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Let the $\operatorname{IVP}$ be given by:

$\dfrac{\operatorname{dy}}{\operatorname{dx}}=y^2+\cos^2 x;x>0;y(0)=0$

Show that the largest interval of existence of the solution predicted by Picard's Theorem is $[0,\frac{1}{2}]$

By Picard's Existence and Uniqueness Theorem;If $f$ is continuous on a domain $D$ and $f$ satisfies Lipschitz condition on $D$ .If $R=\{|x-x_0|\leq a;|y-y_0|\leq b\}$ lies in $D$ and $M=\sup |f(x,y)|,h=\min\{a,\frac{b}{M}\}$. Then the $\operatorname{IVP}$ has a unique solution on the interval $|x-x_0|\leq h$.

Obviously the Lipscitz Condition is satisfied here.But I can't find the rectangle which is needed to apply the theorem.How should I do it?

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In your case $x_0 = y_0 = 0$. For any $a > 0, b > 0$, the function $f(x, y) = y^2 + \cos^2 x$ is defined and Lipschitz continous on the rectangle $R=\{|x|\leq a, |y|\leq b\}$.

On this rectangle, $$|f(x, y)| \le |y|^2 + |\cos x|^2 \le b^2 + 1$$ with equality for $x = 0$ and $y = b$, so the supremum is $M = b^2 + 1$ and therefore $$h= \min\{a,\frac{b}{b^2 + 1}\}$$

The Picard existence theorem states that the IVP has a (unique) solution on the interval $[-h, h]$, or – if you restrict the problem to $x \ge 0$ – on $[0, h]$.

The task is now to choose $a$ and $b$ such that $h$ becomes as large as possible. From the AM-GM inequality is follows that $$ b = \sqrt{1 \cdot b^2} \le \frac{b^2 + 1}2 \Longrightarrow \frac{b}{b^2 + 1} \le \frac 12 $$ with equality for $b=1$.

It follows that $h \le \frac 12$ for any choice of $a, b$, and $h = \frac 12$ for $a = \frac 12, b = 1$. So $h = \frac 12$ is the largest value that can be obtained by this method.

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    $\begingroup$ Great answer +1 $\endgroup$ – Learnmore Dec 9 '15 at 13:31
  • $\begingroup$ I been tying to figure out why I have a theorem in my text stating that the maximal intervall is never closed. This seems not to be the case here. You heard about such statements? $\endgroup$ – user1 Aug 14 '18 at 20:04
  • $\begingroup$ @user1: This question is about the "largest interval ... predicted by Picard's Theorem," not about the maximal interval in which the solution actually exists. $\endgroup$ – Martin R Aug 14 '18 at 20:06
  • $\begingroup$ oh, I thought thats what that meant. Regardsless, the intervall that we obtain via global existance on rectangles is never maximal? $\endgroup$ – user1 Aug 14 '18 at 20:08
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    $\begingroup$ @user1: Sorry, I don't get what you mean. Picard's theorem states that a solution exists on some (closed) interval. That does not exclude that a solution actually exists on a larger interval. $\endgroup$ – Martin R Aug 14 '18 at 20:11
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Hint: You could choose $M = b^2 + 1$. Then, try to find $a,b$ which maximizes $h = \min\{a, b/(b^2+1)\}$.

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  • $\begingroup$ Thanks for fixing my typo! $\endgroup$ – Martin R Dec 9 '15 at 13:16

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