 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-1-course-of-the-university-of-oxford/2013/differential-equations/7/b/solution

Each denominator is at least one and positive, so each component of

F

lies in

(0, 1]

. Taking the maximum norm therefore gives

∥ F (x, Z) ∥_{\infty} = max {\frac{1}{∣ z _{2} ∣ + x ^{2} + 1}, \frac{1}{∣ z _{1} ∣ + 1}} \leq 1.

This is a global bound, so in particular the constant

M = 1

is valid throughout the specified cube.

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