathcal{V}_{2}+\mathcal{V}_{3}+\mathcal{V}_{4}+\mathcal{V}_{5})=0

Since V is a continuous function of $u_{m}$ , we know that there is at least one $u_{m}<0$ such that $\mathcal{V}(u_{m})=V$ ∎

By using this theorem we can get the existence of the solution function. However, for its uniqueness it is not easy to prove. Although we know that $V_{1}$ and $V_{2}$ are monotone increasing function of $u_{m}$ . Also, using the computation in the previous theorem and the previous subsection, we know that $u_{1}-u_{2}$ , $r_{1}$ , and $r_{2}$ are increasing functions of $v_{m}$ . However, the monotonicity of $x_{0}$ , $r_{1}-x_{0}$ , and $r_{2}+x_{0}$ is still unknown. We then need to figure out how to show the monotonicity.

Before proving the final theorem, we first need to prove a lemma which is used to derive the monotonicity of $\mathcal{V}_{1}$ and $\mathcal{V}_{2}$ in Theorem 2.6.

lemma 2.9.

We have the following inequalities

{}a(\psi)=\int_{0}^{\psi}\frac{\cos\gamma d\gamma}{(u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\gamma))^{\frac{3}{2}}}\geq\int_{0}^{\psi}\frac{\cos\gamma d\gamma}{(u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi))^{\frac{3}{2}}}\geq 0

(67)

for any $\psi\in(0,\psi_{1})$ . And

b(\psi)=\int_{0}^{\psi}\frac{\cos\gamma d\gamma}{(u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\gamma))^{\frac{1}{2}}}\geq\int_{0}^{\psi}\frac{\cos\gamma d\gamma}{(u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi))^{\frac{1}{2}}}\geq 0

Proof.

To prove that the integral is non-negative we only need to use the fact that:

\int_{0}^{\psi}\frac{\cos\gamma d\gamma}{(u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi))^{\frac{3}{2}}}=\frac{\sin\psi}{(u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi))^{\frac{3}{2}}}\geq 0

we have the last inequality because of the fact that $\psi\in(0,\pi)$ . Then we only need to prove that:

{}\int_{0}^{\psi}\frac{\cos\gamma d\gamma}{(u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\gamma))^{\frac{3}{2}}}\geq\int_{0}^{\psi}\frac{\cos\gamma d\gamma}{(u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi))^{\frac{3}{2}}}

(68)

We notice that the inequality (68) has a similar form as the inequality (51) which is proved in Lemma 2.5. In fact, we can use the same idea to split the integral into three parts and perform a similar computation as we did in Lemma 2.5. Then we can get the result we want. The proof of function be is the same.

∎

Theorem 2.10.

When $[[\gamma]]<0$ and $\theta_{1}\geq\theta_{2}$ , $\mathcal{V}_{c}=\mathcal{V}_{3}+\mathcal{V}_{4}+\mathcal{V}_{5}$ is a monotone increasing function of $u_{m}$

Proof.

By using the equation (60),(61) and (62) We can get:

\mathcal{V}_{c}=\mathcal{V}_{3}+\mathcal{V}_{4}+\mathcal{V}_{5}=\frac{1}{2}(r_{1}-x_{m})^{2}\tan\theta_{1}+\frac{1}{2}(r_{2}+x_{m})^{2}\tan\theta_{2}+x_{m}(u_{1}-u_{2})

We then plug equation (37) into the equation above and obtain:

$\displaystyle\mathcal{V}_{c}=$	$\displaystyle\frac{1}{2}(r_{1}-\frac{r_{1}\tan\theta_{1}-r_{2}\tan\theta_{2}-(u_{1}-u_{2})}{\tan\theta_{1}+\tan\theta_{2}})^{2}\tan\theta_{1}$
	$\displaystyle+\frac{1}{2}(r_{2}+\frac{r_{1}\tan\theta_{1}-r_{2}\tan\theta_{2}-(u_{1}-u_{2})}{\tan\theta_{1}+\tan\theta_{2}})^{2}\tan\theta_{2}$
	$\displaystyle+(\frac{r_{1}\tan\theta_{1}-r_{2}\tan\theta_{2}-(u_{1}-u_{2})}{\tan\theta_{1}+\tan\theta_{2}})(u_{1}-u_{2})$
$\displaystyle=$	$\displaystyle\frac{1}{2}\frac{(r_{1}\tan\theta_{2}+r_{2}\tan\theta_{2}+(u_{1}-u_{2}))^{2}}{(\tan\theta_{1}+\tan\theta_{2})^{2}}\tan\theta_{1}+\frac{1}{2}(\frac{(r_{2}\tan\theta_{1}+r_{1}\tan\theta_{1}-(u_{1}-u_{2}))^{2}}{(\tan\theta_{1}+\tan\theta_{2})^{2}})\tan\theta_{2}$
	$\displaystyle+(\frac{r_{1}\tan\theta_{1}-r_{2}\tan\theta_{2}-(u_{1}-u_{2})}{\tan\theta_{1}+\tan\theta_{2}})(u_{1}-u_{2})$	(69)

Using a rearrangement of these terms in the right hand side we have:

$\displaystyle\eqref{equ:2.2.113}=$	$\displaystyle\frac{1}{2}\frac{(r_{1}\tan\theta_{2}+r_{2}\tan\theta_{2})^{2}}{(\tan\theta_{1}+\tan\theta_{2})^{2}}\tan\theta_{1}+\frac{1}{2}\frac{(r_{1}\tan\theta_{1}+r_{2}\tan\theta_{1})^{2}}{(\tan\theta_{1}+\tan\theta_{2})^{2}}\tan\theta_{2}$
	$\displaystyle+(u_{1}-u_{2})(r_{1}+r_{2})\frac{\tan\theta_{1}\tan\theta_{2}}{(\tan\theta_{1}+\tan\theta_{2})^{2}}-(u_{1}-u_{2})(r_{1}+r_{2})\frac{\tan\theta_{1}\tan\theta_{2}}{(\tan\theta_{1}+\tan\theta_{2})^{2}}$
	$\displaystyle+\frac{1}{2}\frac{(u_{1}-u_{2})^{2}\tan\theta_{1}}{(\tan\theta_{1}+\tan\theta_{2})^{2}}+\frac{1}{2}\frac{(u_{1}-u_{2})^{2}\tan\theta_{2}}{(\tan\theta_{1}+\tan\theta_{2})^{2}}$
	$\displaystyle+(\frac{r_{1}\tan\theta_{1}-r_{2}\tan\theta_{2}-(u_{1}-u_{2})}{\tan\theta_{1}+\tan\theta_{2}})(u_{1}-u_{2})$
$\displaystyle=$	$\displaystyle\frac{1}{2}\frac{(r_{1}+r_{2})^{2}}{\tan\theta_{1}+\tan\theta_{2}}\tan\theta_{1}\tan\theta_{2}+\frac{1}{2}\frac{(u_{1}-u_{2})^{2}}{(\tan\theta_{1}+\tan\theta_{2})}$
	$\displaystyle+(u_{1}-u_{2})(\frac{r_{1}\tan\theta_{1}-r_{2}\tan\theta_{2}}{\tan\theta_{1}+\tan\theta_{2}})-\frac{(u_{1}-u_{2})^{2}}{\tan\theta_{1}+\tan\theta_{2}}$
$\displaystyle=$	$\displaystyle\frac{1}{2}\frac{(r_{1}+r_{2})^{2}}{\tan\theta_{1}+\tan\theta_{2}}\tan\theta_{1}\tan\theta_{2}+(u_{1}-u_{2})(\frac{r_{1}\tan\theta_{1}-r_{2}\tan\theta_{2}}{\tan\theta_{1}+\tan\theta_{2}})-\frac{1}{2}\frac{(u_{1}-u_{2})^{2}}{\tan\theta_{1}+\tan\theta_{2}}$	(70)

To move on to the next strep, we need introduce following result which we will prove after we prove the main part of this theorem:

{}\frac{1}{\tan\theta_{1}+\tan\theta_{2}}(r_{1}\tan\theta_{1}-(u_{1}-u_{2}))

(71)

is a positive monotone increasing function of $u_{m}$ . Using this result, we then transform equation (70) into:

$\displaystyle\eqref{equ:2.2.38}=$	$\displaystyle\frac{1}{2}\frac{r_{1}^{2}+r_{2}^{2}}{\tan\theta_{1}+\tan\theta_{2}}\tan\theta_{1}\tan\theta_{2}+\frac{r_{1}r_{2}}{\tan\theta_{1}+\tan\theta_{2}}\tan\theta_{1}\tan\theta_{2}-(u_{1}-u_{2})\frac{r_{2}\tan\theta_{2}}{\tan\theta_{1}+\tan\theta_{2}}$
	$\displaystyle+(u_{1}-u_{2})\frac{r_{1}\tan\theta_{1}}{\tan\theta_{1}+\tan\theta_{2}}-\frac{1}{2}\frac{(u_{1}-u_{2})^{2}}{\tan\theta_{1}+\tan\theta_{2}}$
$\displaystyle=$	$\displaystyle\frac{1}{2}\frac{r_{1}^{2}+r_{2}^{2}}{\tan\theta_{1}+\tan\theta_{2}}\tan\theta_{1}\tan\theta_{2}+\frac{r_{2}\tan\theta_{2}}{\tan\theta_{1}+\tan\theta_{2}}(r_{1}\tan\theta_{1}-(u_{1}-u_{2}))$
	$\displaystyle+\frac{u_{1}-u_{2}}{\tan\theta_{1}+\tan\theta_{2}}(r_{1}\tan\theta_{1}-\frac{1}{2}(u_{1}-u_{2}))$	(72)

For the first term in (72), it is monotonically increasing since $r_{1}^{2}$ and $r_{2}^{2}$ are monotonically increasing function of $u_{m}$ (from (64) and (65)). Also, we will show that $\frac{\tan\theta_{1}\tan\theta_{2}}{\tan\theta_{1}+\tan\theta_{2}}>0$ in Lemma 2.11. So,

\frac{1}{2}\frac{r_{1}^{2}+r_{2}^{2}}{\tan\theta_{1}+\tan\theta_{2}}\tan\theta_{1}\tan\theta_{2}

is a monotonically increasing function of $u_{m}$ .

For the second term in (72), we already know that $r_{2}$ is increasing monotonically by using the remark after Theorem 2.6. In addition, $\frac{1}{\tan\theta_{1}+\tan\theta_{1}}(r_{1}\tan\theta_{1}-(u_{1}-u_{2}))$ is a positive monotone increasing function using Lemma 2.11. So:

\frac{r_{2}\tan\theta_{2}}{\tan\theta_{1}+\tan\theta_{2}}(r_{1}\tan\theta_{1}-(u_{1}-u_{2}))

is a monotone increasing function of $u_{m}$ since a positive increasing function times another positive increasing function is still monotone increasing.

Finally, for the third term, it is monotone increasing since $u_{1}-u_{2}$ is positive monotone increasing and

\frac{1}{\tan\theta_{1}+\tan\theta_{2}}(r_{1}\tan\theta_{1}-\frac{1}{2}(u_{1}-u_{2}))

is also positive and increasing by using Lemma 2.11. So the third term is increasing.

In conclusion, we should have (72) is an increasing function of $u_{m}$ . Thus $\mathcal{V}_{c}$ is an increasing function of $u_{m}$ . ∎

Then we only need to show the lemma.

lemma 2.11.

We assume that $\theta_{1}>\theta_{2}$ and $[[\gamma]]<0$ . Then the functions $g_{1}$ and $g_{2}$ below

{}g_{1}(u_{m})=\frac{1}{\tan\theta_{1}+\tan\theta_{2}}(r_{1}\tan\theta_{1}-(u_{1}-u_{2}))

(73)

{}g_{2}(u_{m})=\frac{1}{\tan\theta_{1}+\tan\theta_{2}}(r_{1}\tan\theta_{1}-\frac{1}{2}(u_{1}-u_{2}))

(74)

and:

{}u_{1}-u_{2}

(75)

are all positive monotone increasing functions of $u_{m}$ .

Proof.

We divide our problem into two cases: The first case where $\theta_{1}\in(0,\frac{\pi}{2})$ , the second case where $\theta_{1}\in(\frac{\pi}{2},\pi)$ .

Case 1 $\theta_{1}\in(0,\frac{\pi}{2})$ :

Based on our assumption that $\theta_{1}\geq\theta_{2}$ , we should have $\tan\theta_{1}>0$ and $\tan\theta_{2}>0$ . So $\frac{\tan\theta_{1}}{\tan\theta_{1}+\tan\theta_{2}}>0$ . We split our proof into three steps.

Step 1: (75) is positive and monotonically increasing.

We compute the derivative of $u_{1}-u_{2}$ which is defined as in the (27) and (28):

$\displaystyle\frac{d(u_{1}-u_{2})}{du_{m}}=$	$\displaystyle\frac{d(-\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1})}+\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{2})})}{du_{m}}$
$\displaystyle=$	$\displaystyle\frac{u_{m}}{\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{2})}}-\frac{u_{m}}{\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1})}}$
$\displaystyle=$	$\displaystyle u_{m}\frac{2(\cos\psi_{2}-\cos\psi_{1})}{(\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{2})}+\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1})})}\times$
	$\displaystyle\frac{1}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{2})}\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1})}}$	(76)

Since $\theta_{1}>\theta_{2}$ , we can obtain:

{}\cos\psi_{2}=\cos(\gamma+\frac{\pi}{2}-\theta_{2})=\sin(\theta_{2}-\gamma)<\sin(\theta_{1}-\gamma)=\cos(\gamma+\frac{\pi}{2}-\theta_{1})=\cos\psi_{1}

(77)

where we use the definition of $\psi_{1}$ and $\psi_{2}$ (6) and (7). Therefore, we have:

{}u_{m}(\cos\psi_{2}-\cos\psi_{1})\geq 0

(78)

We can then plug (78) back into the equation (76) and we can obtain that:

\eqref{equ:2.2.46}\geq 0

which finishes the first step. In fact, this proof also works for Case 2.

Step 2: Computing the first derivative of $g_{1}$ and $g_{2}$ with respect to $u_{m}$ .

We take the derivative of function $g_{1}$ with respect to $u_{m}$ and get:

	$\displaystyle\frac{dg_{1}(u_{m})}{du_{m}}$	$\displaystyle=\frac{1}{\tan\theta_{1}+\tan\theta_{2}}\frac{d(r_{1}\tan\theta_{1}-(u_{1}-u_{2}))}{du_{m}}$
		$\displaystyle=\frac{\tan\theta_{1}}{\tan\theta_{1}+\tan\theta_{2}}\frac{d}{du_{m}}r_{1}-\frac{1}{\tan\theta_{1}+\tan\theta_{2}}\frac{d(u_{1}-u_{2})}{du_{m}}$		(79)

We first use the equation (31) to compute the first term in (79):

	$\displaystyle\frac{dr_{1}\tan\theta_{1}}{du_{m}}$	$\displaystyle=\frac{d}{du_{m}}\tan\theta_{1}\int_{0}^{\psi_{1}}\frac{\cos\psi d\psi}{g\sqrt{u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}}$
		$\displaystyle=-u_{m}\tan\theta_{1}\int_{0}^{\psi_{1}}\frac{\cos\psi d\psi}{{g(u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi))^{\frac{3}{2}}}}$		(80)

Noticing that $u_{m}<0$ and the integral in equation (79) is positive by Lemma 2.9, this derivative is a positive number. Then we use Lemma 2.9 again to show:

{}\frac{dr_{1}\tan\theta_{1}}{du_{m}}\geq-u_{m}\tan\theta_{1}\int_{0}^{\psi_{1}}\frac{\cos\psi d\psi}{{g(u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1}))^{\frac{3}{2}}}}

(81)

For the second term in (79), we use the relation (77) again in the equation (76) and then obtain the following result:

{}\frac{d(u_{1}-u_{2})}{du_{m}}\leq u_{m}\frac{2}{g}\frac{\cos\psi_{2}-\cos\psi_{1}}{2(u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1}))^{\frac{3}{2}}}=\frac{u_{m}}{g}\frac{\cos\psi_{2}-\cos\psi_{1}}{(u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1}))^{\frac{3}{2}}}

(82)

We then combine equation (82) and $\eqref{equ:2.2.44}$ together to obtain:

$\displaystyle\frac{dg(u_{m})}{du_{m}}$	$\displaystyle=\frac{1}{\tan\theta_{1}+\tan\theta_{2}}\frac{d(r_{1}\tan\theta_{1}-(u_{1}-u_{2}))}{du_{m}}$
	$\displaystyle\geq\frac{1}{\tan\theta_{1}+\tan\theta_{2}}(-u_{m}\tan\theta_{1}\int_{0}^{\psi_{1}}\frac{\cos\psi d\psi}{g(u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1}))^{\frac{3}{2}}}-\frac{u_{m}}{g}\frac{\cos\psi_{2}-\cos\psi_{1}}{(u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1}))^{\frac{3}{2}}})$
	$\displaystyle=\frac{1}{\tan\theta_{1}+\tan\theta_{2}}(-u_{m}\tan\theta_{1}\sin\psi_{1}\frac{1}{g}\frac{1}{(u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1}))^{\frac{3}{2}}}-\frac{u_{m}}{g}\frac{\cos\psi_{2}-\cos\psi_{1}}{(u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1}))^{\frac{3}{2}}})$
	$\displaystyle=\frac{1}{\tan\theta_{1}+\tan\theta_{2}}(-\frac{u_{m}}{g}\frac{1}{(u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1}))^{\frac{3}{2}}}(\tan\theta_{1}\sin\psi_{1}-(\cos\psi_{1}-\cos\psi_{2})))$	(83)

Using the fact that $u_{m}<0$ , we have:

\frac{1}{\tan\theta_{1}+\tan\theta_{2}}(-\frac{u_{m}}{g}\frac{1}{(u_{m}^{2}+\frac{2\sigma}{g}(1-\cos\psi_{1}))^{\frac{3}{2}}})>0

Therefore, we only need to care about the sign of the term $\tan\theta_{1}\sin\psi_{1}-(\cos\psi_{1}-\cos\psi_{2})$ in (83). We call it $k(\gamma)$ . So now we have the formula of the derivative and only need to discuss the sign of a term which only depends on the inclined angles and contact angles.

Step 3: $k(\gamma)$ is positive .

Using the definition of $\psi_{1}$ and $\psi_{2}$ (33) and (34), we have:

	$\displaystyle k(\gamma)$	$\displaystyle=\tan\theta_{1}\sin\psi_{1}-(\cos\psi_{1}-\cos\psi_{2})$
		$\displaystyle=\tan\theta_{1}\sin(\frac{\pi}{2}-\theta_{1}+\gamma)-(\cos(\frac{\pi}{2}-\theta_{1}+\gamma)-\cos(\frac{\pi}{2}-\theta_{2}+\gamma))$
		$\displaystyle=\tan\theta_{1}\cos(\theta_{1}-\gamma)-(\sin(\theta_{1}-\gamma)-\sin(\theta_{2}-\gamma))$
		$\displaystyle\geq\tan\theta_{1}\cos(\theta_{1}-\gamma)-(\sin(\theta_{1}-\gamma)+\sin(\gamma))=p(\gamma)$

Noticing that when $\gamma=0$ , we have:

p(0)=\sin\theta_{1}-\sin\theta_{1}+\sin 0=0

Then we compute the derivative of function $p(\gamma)$ with respect to $\gamma$ :

	$\displaystyle p^{\prime}(\gamma)$	$\displaystyle=\tan\theta_{1}\sin(\theta_{1}-\gamma)+\cos(\theta_{1}-\gamma)-\cos\gamma$
		$\displaystyle=\tan\theta_{1}(\sin\theta_{1}\cos\gamma-\cos\theta_{1}\sin\gamma)+\cos\theta_{1}\cos\gamma+\sin\theta_{1}\sin\gamma-\cos\gamma$
		$\displaystyle=\frac{\sin^{2}\theta_{1}}{\cos\theta_{1}}\cos\gamma-\sin\theta_{1}\sin\gamma+\cos\theta_{1}\cos\gamma+\sin\theta_{1}\sin\gamma-\cos\gamma$
		$\displaystyle=\frac{\sin^{2}\theta_{1}}{\cos\theta_{1}}\cos\gamma+\cos\theta_{1}\cos\gamma-\cos\gamma$
		$\displaystyle=\frac{1}{\cos\theta_{1}}\cos\gamma-\cos\gamma\geq 0\leavevmode\nobreak\ for\leavevmode\nobreak\ \gamma\in(0,\frac{\pi}{2})$

The last inequality is due to the fact that $\gamma\in(0,\frac{\pi}{2})$ when $[[\gamma]]<0$ . So finally we get:

k(\gamma)\geq p(\gamma)\geq p(0)=0

which implies that (73) is increasing monotonically.

For (74), we know that:

g_{2}(u_{m})=g_{1}(u_{m})+\frac{1}{2}\frac{1}{\tan\theta_{1}+\tan\theta_{2}}(u_{1}-u_{2})

Since $g_{1}(u_{m})$ and $u_{1}-u_{2}$ are both increasing functions of $u_{m}$ , $g_{2}$ must be an increasing function.

Finally, we only need to show the positivity. We use the following fact:

\lim_{u_{m}\rightarrow-\infty}r_{1}=\lim_{u_{m}\rightarrow-\infty}(u_{1}-u_{2})=0

which implies that

{}\lim_{u_{m}\rightarrow-\infty}g_{1}(u_{m})=\lim_{u_{m}\rightarrow-\infty}g_{2}(u_{m})=0

(84)

Then the positivity of $g_{1}$ and $g_{2}$ follows from (84) and the monotonicity of the function $g_{1}$ and $g_{2}$ .

Case 2 $\theta_{1}\in(\frac{\pi}{2},\pi-\theta_{2})$ :

In this case, we observe that

\tan\theta_{1}\leavevmode\nobreak\ \leavevmode\nobreak\ \operatorname{and}\leavevmode\nobreak\ \leavevmode\nobreak\ \tan\theta_{1}+\tan\theta_{2}

are both negative due to the assumption that $\frac{\pi}{2}<\theta_{1}<\pi-\theta_{2}$ . So $\frac{\tan\theta_{1}}{\tan\theta_{1}+\tan\theta_{2}}$ is positive. However,

r_{1}\tan\theta_{1}-(u_{1}-u_{2})

is negative since $u_{1}>u_{2}$ , $r_{1}>0$ and $\tan\theta_{1}<0$ . So $g_{1}(u_{m})$ should be positive and so do $g_{2}(u_{m})$ .

Using the Remark after Theorem 2.6 and Step 1 in Case 1 we have $r_{1}$ and $u_{1}-u_{2}$ are monotone increasing function of $u_{m}$ . So

r_{1}\tan\theta_{1}-(u_{1}-u_{2})\leavevmode\nobreak\ \operatorname{and}\leavevmode\nobreak\ r_{1}\tan\theta_{1}-\frac{1}{2}(u_{1}-u_{2})

are monotone decreasing. Therefore, $g_{1}(u_{m})$ and $g_{2}(u_{m})$ should be monotone increasing since $\frac{1}{\tan\theta_{1}+\tan\theta_{2}}$ is negative and we have the lemma proved. ∎

2.2.2 $\theta_{1}=\frac{\pi}{2}$

When $\theta_{1}=\frac{\pi}{2}$ , we should have $\mathcal{V}=\mathcal{V}_{1}+\mathcal{V}_{2}+\mathcal{V}_{4}+\mathcal{V}_{5}$ . From Theorem 2.6, we already have the monotonicity of $\mathcal{V}_{1}$ and $\mathcal{V}_{2}$ . Then we have the following theorem.

Theorem 2.12.

When $\theta_{1}=\frac{\pi}{2}$ and $[[\gamma]]<0$ , we have $\mathcal{V}_{4}$ and $\mathcal{V}_{5}$ monotone increasing functions of $u_{m}$ .

Proof.

First we show the monotonicity for $\mathcal{V}_{4}$ :

\mathcal{V}_{4}=\frac{1}{2}(r_{2}+x_{m})^{2}\tan\theta_{2}

By using Theorem 2.4 we know that $x_{m}=r_{1}$ . So:

\mathcal{V}_{4}=\frac{1}{2}(r_{2}+r_{1})^{2}\tan\theta_{2}

is monotone increasing since $r_{1}$ and $r_{2}$ are both positive and monotone increasing.

For $\mathcal{V}_{5}$ , we can show that:

\mathcal{V}_{5}=x_{m}(u_{1}-u_{2})=r_{1}(u_{1}-u_{2})

is monotone increasing since $r_{1}$ and $u_{1}-u_{2}$ are both positive and monotone increasing. So the theorem is proved

∎

Then we can show the following theorem:

Theorem 2.13.

When $\theta_{1},\theta_{2}\in(0,\pi)$ and $\theta_{1}\in(0,\pi-\theta_{2})$ and $[[\gamma]]\leq 0$ , $\mathcal{V}(u_{m})$ is a monotonically increasing function of $u_{m}$ .

In conclusion, we have the following theorem:

Theorem 2.14.

For any given constant V, there exist a $u_{m}$ and a solution to the equation system (10) such that

\mathcal{V}(\rho(\theta))=\mathcal{V}(u_{m})=V

Furthermore, $\mathcal{V}(u_{m})$ is a monotone increasing function of $u_{m}$ when $[[\gamma]]<0$ . Therefore, for any given V, there exists a $u_{m}$ such that $\mathcal{V}(u_{m})=V$ . And when $[[\gamma]]<0$ , the choice of $u_{m}$ and the solution function is unique.

When $\psi_{1}<0$ and $\psi_{2}>0$ , the function curve reaches a minimum at a point $(x_{0},u_{0})$ . Then we use a similar computation as in the previous case; we can also prove the existence. However, the uniqueness is still unknown in this case.

We then have the following theorem:

Theorem 2.15.

The function we constructed in Theorem 2.14 is exactly the solution function of the system (3) after the shift $(x,y)\rightarrow(x,y+\frac{P_{0}}{g})$

Proof.

Using Theorem 2.14, we now have a solution $(x(\psi),y(\psi))$ to (10) and $u_{m}$ for any given V. We then use equation (39) in Theorem 2.3 to obtain the value of $P_{0}$ .

After the shift $(x,y)\rightarrow(x,y+\frac{P_{0}}{g})$ , the curve $(x(\psi),y(\psi)+\frac{P_{0}}{g})$ exactly solves the first two equations (9) since they are the solution of the system (10). The only problem is whether it satisfies the boundary condition. In other words, we need to prove that after the shift $(x,y)\rightarrow(x,y+\frac{P_{0}}{g})$ , the two lines $\theta=\theta_{2}$ and $\theta=\pi-\theta_{1}$ are the given two walls. Since the incline angles of the two walls are $\theta_{1}$ and $\theta_{2}$ , we only need to show that the original point of the polar coordinate coincides with the meeting points of the two walls after the shift. This statement is true since this is how we use $u_{m}$ to represent $P_{0}$ in Theorem 2.3. So we have the theorem proved. ∎

3 The second case: $\psi_{1}\cdot\psi_{2}\geq 0$

In this section, we discuss the case where $\psi_{1}$ and $\psi_{2}$ have the same sign. We begin with discussing a special case before we derive the theorem regarding the shape of the surface curve. And then we can follow similar methods as in the first case to derive the result we want.

3.1 A special case where $\theta_{1}=\frac{\pi}{2}$ and $\theta_{2}=0$

We also want to first discuss the shape of the function. But it is more complicated than the discussion in Section 2. So we begin with discussing a special case:

Theorem 3.1.

When $\theta_{1}=\frac{\pi}{2}$ and $\theta_{2}=0$ , we assume that $[[\gamma]]>0$ and the contact angle $\gamma\in(-\frac{\pi}{2},-\frac{\pi}{4})$ , we assume that we have a steady state in this case. Then the slope angle $\psi$ (defined by equation (5)) of the curve is neither a monotone increasing function nor a monotone decreasing function of x when the total volume V is large enough.

Proof.

We first consider the problem in Cartesian coordinates whose origin point is at the point where the two different walls connect with each other. Then we have the equation for the free surface (which is the first equation in (18)):

{}gu-\sigma\mathcal{H}=P_{0}

(85)

where:

\mathcal{H}=\partial_{\theta}(\frac{\partial_{x}u(x)}{\sqrt{1+|\partial_{x}u|^{2}}})

is the mean curvature of the curve.

When $P_{0}$ is less than zero, we integrate both sides of the equation (85) from $0$ to $x_{2}$ to obtain the following inequality:

gV+\sigma\sin(\psi_{1})-\sigma\sin(\psi_{2})=x_{2}P_{0}

Since we assume V is big enough, we have the left-hand side of the equation is greater than zero. However, the term on the right-hand side of the equation is non-positive by using the assumption that $P_{0}\leq 0$ . So $P_{0}$ should be greater or equal to zero.

Since $u(x_{2})=0$ , we can derive the following relation from (85):

-\sigma\partial_{x}(\sin\psi)|_{x=x_{2}}=P_{0}>0

which implies that:

\partial_{x}(sin\psi)|_{x=x_{2}}<0

So if $\psi$ is a monotone function of $x$ , then it should be monotone decreasing. But this can not be true since:

{}\sin(\psi)|_{x=0}=\sin\psi_{1}=\sin(\gamma)<\sin(-\frac{\pi}{2}-\gamma)=\sin\psi_{2}=\sin(\gamma)|_{x=x_{2}}

(86)

where we used the definition of $\psi_{1}$ and $\psi_{2}$ ((6) and (7)). We also used the assumption $\gamma<-\frac{\pi}{4}$ to obtain the inequality $-sin(\gamma)<-sin(\frac{\pi}{2}-\gamma)$ .From this, we know that $\psi$ can not be monotone decreasing. Therefore, when V is large enough, $\psi$ is not a global legal parameter for the boundary curve. ∎

Theorem 3.1 implies that the shape of the curve differs when $[[\gamma]]>0$ and $[[\gamma]]<0$ . We now aim to derive a theorem for this special case, which may provide insight into the curve’s shape in the general case.

Theorem 3.2.

The shape of the function

When $\theta_{1}=\frac{\pi}{2}$ and $\theta_{2}=0$ , we assume that the contact angle $\gamma\in(-\frac{\pi}{2},-\frac{\pi}{4})$ , and we also assume that the steady surface exists.

1: When $gV+\sigma sin(\gamma)-\sigma sin(-\frac{\pi}{2}-\gamma)\leq 0$ , $\psi$ is a monotone increasing of x.

2:When $gV+\sigma sin(\gamma)-\sigma sin(-\frac{\pi}{2}-\gamma)>0$ . $\psi$ reaches to a maximal value $\psi_{3}$ such that $\psi_{3}\in(max(\gamma-\frac{\pi}{2},-\gamma)$ .

Proof.

proof of (1): We integrate both sides of the equation (85) from 0 to $x_{2}$ and get

gV+\sigma\sin(\frac{\pi}{2}-\gamma)-\sigma\sin(\gamma)=x_{2}P_{0}

Using the condition $gV+\sigma sin(\phi)-\sigma sin(\frac{\pi}{2}-\phi)\leq 0$ , we have $P_{0}\leq 0$ . Then the equation (85) becomes:

\sigma\partial_{x}sin\psi=gv(x)-P_{0}>0\leavevmode\nobreak\ for\leavevmode\nobreak\ any\leavevmode\nobreak\ x

So we know that $\psi$ is an increasing function of x and it is a legal global parameter for the curve.

proof of (2): From theorem 3.1, we know $P_{0}\geq 0$ when $gV+\sigma sin(\phi)-\sigma sin(\frac{\pi}{2}-\phi)>0$ then $\gamma$ is not monotone decreasing or increasing and it can reach a maximal value at some point since it is continuous. So we only need to prove that this maximum point $\psi_{3}<0$ .

Step 1 $\psi_{3}\leq 0$

We use a contradiction argument. We suppose that $\psi_{3}>0$ . Then we want to first prove the following property:

\sigma\frac{d}{dx}sin(\psi)|_{x=0}=gv(x_{1})-P_{0}>0

We use a contradiction argument and first assume that:

{}\sigma\frac{d}{dx}sin(\psi)|_{x=0}=gv(x_{1})-P_{0}\leq 0

(87)

We then take derivative with respect to x on the left hand side of the equation (85) to obtain:

{}\sigma\frac{d^{2}}{dx^{2}}sin(\psi)|_{x=0}=g\partial_{x}v|_{x=0}=g\tan\psi|_{x=0}=gtan(-\gamma)<0

(88)

which implies that:

\sigma\frac{d^{2}}{d^{2}x}sin(\psi)<0

and:

\sin\psi(x)<0

for x near zero. We choose $x_{0}$ such that:

x_{0}=\min\{x:\frac{d^{2}}{dx^{2}}\sin\psi=0\}

This is well-defined since $\psi$ is not a monotone decreasing function of x from the assumption. From the computation in equation (88) we should have $\tan\psi=0$ when $x=x_{0}$ . So $\sin\psi(x_{0})>\sin\psi(0)$ . We combine this with the assumption (87) to show that $\frac{d}{dx}(\sin\psi)=0$ at some $x_{1}\in(x,x_{0})$ . However, by using the definition of $x_{0}$ , we know that $\frac{d^{2}}{dx^{2}}(\sin\psi)<0$ when $0<x<x_{0}$ . So $\frac{d}{dx}(\sin\psi)<0$ for any $x\in(0,x_{0})$ . This contradicts to the existence of $x_{1}$ . So (87) is not true.

Therefore, we have the property that:

{}\sigma\frac{d}{d\psi}\sin\psi|_{x=0}=gv(0)-P_{0}>0

(89)

Suppose that $\psi_{3}>0$ is a maximum value, we have:

\sigma\frac{d}{d\psi}\sin\psi|_{x=x_{3}}=gv(x_{3})-P_{0}=0

We then use (88) to show:

{}\sigma\frac{d^{2}}{d\psi^{2}}sin(\psi)|_{x=x_{3}}=g\tan\psi_{3}>0

(90)

Since $\psi_{3}$ is the maximum we should have:

\sigma\frac{d^{2}}{d\psi^{2}}sin(\psi)|_{\psi=\psi_{3}}=g\tan\psi_{3}\leq 0

which contradicts to (90). So we have $\psi_{3}\leq 0$ .

Step 2: $\psi_{3}\neq 0$ .

We use a contradiction argument again. We suppose that $\psi_{3}=0$ then we can know from the first and second equation of the system (9) we can know that:

{}\frac{d\psi}{dx}=\frac{gv-P_{0}}{\sigma\cos\psi}

(91)

{}\frac{dv}{d\psi}=\frac{\sigma\sin\psi}{gv-P_{0}}

(92)

Then we we integrate both sides of the equation (92) from $\psi$ to $\psi_{3}$ to show:

{}v=\frac{P_{0}\pm\sqrt{P_{0}^{2}-2g\sigma(cos\psi-C)}}{g}

(93)

Where $C$ is just a constant such that $u(\psi_{3})=\frac{P_{0}}{g}$ .

We plug the equation (93) back to the ODE (91) and get:

{}\frac{d\psi}{dx}=\frac{\sqrt{P_{0}^{2}-2g\sigma(cos\psi-C)}}{\sigma\cos\psi}=F(\psi)

(94)

Taking derivative on function F with respect to $\psi$ near $\psi_{3}$ , we can show that:

{}\frac{dF(\psi)}{d\psi}=-\frac{\sin\psi\sqrt{P_{0}^{2}-2g\sigma(cos\psi-C)}}{\sigma\cos^{2}\psi}+\frac{d\sqrt{P_{0}^{2}-2g\sigma(\cos\psi-C)}}{d\psi}

(95)

For the first term of equation (95), it is a smooth function and converges to zero when $\psi$ converges $\psi_{3}=0$ , which means that this function is bounded in a neighborhood of $\psi_{3}=0$ . Then we look at the second term.

{}\frac{d\sqrt{P_{0}^{2}-2g\sigma(cos\psi-C)}}{d\psi}=-\frac{1}{2}\frac{2g\sigma\sin\psi}{\sqrt{P_{0}^{2}-2g\sigma(\cos\psi-C)}}

(96)

Then we take Taylor expansion of the right hand side of equation (96) near $\psi=0$ to show:

	$\displaystyle\frac{d\sqrt{P_{0}^{2}-2g\sigma(\cos\psi-C)}}{d\psi}$	$\displaystyle=-\frac{1}{2}\frac{2g\sigma\sin\psi}{\sqrt{P_{0}^{2}-2g\sigma(\cos\psi-C)}}$
		$\displaystyle=-g\sigma\frac{\psi+O(\psi^{2})}{\sqrt{2g\sigma}\sqrt{\frac{P_{0}^{2}}{2g\sigma}+C-\cos\psi}}$		(97)

We know from the definition of constant C:

\frac{P_{0}^{2}}{2g\sigma}+C-\cos\psi_{3}=0

So we can have the Taylor expansion of the denominator which appears in (97):

\sqrt{\frac{P_{0}^{2}}{2g\sigma}+C-\cos\psi}=\sqrt{1-\cos\psi}=\frac{\sqrt{2}}{2}\psi+O(\psi^{2})

We plug it back into the derivative term (97) and get:

$\displaystyle\eqref{equ:3.1.12}$	$\displaystyle=-g\sigma\frac{\psi+O(\psi^{2})}{\sqrt{2g\psi}\sqrt{\frac{P_{0}^{2}+C}{2g\psi}-\cos\psi}}$
	$\displaystyle=-g\sigma\frac{\psi+O(\psi^{2})}{\sqrt{2g\sigma}\sqrt{1-\cos\psi}}$
	$\displaystyle=-g\sigma\frac{\psi+O(\psi^{2})}{\sqrt{g\sigma}\psi+O(\psi^{2})}$
	$\displaystyle=-\sqrt{g\sigma}(1+O(\psi))$	(98)

So for our ODE (94):

\frac{d\psi}{du}=\frac{\sqrt{P_{0}^{2}-2g\sigma(\cos\psi-C)}}{\sigma\cos\psi}=F(\psi)

We know from (95) and (98) that $F(\psi)$ has bounded first derivative and thus is a Lipschitz function near the point $\psi=\psi_{3}=0$ . So by using the uniqueness of ODE we have $\psi=0$ is the unique solution of the ODE, which contradicts our assumption. ∎

3.2 General case when $\theta_{1}\leq\frac{\pi}{2}$

After discussing this special case, we want to discuss the general case. We first need to assume that $0\leq\theta_{2}\leq\theta_{1}\leq\frac{\pi}{2}$ . From the previous subsection, we may infer that $\psi$ can serve as a valid global parameter only when $V$ is small. This makes the analysis more straightforward in such cases, as we can apply a similar approach to the first case. Therefore, our primary focus should be on the scenario where $V$ is large.

The idea:

From Section 3.1, we know that $\psi$ might not be a valid parameter for the boundary curve and attains its maximum value $\psi_{m}$ at a point $(x_{m},v_{m})=(x(\theta_{m}),v(\theta_{m}))$ . After applying the shift $(x,y)\rightarrow(x,y-\frac{P_{0}}{g})$ . We should have $u_{m}=v_{m}-\frac{P_{0}}{g}=0$ . Our goal is to use $\psi_{m}$ to express the coordinates of two contact points. See Figure 7.

In addition, the curve seems to be a function of x which intrigues us to integrate both sides of the first equation in the system (18) from $x_{1}$ to $x_{2}$ and get the following:

{}P_{0}=\frac{g{\mathcal{V}}(\rho(\theta))+\frac{1}{2}gx_{1}^{2}\tan\theta_{1}+\frac{1}{2}gx_{2}^{2}\tan\theta_{2}-\sigma\sin(\psi_{2})+\sigma\sin(\psi_{1})}{x_{2}-x_{1}}

(99)

Where $\mathcal{V}(\rho(\theta))$ is defined by the equation (2) and can be rewritten as the following equation in Cartesian coordinates:

{}\mathcal{V}(\rho(\theta))=\int_{x_{1}}^{x_{2}}v(x)dx-\frac{1}{2}(v(x_{1}))^{2}\frac{1}{\tan\theta_{1}}-\frac{1}{2}(v(x_{2}))^{2}\frac{1}{\tan\theta_{2}}

(100)

It is the total volume enclosed by the boundary curve and two walls. Then we only need to determine $P_{0}$ to make equation $\eqref{equ:3.3.14}$ the relation between $\mathcal{V}$ and $\psi_{0}$ . Our approach is to first show that x serves as a global variable for the curve and then try to use the specially chosen point to express $P_{0}$ .

We begin with deriving a theorem concerning the shape of the curve. We assume that V is large enough such that $\psi$ is not a legal global parameter to parameterize the curve. Under this assumption, we establish the following lemma and theorems:

lemma 3.3.

When $0\leq\theta_{2}\leq\theta_{1}\leq\frac{\pi}{2}$ and the sign of $\psi_{1}$ is the same as the sign of $\psi_{2}$ , then we have $\psi_{1}<0$ and $[[\gamma]]>0$ .

Proof.

First we need to prove that $[[\gamma]]>0$ in this case. By definition we have:

\psi_{1}=\frac{\pi}{2}-\theta_{1}+\gamma

and:

\psi_{2}=-\gamma-\frac{\pi}{2}+\theta_{2}

Then if $[[\gamma]]\leq 0$ we should have $\gamma\in(0,\frac{\pi}{2})$ by using the equation (8). Since $\theta_{1}\in(0,\frac{\pi}{2})$ , we should have $\psi_{1}=\frac{\pi}{2}-\theta_{1}+\gamma\geq 0$ and $\psi_{2}=-\gamma-\frac{\pi}{2}+\theta_{2}<0$ . So $\psi_{1}$ and $\psi_{2}$ have different signs, which contradicts our assumption. So $[[\gamma]]>0$

To prove the theorem, we suppose that $\psi_{1}=\frac{\pi}{2}-\theta_{1}+\gamma>0$ . Then $-\gamma<\frac{\pi}{2}-\theta_{1}$ . Since $\theta_{2}\leq\theta_{1}\leq\frac{\pi}{2}$ , we should have $-\gamma<\frac{\pi}{2}-\theta_{1}\leq\frac{\pi}{2}-\theta_{2}$ , which means that $\psi_{2}=-\gamma-\frac{\pi}{2}+\theta_{2}<0$ . So $\psi_{1}$ and $\psi_{2}$ have different signs, which contradicts our assumption again.

So we should have $\psi_{1}$ and $\psi_{2}$ are both less than zero. ∎

By using Lemma 3.3 we can get a theorem about the shape of the curve:

Theorem 3.4.

We suppose that $0\leq\theta_{2}\leq\theta_{1}\leq\frac{\pi}{2}$ and that the boundary curve u is not a function of $\psi$ . We also assume that there exists a solution function for our equation system (10). Then $\psi(\theta)$ increases from $\psi_{1}=\psi(\pi-\theta_{1})$ to a negative maximum $\psi_{m}=\psi(\theta_{m})$ . And then decreases from the maximal point to $\psi(\theta_{2})$ . Also, the curve is a graph of x.

Proof.

Since $\psi_{1}<0$ , the curve is a function of x near $\theta=\pi-\theta_{1}$ . We can use the first equation in the system (18). It can be turned into the following form:

{}\partial_{x}(\sin\psi)=gu

(101)

We first suppose that $u_{1}\leq 0$ . Then $\sin\psi$ is a decreasing function in a small neighborhood of $(x_{1},u_{1})$ . Since $\psi_{1}<0$ , we can easily show $\psi<\psi_{1}<0$ when $x\in(x_{1},x_{1}+\delta)$ . Then we can repeat the discussion for the point $(x_{1}+\delta,\psi(x_{1}+\delta))$ and prove that $\psi(x)<\psi_{1}$ for any $x\in(x,x+2\delta)$ . Finally, we can prove that $\psi(x)<\psi_{1}$ for any $x\in(x_{1},x_{3})$ where $x_{3}$ is a point such that $\psi(x_{3})=-\frac{\pi}{2}$ . This part of the curve cannot touch the right wall due to the assumption that $\psi$ is not a global variable.

Similarly to Theorem 2.1, we can extend the solution function to $\psi\in(-\frac{\pi}{2},-\frac{3}{2}\pi)$ and similarly to $\psi\in(-\frac{3}{2}\pi,-2\pi-\psi_{1})$ . Then using a similar discussion as in Theorem 2.1, we can know that the curve will self-intersect or pass through the left wall, leading to a contradiction.

Thus, we must have $u_{1}>0$ and $\psi$ increases to the maximal value $\psi_{m}$ . Since $\psi_{m}$ is the maximum value, we obtain $u_{m}=0$ . We also know from Theorem 3.2 that $\psi_{m}<0$ (Although we only proved Theorem 3.2 in the special case, the proof also works for the general case). Examining equation (101), we see that $\psi$ decreases when $u(x)<0$ , implying that $\psi$ decreases monotonically to $\psi_{2}$ as a function of x. This completes the proof. Moreover, we establish $\psi\in(-\frac{\pi}{2},0)$ for any $\theta$ , confirming that the curve represents a graph of x.

∎

So from Theorem 3.4, we can divide the surface curve into two segments: the first extending from $x_{1}$ to $x_{m}$ and the second from $x_{m}$ to $x_{2}$ . See the figure 7. And the system (18) provides a global description of the curve. And we use $v(x)$ to represent the curve and use $u(x)$ to represent the curve after shifting. By making the separation, we establish that each segment of the curve can be parameterized by $\psi$ , which is defined by (5), allowing us to use the equation (11) for each part.

Let $(x_{m},u_{m})$ be the point where the slope angle reaches its maximum, denoted by $\psi_{m}$ . Since it is the maximum value, we must have $gu_{m}=0$ , implying that the coordinate of the special point is $(x_{m},0)$ with maximum angle $\psi_{m}$ . The remainder of this chapter focuses on using $\psi_{m}$ to represent the unknown parameters $P_{0}$ , $x_{1}$ , and $x_{2}$ . We first prove the following theorem in terms of the coordinates of the two contact points:

Theorem 3.5.

We have the following formula for the coordinates of two contact points:

$\displaystyle u_{1}$	$\displaystyle=\sqrt{\frac{2\sigma}{g}(\cos\psi_{m}-\cos\psi_{1})}$	(102)
$\displaystyle u_{2}$	$\displaystyle=-\sqrt{\frac{2\sigma}{g}(\cos\psi_{m}-\cos\psi_{2})}$	(103)
$\displaystyle x_{1}$	$\displaystyle=x_{m}-\int_{\psi_{1}}^{\psi_{m}}\frac{\cos\gamma}{g\sqrt{\frac{2\sigma}{g}(\cos\psi_{m}-\cos\gamma)}}d\gamma$	(104)
$\displaystyle x_{2}$	$\displaystyle=x_{m}-\int_{\psi_{m}}^{\psi_{2}}\frac{\cos\gamma}{g\sqrt{\frac{2\sigma}{g}(\cos\psi_{m}-\cos\gamma)}}d\gamma$	(105)
$\displaystyle r(\psi)$	$\displaystyle=\|x(\psi)-x_{m}\|=\|x_{m}-\int_{\psi_{m}}^{\psi}\frac{\cos\gamma}{g\sqrt{\frac{2\sigma}{g}(\cos\psi_{m}-\cos\gamma)}}d\gamma\|$	(106)

Proof.

Since each segment of the curve can be parameterized by $\psi$ , the computation is the same as Theorem 2.2. We just use the equation (10) and the corresponding boundary condition to do these computations. ∎

Unlike the previous section, we now separate the curve into two segments, each of which may exhibit different properties. We need to prove that the second part does not intersect the left wall, which implies that $x_{2}>0$ .

lemma 3.6.

For any $x\in(x_{m},x_{2})$ , the curve $u(x)$ does not intersect the left wall.

Proof.

Using the construction, we know that the maximal point $(x_{m},u_{m})$ should be right to the left wall. So we only need to prove that $\psi(x)\geq-\theta_{1}$ when $x\in(x_{m},x_{2})$ .

By Theorem 3.4 we know that $\psi$ is a monotonically decreasing function of x. Therefore, it suffices to prove that $\psi_{2}\geq-\theta_{1}$ . From the definition of $\psi_{2}$ ((7)), we have $\psi_{2}=\gamma-\frac{\pi}{2}+\theta_{2}$ and $\psi_{1}=\frac{\pi}{2}-\theta_{1}-\gamma$ . Using Lemma 3.3, we know that $\psi_{1}<0$ and $\psi_{2}<0$ , which allows us to show:

\frac{\pi}{2}-\theta_{1}<\gamma<\frac{\pi}{2}-\theta_{2}

which means that :

\gamma-\frac{\pi}{2}+\theta_{2}>\frac{\pi}{2}-\theta_{1}-\frac{\pi}{2}+\theta_{2}=\theta_{2}-\theta_{1}\geq-\theta_{1}

So we have the theorem proved. ∎

Now we have two parameters $x_{m}$ and $\psi_{m}$ , we need to find the relation between these two parameters. We can also use the same idea as Section 2 to compute $x_{m}$ and get:

{}x_{m}=\frac{r_{1}\tan\theta_{1}-r_{2}\tan\theta_{2}-(u_{1}-u_{2})}{\tan\theta_{1}+\tan\theta_{2}}\leavevmode\nobreak\ \operatorname{when}\leavevmode\nobreak\ \theta_{1}\neq\frac{\pi}{2}

(107)

{}x_{m}=r_{1}\leavevmode\nobreak\ \operatorname{when}\leavevmode\nobreak\ \theta_{1}=\frac{\pi}{2}

(108)

Here $r_{1}=x_{m}-x_{1}$ and $r_{2}=x_{2}-x_{m}$ . For the special case $\theta_{1}=\frac{\pi}{2}$ , the computation is straightforward and follows a similar approach to Subsection 2.2.2. Therefore, we assume that $\theta_{1}\neq\frac{\pi}{2}$ in the following proof.

We can then compute $P_{0}$ by determining the y component of the connect point O (the point where two walls meet each other). After the shifting $(x,y)\rightarrow(x,y-\frac{P_{0}}{g})$ , we know that the y component of the point O is $y_{0}=-\frac{P_{0}}{g}$ . Also, using the same computation as in Theorem 2.3, we can show $y_{0}=-((r_{2}+x_{m})\tan\theta_{2}-u_{2})$ . We can obtain $P_{0}=g((r_{2}+x_{m})\tan\theta_{2}-u_{2})$ . Then we use this expression substituting $P_{0}$ in the equation (99) and obtain:

{}g{\mathcal{V}}(x_{m},\psi_{m})=(x_{2}-x_{1})g((r_{2}+x_{m})\tan\theta_{2}-u_{2})-\frac{1}{2}gx_{1}^{2}\tan\theta_{1}-\frac{1}{2}gx_{2}^{2}\tan\theta_{2}+\sigma\sin\psi_{2}-\sigma\sin\psi_{1}

(109)

Now we have the expression for the functional $\mathcal{V}$ and we want to simplify the representation (109). To achieve this goal, we plug equation (107) into this equation and then we obtain:

$\displaystyle g{\mathcal{V}}(\psi_{m})=$	$\displaystyle(x_{2}-x_{1})g(\frac{r_{2}\tan\theta_{1}+r_{1}\tan\theta_{1}}{\tan\theta_{1}+\tan\theta_{2}}-(u_{1}-u_{2})\frac{\tan\theta_{2}}{\tan\theta_{1}+\tan\theta_{2}}-u_{2})$
	$\displaystyle-\frac{1}{2}gx_{1}^{2}\tan\theta_{1}-\frac{1}{2}gx_{2}^{2}\tan\theta_{2}+\sigma\sin\psi_{2}-\sigma\sin\psi_{1}$
$\displaystyle=$	$\displaystyle g((r_{2}+r_{1})\frac{\tan\theta_{2}\tan\theta_{1}}{\tan\theta_{1}+\tan\theta_{2}}-u_{1}\frac{\tan\theta_{2}}{\tan\theta_{1}+\tan\theta_{2}}-u_{2}\frac{\tan\theta_{1}}{\tan\theta_{1}+\tan\theta_{2}})(r_{1}+r_{2})$
	$\displaystyle-\frac{1}{2}g(r_{1}-x_{m})^{2}\tan\theta_{1}-\frac{1}{2}g(r_{2}+x_{m})^{2}\tan\theta_{2}+\sigma\sin\psi_{2}-\sigma\sin\psi_{1}$
$\displaystyle=$	$\displaystyle g((r_{2}+r_{1})\frac{\tan\theta_{2}\tan\theta_{1}}{\tan\theta_{1}+\tan\theta_{2}}-u_{1}\frac{\tan\theta_{2}}{\tan\theta_{1}+\tan\theta_{2}}-u_{2}\frac{\tan\theta_{1}}{\tan\theta_{1}+\tan\theta_{2}})(r_{1}+r_{2})$
	$\displaystyle-\frac{1}{2}g\frac{\tan\theta_{1}\tan\theta_{2}}{\tan\theta_{1}+\tan\theta_{2}}(r_{1}^{2}+r_{2}^{2})-\frac{1}{2}g\frac{1}{(\tan\theta_{1}+\tan\theta_{2})}(u_{1}-u_{2})^{2}+\sigma\sin\psi_{2}-\sigma\sin\psi_{1}$
$\displaystyle=$	$\displaystyle g(\frac{1}{2}(r_{2}+r_{1})\frac{\tan\theta_{2}\tan\theta_{1}}{\tan\theta_{1}+\tan\theta_{2}}-u_{1}\frac{\tan\theta_{2}}{\tan\theta_{1}+\tan\theta_{2}}-u_{2}\frac{\tan\theta_{1}}{\tan\theta_{1}+\tan\theta_{2}})(r_{1}+r_{2})$
	$\displaystyle+\frac{1}{2}g(r_{1}+r_{2})^{2}\frac{\tan\theta_{1}\tan\theta_{2}}{\tan\theta_{1}+\tan\theta_{2}}-\frac{1}{2}g\frac{\tan\theta_{1}\tan\theta_{2}}{\tan\theta_{1}+\tan\theta_{2}}(r_{1}^{2}+r_{2}^{2})$
	$\displaystyle-\frac{1}{2}g\frac{1}{(\tan\theta_{1}+\tan\theta_{2})}(u_{1}-u_{2})^{2}+\sigma\sin\psi_{2}-\sigma\sin\psi_{1}$
$\displaystyle=$	$\displaystyle g(\frac{1}{2}(r_{2}+r_{1})\frac{\tan\theta_{2}\tan\theta_{1}}{\tan\theta_{1}+\tan\theta_{2}}-u_{1}\frac{\tan\theta_{2}}{\tan\theta_{1}+\tan\theta_{2}}-u_{2}\frac{\tan\theta_{1}}{\tan\theta_{1}+\tan\theta_{2}})(r_{1}+r_{2})$
	$\displaystyle+gr_{1}r_{2}\frac{\tan\theta_{1}\tan\theta_{2}}{\tan\theta_{1}+\tan\theta_{2}}-\frac{1}{2}g\frac{1}{(\tan\theta_{1}+\tan\theta_{2})}(u_{1}-u_{2})^{2}+\sigma\sin\psi_{2}-\sigma\sin\psi_{1}$	(110)

We call the right-hand side of (110) $f(\psi_{m})$ . Then we have the following theorem:

Theorem 3.7.

If we view $r_{1}$ and $r_{2}$ as functions of $\psi_{m}$ , then the function f defined below converges to $+\infty$ when $\psi_{m}\rightarrow 0$ .

	$\displaystyle f(\psi_{m})=$	$\displaystyle g(\frac{1}{2}(r_{2}+r_{1})\frac{\tan\theta_{2}\tan\theta_{1}}{\tan\theta_{1}+\tan\theta_{2}}-u_{1}\frac{\tan\theta_{2}}{\tan\theta_{1}+\tan\theta_{2}}-u_{2}\frac{\tan\theta_{1}}{\tan\theta_{1}+\tan\theta_{2}})(r_{1}+r_{2})$
		$\displaystyle+gr_{1}r_{2}\frac{\tan\theta_{1}\tan\theta_{2}}{\tan\theta_{1}+\tan\theta_{2}}-\frac{1}{2}g\frac{1}{(\tan\theta_{1}+\tan\theta_{2})}(u_{1}-u_{2})^{2}+\sigma\sin\psi_{2}-\sigma\sin\psi_{1}$		(111)

Proof.

Step 1 $r_{1}$ and $r_{2}$ converge to $+\infty$ when $\psi_{m}\rightarrow 0$ .

Using equation (104) we know that:

r_{1}=\int_{\psi_{1}}^{\psi_{m}}\frac{\cos\gamma}{g\sqrt{\frac{2\sigma}{g}(\cos\psi_{m}-\cos\gamma)}}d\gamma

When $\psi_{m}=0$ , we can use Taylor expansion and get:

\cos\gamma=1-\frac{1}{2}\gamma^{2}+O((\gamma)^{3})

We apply this expansion to the integral above:

{}r_{1}(0)\geq\int_{\psi_{1}}^{0}\frac{1}{\sqrt{2\sigma g}}\frac{\cos\gamma}{-\gamma}d\gamma=+\infty

(112)

From a simple observation, we know that $r_{1}$ is a increasing function of $\psi_{m}$ when $\psi_{m}\in(\psi_{1},0)$ . Therefore, using Fatou’s lemma, we can show that:

\infty=\int_{\psi_{1}}^{0}\frac{\cos\gamma}{g\sqrt{\frac{2\sigma}{g}(1-\cos\gamma)}}d\gamma\leq\lim_{\psi_{m}\rightarrow 0}\int_{\psi_{1}}^{\psi_{m}}\frac{\cos\gamma}{g\sqrt{\frac{2\sigma}{g}(\cos\psi_{m}-\cos\gamma)}}d\gamma

which is equivalent to the fact that:

\lim_{\psi_{m}\rightarrow 0}r_{1}=+\infty

And we can use the same method to prove that $r_{2}\rightarrow+\infty$ when $\psi_{m}\rightarrow 0$ , which completes the proof in step 1

Step 2 The behavior of $u_{1}$ and $u_{2}$ as functions of $\psi_{m}$ .

Using (102) and (103), we can obtain that $u_{1}\rightarrow\sqrt{\frac{2\sigma}{g}(1-\cos\psi_{1})}$ when $\psi_{m}\rightarrow 0$ . and $u_{2}\rightarrow\sqrt{\frac{2\sigma}{g}(1-\cos\psi_{2})}$ when $\psi_{m}\rightarrow 0$ . Both of $u_{1}$ and $u_{2}$ are bounded no matter what $\psi_{m}$ is.

Step 3: The asymptotic behavior of f.

We examining equation (111). For the first row we have:

g(\frac{1}{2}(r_{2}+r_{1})\frac{\tan\theta_{2}\tan\theta_{1}}{\tan\theta_{1}+\tan\theta_{2}}-u_{1}\frac{\tan\theta_{2}}{\tan\theta_{1}+\tan\theta_{2}}-u_{2}\frac{\tan\theta_{1}}{\tan\theta_{1}+\tan\theta_{2}})(r_{1}+r_{2})\rightarrow+\infty

whe $\psi_{m}\rightarrow 0$ . This is because

\frac{1}{2}(r_{2}+r_{1})\frac{\tan\theta_{2}\tan\theta_{1}}{\tan\theta_{1}+\tan\theta_{2}}-u_{1}\frac{\tan\theta_{2}}{\tan\theta_{1}+\tan\theta_{2}}-u_{2}\frac{\tan\theta_{1}}{\tan\theta_{1}+\tan\theta_{2}}\rightarrow+\infty

by using step 1 and step 2 and:

r_{2}+r_{1}\rightarrow+\infty

by using step 2.

For the second row in the equation in (111) we can show the following result:

{}gr_{1}r_{2}\frac{\tan\theta_{1}\tan\theta_{2}}{\tan\theta_{1}+\tan\theta_{2}}-\frac{1}{2}g\frac{1}{(\tan\theta_{1}+\tan\theta_{2})}(u_{1}-u_{2})^{2}+\sigma\sin\psi_{2}-\sigma\sin\psi_{1}\rightarrow+\infty

(113)

due to the fact that the first term in equation (113) converges to $\infty$ when $\psi_{m}\rightarrow 0$ . The other terms are bounded for any $\psi_{m}\in(\psi_{1},0)$ .

In conclusion, we have $f(\psi_{m})\rightarrow+\infty$ when $\psi_{m}\rightarrow 0$

∎

So from the previous theorem we know that when $\psi_{m}\rightarrow 0$ , the total volume $\mathcal{V}(\psi_{m})$ enclosed by the boundary curve and two walls converges to $+\infty$ . Then we want to study the minimum bound of the total volume. We have the following theorem:

Theorem 3.8.

The volume functional ${\mathcal{V}}(\psi_{m})$ is bounded from below by a minimum volume $V_{m}$ which is defined by:

V_{m}:=\min_{\psi_{m}\in(\max(\psi_{1},\psi_{2}),0)}\mathcal{V}(\psi_{m})

$V_{m}$ is positive when $\psi_{1}\neq\psi_{2}$

Proof.

We also divide the area into several parts. We call the area enclosed by $y=u_{2}$ and two walls $\Omega_{1}$ and the corresponding area ${\mathcal{V}}_{1}(\psi_{m})$ . Then we have the total volume ${\mathcal{V}}(\psi_{m})\geq{\mathcal{V}}_{1}(\psi_{m})$ . See the figure 7 :

From a simple geometry relation, we obtain:

{\mathcal{V}}_{1}(\psi_{m})=\frac{1}{2}(u_{2}+\frac{P_{0}}{g})^{2}(\frac{1}{\tan\theta_{1}}+\frac{1}{\tan\theta_{2}})

Then plug the equation (103) and $P_{0}=g((r_{2}+x_{m})\tan\theta_{2})-u_{2}$ into the equation above and get:

	$\displaystyle{\mathcal{V}}_{1}$	$\displaystyle=\frac{1}{2}(r_{2}+x_{m})^{2}\tan\theta_{2}(\frac{1}{\tan\theta_{1}}+\frac{1}{\tan\theta_{2}})$
		$\displaystyle=\frac{1}{2}(\frac{r_{1}\tan\theta_{1}+r_{2}\tan\theta_{1}-(u_{1}-u_{2})}{\tan\theta_{1}+\tan\theta_{2}})^{2}\tan\theta_{2}(\frac{1}{\tan\theta_{1}}+\frac{1}{\tan\theta_{2}})$

using Lemma 3.6, we know that $r_{2}+x_{m}=x_{2}>0$ , which implies that

r_{1}\tan\theta_{1}+r_{2}\tan\theta_{1}-(u_{1}-u_{2})\geq 0

for any maximal angle $\psi_{m}$ . And it equals zero if and only if $x_{1}=0$ , which means that $(x_{1},u_{1})$ is the original point. Then we plug $x_{1}=r_{1}=0$ and $u_{1}=u_{2}$ into the equation (107) to obtain that:

0=x_{m}=\frac{-r_{2}\tan\theta_{2}}{\tan\theta_{1}+\tan\theta_{2}}

which implies that $r_{2}=0$ . We then use (106) to obtain that:

{}0=\int_{\psi_{1}}^{\psi_{2}}\frac{\cos\gamma}{g\sqrt{\frac{2\sigma}{g}(\cos\psi_{m}-\cos\gamma)}}d\gamma

(114)

The right-hand side of (114) is positive when $\psi_{1}\neq\psi_{2}$ . Therefore $\mathcal{V}(\psi_{1})$ positive when $\psi_{1}\neq\psi_{2}$ . And $\mathcal{V}_{1}(\psi_{m})$ is positive when $\psi\in(\psi_{1},0)$

Furthermore, since $\psi_{m}$ is the maximum angle, we have $\psi_{m}\in[\psi_{1},0)$ . We already know that $r_{2}+x_{m}\rightarrow+\infty$ when $\psi_{m}\rightarrow 0$ which implies that $\lim_{\psi_{m}\rightarrow 0}\mathcal{V}_{1}(\psi_{m})=+\infty$ We then conclude that ${\mathcal{V}}_{1}$ is bounded below by some constant. Since ${\mathcal{V}}\geq{\mathcal{V}}_{1}$ , we have ${\mathcal{V}}$ bounded below by a positive constant $V_{m}$ . And $V_{m}$ is positive when $\psi_{1}\neq\psi_{2}$

∎

Then we only need to consider the case where V is small, such that $V\leq V_{m}$ . In the previous calculation, we assumed that the maximal angle $\psi_{3}$ is not on the boundary. Therefore, we only need to discuss the scenario in which $\psi(x)$ reaches its maximum at the boundary and is either a monotonous increasing function or a monotone decreasing function. So, $\psi$ serves as a legal parameter for the boundary curve in this case.

The remaining computation is similar to Section 2. We proceed with the assumption that the boundary curve meets the left wall at the point $(x_{1},u_{1})$ . We then derive the following expression for the other contact point:

Theorem 3.9.

We let $(x_{1},u_{1})$ and $(x_{2},u_{2})$ be coordinates of two contact points. We have the following results:

1: When $\psi_{1}\geq\psi_{2}$ , then $\psi(x)$ is a monotone decreasing function of x and we have their expressions as follows:

	$\displaystyle x_{2}$	$\displaystyle=x_{1}+\int_{\psi_{2}}^{\psi_{1}}\frac{\cos\gamma d\gamma}{g\sqrt{u_{1}^{2}+\frac{2\sigma}{g}(\cos\psi_{1}-\cos\psi_{2})}}$		(115)
	$\displaystyle u_{2}$	$\displaystyle=\sqrt{u_{1}^{2}+\frac{2\sigma}{g}(\cos\psi_{1}-\cos\psi_{2})}$		(116)

2: When $\psi_{2}\geq\psi_{1}$ , then $\psi(x)$ is a monotone increasing function of x and we have expressions for $x_{2}$ and $u_{2}$ as follows:

	$\displaystyle x_{2}$	$\displaystyle=x_{1}+\int_{\psi_{1}}^{\psi_{2}}\frac{\cos\gamma d\gamma}{g\sqrt{u_{1}^{2}+\frac{2\sigma}{g}(\cos\psi_{1}-\cos\psi_{2})}}$		(117)
	$\displaystyle u_{2}$	$\displaystyle=\sqrt{u_{1}^{2}+\frac{2\sigma}{g}(\cos\psi_{1}-\cos\psi_{2})}$		(118)

Proof.

The proof is the same as Theorem 3.5 and Theorem 2.2 ∎

Then for the remaining part of the discussion we suppose that $\psi_{2}\leq\psi_{1}<0$ . If $\psi_{1}<\psi_{2}$ , we can use similar computation. See the figure 8

Just like the proof of the equation (107), we have:

{}\tan\theta_{1}(-x_{1})=x_{2}\tan\theta_{2}+(u_{1}-u_{2})

(119)

We then plug $x_{2}=x_{1}+r_{2}$ into the equation (119). and then obtain:

{}x_{1}=\frac{-r_{2}\tan\theta_{2}+(u_{2}-u_{1})}{\tan\theta_{1}+\tan\theta_{2}}

(120)

Next, we express the total volume in terms of $x_{1}$ and $u_{1}$ . We divide the total volume into two regions: the first one $\Omega_{1}$ , is the area enclosed by the boundary curve, the left wall, and $u=u_{2}$ ; the second: $\Omega_{2}$ is the area bounded by two walls and $u=u_{2}$ . See the figure 8. The corresponding volumes of these regions are denoted by ${\mathcal{V}}_{1}$ and ${\mathcal{V}}_{2}$ , respectively.

We can use the same computation in Section 2 to obtain the following results:

$\displaystyle{\mathcal{V}}_{1}(u_{1})$	$\displaystyle=\mathcal{V}_{1}(x_{1}(u_{1}),u_{1})$
	$\displaystyle=\int_{\psi_{1}}^{\psi_{2}}\int_{\psi}^{\psi_{1}}\frac{\cos\gamma}{g\sqrt{u_{1}^{2}+\frac{2\sigma}{g}(1-\cos\gamma)}}d\gamma\frac{\sin\psi}{g\sqrt{u_{1}^{2}+\frac{2\sigma}{g}(1-\cos\psi)}}d\psi-\frac{1}{2\tan\theta_{1}}(u_{2}-u_{1})^{2}$	(121)
$\displaystyle{\mathcal{V}}_{2}(u_{1})$	$\displaystyle=\mathcal{V}_{2}(x_{1}(u_{1}),u_{1})=\frac{1}{2}(\frac{1}{\tan\theta_{1}}+\frac{1}{\tan\theta_{2}})(u_{2}-(u_{1}+x_{1}\tan\theta_{1}))$	(122)

Then we plug equation (120) into equation (122) and we can show that:

{\mathcal{V}}_{2}=\frac{1}{2}\frac{1}{(\tan\theta_{1}+\tan\theta_{2})}(\frac{1}{\tan\theta_{1}}+\frac{1}{\tan\theta_{2}})(r_{2}\tan\theta_{2}+(u_{2}-u_{1})\tan\theta_{2})

Based on the computation above, we can now express the total volume ${\mathcal{V}}={\mathcal{V}}_{1}+{\mathcal{V}}_{2}$ in terms of $u_{1}$ . We combine our ODE (19):

\partial_{x}(\sin\psi)=gu

with the result in Theorem 3.9 that $\psi(x)$ is monotonically decreasing to obtain:

gu_{1}=\partial_{x}(\sin\psi)(x_{1})\leq 0

which implies that:

u_{1}\leq 0

Thus, we determine the possible range of values for $u_{1}$ and proceed to establish the following theorem.

Theorem 3.10.

The total volume ${\mathcal{V}}(u_{1})$ is a bounded function of $u_{1}$ when $u_{1}\leq 0$ . And the upper bound $V_{1}$ is bigger than or equal to $V_{m}$ .

Proof.

We first prove that $\mathcal{V}$ is a bounded function of $u_{1}$ . We notice that it is a continuous function of $u_{1}$ . Also when $u_{1}\rightarrow-\infty$ , (115) and (116) imply that:

{}r_{2}=x_{2}-x_{1}=\int_{\psi_{2}}^{\psi_{1}}\frac{\cos\gamma d\gamma}{g\sqrt{u_{1}^{2}+\frac{2\sigma}{g}(\cos\psi_{1}-\cos\psi_{2})}}\rightarrow 0\leavevmode\nobreak\ \operatorname{when}\leavevmode\nobreak\ u_{1}\rightarrow-\infty

(123)

{}u_{2}-u_{1}\rightarrow 0\leavevmode\nobreak\ \operatorname{when}\leavevmode\nobreak\ u_{1}\rightarrow-\infty

(124)

Then by equation (120), we have:

{}x_{1}=\frac{-r_{2}\tan\theta_{2}+(u_{2}-u_{1})}{\tan\theta_{1}+\tan\theta_{2}}\rightarrow 0\leavevmode\nobreak\ \operatorname{when}\leavevmode\nobreak\ u_{1}\rightarrow-\infty

(125)

Based on the expression for $\mathcal{V_{1}}$ and $\mathcal{V}_{2}$ ((121) and (122)), we obtain the following result by combining(123),(124) and (125) together:

\mathcal{V}_{1}\rightarrow 0\leavevmode\nobreak\ \operatorname{when}\leavevmode\nobreak\ u_{1}\rightarrow-\infty

and:

\mathcal{V}_{2}\rightarrow 0\leavevmode\nobreak\ \operatorname{when}\leavevmode\nobreak\ u_{1}\rightarrow-\infty

The above two asymptotic results mean that:

\mathcal{V}=\mathcal{V}_{1}+\mathcal{V}_{2}\rightarrow 0\leavevmode\nobreak\ \operatorname{when}\leavevmode\nobreak\ u_{1}\rightarrow-\infty

Based on the discussion above, when $u_{1}\leq 0$ , $\mathcal{V}$ is a bounded function of $u_{1}$ .(Here we only need to care about the point $u_{1}=0$ . Based on our previous discussion, the integral in the representation (121) is infinity only when $\psi_{1}=0$ . So the volume functional is bounded at this point since $\psi_{1}<0$ .)

We then only need to show that the upper bound $V_{1}$ is bigger than or equals $V_{m}$ . We notice that when $u_{1}=0$ , the curve is the same as the curve in Theorem 3.8 when $\psi_{m}=\psi_{1}$ . So we have:

V_{1}\geq\lim_{u_{1}\rightarrow 0}{\mathcal{V}}(u_{1})={\mathcal{V}}(\psi_{1})\geq V_{m}

which finishes the proof. ∎

We then combine Theorem 3.8 and Theorem 3.10 to get:

Theorem 3.11.

When $\theta_{1}$ and $\theta_{2}$ are both acute angles, we have at least one steady state for any total volume V.

We also have two constant $V_{1}$ and $V_{m}$ which depend on $\theta_{1}$ , $\theta_{2}$ , $\sigma$ and $[[\gamma]]$ such that:

•

When $V\geq V_{m}$ , we can find a $\psi_{m}$ with a solution function of system (3) which reaches its maximal slope angle $\psi$ in the interval $(\psi_{2},\psi_{1})$ (or $(\psi_{1},\psi_{2})$ if $\psi_{1}\leq\psi_{2}$ ).
•

When $V\leq V_{1}$ , we can find a $u_{1}$ with a solution function of system (3) which reaches its maximal slope angle on the boundary.
•

When $V_{1}>V_{m}$ , the solution is not unique.

3.3 General case when $\theta_{1}>\frac{\pi}{2}$

In this case, the only difference compared with Subsection 3.2 is that $\psi_{1}$ and $\psi_{2}$ can have the same sign even when $[[\gamma]]<0$ .

By definition, we have $\psi_{1}=-\theta_{1}+\frac{\pi}{2}+\gamma$ and $\psi_{2}=\theta_{2}-\frac{\pi}{2}-\gamma$ . Then we can observe that $\psi_{2}<-\frac{\pi}{2}$ when $\gamma>\theta_{2}$ , which means that the boundary curve may no longer be a function of x. Therefore, we cannot use equation (99) to derive the equation for $P_{0}$ . We need to use the method in Section 2 to compute the total volume using the coordinates of a special point. This is the only difference of this subsection compared with Subsection 3.2.

In this section, we only discuss the case $\gamma>\theta_{2}$ . For the case $\gamma\leq\theta_{2}$ , we can just follow the method in Subsection 3.2. We also assume that $V\geq V_{m}$ for a constant $V_{m}$ to be determined such that $\psi$ attains its maximum in the interior. And we assume that $\psi(x)$ reaches its maximal point at $(x_{m},0)$ with the maximal angle $\psi_{m}$ .

We suppose that the curve meets two walls at two contact points $(x_{1},v_{1})$ and $(x_{2},v_{2})$ . After the shift, they become $(x_{1},u_{1})$ and $(x_{2},u_{2})$ . See Figure 9. Then we have the computation (It is the same as the computation in Section two):

	$\displaystyle x_{1}$	$\displaystyle=x_{m}-\int_{\psi_{1}}^{\psi_{m}}\frac{\cos\gamma}{g\sqrt{\frac{2\sigma}{g}(\cos\psi_{m}-\cos\gamma)}}d\gamma$		(126)
	$\displaystyle u_{1}$	$\displaystyle=\sqrt{\frac{2\sigma}{g}(\cos\psi_{m}-\cos\psi_{1})}$		(127)

	$\displaystyle x_{2}$	$\displaystyle=x_{m}+\int_{\psi_{2}}^{\psi_{m}}\frac{\cos\gamma}{g\sqrt{\frac{2\sigma}{g}(\cos\psi_{m}-\cos\gamma)}}d\gamma$		(128)
	$\displaystyle u_{2}$	$\displaystyle=-\sqrt{\frac{2\sigma}{g}(\cos\psi_{m}-\cos\psi_{2})}$		(129)

We also use the same method as in the previous Subsection to derive the relation between $x_{m}$ and $\psi_{m}$ . We should have:

-x_{1}\tan\theta_{1}-x_{2}\tan\theta_{2}=u_{1}-u_{2}

which implies that:

{}x_{m}=\frac{r_{1}\tan\theta_{1}-r_{2}\tan\theta_{2}-(u_{1}-u_{2})}{\tan\theta_{1}+\tan\theta_{2}}

(130)

Then we use $x_{m}$ and $\psi_{m}$ to express the total volume. By using geometric computation, we have:

$\displaystyle\mathcal{V}(\psi_{m})=$	$\displaystyle(r_{1}u_{1}-\int_{\psi_{1}}^{\psi_{m}}\int_{\psi}^{\psi_{m}}\frac{\cos\gamma}{g\sqrt{\frac{2\sigma}{g}(\cos\psi_{m}-\cos\gamma)}}d\gamma\frac{-\sin\psi}{g\sqrt{\frac{2\sigma}{g}(\cos\psi_{m}-\cos\psi)}}d\psi)$
	$\displaystyle+\int_{\psi_{2}}^{\psi_{m}}\int_{\psi}^{\psi_{m}}\frac{\cos\gamma}{g\sqrt{\frac{2\sigma}{g}(\cos\psi_{m}-\cos\gamma)}}d\gamma\frac{-\sin\psi}{g\sqrt{\frac{2\sigma}{g}(\cos\psi_{m}-\cos\psi)}}d\psi$
	$\displaystyle+\frac{1}{2}(u_{1}+u_{2})^{2}\frac{1}{-\tan\theta_{1}}+(-u_{2})r_{1}+\frac{1}{2}(u_{2}-u_{1}-x_{1}\tan\theta_{1})^{2}(\frac{1}{\tan\theta_{1}}+\frac{1}{\tan\theta_{2}})$	(131)

Then we need to prove the following theorem about the total volume:

Theorem 3.12.

When $\psi_{m}\rightarrow 0$ , we have ${\mathcal{V}}(\psi_{m})\rightarrow+\infty$ .

Proof.

Using the computation in Theorem 3.7, we know that:

\int_{\psi}^{\psi_{m}}\frac{\cos\gamma}{g\sqrt{\frac{2\sigma}{g}(\cos\psi_{m}-\cos\gamma)}}d\gamma\rightarrow+\infty

when $\psi_{m}\rightarrow 0$ for any $\psi\in(\psi_{2},\psi_{m})$ . So we should have:

\mathcal{V}(\psi_{m})\geq\int_{\psi_{2}}^{\psi_{m}}\int_{\psi}^{\psi_{m}}\frac{\cos\gamma}{g\sqrt{\frac{2\sigma}{g}(\cos\psi_{m}-\cos\gamma)}}d\gamma\frac{-\sin\psi}{g\sqrt{\frac{2\sigma}{g}(\cos\psi_{m}-\cos\psi)}}d\psi\rightarrow+\infty

when $\psi_{m}\rightarrow 0$ , which means that:

\mathcal{V}(\psi_{m})\rightarrow+\infty

when $\psi_{m}\rightarrow 0$ , which finishes the proof. ∎

Just like the previous subsection, we want to show that ${\mathcal{V}}(\psi_{m})$ has a lower bound $V_{m}$ .

Theorem 3.13.

The energy functional $\mathcal{V}(\psi_{m})$ has a positive lower bound $V_{m}$ .

Proof.

Using equation (131), we know that:

{}V(\psi_{m})\geq\int_{\psi_{2}}^{\psi_{m}}\int_{\psi}^{\psi_{m}}\frac{\cos\gamma}{g\sqrt{\frac{2\sigma}{g}(\cos\psi_{m}-\cos\gamma)}}d\gamma\frac{-\sin\psi}{g\sqrt{\frac{2\sigma}{g}(\cos\psi_{m}-\cos\psi)}}d\psi

(132)

Using the definition of $\psi_{1}$ and $\psi_{2}$ ( (6) and (7)), we know that $\psi_{1}=-\theta_{1}+\frac{\pi}{2}+\gamma$ and $\psi_{2}=\theta_{2}-\frac{\pi}{2}-\gamma$ . First we can show that $\psi_{1}\in(-\frac{\pi}{2},0)$ since

{}0>\psi_{1}=-\theta_{1}+\frac{\pi}{2}+\gamma\geq-\theta_{1}+\frac{\pi}{2}+\theta_{2}>-\pi+\frac{\pi}{2}=-\frac{\pi}{2}

(133)

We also have the following computation:

{}\psi_{m}\geq\psi_{1}=-\theta_{1}+\frac{\pi}{2}-\gamma=\theta_{2}-\theta_{1}-\psi_{2}>-\pi-\psi_{2}

(134)

We can use (133) and (134) to show that $\cos\psi_{m}>\cos\psi_{2}$ . Therefore, we obtain that:

{}\int_{\psi}^{\psi_{m}}\frac{\cos\gamma}{g\sqrt{\frac{2\sigma}{g}(\cos\psi_{m}-\cos\gamma)}}d\gamma\geq\int_{\psi}^{\psi_{m}}\frac{\cos\gamma}{g\sqrt{\frac{2\sigma}{g}(\cos\psi_{m}+1)}}d\gamma=\frac{\sin\psi_{m}-\sin\psi}{g\sqrt{\frac{2\sigma}{g}(\cos\psi_{m}+1)}}

(135)

The proof of (135) is the same as the proof in Lemma 2.9. We do not write down the details here. Since $\psi\in(\psi_{2},\psi_{m})$ , we have $\sin\psi_{m}>\sin\psi$ . We combine (135) with the fact that $\sin\psi_{m}>\sin\psi$ and plug them back into the equation (132) to get:

	$\displaystyle\mathcal{V}(\psi_{m})$	$\displaystyle\geq\int_{\psi_{2}}^{\psi_{m}}\frac{\sin\psi_{m}-\sin\psi}{g\sqrt{\frac{2\sigma}{g}(\cos\psi_{m}+1)}}\frac{-\sin\psi}{g\sqrt{\frac{2\sigma}{g}(\cos\psi_{m}-\cos\psi)}}d\psi$
		$\displaystyle>\frac{1}{2\sigma g(\cos\psi_{m}+1)}\int_{\psi_{2}}^{\psi_{m}}-\sin\psi\sin\psi_{m}+\sin^{2}\psi d\psi$
		$\displaystyle>\frac{1}{4\sigma g}(-\sin\psi_{m}(\cos\psi_{m}-\sin\psi_{2})+\int_{\psi_{2}}^{\psi_{m}}\sin^{2}\psi d\psi)$
		$\displaystyle>\frac{1}{4\sigma g}\int_{\psi_{2}}^{\psi_{1}}\sin^{2}\psi d\psi>C(\psi_{1},\psi_{2})>0$

This inequality combined with Theorem 3.12 imply that that ${\mathcal{V}}(\psi_{m})$ is a positive continuous function of $\psi_{m}\in[\psi_{1},0)$ . Also when $\psi_{m}\rightarrow 0$ , ${\mathcal{V}}\rightarrow+\infty$ . So $\mathcal{V}(\psi_{m})$ is bounded from below. ∎

Then we want to compute the case when $V$ is small. Also we assume that $\psi_{1}>\psi_{2}$ . We suppose that the curve intersects the left wall at the contact point $(x_{1},u_{1})$ . Then we have the representation $x_{2}$ and $u_{2}$ the same as equation (115) and equation (116).

Actually, the representation for $x_{1}$ and total volume $\mathcal{V}(u_{1})$ is the same as equation (120), equation (121) and (122). Then we have the theorem:

Theorem 3.14.

The total volume ${V}=\mathcal{V}_{1}+\mathcal{V}_{2}$ is a function of $u_{1}$ bounded from above when $u_{1}<0$ by a positive constant $V_{a}$ . And $V_{a}\geq V_{m}$ .

Proof.

The proof is the same as Theorem 3.10. ∎

Finally we combine Theorem 3.11, Theorem 3.13, and Theorem 3.14 together to get:

Theorem 3.15.

When $\psi_{1}$ and $\psi_{2}$ have the same sign, we have the existence of the solution function of the equation system (3)

However, unlike the case in Section 2, we cannot show the uniqueness of the steady state in this case.

ACKNOWLEDGEMENTS

The author thanks Yan Guo for numerous comments. His mentorship and constructive feedback significantly contribute to the development of this work.

This work is supported in part by NSF Grant DMS-2405051.

References

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The steady state of the inclined problem

Abstract

1 Introduction

Theorem 1.1.

Theorem 1.2.

Theorem 1.3.

2 The first case: ψ1⋅ψ2<0\psi_{1}\cdot\psi_{2}<0

2.1 The shape of the curve

Theorem 2.1.

Proof.

2.2 The existence of the solution function

Theorem 2.2.

Proof.

Theorem 2.3.

Proof.

lemma 2.4.

Proof.

lemma 2.5.

Proof.

2.2.1 θ1≠π2\theta_{1}\neq\frac{\pi}{2}

Theorem 2.6.

Proof.

Theorem 2.7.

Proof.

Theorem 2.8.

Proof.

lemma 2.9.

Proof.

Theorem 2.10.

Proof.

lemma 2.11.

Proof.

2.2.2 θ1=π2\theta_{1}=\frac{\pi}{2}

Theorem 2.12.

Proof.

Theorem 2.13.

Theorem 2.14.

Theorem 2.15.

Proof.

3 The second case: ψ1⋅ψ2≥0\psi_{1}\cdot\psi_{2}\geq 0

3.1 A special case where θ1=π2\theta_{1}=\frac{\pi}{2} and θ2=0\theta_{2}=0

Theorem 3.1.

Proof.

Theorem 3.2.

Proof.

3.2 General case when θ1≤π2\theta_{1}\leq\frac{\pi}{2}

lemma 3.3.

Proof.

Theorem 3.4.

Proof.

Theorem 3.5.

Proof.

lemma 3.6.

Proof.

Theorem 3.7.

Proof.

Theorem 3.8.

Proof.

Theorem 3.9.

Proof.

Theorem 3.10.

Proof.

Theorem 3.11.

3.3 General case when θ1>π2\theta_{1}>\frac{\pi}{2}

Theorem 3.12.

Proof.

Theorem 3.13.

Proof.

Theorem 3.14.

Proof.

Theorem 3.15.

References

2 The first case: $\psi_{1}\cdot\psi_{2}<0$

2.2.1 $\theta_{1}\neq\frac{\pi}{2}$

2.2.2 $\theta_{1}=\frac{\pi}{2}$

3 The second case: $\psi_{1}\cdot\psi_{2}\geq 0$

3.1 A special case where $\theta_{1}=\frac{\pi}{2}$ and $\theta_{2}=0$

3.2 General case when $\theta_{1}\leq\frac{\pi}{2}$

3.3 General case when $\theta_{1}>\frac{\pi}{2}$