 Solution

ID: past-exam-of-the-differential-equations-and-applications-course-of-the-university-of-oxford/2010/b/6/a/solution

For

η = x / y^{β}

the chain rule gives

u_{y} = - β η f^{'} / y

u_{x} = y^{- β} f^{'}

and

(u u_{x})_{x} = y^{- 2 β} [f f^{''} + (f^{'})^{2}]

y

gives

- β η f^{'} = y^{1 - 2 β} [f f^{''} + (f^{'})^{2}] + \frac{1}{2} f .

A nontrivial scale-invariant balance therefore selects

β = 1/2

and yields

f f^{''} + (f^{'})^{2} + \frac{1}{2} η f^{'} + \frac{1}{2} f = 0.

Its left side is the derivative of

f f^{'} + η f /2

f (0) = 1

and

f^{'} (0) = 0

, so the integration constant is zero:

f (f^{'} + η /2) = 0.

On the positive branch before the first zero, divide by

f

and integrate, obtaining

f = 1 - η^{2} /4

. Its first positive zero is

η_{F} = 2

. Hence

u (x, y) = 1 - \frac{x ^{2}}{4 y} (0 \leq x \leq 2 y), x_{F} (y) = 2 y .

A direct check gives

u_{y} = x^{2} / (4 y^{2})

and

(u u_{x})_{x} + u / (2 y) = x^{2} / (4 y^{2})

. The zero extension past this front is also a natural weak compact-support continuation:

u

is continuous and the degenerate diffusive flux

u u_{x}

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