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Suppose $r>1$ and set $q=\log r/(r-1)$. Substitution of the positive equilibrium into the Jacobian matrix givesPart (b) showed $f(x)<0$ whenever $x\ne 1$. In particular,gives $\log r<r-1$, whereasgives $\det J_{\ast }>1$. The latter already violates the Jury stability criterion condition $1-\det J_{\ast }>0$. More directly, the product of the two eigenvalues is greater than one, so at least one has modulus greater than one. Therefore the positive equilibrium is unstable. This is the instability of the Nicholson-Bailey equilibrium; parasitoid feedback in this simple model does not stabilize coexistence.

For the adult-survival annual pulse-breeding predator-prey model, a positive annual fixed point exists when $r>1$ and $0<s<1$:At this point the Jacobian matrix isIf $0<(1-s)\log r<4$, the two eigenvalues are a complex conjugate pair of moduli one. The annual samples show linearly neutral seasonal oscillations. If $(1-s)\log r>4$, they form a reciprocal negative real pair and the fixed point is a saddle fixed point of a map. Equality at four is a degenerate boundary needing nonlinear analysis. For replacement rather than surviving adults the same calculation gives trace $2-\log r$ and determinant one.

The absence of an attracting annual cycle is stronger than a linear observation. Put $u=\log X$, $v=\log Y$. The update becomesIt is the composition of two nonlinear shear maps, each of determinant one: an area-preserving seasonal predator-prey map. No isolated positive periodic orbit has an open attracting basin. Neutral linear oscillations alone also do not prove nonlinear stability at every resonance. In particular, this density-independent model does not select an asymptotically stable oscillation amplitude.

The Nicholson-Bailey model instead counts parasitoid offspring from attacked hosts:Its positive equilibrium has $Y_{\ast }=\log r/\kappa$ and $X_{\ast }=Y_{\ast }/[c(1-1/r)]$. Its Jacobian matrix hasSince the determinant exceeds one while the positive trace is below two, the eigenvalues are complex with moduli greater than one: instability of the Nicholson-Bailey equilibrium produces outward oscillations near it. The two maps share an exponential prey-survival factor but have different reproduction laws and stability properties.

There is always the fixed point $(0,0)$. For $P>0$, the first fixed-point equation gives $e^{-P}=1/r$, and the second gives $P=N(1-1/r)$. Thus for $r>1$ there is exactly one positive fixed point,For $r=1$, every point $(N,0)$, $N\ge 0$, is fixed. There are no other nonnegative fixed points.

At the origin the Jacobian matrix has eigenvalues $r,0$. It is asymptotically stable for $0<r<1$ and a saddle for $r>1$. For $r<1$, the bounds $N_{t+1}\le r{N}_t$ and $P_{t+1}\le N_t$ also prove global extinction in the nonnegative quadrant.

At the positive fixed point,Here $\log r<r-1$ and $\log r>1-1/r$ for $r>1$. The discriminant is negative because the positive trace is less than two and determinant exceeds one. The conjugate multipliers have modulus $\sqrt{N_{\ast }}>1$, so the positive fixed point is always unstable. This is the instability of the Nicholson-Bailey equilibrium.

For completeness, on the $r=1$ fixed line,For $0\le N<1$, linear perturbations are bounded but have a neutral component along the fixed line; there is no asymptotic stability of an individual point. For $N>1$ there is transverse instability. At $N=1$, the repeated unit eigenvalue has a nontrivial Jordan block, so powers have an unbounded off-diagonal entry and the linearized system is unstable. This degenerate case is not covered by a distinct-eigenvalue assumption.