variational.pdf
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16.21 Techniques of Structural Analysis andDesign
Spring 2005Unit #9 - Calculus of Variations
Raul Radovitzky
March 28, 2005
Let u be the actual configuration of a structure or mechanical system. usatisfies the displacement boundary conditions: u = u on Su. Define:
u = u + v, where: : scalar
v : arbitrary function such that v = 0 o n SuWe are going to define v as u, thefirst variationofu:
u = v (1)
Schematically:
u1
u2
u
a b
u(a)
u(b)
v
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As a first property of thefirst variation:du du dv= + dx dx dx
so we can identify dvdx
with the first variation of the derivative ofu:du dv =
dx dx
But:
dv dv d
dx=
dx=
dx(u)
We conclude that:
du d = (u)
dx dx
Consider a function of the following form:
F= F(x, u(x), u(x))
It depends on an independent variable x, another function ofx (u(x)) and itsderivative (u(x)). Consider the change in F, when u (therefore u) changes:
F= F(x, u + u, u+ u) F(x, u, u)= F(x, u + v, u+ v) F(x, u, u)
expanding in Taylor series:
F F 1 2F 1 2FF= F+ v + v+
u2(v)2 + (v)(v) + F
u u 2! 2! uuF F
= v + v+ h.o.t.u u
First total variation of F:
FF= lim
0 F(x, u + v, u+ v) F(x, u, u)= lim
0 Fv + Fv F F= lim u u = v + v 0 u u
F FF= u + u
u u
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Note that:
dF(x, u + v, u+ v)F= d =0
since:
dF(x, u + v, u+ v) F(x, u + v, u+ v)=
d uevaluated at = 0
F(x, u + v, u+ v)v + v
u
dF(x, u + v, u+ v)=F(x, u, u)v + F(x, u, u)v
d =0 u uNote analogy with differential calculus.
(aF1 + bF2) = aF1 + bF2 linearity
(F1F2) = F1F2 + F1F2
etc
The conclusions for F(x, u, u) can be generalized to functions of severaluiindependent variables xi and functions ui, xj :
F xi, ui,uixj
We will be making intensive use of these properties of the variational operator:
d d dv du(u) = (v) = =
dx dx dx dx
udx = vdx = vdx = udx
Concept of a functional
b
F(x, u(x), u
(x))dxI(u) =a
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First variation of a functional:
I= F(x, u(x), u(x))dx = F(x, u(x), u(x)) dx
F FI= u + u dx
u u
Extremum of a functional
u0 is the minumumof a functional if:
I(u) I(u
0
)uA necessary condition for a functional to attain an extremum at u0 is:
dII(u0) = 0, or (u0 + v,u0
+ v) = 0d =0
Note analogy with differential calculus. Also difference since here we requiredF = 0 at = 0.d
b
F FI= u + u dx
u u
a
Integrate by parts the second term to get rid ofu.
b F d F d F I= u + u u dx
u dx u dx ua b F bF d F = udx + u
u
dx u u aaRequire u to satisfy homogeneous displacement boundary conditions:
u(b) = u(a) = 0
Then: b F d F I= udx = 0,
u
dx ua
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u that satisfy the appropriate differentiability conditions and the homoge-
neous essential boundary conditions. Then:
F d F = 0
u
dx u
These are the Euler-Lagrange equations corresponding to the variationalproblem of finding an extremum of the functional I.
Natural and essential boundary conditions A weaker condition onu also allows to obtain the Euler equations, we just need:
F bu = 0u a
which is satisfied if:
u(a) = 0 and u(b) = 0 as before u(b) = 0 and F(b) = 0
uFu(a) = 0 and u(b) = 0
F(a) = 0 and F(b) = 0u u
Essential boundary conditions: u = 0, or u = u0 on SuSuNatural boundary conditions: F = 0 on S.u
Example: Derive Eulers equation corresponding to the total po-tential energy functional = U+ V of an elastic bar of length L, Youngsmodulus E, area of cross section A fixed at one end and subject to a load Pat the other end. L EA du2
(u) = dxPu(L)2 dx0
Compute the first variation:
EA du du = 2 dxPu(L)
2
dx dx
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Integrate by parts
d du d du = EA u EA u dxPu(L)dx dx
dx dx L du Ld du
= u EA dx + EA u0
Pu(L)0 dx dx dx
Setting = 0, u / u(0) = 0:d du
EA = 0dx dx
du
P= EA
dx L
Extension to more dimensions
I= F(xi, ui, ui,j)dV V F FI= ui + ui,j dV
V ui ui,j
F F F
= ui + ui ui dVV ui xj ui,j
xj ui,j
Using divergence theorem:
F F F I= uidV+ uinjdS
V ui
xj ui,j Sui,j
Extremum of functional Iis obtained when I= 0, or when:
F F = 0 , and
ui
xj ui,j
ui = 0 on Su
Fui,j
nj onSSu = StThe boxed expressions constitute the Euler-Lagrange equations correspond-ing to the variational problem of finding an extremum of the functional I.
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