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The second equation gives $y=x(1-x)$. The first then says either $y=0$, giving the persistent equilibrium points $(0,0)$ and $(1,0)$, orgiving $x_{\pm }=((\beta +1)\pm \sqrt{(\beta +1{)}^2+4\mu })/2$ and $y_{\pm }=x_{\pm }(1-x_{\pm })$. These two points exist for $\mu \ge -(\beta +1{)}^2/4$.

At the origin the Jacobian matrix has trace $-1$ and determinant $-\mu$, so it is stable for $\mu <0$ and a saddle point for $\mu >0$. At $(1,0)$ the trace is $-1$ and determinant $\mu +\beta$, so it is stable for $\mu >-\beta$ and a saddle for $\mu <-\beta$. Stationary bifurcations occur atwith the first two coinciding when $\beta =1$.

For $F(t)=\mathrm{sech}\alpha \sinh t-t$, the relevant saddle point is $t=\alpha$, where $F(\alpha )=\tanh \alpha -\alpha$ and $F^{\prime \prime }(\alpha )=\tanh \alpha$. The method of steepest descent gives

Write the phase and amplitude asThe saddle point relevant to the contour is $z=0$, since ${\varphi }^{\prime }(z)=\sinh z$ andThe local steepest-descent direction is therefore vertical: with $z=iy$,

The given contour approaches the lines $\mathrm{Im}z=-\pi$ at its left end and $\mathrm{Im}z=\pi$ at its right end. The poles of $f$ are at $z=\pm 4i$, outside the closed strip $|\mathrm{Im}z|\le \pi$. The Cauchy integral theorem therefore permits deformation to the contour consisting of the lower horizontal ray from $-\infty -i\pi$ to $-i\pi$, the vertical segment from $-i\pi$ to $i\pi$, and the upper horizontal ray from $i\pi$ to $\infty +i\pi$. On either horizontal ray,so those two integrals are exponentially small, of order $e^{-x}{x}^{-1/2}$.

The vertical segment, oriented upward, contributesIts real phase has a unique maximum at $y=0$. Sincethe simple-saddle contribution in steepest descent, equivalently the local Gaussian integral, givesThe factor $i$ comes from the upward tangent $dz=idy$ of the deformed contour.

The exponential phase isIts saddle points satisfyFor positive real $\lambda$, the original positive-real contour passes through $s=\lambda$; the singularity at $s=0$ separates it from the other saddle. At the contributing saddle,Set $s=\lambda +{\lambda }^{1/2}t$. ThenThe method of steepest descent therefore gives

Write $w=z-1=X+iY$. The phase and its derivative areso the saddle points arewith $\varphi (z_{+})=2i$ and $\varphi (z_{-})=-2i$. The contour geometry can be drawn fromThe stationary-phase level $\mathrm{Re}\varphi =0$ consists of $X=0$ and $X^2-3Y^2+3=0$, meeting at the saddles. The steepest curves through $z_{\pm }$ are the levels $\mathrm{Im}\varphi =\pm 2$. Far away, sectors with $\cos (3\arg w)>0$ are exponential hills and those with $\cos (3\arg w)<0$ are valleys.

The stated contour deforms through the upper saddle. If $z=z_{+}+s$, thenThe descent tangent has $s=e^{i\pi /4}t$, because $3i{s}^2=-3t^2$. The simple-saddle contribution in steepest descent is therefore

When the contour begins at $z=1$, deform it first from the endpoint into the decaying negative-real direction and then onto the same upper-saddle descent path. Near the endpoint, $s=z-1$ and $\varphi =3s+O(s^3)$, so the endpoint contribution in steepest descent isAdding the saddle and endpoint pieces gives

Write the Schläfli contour integral for Legendre polynomials asThe saddle points satisfyFor $\mu =\cos \theta$, they are $z_{\pm }=e^{\pm i\theta }$. Deforming the contour through both conjugate saddles and using the local Gaussian contributions from the method of steepest descent gives conjugate exponentials. Their sum is the Debye asymptotic for Legendre polynomialsBoth saddles are required for the real oscillatory answer.

A harmonic line force localized at one point of a two-dimensional sheet gives $\hat{\eta }=F/D$ and $\Delta =\gamma D$ in the Fourier transform. The outgoing upper pressure is the displayed amplitude. Dividing by $c_0^2$ converts it to acoustic density perturbation. The method of steepest descent gives cylindrical sound from the saddle point, while crossed zeros of the dispersion relation give guided or leaky modal residues. Retaining $k_0=\omega /c_0$ is essential for dimensionally consistent spreading and prefactors.

For $e^{-\lambda z^2}$, the saddle point descent is the real axis and ascent is the imaginary axis. A descent path from a nonzero endpoint $b$ has $z(s{)}^2=b^2+s$, $s\ge 0$, with the continuous square-root branch from $b$. It ends at positive or negative real infinity according to the sign of $\mathrm{Re}b$. The oriented outward endpoint contribution in steepest descent is $e^{-\lambda b^2}/(2\lambda b)$.

On a ray $x/t=V$, a localized wave packet is governed by a physically selected saddle point satisfying $d\omega /dk=V$ in its analytic dispersion relation. The exponent's real growth is $\mathrm{Im}({\omega }_{\star }-V{k}_{\star })$. The fixed-frame case $V=0$ gives the absolute growth rate. This ray formulation makes the frame dependence of convective hydrodynamic instability explicit.

The completed-square form gives global minima at $(1,-1)$ and $(-1,1)$, both with value $-1$, and a saddle point at $(0,0)$ with value zero. Negative levels split into two closed components; the zero contour joins their lobes at the saddle. The gradient flow $\dot{x}=-V_x,\dot{y}=-V_y$ decreases $V$ by the gradient-flow dissipation identity. A negative initial level traps each trajectory on its original side of $x=0$, because $V(0,y)\ge 0$.

For $m\ne 0$ and positive $\beta$, the Polonyi model has a unique zero-energy global minimum at $\beta =2-\sqrt{3}$, $⟨z⟩=\sqrt{3}-1$. Writing $u=\mathrm{Re}z-(\sqrt{3}-1)$ and $y=\mathrm{Im}z$, its potential bracket is $(u^2+y^2+\sqrt{3}u{)}^2+(2\sqrt{3}-3)u^2+(4-2\sqrt{3})y^2$, a sum of nonnegative terms. The real and imaginary curvatures are $4\sqrt{3}$ and $8-4\sqrt{3}$. The other zero-energy stationary point with $\beta =2+\sqrt{3}$ and $z=-\sqrt{3}-1$ is a saddle point. The stable vacuum has nonzero supergravity auxiliary field, so it has supersymmetry breaking even with zero cosmological constant. The $m=0$ branch does not fix these values.