Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 39C / c

Table of contents
- Solution c

Solution

The simultaneous diagonalization from part (a) applies to both

B

and

C

. With

θ_{k} = kπh

,

Q^{- 1} BQ = D = diag (d_{1}, \dots, d_{m}), d_{k} = - \frac{10}{3} + \frac{4}{3} cos θ_{k},

and

Q^{- 1} CQ = E = diag (e_{1}, \dots, e_{m}), e_{k} = \frac{2}{3} + \frac{1}{3} cos θ_{k} .

Multiplying the

j

th block equation by

Q^{- 1}

v_{j} = Q^{- 1} u_{j}

and

c_{j} = Q^{- 1} b_{j}

gives

E v_{j - 1} + D v_{j} + E v_{j + 1} = c_{j},

with the absent boundary terms omitted. This is exactly system (3).

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