 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-3/38b/b/iv/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

F (η) = f^{'} (η) = \frac{5 - c o s h z}{1 + c o s h z}, z = 2 η + c .

C = cosh z

, direct differentiation gives

F^{''} (η) = \frac{12 ( C - 2 )}{( 1 + C ) ^{2}} .

On the other hand,

1 - F^{2} = \frac{( 1 + C ) ^{2} - ( 5 - C ) ^{2}}{( 1 + C ) ^{2}} = \frac{12 ( C - 2 )}{( 1 + C ) ^{2}} .

Thus

F^{''} = 1 - F^{2}

, which is precisely

f^{'''} = 1 - (f^{'})^{2}

. The far-field condition is automatic because

F \to - 1

. No slip requires

0 = F (0) = \frac{5 - c o s h c}{1 + c o s h c},

c = \pm arcosh 5 .

For

c = + arcosh 5

F

decreases directly from

0

- 1

, matching flow everywhere toward the sink. For the negative root,

F

initially becomes positive and creates a reverse-flow region next to the wall. The physically likely choice is therefore

c = + arcosh 5 .

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