 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-336/2/b/solution

X = ϵ x

. The equation becomes

ϵ^{2} y_{XX} + k (X)^{2} y = 0, k (X) = 1 + X^{3} .

Since

k

is positive for every

X \geq 0

y \sim \frac{1}{k ( X )} [C_{+} e^{i ϵ^{- 1} \int_{0}^{X} k (S) d S} + C_{-} e^{- i ϵ^{- 1} \int_{0}^{X} k (S) d S}] .

The condition

y (0) = 0

selects a sine, and

y_{x} (0) = 1

fixes its coefficient. Since

\frac{1}{ϵ} \int_{0}^{X} (1 + S^{3}) d S = x + \frac{ϵ ^{3} x ^{4}}{4},

the result is

y (x) \sim \frac{1}{1 + ( ϵ x ) ^{3}} sin (x + \frac{ϵ ^{3} x ^{4}}{4}) .

The WKB validity measure

ϵ \frac{∣ k ^{'} ( X ) ∣}{k ( X ) ^{2}} = ϵ \frac{3 X ^{2}}{( 1 + X ^{3} ) ^{2}}

is uniformly

O (ϵ)

and tends to zero at both ends of

X \geq 0

. With no turning point, the approximation is therefore uniformly valid on the entire stated half-line.

 New to topics? Read the docs here!