(1/2)y'' + y^2 = 1

How would you attack this equation?

(1/2)y'' + y^2 = 1

I'm trying to find an approximate solution for the trajectory of a ball acted on by constant gravity acceleration and quadratic drag (proportional to the speed squared), and it turns out the speed squared satisfies an equation that is equivalent to this one after non-dimensionalizing. Any kind of solution would be nice, whether it is good for large x, small x, or whatever.

I was thinking that for x << 1, the equation would be approximately (1/2)y'' + y^2 = 0, in which case y = -3/x^2 is a solution. But then I don't know how to proceed from there to a better approximation. For x >> 1, I was thinking y^2 = 1 would be a first approximation, but again I don't know how to get any farther than that.

thanks

I'd multiply through by y', then note that you can integrate each of the three terms individually to get a 1st order diff eqn.

Maybe you probably meant (y')2, in which case try writing as y'' / ( 1 - (y')2)=2 and integrate.

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I'd multiply through by y', then note that you can integrate each of the three terms individually to get a 1st order diff eqn.

Maybe you probably meant (y')2, in which case try writing as y'' / ( 1 - (y')2)=2 and integrate.

No, y2 was what I meant. I forgot to mention that I had tried integrating it once as you suggested above, but I wasn't able to do the resulting integral

$$\int\frac{dy}{\sqrt{c+y-\frac{y^3}{3}}}$$

Is there a way to integrate that?

thanks

$$\int\frac{dy}{\sqrt{c+y-\frac{y^3}{3}}}$$

Is there a way to integrate that?

I'm no expert on this, but it looks like an elliptic integral to me. I don't think you can do it explicitly.

to quote http://en.wikipedia.org/wiki/Elliptic_integral" [Broken]

In general, elliptic integrals cannot be expressed in terms of elementary functions. Exceptions to this general rule are when P has repeated roots, or when R(x,y) contains no odd powers of y. However, with the appropriate reduction formula, every elliptic integral can be brought into a form that involves integrals over rational functions and the three canonical forms (i.e. the elliptic integrals of the first, second and third kind).

and, c + y - y3/3 won't in general have repeated roots.

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I'm no expert on this, but it looks like an elliptic integral to me. I don't think you can do it explicitly.

OK, that helps. Perhaps I could look up an approximation to elliptic integrals, and then invert it to get y = y(x).

gel said:
to quote wikipedia

In general, elliptic integrals cannot be expressed in terms of elementary functions. Exceptions to this general rule are when P has repeated roots, or when R(x,y) contains no odd powers of y. However, with the appropriate reduction formula, every elliptic integral can be brought into a form that involves integrals over rational functions and the three canonical forms (i.e. the elliptic integrals of the first, second and third kind).

and, c + y - y3/3 won't in general have repeated roots.

Excellent. I'll take a look at that wikipedia article. Thank you.

Also, how do you write an inline fraction, such as 1/2, so that the 1 is on top of the 2?

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Well darn. I went back and checked my derivation of the d.e. and realized I had a minus sign wrong. The correct equation is (I pretty sure this time ),

$$\frac{1}{2}y'' + 2y^{\frac{1}{2}}y' + y^2 = 1$$

So I guess I'll get to work on this one...

Suppose a solution y(x) is invertible locally, then we can define $$v(y)=y'(x(y))$$, where x(y) is the inverse of y(x). Substituting this in your d.e., you get

$$\frac{1}{2} v \dot{v} + 2 y^{1/2} v + y^2 = 1,$$

where $$\dot{v}$$ denotes $$dv/dy$$. This is a first order d.e. which is easier to solve. Try substituting $$w=v/\sqrt{y}$$ to get rid of the square root. If you have found the solution v(y), you can find y(x) by integrating

$$x-x_0 = \int_{y_0}^y \frac{dy}{v(y)}$$

and then solving the equation for y. Good luck!

......or use Mathematica . It gives two solution: $$y=\tan^2 (x-x_0)$$ and $$y=\coth^2(x-x_0)$$. But you should check (and find the general solution).

Suppose a solution y(x) is invertible locally, then we can define $$v(y)=y'(x(y))$$, where x(y) is the inverse of y(x). Substituting this in your d.e., you get

$$\frac{1}{2} v \dot{v} + 2 y^{1/2} v + y^2 = 1,$$

where $$\dot{v}$$ denotes $$dv/dy$$. This is a first order d.e. which is easier to solve. Try substituting $$w=v/\sqrt{y}$$ to get rid of the square root. If you have found the solution v(y), you can find y(x) by integrating

$$x-x_0 = \int_{y_0}^y \frac{dy}{v(y)}$$

and then solving the equation for y. Good luck!

......or use Mathematica . It gives two solution: $$y=\tan^2 (x-x_0)$$ and $$y=\coth^2(x-x_0)$$. But you should check (and find the general solution).

That is an excellent suggestion for clearing the square root. Thank you. I had found the solution y = tanh2(x-x0) by guessing and checking. What led me to guess it was the special case of a ball falling straight down (angle the velocity makes with the vertical is zero or pi, depending on how you define it) acted on by quadratic drag. In that case the non-dimensionalized speed is u = tanh(x-x0), and y = u2 in the above d.e.

The problem is, I don't know how to find a general solution for a non-linear d.e. Is there even a method of knowing how many solutions such an equation has?

Perhaps it would be better to work directly with the d.e. for the nondimensionalized velocity, u?

u(u'' + 2uu') + (u' + u2)2 = 1

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..............................................
this eqn. comes from the pair of equations

u(u'' + 2uu') = sin2A
(u' + u2)2 = cos2A

where A(x) is the angle of the velocity vector with the vertical, as a function of the nondimensionalized time x.
................................................

That way the special case of (u' + u2)2 = 1, which implies u'' + 2uu' = 0, would be apparent. But I'm guessing there are more solutions than that.

I will try out your suggestion. Thanks :)

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