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Consider a mass $m$ is placed on a horizontal level surface and attached to a spring, whose other end is attached to a vertical wall. The mass moves in a viscous medium, where the resistance force acting on it is proportional to the velocity: $\vec{F}_p=-\gamma\vec{\nu}$, where $\gamma$ is a known positive constant. Due to aging, the spring stiffness coefficient decreases exponentially over time: $k\left ( t\right ) = k_0\mathrm{e}^{-\alpha t}$, where $k_0$ and $\alpha$ are other known positive constants. Write down the differential equation describing the motion of the equilibrium position derived mass (in the one-dimensional case) and find its general solution $x(t)$, expressing it in terms of Bessel functions. Hint: Use the substitution $s=\mathrm{e}^{-\alpha t/2}.$

I tried applying Newton's second law and got $m\ddot{x}=-k_0e^{-\alpha t}x-\gamma\dot{x} \implies \ddot{x}=-\frac{k_0}me^{-\alpha t}x-\frac\gamma m\dot{x}$, if correct. Then I denote $\begin{aligned}&\omega_0=\sqrt{k_0/m}. , \beta=\frac\gamma m\end{aligned}$, and get $\ddot{x}=-\omega_0^2e^{-\alpha t}x-\beta\dot{x}$.

I don't have ideas on how I apply the substitution and where to use the Bessel functions? Any help?

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  • $\begingroup$ By "use the substitution $s=e^{-\alpha t/2}$ they mean that you can think of $x$ as a function not of $t$ but of $s$. (Note that, so long as $\alpha>0$, $s$ is a strictly decreasing of $t$. Hence for any given $t$ there's a well-defined $s$ and vice-versa. Otherwise, this kind of substitution in the dependent variable would make little sense.) The chain rule (and some messy algebra) then allows one to express the time derivatives in terms of $s$-derivatives. This yields a new (and hopefully tractable) differential equation. $\endgroup$ Apr 15 at 21:11
  • $\begingroup$ Yes thank you I made a fix $\endgroup$
    – gujaral
    Apr 15 at 21:42

1 Answer 1

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Applying the substitution $s = \mathrm{e}^{-\alpha t / 2}$ in the hint transforms the differential equation to $$s^2 x_{ss} + \left(1-\frac{2 \beta}{\alpha}\right) s x_s + \frac{4 \omega_0^2}{\alpha^2} s^2 x = 0,$$ which, except for the coefficients of the first- and zeroth-order terms, is already the Bessel equation.

Hint We can adjust the zeroth-order term by changing variables again by setting $r = \frac{2 \omega_0}{\alpha} s$, which simplifies the equation to $$r^2 x_{rr} + \left(1-\frac{2 \beta}{\alpha}\right) r x_r + r^2 x = 0 .$$ To adjust the first-order term, change the dependent variable with the ansatz $x = r^\lambda y$, which transforms the equation to $$r^2 y_{rr} + \left(1 + 2 \left(\lambda - \frac{\beta}{\alpha}\right)\right) r y_r + \left(r^2 + \lambda^2 - \frac{2 \lambda \beta}{\alpha} \right) y = 0 .$$ For an appropriate choice of $\lambda$ this equation becomes the modified Bessel equation.

Fixing $\lambda = \frac{\beta}{\alpha}$ yields $$r^2 y_{rr} + r y_{r} + \left(r^2 - \frac{\beta^2}{\alpha^2}\right) y = 0 ,$$ which is the modified Bessel equation with parameter $\frac{\beta}{\alpha}$, so the general solution in $y$ is $$y(r) = A J_{\frac\beta\alpha}(r) + B Y_{\frac\beta\alpha}(r) ,$$ where $J_\bullet$ and $Y_\bullet$ are the modified Bessel functions of the first and second kinds, respectively.

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  • $\begingroup$ Thank you, did you use $\omega_0=\sqrt{k_0/m}$ here? I also have some trouble writing your first equation, can you maybe write an intermediate step? I possibly got $\ddot{x}=\frac d{dt}\left(-\frac\alpha2sy^{\prime}(s)\right)=-\frac\alpha2\frac{ds}{dt}y^{\prime}(s)-\frac{\alpha}2sy^{\prime\prime}(s)$ wrong or something $\endgroup$
    – gujaral
    Apr 15 at 21:41
  • $\begingroup$ This answer used $\omega_0$ as originally defined in the question statement, but I've adjusted the answer to reflect the updated version of the question/the more standard usage. Regard $x$ as a function of $s$ and use the chain rule, $\frac{d}{dt} x(s(t)) = x_s \frac{ds}{dt}$. $\endgroup$ Apr 15 at 21:57

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