Question about saddle-node bifurcation in a $2$D-system of vector fields

bifurcationnonlinear dynamics

Consider a two dimensional continuously differentiable vector fields
in $\mathbb{R}^2$ $$\dot{u}=a(1-u)-uv^2$$ $$\dot{v}=uv^2-(a+k)v$$ where
$a,k>0$ are parameters. Show that saddle-node bifurcation occurs at
$\displaystyle k=-a\pm \frac{\sqrt{a}}{2}$.

A saddle-node bifurcation is defined at a point if two equilibrium points annihilate each other upon changing the parameter values $a,k$. The given system has two stable equilibrium points $(0,0)$ and $(1,0)$ since their Jacobian is $\begin{pmatrix} -a & 0 \\ 0 & -(a+k) \end{pmatrix}$. Now we need to vary the parameters $a,k$ in such a way that these two equilibrium points annihilate each other. But how to do this in $2$-dimensional system I don't understand. Any help is appreciated.

Best Answer

The equilibrium points are at

$$ \left[ \begin{array}{ccc} n & u & v \\ 1 & 1 & 0 \\ 2 & \frac{a+\sqrt{a \left(-4 a^2-8 k a+a-4 k^2\right)}}{2 a} & \frac{a-\sqrt{a \left(a-4 (a+k)^2\right)}}{2 (a+k)} \\ 3 & \frac{a-\sqrt{a \left(-4 a^2-8 k a+a-4 k^2\right)}}{2 a} & \frac{a+\sqrt{a \left(a-4 (a+k)^2\right)}}{2 (a+k)} \\ \end{array} \right] $$

The equilibrium points $\{2,3\}$ collapse (have the same coordinates) at $k = -a \pm\frac{\sqrt{a}}{2}$: thus for $k = -a+\frac{\sqrt{a}}{2}$ they collapse in $\{\frac 12,\sqrt{a} \}$ and for $k = -a-\frac{\sqrt{a}}{2}$ in $\{\frac 12,-\sqrt{a} \}$

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