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Jim Lambers
MAT419/519
Summer Session 2011-12
Lecture 11 Notes
These notes correspond to Section 3.4 in the text.
Broyden’s Method
One of the drawbacks of using Newton’s Method to solve a system of nonlinear equations g(x) = 0
is the computational expense that must be incurred during each iteration to evaluate the partial
derivatives of g at x(k), and then solve a system of linear equations involving the resulting Jacobian
matrix. The algorithm does not facilitate the re-use of data from previous iterations, and in some
cases evaluation of the partial derivatives can be unnecessarily costly.
Analternative is to modify Newton’s Method so that approximate partial derivatives are used,
since the slightly slower convergence resulting from such an approximation is offset by the improved
efficiency of each iteration. However, simply replacing the analytical Jacobian matrix J (x) of g
g
with a matrix consisting of finite difference approximations of the partial derivatives does not do
muchtoreduce the cost of each iteration, because the cost of solving the system of linear equations
is unchanged.
However, because Jg(x) consists of the partial derivatives evaluated at an element of a conver-
gent sequence, intuitively Jacobian matrices from consecutive iterations are “near” one another in
some sense, which suggests that it should be possible to cheaply update an approximate Jacobian
matrix from iteration to iteration, in such a way that the inverse of the Jacobian matrix, which is
what is really needed during each Newton iteration, can be updated efficiently as well.
This is the case when an n×n matrix B has the form
B=A+u⊗v,
where u and v are given vectors in Rn, and u⊗v is the outer product of u and v, defined by
u1v1 u1v2 · · · u1vn
u2v1 u2v2 · · · u2vn
u⊗v=uvT = .
. . .
. .. .
. .
u v u v · · · u v
n 1 n 2 n n
This modification of A to obtain B is called a rank-one update. This is because u⊗v has rank one,
since every column of u⊗v is a scalar multiple of u. To obtain B−1 from A−1, we note that if
Ax=u,
1
then
Bx=(A+u⊗v)x=Ax+uvTx=u+u(v·x)=(1+v·x)u,
which yields
B−1u= 1 A−1u.
−1
−11+v·A u
Onthe other hand, if x is such that v · A x=0,then
−1 −1 −1 T −1 −1
BA x=(A+u⊗v)A x=AA x+uv A x=x+u(v·A x)=x,
which yields
−1 −1
B x=A x.
This takes us to the following more general problem: given a matrix C, we wish to construct a
matrix D such that the following conditions are satisfied:
• Dw=z,for given vectors w and z
• Dy=Cy,if y is orthogonal to a given vector g.
−1 −1 −1 −1 −T
In our application, C = A , D = B , w = u, z = 1/(1+v·A u)A u, and g = A v.
To solve this problem, we set
D=C+(z−Cw)⊗g.
g·w
Then, if g · y = 0, the second term in the definition of D vanishes, and we obtain Dy = Cy, but
in computing Dw, we obtain factors of g·w in the numerator and denominator that cancel, which
yields
Dw=Cw+(z−Cw)=z.
Applying this definition of D, we obtain
h 1 −1 −1 i −1
−1 A u−A u ⊗v A −1 −1
B−1 =A−1+ 1+v·A u =A−1−A (u⊗v)A .
−1 −1
v·A u 1+v·A u
This formula for the inverse of a rank-one update is known as the Sherman-Morrison Formula.
We now return to the problem of approximating the Jacobian of g, and efficiently obtaining
its inverse, at each iterate x(k). We begin with an exact Jacobian, D = J (x(0)), and use D to
0 g 0
compute the first iterate, x(1), using Newton’s Method as follows:
(0) −1 (0) (1) (0) (0)
d =−D g(x ), x =x +d .
0
Then, we use the approximation
g(x )−g(x )
g′(x ) ≈ 1 0 .
1 x −x
1 0
Generalizing this approach to a system of equations, we seek an approximation D1 to Jg(x(1)) that
has these properties:
2
• D (x(1) −x(0)) = g(x(1))−g(x(0)) (the Secant Condition)
1
• If z · (x(1) − x(0)) = 0, then D z = J (x(0))z = D0z.
1 g
It follows from previous discussion that
(0)
y −D d
0 0 (0)
D =D + ⊗d ,
1 0 (0) (0)
d ·d
where
(0) (1) (0) (0) (1) (0)
d =x −x , y =g(x )−g(x ).
(0) (1)
However, it can be shown (Chapter 3, Exercise 15) that y − D d =g(x ), which yields the
0 0
simplified formula
1 (1) (0)
D =D + g(x )⊗d .
1 0 (0) (0)
d ·d
Once we have computed D−1, we can apply the Sherman-Morrison formula to obtain
0
−1 1 (1) (0) −1
D (0) (0) g(x ) ⊗d D
−1 −1 0 d ·d 0
D = D −
1 0 (0) −1 1 (1)
1+d ·D (0) (0) g(x )
0 d ·d
−1 (1) (0) −1
D g(x )⊗d D
= D−1− 0 0
0 (0) (0) (0) −1 (1)
d ·d +d ·D g(x )
0
(0) (0) −1
= D−1− (u ⊗d )D0 ,
0 (0) (0) (0)
d ·(d +u )
(0) −1 (1) (1)
where u =D g(x ). Then, as D is an approximation to J (x ), we can obtain our next
0 1 g
iterate x(2) as follows:
D d(1) = −g(x(1)), x(2) = x(1) + d(1).
1
Repeating this process, we obtain the following algorithm, which is known as Broyden’s Method:
Choose x(0)
D =J (x(0))
0 g
(0) −1 (0)
d =−D g(x )
0
(1) (0) (0)
x =x +d
k = 0
while not converged do
u(k) = D−1g(x(k+1))
k
(k) (k) (k)
ck = d · (d +u )
−1 −1 1 (k) (k) −1
D =D − [u ⊗d ]D
k+1 k ck k
k = k +1
3
d(k) = −D−1g(x(k))
k
(k+1) (k) (k)
x =x +d
end
NotethatitisnotnecessarytocomputeD fork ≥ 1; onlyD−1 isneeded. Itfollowsthatnosystems
k k
of linear equations need to be solved during an iteration; only matrix-vector multiplications are
required, thus saving an order of magnitude of computational effort during each iteration compared
to Newton’s Method.
Example We consider the system of equations g(x) = 0, where
2 2 2
x2+y2+z −3
g(x,y,z) = x +y −z−1 .
x+y+z−3
We will begin with one step of Newton’s Method to solve this system of equations, with initial
(0) (0) (0) (0)
guess x =(x ,y ,z )=(1,0,1).
As computed in a previous example,
2x 2y 2z
Jg(x,y,z) = 2x 2y −1 .
1 1 1
Therefore, the Newton iterate x(1) is obtained by solving the system of equations
J (x(0))(x(1) − x(0)) = −g(x(0)),
g
(0) (0)
or, equivalently, D d =−g(x )where
0
(0) (0) (0) (1) (0)
(0) 2x 2y 2z (0) x −x
D =J (x )= (0) (0) , d = y(1) −y(0) ,
0 g 2x 2y −1 (1) (0)
1 1 1 z −z
(0) 2 (0) 2 (0) 2
(0) (x ) +(y ) +(z ) −3
g(x )= (0) 2 (0) 2 (0) .
(x ) +(y ) −z −1
(0) (0) (0)
x +y +z −3
(0) (0) (0)
Substituting (x , y , z ) = (1,0,1) yields the system
2 0 2 x(1) −1 1
2 0 −1 y(1) = 1 .
1 1 1 z(1) − 1 1
4
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