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= 0F(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= 0F(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
dλF(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= 0F(u+λv) =∫Rn[
∇u·∇v+f′(u)v]
dx=(by the Gauss formulas( 3. 2. 36 ))=∫Rn[
−div(∇u)+f′(u)]
vdx.