For the directly forced cubic amplitude equation, a saddle-node bifurcation requires and . The right side has its unique maximum at , giving a cusp bifurcation at . The stability matrix has trace and determinant . A Hopf bifurcation requires , and . These imply ; equality is a double-zero linear degeneracy rather than an ordinary Hopf bifurcation.
Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 60 2 i Solution Created 2026-10-03 Updated 2026-10-14
Write . A steady complex amplitude obeys , soThus every positive intensity root gives exactly one steady phase, and is impossible. The cubic intensity polynomial is . Its positive steady branch can be parameterized completely asThe two formulae join smoothly as a single curve at the maximum radius ; this joining point is not a saddle-node bifurcation. A turning point with respect to satisfies . Subtracting the equilibrium and derivative conditions givesThese points lie on . Since , its maximum is . Therefore there is exactly one distinct steady state for every when . One can also see uniqueness directly: is increasing and is decreasing when no turning points exist, and their respective ranges meet only at their common endpoint. At the equality there is a cusp bifurcation with a triple intensity root at .
When , the two turning intensities satisfy . Define and ; the curve has a local minimum at the former and a local maximum at the latter, with . There are three distinct steady intensities for , two distinct ones at an endpoint, and one otherwise. The bifurcation diagrams below show the folded small- curves and the unfurled large- curve, together with the linear stability calculated next. Both ends of the full branch have as , while its middle reaches the finite maximum .