# Non linear Power Series ODE

1. May 15, 2012

### Euler1707

1. The problem statement, all variables and given/known data

I have not had luck in finding a solution that describes an object falling. Forces include gravitational force which is constant and a vicous force directly proportional to the cube of the velocity. I am supposed to find v as a function of time.

2. Relevant equations

v' + a(v)^3 = b where a and b are constants.

3. The attempt at a solution

I have a lot of paper crumbled on the floor. I might add that I am just doing this for fun. I am practicing and learning some math. It is my hobby. :)

Last edited: May 15, 2012
2. May 15, 2012

### Steely Dan

If you are attempting a power series solution, why don't you tell us what your guess for the series is, and what the ODE looks like after substituting in the series?

3. May 15, 2012

### Euler1707

Well,

I have attempted to guess

v = $\Sigma$c$_{n}$t $^{n}$

But my problem is the cubed term. Series solutions seem to only be easy when dealing with polynomial coefficients, but this time the dependent variable is causing the non-linearity of the ODE.

I have ofcourse substituted the derivative terms of the power series but I do not know how to cube the series in order to then find the coefficients. Cubing the series would result in a mess.

Thanks for any input.

By the way this problem is from an OLD book in mechanics. Very cool read if anyone is interested

Slater and Frank Mechanics 1947.

4. May 15, 2012

### Dickfore

Your eqution should be:
$$m \, \dot{v} = m \, g - k \, v^3$$

It is a 1st order ODE in velocity! The one you have is 2nd order. If you introduce dimensionless variables for time and velocity:
$$t = a \, x, v = b \, y, \ y = y(x)$$
then, the equation reduces to:
$$m \, \frac{b}{a} \, y'(x) = m \, g - k \, b^3 \, y^3$$
Let us choose a, and b so that:
$$m \, \frac{b}{a} = m \, g = k \, b^3$$
$$a = \left( \frac{m}{k \, g^2} \right)^{\frac{1}{3}}, \ b = \left( \frac{m \, g}{k} \right)^{\frac{1}{3}}$$
then the ODE simplifies in form:
$$y' = 1 - y^3$$
This equation is with separable variables:
$$\frac{dy}{1 - y^3} = dx$$
Integrating:
$$\int_{y_0}^{y}{\frac{dy'}{1 - (y')^3}} = x, \ y(x = 0) = y_0$$
Do the partial fraction decomposition:
$$\frac{1}{1 - (y')^3} = \frac{1}{(1 - y') (1 + y' + (y')^2)} = \frac{A}{1 - y'} + \frac{B \, y' + C}{1 + y' + (y')^2}$$
$$1 = A (1 + y' + (y')^2) + (B \, y' + C)(1 - y')$$
$$1 = (A + C) + (A + B - C) \, y' + (A - B) \, (y')^2$$
$$\left\lbrace\begin{array}{lcl} A + C & = & 1 \\ A + B - C & = & 0 \\ A - B & = & 0 \end{array} \right. \Leftrightarrow A = \frac{1}{3} \, B = \frac{1}{3} \, C = \frac{2}{3}$$
Thus, you have the integrals:
$$\frac{1}{3} \, \int_{y_0}^{y}{\frac{dy'}{1 - y'}} + \frac{1}{3} \, \int_{y_0}^{y}{dy' \, \frac{y' + 2}{1 + y' + (y')^2}} = x$$
Do the integrals (assuming $0 \le y, \ y_0 < 1$

5. May 15, 2012

### Euler1707

Thank you

Your solution is very elegant. Although the book calls for a power series solution ( at least first 4 terms). I was reviewing from my ODE books but like always math books turn out not to have the thing you are looking for. hehe

Nevertheless thanks for taking the time to solve it.

Also yeah, I did realized that I got carried away with the primes. Ofcourse the physical description was enough to figure it out.

6. May 15, 2012

### Euler1707

Oh, I think I might have found a better method that involves a polynomial solution. Maybe just by iteration using the Taylor series. The problem does not say anything about I.C s but I cannot think of anything else. xD