Page 101 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

SOLUTIONS USING FOURIER SERIES AND INTEGRALS 91
solution of the form

λx

2
B n exp(−λ t)sin (5.98)
c
for a region of x on the interval (0, c ). Thus, for x on the interval 0 ≤ x ≤∞ we have
∞
2
B(α) = f (ς)sin ας dς (5.99)
π
ς=0

and the solution is

∞ ∞

2
2
u(x, t) = exp(−λ t)sin(λx) f (s )sin(λs )ds d α (5.100)
π
λ=0 s =0
Noting that
2sin α s sin α x = cos α (s − x) − cos α(s + x) (5.101)

and that

∞
1 π b
2
2
exp(−γ α)cos(γ b)dγ = exp − (5.102)
2 α 4α
0
we have
∞
1 (s − x) 2 (s + x) 2
u(x, t) = √ f (s ) exp − − exp − ds (5.103)
2 πt 4t 4t
0
2 (s −x) 2
Substituting into the first of these integrals σ = and into the second integral
4t
(s + x) 2
2
σ = (5.104)
4t
∞
1 √ −σ 2
u(x, t) = √ f (x + 2σ t)e d σ
π
√
−x/2 t
∞
1 √ −σ 2
− √ f (−x + 2σ t)e d σ (5.105)
π
√
x/2 t