# Solving higher order ODE as system of first order

1. Aug 11, 2014

### Maylis

For this problem, I am stuck on the actual system. I don't see what substitution I can make, and the fact that $u(v)$ is a piece-wise function is tripping me up. How the heck do I approach this?? This doesn't look like a standard problem at all.

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2. Aug 12, 2014

### HallsofIvy

Staff Emeritus
Actually, that's very much a standard problem- except perhaps that the function on the right, u, is much simpler than usual- it only has two values. No, you do not let "$v=\theta$" First, it isn't the independent variable you want to change to reduce the order and second, setting one variable equal to a new variable is just changing its name- not really changing anything.

Instead, let the new variable be equal to the derivative of $\theta$: $\phi= d\theta/dt$ so that the second derivative of $\theta$ is the first derivative of $\phi$- $d^2\theta/dt^2= d\phi/dt$. Now what does $d^2\theta/dt^2= v(\theta)$ become?

3. Aug 12, 2014

### Maylis

okay,
Are these two my new system of two first order equations? They don't have to be derivatives with respect to the same variable?
$\phi = \frac{d{\theta}}{dt}$
$\frac{d^2{\theta}}{dt^2} = \frac{d{\phi}}{dt}$

Therefore,
$\frac{d^2{\theta}}{dt^2} = u(\int \phi{dt})$ ???

How do I specify the new initial conditions

Last edited: Aug 12, 2014
4. Aug 12, 2014

### Zondrina

Specifically:

$\frac{d \phi}{dt} = u(\int \phi(t) dt)$

Really there are two first order equations here. Do you see them?

5. Aug 12, 2014

### Maylis

I'm afraid I don't see them

6. Aug 12, 2014

### Zondrina

You are given a specified piecewise function. It only has two values.

7. Aug 12, 2014

### Maylis

So does that mean the two equations are

$\frac {d{\phi}}{dt} = -1$ if $u(\int \phi{dt}) \ge 0$
$\frac {d{\phi}}{dt} = 1$ if $u(\int \phi{dt}) \lt 0$.

and if so, now how do I determine the new conditions?

Last edited: Aug 12, 2014
8. Aug 12, 2014

### HallsofIvy

Staff Emeritus
No, you put "$\phi$" in the wrong place!
$\frac{d\phi}{dt}= u(\theta)$

Your initial conditions were originally $\theta(5)= \pi/2$ and $\theta'(5)= \pi/10[/tex] so in terms of [itex]\theta$ and $\phi[/tex] would be [itex]\theta(5)= \pi/2$, $\phi(5)= \pi/10$.

9. Aug 12, 2014

### Maylis

HallsOfIvy,

So does that mean my two equation system is

$\frac {d{\phi}}{dt} = -1$ if $\theta \ge 0$

$\frac {d{\phi}}{dt} = 1$ if $\theta \lt 0$ ??

with initial condition $\phi(5) = \pi/10$ and $\dot{\theta}(5) = \pi/2$?? But then only one condition is in terms of $\phi$, and the other in terms of $\dot{\theta}$, is there a way to make the second one in terms of $\phi$?

Last edited: Aug 12, 2014