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I have this question asked in an exam:

The solution to $$ \left\{ \begin{aligned} \frac{dy}{dx} &= y^2+x^2 \quad , x \gt 0 \\ y(0) &= 2 \end{aligned} \right. $$ exists in the interval $[0, L_0)$ and the maximal interval of existence of $$ \left\{ \begin{aligned} \frac{dz}{dx} &= z^2 \quad , x\gt 0 \\ z(0) &= 1 \end{aligned} \right.$$ is $[0,L_1) $.

I have to find the values of $L_0$ and $L_1$.

In the second case the conditions of Existence and Uniqueness theorem are satisfied in every rectangle $$|x| \le a, \quad |z-1| \le b$$ about the point $(0,1)$.Now if $M $ is maximum of $z^2$ and $h= min(a,b/M) $ then the IVP possess a solution on $|x| \le h$.

After calculating I get $h=1/4$ , so will I conclude that $L_1 = 1/4$ ?

Also in the 1st case I am unable to find $L_0$.

Any help. Thank you.

hiren_garai
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1 Answers1

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The second equation is easy to solve in closed form, and you can see from that closed form where the solution goes to $\infty$, giving you $L_1$.

The first equation has a complicated closed-form solution in terms of Bessel functions, which nobody would expect you to find on an exam. Fortunately it doesn't ask for the maximal interval of existence in this case, just an interval of existence. So you can choose some convenient $R > 2$ and find some $L_0$ for which you can prove that the solution will stay in the interval $[2, R]$ for $0 \le x < L_0$. Thus if you take $R = 3$, $y' = y^2 + x^2 < 10$ if $2 \le y \le 3$ and $0 < x < 1$, and so the solution is guaranteed to exist at least on $[0, 1/10)$.

Robert Israel
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  • Actually I am not familiar with closed form solution,so it's hard to make out your answer, will you please give any other way solving. @RobertIsrael – hiren_garai Sep 07 '17 at 13:37
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    "Closed form" just means an explicit formula for the solution: $y = \ldots$. – Robert Israel Sep 07 '17 at 18:59
  • Oh! I got the closed form solution ! But how the solution is guaranteed on $[0,1/10)$ ? , Will you explain please. – hiren_garai Sep 08 '17 at 15:35
  • In order for the solution to go to $\infty$, it must first get to $y=3$. While in the rectangle $0 \le x \le 1$, $2 \le y \le 3$ the slope $y' \le 10$, so $y(1/10) \le 2 + 10/10 = 3$. – Robert Israel Sep 08 '17 at 19:16