# An interesting Nonlinear Differential Equation

• I
• Isaac0427
In summary: This comes from taking the derivative of y' with respect to y. The derivative of y' can also be written as y'' or d^2y/dt^2.
Isaac0427
TL;DR 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.

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

fresh_42 said:
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.

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

Isaac0427 said:
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.

This is of the form $y'' = f(y)$ with $f(y) = - K/y^2$, so multiplying by $y'$ and integrating is the first step: $$y' = \pm \sqrt{C + \frac{2K}{y}}.$$ Getting any further requires that you be able to solve $$\int \sqrt{\frac{y}{Cy + 2K}}\,dy = t$$ 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 $$r'' = \frac{-K}{r^2}$$ will yield $$y'' = - \frac{K}{R^2}\left(1 + \frac{y}{R}\right)^{-2}$$ 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 $$y'' = - \frac{K}{R^2}\left(1 - \frac{2y}{R}\right)$$ which is a second-order non-homogenous linear ODE with constant coefficients.

Isaac0427
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.

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:

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}]

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:

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}]

Last edited:
Isaac0427
$$L=\frac{1}{2}y'^2-V,\quad V=-\frac{K}{y}$$
1-dimensional motion in Coulomb's field

epenguin
Or gravitational - inverse square law? Solutions well known

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

wrobel said:
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.

bigfooted said:
where does the nonlinear term come from if there is no air resistance?
Force varies with distance from the center of Earth.
Fred Wright said:
and observe,
y¨=y˙dy˙dy
so that
Where does this come from? I may just be missing something obvious here.

## What is a nonlinear differential equation?

A nonlinear differential equation is a mathematical equation that involves both the dependent variable and its derivatives in a nonlinear way. This means that the rate of change of the dependent variable is not directly proportional to the value of the dependent variable itself.

## Why are nonlinear differential equations important?

Nonlinear differential equations are important because they can model a wide range of real-world phenomena that cannot be accurately described by linear equations. They are used in fields such as physics, engineering, biology, and economics to understand and predict complex systems.

## How do you solve a nonlinear differential equation?

Solving a nonlinear differential equation can be a challenging task and often requires advanced mathematical techniques. In some cases, an analytical solution may not be possible, and numerical methods must be used. These methods involve approximating the solution at discrete points using algorithms and computer simulations.

## What are some applications of nonlinear differential equations?

Nonlinear differential equations have numerous applications in various fields. They are used to model population growth, chemical reactions, fluid dynamics, electrical circuits, and many other phenomena. They are also used in control systems, optimization problems, and machine learning algorithms.

## What are some techniques for analyzing nonlinear differential equations?

There are several techniques for analyzing nonlinear differential equations, including phase plane analysis, stability analysis, and bifurcation analysis. These methods involve studying the behavior of the solutions of the equations and can provide insights into the long-term behavior of the system.

• Differential Equations
Replies
1
Views
1K
• Differential Equations
Replies
6
Views
2K
• Differential Equations
Replies
1
Views
1K
• Differential Equations
Replies
11
Views
2K
• Differential Equations
Replies
5
Views
1K
• Differential Equations
Replies
5
Views
2K
• Differential Equations
Replies
1
Views
694
• Differential Equations
Replies
25
Views
2K
• Differential Equations
Replies
2
Views
1K
• Differential Equations
Replies
6
Views
3K