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

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

• What do you mean "...we should only see the interval of validity..."? We can see three possible intervals of validity. The question is which one this particular solution of the equation belongs to. If you had a different initial condition, then the particular solution of the same equation might belong to one of the other intervals. – NickD Jan 10 '18 at 18:31

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.

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.

• why can't we define the interval of definiton as a union of intervals where the solution is defined ? – onurcanbektas Feb 28 '18 at 8:16

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.