Transform $dx/dt = 1 + r + x^2$ into normal form $dX/dt = R +- X^2$ (saddle node bifurcation)

Question

I am trying to find a substitution $X(x,r)=?$ and $R(x,r)=?$ to allow me to transform

$dx/dt = 1 + r + x^2$

into

$dX/dt = R +- X^2$ (normal form for saddle node bifurcation)

I'm sure it's a simple substitution, but I just can't see it for the life of me.

Thanks!

Answer 1

The solution was in fact to complete the square which gives:

$dx/dt = 1 - (r/2) + (x+r/2)^2$

then let X=x+r/2, which means dX/dt = dx/dt, and then let R=1-(r/2). Substituting then gives:

$dX/dt = R + X^2$

which is the normal form for a saddle node

Answer 2

We are working with the equation: $$\frac{dx}{dt}=r+1+x^2$$ I believe the answer is connected to the total derivative of the substitution function. For example we can simplify: $$\frac{dX(x,r)}{dt}=R(x,r)\pm X(x,r)^2$$ Into $$\frac{\partial X}{\partial x}\frac{dx}{dt}+\frac{\partial X}{\partial r}\frac{dr}{dt}=R(x,r)\pm X(x,r)^2$$ Therefore, we match the left side to the first equation and state that: $$\frac{\partial X}{\partial r}=0\rightarrow X(x,r)=X(x)$$ Therefore we rearrange the equation to: $$\frac{dX}{dx}\frac{dx}{dt}=r+1+x^2\rightarrow\frac{dx}{dt}=\frac{R(x,r)}{X'(x)}\pm\frac{X(x)^2}{X'(x)}$$ By matching expressions, we obtain two ODEs for the functions: $$\frac{X(x)^2}{X'(x)}=\pm(1+x^2)\text{ }\&\text{ }R(x,r)=r\cdot\frac{dX}{dx}$$ With the solution of: $$X(x(t))=\frac{1}{C\mp\tan^{-1}x(t)}\text{ and }R(x,r)=r\cdot\pm\frac{1}{\left(C\mp\tan^{-1}x(t)\right)^2\left(1+x(t)^2\right)}$$