# Existence Theorem in ODE doubt in understanding the theorem [closed]

Theorem 1. (Existence theorem): Suppose that $f(x, y)$ is a continuous function in some region

$$R = \{(x, y) : |x − x_0| ≤ a, |y − y_0| ≤ b\}\qquad a, b > 0$$

Since $f$ is continuous in a closed and bounded domain, it is necessarily bounded in $R$, i.e., there exists $K > 0$ such that $|f(x, y)| \leq K$ $\forall (x, y) \in R$. Then the IVP $$y'=f(x,y), \qquad y(x_0)=y$$ has at least one solution $y = y(x)$ defined in the interval $|x − x_0| \leq \alpha$ where $α = \min(a , b/K )$.

Why is the interval for finding solution $x=\min(a,b/K)$? Particularly why is there $b/k$ and not just $b$?

## closed as off-topic by Namaste, user99914, mau, Ethan Bolker, José Carlos SantosMar 22 '18 at 23:21

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Community, mau, Ethan Bolker, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.

• Have you looked at the proof of the theorem? – Alex R. Mar 22 '18 at 17:31
• Take $y' \equiv 2$, $y(0) = 0$, $a = b = 1$, and look what happens. – user539887 Mar 22 '18 at 22:03