I have a function in terms of a vector x, and a parameter a, say F(x,a). I also know that there exists a smooth unique parametrized curve x(a) such that F(x(a),a) = 0 for all a. I want to know if the Jacobian matrix of F differentiated with respect to x is nonsingular when evaluated on the curve x(a).
In other words, if we have a dynamical system can we say that local solvability of an equilibrium guarantees that stability cannot change assuming a Hopf bifurcation does not take place?
The following StackExchange links gives an overview, probably you'd benefit from that:
http://math.stackexchange.com/questions/171720/converse-to-inverse-function-theorem
http://math.stackexchange.com/questions/501505/prove-local-converse-to-the-implicit-function-theorem
May be next notes will be useful to you
www.math.washington.edu/.../impft_notes.pdf
Dear Jason,
Consider simple situation F maps R^2 into R. If locally around the origin 0 we can solve F(x,y)=0 is a graph y=f(x), then take F(x,y)=f(x)-y in a nbg of 0. Since F’_x=f’(x) and F’_y=-1, then rank F=1. We can generalize this to the setting in which F maps Rn into Rm where m=n-k.
Suppose that M ⊂ R n there exists an open neighborhood U of 0 such M∩U is the graph of a smooth map f which maps a k-dim ngb V k of 0 (which belongs
R k ) into R m where n=k+m. Abusing notation we write y=f(x) and x=(x_1,…x_k) and y=(y1,…,y m).
The function F(x,y)=f(x) –y has rank m at origin.
If I understand this problem and if this answer is related to your question, we can give more details.
See http://math.stackexchange.com/questions/501505/prove-local-converse-to-the-implicit-function-theorem/1400620#1400620
The Jacobian matrix of F with respect to x may be singular.
Let F(x,a) = (x-a)^2.
The only solution of equation F(x(a),a)=0 is the curve x(a)=a, and the derivative of F with respect to x on the curve vanishes.
For an example in several variables, take F(x_1,...x_n,a) = ((x1-a)^2,...(x_n-a)^2), then x(a)=(a,...a) and the Jacobian matrix of F with respect to x vanishes on the curve.
Let F(x,a) = (x-a)(x^2+a^2).
We have only the curve x(a)=a; the derivative of F with respect to x is x^2+a^2+2x(x-a), hence F_x(a,a) = 2a^2 i.e. F_x(a,a)=0 if and only of a=0.
Thank you Giovanni. I deleted my original questions because I found a counterexample myself.
Consider simple form of the implicit function theorem:
Suppose that (i): F maps a nbg V in R^2 into R , F(0,0)=0 and that F’_x(x,a) different from 0 in V .
Then (ii) there is b>0 such that the set of points F(x,a)= 0 in Q= [-b,b]^2, is graph of a function x=x(a), -b
Dear Jason,
First of all, I think that your problem should be reformulated. If F is a scalar valued function, then (F_x)( , )=F(x(a),a) is not a matrix, but a vector field, depending on a. So, because of
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