826
Appendix: All Solutions to the Problems
The Euler equation of the functional is
2y − 2y
= 0
or
y
− y = 0
The general solution of the equation is
y = c 1 e
x
+ c 2 e
−x
From the boundary condition y(0) = 0, we get c 2 = −c 1 . Thus the extremal
curve is y = c sinh x, the derive is y
= c 1 (e
x
+ e
−x
) = c cosh x.
On the right boundary, the extremal curve, the extremal curve and the boundary
curve intersect, the intersection point relation is
y 1 = c sinh x 1 = e
2x 1
Substituting the extremal curve and the intersection point relation into the natural
boundary condition, we obtain
sinh
2 x 1 − cosh
2 x 1 + 4 cosh x 1 sinh x 1 = 0
Using the relation sinh
2 x 1 + 1 = cosh
2 x 1 , the above equation can be changed to
4 cosh x 1 sinh x 1 − 1 = 0
Expanding the above equation and operating
(e
x 1 + e
−x 1 )(e
x 1 − e
−x 1 ) − 1 = e
2x 1 − 1 − e
−2x 1 = 0
The above equation is multiplied by e
2x 1 and put y = e
2x 1 , then
y
2
− y − 1 = 0
The solution of the equation is
y =
1 +
√
5
2
= e
2x 1
Solve for x 1
x 1 =
1
2
ln
1 +
√
5
2