Example 6.5.11.
As a second example, suppose
\begin{align*}
\frac{dx}{dt} & = -x^2 \cos y \sin x\\
\frac{dy}{dt} & = x^2 \cos x \sin y + 2x(\sin x \sin y - 1).
\end{align*}
Since
\begin{equation*}
\frac{\partial f}{\partial x} = x^2 \cos x \cos y - 2x \cos y \sin x = -
\frac{\partial g}{\partial y}
\end{equation*}
our system is Hamiltonian. To find a Hamiltonian, we compute
\begin{equation*}
H(x, y) = \int f(x, y) \, dy + \phi(x) = \int (-x^2 \cos y \sin x) \, dy +
\phi(x) = -x^2 \sin x \sin y + \phi(x).
\end{equation*}
Thus,
\begin{align*}
\phi'(x) & = - g(x, y) - \frac{\partial}{\partial x} \int f(x, y) \,
dy\\
& = -(x^2 \cos x \sin y + 2x(\sin x \sin y - 1)) - ( 2x \sin x \sin y
- x^2 \cos x \sin y)\\
& = 2x,
\end{align*}
and \(\phi(x) = x^2 + C\text{.}\) Letting \(C = 0\text{,}\) our Hamiltonian function is
\begin{equation*}
H(x, y) = x^2 - x^2 \sin x \sin y = x^2(1 - \sin x \sin y).
\end{equation*}
Solution curves for this system are given in Figure 6.5.12.
solution curves as level sets on a direction field of slope arrows