1 First solution: Restriction on initial condition. No resonance

The Van Der Pol equation is

\[ \ddot {x}-\alpha \left ( 1-x^{2}\right ) \dot {x}+x=0 \]

Initial conditions are \(x(0)=\varphi \) and \(\dot {x}(0)=\xi \).

If \(\alpha \lll 1\), then the equation becomes \(\ddot {x}_{0}+x_{0}=0\) which has an exact solution. We perturb the exact solution for \(\ddot {x}_{0}+x_{0}=0\) to approximate the solution of the nonlinear equation \(\ddot {x}-\alpha \left ( 1-x^{2}\right ) \dot {x}+x=0\) in terms of perturbation parameter \(\alpha \) Let the solution of the Van Der Pol equation be \(x(t)\) and the solution to the exact equation \(\ddot {x}_{0}+x_{0}=0\) be \(x_{0}(t)\), then

\[ x(t)=x_{0}(t)+\alpha x_{1}(t)+\alpha ^{2}x_{2}(t)+\cdots \]

First order approximation results in

\[ x(t)=x_{0}(t)+\alpha x_{1}(t) \]

Now we need to determine \(x_{0}\left ( t\right ) \) and \(x_{1}(t)\). Substituting (2) into (1) gives

\[ \ddot {x}_{0}+\alpha \ddot {x}_{1}-\alpha \dot {x}_{0}+\alpha x_{0}^{2}\dot {x}_{0}+\alpha ^{3}x_{1}^{2}\dot {x}_{0}+2\alpha ^{2}x_{1}x_{0}\dot {x}_{0}-\alpha ^{2}\dot {x}_{1}+\alpha ^{2}x_{0}^{2}\dot {x}_{1}+\alpha ^{4}x_{1}^{2}\dot {x}_{1}+2\alpha ^{3}x_{1}x_{0}\dot {x}_{1}+x_{0}+\alpha x_{1}=0 \]

Collecting terms with same power of \(\alpha \) gives

\[ \alpha ^{0}\left ( \ddot {x}_{0}+x_{0}\right ) +\alpha \left ( \ddot {x}_{1}-\dot {x}_{0}+x_{0}^{2}\dot {x}_{0}+x_{1}\right ) +\alpha ^{2}\left ( x_{0}^{2}\dot {x}_{1}+2x_{1}x_{0}\dot {x}_{0}-\dot {x}_{1}\right ) +\alpha ^{3}\left ( x_{1}^{2}\dot {x}_{0}+2x_{1}x_{0}\dot {x}_{1}\right ) +\alpha ^{4}x_{1}^{2}\dot {x}_{1}=0 \]

Setting terms which multiply by higher power \(\alpha \) to zero and since it is assumed that \(\alpha \) is very small therefore

\[ \left ( \ddot {x}_{0}+x_{0}\right ) +\alpha \left ( \ddot {x}_{1}-\dot {x}_{0}+x_{0}^{2}\dot {x}_{0}+x_{1}\right ) =0 \]

For the LHS to be zero implies that

\[ \ddot {x}_{0}+x_{0}=0 \]

And

\[ \ddot {x}_{1}-\dot {x}_{0}+x_{0}^{2}\dot {x}_{0}+x_{1}=0 \]

Equation (3) is solved for \(x_{0}\) and (4) is solved for \(x_{1}\) in order to find the solution \(x(t)\). Solution to (3) is

\[ x_{0}(t)=A_{0}\cos t+B_{0}\sin t \]

Assuming initial conditions for \(x_{1}(t)\) are zero, then initial conditions for \(x_{0}(t)\) can be taken to be those given for \(x(t)\). Solving for \(A_{0},B_{0}\) gives

\[ x_{0}(t)=\varphi \cos t+\xi \sin t \]

Substituting the solution just found (and its derivative) into (4) gives

\begin{align*} \ddot {x}_{1}+x_{1} & =-\varphi \sin t+\xi \cos t-\xi ^{3}\cos t\sin ^{2}t-2\xi ^{2}\varphi \cos ^{2}t\sin t+\xi ^{2}\varphi \sin ^{3}t\\ & -\xi \varphi ^{2}\cos ^{3}t+\allowbreak 2\xi \varphi ^{2}\cos t\sin ^{2}t+\varphi ^{3}\cos ^{2}t\sin t \end{align*}

Using \(\sin t\cos ^{2}t=\frac {1}{4}\left ( \sin t+\sin 3t\right ) \) and \(\cos t\sin ^{2}t=\frac {1}{4}(\cos t-\cos 3t)\) the above can be simplified to

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

Using \(\xi \sin ^{3}t-\varphi \cos ^{3}t=\frac {1}{4}\left ( -3\varphi \cos t-\varphi \cos 3t+3\xi \sin t-\xi \sin 3t\right ) \) the above can be simplified further to

\begin{align} \ddot {x}_{1}+x_{1} & =\left ( -\varphi -\frac {\xi ^{2}\varphi }{2}+\frac {\varphi ^{3}}{4}+\frac {3\xi ^{2}\varphi }{4}\right ) \sin t+\left ( \xi -\frac {\xi ^{3}}{4}+\frac {\allowbreak \xi \varphi ^{2}}{2}-\frac {3}{4}\xi \varphi ^{2}\right ) \cos t\nonumber \\ & +\left ( \frac {\xi ^{3}}{4}-\frac {\allowbreak \xi \varphi ^{2}}{2}-\frac {\xi \varphi ^{2}}{4}\right ) \cos 3t+\left ( -\frac {\xi ^{2}\varphi }{2}+\frac {\varphi ^{3}}{4}-\frac {\xi ^{2}\varphi }{4}\right ) \sin 3t\tag {5}\end{align}

The above is the equation we will now solve for \(x_{1}\left ( t\right ) \), which has four forcing functions, hence four particular solutions. two of these particular solutions will cause a complete solution blows up as time increases (the first two in the RHS shown above). If we however restrict the initial conditions such that

\[ \left ( -\varphi -\frac {\xi ^{2}\varphi }{2}+\frac {\varphi ^{3}}{4}+\frac {3\xi ^{2}\varphi }{4}\right ) =0 \]

And

\[ \left ( \xi -\frac {\xi ^{3}}{4}+\frac {\allowbreak \xi \varphi ^{2}}{2}-\frac {3}{4}\xi \varphi ^{2}\right ) =0 \]

Then those forcing function will vanish. The above two equation result in the same restriction, which can be found to be

\[ 4=\xi ^{2}+\varphi ^{2}\]

By restricting \(\xi ^{2}+\varphi ^{2}\) to be exactly \(4\), then the solution we obtain for \(x_{1}\left ( t\right ) \) from (5) will not blow up. Hence (4) can now be rewritten without the two forcing functions which caused resonance resulting in

\[ \ddot {x}_{1}+x_{1}=\xi \left ( \frac {\xi ^{2}-3\varphi ^{2}}{4}\right ) \cos 3t+\varphi \left ( \frac {\varphi ^{2}-3\xi ^{2}}{4}\right ) \sin 3t \]

But \(\varphi ^{2}=4-\xi ^{2}\) and \(\xi ^{2}=4-\varphi ^{2}\), hence the above becomes after some simplification

\begin{equation} \ddot {x}_{1}+x_{1}=\xi \left ( \xi ^{2}-3\right ) \cos 3t+\varphi \left ( \varphi ^{2}-3\right ) \sin 3t\tag {6}\end{equation}

The homogeneous solution to the above is

\[ x_{1,h}=c_{1}\cos t+c_{2}\sin t \]

And we guess two particular solutions \(x_{1,p1}=a_{1}\cos 3t+a_{2}\sin 3t\) and \(x_{1,p2}=d_{1}\cos 3t+d_{2}\sin 3t\). By substituting each of these particular solutions into (6) one at a time, and comparing coefficients, we can determine \(a_{1},a_{2},d_{1},d_{2}\). This results in

\[ x_{1,p1}=\frac {\xi \left ( 3-\xi ^{2}\right ) }{8}\cos 3t \]

Similarly,

\[ x_{1,p2}=\frac {\varphi \left ( 3-\varphi ^{2}\right ) }{8}\sin 3t \]

Hence the solution to (6) becomes

\begin{align} x_{1}\left ( t\right ) & =x_{1,h}+x_{1,p1}+x_{1,p2}\nonumber \\ & =\left ( c_{1}\cos t+c_{2}\sin t\right ) +\frac {\xi \left ( 3-\xi ^{2}\right ) }{8}\cos 3t+\frac {\varphi \left ( 3-\varphi ^{2}\right ) }{8}\sin 3t\tag {7}\end{align}

Now applying initial conditions, which are \(x_{1}\left ( 0\right ) =0\) and \(\dot {x}_{1}\left ( 0\right ) =0\), to (7) and determining \(c_{1}\) and \(c_{2}\) results in

\[ x_{1}\left ( t\right ) =\frac {\xi \left ( \xi ^{2}-3\right ) }{8}\cos t+\frac {3}{8}\varphi \left ( \varphi ^{2}-3\right ) \sin t+\frac {\xi \left ( 3-\xi ^{2}\right ) }{8}\cos 3t+\frac {\varphi \left ( 3-\varphi ^{2}\right ) }{8}\sin 3t \]

Therefore we now have the final perturbation solution, which is

\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 {\xi \left ( \xi ^{2}-3\right ) }{8}\cos t+\frac {3}{8}\varphi \left ( \varphi ^{2}-3\right ) \sin t+\frac {\xi \left ( 3-\xi ^{2}\right ) }{8}\cos 3t+\frac {\varphi \left ( 3-\varphi ^{2}\right ) }{8}\sin 3t\right ) \end{align*}

Where \(x\left ( 0\right ) =\varphi \) and \(\dot {x}\left ( 0\right ) =\xi \) and the above solution is valid under the restriction that

\[ x^{2}\left ( 0\right ) +\dot {x}^{2}\left ( 0\right ) =4 \]

We notice that the above restriction implies that both \(x\left ( 0\right ) \) and \(\dot {x}\left ( 0\right ) \) have to be less than \(4\) to avoid getting a quantity which is complex. This implies that his solution is valid near the center of the phase plane only. In addition, it is valid only for small \(\alpha \). To illustrate this solution, We show the phase plot which results from the above solution, and also plot the solution \(x\left ( t\right ) \). Selecting \(x\left ( 0\right ) =1.5\) and \(\dot {x}\left ( 0\right ) =\sqrt {4-1.5^{2}}=1.3229\)

Figure 1: No resonance