Can we solve Newton's second law \begin{equation} m\frac{d^2}{dt^2}x(t)=F(x(t),t), \label{eq:1} \end{equation} a second-order ODE, without resorting to Euler's method which re-casts the problem as two coupled, first-order ODEs?

Here, we employ the initial condition $x(t_0)=x_0$ and $v(t_0)=v_0$, and restrict our analysis to one dimension. Also, we assume for now that the force $F$ does not depend on the velocity $v$.

We can divide equation \eqref{eq:1} by the mass $m$ and write instead $$ \frac{d^2}{dt^2}x(t)=a(x(t),t), $$ where $a(x,t)$ is the acceleration.

**Example 1:**
\begin{equation}
\frac{d^2x}{dt^2}=-x+x^3+0.1\cos(t),
\label{eq:2}
\end{equation}
with the initial conditions $x(t_0)=0$ and $v(t_0)=0$.

In a fairly crude way, we now approximate the second derivative of $x(t)$ with respect
to time, i.e. the acceleration, by
\begin{equation}
\begin{aligned}
\frac{d^2x}{dt^2}\bigg|_{t=t_n}=\frac{dv}{dt}\bigg|_{t=t_n}&
\approx&\frac{v(t_n+h/2)-v(t_n-h/2)}{h} \\
&\approx&\frac{
\frac{x(t_n+h)-x(t_n)}{h}
-\frac{x(t_n)-x(t_n-h)}{h}
}{h},
\end{aligned}
\label{eq:3}
\end{equation}
where we used the **central difference approximation(s)**
$$
\frac{dv}{dt}\bigg|_{t=t_n}\approx\frac{v(t_n+h/2)-v(t_n-h/2)}{h}
$$
with
$$v(t_n+h/2)\approx \frac{x(t_n+h)-x(t_n)}{h},\\
v(t_n-h/2)\approx \frac{x(t_n)-x(t_n-h)}{h}.$$

The above formulation follows the usual discretization $$ t_n=t_0+n\cdot h~~~\mathrm{with}~~~n=0,1,2,3,...,N. $$ Again, this suggests the following abbreviation: $x_n=x(t_n)$.

With this in mind, substitution of expression \eqref{eq:3} into equation \eqref{eq:2} and subsequent multiplication by $h^2$ yields $$ x_{n+1}-2x_n+x_{n-1}=h^2\left[-x_n+x_n^3+0.1\cos(t_n)\right], $$ where we evaluated equation \eqref{eq:2} at $t=t_n$. This results in the following recursive formulation of the solution \begin{equation} x_{n+1}=2x_n-x_{n-1}+h^2\left[-x_n+x_n^3+0.1\cos(t_n)\right] \label{eq:4} \end{equation} with $x_0=0$ and $v_0=0$.

More rigorously, one can derive the generalized version of the recursive formula \eqref{eq:4}, namely

\begin{equation} \boxed{x_{n+1}=2x_n-x_{n-1}+h^2 a(x_n,t_n),} \label{eq:5} \end{equation}by use of Taylor expansions about $x(t_n)$ (see Appendix). The Taylor-expansion method tells us that the local error we make in using the approximation \eqref{eq:5}, scales like $h^4$. This is, perhaps, surprisingly good.

The method \eqref{eq:5} is called **Verlet integration**.
Often, it includes an acceleration term which does not depend explicitly on time:
$a=a(x(t))$.

The remaining problem is now to use equation \eqref{eq:4} at $t=t_0$, i.e. at $n=0$, since we do not know the value of $x$ for $t<0$, in particular $x_{-1}$. Here, one can estimate the first value $x_1$ after the starting point $t_0$ by Taylor expansion of $x(t)$ at $t=t_0$: $$ x_1=x_0+h v_0 + \frac{h^2}{2}a(x_0,t_0)+\mathcal{O}(h^3). $$ In our example, this becomes ($x_0=0$, $v_0=0$, $t_0=0$) \begin{equation} x_1=x_0+h v_0 + \frac{h^2}{2}a(x_0,t_0)=\frac{h^2}{20}. \label{eq:6} \end{equation} Thereafter, we can stick to the relation \eqref{eq:4} to find all other $x_n$.

(__Note__: The error of the approximation \eqref{eq:6} scales like $h^3$.)

Let's use \eqref{eq:4} and \eqref{eq:6} to solve \eqref{eq:2}:

In [2]:

```
%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
# Set common figure parameters
newparams = {'figure.figsize': (16, 6), 'axes.grid': True,
'lines.linewidth': 1.5, 'lines.linewidth': 2,
'font.size': 14}
plt.rcParams.update(newparams)
```

In [3]:

```
N = 100000 # number of steps
h = 0.001 # step size
t = np.zeros(N+1)
x = np.zeros(N+1)
# initial values
t_0 = 0
x_0 = 0
t[0] = t_0
x[0] = x_0
for n in range(N):
# Second grid point:
if n==0:
t[n+1] = h
x[n+1] = h**2/20.0
# Verlet integration
else:
t[n+1] = t[n] + h
x[n+1] = 2.0*x[n] - x[n-1] + h**2*(-x[n]+x[n]**3+0.1*np.cos(t[n]))
# Plot the solution
plt.plot(t,x)
plt.ylabel(r'$x(t)$')
plt.xlabel(r'$t$')
plt.grid();
```

Here we have chosen $h=0.001$ and $N=100\,000$, and so $t_N=100$. In the plot of $x(t)$, the discrete points have been connected by straight lines.

What if the force, and hence the acceleration, also depends on the velocity $v$? In other words, let's look at $$ \frac{d^2}{dt^2}x(t)=a(x(t),v(t),t), $$ with the initial conditions $x(t_0)=x_0$ and $v(t_0)=v_0$.

We approximate the left-hand side at $t=t_n$ just like before, and we approximate the right-hand side by $$ a(x_n,v_n,t_n)\approx a\left( x_n, \frac{x_{n+1}-x_{n-1}}{2h}, t_n \right), $$ where we used a central difference approximation for $v_n$ again: $$ \frac{dv}{dt}\bigg|_{t=t_n}\approx\frac{v(t_n+h)-v(t_n-h)}{2h}. \label{eq:7} $$

However, the error of the approximation \eqref{eq:7} scales like $h^2$, thereby reducing the overall accuracy of the method.

Again, we estimate $$ x_1=x_0+h v_0 + \frac{h^2}{2}a(x_0,v_0,t_0). $$ Thereafter, we now use $$ x_{n+1}=2x_n-x_{n-1}+h^2 a\left( x_n, \frac{x_{n+1}-x_{n-1}}{2h}, t_n \right) . $$ Note that we need to solve this equation for $x_{n+1}$ which is not always possible in analytical form, i.e. in closed form, if the problem is nonlinear.

Let us focus on a problem which is linear in $v$.

**Example 2**:
$$
\frac{d^2x}{dt^2}=-v-x^3
$$
with the initial conditions $x_0=x(0)=10$ and $v_0=v(0)=0$.
Using our above formalism, this reads in discretized
form
\begin{align*}
x_{n+1}&=2x_n-x_{n-1}+h^2 \left( -\frac{x_{n+1}-x_{n-1}}{2h}-x_n^3 \right) \\
\Rightarrow x_{n+1}&=\frac{2x_n-(1-h/2)\,x_{n-1}-h^2 x_n^3}{1+h/2}.
\end{align*}
When combined with $x_0=10$ and
$$
x_1=x_0+h v_0 + \frac{h^2}{2}a(x_0,v_0,t_0)=10-500 h^2,
$$
this determines the solution of the problem. This is implemented in the code below. In the code we choose $h=0.001$ and
$N=3\,000$, and so $t_N=3.0$

In [4]:

```
N = 3000 # number of steps
h = 0.001 # step size
t = np.zeros(N+1)
x = np.zeros(N+1)
# initial values
t_0 = 0
x_0 = 10
t[0] = t_0
x[0] = x_0
for n in range(N):
# Second grid point:
if n==0:
t[n+1] = h
x[n+1]= 10.0-500.0*h**2
# Verlet integration
else:
t[n+1] = t[n] + h
x[n+1] = (2.0*x[n]-(1.0-h/2)*x[n-1]-h**2*x[n]**3)/(1+h/2.0)
# Plot the solution
plt.plot(t,x)
plt.ylabel(r'$x(t)$')
plt.xlabel(r'$t$')
plt.grid();
```

We have seen that there is more than one method (Euler's method) to solve ODEs. Verlet integration is often used to solve Newton's second law.

- It provides higher accuracy than Euler's method.
- It is more stable than Euler's method and
- might require a nonlinear solve for $x_{n+1}$.
- It can be extended in a straightforward manner to three-dimensional motion.

Given the dynamics of $x(t)$ at $t=t_n$, let us approximate the values $x_{n-1}=x(t_n-h)$ and $x_{n+1}=x(t_n+h)$ by use of Taylor series at $x_{n}=x(t_n)$: \begin{eqnarray*} x(t_n-h)&=&x(t_n)+v(t_n) \cdot (-h)+\frac{a(t_n)}{2}\cdot (-h)^2+\frac{b(t_n)}{6}\cdot (-h)^3+\mathcal{O}(h^4), \\ x(t_n+h)&=&x(t_n)+v(t_n)\,h+\frac{a(t_n)}{2}h^2+\frac{b(t_n)}{6}h^3+\mathcal{O}(h^4), \end{eqnarray*} with \begin{align*} v(t_n)=\frac{dx}{dt}\bigg|_{t=t_n}, \quad a(t_n)=\frac{d^2x}{dt^2}\bigg|_{t=t_n}, \quad \text{and} \quad b(t_n)=\frac{d^3x}{dt^3}\bigg|_{t=t_n}. \end{align*} Adding these two equations and subsequent re-arranging of terms yields $$ x(t_n+h)=2x(t_n)-x(t_n-h)+a(t_n)\,h^2+\mathcal{O}(h^4) $$ or, in different notation, $$ x_{n+1}=2 x_n-x_{n-1}+h^2\,a(t_n)+\mathcal{O}(h^4). $$ This is our Verlet formula \eqref{eq:5} if we drop the higher-order terms $\mathcal{O}(h^4)$ and write $a(x(t_n),t_n)=a(x_n,t_n)$ instead of simply $a(t_n)$.