846
Appendix: All Solutions to the Problems
From the boundary conditions x(0) = x 0 , x(t 1 ) = x 1 , y(0) = y 0 , y(t 1 ) = y 1 , we
obtain
c 1 = 12
(y 1 + y 0 )t 1 − 2(x 1 − x 0 )
t
3
1
, c 2 = 4
3(x 1 − x 0 ) − t 1 (y 1 + 2y 0 )
t
2
1
c 3 = y 0 , c 4 = x 0
Thus, the expressions of the constraint conditions are
x =
(y 1 + y 0 )t 1 − 2(x 1 − x 0 )
t
3
1
t
3
+
3(x 1 − x 0 ) − t 1 (y 1 + 2y 0 )
t
2
1
t
2
+ y 0 t + x 0
y = 3
(y 1 + y 0 )t 1 − 2(x 1 − x 0 )
t
3
1
t
2
+ 2
3(x 1 − x 0 ) − t 1 (y 1 + 2y 0 )
t
2
1
t + y 0
The extremal function is
u = 6
(y 1 + y 0 )t 1 − 2(x 1 − x 0 )
kt
3
1
t + 2
3(x 1 − x 0 ) − t 1 (y 1 + 2y 0 )
kt
2
1
Solution 2 It follows from the constraint conditions that u = x 1
, substituting the
relation into the functional, there is
J =
t 1
0
x
2
1 dt
The Euler equation of the functional is
2
d
4 x 1
dt 4 = 0
Integrating four times, we get
x 1 = c 1 t
3
+ c 2 t
2
+ c 3 t + c 4
From the boundary conditions x(0) = x 0 , x(t 1 ) = x 1 , y(0) = y 0 , y(t 1 ) = y 1 , we
obtain
c 1 =
(y 1 +y 0 )t 1 −2(x 1 −x 0 )
t
3
1
, c 2 =
3(x 1 −x 0 )−t 1 (y 1 +2y 0 )
t
2
1
c 3 = y 0 , c 4 = x 0
Thus the expressions of the constraint conditions are
x =
(y 1 + y 0 )t 1 − 2(x 1 − x 0 )
t
3
1
t
3
+
3(x 1 − x 0 ) − t 1 (y 1 + 2y 0 )
t
2
1
t
2
+ y 0 t + x 0