3.4. VARIATIONS WITHDIVA-FREE CONSTRAINTS 145
whereδFis the derivative operator ofF.
Given a variational problem, it is important to computeδFfor a given functionalF.
Hereafter we give a brief introduction of general methods tocompute the derivative operators
fromF.
LetFbe the functional given by (3.4.1), andX∗be the dual space ofX. The derivative
operatorδF(u)ofFatu∈Xis a linear functional onX, i.e.δF(u)∈X∗for eachu∈X. In
other words,δFis a mapping fromXtoX∗:
(3.4.4) δF:X→X∗.
Denote〈·,·〉the product betweenXandX∗, i.e.
〈·,·〉:X×X∗→R.
Then the derivative operatorδFin (3.4.4) satisfies the relation:
(3.4.5) 〈δF(u),v〉=
d
dλ
∣
∣
∣
λ= 0
F(u+λv) foru,v∈X,
whereλ∈R^1 is real number.
Based on (3.4.5), it is easy to see that the minimal pointusatisfying (3.4.2) is a solution
of the variational equation (3.4.3). In fact, by (3.4.2) for any givenv∈Xthe functionf(λ) =
F(u+λv)is minimal atλ=0:
d f( 0 )
dλ
= 0 ⇒
d
dλ
∣
∣
∣
λ= 0
F(u+λv) = 0 ∀v∈X.
It follows then by (3.4.5) that
〈δF(u),v〉= 0 ∀v∈X,
which means thatusatisfies (3.4.3).
In the following, we give a simple example to show how to computeδFusing formula
(3.4.5).
LetX=H^1 (Rn), andF:H^1 (Rn)→R^1 be given by
(3.4.6) F(u) =
∫
Rn
[
1
2
|∇u|^2 +f(u)
]
dx foru∈H^1 (Rn).
We see that
d
dλ
F(u+λv) =
d
dλ
∫
Rn
[
1
2
|∇u+λ∇v|^2 +f(u+λv)
]
dx
=
∫
Rn
[
(∇u+λ∇v)·∇v+f′(u+λv)v
]
dx.
Hence we have
d
dλ
∣
∣
∣
λ= 0
F(u+λv) =
∫
Rn
[
∇u·∇v+f′(u)v
]
dx
=(by the Gauss formulas( 3. 2. 36 ))
=
∫
Rn
[
−div(∇u)+f′(u)
]
vdx.