738 CHAPTER 22. THE DERIVATIVE

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 )= 0 whichis valid for small λ ,x,y, this in addition to the zero solution. Note that for small nonzeroλ ,−1+ sin(λ )

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

2

3! for small λ from the power seriesfor sin .

In the next example, the same procedure gives a solution to a problem f((x,y) ,λ ) = 0such that for small λ , (x,y) is a function of λ which is nonzero and f((0,0) ,λ ) = 0. Thusfor small λ , there are two solutions to the nonlinear system of equations.

Example 22.12.3 Let

f((x,y) ,λ ) =(

x+ xy+ y2 + xsin(λ )x+ y2− x2 + xλ

)In this case f((0,0) ,λ ) = 0 even though λ might not be 0. The Lyapunov Schmidt proce-dure will be used to show that there are nonzero solutions x(λ ) ,y(λ ) such that

f ((x(λ ) ,y(λ )) ,λ ) = 0

At origin,

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

1 01 0

)Thus X1 = span(e1) and X2 = span(e2). Then Y1 = span(e1 + e2) and Y2 = span(e1− e2) .Also D1f((0,0) ,0) is one to one on X1 and its range is Y1. Then let

Q(

α

β

)=

(α+β

2α+β

2

)=

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

)(α

β

)

738 CHAPTER 22. THE DERIVATIVEThis reduces toex tay taty + 2A +7 sind _ 05x + 5xy+xt+y?+5A44sinA 012.1 1 Loxx + syx— 5A + 5 sind _ 05x" — gyx+5A—4sind 0Note how in both the top and the bottom, there is only one equation and one can solvefor x in terms of y,A 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(A)? This would involve trying to solve for y as a function of A in the following wherex(y,A) comes from the first equations.I 4 1 1 1g5° (yA) + 5yxQ,A)— 54+ 5 sind =0Tf you can do this, then you would have found (x,y) as a function of 4 for small 2.In this example, in the top equation, at (0,0,0) ,x, = 0. Also xy =—1 sox(y,A) ®—Aother than higher order terms for small y, 2. Then in the bottom equation, for all variablesvery small, you would have A” +y(—A) —A +sin(A) =0, y(A) =—1+ sin) +A at leastapproximately. Thus it seems there is a nonzero solution to the equation f (x,y, A) = 0 whichis valid for small A,x,y, this in addition to the zero solution. Note that for small nonzeroA,-1+ sin) +A #0. It equals approximately A — 4 for small A from the power seriesfor sin.In the next example, the same procedure gives a solution to a problem f((x,y),A) =0such that for small 2, (x,y) is a function of A which is nonzero and f((0,0),A) =0. Thusfor small J, there are two solutions to the nonlinear system of equations.Example 22.12.3 Let+xy+y? +xsin (AF((sy).ay= (Ee ee )In this case £((0,0) ,A) = 0 even though A might not be 0. The Lyapunov Schmidt proce-dure will be used to show that there are nonzero solutions x(A) ,y(A) such thatf((x(A),¥(A)) A) =0At origin,Dif((0.0).0) = ( ro )Thus X; = span (e;) and Xz = span (ez). Then Y; = span (e; + e2) and Y; = span (e; — 2).Also D;f((0,0) ,0) is one to one on X, and its range is Y,. Then let0 )-( a8 )-(13 1B )(8)