Appendix: All Solutions to the Problems
807
or
y
+ y = −3 sin 2x
The general solution of the function is y = c 1 cos x + c 2 sin x + sin 2x. From the
boundary conditions y(0) = 0, y
π
4
= 1, we get c 1 = c 2 = 0, hence the extremal
curve is y = sin 2x.
The Weierstrass function is
E = y
2
− (y
+ u)
2
+ 6y sin 2x − (y
2
− y
2
+ 6y sin 2x) + u2y
= −u
2
< 0
Therefore y = sin 2x reaches the strong maximum.
3.13 J [y] =
x 1
0
dx
y , y(0) = 0, y(x 1 ) = y 1 , x 1 > 0, y 1 > 0.
Solution F =
1
y , the Euler equation of the functional is
d
dx
−
1
y 2
= 0
The first integral is
dx
dy
=
1
c 1
The general solution of the equation is y = c 1 x +c 2 . Of course, since the integrand
is only the function of y
, the above result can also be directly obtained. From the
boundary conditions y(0) = 0, y(x 1 ) = y 1 , x 1 > 0, y 1 > 0, we get c 1 =
y 1
x 1
, c 2 = 0.
The extremal curve is y =
y 1
x 1
x. The Weierstrass function is
E =
1
(y + u) 2 −
1
y 2 +
2u
y 3 =
u
2
(3y
+ 2u)
(y + u) 2 y 3
When 3y
+ 2u > 0, E > 0, when 3y
+ 2u < 0, E < 0, so the sign of E
can not be determined, the extremal curve can not reach the strong extremum. But
F y y =
2
y 3 , along the extremal curve y =
y 1
x 1
x, there is F y y =
2x
3
1
y
3
1
> 0, so the
functional can reach the weak minimum on the extremal curve y =
y 1
x 1
x.
3.14 J [y] =
x 1
0
dx
y 2 , y(0) = 0, y(x 1 ) = y 1 , x 1 > 0, y 1 > 0.
Solution F =
1
y 2 , the Euler equation of the functional is
d
dx
−
2
y 3
= 0