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I want to solve the differential equation$$\begin{cases}x''=\frac{2x}{x^2-1}\\x'(0)=0\\x(0)=x_0\end{cases}$$

This is what I have done so far. I have not studied differential equation much, and introducing the function $s$ below is just a trick that I learned, but I'm not sure why/if it works.

Let $s:=x'$. Then $$\frac{d^2x}{dt^2}=\frac{ds}{dt}=\frac{ds}{dx}\frac{dx}{dt}=\frac{ds}{dx}s$$ so the above equation becomes $$s\frac{ds}{dx}=\frac{2x}{x^2-1}$$ $$s\,ds=\frac{2x}{x^2-1}\,dx$$$$\int s\,ds=\int\frac{2x}{x^2-1}\,dx+C$$ $$s^2=\log\lvert x^2-1\rvert+C$$$$x'=\omega\sqrt{\log\lvert x^2-1\rvert+C},\,\omega:\mathbb R_+\mapsto\{-1,1\}$$ and with $x'(0)=0$, $x(0)=x_0$ we have $$x'=\omega\sqrt{\log\bigg\lvert\frac{x^2-1}{x_0^2-1}\bigg\rvert}$$ How can I continue to solve for $x(t)$? And am I justified in introducing $s$ and treat the notation like I did to obtain $x'$?

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  • $\begingroup$ Your steps are pretty fine. Just go on as if you are solving a first order differential equation. But maybe there are some integrals that can’t be done by hand. $\endgroup$ – Szeto Aug 15 '18 at 11:01
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Your calculus is correct. $$\frac{dx}{dt}=\omega\sqrt{\ln\bigg\lvert\frac{x^2-1}{x_0^2-1}\bigg\rvert}$$ with $\omega=\pm 1$ . $$t=\pm\int \frac{dx}{\sqrt{\ln\bigg\lvert\frac{x^2-1}{x_0^2-1}\bigg\rvert}}$$ With condition $x(0)=x_0$ : $$t=\int_{x_0}^x \frac{d\xi}{\sqrt{\ln\bigg\lvert\frac{\xi^2-1}{x_0^2-1}\bigg\rvert}}$$ One can show that the integral is convergent for $\xi\to x_0$. That isn't the main trouble.

The integral cannot be expressed in terms of standard functions, a fortiori the inverse function $x(t)$ as well. Numerical calculus is required to fully solve the problem.

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  • $\begingroup$ This problems is just from a thought experminent in physics involving two chambers separated by a moveable wall that is pushed by the gas pressure in the two chambers. Thinking about the problem now I realize $\omega=$sgn$(x')$ must some how be dependent of $t$, but I don't know how... $\endgroup$ – Ludvig Lindström Aug 15 '18 at 11:52
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    $\begingroup$ OK. I realize that $\omega=\pm 1$. So, some details have been corrected in my answer. This doesn't change the conclusion about the non-elementary form of analytic solution. $\endgroup$ – JJacquelin Aug 15 '18 at 13:48
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The procedure is correct. The last equation is separable, and its solution can be written as $$ \int_{x_0}^x\frac{dx}{\sqrt{\log\Big\lvert\dfrac{x^2-1}{x_0^2-1}\Big\rvert}}=x_0+\omega\,t. $$ However, the integral has no closed form in terms of elementary functions (or even special functions, as far as I can tell.)

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