1 5CCM211A & 6CCM211B – outline solutions 4 Part II Setting t = 0 in eqn (0.3) of the Assignment sheet we obtain ψ(x)= ∞ X n=1 A n sin nπx a Using question [1] we hence obtain the stated formula for A n . Differentiating eqn (0.3) on the Assignment w.r.t. t and setting t = 0 the other boundary condition gives φ(x)= ∞ X n=1 B n nπc a sin nπx a and once again question [1] implies the remaining formula. Consider now the specific example given, in which φ(x)=0 and ψ(x)= 3 10 x, if 0 ≤ x ≤ 1 3 3 20 (1 - x), if 1 3 ≤ x ≤ 1 (1) Since φ = 0 we have B n = 0, while since L = 1 the A n are given by A n = 2 Z 1 0 ψ(x) sin (nπx) dx = 3 5 Z 1/3 0 x sin (nπx) dx + 3 10 Z 1 1/3 (1 - x) sin (nπx) dx = - cos(nπ/3) 5nπ + 3 5 sin(nπ/3) 5n 2 π 2 + cos(nπ/3) 5nπ + 3 10 sin(nπ/3) 5n 2 π 2 = 9 10π 2 sin(nπ/3) n 2 . Substituting these values for A n and B n the asserted formula results.
