Let \(S\) be a real-valued function on the \(xy\)-plane. The gradient of \(S\) is
\begin{equation*}
\nabla S = \left( \frac{\partial S}{\partial x}, \frac{\partial S}{\partial y}
\right).
\end{equation*}
The system
\begin{align*}
\frac{dx}{dt} & = f(x, y)\\
\frac{dy}{dt} & = g(x, y)
\end{align*}
is a gradient system if
\begin{align*}
f(x, y) & = \frac{\partial S}{\partial x}\\
g(x, y) & = \frac{\partial S}{\partial y}.
\end{align*}
For example, the system
\begin{align*}
\frac{dx}{dt} & = x - x^3\\
\frac{dy}{dt} & = -y
\end{align*}
is a gradient system, where
\begin{equation*}
S(x, y) = \frac{x^2}{2} - \frac{x^4}{4} - \frac{y^2}{2} + 8.
\end{equation*}
Now, let us see what happens on the solution curves of this gradient system. If \((x(t), y(t))\) is a solution curve,
\begin{equation*}
\frac{dS}{dt} = \frac{\partial S}{\partial x} \frac{dx}{dt} + \frac{\partial
S}{\partial y} \frac{dy}{dt} = (x - x^3)^2 + y^2 \geq 0.
\end{equation*}
Thus, \(S\) increases at the point on the solution curve where the gradient of \(S\) is nonzero. That is, \(S\) increases at every point on the solution curve except at the equilibrium points.