First-order differential equations - why is only the interval of validity with the initial value valid?

Question

I'm reading through Paul's online math notes on differential equations because I'd like to have a basic grasp on the subject (I'm interested in physics ...)

There's something I don't understand about initial value problems. I get that the solution y(x) to a first-order differential equation with a given initial value constraint might not be valid for every x. But what I don't get is that the author claims that we should only see the interval of validity in which the x of the initial value constraints lays as the interval of validity.

I'm getting these claims from this page, at the Example 1 section.

I just see no reason for this claim and I was wondering if someone could offer a mathematical explanation of why we discard all the intervals of validity except for the one in which the initial value lays.

( I could understand it from a physical point of view ).

Thanks very much in advance, Joshua

Answer 1

You can think of the problem of solving a differential equation like $y^\prime = F(x,y)$ given an initial condition $y(x_0)=C$ as the problem of finding:

1. A function $\varphi(x)$ which is smooth ...
2. ...which satisfies the differential equation $\varphi^\prime = F(x,\varphi(x))$
3. ...and satisfies the initial condition $\varphi(x_0) =C$
4. ...and is defined on the largest possible domain.

Because of condition (3), the domain of the solution must contain the initial point $x_0$.

Because of condition (4), we want to extend the domain to the largest possible set containing $x_0$.

Because of condition (1), the solution can't contain any discontinuities, gaps between intervals, or "domain errors" such as divide-by-zero. This means that the domain of the solution must be a contiguous interval where the function is smooth and well-defined throughout.

For this reason, you must exclude from the domain any points where the formal solution has a domain error—and you can't extend the domain interval beyond such points.

In short, you can consider it baked into the definition of what it means to solve a differential equation: to solve a differential equation means to find a smooth solution on the largest possible contiguous interval containing the initial condition.

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To look at it another way, you can consider a simpler differential equation such as $y^\prime = 0$ where $y(0.5) = 0$. One solution which satisfies the differential equation is:

$$\phi(x) = \lfloor x \rfloor$$

because what we've essentially done is taken a bunch of smooth solutions (flat line functions which differ by a constant) and stitched them together. This "solution" satisfies property (2) and also property (3). But it violates the other conditions, and therefore is not considered a proper solution to this equation.

Answer 2

A solution to the differential equation $$ y'=f(x,y) $$ is a function $ y=y(x)$ which satisfies $$ y' = f(x,y(x))$$

Therefore $y(x)$ is a differentiable function which makes it continuous as well.

Now if you have an initial condition and a solution which has a vertical asymptote , the continuity of your function will be lost at the point where the vertical asymptote happens.

Thus you can not cross that point which puts a limit on the interval of definition.

For example $y'= 1+ y^2$ subject to $y(0)=0$ has its solution as $y=tan(x)$ which does not extend beyond $x=\pi /2 $ without loss of continuity.