3
$\begingroup$

Let $f: \mathbb R^n\to \mathbb R^n$ be a $C'$ map with $f(0)=0$. The set $$A=\{(x,\lambda)\in \mathbb R^n\times \mathbb R^1 :f(x)=\lambda x\}$$ contains the set $\{(0,\lambda)\in\mathbb R^n\times \mathbb R^1\}$. We say $\lambda_0$ is a bifurcation point of $A$ if every neighborhood of $(0,\lambda_0)$ contains a point $(x,\lambda)$ of $A$ with $x\ne 0$. Use the implicit function theorem to derive a necessary and sufficient condition that $\lambda_0$ be a bifurcation point.

The only idea I have is to apply the implicit function theorem for the map $F:\mathbb R^{n+1}\to \mathbb R^n, (x,\lambda)\mapsto f(x)-\lambda x$. Then $F(0, \lambda)=0$. The theorem says that if the $n\times n$ matrix of partial derivatives w.r.t. $x_i\ (i=1,\dots, n)$ at $(0,\lambda)$ is non-singular, then there is a neighborhood $U\subset \mathbb R$ of $\lambda$ and a unique $C'$ function $g: U \to \mathbb R^n$ such that $g(\lambda)=0$ and for any point $t\in U$, $F(g(t),t)=0$.

Is this a correct idea to apply the theorem to the map I specified? If not, how should I approach the problem? If yes, I have two questions. First, how to show that the matrix we consider is non-singular? (This matrix is $\operatorname{Jac}_0 f-\lambda I$.) Second, how exactly the condition we need follow from the implicit function theorem? It doesn't say anything about neighborhoods of $(0,\lambda_0)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.