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An interior stationary point has vanishing gradient:For a differentiable function, this is the necessary first-order condition for an interior local minimum or local maximum.

The product rule gives $y^{\prime }=e^x(\cos x-\sin x)$. The exponential function never vanishes, so a stationary point requires $\tan x=1$. In the given interval this means $x=\pi /4$. Substitution givesThe derivative changes from positive to negative there, making it a strict local maximum.

The Hessian matrix has entriesAt each stationary point the mixed entry vanishes, so the signs of the diagonal entries directly classify the local quadratic change:In particular the two local minima are also global minima: the squared term is nonnegative and $\ln (3+y^2)\ge \ln 3$, with equality in both bounds exactly at $(\pm 1,0)$. The local maximum at the origin is not a global maximum, since $f$ grows without bound as $x^2+y^2\to \infty$. The opposite diagonal signs at the other two points establish their saddle point character.

The contour plot is symmetric under either $x\mapsto -x$ or $y\mapsto -y$, because $f$ depends only on their squares. Denote the three critical heights by $m=\frac{7}{2}\ln 3$, $h=\frac{1}{4}+\frac{7}{2}\ln (7/2)$ and $M=1+\frac{7}{2}\ln 3$, where $m<h<M$.

For $m<c<h$, the level set $f=c$ consists of separate closed curves around the two minima. At $c=h$ the branches meet at the two saddle points. For $h<c<M$, there is an inner closed curve surrounding the local maximum and an outer closed curve surrounding all five stationary points. At $c=M$ the inner curve shrinks to the origin; for $c>M$ only the outer closed curve remains. At each saddle point the critical level set crosses in the two directions determined byThe figure plots representative level sets, including the critical saddle height; its second panel adds the gradient directions requested below.

Near each local minimum, the gradient points outward; near each local maximum, it points inward. The five remaining points have saddle directions, determined by their indefinite Hessian matrices.

On the $x$-axis the exact gradient is $(0,x(x^2-1){)}^T$, and on the $y$-axis it is $(y(y^2-1),0{)}^T$. Thus arrows are perpendicular to the axes and reverse direction at the origin and the unit-axis stationary points.

On the unit circle, substitute $x^2+y^2=1$ into the gradient to obtainIt is radial outward in quadrants I and III, radial inward in quadrants II and IV, and zero at the four axis intersections. These arrows, together with the nearby saddle and extremum patterns, are shown in the right panel of the contour figure. The gradient is normal to every regular level curve and points toward increasing values of $f$.

The chain rule and the quotient rule giveThe exponential function and denominator are positive, so the stationary points occur exactly at $x=\pm 1$. Their values areThe derivative is negative for $|x|>1$ and positive for $|x|<1$, so $x=-1$ is a local minimum and $x=1$ a local maximum. They are also global extrema, since $-1/2\le x/(1+x^2)\le 1/2$.

The Hessian matrix is diagonal:Its eigenvalues are $c$ and $d$. Hence the second-derivative test gives a strict local minimum if $c,d>0$, a strict local maximum if $c,d<0$, and a saddle point of a scalar function if $cd<0$. No degenerate case occurs because both are nonzero.

The local minimum is the unique global minimum, since both separated summands are strictly convex functions and tend to $+\infty$ at either end of the real line. Similarly, when both signs are negative there is a unique global maximum.

First take $a\ne 0$, put $A=|a|$, and suppose $k>0$. The force is the negative derivative of the potential energy:The potential energy is an odd function, passes through $U(0)=0$, and approaches zero from above as $x\to -\infty$ and from below as $x\to +\infty$. Its stationary points are $x=\pm A$:Thus the sole stable equilibrium is the local minimum at $x=A$; the local maximum at $x=-A$ is an unstable equilibrium. The requested values at the stable equilibrium areThe force is an even function, positive for $|x|<A$, negative for $|x|>A$, and zero at $x=\pm A$. Its central maximum is $F(0)=k/A^2$. Sinceits two minima occur at $x=\pm \sqrt{3}A$, with $F=-k/(8A^2)$, and $F\to 0^{-}$ at both ends. These values fix the shapes of both requested sketches.

In one dimension, attraction towards the origin means $xF(x)<0$, while repulsion from the origin means $xF(x)>0$. With that explicitly stated convention,A positive force alone means rightward, not necessarily outward. At the origin $F(0)\ne 0$, so it is not an equilibrium; the inward/outward distinction is undefined exactly there.

The sign assumptions matter because the paper calls $a,k$ merely constants. For $k<0$ both curves reverse sign, the stable equilibrium is $x=-A$, its potential energy is $-|k|/(2A)$, and the attraction/repulsion intervals interchange. For $k=0$ the force vanishes everywhere on the nonsingular domain and there is no isolated potential energy minimum. If $a=0$ and $k\ne 0$, then $U=-k/x$ for $x\ne 0$ and $F=-k/x^2$, with no finite equilibrium.

At $N=M$, the two spin occupations per site are $x$ and $1-x$. With $\tau =|t|>0$, the trial energy per site isIts derivative is $e^{\prime }(x)=-4\tau \cos (\pi x)+V[4x-2-(2/\pi )\sin (2\pi x)]$. Since $x=0$ is a boundary point, its local minimum test is one-sided: $e^{\prime }(0)=-4\tau -2V$. At $V=-2\tau$, where this derivative vanishes, its Taylor series begins $e(x)-e(0)=(2{\pi }^2\tau /3)x^3+O(x^4)>0$ for small positive $x$. HenceThe same applies to $x=1$ by spin interchange.

The symmetry $e(x)=e(1-x)$ makes $x=1/2$ a stationary point for every $V$. Here $e^{\prime \prime }(1/2)=4\pi \tau +8V$, so the second-derivative test settles all cases except equality. Writing $y=x-1/2$, at $V=-\pi \tau /2$ the Taylor series is $e(1/2+y)-e(1/2)=({\pi }^3\tau /6)y^4+O(y^6)$. ThusThe minimizing branches interpolate continuously. To find all of them, put $u=\pi (1/2-x)$ on $0<x<1/2$. An interior stationary point obeysThe right side decreases strictly from $-\pi /2$ to $-2$, sinceThe positivity follows from $u-\sin u\cos u={\int }_0^{u}2{\sin }^{2}vdv$. There is therefore exactly one such stationary point on each side of $1/2$ when $-2<V/\tau <-\pi /2$. The signs of $e^{\prime }$ show that both are global minima, whereas $1/2$ is then a local maximum. The complete answer isThe pair merges through a supercritical pitchfork bifurcation at $V/\tau =-\pi /2$. Comparing only $e(0)$ and $e(1/2)$ would miss the intervening ground states with partial spin polarization.

Choose a basis whose first axis is $\mathbf{n}$. The tensor with uniaxial nematic order then has eigenvalues $2Q,-Q,-Q$. Hence $\mathrm{Tr}{S}^2=6Q^2$, $\mathrm{Tr}{S}^3=6Q^3$ and $\mathrm{Tr}{S}^4=18Q^4$. Substitution givesFor $a,c>0$, $Q=0$ remains a local minimum with curvature $12a$. Its nonzero stationary points satisfy $2a+3bQ+12c{Q}^2=0$, soThey first appear at $b_s=-\sqrt{32ac/3}$ as a stationary inflection point; for more negative $b$, $Q_{-}$ is a local maximum and $Q_{+}$ a local minimum. The latter is initially metastable, becomes degenerate with $Q=0$ at $b_t=-\sqrt{12ac}$, and then becomes the global minimum as $b$ decreases further. This is a first-order transition in a cubic-quartic Landau potential: the order changes discontinuously. The saddle-node bifurcation creating a local minimum is not the coexistence condition. The isotropic minimum never loses local stability in this fixed positive-$a$ scan.

The illustrative plots set $a=c=1$ and show $f/6$. A cubic invariant is permitted because head-tail reversal leaves $S$ unchanged; it does not implement $S\mapsto -S$. The physical range of a uniaxial nematic order parameter imposes $-1/2\le Q\le 1$; the usual small-order-parameter Landau result assumes the displayed minimum lies in that interval. The unrestricted quartic graphs express the intended Landau model, with the effect of enforcing the bound stated in part (f).^[ref]

Landau description. Introduce a macroscopic order parameter and expand an appropriate free energy analytically in that parameter, retaining terms permitted by the physical symmetries. Stable equilibrium minimizes this free energy globally; a local minimum with larger free energy is metastable. Coefficient changes with temperature can exchange the preferred minima. A continuous emergence from zero can produce a continuous transition, while competing separated minima can produce a first-order transition. Stability requires a positive highest retained even-power term. This local mean-field description neglects spatial fluctuations and nucleation dynamics unless gradient and dynamical terms are added; near a critical point those omissions can matter.

For this temperature-dependent cubic-quartic Landau free energy, abbreviate $B=a/(12\pi )>0$ and $A(T)=c(T^2-T_0^2)$, so $F=A{\varphi }^2/2-BT{\varphi }^3+\lambda {\varphi }^4/4$ and $\gamma =2B^2/(\lambda c)$.

Local behavior at the origin. To determine constrained order-parameter boundary stability, note that $F^{\prime }(0)=0$ and $F^{\prime \prime }(0)=c(T^2-T_0^2)$, the origin is a local minimum for $T>T_0$ and a local maximum for $0\le T<T_0$. At $T=T_0$, the quadratic term vanishes, but $F=-B{T}_0{\varphi }^3+O({\varphi }^4)<0$ for small positive $\varphi$. ThereforeThe equality case uses the stipulated domain $\varphi \ge 0$. On an unrestricted real line, the nonzero cubic term would instead make it a stationary inflection point.

Global stability in both inline cases. For $\varphi >0$, complete the square in $F/{\varphi }^2$:The quartic coefficient guarantees a bounded global minimum. A negative value exists precisely when the minimum of this quadratic factor is negative. Hence case a, $\gamma >1$, has a nonzero globally stable phase with $F<0$ for every physical temperature. Case b, $0<\gamma <1$, has such a phase exactly forAt $T=0$, the square's centre is at the boundary, but its negative constant still gives $F<0$ for small positive $\varphi$, so this endpoint is included.

At $T=T^{\ast }$,The two global minima $\varphi =0,{\varphi }^{\ast }$ have equal free energy, and $F^{\prime \prime }({\varphi }^{\ast })=\lambda ({\varphi }^{\ast }{)}^2/2>0$. The coexistence barrier of a cubic-quartic Landau transition has its maximum at ${\varphi }^{\ast }/2$, with barrier height $\lambda ({\varphi }^{\ast }{)}^4/64$.

For locating the extrema away from coexistence,Retain only real positive roots; when both are positive, ${\varphi }_{-}$ is a local maximum and ${\varphi }_{+}$ a local minimum. For $T<T_0$, only ${\varphi }_{+}$ is positive. If $\gamma <8/9$, the pair first appears on cooling at $T_{\mathrm{sp}}=T_0/\sqrt{1-9\gamma /8}>T^{\ast }$; if $8/9\le \gamma <1$, a higher-free-energy nonzero minimum already exists at arbitrarily high temperature. These local-stability limits must not be mistaken for the global coexistence condition. The boundary $\gamma =1$ also favors a nonzero minimum at every finite temperature, but has no finite $T^{\ast }$.

Cooling and transition temperature. For $\gamma <1$, equilibrium starts at the zero minimum at high temperature and switches to the nonzero minimum at $T^{\ast }$. The jump is finite, $0\to {\varphi }^{\ast }>0$, and there is a barrier between the coexisting states. This is a first-order transition. The latent heat of a temperature-dependent cubic-quartic Landau transition follows from differentiating the minimized free energy: the heat released on cooling isIf $F$ is an energy density rather than total energy, $L$ is likewise a density. An actual system can supercool below $T^{\ast }$ before nucleation, so $T^{\ast }$ is the equilibrium transition temperature.

For thermal symmetry nonrestoration in a cubic-quartic model, with $\gamma >1$, the equilibrium global minimum is nonzero already at arbitrarily high temperature. Following it does not force a transition. A system initially prepared in the metastable zero minimum can still leave it by nucleation while $T>T_0$; that minimum loses local stability at $T_0$. This is the sense in which a change is possible rather than inevitable. There is no equilibrium coexistence temperature for $\gamma >1$ in this model. A kinetic transition can depend on preparation and nucleation dynamics; $T_0$ is the limiting loss of local stability of the zero state, not a second coexistence temperature. Continuing $T^{\ast }$ to an imaginary value does not define a physical transition temperature.

Mass window in the specified electroweak phase transition model. For the mass window in a cubic electroweak transition model, put $x=m_H^2/m_0^2>0$. The supplied ratios implyThe two strict conditions therefore requireSince $x(A+x)$ is increasing for $x>0$, the finite window isUsing the specified $A$ and $m_0$ givesAt $m_H=134$ u, $x=0.3581801$, $\gamma \simeq 0.0030033<1$ but ${\varphi }^{\ast }/T^{\ast }\simeq 0.1481<1$. Thus 134 u lies outside the required window, failing the order-parameter-strength condition. This does not remove the polynomial model's finite first-order jump; it fails the additional requirement ${\varphi }^{\ast }>T^{\ast }$. The result is a bound within the specified simplified model, not a model-independent Higgs-mass bound.

Here a dense $100000$-by-$100000$ matrix has $10^{10}$ entries, about $80$ GB even with eight-byte storage per entry. Use limited-memory BFGS on the negative of the objective, or equivalently its positive-definite approximate-metric form of gradient ascent. It stores only a fixed small number $\ell$ of vector pairs, costing $O(\ell d)$ memory rather than $O(d^2)$.

Let $G(x)={\sum }_{i=1}^{100}{e}^{e^{-\Vert x-a_i{\Vert }^2}}$ and put $c_i(x)=e^{e^{-\Vert x-a_i{\Vert }^2}}{e}^{-\Vert x-a_i{\Vert }^2}$. Direct use of the chain rule givesIf $H_n$ denotes the positive-definite inverse-Hessian matrix approximation for $-G$, the iteration iswith a line search increasing $G$ and the usual safeguarded limited-memory updates. No exact Hessian matrix is required. With fixed memory and the usual bounded-metric and line search hypotheses, expect local linear convergence near a nondegenerate strict maximum; fixed-memory limited-memory BFGS does not generally inherit the superlinear theorem for full-memory BFGS.

Unlike the preceding objective, $G$ is not generally a concave function. Thus this is a method for finding a local maximum; comparison of different converged candidates or a global search is needed to certify a global maximum for arbitrary centres. The symbolic problem supplies no assumption that would justify an unconditional global-convergence claim.^[ref]

The stationary point equations give $y=x$ and $x^3=x$, hence $(0,0),(1,1),(-1,-1)$. At $(0,0)$ the Hessian matrix has negative determinant $-16$, so $(0,0)$ is a saddle point. At each of $(1,1)$ and $(-1,-1)$, the Hessian matrix has positive leading minors $12$ and $32$, proving a strict local minimum. More strongly,Thus $(1,1)$ and $(-1,-1)$ are exactly the global minima, with value $3$. There are no local maxima, since any differentiable interior local maximum would be a stationary point, and all three have been classified.

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.

Rolle's theorem is the equal-endpoint case of the mean value theorem. Let $a<b$ and let the real-valued function $f:[a,b]\to \mathbb{R}$ be continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$. If $f(a)=f(b)$, there exists $c\in (a,b)$ with $f^{\prime }(c)=0$. Geometrically, a tangent line somewhere in the interior is horizontal, like the secant line joining the endpoints. ^[ref]

For the polynomial $f(x)=x(x-2)$ on $[0,2]$, both endpoint values are zero and the derivative is $f^{\prime }(x)=2x-2$. Thus $c=1$, where $f(1)=-1$, and the horizontal tangent line is $y=-1$. The horizontal secant line is $y=0$: equal endpoint values do not force the interior tangent line to pass through those endpoints.

The absolute value function $f(x)=|x|$ on $[-1,1]$ exposes the importance of differentiability. It is continuous and $f(-1)=f(1)=1$, so the secant line has slope zero. However,and the derivative does not exist at zero because the one-sided derivatives disagree. Every existing derivative in the interior is nonzero. The corner fails a hypothesis of both Rolle's theorem and the mean value theorem, so neither conclusion is guaranteed.

For a short proof, the extreme value theorem supplies a greatest and a least value on $[a,b]$. If $f$ is a constant function, every interior point works. Otherwise at least one of these values differs from the common endpoint value, so it is attained at an interior extremum. At an interior local maximum or local minimum, differentiability forces the derivative to be zero: the difference quotients from the two sides have opposite weak signs and the same limit.

This turns repeated values into information about derivatives. Two distinct roots of a polynomial force a root of its derivative between them; iterating this idea gives the Rolle root count with multiplicities. More generally, a differentiable function with nonzero derivative throughout a real interval cannot repeat a value there, hence is injective. The beauty is that knowing only two equal values guarantees an exact local feature somewhere between them.

A saddle point of a differentiable real-valued function is a critical point which is neither a local maximum nor a local minimum. For example $f(x,y)=x^2-y^2$ has a saddle at the origin. For a nonconstant holomorphic function, critical points of its real part supply the complex-phase application used by the method of steepest descent.

Let $f$ have continuous second partial derivatives near an interior critical point $x_{\ast }$ of an open subset of ${\mathbb{R}}^n$, with $\nabla f(x_{\ast })=0$. The Hessian matrix gives the following sufficient conditions:

A positive semidefinite or negative semidefinite Hessian matrix with a zero eigenvalue does not settle the classification. The higher-order test for separated extrema gives two quartic examples with exactly the same zero Hessian. These conditions apply to an interior critical point; a boundary optimum or a constrained optimization problem requires attention to the allowed directions. ^[ref]

The proof outline is the Taylor expansionFor a positive-definite matrix, the smallest eigenvalue supplies a positive bound proportional to $\Vert h{\Vert }^2$, which dominates the remainder for small enough $h$. Negate the function for the local maximum case. For opposite-sign eigenvalues, choose the two corresponding eigenvectors: the leading terms have opposite signs along their lines.

There are also necessary conditions. At an interior local minimum, the Hessian matrix is positive semidefinite; at an interior local maximum, it is negative semidefinite. Restriction to each straight line reduces this to the one-variable second derivative condition. Since the Laplacian is the trace of the Hessian, it is nonnegative at such a local minimum and nonpositive at such a local maximum. The sign of the Laplacian alone is insufficient: $x^2-2y^2$ has negative Laplacian but a saddle point of a scalar function at the origin.

For a twice continuously differentiable real function, define the Newton map in optimization $N(p)=p-H(p{)}^{-1}\nabla f(p)$ wherever its Hessian matrix $H$ is invertible. Every stationary point in this domain is a fixed point, because its gradient is zero. This includes a local maximum and a saddle point of a scalar function, as well as a local minimum. Thus Newton's method in optimization needs a separate minimum classification or a method for ensuring descent when it is used for minimization problem.

For example, $f(x,y)=(x+y^2{)}^2+(x^2+y{)}^2$ has a stationary point at $(-1/2,-1/2)$, with Hessian matrixIts eigenvalues are $-1$ and $7$, so it is a saddle point of a scalar function, but starting Newton's method in optimization there gives a constant sequence. Its value is $1/8$, whereas the global minimum zero is attained at $(0,0)$ and $(-1,-1)$.

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.

A function has a strict local maximum at $t$ if some neighbourhood contains no other point $s$ with $f(s)\ge f(t)$. Unlike a non-strict local maximum, equality at another sufficiently nearby point is excluded. Use neighbourhoods relative to the domain, including when $t$ is an endpoint.