2 Second solution. No restriction on initial conditions. Resonance present in the solution

The same initial steps are repeated as in the first solution, up the equation to solve for \(x_{1}\left ( t\right ) \)

\begin{align} \ddot {x}_{1}+x_{1} & =\varphi \left ( \frac {\varphi ^{2}+\xi ^{2}}{4}-1\right ) \sin t+\xi \left ( 1-\frac {\xi ^{2}-\varphi ^{2}}{4}\right ) \cos t\tag {8}\\ & +\xi \left ( \frac {\xi ^{2}-3\varphi ^{2}}{4}\right ) \cos 3t+\varphi \left ( \frac {\varphi ^{2}-3\xi ^{2}}{4}\right ) \sin 3t\nonumber \end{align}

There exists four particular solutions relating to the above four forcing functions

\begin{align*} f_{1}\left ( t\right ) & =\varphi \left ( \frac {\varphi ^{2}+\xi ^{2}}{4}-1\right ) \sin t\\ f_{2}\left ( t\right ) & =\xi \left ( 1-\frac {\xi ^{2}-\varphi ^{2}}{4}\right ) \cos t\\ f_{3}\left ( t\right ) & =\xi \left ( \frac {\xi ^{2}-3\varphi ^{2}}{4}\right ) \cos 3t\\ f_{4}\left ( t\right ) & =\varphi \left ( \frac {\varphi ^{2}-3\xi ^{2}}{4}\right ) \sin 3t \end{align*}

For each of the above, we guess a particular solution, and determine the solution parameters. We guess \(x_{1,p1}=t\left ( c_{1}\sin t+c_{2}\cos t\right ) \),  \(x_{1,p2}=t\left ( c_{1}\sin t+c_{2}\cos t\right ) \), \(x_{1,p3}=c_{1}\sin 3t+c_{2}\cos 3t\), and \(x_{1,p4}=c_{1}\sin 3t+c_{2}\cos 3t\). By substituting each of the above into (8) one at a time and comparing coefficients, we arrive at the following solutions

\begin{align*} x_{1,p1} & =t\left ( \frac {1}{2}\varphi \left ( 1-\frac {\varphi ^{2}+\xi ^{2}}{4}\right ) \cos t\right ) \\ x_{1,p2} & =t\left ( \frac {1}{2}\xi \left ( 1-\frac {\xi ^{2}-\varphi ^{2}}{4}\right ) \sin t\right ) \\ x_{1,p3} & =\frac {1}{8}\xi \left ( \frac {\xi ^{2}-\varphi ^{2}}{4}-1\right ) \cos 3t\\ x_{1,p4} & =\frac {1}{8}\varphi \left ( \frac {3\xi ^{2}-\varphi ^{2}}{4}\right ) \sin 3t \end{align*}

\begin{align*} x_{1}\left ( t\right ) & =x_{1,h}+x_{1,p4}+x_{2,p4}+x_{3,p4}+x_{4,p4}\\ & =\left ( c_{1}\cos t+c_{2}\sin t\right ) +t\left ( \frac {1}{2}\varphi \left ( 1-\frac {\varphi ^{2}+\xi ^{2}}{4}\right ) \cos t\right ) +t\left ( \frac {1}{2}\xi \left ( 1-\frac {\xi ^{2}-\varphi ^{2}}{4}\right ) \sin t\right ) \\ & +\frac {1}{8}\xi \left ( \frac {\xi ^{2}-\varphi ^{2}}{4}-1\right ) \cos 3t+\frac {1}{8}\varphi \left ( \frac {3\xi ^{2}-\varphi ^{2}}{4}\right ) \sin 3t \end{align*}

Now applying initial conditions which are \(x_{1}\left ( 0\right ) =0\) and \(\dot {x}_{1}\left ( 0\right ) =0\) and determining \(c_{1}\) and \(c_{2}\) results in

\begin{align*} x_{1}\left ( t\right ) & =\left ( -\frac {1}{32}\xi ^{3}+\frac {1}{32}\xi \varphi ^{2}+\frac {1}{8}\xi \right ) \cos t-\frac {1}{32}\varphi \left ( 5\xi ^{2}-7\varphi ^{2}+16\right ) \sin t\\ & +t\left ( \frac {1}{2}\varphi \left ( 1-\frac {\varphi ^{2}+\xi ^{2}}{4}\right ) \cos t\right ) +t\left ( \frac {1}{2}\xi \left ( 1-\frac {\xi ^{2}-\varphi ^{2}}{4}\right ) \sin t\right ) \\ & +\frac {1}{8}\xi \left ( \frac {\xi ^{2}-\varphi ^{2}}{4}-1\right ) \cos 3t+\frac {1}{8}\varphi \left ( \frac {3\xi ^{2}-\varphi ^{2}}{4}\right ) \sin 3t \end{align*}

We can now write the final solution for \(x\left ( t\right ) \) as

\begin{align*} x\left ( t\right ) & =x_{0}\left ( t\right ) +\alpha x_{1}\left ( t\right ) \\ & =\varphi \cos t+\xi \sin t+\\ & \alpha \left ( -\frac {1}{32}\xi ^{3}+\frac {1}{32}\xi \varphi ^{2}+\frac {1}{8}\xi \right ) \cos t-\alpha \frac {1}{32}\varphi \left ( 5\xi ^{2}-7\varphi ^{2}+16\right ) \sin t\\ & +\alpha t\left ( \frac {1}{2}\varphi \left ( 1-\frac {\varphi ^{2}+\xi ^{2}}{4}\right ) \cos t\right ) +\alpha t\left ( \frac {1}{2}\xi \left ( 1-\frac {\xi ^{2}-\varphi ^{2}}{4}\right ) \sin t\right ) \\ & +\alpha \frac {1}{8}\xi \left ( \frac {\xi ^{2}-\varphi ^{2}}{4}-1\right ) \cos 3t+\alpha \frac {1}{8}\varphi \left ( \frac {3\xi ^{2}-\varphi ^{2}}{4}\right ) \sin 3t \end{align*}

To illustrate this solution, we show the phase plot which results from the above solution, and also plot the solution \(x\left ( t\right ) \). We select \(x\left ( 0\right ) =1.5\) and \(\dot {x}\left ( 0\right ) =1.3229\). We note that the same initial conditions are used as in the earlier solution to compare both solutions.

Figure 2: Rresonance

We see clearly the effect of including resonance particular solutions. The limit cycle boundary is increasing with time.

The effect becomes more clear if we plot the solution \(x\left ( t\right ) \) itself, and compare it to the solution earlier with no resonance. Below, We show \(x(t)\) from both solutions. We clearly see the effect of including the resonance terms. This is the solution with no resonance terms

Figure 3: No rresonance terms

This is the solution with resonance terms

Figure 4: Rresonance terms

In the second solution (with resonance), there is no restriction on initial conditions. Hence we can for example look at the solution starting from any \(x\left ( 0\right ) \) and \(\dot {x}\left ( 0\right ) \). This is the phase plot for \(x\left ( 0\right ) =5\) and \(\dot {x}\left ( 0\right ) =2\). This would not be possible using the first solution, due to the restriction on \(x\left ( 0\right ) ^{2}\) \(+\) \(\dot {x}\left ( 0\right ) ^{2}\) having to be equal to \(4\)

Figure 5: phase plot