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So, I have this pair of differential equations:

$$\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + \frac{2g(H/L)x}{4(H/L)^2x^2+1} = 0$$ $$\frac{\mathrm{d}^2y}{\mathrm{d}t^2} + \frac{4g(H/L)y}{4(H/L)y+1} = 0$$ I am looking for two functions, $x(t),\ y(t)$ and the initial conditions are:

$$x(0) = L,\ y(0) = H,\ x'(0) = 0,\ y'(0)=0$$

I got these by applying Newton's second law to a specific problem.

I don't have the necessary background to solve this equation but am really curious as to what the solution would look like.

If it couldn't be solved analytically, I would love to see a numerical solution.

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  • $\begingroup$ Are $L,H$ and $g$ constants? $\endgroup$ – Arthur Jan 22 '18 at 7:20
  • $\begingroup$ Yes. $L, H$ represent initial $x, y$ of the particle. And $g$ is gravitational acceleration near the earth. $\endgroup$ – Anonymous Jan 22 '18 at 7:22
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    $\begingroup$ First of all, check that both equations are uncoupled, so you can treat each one as a second order ordinary differential equation. I mean, if what you wrote is ok, you have two separate problems there, not an actual "pair" or "system". $\endgroup$ – Alejandro Nasif Salum Jan 22 '18 at 7:29
  • $\begingroup$ Yes, I just meant, there are two. And you're right. Mathematically, they're two different problems. $\endgroup$ – Anonymous Jan 22 '18 at 7:30
  • $\begingroup$ If it couldn't be solved analytically, I would love to see a numerical solution. $\endgroup$ – Anonymous Jan 22 '18 at 8:22
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$$\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + \frac{2g(H/L)x}{4(H/L)^2x^2+1} = 0$$ $2\frac{\mathrm{d}^2x}{\mathrm{d}t^2}\frac{\mathrm{d}x}{\mathrm{d}t} + \frac{4g(H/L)x}{4(H/L)^2x^2+1}\frac{\mathrm{d}x}{\mathrm{d}t} = 0$

$\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \frac{gL}{2H}\ln\big|4(H/L)^2x^2+1\big| = c_1$

$\frac{\mathrm{d}x}{\mathrm{d}t} = \pm\sqrt{-\frac{gL}{2H}\ln\big|4(H/L)^2x^2+1\big| + c_1}$ $$t(x)=\pm\int \frac{\mathrm{d}x}{\sqrt{-\frac{gL}{2H}\ln\big|4(H/L)^2x^2+1\big| + c_1}} +c_2$$ As far as I know, there is no closed form for this integral in terms of a finite number of standard functions. A-fortiori, there is no closed form for the inverse function $x(t)$.

$$\frac{\mathrm{d}^2y}{\mathrm{d}t^2} + \frac{4g(H/L)y}{4(H/L)y+1} = 0$$ $2\frac{\mathrm{d}^2y}{\mathrm{d}t^2}\frac{\mathrm{d}y}{\mathrm{d}t} + \frac{8g(H/L)y}{4(H/L)y+1}\frac{\mathrm{d}y}{\mathrm{d}t} = 0$

$\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2 + \frac{g}{4H}\left(4Hy -L\ln\big|4Hy+L\big| \right)= c_1$

$\frac{\mathrm{d}y}{\mathrm{d}t}= \pm\sqrt{-\frac{g}{4H}\left(4Hy -L\ln\big|4Hy+L\big| \right)+ c_1}$ $$t(y)=\pm\int \frac{\mathrm{d}y}{\sqrt{-\frac{g}{4H}\left(4Hy -L\ln\big|4Hy+L\big| \right)+ c_1}}+c_2$$ Same comment than for the above integral.

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  • $\begingroup$ But the solution does exist, meaning I can still numerically plot it, right? It just doesn't have a standard name. $\endgroup$ – Anonymous Jan 22 '18 at 8:41
  • $\begingroup$ Of course, the solution exists, but it cannot be expressed on closed form with the a finite number of standard functions. It can be expressed on implicit form with integrals as shown above. For approximate solution, you can try series expansion (probably arduous), or numerical integration of the integrals, or numerical method of direct solving ODEs (certainly the easier way). $\endgroup$ – JJacquelin Jan 22 '18 at 9:14
  • $\begingroup$ From the answer provided, can you "see" what the shape would look like, I am just asking to know for sure, if I am on the right path. $x(t)$ should start from $H$ (given in the initial condition) and then decrease with $t$ (from my intuition, if I modeled the problem, correctly). $\endgroup$ – Anonymous Jan 22 '18 at 9:20
  • $\begingroup$ OK. But don't forget that the behaviour depends of not only of one initial condition $x(0)=L$, but of two initial conditions. The second isn't specified in the wording of the question. So, $x(t)$ in not fully determined and we cannot say what will be exactly the behaviour in all cases. For example if the second initial condition is $x'(0)=c_0>0$ the function $x(t)$ will not be always decreasing. $\endgroup$ – JJacquelin Jan 22 '18 at 9:37
  • $\begingroup$ I am sorry I forgot to mention. I'll add it in the question as well, $x'(0) = 0,\ y'(0)=0$ $\endgroup$ – Anonymous Jan 22 '18 at 9:39

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