12.2 Calculus of Variations 669
12.2.2 Problem of Calculus of Variations
A simple problem in the theory of the calculus of variations with no constraints can
be stated as follows:
Find a functionu(x)that minimizes the functional (integral)
A=
∫x 2
x 1
F(x, u, u′, u′′) dx (12.1)
whereAandFcan be called functionals (functions of other functions). Herexis the
independent variable,
u=u(x), u′=
du(x)
dx
, and u′′=
d^2 u(x)
dx^2
In mechanics, the functional usually possesses a clear physi cal meaning. For example,
in the mechanics of deformable solids, the potential energy (π) plays the role of the
functional (πis a function of the displacement componentsu, v, andw, which, in turn,
are functions of the coordinatesx, y, andz).
The integral in Eq. (12.1) is defined in the region or domain [x 1 , x 2 ]. Let the values
ofube prescribed on the boundaries asu(x 1 )=u 1 and u(x 2 )=u 2. These are called
theboundary conditionsof the problem. One of the procedures that can be used to
solve the problem in Eq. (12.1) will be as follows:
1.Select a series of trial or tentative solutionsu(x)for the given problem and
express the functionalAin terms of each of the tentative solutions.
2.Compare the values ofAgiven by the different tentative solutions.
3.Find the correct solution to the problem as that particular tentative solution
which makes the functionalAassume an extreme or stationary value.
The mathematical procedure used to select the correct solution from a number of
tentative solutions is called the calculus of variations.
Stationary Values of Functionals. Any tentative solutionu(x)in the neighborhood
of the exact solutionu(x)may be represented as (Fig. 12.1)
u(x) =u(x)+δu(x)
tentative exact variation
solution solution ofu
(12.2)
The variation inu(i.e.,δu) is defined as an infinitesimal, arbitrary change inufor a
fixed value of the variablex(i.e., forδx=0). Hereδis called thevariational operator
(similar to the differential operatord). The operation of variation is commutative with
both integration and differentiation, that is,
δ
(∫
F dx
)
=
∫
(δF ) dx (12.3)
δ
(
du
dx
)
=
d
dx
(δu) (12.4)
