Topics In Analysis

20.12 Lyapunov Schmidt Procedure

You have f : X × Λ → Y where here X,Λ are Banach spaces. Suppose

∈ X × Λ and f

= 0. Then if D₁f

⁻¹ is in ℒ

, the implicit function theorem says that there exists x

a C^p function such that locally f

= 0. So what if D₁f

fails to be one to one? Sometimes this case is also considered. It may be that D₁f

is one to one on some subspace and other nice things happen. In particular, suppose the following.

Letting X₂ ≡ kerD₁f

assume

X = X \oplus X , dim (X ) < \infty 1 2 2

where X₁ is a closed subspace. Thus D₁f

is one to one on X₁. We let

Y1 = D1f (0,0) (X1 )

and suppose that Y = Y ₁ ⊕ Y ₂ where dim

< ∞,Y ₁ also a closed subspace.

X1 D1f\to(0,0)Y1 = D1f (0,0)(X1 ), Y1 closed Y = Y1 \oplusY2, dim (Y2) < \infty < \infty

By the open mapping theorem, D₁f

⁻¹ is also continuous.

Let Q be a continuous projection onto Y ₁ which is assumed to exist² so that

is a projection onto Y ₂. Then the equation f

= 0 can be written as the pair

Qf (x,λ) = 0 (I - Q)f (x,λ) = 0

Consider the top. For x = x₁ + x₂ where x_i ∈ X_i, this is

Qf (x1 + x2,λ) = 0

Then if g

= Qf

, one has g : X₁ × X₂ × Λ → Y ₁

D1g (x1,x2,λ)h = D1Qf (x1 + x2,λ)h, h \in X1.

Thus D₁g

⁻¹ is continuous by the open mapping theorem (D₁f

is one to one on X₁), and by the implicit function theorem, there is a solution to

Qf (x1 + x2,λ) = 0

for x₁ = x₁

. (Note how it is important that X₁ and Y ₁ be Banach spaces.) Then the other equation yields

(I - Q)f (x1(x2,λ) + x2,λ) = 0

and so for fixed λ, this is a finite set of equations of a variable in a finite dimensional space.

This depends on being able to write X = X₁ ⊕ X₂ where X₁ is closed, X₂ = kerD₁f

, a similar situation for Y = Y ₁ ⊕Y ₂. So when does this happen? Are there conditions on D₁f

which will cause it to occur?

There are such conditions. For example, D₁f

could be a Fredholm operator defined in Definition 15.6.7. The following are some easy examples in which all that nonsense about things being finite dimensional and part of a direct sum does not need to be considered.

Example 20.12.1 Say X = ℝ² and Λ = ℝ. Let f

= x + xy + y² + λ. Then

D1f (0,0,0) = (1,0)

this 1 × 2 matrix mapping ℝ² to ℝ. Thus X₂ =

^T : α ∈ ℝ and X₁ =

^T : α ∈ ℝ. In this case, Y ₁ = ℝ and so Q = I. Thus the above reduces to the single equation

f ((α,0)+ (0,β),λ) = 0

and so since D₁f

is one to one, x₁ =

= x₁

. Of course this is completely obvious because if you consider f in the natural way as a function of three variables, then the implicit function theorem immediately gives x = x

which is essentially the same result. We just write

in place of α. The first independent variable is a function of the other two.

Example 20.12.2 Here is another easy example. f : ℝ² × ℝ → ℝ²

( x+ xy+ y2 + sin(λ) ) f (x,y,λ) = x+ y2 - x2 + λ

Then

( ) D1f (x,y,λ) = 1+ y x + 2y 1- 2x 2y

( ) D f ((0,0),0) = 1 0 1 1 0

Then

{ ( ) } X = kerD f ((0,0),0) = 0 : β \in ℝ 2 1 β

and X₁ =

and clearly D₁f

is indeed one to one on X₁. D₁f

= Y ₁. In this case, let

( α ) ( α+2β-) ( 1∕2 1∕2 ) ( α ) Q β = α+β- = 1∕2 1∕2 β 2

. Thus the equations are

Qf (x,λ ) = 0 (I - Q) f (x,λ ) = 0

This reduces to

( - 1x2 + 1xy+ x + y2 + 1λ+ 1sin λ ) ( 0 ) -2 1x2 + 21xy+ x + y2 + 21λ+ 21sin λ = 0 2 ( 21 2 1 12 12 ) ( ) 2x1 +2 21yx- 2λ1 + 2 s1in λ = 0 - 2x - 2yx + 2λ- 2 sin λ 0

Note how in both the top and the bottom, there is only one equation and one can solve for x in terms of y,λ near

which is what the above general argument shows. Of course you can see this directly using the implicit function theorem. Then can you solve for y = y

? This would involve trying to solve for y as a function of λ in the following where x

comes from the first equations.

1x2(y,λ)+ 1 yx(y,λ)- 1λ + 1sinλ = 0 2 2 2 2

If you can do this, then you would have found

as a function of λ for small λ.

In this example, in the top equation, at

,x_y = 0. Also x_λ = −1 so x

≈−λ other than higher order terms for small y,λ. Then in the bottom equation, for all variables very small, you would have λ² + y

− λ + sin

= 0, y

= −1 +

+ λ at least approximately. Thus it seems there is a nonzero solution to the equation f

= 0 which is valid for small λ,x,y, this in addition to the zero solution. Note that for small nonzero λ,−1 +

+ λ≠0. It equals approximately λ−

for small λ from the power series for sin.

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

= 0 such that for small λ,

is a function of λ which is nonzero and f

= 0. Thus for small λ, there are two solutions to the nonlinear system of equations.

Example 20.12.3 Let

( x+ xy + y2 + xsin(λ) ) f ((x,y),λ ) = x + y2 - x2 + xλ

In this case f

= 0 even though λ might not be 0. The Lyapunov Schmidt procedure will be used to show that there are nonzero solutions x

such that f

= 0.

At origin,

( 1 0 ) D1f ((0,0),0) = 1 0

Thus X₁ = span

and X₂ = span

. Then Y ₁ = span

and Y ₂ = span

. Also D₁f

is one to one on X₁ and its range is Y ₁. Then let

( α ) ( α+β-) ( 1∕2 1∕2 ) ( α ) Q β = α2+β- = 1∕2 1∕2 β 2

( ) (I - Q ) = 1∕2 - 1∕2 - 1∕2 1∕2

Then Qf = 0 is yields the equation

1 1 1 1 2 2 x+ 2x λ+ 2x sinλ + 2xy - 2x +y = 0

Also

f = 0 yields the equation

1 1 1 1 -xsinλ- -xλ+ -xy+ - x2 = 0 2 2 2 2

Now consider x_y and x_λ at

from the first equation. Both of these are easily seen to be 0. Now consider x_yy. After some computations, this is seen to be x_yy = −2. Similarly, x_yλ

= 0,x_λλ

= 0 also. Thus up to terms of degree 3,

1 x (y,λ) = - y2 = 2 (- 2)y2

Place this in the bottom equation.

1y2λ - 1y2sin λ- 1 y3 + 1y4 = 0 2 2 2 2

Now the idea is to find y = y

, hopefully nonzero. Divide by y² and multiply by 2.

y2 - y + λ- sin λ = 0

Then for small λ this is approximately equal to

3 y2 - y+ λ- = 0 6

Then a solution for y for small λ is

\circ ------- 1 + 1 - 2λ3 y = ---------3-- 2

Of course there is another solution as well, when you replace the + with a minus sign. This is the one we want because when λ = 0 it reduces to y = 0. This shows that there exist solutions to the equations f

= 0 which for small λ are approximately

( \circ ----2-3) (x (λ),y(λ)) = (- y2, 1---1---3λ-) 2

In terms of λ very small,

( \circ ---2---) (1 3 1\sqrt -\circ -----3- 1 1----1--3λ3-) (x (λ) ,y (λ )) = 6 λ + 6 3 3- 2λ - 2, 2

Using a power series in λ to approximate these functions, this reduces to

( 1 6 1 3 1 6 1 9) (x(λ),y(λ)) = - 36λ ,6λ + 36λ + 108-λ

where higher order terms are neglected. Thus there exist other solutions than the zero solution even though λ may be nonzero. Note that in this example, f

= 0.