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The gradient vanishes at $0$. Decompose any vector as $x=tb+w$, with $w\perp b$. Symmetry of $A$ and $Ab=\frac{1}{2}b$ imply $b^{T}Aw=(Ab{)}^{T}w=0$. Since $\Vert b\Vert =1$,If $d\ge 2$, choose a nonzero $w\perp b$. Arbitrarily close to zero, the direction $b$ gives negative values and the direction $w$ gives positive values. Thus $0$ is a saddle point for $d\ge 2$. Equivalently, the Hessian matrix has one negative eigenvalue and all its other eigenvalues positive.

The dimension matters: if $d=1$, the orthogonal complement is zero-dimensional and $g(tb)=-t^2/2$. In that case $0$ is the unique global maximum. This edge case is not excluded by the stated hypotheses. The decomposition is a rank-one downdate along an eigenvector.

At $x=0$, the gradient from the preceding part vanishes and the Hessian matrix is zero. The second-order test is therefore inconclusive. Instead inspect the eigenline:For arbitrarily small positive and negative $t$, these values have opposite signs. Thus $0$ is neither a local maximum nor a local minimum. In dimension at least two it is a degenerate saddle point; in one dimension it is a stationary inflection point. Looking only at the zero Hessian matrix would not determine this classification.

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 $e^{-\lambda b^2}/(2\lambda b)$.

For $\dot{Y}=-Y+\gamma /(1+Z^2)$ and $\dot{Z}=-Z+\gamma /(1+Y^2)$, subtraction of the equilibrium conditions gives $(Y-Z)(1-YZ)=0$. The symmetric state solves $x^3+x=\gamma$. Two asymmetric positive states exist precisely for $\gamma >2$: $Y_{\pm }=(\gamma \pm \sqrt{{\gamma }^2-4})/2$, $Z_{\pm }=Y_{\mp }$. Their Jacobian matrix has eigenvalues $-1\pm 2/\gamma$, both negative; the symmetric state is then a saddle point. For $0<\gamma <2$ only the attracting symmetric state remains.

For $\dot{x}=\alpha x-y+y^3$, $\dot{y}=-x$, the energy $H=x^2-y^2+y^4/2$ satisfies $\dot{H}=2\alpha x^2$. At $\alpha =0$, its compact level sets give two families of closed periodic orbits around $(0,\pm 1)$ and an outer family around both; $H=0$ is a figure-eight homoclinic orbit pair through the saddle point at the origin. The two minima of $H$ are center equilibria for zero damping, stable foci for small negative $\alpha$, and unstable foci for small positive $\alpha$. For $\alpha \ne 0$, strict monotonicity along nonconstant solutions excludes periodic orbits and homoclinic orbits.

The stationary points of this polynomial are the origin and $(a,a)$, $(-a,-a)$. Its Hessian matrix at the origin has eigenvalues $\pm 4a^2$, so the origin is a saddle point. At each of the other points the eigenvalues are $8a^2$ and $16a^2$, giving strict local minima. The identityshows that they are also the only global minima, with value $-2a^4$.

For $-2a^4<c<0$, the level set $g=c$ has two closed components, one surrounding each minimum. The zero level set has two lobes meeting at the origin with the coordinate axes as tangent directions. For $c>0$, there is one closed outer component. These statements also follow in polar coordinates from the quadratic equation in $r^2$, together with the absence of other stationary points.

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.

Suppose $h$ is differentiable and $F^{\prime }>0$ on its range. The multivariable chain rule shows that $F\circ h$ has exactly the same stationary points as $h$. Since $F$ is strictly increasing, the type of every local maximum, local minimum, and saddle point is preserved. If both functions are twice differentiable, then at a stationary pointso the Hessian matrix signature is unchanged. The hypothesis $F^{\prime }>0$ matters for the converse about stationary points: a strictly increasing function such as $F(s)=s^3$ can have a zero derivative and create new stationary points in a composition.

For the auxiliary-field convention with cost $M^2/(2U)$ and coupling $-M\bar{\psi }{\sigma }_z\psi$, the homogeneous saddle point equation equates M to U times spin polarization per site. M has energy units. It is not itself a magnetic moment, and its coefficient changes if the contact interaction or density-of-states spin convention is changed. The exact three-component decoupling must be distinguished from selecting one constant spin direction as a mean-field approximation.

This real planar dynamical system has a saddle point at $(-1,0)$ and a repelling weak focus at $(1,0)$. The latter has purely imaginary linearized eigenvalues $\pm 2i$, so its linearization has closed elliptical orbits. Nonlinear terms nevertheless produce outward drift.

Set $X=2(x-1)$, $Y=y$, $\tau =2t$ and $r^2=X^2+Y^2$. The cubic expansion isThe linear rotation acts on scalar functions by $L=Y{\partial }_X-X{\partial }_Y$. For $H_2=r^2/2$, its cubic derivative is $X{Y}^2-3X^{2}Y/4-Y^3/2$. The homogeneous correction $H_3=7X^3/12+X{Y}^2/2+Y^3/3$ satisfies $L{H}_3$ equal to the negative of that derivative. The remaining quartic derivative of $H_2+H_3$ is $3X^2{Y}^2/2-X^{3}Y$. Withdirect differentiation givesfor all sufficiently small nonzero $r$. Meanwhile $H_2+H_3+H_4=r^2/2+O(r^3)>0$. In reversed time it is a strict Lyapunov function, proving asymptotic stability of the reversed system and hence repulsion forward in time. The angular speed stays close to $-1$, so the nearby nonzero trajectories spiral rather than approaching along a fixed direction. This explicit calculation illustrates why a centre in the linearization need not be a nonlinear centre.