
602 SECOND-ORDER HYPERBOLIC EQUATIONS WITH ONE SPACE VARIABLE
leads to the equation ∂
zy
u = 0. Thus, the general solution of the original equation has the
form
w = (ax + b)[f (z) + g(y)],
where f = f(z) and g = g(y) are arbitrary functions.
⊙ Literature: N. H. Ibragimov (1994).
6.
∂
2
w
∂t
2
= (a
2
− x
2
)
2
∂
2
w
∂x
2
+ Φ(x, t).
Domain: −l ≤ x ≤ l. First boundary value problem.
The following conditions are prescribed:
w = f (x) at t = 0 (initial condition),
∂
t
w = g(x) at t = 0 (initial condition),
w = 0 at x = l (boundary condition ),
w = 0 at x = −l (boundary condition).
Solution for 0 < l < a:
w(x, t)=
∂
∂t
Z
l
−l
f(ξ)G(x, ξ, t) dξ+
Z
l
−l
g(ξ)G(x, ξ, t) dξ+
Z
t
0
Z
l
−l
Φ(ξ, τ )G(x,