Discussion
For the initial value problem
\[ y^{1} = y^{2} + \cos^{2} x,\ y(0) = 0 \]
the interval of existence of solution in rectangle
\[ \begin{aligned} R : \{(x, y) \mid &0 \le x \le a,\ |y| \le b, \\ &a \gt 1/2,\ b \gt 0\} \end{aligned} \]
is :
Answer: C
The interval of existence is [0, 1/2]. (The paper writes \(y^{1}\) for \(y'\).)
Picard's theorem on the rectangle \(0 \le x \le a\), \(|y| \le b\) guarantees a solution on \([0, h]\) with \(h = \min(a, b/M)\), where \(M\) bounds \(|f|\) on the rectangle. Here
\[ \begin{aligned} M &= \max |y^{2} + \cos^{2} x| = b^{2} + 1 \\ \frac{b}{M} &= \frac{b}{b^{2} + 1} \end{aligned} \]
The function \(b/(b^{2} + 1)\) is largest at \(b = 1\), where it equals 1/2. Since \(a \gt 1/2\), the best guaranteed interval is \(h = \min(a, 1/2) = 1/2\), that is [0, 1/2].
The intervals [0, 1] and [0, 2] would need \(b/M\) larger than any value it can take, and \([0, 1/\sqrt{2}]\) does not come from the maximisation.
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