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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2007/paper-1/8/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2007 / Paper 1 / 8 / Solution by

Codex  0 Created 2026-10-18 Updated 2026-10-20

Choose the left homogeneous solution

v (x) = sinh x

and the right one

w (x) = sinh (1 - x)

. Their Wronskian is

v w^{'} - v^{'} w = - sinh 1

. The Dirichlet modified Helmholtz Green function on an interval, with operator sign

D_{x}^{2} - 1

, is therefore

G (x, ξ) = - \frac{s i n h ( x _{<} ) s i n h ( 1 - x _{>} )}{s i n h 1}, x_{<} = min (x, ξ), x_{>} = max (x, ξ) .

It vanishes at both endpoints, is continuous at

x = ξ

, and its first derivative jumps by

+ 1

(D_{x}^{2} - 1) G = δ (x - ξ)

. The Green-function representation then gives

u (x) = - \frac{s i n h ( 1 - x )}{s i n h 1} \int_{0}^{x} e^{ξ} sinh ξ d ξ - \frac{s i n h x}{s i n h 1} \int_{x}^{1} e^{ξ} sinh (1 - ξ) d ξ .

The two integrals are

(e^{2 x} - 1) /4 - x /2

and

e (1 - x) /2 - (e - e^{2 x - 1}) /4

, respectively. Substituting them and simplifying gives

u (x) = \frac{1}{2} [x e^{x} - \frac{e sinh x}{sinh 1}] .

Direct differentiation verifies

u^{''} - u = e^{x}

and both zero boundary values. The negative kernel also explains why the solution is negative inside the interval despite the positive right-hand side.

Figure 1.
A negative Dirichlet Green kernel produces a negative response to positive forcing
. Left: three source locations give continuous kernels with derivative jump $+ 1$ . Right: the integrated solution of $u^{''} - u = e^{x}$ returns to zero at both boundaries. Details.

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