
612 SECOND-ORDER HYPERBOLIC EQUATIONS WITH ONE SPACE VARIABLE
6
◦
. Domain: 0 ≤ x ≤ l. Third boundary value problem.
The following conditions are prescribed:
w = f
0
(x) at t = 0 (initial condition),
∂
t
w = f
1
(x) at t = 0 (initial condition),
∂
x
w − s
1
w = g
1
(t) at x = 0 (boundary condition),
∂
x
w + s
2
w = g
2
(t) at x = l (boundary condition).
The solution w(x, t) is determined by the formula in Item 5
◦
with
G(x, ξ, t) = exp
−
1
2
kt
∞
X
n=1
y
n
(x)y
n
(ξ) sin
t
p
a
2
µ
2
n
− p
B
n
p
a
2
µ
2
n
− p
, p = b +
1
4
k
2
,
y
n
(x) = cos(µ
n
x) +
s
1
µ
n
sin(µ
n
x), B
n
=
s
2
2µ
2
n
µ
2
n
+ s
2
1
µ
2
n
+ s
2
2
+
s
1
2µ
2
n
+
l
2
1 +
s
2
1
µ
2
n
.
Here, the µ
n
are positive roots of the transcendental equation
tan(µl)
µ
=
s
1
+ s
2
µ
2
− s
1
s
2
.
If the inequali ...