Euler Equation

1 The fundamental problem of the calculus of variation1 is to find a function u(x) such that ib n= / F (x,u,u)dx (3.1)

J a is stationary. Or,

where S 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 n.

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. C.1.

where e is a small parameter, and Su(x) is the variation of u(x)

1 Differential calculus involves a function of one or more variable, whereas variational calculus involves a function of a function, or a functional.

Euler Equation

Figure C.1: Variational and Differential Operators and n(x) is twice differentiable, has undefined amplitude, and n(a) = n(b) = 0. We note that u coincides with u if e = 0

6 The variational operator S and the differential calculus operator d have clearly different meanings. du is associated with a neighboring point at a distance dx, however Su is a small arbitrary change in u for a given x (there is no associated Sx).

7 For boundaries where u is specified, its variation must be zero, and it is arbitrary elsewhere. The variation Su of u is said to undergo a virtual change.

8 To solve the variational problem of extremizing n, we consider

Since u ^ u as e ^ 0, the necessary condition for n to be an extremum is d$(e)

10 From Eq. 3.3 and applying the chain rule with e = 0, u = u, we obtain d$(e)

11 It can be shown (through integration by part and the fundamental lemma of the calculus of variation) that this would lead to _

12 This differential equation is called the Euler equation associated with n and is a necessary condition for u(x) to extremize n.

13 Generalizing for a functional n which depends on two field variables, u = u(x, y) and v = v(x, y)

n= / F (x,y,u,v,utX,uty ,v,x,v,y, •••,v,yy) dxdy (3.9)

There would be as many Euler equations as dependent field variables dF du dF dv d dF dx du>x d dF dx dv x d dF dy duy d dF dy dvy d2

dy2 dvy dxdy du,Xy dy2 du,yy

14 We note that the Functional and the corresponding Euler Equations, Eq. 3.1 and 3.8, or Eq. 3.9 and 3.10 describe the same problem.

15 The Euler equations usually correspond to the governing differential equation and are referred to as the strong form (or classical form).

16 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).

17 Euler equations are differential equations which can not always be solved by exact methods. 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.

18 Finite Element formulation are based on the weak form, whereas the formulation of Finite Differences are based on the strong form.

19 Finally, we still have to define Sn

As above, integration by parts of the second term yields

20 We have just shown that finding the stationary value of n by setting ¿n = 0 is equivalent to finding the extremal value of n by setting equal to zero.

2i Similarly, it can be shown that as with second derivatives in calculus, the second variation S2n can be used to characterize the extremum as either a minimum or maximum.

22 Revisiting the integration by parts of the second term in Eq. 3.7, we obtain

rb .d—



t d d— d

Ja nd—dx =



We note that

1. Derivation of the Euler equation required n(a) = n(b) = 0, thus this equation is a statement of the essential (or forced) boundary conditions, where u(a) = u(b) = 0.

2. If we left n arbitrary, then it would have been necessary to use dF = 0 at x = a and b. These are the natural boundary conditions.

23 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.

These boundary conditions were already introduced, albeit in a less formal way, in Table 9.1.

24 Table C.1 illustrates the boundary conditions associated with some problems


Axial Member

Flexural Member

Distributed load

Distributed load

Differential Equation

AE dxU + q = 0

E & - q = 0




Essential B.C. [0,m - 1]



Natural B.C. [m, 2m - 1]

dx2 dx3

or M = EIw xx and V = EIw xxx

Table C.1: Essential and Natural Boundary Conditions

The total potential energy n of an axial member of length L, modulus of elasticity E, cross sectional area A, fixed at left end and subjected to an axial force P at the right one is given by n = fL EA(du\ 2 dx - Pu(L) (3.14)

Determine the Euler Equation by requiring that n be a minimum. Solution:

Solution I The first variation of n is given by

Integrating by parts we obtain sn dx du dx dx du



— Su




EA —

Idu dx

The last term is zero because of the specified essential boundary condition which implies that âu(0) = 0. Recalling that â in an arbitrary operator which can be assigned any value, we set the coefficients of âu between (0, L) and those for 5u at x = L equal to zero separately, and obtain

Euler Equation:

Natural Boundary Condition:

Solution II We have

(note that since P is an applied load at the end of the member, it does not appear as part of F(x, u, u') To evaluate the Euler Equation from Eq. 3.8, we evaluate dF dF

Thus, substituting, we obtain dF d dF

dx dx

Example C-2: Flexure of a Beam

The total potential energy of a beam is given by

Derive the first variational of n. Solution:

Extending Eq. 3.11, and integrating by part twice rL ( dF ç ,, , dF

(EIw''Sw')\L - [(EIw'')'öw' - pôw] dx (3.23-c) Jo

(EIw''ôw')\L - [(EIw'')'ôw]\L + / [(EIw'')'' + p] ôwdx = 0 (3.23-d)

(EIw'')'' = -p for all x which is the governing differential equation of beams and

0 0

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