Help with transforming a second order ode into a system of first order ode's

Question

I have the following equation:

\begin{align*} \frac{d^2\theta}{dt}=\alpha(\theta-1)+\beta(\theta-1)^3-\gamma\frac{d\theta}{dt} \tag{1} \end{align*}

Where $\alpha, \beta, \gamma \in \mathbb{R}$.

This is my solution in attempting to convert (1) into a system of first order ODE's:

\begin{align*} \frac{d\theta}{dt}&=\phi \tag{2} \\ \\ \frac{d\phi}{dt} &= \alpha(\theta-1)+\beta(\theta-1)^3-\gamma\phi \tag{3} \end{align*}

Is the above system correct? Also in equation (3) is it okay to have the $\theta$ term?

Answer 1

We are given:

\begin{align*} \frac{d^2\theta}{dt}=\alpha(\theta-1)+\beta(\theta-1)^3-\gamma\frac{d\theta}{dt} \tag{1} \end{align*}

Where $\alpha, \beta, \gamma \in \mathbb{R}$.

We can proceed as follows:

• $x_1 = \theta \implies x_1' = \theta' = x_2$
• $x_2 = \theta' \implies x_2' = \theta'' = \alpha(\theta-1)+\beta(\theta-1)^3-\gamma\frac{d\theta}{dt} = \alpha(x_1-1) + \beta(x_1 - 1)^3 -\gamma ~x_2$

So, our new system is:

$$\begin{align} x_1' & = x_2 \\ x_2' & = \alpha(x_1-1) + \beta(x_1 - 1)^3 -\gamma ~x_2 \end{align}$$

Note, given initial conditions, similarly follow as:

• $\theta(t_0) = a \implies x_1(t_0) = a$
• $\theta'(t_0) = b \implies x_2(t_0) = b$

Answer 2

Yes it's fine. Here, you have taken the states of the system as $\{x_1 = \theta, x_2 = \phi\}$ and expressed

$\dot{x_1} = f(x_1,x_2)$, $\dot{x_2} = g(x_1, x_2)$