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$f$ is locally Lipschitz in $y$ if for every $(t_0, y_0) \in (c,d) \times U$, there exists a neighborhood V of $(t_0, y_0)$, (i.e. $V = \{f(t,y) \in(c,d) \times U : ||t-t_0|<a \ \text{and} \ |y-y_0|\leq b\}$) and a constant K = K(V) such that $||f(t,x)-f(t,y)||\leq K||x-y||$ for any $(t,x),(t,y) \in V$

Existence Theorum:

Consider the IVP $y'=f(t,y), \ y(t_0) = y_0 $ (1)

If f is continuous on $(c,d) \times U$ and locally Lipschitz then the IVP (1) had a unique local solution. More precisely, there exists a neighborhood $\Omega$ of $(t_0, y_0)$, that is $\Omega=\{(t,y)\in (c,d) \times U:|t-t_0|\leq a, |y-y_0|\leq b\}$ such that $f$ is Lipschitz in y with Lipschitz constant K on $\Omega$ and let M be a number such that $||f(t,y)||\leq M$ for $(t,y)\in \Omega$. Choose $0<\alpha<\min[\frac{1}{K},\frac{b}{M},a]$. Then there exists a unique solution of the IVP (1) valid on $[t_0-\alpha , t_0 +\alpha ]$.

QUESTION) The function $f(y) = 1 + y^2$ is locally Lipschitzian. Consider the IVP $y' = 1 + y^2 ,y(0) = 0$:

(a) Using the rectangle in the hypothesis of the local existence and uniqueness Theorem, compute $\alpha$ in terms of a, b, M, and the Lipschitz constant of $f$.

(b) Is it possible for your found in part (a) to be greater than $\frac{\pi}{2}$? Justify your answer. [Do not compute explicitly the solution of the IVP.

For this question, I was unable to start on part (a).

Part (b), I realize that the solution is $y=tan(x)$, so crossing $\frac{\pi}{2}$ would mean that the function is not continuous because that is how the tan function behaves. But I do not know how to show that without computing IVP because I couldn't figure out part (a).

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  • $\begingroup$ The steps in (a): First, find $M$, the maximum of $1+y^2$ on the interval $[-b,b]$. Can you do this? Then, find $K$ by using the fact that the Lipschitz constant of $1+y^2$ is just the maximum of the absolute value of its derivative on the interval $[-b,b]$. Can you find the value of this maximum? Then $\alpha$ comes out of the statement of the theorem. $\endgroup$ – user53153 Feb 24 '13 at 15:25
  • $\begingroup$ Ok so I did this: $f(t,y)=1+y^2\leq 1+b^2=M$. Also, $|f(x)-f(y)|=|x^2-y^2|=|x+y|.|x-y|\leq 2b|x-y|$. So our function is locally lipschitz with K=2b. But our $\alpha$ must be less than $\min[\frac{1}{K},\frac{b}{M},a]$. This implies that $\alpha$ must be less than $\min[ \frac{1}{2b},\frac{b}{1+b^2},a]$. How do i move forward from here? Thanks for the reply. $\endgroup$ – user52932 Feb 24 '13 at 23:22
  • $\begingroup$ Good. Notice we don't really get a "canonical" value of $\alpha$; we are just told by the Theorem that it's okay to use $\alpha$ as long as $0<\alpha<\min[ \frac{1}{2b},\frac{b}{1+b^2},a]$. This is the answer to (a): any $\alpha$ in this range works. // Now on to b): does our inequality for $\alpha$ allow it to be more than $\pi/2$? Well, neither $\alpha<\frac{1}{2b}$ nor $\alpha<a$ seem to prohibit such scenario outright, since we don't know what $a$ and $b$ are. But look at $\alpha<\frac{b}{1+b^2}$ ... the right side here is never very large, for any $b$. Try to make this precise. $\endgroup$ – user53153 Feb 24 '13 at 23:54
  • $\begingroup$ So we can take derivative of $\frac{b}{1+b^2}$. This gives us $\frac{1-b^2}{(b^2+1)^2}$. This implies that the maximum value of this function is at b=1 and min is at b=-1 (when we set derivative = 0). Hence, our $\alpha$ must be between -1 and 1. And $\frac{\pi}{2}$ is greater than 1.5. Is this the correct answer? $\endgroup$ – user52932 Feb 25 '13 at 0:12
  • $\begingroup$ Not quite. For one thing, $b$ is positive by the logic of the problem (you see $|y-y_0|\le b$ there), so there is no need to check negative values of $b$. More importantly, the relevant number here is the maximum of $\frac{b}{1+b^2}$, not where it is attained. This maximum is $\frac12$. Hence, the conclusion is that $\alpha<\frac12$. $\endgroup$ – user53153 Feb 25 '13 at 0:15
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Summary of the discussion in comments.

(a) By definition, $M = \max (1+y^2)$ on the interval $[−b,b]$. The Lipschitz constant $K$ is the maximum of $|(1+y^2)'|$ on the same interval. Hence, $\alpha<\min(1/(2b), b/(1+b^2),a)$.

(b) Since $b/(1+b^2)\le 1/2$ for any $b\ge 0$, any $\alpha$ allowed to us by the theorem must be less than $1/2$.

The value of $\alpha$ is the size of the interval in which the Picard-Lindelöf theorem guarantees the existence and uniqueness. The solution actually exists for $|x|<\pi/2$, beyond its warranty period. Like rovers on Mars.

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