Appendix: All Solutions to the Problems
787
By y(0) = 0, we get c 1 = 0. From y
(0) = 0, we get c 2 +c 3 = 0. From y
π
2
= 1,
we get c 3 + c 4
π
2
= 1. From y
π
2
=
π
2
, we get −c 2
π
2
+ c 4 =
π
2
. It can work out
c 2 = −1, c 3 = 1, c 4 = 0. Substituting them into the above equation, we obtain
y = sin x − x cos x
2.62 In the curves through A(0, 0) and B(1, 0), find the curve satisfying the boundary
conditions y
(0) = a, y
(1) = b, and make the functional J [y] =
1
0 y
2 dx obtain
extremum.
Solution The Euler–Poisson equation of the functional is
d
2
dx 2 F y = 2y
(4)
= 0
Integrating four times, we get
y = c 1 x
3
+ c 2 x
2
+ c 3 x + c 4
From the boundary conditions y(0) = 0, y(1) = 0, y
(0) = a, y
(1) = b, we
get c 4 = 0, c 3 = a, c 2 = −(2a + b), c 1 = a + b, thus, the extremal curve is
y = (a + b)x
3
− (2a + b)x
2
+ ax.
2.63 Find the extremal curve of the functional J [y] =
x 1
x 0
y
k dx, where, k = 0.
Solution Because the integrand F = y
k does not contain y and y
, the Euler
equation of the functional is
d
2
d
2 x
ky
k−1
= 0
or
y
= (c 11 x + c 21 )
1
k−1
Integrating two times, we get
y =
(k − 1) 2
k(2k − 1)
(c 11 x + c 21 )
2k−1
k−1 + c 3 x + c 4 = (c 1 x + c 2 )
2k−1
k−1 + c 3 x + c 4
k = 1, k =
1
2
When k =
1
2
, F y = −
1
2
y
−
1
2 , the Euler equation is
d
2
dx 2
−
1
2
y
−
1
2
= 0