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In the twenty third Feynman lecture, the solution of the following differential equation is discussed:

$$ \frac{d^2 x}{dt^2} + \frac{kx}{m} = \frac{F}{m}$$

AFter 'complexifying' this differential equation, he gets:

$$ \frac{d^2 x}{dt^2} + \frac{kx}{m} = \frac{\hat{F} e^{iwt} }{m}$$

And then it is written that we can write:

$$ x = |x| e^{i \omega t}$$

Assuming $x$ is a complex number and this leads to:

$$ \frac{dx}{dt} = i \omega x$$

However, the above result assumes that $|x|$ is constant, how do we rigorously justify this assumption?

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  • $\begingroup$ Related : Need help understanding an equation of motion for a pendulum. $\endgroup$ – Frobenius Mar 24 at 21:00
  • $\begingroup$ I'm not sure understand what you're looking for vis-à-vis a 'rigorous justification'. Isn't it just that one is seeking solutions of such form in order to be able to express a general solution over such a basis? $\endgroup$ – Alfred Centauri Mar 24 at 23:33
  • $\begingroup$ Hmm , the concept of solution basis and other's feels outside the text which I have linked. @AlfredCentauri $\endgroup$ – Buraian Mar 25 at 7:08
  • $\begingroup$ No effort then to clarify what it is that you're looking for? $\endgroup$ – Alfred Centauri Mar 26 at 2:48
  • $\begingroup$ I mean, how exactly do you think I can clarify the question? If you see the text of Feynman, he explains without bringing up the concepts you've mentioned (however it is unclear to me) $\endgroup$ – Buraian Mar 26 at 5:28
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$x$ and $F$ can each be expressed as a Fourier integral: $$x(t)=\int x(\omega)e^{i\omega t} d\omega$$ $$F(t)=\int F(\omega)e^{i\omega t} d\omega$$ This of course assumes that $F(t)$ is square integrable ($L^2$ Hilbert space).

The rest is just substitution into the equations of motion and comparison of coefficients (which is possible due to the perpendicularity of the Fourier basis functions and linearity of the equations of motion).

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    $\begingroup$ It must also be noted that this per-Fourier-coefficient analysis works only thanks to linearity of the equation. $\endgroup$ – Ruslan Mar 24 at 21:54
  • $\begingroup$ Absolutely true. Thanks! $\endgroup$ – oliver Mar 24 at 22:08

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