21.7. LYAPUNOV SCHMIDT PROCEDURE 565

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

This reduces to (− 1

2 x2 + 12 xy+ x+ y2 + 1

2 λ + 12 sinλ

− 12 x2 + 1

2 xy+ x+ y2 + 12 λ + 1

2 sinλ

)=

(00

)( 1

2 x2 + 12 yx− 1

2 λ + 12 sinλ

− 12 x2− 1

2 yx+ 12 λ − 1

2 sinλ

)=

(00

)Note how in both the top and the bottom, there is only one equation and one can solvefor x in terms of y,λ near (0,0,0) which is what the above general argument shows. Ofcourse you can see this directly using the implicit function theorem. Then can you solve fory = y(λ )? This would involve trying to solve for y as a function of λ in the following wherex(y,λ ) comes from the first equations.

12

x2 (y,λ )+12

yx(y,λ )− 12

λ +12

sinλ = 0

If you can do this, then you would have found (x,y) as a function of λ for small λ .

In this example, in the top equation, at (0,0,0) ,xy = 0. Also xλ =−1 so x(y,λ )≈−λ

other than higher order terms for small y,λ . Then in the bottom equation, for all variablesvery small, you would have λ

2 +y(−λ )−λ + sin(λ ) = 0, y(λ ) =−1+ sin(λ )λ

+λ at leastapproximately. Thus it seems there is a nonzero solution to the equation f (x,y,λ ) = 0which is valid for small λ ,x,y, this in addition to the zero solution. Note that for smallnonzero λ ,−1 + sin(λ )

λ+ λ ̸= 0. It equals approximately λ − λ

2

3! for small λ from thepower series for sin .

In the next example, the same procedure gives a solution to a problem

f ((x,y) ,λ ) = 0

such that for small λ , (x,y) is a function of λ which is nonzero and

f ((0,0) ,λ ) = 0

Thus for small λ , there are two solutions to the nonlinear system of equations.