Solving Advanced ODEs: Help Needed!

[kitsch_22]

Hello and thanks in advance for anyone who can help at all. I have two problems that have stumped me.. I'm in an advanced ODE class. Here they are:

1) Consider the first order ODE f_a(x) where a is a parameter; let f_a(x0) = 0
for some solution x0 and also let f'_a(x0) != 0. Prove that the ODE
f_a+e(x) has an equlibrium point x0(e) where e -> x0(e) is a smooth function satisfying x0(0) = x0 for e sufficiently small.

2) Consider the system X' = F(X) where X is in R_n. Suppose F has an equilbrium point at X0. Show that there exists a change of coordinates that moves X0 to the origin and converts the system to X' = AX + G(X) where A is an nxn matrix which is the canonical form of DF_X0 and where G(X) satifies

lim (|G(X)| / |X|) = 0.
|X|->0

I am so lost on these...can anyone help pleeeeeeeeeease?

Michelle

 

[Reb]

Ok let's see...

For 1), you are given the ODE with the parameter shifted by a small [tex]\epsilon[/tex], and you are required to show that this new ODE with [tex]\epsilon[/tex] will have an equilibrium, which is "close" to the original one. Since [tex]f_a[/tex] has nonzero gradient, continuity implies that [tex]f'_{a+\epsilon}[/tex] will also be nonzero for small [tex]\epsilon[/tex]. Invoking the existence theorem, there is a smooth equilibrium that depends on [tex]\epsilon[/tex]. Call this [tex]x_0(\epsilon,\cdot)[/tex]. By the dependence on parameters theorem, [tex]x_0(\epsilon,\cdot)\rightarrow x_0(\cdot)[/tex] as [tex]\epsilon\rightarrow0[/tex].


For 2), note that [tex]DF_{x_0}[/tex] being nonzero, implies that [tex]F[/tex] is a local diffeomorphism in a neighbouhood of [tex]x_0[/tex]. This grants us the validity of a local change of variables to [tex]y=F(x)[/tex]. Under [tex]F[/tex], the equilibrium is mapped to the origin. For the last part, note that [tex]F^{-1}[/tex] will have a similar Taylor expansion as [tex]F[/tex], and that a Taylor expansion for [tex]F[/tex] gives [tex]F(x)=F(x_0)+DF_{x_0}(x)+G(x)=0+A\cdot x+G(x)[/tex], where [tex]G[/tex] will contain higher order terms than [tex]|x|[/tex], and so [tex]G(x)=0(|x|)[/tex].