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i have the following theorem:

Let the Cauchy problem $$ y'=f(x,y),~~ y(x_0)= y_0 $$ in $$ R=\{(x,y) \in \mathbb{R}^2: |x-x_0| \leq a, |y-y_0| \leq b\} $$ If $f$ is continuous and bounded in $R$ such as $\forall (x,y) \in R: |f(x,y)| \leq M$ and if $\dfrac{\partial f}{\partial y}$ is continuous and bounded in $R$ then the Cauchy problem for $f,x_0,y_0$ admits a unique solution in the interval $|x-x_0| \leq \alpha$ with $\alpha = \min(a, b/M)$.

My question is why the specific choice $\alpha= \min(a,b/M)$? For exemple why we don't take $\alpha= \min(a,b)$?

Thank's in advance.

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  • $\begingroup$ Possible duplicate of Existence Theorem in ODE doubt in understanding the theorem (closed as off-topic). $\endgroup$ – user539887 Apr 17 '18 at 12:05
  • $\begingroup$ the post who show me are closed withut answer. Can you help me please $\endgroup$ – user487908 Apr 17 '18 at 12:13
  • $\begingroup$ Look at my comment there. $\endgroup$ – user539887 Apr 17 '18 at 12:18
  • $\begingroup$ Yes i read it, but it's an particular case. My question is in general why we choose $\alpha= \min(a,b/M)$ why $b/M$ exactly? Please $\endgroup$ – user487908 Apr 17 '18 at 12:22
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    $\begingroup$ Take $a = b = 1$, $x' \equiv A$, where $A > 1$. Then $M = A$. $\alpha$ must be $\le a = 1$. But if you take $\alpha > b/M = 1/A$ then the graph of the solution, that is, of the function $x(t) = A t$, $t \in [-\alpha, \alpha]$, sticks out: for $t \in (1/A, \alpha)$ the value of the function is $> b$. And this is an example showing that $\alpha = \min(a, b/M)$ is the best choice. $\endgroup$ – user539887 Apr 17 '18 at 12:52
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An image may say more than a thousand words:

enter image description here

With the given data and assumptions you can only make claims on the part of a solution that is inside the rectangle $R$. As to what part of the solution, if it exists, can be guaranteed to be in $R$, you only know that $$\|y(x)-y_0\|\le \int_{x_0}^x\|f(s,y(x))\|\,dx\le M\,|x-x_0|.$$ For this "butterfly" bound to restrict the solution to $R$ you need to restrict $x$ so that $M\,|x-x_0|\le b$ in addition to the restriction $|x-x_0|\le a$. Thus $$ |x-x_0|\le α=\min(a,b/M). $$

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  • $\begingroup$ Thank you so much $\endgroup$ – user487908 Apr 17 '18 at 16:06

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