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Assume $f$ and $g$ are continuously differentiable near the point. At a regular feasible point, meaning $g(x)=0$ and $\nabla g(x)\ne 0$, a differentiable constrained local extremum satisfies the Lagrange multiplier equationsThey arise by differentiating the optimization Lagrangian $L(x,\lambda )=f(x)-\lambda g(x)$ with respect to all coordinates and the multiplier. Geometrically, the objective has zero derivative along the tangent space of the constraint surface, so its gradient must lie in the normal direction.

Solve these equations for candidates, examine feasible points where the regularity condition fails separately, and compare objective values or use a second-order test restricted to feasible directions. The equations locate stationary points; they do not by themselves establish which are minima or whether a global minimum is attained.

Here the constraint surface $x^4+2y^2=3$ is compact and regular, so a global minimum exists and satisfies the Lagrange multiplier equationsKeep the zero-coordinate cases. If $y=0$, the two points are $(\pm 3^{1/4},0)$, with value $3\sqrt{3}$. If $x=0$, the two points are $(0,\pm \sqrt{3/2})$, with value $27/8$. If both coordinates are nonzero, the equations give $x^2=y^2=t>0$. The constraint becomes $t^2+2t=3$, whose only positive solution is $t=1$. These are the four points $(\pm 1,\pm 1)$, all with value $5/2$.

These cases exhaust the stated eight constrained stationary points. Comparing their values givesFor additional classification, substituting $u=x^2\in [0,\sqrt{3}]$ gives $F(u)=u^3+\frac{3}{8}(3-u^2{)}^2$, withIt decreases from $u=0$ to $u=1$ and increases from $u=1$ to $u=\sqrt{3}$. Thus the two $y$-axis points are constrained local maxima, the two $x$-axis points are global maxima, and the four mixed-coordinate points are global minima.

Both terms $x^6$ and $\frac{3}{2}{y}^4$ are nonnegative, and they vanish simultaneously only at $(0,0)$. That point satisfies $x^4+2y^2\le 3$. Consequentlyand the global minimum is unique. No Lagrange multiplier calculation is needed: nonnegativity gives a lower bound attained at a feasible point.

The eigenvector identity $Ab=\frac{1}{2}b$ gives $\nabla f(-b)=-2Ab+b=0$. Since $H_f=2A$ is a positive-definite matrix, this is a strict local minimum. A global certificate follows by completing the quadratic form:Positive definiteness makes the first term nonnegative, with equality only at $x=-b$. Thus $-b$ is the unique global minimum, with value $-1/2$.

Substitution givesIn particular, $1$ is a stationary point. For $x$ sufficiently close to $1$, Taylor's theorem with fifth-order remainder giveswhere $\xi$ lies between $1$ and $x$. The fifth derivative is bounded near $1$, so the remainder is $O((x-1{)}^5)$ and is smaller in magnitude than half the positive quartic term for sufficiently small nonzero $x-1$. Therefore $x=1$ is a strict local minimum, despite the vanishing second derivative.

An exact algebraic check gives the stronger global conclusion:Equality holds only at $x=1$, so it is the unique global minimum on the actual domain.

In any finite-dimensional Euclidean space, this extension is a smooth function and all its partial derivatives at the origin vanish. Away from zero, repeated differentiation gives finite sums of coordinate polynomials divided by powers of $|x|$, multiplied by $e^{-1/|x{|}^2}$. Every such term is bounded in absolute value by $C|x{|}^{-N}{e}^{-1/|x{|}^2}$ near zero. Since Gaussian decay dominates every power, this bound tends to zero faster than any prescribed positive power of $|x|$. Induction using difference quotients then proves smoothness and zero derivatives of every order at zero.

The origin is nevertheless a strict global minimum, because $F(x)>0$ for $x\ne 0$. All its Taylor series coefficients vanish, so the function is not real analytic at zero. Adding a constant preserves the stationary minimum but changes the zeroth-order value. An expression written only for $x\ne 0$ has no stationary point at the excluded origin until an extension is specified.

If a free real scalar field's three-momentum ultraviolet cutoff is identified with the absolute vacuum expectation value of another scalar field, the displayed scalar potential is even and has its unique global minimum at zero. It grows as ${\phi }^4/(16{\pi }^2)$ for $|\phi |\gg m$, but as $m|\phi {|}^3/(12{\pi }^2)$ for $|\phi |\ll m$ with $m>0$. The positive derivative with respect to $|\phi |$ proves that this contribution alone does not produce spontaneous symmetry breaking. This cutoff identification is a model assumption, distinct from deriving a complete effective potential for a dynamical field.

For the Landau free energy $G(\varphi )={\varphi }^6/6+g{\varphi }^4/4+\epsilon {\varphi }^2/2$, positive $g$ gives a continuous phase transition at $\epsilon =0$. For negative $g$, phase coexistence between the disordered global minimum $\varphi =0$ and the ordered global minima occurs on $\epsilon =3g^2/16$, with ${\varphi }^2=-3g/4$. Indeed, on that locus the polynomial is $G(\varphi )={\varphi }^2({\varphi }^2+3g/4{)}^2/6\ge 0$, so these three zeros are global minima. For $g<0$, the disordered and ordered spinodal points are respectively $\epsilon =0$ and $\epsilon =g^2/4$; equality of stationary free energies determines phase coexistence, while loss of a local minimum determines a spinodal point.

Simulated annealing accepts worse feasible solutions to escape a local minimum. Minimize an objective function $E(x)$, analogous to energy. Effective temperature $T>0$ has the same units as $E$, absorbing the Boltzmann constant; it controls exploration, rather than a material's thermodynamic temperature.^[ref]

Given current feasible point $x$, draw $y$ from a proposal distribution $q(x,y)$. For a symmetric proposal distribution, $q(x,y)=q(y,x)$, use the Metropolis–Hastings acceptance probabilityAccept with this conditional probability; on rejection retain $x$. Every downhill or equal-cost proposal is accepted. Draw a fresh uniform random variable $U$ on $(0,1)$ and accept if $U<a_T(x,y)$. The off-diagonal transition probability $P_T(x,y)=q(x,y)a_T(x,y)$ includes proposing $y$.^[ref]

For a maximization problem with objective function $f$, set $E=-f$: every improving proposed neighbor is accepted, and a worsening move of loss $\Delta =f(x)-f(y)>0$ has conditional probability $\exp (-\Delta /T)$ of acceptance, under the same symmetry assumption. Likelihood-power annealing uses precisely this sign convention for a log-likelihood. Retain the best feasible solution seen, since the current state can worsen even after a global minimum has been visited.

If the proposal distribution is asymmetric, use the full Metropolis–Hastings acceptance probability, for $q(x,y)>0$,A missing reverse proposal gives zero acceptance. Even a downhill proposal need not be accepted with probability one in this case. Choosing uniformly among a variable number of neighbors is generally asymmetric: $q(x,y)=1/d(x)$ and $q(y,x)=1/d(y)$ give the correction $d(x)/d(y)$, where $d$ is vertex degree.^[ref]

The useful idea is to separate exploring from insisting on improvement: an uphill move can open a route to a better feasible solution. Its usefulness depends on the proposal distribution, the scale of the objective function, and the cooling schedule for simulated annealing. Finite practical schedules carry no general guarantee of finding a global minimum.

For an ideal soap film with equal constant surface tension $\gamma$ on both faces, the effective tension is $\tau =2\gamma$. Choose a unit normal $N$ from the minus side to the plus side, and $\Delta p=p_{-}-p_{+}$. Translating the Young–Laplace equation to $S=-dN$ givesA single interface uses $\tau =\gamma$. An outward-oriented spherical soap bubble has $H=-1/R$, giving $\Delta p=4\gamma /R$. The positive-convex curvature sum in the Young–Laplace equation article is $-2H$. ^[ref]

An open film with the same resting gas on both sides has zero pressure jump. Neglecting film weight, soap film pressure balance gives $H=0$: a minimal surface. A pressurized bubble instead has nonzero mean curvature. These statements assume equilibrium, constant surface tension, and negligible normal viscous stress. Surfactant gradients or motion require the fuller normal stress balance. ^[ref]

The variational outline is short. With fixed boundary, a normal displacement $fN$ gives $\delta A=-2{\int }_{M}HfdA$ by the first variation of area formula. For a closed interface oriented outward, $\delta V={\int }_{M}fdA$, so stationarity of $\tau A-\Delta pV$ gives $-2\tau H-\Delta p=0$. This is a first variation calculation; testing for a stable minimal surface needs the second variation, and deciding a global minimum needs a comparison of competitors.

A stable minimal surface has nonnegative second variation of surface area for every admissible normal variation $fN$ fixing the boundary. For a two-sided minimal surface in three-dimensional Euclidean space, the quadratic form isHere ${\nabla }_{M}f$ is the gradient for the induced Riemannian metric. The gradient term penalizes rapid changes in the displacement; the principal curvature term can make the second variation negative. A plane has only the nonnegative gradient term. Nonnegativity alone, especially with zero modes, does not prove a strict local minimum or a global minimum.

For a finite catenoid between fixed rings, the answer depends on the ring separation, waist radius, and allowed variations. The axisymmetric second variation of a catenoid tests a specified restricted class. Comparing with two disconnected disks tests a different class of competitors, including a change in topology. Consequently, a stationary catenoid can have greater surface area than two disks without immediately collapsing under every small perturbation.

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 $f$ is bounded below, $0\prec B_k⪯MI$ for one fixed $M$, $B_k{s}_k=-\nabla f(x_k)$, and $x_{k+1}=x_k+{\alpha }_k{s}_k$ satisfies the Armijo condition with ${\alpha }_k\ge C>0$. ThenA telescoping sum proves the displayed convergence of a nonnegative series and hence $\nabla f(x_k)\to 0$. A uniform positive lower bound on the eigenvalues of $B_k$ is unnecessary for this conclusion. The conclusion is first-order stationarity; without further assumptions it does not assert convergence of the iterates or attainment of a global minimum.

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 $(-1,-1)$.

Stochastic gradient descent updates parameters using an unbiased or approximate gradient computed from one observation or a mini-batch rather than the complete data set.

For a three-dimensional zero-temperature parabolic Fermi gas, let $\zeta =(N_{\uparrow }-N_{\downarrow })/N$ and $E_F={\hslash }^2(3{\pi }^{2}N/V{)}^{2/3}/(2m)$. Suppose the interaction energy per electron is $U{N}_{\uparrow }{N}_{\downarrow }/N^2=U(1-{\zeta }^2)/4$. The total energy per electron isIts Taylor expansion has quadratic coefficient $E_F/3-U/4$ and positive quartic coefficient $E_F/81$. Hence the unpolarized state loses stability at $U_c=4E_F/3\propto (N/V{)}^{2/3}$. This $U$ has units of energy. It differs from a contact coupling $g$ defined by interaction-energy density $g{n}_{\uparrow }{n}_{\downarrow }$, for which $g=U/n$ and $g_c=4E_F/(3n)$. Normalization must be fixed before comparing critical couplings. For $U\le 4E_F/3$, convexity of $x^{-1/3}$ implies $f^{\prime \prime }(\zeta )=(E_F/3)[(1+\zeta {)}^{-1/3}+(1-\zeta {)}^{-1/3}]-U/2\ge 0$, so the unpolarized minimum is a global minimum; a first-order polarized minimum does not pre-empt this threshold in the stated model.

For $\varphi \ge 0$ and positive $c,B,\lambda ,T_0$, define $\gamma =2B^2/(\lambda c)$. Completing the square givesFor $0<\gamma <1$, phase coexistence occurs at $T^{\ast }=T_0/\sqrt{1-\gamma }$ with a discontinuous minimum at ${\varphi }^{\ast }=2B{T}^{\ast }/\lambda$. For $\gamma \ge 1$, a nonzero global minimum has lower free energy at every finite temperature. Following the global equilibrium branch then does not force a restoration transition.

A free energy is a thermodynamic potential that compares possible states under specified environmental constraints. For a closed system of fixed composition, treated as a simple compressible system with only pressure-volume work, stable thermodynamic equilibrium minimizes Helmholtz free energy at fixed temperature and volume, or Gibbs free energy at fixed temperature and pressure. The choice follows the experiment: a rigid sealed vessel and a sealed cylinder with a freely moving loaded piston impose different constraints. Both may exchange heat with a thermal reservoir. See W. C. Carter, MIT thermodynamics notes, sections 2-1.14 and 2-1.19.

The useful insight is that internal energy alone cannot rank these alternatives: exchanging heat also changes the surroundings' entropy. Subtracting $TS$ packages that accounting into a property of the chosen system. Adding $pV$ additionally accounts for displacing the surroundings at fixed pressure. A free energy is a state function; supplied heat describes a transfer during a process. Under the stated constraints, supplied heat is $Q=\Delta U$ at fixed volume or $Q=\Delta H$ at fixed pressure, whereas neither $\Delta F$ nor $\Delta G$ generally equals $Q$.

A lower free energy identifies a thermodynamic preference, without specifying a timescale. A barrier can trap a system in a local minimum, producing metastability although a lower global minimum exists. The enzyme example illustrates the corresponding distinction for chemical reactions: changing the route can change the reaction rate while preserving the endpoint Gibbs free energy difference and chemical equilibrium.

The trust-region subproblem minimizes a quadratic function on a closed Euclidean ball. Even if $H$ has a negative eigenvalue, the compactness of this ball guarantees a global minimum. An exact answer is often unnecessary: a fixed fraction of the Cauchy decrease bound supplies a useful inexpensive standard.

For $F=F_{\mathrm{reg}}+a{m}^2/2+c{m}^6/6$, $c>0$, $a=a_0(T-T_c)$, the ordered global minimum gives $F-F_{\mathrm{reg}}=-(-a{)}^{3/2}/(3\sqrt{c})$ and $C_{\mathrm{sing},-}=T{a}_0^2/(4\sqrt{c}\sqrt{-a})$. On the disordered side the minimum is exactly $m=0$, so the polynomial's heat capacity is zero. The tricritical scaling label $\alpha =1/2$ can be used with zero upper-side amplitude, but no nonzero upper-side power law follows from this saddle alone. A smooth background such as $F_{\mathrm{reg}}=-C_{0}T\log (T/T_{\ast })$ gives a finite upper-side heat capacity $C_0$ without changing the phase diagram.