3.5 Sufficient Conditions of Extrema of Functionals
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in the Legendre strong condition (3.5.3), then the functional (3.5.1) at least can get
a weak maximum.
Example 3.5.6 Discuss the extremum of the functional J [y] =
x 1
0
1+y 2
2gy
dx, the
boundary conditions are y(0) = 0, y(x 1 ) = y 1 .
Solution This is the brachistochrone problem. Solve for that the family of extremal
curves satisfying the boundary condition y(0) = 0 is the cycloid
⎧
⎪ ⎨
⎪ ⎩
x =
c
2
(θ − sin θ)
y =
c
2
(1 − cos θ)
where, the constant c can be determined by another boundary condition y(x 1 ) = y 1 .
Let c = 2a, then the extremal curve satisfying the given boundary condition is
x = a(θ − sin θ)
y = a(1 − cos θ)
When 0 < x 1 < 2π a, the above bunch of cycloids forms the extremal curve
field taking the coordinate origin (0, 0) as the center, and including extremal curve
satisfying the two given boundary conditions, namely the Jacobi condition holds.
Moreover for any y
, there is
F y y =
1
√
2gy(1 + y 2 )
3
2
> 0
Therefore, when 0 < x 1 < 2πa, on the cycloid, the given functional gets a strong
minimum.
Example 3.5.7 Let the functional J [y] =
2
0 (e
y
+ a)dx, where a is an arbitrary
real constant, the boundary conditions are y(0) = 1, y(2) = 3. Discuss its extremal
situation.
Solution Since the integrand of the functional is only the function of y
, so the
extremal curve is a family of straight lines y = c 1 x + c 2 , the extremal curve
conforming to the boundary conditions is y = x + 1, it can be included in the
extremal curve field of the family of extremal curves y = c 1 x + c 2 . For any y
, there
is F y y = e
y
> 0, thus, the functional gets a strong minimum on the extremal curve
y = x + 1.
Example 3.5.8 Making use of the Legendre condition to discriminate whether the
functional J [y] =
2
1 (x y
4
− 2yy
3
)dx has extremum, the boundary conditions are
y(1) = 0, y(2) = 1.