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So I have just started taking a course in bifurcations and finding some parts of the course rather tricky when I don't think they should be. I was wondering if anyone could explain this question.

Find a transformation putting each of the f(r, x) in normal form for a saddle-node bifurcation in a vicinity of the bifurcation point.

$$x'=1+r/2+x^2$$

From what I've researched, the normal form is the simplest form of expressing a saddle node bifurcation which is $x'=r+x^2$ and so the question is asking me to put the equation into normal form.

This is what I have done,

I have completed the square and got the equation into:

$$(x+r/4)^2+1-r^2/16$$

Thus, I let $X=(x+r/4)$ and $R=(1-r^2/16)$ and therefore can express it as:

$$X'=X^2+R$$

From here I am very confused because I believe that I need to incorporate the use of critical points but I am not sure. If someone could explain the procedure(?) in finding the normal form for a saddle node bifurcation, it would help clarify things so much for me

user415903
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    Why not to introduce a new parameter $\alpha=1+r/2$ to get $\dot x=\alpha+x^2$ and you are done. – Artem Mar 06 '18 at 23:56
  • You completed the square wrong. $(x+r/4)^2 = x^2+xr/2+r^2/16$ ; notice the factor of $x$ with $r/2$ . – mr_e_man Apr 26 '18 at 00:36

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