282
4 Problems with Variable Boundaries
where, y i ∈ C
2n i [x 0 , x 1 ], i = 1, 2, . . . , r , F ∈ C max(n 1 ,n 2 ,...,n r )+1 , the admissible
curve y i = y i (x) is fixed at the left endpoint A(x 0 , y i0 ), it is variable at the right
endpoint B(x 1 , y i1 ). The extremal function of the functional (4.3.36) must satisfy
the Euler-Poisson equations
F y i +
n i
k=1
(−1)
k
d
k F y
(k)
i
dx k = 0 (i = 1, 2, . . . , r )
(4.3.37)
Generally, the every one of Euler-Poisson Eqs. (4.3.37) is an ordinary differential
equation of 2 n i -th order, its general solution has 2 n i arbitrary constants. Because
the right endpoint B is variable, x 1 is undetermined, therefor 2(n 1 +n 2 +· · ·+n r )+1
arbitrary constants need to be determined in all. Because the left endpoint A is fixed,
n 1 + n 2 + · · · + n r arbitrary constants of the general solution of Euler-Poisson
equations can be determined by the boundary conditions y i (x 0 ) = y i0 , y
i (x 0 ) = y
i0 ,
y
i (x 0 ) = y
i0 , …, y
(n i −1)
i
(x 0 ) = y
(n i −1)
i0
. In order to determine the unique solution, it
is necessary to find out other n 1 + n 2 + · · · + n r + 1 equations to determine the rest
n 1 +n 2 +· · ·+n r +1 undetermined constants, the n 1 +n 2 +· · ·+n r +1 equations can
be obtained by means of the necessary condition δ J = 0 of the functional obtaining
the extremum. Taking after the preceding analysis methods, there is
δ J =
⎧
⎨
⎩
F − y
i
⎡
⎣ F y
i
+
ni −1
k=1
(−1)
k
d k F y
(k+1)
i
dx k
⎤
⎦ − y
i
⎡
⎣ F y
i
−
ni −2
k=1
(−1)
k
d k F y
(k+2)
i
dx k
⎤
⎦ − · · ·
− y
( j)
i
⎡
⎣ F y
( j)
i
+
ni − j
k=1
(−1)
k
d k F y
(k+ j)
i
dx k
⎤
⎦ − · · · − y
(ni −1)
i
F
y
(n i −1)
i
−
d F
y
(n i )
i
dx
− y
(ni )
i
F
y
(n i )
i
x=x1
δx 1
+
⎡
⎣ F y
i
+
ni −1
k=1
(−1)
k
d k F y
(k+1)
i
dx k
⎤
⎦
x=x1
δy i1 +
⎡
⎣ F y
i
+
ni −2
k=1
(−1)
k
d k F y
(k+2)
i
dx k
⎤
⎦
x=x1
δy
i1 + · · ·
+
⎡
⎣ F y
( j)
i
+
ni − j
k=1
(−1)
k
d k F y
(k+ j)
i
dx k
⎤
⎦
x=x1
δy
( j−1)
i1
+ · · ·
+
F
y
(n i −1)
i
−
d F
y
(n i )
i
dx
x=x1
δy
(ni −2)
i1
+ F
y
(n i )
i
x=x1
δy
(ni −1)
i1
+
x1
x0
⎡
⎣ F yi +
ni
k=1
(−1)
k
d k F y
(k)
i
dx k
⎤
⎦ δy i dx = 0
(4.3.38)
4 Problems with Variable Boundaries
where, y i ∈ C
2n i [x 0 , x 1 ], i = 1, 2, . . . , r , F ∈ C max(n 1 ,n 2 ,...,n r )+1 , the admissible
curve y i = y i (x) is fixed at the left endpoint A(x 0 , y i0 ), it is variable at the right
endpoint B(x 1 , y i1 ). The extremal function of the functional (4.3.36) must satisfy
the Euler-Poisson equations
F y i +
n i
k=1
(−1)
k
d
k F y
(k)
i
dx k = 0 (i = 1, 2, . . . , r )
(4.3.37)
Generally, the every one of Euler-Poisson Eqs. (4.3.37) is an ordinary differential
equation of 2 n i -th order, its general solution has 2 n i arbitrary constants. Because
the right endpoint B is variable, x 1 is undetermined, therefor 2(n 1 +n 2 +· · ·+n r )+1
arbitrary constants need to be determined in all. Because the left endpoint A is fixed,
n 1 + n 2 + · · · + n r arbitrary constants of the general solution of Euler-Poisson
equations can be determined by the boundary conditions y i (x 0 ) = y i0 , y
i (x 0 ) = y
i0 ,
y
i (x 0 ) = y
i0 , …, y
(n i −1)
i
(x 0 ) = y
(n i −1)
i0
. In order to determine the unique solution, it
is necessary to find out other n 1 + n 2 + · · · + n r + 1 equations to determine the rest
n 1 +n 2 +· · ·+n r +1 undetermined constants, the n 1 +n 2 +· · ·+n r +1 equations can
be obtained by means of the necessary condition δ J = 0 of the functional obtaining
the extremum. Taking after the preceding analysis methods, there is
δ J =
⎧
⎨
⎩
F − y
i
⎡
⎣ F y
i
+
ni −1
k=1
(−1)
k
d k F y
(k+1)
i
dx k
⎤
⎦ − y
i
⎡
⎣ F y
i
−
ni −2
k=1
(−1)
k
d k F y
(k+2)
i
dx k
⎤
⎦ − · · ·
− y
( j)
i
⎡
⎣ F y
( j)
i
+
ni − j
k=1
(−1)
k
d k F y
(k+ j)
i
dx k
⎤
⎦ − · · · − y
(ni −1)
i
F
y
(n i −1)
i
−
d F
y
(n i )
i
dx
− y
(ni )
i
F
y
(n i )
i
x=x1
δx 1
+
⎡
⎣ F y
i
+
ni −1
k=1
(−1)
k
d k F y
(k+1)
i
dx k
⎤
⎦
x=x1
δy i1 +
⎡
⎣ F y
i
+
ni −2
k=1
(−1)
k
d k F y
(k+2)
i
dx k
⎤
⎦
x=x1
δy
i1 + · · ·
+
⎡
⎣ F y
( j)
i
+
ni − j
k=1
(−1)
k
d k F y
(k+ j)
i
dx k
⎤
⎦
x=x1
δy
( j−1)
i1
+ · · ·
+
F
y
(n i −1)
i
−
d F
y
(n i )
i
dx
x=x1
δy
(ni −2)
i1
+ F
y
(n i )
i
x=x1
δy
(ni −1)
i1
+
x1
x0
⎡
⎣ F yi +
ni
k=1
(−1)
k
d k F y
(k)
i
dx k
⎤
⎦ δy i dx = 0
(4.3.38)