2
$\begingroup$

There was this question asked in a competitive examination , the solution of which is very confusing to me.

The number of differentiable functions $y:(-\infty, \infty) \rightarrow [0, \infty)$ satisfying $ y' = 2 \sqrt y $ , $y(0)=0$ is

The answer says there exists infinitely many solutions but I could get only one by solving the differential equation.

Can someone help me figuring this out ?

$\endgroup$
2

1 Answer 1

5
$\begingroup$

My answer is treating $y'=2\sqrt{|y|}$ instead of $y'=2\sqrt{y}$. In my experience the problem I consider is the more common way this exercise is stated. If you actually meant $y'=2\sqrt{y}$ and are restricting attention to nonnegative solutions, then you can recover what you want by setting $a=-\infty$ at the end of the discussion below.

The "obvious" solution to this IVP is the one you get by separation of variables. This is just $y=x |x|$, i.e. $y=\begin{cases} x^2 & x \geq 0 \\ -x^2 & x<0 \end{cases}$.

But separation of variables more or less assumes you don't divide by zero, certainly that you don't divide by zero on an interval. The other solutions to this IVP do have the RHS equal to zero on an interval.

Specifically, what happens is that these solutions can start going down from zero as you move to the left anywhere, and can start moving up from zero as you move to the right anywhere as well. Once you move away from zero even a little bit, separation of variables is valid again, and so you can solve the problem off the interval on which $y$ is zero.

Mathematically the result of the above is

$$y(x)=\begin{cases} -(x-a)^2 & x< a \\ 0 & a \leq x \leq b \\ (x-b)^2 & x>b \end{cases}$$

where $-\infty \leq a \leq b \leq \infty$. The solutions with $y(0)=0$ are those with $0 \in [a,b]$.

This is a rather standard exercise in real analysis, posed to illustrate what can happen when the hypotheses of the Picard-Lindelof theorem are violated.

$\endgroup$
8
  • $\begingroup$ For the $x<a$ piece, should that be a $+$ rather than $-$? ie. $+(x-a)^2$ $\endgroup$
    – Sal
    May 24 at 15:49
  • $\begingroup$ @Sal No, but I did make a mistake at the very top related to this issue. OP also either made a mistake or at least deviated from what I was expecting by writing $\sqrt{y}$ instead of $\sqrt{|y|}$. $\endgroup$
    – Ian
    May 24 at 15:49
  • $\begingroup$ (+1) Very nice! I misread the question the first time :) $\endgroup$ May 24 at 15:50
  • $\begingroup$ @Ian Possibly I miss something obvious, but it seems that $-(x-a)^2$ is negative while the question asks for $y\geq 0$, and also if we want to verify that the $x<a$ part is a solution of the DE, we need to take the square root of $y<0$ $\endgroup$
    – Sal
    May 24 at 15:53
  • $\begingroup$ @Sal Yeah, I answered a slightly different question which generalizes the answer to this actual question. See revised first paragraph. $\endgroup$
    – Ian
    May 24 at 15:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.