 Solution

ID: past-exam-of-the-topics-in-applied-mathematics-course-of-the-university-of-oxford/2011/a/2/c/i/solution

Y = U^{q /2}

and

k = 2 b / q

. The proposed profile with

a, b > 0

gives

U^{'} = - k U (1 - Y), U^{''} = k^{2} U (1 - Y) [1 - (1 + q /2) Y] .

Substituting into

U^{''} + c U^{'} + U (1 - U^{q}) = 0

and dividing in the interior by

U (1 - Y)

yields

k^{2} [1 - (1 + q /2) Y] - c k + 1 + Y = 0.

Equating the constant and

Y

k = \frac{2}{q + 2}, b = \frac{q}{2 ( q + 2 )}, c = \frac{q + 4}{2 ( q + 2 )} .

U (z) = (1 + a e^{q z / 2 (q + 2)})^{- 2/ q}, a > 0.

The free positive

a

translates the front. It connects one on the left to zero on the right and is strictly decreasing. At

q = 1

, its speed is

5/ 6

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