 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-336/3/a/solution

y = y_{0} + ϵ y_{1} + \dots, ω^{2} = ω_{0}^{2} + ϵ ω_{1}^{2} + \dots .

The reduced leading equation is

y_{0}^{''} + ω_{0}^{2} y_{0} = 0.

For the symmetric fundamental mode,

y_{0} = a cos (ω_{0} x)

. The reduced second-order problem can impose the displacement conditions

y_{0} (\pm 1) = 0

but not both clamped-slope conditions. The smallest positive root is

ω_{0} = \frac{π}{2}, y_{0} = a cos \frac{π x}{2} .

Near the right endpoint introduce the stretched coordinate

X_{+} = \frac{1 - x}{ϵ},

and near the left endpoint use

X_{-} = (1 + x) / ϵ

. The leading outer solution is

O (ϵ)

in either inner region, so put

y = ϵ Y_{0} + \dots

. The dominant inner equation is

Y_{0}^{''''} - Y_{0}^{''} = 0.

At either endpoint, clamping gives

Y_{0} (0) = Y_{0}^{'} (0) = 0

, bounded matching excludes the growing exponential, and the outer behavior requires

Y_{0} \sim \frac{aπ}{2} X_{\pm} (X_{\pm} \to \infty) .

Thus the two leading inner solutions have the same form:

y_{in}^{\pm} = ϵ \frac{aπ}{2} (X_{\pm} - 1 + e^{- X_{\pm}}) .

They show explicitly that each clamped endpoint has a boundary layer of width

O (ϵ)

At order

ϵ

, the outer equation is

y_{1}^{''} + \frac{π ^{2}}{4} y_{1} = - ω_{1}^{2} a cos \frac{π x}{2} .

The symmetric solution is

y_{1} = b cos \frac{π x}{2} - \frac{ω _{1}^{2} a}{π} x sin \frac{π x}{2},

which is the stated resonant particular solution plus a homogeneous normalization term.

x = 1 - ϵ X_{+}

approaches the right endpoint from the outer region,

y_{0} + ϵ y_{1} \sim ϵ [\frac{aπ}{2} X_{+} - \frac{ω _{1}^{2} a}{π}] .

The large-

X_{+}

y_{in}^{+} \sim ϵ [\frac{aπ}{2} X_{+} - \frac{aπ}{2}] .

Matching the constant terms gives

ω_{1}^{2} = \frac{π ^{2}}{2} .

The left endpoint gives the same condition by symmetry. Therefore

ω^{2} \sim \frac{π ^{2}}{4} + \frac{π ^{2}}{2} ϵ + O (ϵ^{2}) .

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