# How does a fixed point of a dynamical system change when the system is changed slightly?

I've come across this problem in the course of my work, and I'm a bit stuck on it.

Suppose I have an $n$-dimensional smooth dynamical system $$\dot{x_i} = f_i(\mathbf{x}),$$ and suppose there is a fixed point $\hat{\mathbf{x}}$, i.e. $f_i(\hat{\mathbf{x}})=0$ for every $i$.

Now let us define a new dynamical system $$\dot{x_i} = f'_i(\mathbf{x}),$$ where $f_i'(\mathbf{x}) = f_i(\mathbf{x}) + \epsilon\delta f_i(\mathbf{x})$, for some set of arbitrary smooth functions $\delta f_i$.

It's possible that the fixed point of the original system was a bifurcation point, so that there will be no corresponding fixed point in the new system, no matter how small we make $\epsilon$. But supposing this isn't the case, the fixed point should just move to a slightly different position. That is, $\hat{\mathbf{x}}' = \hat{\mathbf{x}} + \delta \hat{\mathbf{x}}$, where every $f_i'(\hat{\mathbf{x}}')=0$ and $\delta\hat{\mathbf{x}}$ will be of the same order as $\epsilon$.

What I want is an expression for $\delta\hat{\mathbf{x}}$ (in the small $\epsilon$ limit) in terms of $\epsilon$ and the partial derivatives of the functions $f_i$ and $\delta f_i$ at the original fixed point. It seems like it should be possible, but after going cross-eyed scribbling on the whiteboard I haven't come up with the answer. I'm probably just missing some simple trick - can anyone help me out?

I will in the following do nothing but simple calculus trick. Assume $f'(x)=f(x)+\epsilon g(x)$ where $g(x)$ is arbitrary smooth function. Denote fixed points of two dynamical systems by $x_0,x_0'$ so that $f(x_0)=f'(x_0')=0$ and $\delta x=x_0'-x_0$. Then we have \begin{align}f'(x'_0)&=f(x_0+\delta x)+\epsilon g(x_0+\delta x)\\ &=f(x_0)+Jf(x_0)\delta x+\epsilon\left(g(x_0)+Jg(x_0)\delta x\right)\\ &=Jf(x_0)\delta x+\epsilon g(x_0)\\ &=0\end{align} Hence $\delta x=-\epsilon(Jf)^{-1}(x_0)g(x_0)$ where $Jf$ is Jacob matrix. Note that in the second line, $f(x_0)=0$ and $\epsilon\delta xJg(x_0)$ is infinitesimal compared to $\epsilon$.
• Thanks - I knew it would be simple, I just wasn't looking at it in the right way. In the multivariate case it becomes $\delta \mathbf{x} = -\epsilon J^{-1}\mathbf{g}(\mathbf{x}_0)$, where $J$ is the Jacobian matrix, i.e. $J_{ij} = \partial f_i/\partial x_j$, or $(J^{-1})_{ij} = \partial x_i/\partial f_j$. – Nathaniel Nov 19 '13 at 4:38
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