To find the solution of the nonlinear initial-value problem (1), we consider a
sequence of auxiliary linear initial-value problems and their continuous solutions. The
characteristic start time is shown in Fig. 2.
Mathematical problem (1a):
x
m
c
dt
x
d
1
2
2
−
=
,
1
,
0t
t
,
()
0
0=
x
,
()
0
0
V
dt
dx
=
.
A solution to the problem (1a):
()
t
m
c
c
m
V
t
x
=
1
1
0
sin
,
1
1
2
c
m
t
=
.
Mathematical problem (2a):
x
m
c
dt
x
d
2
2
2
−
=
,
2
1
,t
t
t
,
()
1
0
1
c
m
V
t
x
=
,
()
0
1
=
t
dt
dx
.
A solution to the problem (2a):
()
(
)
−
=
1
2
1
0
cos
t
t
m
c
c
m
V
t
x
,
()
(
)
−
−
=
1
2
1
2
0
sin
t
t
m
c
c
c
V
t
dt
dx
.
Find the instant of time
2
t, when
()
0
2
=
t
x
.
We have the equation:
()
(
)
0
cos
1
2
1
0
=
−
=
t
t
m
c
c
m
V
t
x
, and then we obtain
2
1
2
2
c
m
t
t
+
=
. (3)
When
2
t
t
we have an initial-value problem (3a):
x
m
c
dt
x
d
3
2
2
−
=
,
4
2
,t
t
t
,
()
0
2
=
t
x
,
()
1
1
2
0
2
V
c
c
V
t
dt
dx
=
−
=
. (4)
We have the solution to the problem (3a):
()
)
(
sin
2
3
3
1
t
t
m
c
c
m
V
t
x
−
=
,
).
(
cos
2
3
1
t
t
m
c
V
dt
dx
−
=
(5)
From condition
()
0
4
=
t
x
we find
2
2
4
c
m
t
t
+
=
.
Mathematical problem (4a):
x
m
c
dt
x
d
1
2
2
−
=
,
5
4
,t
t
t
,
()
0
4
=
t
x
,
()
1
4
V
t
dt
dx
−
=
.
Solution to the problem (4a):
()
)
(
sin
4
1
1
1
t
t
m
c
c
m
V
t
x
−
−
=
,
)
(
cos
4
1
1
t
t
m
c
V
dt
dx
−
−
=
,
1
4
5
2
c
m
t
t
+
=
,
()
()
1
2
1
1
2
1
0
1
1
5
2
sin
c
c
t
x
c
c
c
m
V
c
m
V
t
x
=
=
−
=
.
We give a general formula for solving the general initial-value problem:
- 1563 -