An interesting Nonlinear Differential Equation

[Isaac0427]

TL;DR

I have been thinking about the differential equation for the motion of a free falling object without air resistance (on my own, not as homework), which is in the form $y^2y''+K=0$ with $K$ as a positive constant. I'm looking for advice on a first step to solve this equation.

That's pretty much it. If there is a very basic strategy that I am forgetting from ODEs, please let me know, though I don't recall any strategies for nonlinear second order equations.

I've tried looking up "motion of a free falling object" with various specifications to try to get the solution to this equation, but the sources I find still always assume $g$ does not vary with distance. I have also tried looking up "motion of a charge in a coulomb field" but to no avail.

Thank you in advance!

 

[fresh_42]

I looked up the solution and it didn't give any hint how to start, it looks awful.

 

[Isaac0427]

I looked up the solution and it didn't give any hint how to start, it looks awful.

Where does one find the solution? Wolfram Alpha didn't come up with anything.

I could try to reverse engineer it, or in any case check a few things with the solution that I've wanted to look at.

 

[fresh_42]

I used K=7 so it was easy to see where it occurs in the solution:
https://www.wolframalpha.com/input/?i=y^2y''=-7

 

[bigfooted]

where does the nonlinear term come from if there is no air resistance?

 

[pasmith]

Summary:: I have been thinking about the differential equation for the motion of a free falling object without air resistance (on my own, not as homework), which is in the form $y^2y''+K=0$ with $K$ as a positive constant. I'm looking for advice on a first step to solve this equation.

That's pretty much it. If there is a very basic strategy that I am forgetting from ODEs, please let me know, though I don't recall any strategies for nonlinear second order equations.

I've tried looking up "motion of a free falling object" with various specifications to try to get the solution to this equation, but the sources I find still always assume $g$ does not vary with distance. I have also tried looking up "motion of a charge in a coulomb field" but to no avail.

Thank you in advance!

This is of the form y&#039;&#039; = f(y) with f(y) = - K/y^2, so multiplying by y&#039; and integrating is the first step: <br /> y&#039; = \pm \sqrt{C + \frac{2K}{y}}. Getting any further requires that you be able to solve <br /> \int \sqrt{\frac{y}{Cy + 2K}}\,dy = t analytically.

Given your actual question, setting r = R + y, where R is the radius of the Earth and y is height above the surface, in <br /> r&#039;&#039; = \frac{-K}{r^2} will yield <br /> y&#039;&#039; = - \frac{K}{R^2}\left(1 + \frac{y}{R}\right)^{-2} and on the assumption that y/R \ll 1 you can expand the right hand side in binomial series to either O(1) (getting the familiar constant acceleration) or to O(y/R), getting <br /> y&#039;&#039; = - \frac{K}{R^2}\left(1 - \frac{2y}{R}\right) which is a second-order non-homogenous linear ODE with constant coefficients.

 

[Fred Wright]

As an alternate approach to @pasmith consider the DE,
$$
y^2\ddot{y}=-K
$$
and observe,
$$
\ddot{y}=\dot{y}\frac{d\dot{y}}{dy}
$$
so that
$$
\dot{y}\frac{d\dot{y}}{dy}=-\frac{K}{y^2}
$$
and thus,
$$
\int_{\dot{y}_0}^{\dot{y}}\dot{y}d\dot{y}=-K\int_{y_0}^{y}\frac{dy}{y^2}
$$
$$
\frac{1}{2}[\dot{y}^2 - \dot{y}_0^2]=K[\frac{1}{y}-\frac{1}{y_0}]
$$
where $y_0$ is the initial position and $\dot{y}_0$ is the initial velocity. We have,
$$
\frac{dy}{dt}=\sqrt{2}\sqrt{\frac{K}{y} + C}
$$
$$
C=\frac{1}{2}\dot{y_0}^2-\frac{K}{y_0}
$$
$$
t=\int_{y_0}^{y}\sqrt{\frac{y}{2K}}\frac{dy}{\sqrt{\frac{Cy}{K}}+1}
$$
Make the substitution,
$$
u^2=\frac{Cy}{K}
$$
to get,
$$
t=\frac{\sqrt{2}K}{C^{\frac{3}{2}}}\int_{\sqrt{\frac{Cy_0 }{K}}}^{\sqrt{\frac{Cy }{K}}}\frac{u^2du}{\sqrt{u^2+1}}
$$
Integrating by parts we find,
$$
t=\frac{K}{\sqrt{2}C^{\frac{3}{2}}}[u\sqrt{u^2+1}-\sinh^{-1}(u)]|_{\sqrt{\frac{Cy_0}{K}}}^{\sqrt{\frac{Cy}{K}}}

$$
Clearly $y(t)$ can't be expressed in terms of elementary functions, but you can make a plot of y vs. t by plugging in y values and computing t.

 

[aheight]

Here's how I would check it:

In Fred's solution above:
$$(y')^2=2K(1/y-1/y_0)+(y_0')^2$$
I chose the negative root and solved:
$$y'=-\sqrt{2K/y+c},\quad c=(y_0')^2-2K/y_0$$
to obtain:
$$
x(c,K,y)=\frac{2 K \tanh ^{-1}\left(\frac{\sqrt{c+\frac{2 K}{y}}}{\sqrt{c}}\right)}{c^{3/2}}-\frac{y \sqrt{c+\frac{2K}{y}}}{c}
$$
Next, I solved a test IVP numerically in Mathematica with y(0)=10, y'(0)=0 and K=1 and plotted the solution as the gray trace in the plot below:

[CODE title="Mathematica"]mySol = First@
NDSolve[{y[t]^2 y''[t] + 1 == 0, y[0] == 10, y'[0] == 0},
y, {t, 0, 10}]
theTrace[t_] = Evaluate[Flatten[y[t] /. mySol]];
dPlot = Plot[theTrace[t], {t, 0, 10}][/CODE]

Then I superimposed a parametric plot of {x(c,K,y),y} over this solution as the red dashed lines to confirm the analytic solution matches the numeric solution:

[CODE title="Mathematica"]
myf[c_, k_, y_] = Integrate[-1/Sqrt[2 k/y + c], y]
pp1 = ParametricPlot[{myf[-1/5, 1, y], y}, {y, 9, 10},
AspectRatio -> 1, PlotStyle -> {Dashed, Red}]
Show[{dPlot, pp1}][/CODE]

 

[wrobel]

$$L=\frac{1}{2}y'^2-V,\quad V=-\frac{K}{y}$$
1-dimensional motion in Coulomb's field

 

[epenguin]

Or gravitational - inverse square law? Solutions well known

 

[wrobel]

Gravitation can only attract, here attraction or repulsion depend on the sign of K
inverse square law is for force, for the potential just 1/y

 

[epenguin]

Gravitation can only attract, here attraction or repulsion depend on the sign of K
inverse square law is for force, for the potential just 1/y

Ah, it was late at night, I thought something might be not quite right, note the emoticon. OK he said K positive. But anyone who can do it for K negative can for K positive I'd think.

 

[Isaac0427]

where does the nonlinear term come from if there is no air resistance?

Force varies with distance from the center of Earth.

and observe,
y¨=y˙dy˙dy
so that

Where does this come from? I may just be missing something obvious here.