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For $b>0$, the origin is a strict local minimum. The four axis points at distance one have Hessian eigenvalues $-4$ and $2(1-b)$. The four diagonal points with $x^2=y^2=1/(1+b)$ have Hessian eigenvalues $-4$ and $-4(1-b)/(1+b)$. Thus for $b<1$ the diagonal points are the global maxima and the axis points are saddle points of a scalar function; for $b>1$ these roles reverse. In polar coordinates, $f=r^2-\frac{1}{2}q(\theta )r^4$, where $q=1+\frac{1}{2}(b-1){\sin }^2(2\theta )>0$. Maximizing over $r$ gives $1/(2q)$ and proves the global statements. At $b=1$, $f=\frac{1}{2}-\frac{1}{2}(r^2-1{)}^2$, so the entire unit circle is the maximum set.

Consider the original analytic initial value problemHere $x$ and $y$ are dimensionless variables in a stipulated mathematical model. The trajectories below are exact solutions of a differential equation, with no measured data or empirical parameter fit. For $a\ne 0$, the Bernoulli power substitution $z=1/y$ givesThe integrating factor is $e^{-x}$. Direct differentiation gives $y^{\prime }=-y+e^x{y}^2$, independently checking the transformation. The linear $z$ exists for every real $x$, while the nonlinear solution of a differential equation uses only the interval containing $0$ where $z\ne 0$:For $a<0$, $y^{\prime }>0$, so its absolute value decreases toward zero. The lost zero solution of a differential equation is recovered by substitution into the original equation. Since $f(x,y)=-y+e^x{y}^2$ is locally Lipschitz in $y$ everywhere, the Picard-Lindelöf theorem ensures uniqueness through every finite point, including $y=0$. Thus a nonzero trajectory never meets or crosses zero at a finite $x$.

For $a=1/2$, the exact solution of a differential equation is $y=e^{-x}/(2-x)$ on $(-\infty ,2)$. It first decreases, reaches a local minimum $y(1)=e^{-1}$, then increases without bound. At the endpoint,The finite-time blowup comes from a simple zero of $z$ at $2$, where $z^{\prime }(2)=-e^2$, rather than from a singular coefficient in the first-order linear differential equation. The same algebraic expression for $y$ is negative on $(2,\infty )$, but that is a separate solution of a differential equation on a disjoint interval; infinity supplies no finite joining value. For $a=-1$, $y=-e^{-x}/(1+x)$ on $(-1,\infty )$, giving a distinct decaying branch whose finite blowup is instead in the past.

The direction field also explains the positive turning point. Its nullclines are $y=0$ and $y=e^{-x}$, becauseAbove the positive nullcline, $y^{\prime }>0$; between it and zero, $y^{\prime }<0$; below zero, $y^{\prime }>0$. The moving nullcline $y=e^{-x}$ is not a solution of a differential equation: its derivative is $-e^{-x}$, whereas the prescribed slope there is zero. The $a=1/2$ trajectory crosses it at $x=1$, exactly its local minimum.

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 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.

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 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 local minimum is strict when every other point of the domain of a function in some neighbourhood has a larger value. Strictness is local: two separated strict local minima can have the same value, as $x^4-2x^2$ at $\pm 1$ demonstrates. Negating the function gives a strict local maximum.

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.

A wave trough is a local minimum of the wave's displacement.