 Solution

ID: past-exam-of-the-topics-in-applied-mathematics-course-of-the-university-of-oxford/2008/a/1/b/ii/solution

First establish a uniform upper bound. Direct subtraction gives

F (N) - \frac{3}{2} = - \frac{N}{2} (2 N^{2} - 7 N + 7) < 0 (N > 0),

49 - 56 = - 7

. The assumed positivity of

F (N_{0})

therefore places

N_{1}

(0, 3/2)

This interval is forward invariant. On

[0, 1/2]

and on

[1, 3/2]

, the three factors in

P (N)

have nonnegative product, hence

F (N) \geq 1/2

. On

[1/2, 1]

the product is nonpositive, but

∣ P (N) ∣ = (N - \frac{1}{2}) (1 - N) (2 - N) \leq \frac{1}{2} \frac{1}{2} \frac{3}{2} = \frac{3}{8},

F (N) \geq 1/8 > 0

. Together with the upper bound this proves

F ((0, 3/2)) \subset (0, 3/2)

. Induction now gives

N_{t} > 0 for every t \geq 0.

The hypothesis on the first image is essential for initial values outside that invariant interval.

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