Non-linear ODE: initial conditions

[Arjan82]

TL;DR Summary

What are the rules for specifying initial conditions? (How many, which variables, ...)

Say you have the set of coupled, non-linear ODEs as derived in this thread, it has two unknowns $N(t)$ and $\theta(t)$:

$$ N - mg = - m\frac{L}{2}\left(\dot{\theta}^2\cos(\theta) + \ddot{\theta}\sin(\theta)\right)$$
$$ \frac{L}{2}N\sin(\theta) = \frac{1}{12}ml^2\ddot{\theta}$$

What freedom do I have in specifying the initial conditions to make this a fully determined problem? It definitely works if I specify $\theta$ and $\dot{\theta}$ as is shown in the thread mentioned above. But could I also specify $\ddot{\theta}$ and $N$? Or $N$ and $\dot{N}$? or any other combination of the two independent variables and/or their derivatives? What implications does that have?

I suppose I need two initial conditions because I have two equations and two unknowns, correct?

 

[fresh_42]

How many initial conditions you need what for? I assume, in order to determine a unique solution. A solution can be visualized as a flow through a vector field, i.e. it depends on that vector field how many initial conditions you need to determine a unique flow. As a rule of thumb: it are as many as there are derivatives, e.g. $\dot \theta,\ddot \theta.$

A more basic point of view: you need for every integration one initial condition to fix the constant.

I am not sure whether this is always true, as I could imagine vector fields with branches at some point, in which case we would need more fix points. But it will do as a heuristic.

 

[phyzguy]

I look at it this way. If you specify $\theta(0)$ and $\dot{\theta}(0)$, then you can use your two equations to find N(0) and $\ddot{\theta}(0)$. Similarly if you specify N(0) and $\ddot{\theta}(0)$ you can find the other two. But since $\rm \dot N$ doesn't appear, I don't think it is sufficient to specify N(0) and $\rm \dot N(0)$.

 

[Arjan82]

I totally hadn't thought about branching, but I do not necessarily mean a unique solution. Since this is an initial value problem, and I'm only considering models for which the unknowns evolve in time only, I would like to only specify the initial values and let the solution decide which branches to take (are there problems for which this doesn't work?)

I hadn't yet made the link between integration constants and initial conditions, so that is clarifying. This would then also mean that if I had a single equation with a second derivative, I would still need two initial conditions. And, vice versa, if I had multiple equations with equal amount of unknowns where only a single unknown appears as a single derivative I only need to specify a single initial condition, correct?

 

[fresh_42]

This would then also mean that if I had a single equation with a second derivative, I would still need two initial conditions.

$\ddot y = 1$ leads to $y(t)=\frac{1}{2}t^2+v_0t+s_0$ so, yes, basically, yes.

And, vice versa, if I had multiple equations with equal amount of unknowns where only a single unknown appears as a single

first

derivative I only need to specify a single initial condition, correct?

Yes, one integration makes one constant makes one initial condition.

Of course the initial conditions have to fit to the integration you have to perform. It does not help to have initial conditions $y^{(3)}(1)=y^{(4)}(2)=0$ in the above equation $\ddot y = 1$.