Recipe

Find the bifurcation value of a one-parameter family

Use when: you have a family of autonomous equations $\dfrac{dy}{dt}=f(y,c)$ depending on a parameter $c$ , and want the exact value of $c$ where the number of equilibria changes

Goal: the bifurcation value $c^\star$ , plus what happens to the equilibria on each side of it

Taught in Bifurcation of a 1-D autonomous family · extends Find the equilibria of a harvested logistic population

Quick recipe

Solve $f(y,c)=0$ for $y$ , keeping $c$ as a parameter
Count how many real solutions exist, as a function of $c$
Find the exact value where the count changes
Classify the behavior on each side of $c^\star$

1
Set the rate to zero and solve for $y$ in terms of $c$ — $c$ stays a fixed background constant here, it isn't what you're solving for.

Example

For $\dfrac{dy}{dt}=c-y^2$ : setting $c-y^2=0$ gives $y=\pm\sqrt c$ .

Why: equilibria are always found the same way, parameter or not — set the rate of change to zero and solve.
2
Determine, without picking a specific $c$ yet, whether the solution set has two members, one, or none, and how that count depends on $c$ .

Example

$y=\pm\sqrt c$ gives two real, distinct roots when $c>0$ ; exactly one (repeated) root, $y=0$ , when $c=0$ ; and no real root at all when $c<0$ , since a negative number has no real square root.
- two real roots hold for a whole range of $c$ — these are the branches Step 4 classifies.
- the root count changes at one specific value of $c$ — that value is the bifurcation value; go to Step 3 to pin it down exactly.
Why: a bifurcation is exactly the parameter value where this count changes — not merely where the individual roots happen to move.
3
Solve for the $c$ at which the expression under the root (or a quadratic's discriminant) hits zero.

Example

$y=\pm\sqrt c$ has real solutions exactly when $c\ge0$ , with a single repeated root exactly at $c=0$ . $\boxed{c^\star=0}$

⚡ Shortcut

When $f(y,c)=0$ expands into a quadratic in $y$ , skip the square-root reasoning entirely and set that quadratic's discriminant $b^2-4ac=0$ directly. For the harvested logistic family $y^2-10y+10h=0$ from the previous section, the discriminant is $100-40h$ , which hits zero at $h=2.5$ — the same maximum sustainable harvest rate found there, reached here as one discriminant condition instead of tracking two roots as they collide.

Why: the merge point is precisely where two roots collapse into one, and for a quadratic (or an expression like $y^2=c$ ) that's a single algebraic condition, not a search.
4
Pick one $c$ -value on each side of $c^\star$ and determine the sign of $f$ (hence stability and long-term behavior), using the phase-line reasoning from earlier in this chapter.

Example

At $c=4>c^\star=0$ : equilibria at $y=2$ (stable, since $f'(y)=-2y$ is negative there) and $y=-2$ (unstable). At $c=-4<c^\star$ : no equilibria at all, and $f=-4-y^2<0$ for every $y$ , so every trajectory decreases without bound.

Why: the bifurcation value is only useful once you know what qualitatively different behavior it separates — the number alone isn't the answer.

Common mistakes

Solving $f(y,c)=0$ for $c$ instead of $y$ in Step 1 — $c$ is fixed background data until Step 3, not the unknown.
Assuming the merged equilibrium at $c^\star$ is stable or unstable like its neighbors, when a freshly merged equilibrium is generically semistable — attracting from one side, repelling from the other.
Reporting only the number $c^\star$ without saying which two behaviors it separates (two equilibria vs. none, in the harvesting example: survival vs. certain extinction).
Confusing the bifurcation VALUE (the parameter $c^\star$ ) with the equilibrium value where the merge happens (a $y$ -value) — the two answer different questions.