General Variation of a Functional

Transcription

1 4 General Variation of a Functional 4 1

2 Chapter 4: GENERAL VARIATION OF A FUNCTIONAL TABLE OF CONTENTS Page 4.1 General Variation of a Functional Extremals with Corners

3 4.1 GENERAL VARIATION OF A FUNCTIONAL This Chapter treats the following generalizations of the basic-functional extremization problem: variable end points, transversality conditions, and extremals with corners ( broken extremals ) General Variation of a Functional The general variation of a functional occurs when both dependent variables and the limits of integration are allowed to vary. This is common in many applications, especially in time dependent problems. In this Section, we restrict ourselves to the one-dimensional functional of Chapter 6, but allow the end points to vary, see Figure 9.1. Figure 9.1 The variable end-point problem We have x1 J = F(y, y, x) dx (4.1) where now, x 1 may also vary. Gelfand and Fomin (Chapter 13) show that the first variation becomes x1 [ ] δ J = Ehdx + Fy x1 [ ] δy + (F y Fy y1 )δx y 0 x1 [ ] x1 [ ] x1 (4.2) = Ehdx + p δy Hδx, where as usual E = F y (d/dx)f y denotes the left hand side of the Euler equation, and in which we have introduced the so-called canonical variables: the momentum p and the Hamiltonian H. These are defined by p = F y = F y, H = F + y F y = F + y p. (4.3) The Hamiltonian often has the meaning of total energy of a dynamical system if x is time. 4 3

4 Chapter 4: GENERAL VARIATION OF A FUNCTIONAL The Euler equation E = 0 still defines the extremals. If δ, δy 0, δx 1, δy 1 were all independently varied, then the left end conditions would be { δy0 = 0 (essential) or p 0 = 0 (natural) δ = 0 (essential) or H 0 = 0 (natural) (4.4) and at the right end: { δy1 = 0 (essential) or p 1 = 0 (natural) δx 1 = 0 (essential) or H 1 = 0 (natural) (4.5) These would represent four boundary conditions (BCs) for the second order ODE E = 0. That s two too many! It is necessary to constrain the end variations. If y 0 = ϕ( ) and y 1 = ψ(x 1 ), then δy 0 = ϕ ( )δx and δy 1 = ψ (x 1 )δx because the δ s behave like d s. Then, keeping the δx s as independent end variation we have δ J = x1 Ehdx + (ψ p H) 1 δx 1 (ϕp H) 0 δ = 0. (4.6) The conditions (pϕ H) 0 = 0, (pψ H) 1 = 0, (4.7) are called transversality conditions, and are the new natural boundary conditions for the variable end problem. Note that if, say, ϕ 0, p 0 = 0 and we recover the well known natural BC for δ = 0. Similarly if ψ 1 then p 1 = 0. On the other hand if ϕ 0 = 0 the condition becomes H 0 = 0 because δy 0 = 0. If F has the particular form f (x, y) 1 + y 2, then Gelfand and Fomin show that the transversality conditions become (t ϕ ) 0 = 1at the left and (y ψ ) 1 = 1at the right. Then transversality becomes orthogonality and the extremals intersect the end curves at right angles Extremals with Corners Up to now we have assumed that the extremals are C continuous, that is, they have a continuous first derivative. We may enlarge the class of admissible functions, however, by admitting isolated discontinuities or jumps in the derivatives. This new class comprise the so called piecewise C functions. The simplest example occurs in the one dimensional functional J = xb x a F(y, y, x) dx, (4.8) if the solution y = y(x) is allowed to have a corner at a point x = c. Gelfand and Fomin (Chapter 14) call this a broken extremal although the name is not really appropriate (probably an unhappy 4 4

5 4.2 EXTREMALS WITH CORNERS translation from the Russian). See Figure 9.2. The integral (4.8) naturally splits into two: J = Figure 9.2 Extremal with a corner at x = c c a F(y, y, x) dx + b c F(y, y, x) dx (4.9) The first variation yields two pieces of the Euler equation E = 0: one from a to c, and another one from c to b. Each piece gives two constants of integration; hence four boundary conditions are needed. The end conditions at x = a and x = b provide two conditions. Continuity of y(x) at x = c provides one more. The last one comes from the transversality condition at x = c, which may be enunciated as follows: where p c δy c H c δx c = p + c δy c H + c δx c = 0 (4.10) p c = Fy c, p + c = Fy c+, p c = F y Fy c, p + c = F y Fy c+. (4.11) in which c and c+ denote the values at x = c approached from the left and right, respectively. Three cases may now be distinguished. (I) (II) (III) x = c is fixed, that is, the position of the corner is known a priori. Then δx c = 0 and the momentum p must be continuous at x = c because p c δy c = p + c δy c. The Hamiltonian H may jump at x = c. y c is fixed but the position x = c of the jump may vary. Then the Hamiltonian H must be continuous at x = c, because H c δx c = H + c δx c. The momentum p may jump at x = c. Point c is constrained to lie on the curve y = ϕ(x). Then the combination pϕ H must be continuous at x = c. This situation includes (I) and (II) if one sets ϕ and 4 5

6 Chapter 4: GENERAL VARIATION OF A FUNCTIONAL ϕ = 0, respectively. (Noter similarity to the variable end point problem discussed in the previous Section.) The additional transversality condition furnished by one of these cases completes the four needed for the determination of the solution extremal. Remark 4.1. The paragraph of page 63 of Gelfand and Fomin following their Eq. (18) is incorrect in stating that both p and H are continuous at x = c and thus provide the additional two conditions necessary to match the two Euler solution pieces. The correct conditions are the ones stated above. Example 4.1. An end-loaded bar fabricated with two materials. Let u(x) be the axial displacement. Other quantities are defined below: J = 1 2 xb x a EA(u ) 2 dx Pu b = 1 2 E J A J c a b (u ) 2 dx + 1 E 2 S A S (u ) 2 dx Pu b. (4.12) The transversality condition at x = c is that of case (I) because x = c is given a priori. Thus, the momentum p = F/ u = EAu is continuous: E J A J u c = E S A S u c+, (4.13) But u = e is the axial bar strain, Eu = σ, the axial bar stress, and p = Aσ the axial bar force (usually denoted by N). Thus (4.13) says that the axial force is continuous at the interface as is physically obvious from equilibrium considerations. (In fact, that force is constant along the bar.) The case of a moving material interface in which δu C = 0butδx C may vary is not physically relevant to this problem but appears in more general mechanical models. Examples: elastoplasticity or phase changes (melting, solidification, state transition). c 4 6