16.21 Techniques of Structural Analysis and sig Spring 2003 Unit #9 -Calculus of Variations Let u be the actual configuration of a structure or mechanical system. u satisfies the displacement boundary conditions: u=u* on Su. Define =+0 here a: scalar v: arbitrary function such that v=0 on S, e are going to define au as du, the first variation of u L=QU (1) Schematically u(b)-
16.21 Techniques of Structural Analysis and Design Spring 2003 Unit #9 - Calculus of Variations Let u be the actual configuration of a structure or mechanical system. u ∗ satisfies the displacement boundary conditions: u = u on Su. Define: u¯ = u + αv, where: α : scalar v : arbitrary function such that v = 0 on Su We are going to define αv as δu, the first variation of u: δu = αv (1) Schematically: u1 u2 u a b u(a) u(b) v 1
first property of the first variation so we can identify at with the first variation of the derivative of u But d=dr (6) Consider a function of the following form F=F(x,(x),(x) It depends on an independent variable x, another function of x(u(a)and derivative(u()). Consider the change in F, when u(therefore u) chang △F=F(x,u+6,u2+6u)-F(x,u,) F(a,u+ )-F(x expanding in Taylor series OF OF 1 aF 1 aF 2!2(a) 2 dudu (av)(a')+ OF OF +~0t+h.o.t First total variation of F. SF lim =alim aut aute OF a OF OF SF
� As a first property of the first variation: du¯ du dv = + α dx dx ���� dx so we can identify αdv with the first variation of the derivative of u: dx �du dx = α dv dx δ But: dv dx αdv dx d dx α = = (δu) 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 ∂2F 1 ∂2F ΔF = F + αv + ∂u� αv� + ∂u2 (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 limα→0 α �F(x, u + αv, u� + αv� ) − F(x, u, u� )� = α limα→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
SF=a dF(a, u+au, u'+av) dF(a, u+av, u+av) aF(a,u+av, u+au) aF(a, u +av, u+av) evaluated at a=0 dF(c, u+av, u+au) aF(a, u,u aF(, u, u) a=0 Note analogy with differential calculus d(aFi+bF2)=aoF1+boF2 linearity 6(F1F2)=6F1F2+F16F2 etc The conclusions for F(a, u, u) can be generalized to functions of several independent variables i and functions u ”O We will be making intensive use of these properties of the variational operator d d du dar otdr Sudx=aude=a/udx=8 udr Concept of a functional 1(u)=/F(, u(),a'(a)da 3
�� � � 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 differential calculus. δ(aF1 + bF2) = aδF1 + bδF2 linearity δ(F1F2) = δF1F2 + F1δF2 etc The conclusions for F(x, u, u� ) can be generalized to functions of several ∂ui independent variables xi and functions ui, ∂xj : � ∂ui � F xi, ui, ∂xj We will be making intensive use of these properties of the variational operator δ: d d dv �du� = δ � dx(δu) � = dx(αv) = � αdx 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 F(a, u(a), u(n)d /6( F(a, u(a), u()) Su tSu'dx du Extremum of a functional "ulo" is the minumum of a functional if: m)2/(o A necessary condition for a functional to attain an extremum atuo 6I(0)=0 (uo +av, uo+ai 0 Note analogy with differential calculus. Also difference since here we require 出=0ata=0. OF aF Su+ dx Integrate by parts the second term to get rid of du OF d/aF d/OF aF d/aF OF。中b Require du to satisfy homogeneous displacement boundary conditions 6u(b)=6u(a) 6Ⅰ= af d/aF\lOud.r=0
�� � � � � � � � 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 “u0” is the minumum of a functional if: I(u) ≥ I(u0)∀u A necessary condition for a functional to attain an extremum at “u0” is: dI δI(u0) = 0, or dα(u0 + αv, u� 0 + αv� ) � � = 0 α=0 Note analogy with differential calculus. Also difference since here we require dF = 0 at α = 0. dα b�∂F ∂F � � δI = δu + ∂u� δu dx a ∂u 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 a ∂u dx � 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
WOu that satisfy the appropriate differentiability conditions and the homoge- neous essential boundary conditions. Then oF d/OF au dr laul These are the Euler-Lagrange equations corresponding to the variational problem of finding an extremum of the functional I Natural and essential boundary conditions A weaker condition on Ou also allows to obtain the euler equations, we just need OF which is satisfied if du()=0 and Su(b)=0 as before ar (b) ·m(a)=0 and Sut(b)=0 aF (a)=0 and a (b) Essential boundary conditions: ouls=0, or u=uo on Su Natural boundary conditions: on =0 on S Example: Derive Euler's equation corresponding to the total po- tential energy functional II=U+V of an elastic bar of length L, Youngs modulus E, area of cross section A fixed at one end and subject to a load P at the other end EA 2dx Compute the first variation Ea du/du I 6(元)dx-P6u(L) 5
� � � ∀δu that satisfy the appropriate differentiability conditions and the homogeneous essential boundary conditions. Then: ∂F ∂u − d dx �∂F ∂u� � = 0 These are the Euler-Lagrange equations corresponding to the variational problem of finding 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 satisfied 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� Su = 0, or u = u0 on Su Nat ∂F ural boundary conditions: ∂u� = 0 on S. Example: Derive Euler’s equation corresponding to the total potential energy functional Π = U + V of an elastic bar of length L, Young’s modulus E, area of cross section A fixed 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 first variation: EA du �du� δΠ = �2 �2 dxδ dx dx − Pδu(L) 5
grate by pal EA EA d a+ea-d PSu(l 0,V6u/6a(0) dx du P=EA L Extension to more dimensions OF OF a/ aF OF Build Using divergence theorem aF a/ OF OF duidv+ dinds L Extremum of functional I is obtained when SI=0 or when aF 0/ OF Sui=o on s OF The boxed expressions constitute the Euler-Lagrange equations correspond ing to the variational problem of finding an extremum of the functional I 6
� � � � � � � � � � � � � � � � � � � �� � Integrate by parts � d δΠ = dx � L EA − d dx EA δu dx − Pδu(L) du dxδu du dx EA dx + EA L 0 d dx du dx du dx = − δu δu − Pδu(L) 0 Setting δΠ = 0, ∀ δu / δu(0) = 0: d dx � EAdu dx � = 0 du � P = EA � dx L Extension to more dimensions I = F(xi, ui, ui,j )dV V �∂F V dV ∂F δui,j ∂ui,j � ∂F δI = δui + ∂ui = V � ∂F δui + ∂ui � ∂F δui ∂ui,j δui dV ∂ ∂xj ∂ ∂xj − ∂ui,j Using divergence theorem: δI = V � ∂F ∂ui − S � ∂F ∂ui,j δuidV + ∂ ∂xj ∂F δuinjdS ∂ui,j Extremum of functional I is obtained when δI = 0, or when: ∂F � ∂F � ∂ ∂xj − ∂ui,j = 0 , and ∂ui δui = 0 on Su ∂F nj onS − Su = St ∂ui,j The boxed expressions constitute the Euler-Lagrange equations corresponding to the variational problem of finding an extremum of the functional I. 6
