3.6.1 Example: Transformation of a non linear problem into a linear problem

Motion of a pendulum:

PIC
\[s=l\varphi \qquad F_s=-mg\sin\varphi\]

Newton: \begin{eqnarray*}F_s=m\ddot{s}&\rightarrow&\;m\ddot{s}=-mg\sin\varphi\\ l\ddot\varphi+g\sin\varphi=0&\rightarrow&\ddot\varphi+\omega^2\sin\varphi=0 \qquad(\star)\\ &&\qquad\varphi(t)=?\;\mbox{function is looked for ?}\\ \end{eqnarray*}

Eq. (\(\star\)) is extremely complicated since it is non-linear because of the \(\sin\varphi\). For small \(\varphi\):
\(\sin\phi\approx\varphi\;\;\left|\sin\varphi-\varphi\right|\le\frac{\varphi^3}{3!}\approx10^{-3}\) for \(\varphi\le10^\circ\)

\[\rightarrow\;\ddot{f}(t)+\omega^2f(t)=0\quad\frac{d^2\varphi}{dt^2}+\omega^2\varphi(t)=0\;\leftarrow\,\mbox{linear equation}\]

See also exercises:

\[f(t)=A_1e^{i\omega t}+A_2e^{-i\omega t}=\varphi\cos(\omega t)+\frac{\varphi}{\omega}\sin(\omega t)\]


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© J. Carstensen (Math for MS)