Integrating second order ODE and finding fixed points

Question

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

Answer 1

From my understanding, you have the following system:

$x''(t)+\sin(x(t))=C$

for which you would like to find fixed points, where $x''(t)=x'(t)=0$ for all t, such that $x(t)=x_0$ (i.e., x is a constant function which does not change with t).

You have already pointed out that any fixed point satisfies $\sin(x_0)=C$, and it seems that your confusion stems from attempting to integrate the differential equation with respect to x, done as follows:

$v(t)=x'(t)$

$x''(t)=v'(t)=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}=\frac{d}{dx}(\frac{1}{2}v^2)$

$x''(t)+\sin(x(t))=\frac{d}{dx}(\frac{1}{2}v^2)+\sin(x)=C \implies \frac{1}{2}v^2-\cos(x)=Cx+C_2$, where $C_2$ is a constant of integration.

There are two problems with this. First, in our integration, we cannot integrate by x directly, because x is a function of t. For the $\sin(x)$ term, for example, the integral becomes $\int^{t_1}_{t_0} \sin(x(t))x'(t)\mathrm{d}t$, where we then do a u-sub $u=x(t)$ and complete the integral quite simply. However, a fixed point requires $x'(t)=0$, so our entire integral is 0, and in fact the rigorous integrating process yields the equation 0=0, which does not give us any new information about the system. Second, even if the integration were justified, it still would not be useful. We know that $x'(t)=v(t)=0$, which means our integrated expression becomes:

$\frac{1}{2}v^2-\cos(x)=Cx+C_2 \implies -\cos(x_0)=Cx_0+C_2 \implies C_2=-\cos(x_0)-Cx_0$. We have defined a new constant $C_2$, but we have not learned anything new about $x(t)$ or $x_0$.

In conclusion, your book is right in that $\sin(x)=C$ is the only equation you need to fully describe the fixed points of your system.

Answer 2

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 $x_2=0$.