Solving higher order ODE as system of first order

[gfd43tg]

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.

 

[HallsofIvy]

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 "[itex]v=\theta[/itex]" 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 [itex]\theta[/itex]: [itex]\phi= d\theta/dt[/itex] so that the second derivative of [itex]\theta[/itex] is the first derivative of [itex]\phi[/itex]- [itex]d^2\theta/dt^2= d\phi/dt[/itex]. Now what does [itex]d^2\theta/dt^2= v(\theta)[/itex] become?

 

[gfd43tg]

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?
[itex] \phi = \frac{d{\theta}}{dt} [/itex]
[itex] \frac{d^2{\theta}}{dt^2} = \frac{d{\phi}}{dt}[/itex]

Therefore,
[itex]\frac{d^2{\theta}}{dt^2} = u(\int \phi{dt})[/itex] ?

How do I specify the new initial conditions

 

[STEMucator]

Specifically:

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

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

 

[gfd43tg]

I'm afraid I don't see them

 

[STEMucator]

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

 

[gfd43tg]

So does that mean the two equations are

[itex] \frac {d{\phi}}{dt} = -1 [/itex] if ##u(\int \phi{dt}) \ge 0##
[itex] \frac {d{\phi}}{dt} = 1 [/itex] if ##u(\int \phi{dt}) \lt 0##.

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

 

[HallsofIvy]

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?
[itex] \phi = \frac{d{\theta}}{dt} [/itex]
[itex] \frac{d^2{\theta}}{dt^2} = \frac{d{\phi}}{dt}[/itex]

No, you put "[itex]\phi[/itex]" in the wrong place!
[itex]\frac{d\phi}{dt}= u(\theta)[/itex]

Therefore,
[itex]\frac{d^2{\theta}}{dt^2} = u(\int \phi{dt})[/itex] ?

How do I specify the new initial conditions

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

 

[gfd43tg]

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##?