Appendix: All Solutions to the Problems
777
Solve for c 2
c 2 =
1
e −1 − e −3 =
e
2
e 1 − e −1
Then determine c 1
c 1 = −
e
−2
e 1 − e −1
Thus the extremal curve is
y = −
e
−2
e 1 − e −1 e
x
+
e
2
e 1 − e −1 e
−x
=
e
2−x
− e
−(2−x)
e 1 − e −1
=
sinh(2 − x)
sinh 1
2.42 Find the extremal curve of the functional J [y] =
1
0 yy
2 dx, the boundary
conditions are y(0) = 1, y(1) =
3
√
4.
Solution Because the integrand does not contain x, the first integral of the Euler
equation is
yy
2
− 2yy
2
= c
or
√ ydy = ±cdx
Integrating we get
y
3
2 = ±c 1 x + c 2 , y =
3
(c 2 ± c 1 x) 2
According to the boundary conditions y(0) = 1, y(1) =
3
√
4, we get c 2 = 1,
c 1 = 1, or c 2 = −1, x 1 = 3. Thus the extremal curve is y =
3
(x + 1) 2 和y =
3
(3x − 1) 2 .
2.43 Find the extremal curve of the functional J [y] =
1
0 y
2 y
2 dx, the boundary
conditions are y(0) = 0, y(1) = 1.
Solution Because the integrand does not contain x, the first integral of the Euler
equation is
y
2 y
2
− 2y
2 y
2
= −c
2
or