2.3 Variations of the Simplest Functionals and Necessary Conditions …
105
Table 2.1 The corresponding
relations of the functional and
function
Function
Functional
Function f (x)
Functional J [y(x)]
Variable y = f (x)
Variable J = J [y(x)]
Independent variable x
Function y(x)
Increment x of independent
variable x
Variation δy of the function
y(x)
Differential dy of the function Variation δ J of the functional
Example 2.3.2 Verify J [y] =
x 1
x 0
[ p(x)y + q(x)y
]dx is a linear functional, where,
p(x) and q(x) are known functions of x.
Solution Since
J [cy] =
x 1
x 0
[ p(x)cy + q(x)cy
]dx = c
x 1
x 0
[ p(x)y + q(x)y
]dx = c J [y]
J [y 1 + y 2 ] =
x 1
x 0
[ p(x)(y 1 + y 2 ) + q(x)(y 1 + y 2 )
]dx
=
x 1
x 0
[ p(x)y 1 + q(x)y
1 ]dx +
x 1
x 0
[ p(x)y 2 + q(x)y
2 ]dx
= J [y 1 ] + J [y 2 ]
so J [y] is a linear functional.
Example 2.3.3 Verify J [y, z] =
x 1
x 0
f (x)yzdx is a symmetric bilinear functional,
where, f (x) is the known function of x.
Solution Since
J [y, z] =
x 1
x 0
f (x)yzdx =
x 1
x 0
f (x)zydx = J [z, y]
J [a 1 y 1 + a 2 y 2 , z] =
x1
x0
f (x)(a 1 y 1 + a 2 y 2 )zdx =
x1
x0
f (x)a 1 y 1 zdx +
x1
x0
f (x)a 2 y 2 zdx
= a 1
x1
x0
f (x)y 1 zdx + a 2
x1
x0
f (x)y 2 zdx = a 1 J [y 1 , z] + a 2 J [y 2 , z]
J [y, b 1 z 1 + b 2 z 2 ] =
x1
x0
f (x)y(b 1 z 1 + b 2 z 2 )dx =
x1
x0
f (x)yb 1 z 1 dx +
x1
x0
f (x)yb 2 z 2 dx
= b 1
x1
x0
f (x)yz 1 dx + b 2
x1
x0
f (x)yz 2 dx = b 1 J [y, z 1 ] + b 2 J [y, z 2 ]
so J [y] is a symmetric bilinear functional.