22.12. LYAPUNOV SCHMIDT PROCEDURE 737

Example 22.12.1 Say X = R2 and Λ = R. Let f (x,y,λ ) = x+ xy+ y2 +λ . Then

D1 f (0,0,0) = (1,0)

this 1×2 matrix mapping R2 to R. Thus X2 = (0,α)T : α ∈ R and X1 = (α,0)T : α ∈ R.In this case, Y1 = R and so Q = I. Thus the above reduces to the single equation

f ((α,0)+(0,β ) ,λ ) = 0

and so since D1 f (0,0,0) is one to one, x1 = (α,0) = x1 ((0,β ) ,λ ) . Of course this iscompletely obvious because if you consider f in the natural way as a function of threevariables, then the implicit function theorem immediately gives x = x(y,λ ) which is essen-tially the same result. We just write (α,0) in place of α . The first independent variable isa function of the other two.

Example 22.12.2 Here is another easy example. f : R2×R→ R2

f(x,y,λ ) =(

x+ xy+ y2 + sin(λ )x+ y2− x2 +λ

)Then

D1f(x,y,λ ) =(

1+ y x+2y1−2x 2y

)So

D1f((0,0) ,0) =(

1 01 0

)Then

X2 = kerD1f((0,0) ,0) ={(

0β

): β ∈ R

}and X1 =

{(α

0

): α ∈ R

}and clearly D1f((0,0) ,0) is indeed one to one on X1.

D1f(0,0)(X1) =

{(yy

): y ∈ R

}= Y1

In this case, let

Q(

α

β

)=

(α+β

2α+β

2

)=

(1/2 1/21/2 1/2

)(α

β

)

so (I−Q) =

(1/2 −1/2−1/2 1/2

). Thus the equations are

Qf(x,λ ) = 0(I−Q) f(x,λ ) = 0