20.2: Resonance near Double the Natural Frequency

From the above argument, the place to look for resonance is close to \(\Omega=2 \omega_{0}\). Landau takes

\(\begin{equation}
\ddot{x}+\omega_{0}^{2}\left[1+h \cos \left(2 \omega_{0}+\varepsilon\right) t\right] x=0
\end{equation}\)

and, bearing in mind that we’re looking for oscillations close to the natural frequency, puts in

\begin{equation}
x=a(t) \cos \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t+b(t) \sin \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t
\end{equation}

with \(a(t), b(t)\) slowly varying.

It’s important to realize that this is an approximate approach. It neglects nonresonant frequencies which must be present in small amounts, for example

\begin{equation}
\cos \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t \cos \left(2 \omega_{0}+\varepsilon\right) t=\frac{1}{2} \cos 3\left(\omega_{0}+\frac{1}{2} \varepsilon\right) t+\frac{1}{2} \cos \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t
\end{equation}

and the \(3\left(\omega_{0}+\frac{1}{2} \varepsilon\right)\) term is thrown away.

And, since the assumption is that \(a(t), b(t)\) are slowly varying, their second derivatives are dropped too, leaving just

\begin{equation}
\begin{array}{l}
\ddot{x}=-2 \dot{a}(t) \omega_{0} \sin \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t-a(t)\left(\omega_{0}^{2}+\omega_{0} \varepsilon\right) \cos \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t \\
+2 \dot{b}(t) \omega_{0} \cos \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t-b(t)\left(\omega_{0}^{2}+\omega_{0} \varepsilon\right) \sin \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t
\end{array}
\end{equation}

This must equal

\begin{equation}
-\omega_{0}^{2}\left[1+h \cos \left(2 \omega_{0}+\varepsilon\right) t\right]\left[a(t) \cos \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t+b(t) \sin \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t\right]
\end{equation}

Keeping only the resonant terms, we take \(\cos \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t \cdot \cos \left(2 \omega_{0}+\varepsilon\right) t=\frac{1}{2} \cos \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t\) and \(\sin \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t \cdot \cos \left(2 \omega_{0}+\varepsilon\right) t=-\frac{1}{2} \sin \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t\)

so this expression becomes

\begin{equation}
\begin{array}{l}
-\omega_{0}^{2}\left[1+h \cos \left(2 \omega_{0}+\varepsilon\right) t\right]\left[a(t) \cos \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t+b(t) \sin \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t\right] \\
=-\omega_{0}^{2}\left[a(t) \cos \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t+b(t) \sin \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t+\frac{1}{2} h a(t) \cos \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t-\frac{1}{2} h b(t) \sin \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t\right]
\end{array}
\end{equation}

The equation becomes:

\begin{equation}
\begin{aligned}
\ddot{x}=&-2 \dot{a}(t) \omega_{0} \sin \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t-a(t)\left(\omega_{0}^{2}+\omega_{0} \varepsilon\right) \cos \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t \\
&+2 \dot{b}(t) \omega_{0} \cos \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t-b(t)\left(\omega_{0}^{2}+\omega_{0} \varepsilon\right) \sin \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t \\
=&=\omega_{0}^{2}\left[a(t) \cos \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t+b(t) \sin \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t+\frac{1}{2} h a(t) \cos \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t-\frac{1}{2} h b(t) \sin \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t\right]
\end{aligned}
\end{equation}

The zeroth-order terms cancel between the two sides, leaving

\begin{equation}
\begin{array}{l}
-2 \dot{a}(t) \omega_{0} \sin \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t-a(t) \omega_{0} \varepsilon \cos \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t+2 \dot{b}(t) \omega_{0} \cos \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t-b(t) \omega_{0} \varepsilon \sin \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t \\
=-\omega_{0}^{2}\left[\frac{1}{2} h a(t) \cos \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t-\frac{1}{2} h b(t) \sin \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t\right]
\end{array}
\end{equation}

Collecting the terms in \(\sin \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t, \quad \cos \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t\)

\begin{equation}
-\left(2 \dot{a}+b \varepsilon+\frac{1}{2} h \omega_{0} b\right) \omega_{0} \sin \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t+\left(2 \dot{b}(t)-a \varepsilon+\frac{1}{2} h \omega_{0} a\right) \omega_{0} \cos \left(\omega_{0}+\frac{1}{2} \varepsilon\right) t=0
\end{equation}

The sine and cosine can’t cancel each other, so the two coefficients must both be identically zero. This gives two first order differential equations for the functions \(a(t), b(t)\), and we look for exponentially increasing functions, proportional to \(\begin{equation}
a(t)=a e^{s t}, \quad b(t)=b e^{s t}
\end{equation}\), which will be solutions provided

\begin{equation}
\begin{array}{l}
s a+\frac{1}{2}\left(\varepsilon+\frac{1}{2} h \omega_{0}\right) b=0 \\
\frac{1}{2}\left(\varepsilon-\frac{1}{2} h \omega_{0}\right) a-s b=0
\end{array}
\end{equation}

The amplitude growth rate is therefore

\begin{equation}
s^{2}=\frac{1}{4}\left[\left(\frac{1}{2} h \omega_{0}\right)^{2}-\varepsilon^{2}\right]
\end{equation}

Parametric resonance will take place if \(s\) is real, that is, if

\begin{equation}
-\frac{1}{2} h \omega_{0}<\varepsilon<\frac{1}{2} h \omega_{0}
\end{equation}

a band of width \(h \omega_{0} \text { about } 2 \omega_{0}\)