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.
An image may say more than a thousand words:
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 $$ |x-x_0|\le α=\min(a,b/M). $$