Solution. Equilibria satisfy \(x^2 - \mu x + h = 0\), so \(x = \frac 12(\mu \pm \sqrt {\mu ^2 - 4h})\), real for \(|\mu | \geq 2\sqrt h\). At \(\mu = \pm 2\sqrt h\) there are saddle-node bifurcations. For \(\mu > 2\sqrt h\) the larger root is stable (\(f' = \mu - 2x < 0\) there) and the smaller unstable; for \(\mu < -2\sqrt h\) both roots are negative. The harvest \(h\) has turned the transcritical bifurcation of \(x' = \mu x - x^2\) into a pair of saddle-nodes: a small perturbation destroys a non-generic bifurcation.

Solution. \(x = 0\), and \(x^2 = \frac 12(1 \pm \sqrt {1 + 4\mu })\). The origin is stable for \(\mu < 0\) and unstable for \(\mu > 0\), with a subcritical pitchfork at \(\mu = 0\) (the branches \(x^2 = \frac 12(1 - \sqrt {1 + 4\mu })\), unstable, exist for \(-\frac 14 < \mu < 0\)). The outer branches \(x^2 = \frac 12(1 + \sqrt {1 + 4\mu })\) are stable and exist for \(\mu > -\frac 14\), joining the inner ones in saddle-nodes at \(\mu = -\frac 14\). As \(\mu \) increases through \(0\) the state jumps from \(0\) to the outer branch, \(x^2 = 1\); decreasing \(\mu \) again, it stays on the outer branch until \(\mu = -\frac 14\), where it drops back to \(0\). The interval \(-\frac 14 < \mu < 0\), in which two stable states coexist, is the hysteresis loop.

Solution. At \((a, b/a)\), \(J = \begin {bmatrix} -(b + 1) + 2xy & x^2\\ b - 2xy & -x^2\end {bmatrix} = \begin {bmatrix} b - 1 & a^2\\ -b & -a^2\end {bmatrix}\), since \(2xy = 2b\). The eigenvalues are complex when \((\tr J)^2 < 4a^2\), with real part \(\alpha = \frac 12\tr J = \frac 12(b - 1 - a^2)\). So \(\alpha = 0\) at \(b = 1 + a^2\), where the eigenvalues are \(\pm ia\), and \(\alpha '(b) = \frac 12\).

Solution. Differentiating, \[ \tfrac 12\dot W = \frac {x}{\sigma }\,\sigma (y - x) + y\bigl (x(\rho - z) - y\bigr ) + z(xy - \beta z) = -x^2 + (1 + \rho )xy - y^2 - \beta z^2 , \] the \(xyz\) terms cancelling. The quadratic form \(x^2 - (1 + \rho )xy + y^2\) is positive definite if \((1 + \rho )^2 < 4\), that is \(\rho < 1\). So \(\dot W < 0\) except at the origin, \(W\) is radially unbounded, and every solution tends to \(0\).

Solution. \(x = y\), \(z = \rho - 1 = 27\), \(x^2 = \beta z = 72\), so \(C_\pm = (\pm 8.485, \pm 8.485, 27)\). Then \(\rho _H = 10(10 + \frac 83 + 3)/(10 - \frac 83 - 1) = 10(15.667)/6.333 = 24.74\). Since \(28 > 24.74\), both \(C_\pm \) are unstable at the standard parameters, as are the origin and every equilibrium, while all orbits remain bounded.

Solution. With \(F(x) = rx(1 - x)\), \(F'(x^*) = r(1 - 2x^*) = 2 - r\), of modulus less than \(1\) for \(1 < r < 3\). At \(r = 3\), \(F'(x^*) = -1\): a multiplier crosses \(-1\), the discrete analogue of a Floquet multiplier \(-1\) in Theorem 7.3.2(iii), and a cycle of twice the period is born (a flip or period-doubling bifurcation). Solving \(F(F(x)) = x\) and removing the fixed points gives the 2-cycle \(x = \frac {r + 1 \pm \sqrt {(r + 1)(r - 3)}}{2r}\), real for \(r > 3\).