16.21 Techniques of Structural Analysis and Design Spring 2003 Unit #9 - Calculus of Variations Let u be the actual con?guration of a structure or mechanical system. u ? satis?es the displacement boundary conditions: u = u on S u . De?ne: ˉu = u + αv, where: α : scalar v : arbitrary function such that v = 0 on S u We are going to de?ne αv as δu, the ?rst variation of u: δu = αv (1) Schematically: u1 u2u a b u(a) u(b) v 1 ? As a ?rst property of the ?rst variation: dˉu du dv = + α dx dx ???? dx so we can identify α dv with the ?rst variation of the derivative of u: dx ? du dx = α dv dx δ But: dv dx αdv dx d dx (δu)α = = We conclude that: ? du ? d δ = dx dx (δu) Consider a function of the following form: F = F (x,u(x),u ? (x)) It depends on an independent variable x, another function of x (u(x)) and its derivative (u ? (x)). Consider the change in F , when u (therefore u ? ) changes: ΔF = F (x,u + δu,u ? + δu ? ) ? F (x,u,u ? ) = F (x,u + αv,u ? + αv ? ) ? F (x,u,u ? ) expanding in Taylor series: ?F ?F 1 ? 2 F 1 ? 2 F ΔF = F + αv + ?u ? αv ? + ?u 2 (αv) 2 + ?u?u ? (αv)(αv ? ) + ···? F ?u 2! 2! ?F ?F = αv + ?u ? αv ? + h.o.t. ?u First total variation of F: ? ΔF ? δF = α lim α→0 α ? F (x,u + αv,u ? + αv ? ) ? F (x,u,u ? ) ? = α lim α→0 α ? ?F αv + ?F = α lim ?u ?u ? αv ? ? = ?F αv + ?F ? ?u ? αv α→0 α ?u δF = ?F ?u δu + ?F ?u ? δu ? 2 ? ? ? ? Note that: dF (x,u + αv,u ? + αv ? ) ? ? δF = α ? dα α=0 since: dF (x,u + αv,u ? + αv ? ) ?F (x,u + αv,u ? + αv ? ) ?F (x,u + αv,u ? + αv ? ) ? = v + dα ?u ?u ? evaluated at α = 0 dF (x,u + αv,u ? + αv ? ) ? ? ?F (x,u,u ? ) ?F (x,u,u ? ) ? ? = v + v dα α=0 ?u ?u ? Note analogy with di?erential calculus. δ(aF 1 + bF 2 ) = aδF 1 + bδF 2 linearity δ(F 1 F 2 ) = δF 1 F 2 + F 1 δF 2 etc The conclusions for F (x,u,u ? ) can be generalized to functions of several ?u i independent variables x i and functions u i , ?x j : ? ?u i ? F x i ,u i , ?x j We will be making intensive use of these properties of the variational operator δ: d d dv ? du ? = δ ? dx (δu) ? = dx (αv) = ? α dx dx δudx = αvdx = α vdx = δ udx Concept of a functional b I(u) = F (x,u(x),u ? (x))dx a 3 v ? ? ? ? ? ? ? ? ? First variation of a functional: δI = δ F (x,u(x),u ? (x))dx = δ F (x,u(x),u ? (x)) dx ? ? ?F ?F ? ? δI = δu + dx ?u ?u ? δu Extremum of a functional “u 0 ” is the minumum of a functional if: I(u) ≥ I(u 0 )?u A necessary condition for a functional to attain an extremum at “u 0 ” is: dI δI(u 0 ) = 0, or dα (u 0 + αv,u ? 0 + αv ? ) ? ? = 0 α=0 Note analogy with di?erential calculus. Also di?erence since here we require dF = 0 at α = 0. dα b? ?F ?F ? ? δI = δu + ?u ? δu dx ?u a Integrate by parts the second term to get rid of δu ? . b? ?F d ? ?F ? d ? ?F ?? δI = δu + ?u ? δu ? δu dx ?u ? dx ?u dx a ? b? ?F d ? ?F ?? ?F ? ? b = ? δudx + ? ?u dx ?u ? ?u ? δu a a Require δu to satisfy homogeneous displacement boundary conditions: δu(b) = δu(a) = 0 Then: ? b? ?F d ? ?F ?? δI = ? δudx = 0, ?u dx ?u ? a 4 ? ? ? ?δu that satisfy the appropriate di?erentiability conditions and the homoge- neous essential boundary conditions. Then: ?F ?u ? d dx ? ?F ?u ? ? = 0 These are the Euler-Lagrange equations corresponding to the variational problem of ?nding an extremum of the functional I. Natural and essential boundary conditions A weaker condition on δu also allows to obtain the Euler equations, we just need: ?F ? b ? = 0 ?u ? δu a which is satis?ed if: ? δu(a) = 0 and δu(b) = 0 as before ?F ? δu(b) = 0 and ?u ? (b) = 0 ?F ? ?u ? (a) = 0 and δu(b) = 0 ?F ?F ? ?u ? (a) = 0 and ?u ? (b) = 0 Essential boundary conditions: δu ? S u = 0, or u = u 0 on S u ?F Natural boundary conditions: ?u ? = 0 on S. Example: Derive Euler’s equation corresponding to the total po- tential energy functional Π = U + V of an elastic bar of length L, Young’s modulus E, area of cross section A ?xed at one end and subject to a load P at the other end. ? L EA ? du ? 2 Π(u) = dx ? Pu(L) 0 2 dx Compute the ?rst variation: EA du ? du ? δΠ = ? 2 ? 2 dx δ dx dx ? Pδu(L) 5 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?? ? Integrate by parts ? d δΠ = dx ? L EA ? d dx EA δu dx ? Pδu(L) du dx δu du dx EA dx + EA L 0 d dx du dx du dx = ? ? Pδu(L) δu δu 0 Setting δΠ = 0, ? δu / δu(0) = 0: d dx ? EA du dx ? = 0 du ? P = EA ? dx L Extension to more dimensions I = F (x i ,u i ,u i,j )dV V ? ?F V dV ?F δu i,j ?u i,j ? ?F δI = δu i + ?u i = V ? ?F δu i + ?u i ? ?F δu i ?u i,j δu i dV ? ?x j ? ?x j ? ?u i,j Using divergence theorem: δI = V ? ?F ?u i ? S ? ?F ?u i,j δu i dV + ? ?x j ?F δu i n j dS ?u i,j Extremum of functional I is obtained when δI = 0, or when: ?F ? ?F ? ? ?x j ? ?u i,j = 0 , and ?u i δu i = 0 on S u ?F n j onS ? S u = S t ?u i,j The boxed expressions constitute the Euler-Lagrange equations correspond- ing to the variational problem of ?nding an extremum of the functional I. 6