From Pilkey and Wunderlich book 9.1.1 Euler Equation
1 The fundamental problem of the calculus of variation1 is to find a function u(x) such that Π =
b
a
F (x, u, u)dx (9.1)
is stationary. Or,
δΠ = 0 (9.2)
where δ indicates the variation
2 We define u(x) to be a function of x in the interval (a, b), and F to be a known function (such as the energy density).
3 We define the domain of a functional as the collection of admissible functions belonging to a class of functions in function space rather than a region in coordinate space (as is the case for a function).
4 We seek the function u(x) which extremizes Π.
1Differential calculus involves a function of one or more variable, whereas variational calculus involves a function of a function, or a functional.
u, u
x=b x x=c
x=a
A B C
dx du
u(x) u(x)
Figure 9.1: Variational and Differential Operators
5 Letting ˜u to be a family of neighbouring paths of the extremizing function u(x) and we assume that at the end points x = a, b they coincide. We define ˜u as the sum of the extremizing path and some arbitrary variation, Fig. 9.1.
˜
u(x, ε) = u(x) + εη(x) = u(x) + δu(x) (9.3) where ε is a small parameter, and δu(x) is the variation of u(x)
δu = ˜u(x, ε)− u(x) (9.4-a)
= εη(x) (9.4-b)
and η(x) is twice differentiable, has undefined amplitude, and η(a) = η(b) = 0. We note that
˜
u coincides with u if ε = 0
6 It can be shown that the variation and derivation operators are commutative
d
dx(δu) = u˜(x, ε)− u(x) δu = u˜(x, ε)− u(x)
d
dx(δu) = δ
du dx
(9.5)
7 Furthermore, the variational operator δ and the differential calculus operator d can be simi-larly used, i.e.
δ(u)2 = 2uδu (9.6-a)
δ(u + v) = δu + δv (9.6-b)
δ
udx
=
(δu)dx (9.6-c)
δu = ∂u
∂xδx +∂u
∂yδy (9.6-d)
Draft
9.1 † Variational Calculus; Preliminaries 9–3 however, they have clearly different meanings. du is associated with a neighbouring point at a distance dx, however δu is a small arbitrary change in u for a given x (there is no associated δx).8 For boundaries where u is specified, its variation must be zero, and it is arbitrary elsewhere.
The variation δu of u is said to undergo a virtual change.
9 To solve the variational problem of extremizing Π, we consider Π(u + εη) = Φ(ε) =
Integration by part (Eq. 5.1 and 5.1) of the second term leads to
b
12 The fundamental lemma of the calculus of variation states that for continuous Ψ(x) in a≤ x≤ b, and with arbitrary continuous function η(x) which vanishes at a and b, then
b
condition for u(x) to extremize Π.
14 Generalizing for a functional Π which depends on two field variables, u = u(x, y) and v = v(x, y)
Π =
F (x, y, u, v, u,x, u,y, v,x, v,y,· · · , v,yy)dxdy (9.15) There would be as many Euler equations as dependent field variables
∂F
∂u −∂x∂ ∂u∂F,x −∂y∂ ∂u∂F,y +∂x∂22∂u∂F
,xx + ∂x∂y∂2 ∂u∂F
,xy + ∂y∂22∂u∂F
,yy = 0
∂F
∂v −∂x∂ ∂v∂F,x −∂y∂ ∂v∂F,y +∂x∂22 ∂F
∂v,xx +∂x∂y∂2 ∂v∂F
,xy +∂y∂22 ∂F
∂v,yy = 0 (9.16)
15 We note that the Functional and the corresponding Euler Equations, Eq. 9.1 and 9.14, or Eq. 9.15 and 9.16 describe the same problem.
16 The Euler equations usually correspond to the governing differential equation and are referred to as the strong form (or classical form).
17 The functional is referred to as the weak form (or generalized solution). This classification stems from the fact that equilibrium is enforced in an average sense over the body (and the field variable is differentiated m times in the weak form, and 2m times in the strong form).
18 It can be shown that in the principle of virtual displacements, the Euler equations are the equilibrium equations, whereas in the principle of virtual forces, they are the compatibility equations.
19 Euler equations are differential equations which can not always be solved by exact meth-ods. An alternative method consists in bypassing the Euler equations and go directly to the variational statement of the problem to the solution of the Euler equations.
20 Finite Element formulation are based on the weak form, whereas the formulation of Finite Differences are based on the strong form.
21 We still have to define δΠ. The first variation of a functional expression is δF = ∂F∂uδu +∂F∂uδu
δΠ = )abδF dx
δΠ =
b
a
∂F
∂uδu +∂F
∂uδu
dx (9.17)
As above, integration by parts of the second term yields δΠ =
b
a
δu
∂F
∂u − d dx
∂F
∂u
dx (9.18)
22 We have just shown that finding the stationary value of Π by setting δΠ = 0 is equivalent to finding the extremal value of Π by setting dΦ(ε)dε (((
ε=0 equal to zero.
23 Similarly, it can be shown that as with second derivatives in calculus, the second variation δ2Π can be used to characterize the extremum as either a minimum or maximum.
Draft
9.1 † Variational Calculus; Preliminaries 9–5 9.1.2 Boundary Conditions24 Revisiting the integration by parts of the second term in Eq. 9.10, we had
b
a
η∂F
∂udx = η ∂F
∂u ((((b
a
− b
a
η d dx
∂F
∂udx (9.19)
We note that
1. Derivation of the Euler equation required η(a) = η(b) = 0, thus this equation is a state-ment of the essential (or forced) boundary conditions, where u(a) = u(b) = 0.
2. If we left η arbitrary, then it would have been necessary to use ∂F∂u = 0 at x = a and b.
These are the natural boundary conditions.
25 For a problem with, one field variable, in which the highest derivative in the governing differential equation is of order 2m (or simply m in the corresponding functional), then we have Essential (or Forced, or geometric) boundary conditions, involve derivatives of order zero (the field variable itself) through m-1. Trial displacement functions are explicitely required to satisfy this B.C. Mathematically, this corresponds to Dirichlet boundary-value problems.
Nonessential (or Natural, or static) boundary conditions, involve derivatives of order m and up. This B.C. is implied by the satisfaction of the variational statement but not explicitly stated in the functional itself. Mathematically, this corresponds to Neuman boundary-value problems.
26 Table 9.1 illustrates the boundary conditions associated with some problems
Problem Axial Member Flexural Member
Distributed load Distributed load Differential Equation AEddx2u2 + q = 0 EIddx4w4 − q = 0
m 1 2
Essential B.C. [0, m− 1] u w,dwdx
Natural B.C. [m, 2m− 1] dudx ddx2w2 and ddx3w3
or σx = Eu,x or M = EIw,xx and V = EIw,xxx
Table 9.1: Essential and Natural Boundary Conditions
Example 9-1: Extension of a Bar
sectional area A, fixed at left end and subjected to an axial force P at the right one is given by
Determine the Euler Equation by requiring that Π be a minimum.
Solution:
Solution I The first variation of Π is given by