1.5 Fundamental Lemmas of the Calculus of Variations
47
When m or n increases, the chosen function class of η(x) becomes small, the
conditions of lemma becomes weak. Especially when taking m = n = 0, there is
the following lemma:
Lemma 1.5.2 Let a function f (x) be continuous in the interval [a, b], an arbitrary
function η(x) is continuous in the interval [a, b], and satisfies the boundary condition
η(a) = η(b) = 0, always makes the integral
b
a
f (x)η(x)dx = 0
(1.5.8)
hold, then there must be f (x) ≡ 0 in the interval [a, b].
Lemma 1.5.3 Let a function f (x, y) be continuous in the closed domain D, both
arbitrary function η(x, y) and its first partial derivative are continuous in the domain
D, and η(x, y) is equal to zero on the boundary line L of D, always make the integral
¨
D
f (x, y)η(x, y)dxdy = 0
(1.5.9)
hold, then there must be f (x, y) ≡ 0 in the domain D.
Proof Let at a point (ξ, ζ ) in the domain D, a function f (x, y) be positive value. It
can be seen from the continuity of the function f (x, y) that there must exist a circle
with center (ξ, ζ ) and radius ρ. In the circle, f (x, y) > 0, and the circle is included
in the domain D. Choosing the function η(x, y) as follows
η(x, y) =
ρ
2
− (x − ξ)
2
+ (y − ζ )
2
(x − ξ)
2
+ (y − ζ )
2
< ρ
2
0
(x − ξ)
2
+ (y − ζ )
2
≥ ρ
2
(1.5.10)
It is not difficult to verify that η(x, y) satisfies all the conditions for Lemma 1.5.3,
but the integral value is positive at the moment, and the circle is included in D, this
contradicts the conditions of Lemma 1.5.3, therefore Lemma 1.5.3 is verified. Quod
erat demonstrandum.
Lemma 1.5.4 Let the function f (x, y) be continuous in the closed domain D, all
the nth partial derivatives of arbitrary function η(x, y) are continuous in the domain
D, L is the border of D, as long as
∂
k
η
∂ x i ∂ y k−i
L
= 0,
k = 0, 1, 2, . . . , n − 1
i = 0, 1, 2, . . . , k
,
always make the integral
¨
D
f (x, y)η(x, y)dxdy = 0
(1.5.11)
47
When m or n increases, the chosen function class of η(x) becomes small, the
conditions of lemma becomes weak. Especially when taking m = n = 0, there is
the following lemma:
Lemma 1.5.2 Let a function f (x) be continuous in the interval [a, b], an arbitrary
function η(x) is continuous in the interval [a, b], and satisfies the boundary condition
η(a) = η(b) = 0, always makes the integral
b
a
f (x)η(x)dx = 0
(1.5.8)
hold, then there must be f (x) ≡ 0 in the interval [a, b].
Lemma 1.5.3 Let a function f (x, y) be continuous in the closed domain D, both
arbitrary function η(x, y) and its first partial derivative are continuous in the domain
D, and η(x, y) is equal to zero on the boundary line L of D, always make the integral
¨
D
f (x, y)η(x, y)dxdy = 0
(1.5.9)
hold, then there must be f (x, y) ≡ 0 in the domain D.
Proof Let at a point (ξ, ζ ) in the domain D, a function f (x, y) be positive value. It
can be seen from the continuity of the function f (x, y) that there must exist a circle
with center (ξ, ζ ) and radius ρ. In the circle, f (x, y) > 0, and the circle is included
in the domain D. Choosing the function η(x, y) as follows
η(x, y) =
ρ
2
− (x − ξ)
2
+ (y − ζ )
2
(x − ξ)
2
+ (y − ζ )
2
< ρ
2
0
(x − ξ)
2
+ (y − ζ )
2
≥ ρ
2
(1.5.10)
It is not difficult to verify that η(x, y) satisfies all the conditions for Lemma 1.5.3,
but the integral value is positive at the moment, and the circle is included in D, this
contradicts the conditions of Lemma 1.5.3, therefore Lemma 1.5.3 is verified. Quod
erat demonstrandum.
Lemma 1.5.4 Let the function f (x, y) be continuous in the closed domain D, all
the nth partial derivatives of arbitrary function η(x, y) are continuous in the domain
D, L is the border of D, as long as
∂
k
η
∂ x i ∂ y k−i
L
= 0,
k = 0, 1, 2, . . . , n − 1
i = 0, 1, 2, . . . , k
,
always make the integral
¨
D
f (x, y)η(x, y)dxdy = 0
(1.5.11)