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DF (x0 )[u] = DF 1 (F 2 (x0 ))[DF 2 (x0 )[u]]




* u in the sense that for any u1 and u2 :
The term DF(x0 )[u] is linear with respect to



DF(x0 )[u1 + u2 ] = DF(x0 )[u1 ] + DF(x0 )[u2 ]
30 MATHEMATICAL PRELIMINARIES




EXAMPLE 2.8: Interpretation of Equation (2.105c)
The chain rule Equation (2.105c) is not easy to interpret. In an at-
tempt to clarify the meaning of this equation consider the approxima-
tion statement,
F (x0 + u) ≈ F (x0 ) + DF (x0 )[u]
In the case where F (x) = F 1 (F 2 (x)) the left-hand side becomes,
F (x0 + u) = F 1 (F 2 (x0 + u))
and using the linearization of F2 this can be written as,
F (x0 + u) ≈ F 1 (F 2 (x0 ) + DF 2 (x0 )[u])
Now linearizing F1 at F2 (x0 ) in the direction of the increment DF2 (x0 )[u]
gives,
F (x0 + u) ≈ F 1 (F 2 (x0 )) + DF 1 (F 2 (x0 ))[DF 2 (x0 )[u]]
Comparing the ¬rst and last equations gives Equation (2.105c)




2.3.4 EXAMPLES OF LINEARIZATION
Algebraic systems of equations. Consider of set of nonlinear algebraic
equations f(x) = [f1 , f2 , . . . , fn ]T with unknowns x = [x1 , x2 , . . . , xn ]T as,



f1 (x1 , x2 , . . . , xn ) = 0
f2 (x1 , x2 , . . . , xn ) = 0
(2.106)
.
.
.
fn (x1 , x2 , . . . , xn ) = 0



The directional derivative of f(x) at a solution estimate x0 in the general
direction u = [u1 , u2 , . . . , un ]T is given by (2.101) as,



d
Df(x0 )[u] = f(x0 + u) (2.107)
d =0
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2.3 LINEARIZATION AND THE DIRECTIONAL DERIVATIVE



The above expression can be evaluated using the standard chain rule for the
partial derivatives of a function of several variables as,
d
Df(x0 )[u] = f(x0 + u)
d =0
n
d(x0,i + ui )
‚f
=
‚xi d
xi =x0,i =0
i=1
n
‚f
= ui
‚xi xi =x0,i
i=1
= K(x0 )u (2.108)
where the tangent matrix K is,
® ‚f1 ‚f1 ‚f1 
...
‚x1 ‚x2 ‚xn
 ‚f2 ‚f2 ‚f2
...

 ‚x ‚x2 ‚xn

1
K= . (2.109)

. .
..
. . .
.
°. . .»
‚fn ‚fn ‚fn
...
‚x1 ‚x2 ‚xn

Consequently the Newton“Raphson iterative scheme becomes,
K(xk ) u = ’f(xk ); xk+1 = xk + u (2.110)
Function minimization. The directional derivative given in (2.101) need
not be necessarily associated with the Newton“Raphson method and can be
equally applied to other purposes. An interesting application is the min-
imization of a functional, which is a familiar problem that often arises in
continuum or structural mechanics. For example consider the total poten-
tial energy for a simply supported beam under the action of a uniformly
distributed load q(x) given as (see Figure 2.6),
2
l l
d2 w(x)
1
V(w(x)) = EI dx ’ q(x)w(x) dx (2.111)
dx2
2 0 0

where w(x) is the lateral de¬‚ection (which satis¬es the boundary conditions
a priori), E is Young™s modulus, I is the second moment of area, and l is
the length of the beam. A functional such as V is said to be stationary
at point w0 (x) when the directional derivative of V vanishes for any arbi-
trary increment u(x) in w0 (x). Consequently, the equilibrium position w0 (x)
satis¬es,
d
DV(w0 (x))[u(x)] = V(w0 (x) + u(x)) = 0 (2.112)
d =0
32 MATHEMATICAL PRELIMINARIES



q(x)




w(x)


FIGURE 2.6 Simply supported beam.




for any function u(x) compatible with the boundary conditions. Note that
w0 (x) is the unknown function in the problem and is not to be confused
with a Newton“Raphson iterative estimate of the solution. Substituting for
V in (2.112) from (2.111) gives,

2
l
d2 (w0 (x) + u(x))
d 1
DV(w0 (x))[u(x)] = EI dx
dx2
d =0 2 0
l
d
q(x)(w0 (x) + u(x))dx = 0 (2.113)

d 0
=0


Hence,

l l
d2 w0 (x) d2 u(x)
DV(w0 (x))[u(x)] = EI dx ’ q(x)u(x) dx = 0
dx2 dx2
0 0
(2.114)

If u(x) is considered to be the virtual displacement δu(x) then the above
equation is easily recognised as the virtual work equation for the beam,
which is an alternative expression of equilibrium.
Linearization of the determinant of a tensor. This example further
illustrates the generality of the concept of the linearization obtained using
the directional derivative. Consider the linearization of the determinant
det S of the second-order tensor S (or square matrix) with respect to an
increment in this tensor U as,

det(S + U ) ≈ det S + D det(S)[U ] (2.115)

where the directional derivative of the determinant can be found by direct
33
2.3 LINEARIZATION AND THE DIRECTIONAL DERIVATIVE



application of Equation (2.111) as,
d
D det(S)[U ] = det(S + U )
d =0

d
det[S(I + S ’1 U )]
=
d =0

d
det(I + S ’1 U )
= det S (2.116)
d =0
In order to proceed, note that the characteristic equation of a matrix B
with eigenvalues »B , »B , and »B is,
1 2 3

det(B ’ »I) = »B ’ » »B ’ » »B ’ » (2.117)
1 2 3

Using this equation with » = ’1 and B = S ’1 U gives,
d S U S U
1 + »S U
’1 ’1 ’1
D det(S)[U ] = det S 1 + »1 1 + »2 (2.118)
3
d =0

where »1 U , »2 U , and »3 U are the eigenvalues of S ’1 U . Using the
S S S
’1 ’1 ’1



standard product rule of di¬erentiation in (2.118) and recalling the de¬nition
and properties of the trace of a tensor introduced in Section 2.2.3 gives the
directional derivative of the determinant of a tensor as,
S U S U S U
’1 ’1 ’1
D det(S)[U ] = det S »1 + »2 + »3
= det S tr(S ’1 U )
= det S (S ’T : U ) (2.119)
Linearization of the inverse of a tensor. Finally, consider the lin-
earization of the inverse of a tensor (or square matrix) S with respect to an
increment in this matrix U as,
(S + U )’1 ≈ S ’1 + D(S ’1 )[U ] (2.120)
where the directional derivative is given as,
d
D(S ’1 )[U ] = (S + U )’1 (2.121)
d =0
Clearly the evaluation of this derivative is far from obvious. A simple pro-
cedure to evaluate this linearization, however, emerges from the product
rule property given in Equation (2.105b). For this purpose, note ¬rst that
the linearization of I, the identity tensor, is the null tensor 0, because I is
independent of the increment U , that is,
D(S ’1 S)[U ] = D(I)[U ] = 0 (2.122)
34 MATHEMATICAL PRELIMINARIES



Consequently, using the product rule gives,
D(S ’1 )[U ]S + S ’1 D(S)[U ] = 0 (2.123)
which, after some simple algebra, leads to,
D(S ’1 )[U ] = ’S ’1 U S ’1 (2.124)

EXAMPLE 2.9: Linearization of det(S ’1 )
An interesting application of the chain rule Equation (2.105c) is ob-
tained by combining the linearizations of the determinant and the in-
verse of a tensor into the linearization of the functional det S ’1 . First
note that the directional derivative of this functional can be obtained
directly by noting that det(S ’1 ) = 1/ det S and using Equation (2.119)
to give,
d
D det(S ’1 )[U ] = det(S + U )’1
d =0

d 1

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