## I. INTRODUCTION As one of the most well-known natural laws, Newton's second law of motion can be expressed in the form of a second-order differential equation. In order to preserve the energy conservation property of the original system, it is very important to use the conservative numerical methods to solve these equations with initial conditions [1-4]. In recent years some conservative numerical methods [1-3] have been developed to solve the following second-order initial value problem, $$ \ddot {x} (t) = f (x (t)), t \in I; x (0) = \alpha ; \dot {x} (0) = \beta , \tag {1.1} $$ where $I = (0,T)$, and $\dot{x} \coloneqq \frac{\mathrm{d}x}{\mathrm{d}t}$, $\ddot{x} \coloneqq \frac{\mathrm{d}^2x}{\mathrm{d}t^2}$. Introduce the potential function $\phi (x)$ of $f(x)$ $$ -\frac{\mathrm{d}\phi}{\mathrm{d}x}=f(x), $$ then the energy conservation equation is derived as [1-2] $$ \frac {1}{2} \left(\frac {\mathrm {d} x}{\mathrm {d} t}\right) ^ {2} + \phi (x) = \frac {1}{2} \beta^ {2} + \phi (\alpha). \tag {1.2} $$ Applying iterative computation, Greenspan [1] proposed two conservative implicit numerical methods to solve problem (1.1). According to the energy conservation equation (1.2), Qin [2] constructed an explicit energy-conserving method by using sign function, Sövegjártó [3] also developed conservative spline methods to solve this problem numerically. In these methods [2-3], the function value at the current temporal node is numerically calculated first, and then the derivative value at the current temporal node is numerically computed, in which the sign function is used to prejudice the sign of the derivative value. We design an energy conservation method(numerical method I), which first calculates the derivative value and then calculates the function value, without using the sign function. Moreover, we use the $C^1$ -continuous piecewise-quadratic functions [5-6] to approximate the true solution of problem (1.1), and construct a $C^1$ -continuous energy conservation numerical method(numerical method II), these approximate solutions are global $C^1$ -smooth on the entire temporal interval. Under a linear force assumption of the initial value problems, a priori error estimate for the numerical method II is deduced, which shows that the method has first-order convergence accuracy. The remainder of the paper is organized as follows. In Section II and Section III, we propose two novel conservative numerical methods. We derive a priori error estimate for numerical method II in Section IV. The numerical experiments demonstrating the promising features of the conservative methods are displayed in Section V. ## II. NUMERICAL METHOD I Let $0 = t_0 < t_1 < \dots < t_N = T$ be a subdivision of $\overline{I}$. Define $$ I _ {n} := (t _ {n}, t _ {n + 1}), h := t _ {n + 1} - t _ {n}, n = 0, \dots , N - 1, $$ denote the numerical solution $x_{n}$ and $v_{n}$ for $x(t_{n})$ and $\dot{x}(t_{n})$, respectively. We construct the following numerical scheme satisfying (1.2) for problem (1.1). $$ v_{n+1} = v_{n} + h f(x_{n}), $$ $$ \phi \left(x _ {n + 1}\right) = \frac {1}{2} \beta^ {2} + \phi (\alpha) - \frac {1}{2} \left(v _ {n + 1}\right) ^ {2}. \tag {2.1} $$ From second equation of (2.1), $x_{n+1}$ can be determined by Newton's method [7], where the initial guess $\tilde{x}_{n+1}$ of $x_{n+1}$ is computed by the following equation, $$ \tilde {x} _ {n + 1} = x _ {n} + h v _ {n} + \frac {h ^ {2}}{2} f (x _ {n}). $$ ## III. NUMERICAL METHOD II We apply the piecewise second-order polynomial $\varphi(t) \in C^{1}(\overline{I})$ to approximate the true solution $x(t)$ of problem (1.1). In each subinterval $I_{n}$, $n = 0, \dots, N - 1$, set $$ \varphi (t) = a _ {n} ^ {2} \left(t - t _ {n}\right) ^ {2} + a _ {n} ^ {1} \left(t - t _ {n}\right) + \varphi_ {n}, t \in I _ {n} \tag {3.1} $$ with $\varphi_{n}:= \varphi(t_{n}), \varphi_{0} = \alpha, a_{0}^{1} = \beta$. And the coefficients $a_{n}^{2}$ and $a_{j}^{1}(j = 1, \dots, N - 1)$ are yet to be computed. Noting the continuity condition $\varphi(t) \in C^{1}(\overline{I})$, we have $$ \varphi_ {n + 1} = a _ {n} ^ {2} h ^ {2} + a _ {n} ^ {1} h + \varphi_ {n}, $$ $$ a_{n+1}^{1} = 2 a_{n}^{2} h + a_{n}^{1}. $$ Inserting (3.2) into (1.2), we get $$ \frac{1}{2} \left(2 a_{n}^{2} h + a_{n}^{1}\right)^{2} + \phi \left(a_{n}^{2} h^{2} + a_{n}^{1} h + \varphi_{n}\right) = \frac{1}{2} \beta^{2} + \phi(\alpha). $$ From (3.3), $a_{n}^{2}$ can be computed by Newton's method, where the initial guess $\tilde{a}_{n}^{2}$ of $a_{n}^{2}$ is determined by $\tilde{a}_{n}^{2} = \frac{f(\varphi_{n})}{2}$. ## IV. ERROR ANALYSIS In this section, we assume $f(x) = -\lambda^2 x$ [8] and deduce the convergent result for numerical method (3.1)-(3.3). On one hand, from (3.2) we have $$ \varphi_{n+1} - \varphi_{n} = \frac{h}{2} \left(a_{n+1}^{1} + a_{n}^{1}\right), $$ and hence, $$ \begin{array}{l} \varphi_ {n + 2} - 2 \varphi_ {n + 1} + \varphi_ {n} = \frac {h}{2} (a _ {n + 2} ^ {1} - a _ {n} ^ {1}) \\= h ^ {2} \left(\frac {a _ {n + 2} ^ {1} - a _ {n + 1} ^ {1}}{2 h} + \frac {a _ {n + 1} ^ {1} - a _ {n} ^ {1}}{2 h}\right) \tag {4.2} \\= h ^ {2} (a _ {n + 1} ^ {2} + a _ {n} ^ {2}). \\\end{array} $$ On the other hand, since $\phi (x) = \frac{\lambda^2x^2}{2}$, applying (3.2), (4.1), and the energy conservation equation (1.2) we obtain $$ a_{n}^{2} = \frac{1}{2h} (a_{n+1}^{1} - a_{n}^{1}) \\= \frac{1}{2h} \frac{(a_{n+1}^{1})^{2} - (a_{n}^{1})^{2}}{a_{n+1}^{1} + a_{n}^{1}} \\= \frac{1}{4} \frac{(a_{n+1}^{1})^{2} - (a_{n}^{1})^{2}}{\varphi_{n+1} - \varphi_{n}} $$ $$ = -\frac{1}{2} \frac{\phi(\varphi_{n+1}) - \phi(\varphi_{n})}{\varphi_{n+1} - \varphi_{n}} \\= -\frac{\lambda^{2}}{4} (\varphi_{n+1} + \varphi_{n}) \\= \frac{1}{4} \left(f_{n+1} + f_{n}\right). \tag{4.3} $$ Together with (4.2)-(4.3), this yields $$ \varphi_{n+2} - 2\varphi_{n+1} + \varphi_{n} = h^{2} \left(\frac{1}{4} f_{n+2} + \frac{2}{4} f_{n+1} + \frac{1}{4} f_{n}\right),\,n = 0,\dots,\,N - 2. $$ It is easy to conclude that method (4.4) is stable, of order 2 [9]. Furthermore, using (3.2) and (4.3) we have $$ \varphi_ {1} = a _ {0} ^ {2} h ^ {2} + a _ {0} ^ {1} h + \varphi_ {0} = (- \frac {\lambda^ {2}}{4} \varphi_ {1} - \frac {\lambda^ {2}}{4} \alpha) h ^ {2} + \beta h + \alpha , $$ i.e., $$ \varphi_ {1} = \frac {4 - \lambda^ {2} h ^ {2}}{4 + \lambda^ {2} h ^ {2}} \alpha + \frac {4 h}{4 + \lambda^ {2} h ^ {2}} \beta , $$ recall the Taylor's formula $$ x (h) = \alpha + \beta h + \frac {1}{2} \ddot {x} (\xi) h ^ {2}, 0 < \xi < h, $$ we get $$ x (h) - \varphi_ {1} = \frac {2 \lambda^ {2} h ^ {2}}{4 + \lambda^ {2} h ^ {2}} \alpha + \frac {\lambda^ {2} h ^ {3}}{4 + \lambda^ {2} h ^ {2}} \beta + \frac {1}{2} \ddot {x} (\xi) h ^ {2} = \mathcal {O} (h ^ {2}), $$ which implies the starting values $\varphi_0, \varphi_1$ have second-order accuracy. Hence, by using Theorem 10.6 of chapter III in [9], we conclude that the convergent order of method (4.4) is 1. Thus, we obtain the following convergent theorem for numerical method II. Theorem 4.1: Let $\varphi(t)$ and $x(t)$ be the functions given by (3.1) and (1.1), respectively. Assume that $f(x) = -\lambda^2 x$, then we have the following error estimates, $$ | x(t_i) - \varphi(t_i) | \leq C h, i = 2, \dots, N, $$ where $C$ is a positive constant. ## V. NUMERICAL TESTS We shall apply our two methods to solve three problems given in [1, 8] and display the error results. $$ E x a m p l e 1 [ 1 ]. \quad \ddot {x} = x ^ {2}, t \in (0, 1); x (0) = 1; \dot {x} (0) = 1. $$ $$ E x a m p l e 2 [ 1 ]. \ddot {x} = - \sin x, t \in (0, 1); x (0) = \pi / 2; \dot {x} (0) = 0. $$ In our numerical experiments, we use numerical method I to calculate Example 1 and Example 2. In Tables 1-2 we give the error results with different $h$, where $\text{Error}:= |x(1) - x_N|$. These results indicate that numerical method I is effective. Table 1: Example 1: Numerical results for numerical method I <table><tr><td>h</td><td>Error</td></tr><tr><td>1/2</td><td>4.31e-1</td></tr><tr><td>1/4</td><td>1.76e-1</td></tr><tr><td>1/8</td><td>5.79e-2</td></tr><tr><td>1/16</td><td>1.66e-2</td></tr></table> $$ \text{Example} 3 [ 8 ]. \ddot{x} = - 1 0 0 x, t \in (0, 1); x (0) = 1; \dot{x} (0) = 0. $$ We apply numerical method II to solve Example 3 with different $h$. The errors and the convergent order are displayed in Table 3, where $\text{Error}:= x(1) - \varphi(1)|$. These results validate the conclusion in Theorem 4.1. Table 2: Example 2: Numerical results for numerical method I Table 3: Example 3: <table><tr><td>h</td><td>Error</td></tr><tr><td>1/2</td><td>1.43e-2</td></tr><tr><td>1/4</td><td>4.32e-3</td></tr><tr><td>1/8</td><td>1.16e-3</td></tr><tr><td>1/16</td><td>3.01e-4</td></tr></table> Numerical results for numerical method II <table><tr><td>h</td><td>Error</td><td>Convergent order</td></tr><tr><td>1/1000</td><td>2.83e-1</td><td>---</td></tr><tr><td>1/2000</td><td>1.39e-1</td><td>1.03</td></tr><tr><td>1/4000</td><td>6.87e-2</td><td>1.02</td></tr></table> ### ACKNOWLEDGMENT

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