 Solution

ID: past-exam-of-the-topics-in-applied-mathematics-course-of-the-university-of-oxford/2007/a/3/ii/b/solution

b = - η a

with

a > 0

\frac{d b}{d a} = \frac{- c b - f ( a )}{b} = - c + \frac{f ( a )}{η a} = - c + \frac{η ^{2} - a ^{2}}{η ( 1 + a ^{2} )} \leq - c + η < - η .

Thus

d b / d a < - η

. Since

a^{'} = b < 0

, comparing slopes alone can reverse one's intuition about crossing direction. More directly, for

H = b + η a

the inward test is

H^{'} = b^{'} + η a^{'} = η (c - η) a - f (a) \geq η (c - 2 η) a > 0,

so the sloping edge also points into

H \geq 0

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