834
Appendix: All Solutions to the Problems
that tan(β − α) =
tan β−tan α
1+tan β tan α
=
ϕ
−y
1+ϕ y , and tan
−
π
4
= −1, thus β − α = −
π
4
,
consequently α = β +
π
4
. Quod erat demonstrandum.
4.22 Using the necessary condition δ J = 0 of a functional obtaining extremum, find
the function that can make the functional J [y] =
1
0 (y
2
− 2x y)dx reach extremum.
The boundary conditions are y(0) = y
(0) = 0, y(1) =
1
120
, y
(1) is not given.
Solution The Euler equation of the functional is
−2x + 2y
(4)
= 0
or y
(4)
= x, integrating four times, we get
y =
x
5
120
+ c 1 x
3
+ c 2 x
2
+ c 3 x + c 4
From the boundary conditions y(0) = y
(0) = 0, we get c 3 = c 4 = 0, and from
y(1) =
1
120
, we get c 1 + c 2 = 0. Taking the first variation to the functional, there is
δ J = 2
1
0
(y
δy
− xδy)dx = 2
1
0
y
dδy
− 2
1
0
xδydx
= 2y
δy
x=1
x=0
− 2
1
0
y
dδy − 2
1
0
xδydx
= 2y
δy
x=1
x=0
− 2y
δy
x=1
x=0
+ 2
1
0
[y
(4)
− x]δydx = 0
Since y
(1) has not been given, we get y
(1) = 0, 0 =
1
6
+ 6c 1 + 2c 2 , solving
simultaneously the equation and the equation c 1 + c 2 = 0, we obtain c 1 = −c 2 =
−
1
24
. Therefore, the extremal curve is y =
x
5
120
+
1
24
(x
2
− x
3
).
4.23 Under the condition that the admissible curve can not pass through the interior
of the circle domain surrounded by the circumference (x − 5)
2
+ y
2
= 9, find the
curve that can make the functional J [y] =
10
0 y
3 dx reach extremum, the boundary
conditions are y(0) = y(10) = 0.
Solution Because F = y
3 only depends on y
, the integral curve of the Euler
equation is
y = c 1 x + c 2
From y(0) = 0, we get c 2 = 0, thus y = c 1 x.
Let y = ϕ(x) = ±
9 − (x − 5) 2 and y = c 1 x + c 2 be contingence at point
(x 2 ,y 2 ), point (x 2 ,−y 2 ), point (x 3 ,y 3 ) and point (x 3 ,−y 3 ), here 0 < x 2 < x 3 < 10,
then