3.2 The Jacobi Conditions and Jacobi Equation
207
According to the expression (3.2.4) and expression (3.2.11), when η
− η
u
u
= 0,
δ
2 J is the same as F y y in sign.
Let the functional J [y(x)] =
x 1
x 0
F(x, y, y
)dx, the points are A(x 0 , y 0 ) and
B(x 1 , y 1 ), y = y(x) is the extremal function of the functional, if u(x) is the solution
of the Jacobi equation, u(x 0 ) = 0, besides x 0 , let the root of the equation u(x) = 0
be x
∗ , then the root x
∗ is called conjugate value of x 0 , and point A c (x
∗
, y(x
∗
)) is
called the conjugate point of point A on the extremal function y = y(x).
Example 3.2.1 Let the family of curves be u = c(x − 1)x, find the conjugate point
of the coordinate origin (0, 0).
Solution Obviously, when x = 0 and x = 1, u = 0, the conjugate point of the
coordinate origin (0, 0) is point (1, 0).
Example 3.2.2 Known the family of curves u = c sinh x, determine whether the
coordinate origin (0, 0) has the conjugate point or not?
Solution Because only when x = 0 for the hyperbola, so that y = 0, and when x
is other any value different from zero, there is all u = 0, therefore the coordinate
origin (0, 0) has not the conjugate point.
As the Jacobi equation is a second order linear homogeneous equation, let its
two linearly independent particular solutions be u 1 (x) and u 2 (x), thus the general
solution is
u(x) = c 1 u 1 (x) + c 2 u 2 (x)
(3.2.12)
where, c 1 and c 2 are arbitrary constants.
From the condition u(x 0 ) = 0 of the solution for the Jacobi equation, we get
c 1 u 1 (x 0 ) + c 2 u 2 (x 0 ) = 0
or
u 1 (x 0 )
u 2 (x 0 )
= −
c 2
c 1
If x
∗ is the root of the equation u(x) = 0, namely x
∗ is abscissa of the conjugate
point A c of A(x 0 , y 0 ), then there is
c 1 u 1 (x
∗
) + c 2 u 2 (x
∗
) = 0
207
According to the expression (3.2.4) and expression (3.2.11), when η
− η
u
u
= 0,
δ
2 J is the same as F y y in sign.
Let the functional J [y(x)] =
x 1
x 0
F(x, y, y
)dx, the points are A(x 0 , y 0 ) and
B(x 1 , y 1 ), y = y(x) is the extremal function of the functional, if u(x) is the solution
of the Jacobi equation, u(x 0 ) = 0, besides x 0 , let the root of the equation u(x) = 0
be x
∗ , then the root x
∗ is called conjugate value of x 0 , and point A c (x
∗
, y(x
∗
)) is
called the conjugate point of point A on the extremal function y = y(x).
Example 3.2.1 Let the family of curves be u = c(x − 1)x, find the conjugate point
of the coordinate origin (0, 0).
Solution Obviously, when x = 0 and x = 1, u = 0, the conjugate point of the
coordinate origin (0, 0) is point (1, 0).
Example 3.2.2 Known the family of curves u = c sinh x, determine whether the
coordinate origin (0, 0) has the conjugate point or not?
Solution Because only when x = 0 for the hyperbola, so that y = 0, and when x
is other any value different from zero, there is all u = 0, therefore the coordinate
origin (0, 0) has not the conjugate point.
As the Jacobi equation is a second order linear homogeneous equation, let its
two linearly independent particular solutions be u 1 (x) and u 2 (x), thus the general
solution is
u(x) = c 1 u 1 (x) + c 2 u 2 (x)
(3.2.12)
where, c 1 and c 2 are arbitrary constants.
From the condition u(x 0 ) = 0 of the solution for the Jacobi equation, we get
c 1 u 1 (x 0 ) + c 2 u 2 (x 0 ) = 0
or
u 1 (x 0 )
u 2 (x 0 )
= −
c 2
c 1
If x
∗ is the root of the equation u(x) = 0, namely x
∗ is abscissa of the conjugate
point A c of A(x 0 , y 0 ), then there is
c 1 u 1 (x
∗
) + c 2 u 2 (x
∗
) = 0