
592 SECOND-ORDER HYPERBOLIC EQUATIONS WITH ONE SPACE VARIABLE
◮ Domain: R
1
≤ r ≤ R
2
. Second boundary value problem.
The following conditions are prescribed:
w = f
0
(r) at t = 0 (initial condition),
∂
t
w = f
1
(r) at t = 0 (initial condition),
∂
r
w = g
1
(t) at r = R
1
(boundary condition),
∂
r
w = g
2
(t) at r = R
2
(boundary condition).
Solution:
w(r, t) =
Z
t
0
Z
R
2
R
1
Φ(ξ, τ )G(r, ξ, t − τ) dξ dτ
+
∂
∂t
Z
R
2
R
1
f
0
(ξ)G(r, ξ, t) dξ +
Z
R
2
R
1
f
1
(ξ)G(r, ξ, t) dξ
− a
2
Z
t
0
g
1
(τ)G(r, R
1
, t −τ) dτ + a
2
Z
t
0
g
2
(τ)G(r, R
2
, t − τ ) dτ.
Here,
G(r, ξ, t)=
3ξ
2
sin
t
√
b
(R
3
2
−R
3
1
)
√
b
+
2ξ
(R
2
−R
1
)r
∞
X
n=1
(1+R
2
2
λ
2
n
)Ψ
n
(r)Ψ
n
(ξ) sin
t
p
a
2
λ
2
n
+b
λ
2
n
R
2
1
+R
2
2
+R
1
R
2
(1+R
1
R
2
λ
2
n
)
p
a
2
λ
2
n
+b
,
Ψ
n
(r)=sin[λ
n
(r−R
1
)]+R
1
λ
n
cos[λ
n
(r−R
1
)] ...