We know that fixed points are such points, where $x'$ = $x''$ = $0$.
Let's say we are given second order ODE $:$ $ x'' + sin(x) = C$
We can rewrite it into system of first order ODEs:
$ \begin{cases} x_1' = x_2 \\ x_2' = C – sin(x_1) \end{cases} $
Now, according to this equation fixed points occur at $sin(x_1) = C$
But if we take the first integral of the system we will find that $ \frac {(x_2)^2}{2} – cos(x_1) = C \cdot x_1 + Constant$
Now, points where $sin(x) = C$ don't satisfy the equation above (phase portrait confirms that). Yet the book claims that the fixed points are arrived from first equation.
Could you help wih my confusion? Thanks
Best Answer
The fixed points occur when $\sin(x_1)=C$ and $x_2=0$. When those conditions are met, $x_1'$ and $x_2'$ are both 0. I'm placing a streamline plot with C=0.5 below.
As in Marcus's answer, I get that $U(x(t))= \frac{ x_2^2(t)}2 - C\, x_1(t) - \cos(x_1(t))$ is constant for any solution $x(t)$ of the differential equation (i.e $U(x)= \frac{ x_2^2}2 - C\, x_1 - \cos(x_1)$ is a first integral of the system). The fixed points of the ODE occur at the critical points of $U(x)$ which occur exactly when $\sin(x_1)=C$ and $x_2=0$.