# Determine the *interval* in which the solution is defined?

The ODE: $y' = (1-2x)y^2$

Initial Value: $y(0) = -1/6$

I've solved the particular solution, which is $1/(x^2-x-6)$. I don't understand what they mean about the solution is defined, because when I graph $1/(x^2-x-6)$, it's only discontinuous at $x = -2, 3$.

What does "Determine the interval in which the solution is defined" mean?

• Good question! But here's a bit of nitpicking about terminology: usually one reserves the word "(dis)continuous" for points in the domain of definition of the function; see for example Rudin's Principles, p. 94. If one follows that tradition, your function would in fact neither be said to be continuous nor discontinuous at $x=-2$ and $x=3$ (simply for the reason that it's not defined there). – Hans Lundmark Jan 30 '11 at 10:31

It makes sense to consider solutions only on intervals that contain the initial time. The "interval where the solution is defined" (I would call it the (maximal) interval of existence) is the maximal interval of all intervals $I$ which contain 0 and there exists a solution on $I$. This maximal interval turns out to be $(-2,3)$ in your case.

So why doesn't it make sense to consider solutions defined on other subsets of $\mathbb{R}$ than intervals, e.g. $(-\infty,-2)\cup (-2,3)\cup (3,\infty)$? One reason is that you wouldn't have uniqueness of solutions, for example, $$\begin{cases}1/(x^2-x-6) & x<3 \cr 0 & x>3\end{cases}$$ would be another "solution". What happens in the other components which don't contain 0 can be rather arbitrary, so one does not allow them.

The solution is not only discontinuous at $x=-2$ and $x=3$, it is also undefined at these points. The maximal interval containing $0$ on which there exists a solution is the open interval $(-2,3)$.

Edit to answer a comment: More generally the ODE around a point $x$ such that $y(x)\ne0$ is equivalent to $y'/y^2=1-2x$, that is, $(1/y)'=(x^2-x)'$. Hence, on every interval where $y$ is defined and not identically zero, there exists a constant $c$ such that $y(x)=1/(x^2-x+c)$. The value of $c$ depends on the initial condition one is given, hence the interval on which $y$ is defined also depends on the initial condition.

In your example, $y(0)=-1/6$, hence $c=-6$, $y(x)=1/(x^2-x-6)$ and the maximal interval of definition around $0$ is bounded by the roots of $x^2-x-6=(x+2)(x-3)$ which are closest to $0$, namely $-2$ and $3$.

But consider another example: if $y(-3)=-1/8$, then $c=-20$, $y(x)=1/(x^2-x-20)$ and the maximal interval of definition around $-3$ is bounded by the roots of $x^2-x-20=(x-5)(x+4)$ which are closest to $-3$, hence this interval is $(-4,+5)$.

The interval can also be semi-infinite or infinite. For instance, if $y(-3)=1/15$, then $c=3$, $y(x)=1/(x^2-x+3)$ and the maximal interval of definition around $-3$ is the real line $(-\infty,+\infty)$. If $y(-3)=1/12$, then $c=0$, $y(x)=1/(x^2-x)$ and the maximal interval of definition around $-3$ is the half line $(-\infty,0)$.

And so on.

• Does this mean that if I was given y(-3)=[something] then the interval would be (-infinity, -2)? What do the inverted square brackets denote ("]-2,3[") – JustcallmeDrago Jan 29 '11 at 23:34
• @JustcallmeDrago: Yes, that's what it would mean. The inverted brackets are equivalent to parentheses: it means the interval from $-2$ to $3$, without including the endpoints. – Arturo Magidin Jan 29 '11 at 23:38
• @Drago I replaced brackets by parenthesis. I also added other examples of initial conditions showing that the interval around $-3$ does not have to be $(-\infty,-2)$. – Did Jan 31 '11 at 7:13