Does a Solution's Interval Limit Its Usefulness Outside That Range?

[thegreenlaser]

We've done a little bit on existence/uniqueness of solutions, and there's one thing that's a little confusing to me. We have a theorem which (paraphrased) says that if you have a linear ODE with an initial value problem, then a solution exists on the largest open interval containing t₀ on which the coefficient functions are continuous. (t being the independent variable). Now, say I'm modelling something with this equation:

\frac{d^2 y}{dt^2} + \frac{1}{t(t-4)}\cdot \frac{dy}{dt} + t^2 y = e^t

and say I know that y(2) = y₀ and y'(2) = y'₀. (so t₀ = 2)

Then, the coefficient functions are all continuous except for at t=0 and t=4. Thus, I know a unique solution to my initial value problem exists on the interval 0 < t < 4, since t₀ = 2 lies on that interval. So, my question is, does this mean that my solution might be useless everywhere else? Does this mean I might get a function that is accurate on the interval 0 < t < 4 but is garbage or undefined everywhere else? I guess I'm just a little unsure what "having a solution on that interval" means.

 

[henry_m]

A time at which a coefficient of the ODE has a pole or is otherwise undefined is called a singular point. Essentially two things could happen at such a point:

(i) The solution holds good at and beyond the singular point and you get to extend your unique solution at least until the next singular point.

(ii) The solution or some of its derivatives blow up so you can't extend your solution beyond the singular point.

As an example of both of these, look at the equation y'+y/[t(t-1)]=0, y(1/2)=1. There are singular points at t=0,1. The solution is y=t/(1-t). At t=0, the solution stays bounded and you can continue it as far as you like in that direction. On the other hand, it blows up at t=1 so you can't extend the solution.

 

[bmxicle]

The theorem only guarantees that for a given initial value, if the 'standard form' functions (like the form you have the ODE in) are continuous for a given initial value, then a unique solution is guaranteed to exist, and the solution is guaranteed to be valid on the interval of continuity that contains the initial value.

This means if you find a solution to the IVP, and the conditions of the theorem are met then that is the only solution to the IVP. The solution can be valid over an interval greater than the continuous interval containing the IVP values, but the existence/uniqueness theorem can't say whether that will be the case.

Also, if you find a solution for an initial value problem at a point of discontinuity, it can be valid, but it is not guaranteed to be unique by the existence/uniqueness theorem.

 

[thegreenlaser]

A time at which a coefficient of the ODE has a pole or is otherwise undefined is called a singular point. Essentially two things could happen at such a point:

(i) The solution holds good at and beyond the singular point and you get to extend your unique solution at least until the next singular point.

(ii) The solution or some of its derivatives blow up so you can't extend your solution beyond the singular point.

As an example of both of these, look at the equation y'+y/[t(t-1)]=0, y(1/2)=1. There are singular points at t=0,1. The solution is y=t/(1-t). At t=0, the solution stays bounded and you can continue it as far as you like in that direction. On the other hand, it blows up at t=1 so you can't extend the solution.

So, essentially, even though the solution y=t/(1-t) is defined for t>1, the numbers you would obtain from that function for t>1 are meaningless for a given IVP where t₀ < 1 ?