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Joined 2026-09-21 Message
Incoming links: Local minimum
Here and . This implementation uses the undamped Newton step, checks a stationary point's curvature, and reports failure on a two-cycle or exhausted iteration limit:
R
newton_min <- function(x0, tol = 1e-10, maxit = 100L) {
fp <- function(x) 4*x^3 - 12*x^2 + 4*x + 12
fpp <- function(x) 12*x^2 - 24*x + 4
x <- x0
trace <- x
for (k in seq_len(maxit)) {
g <- fp(x)
h <- fpp(x)
if (abs(g) <= tol) {
return(list(x = x, converged = h > 0,
reason = if (h > 0) "minimum" else "not a strict minimum",
trace = trace))
}
if (!is.finite(h) || abs(h) <= .Machine$double.eps)
stop("Newton step has zero or invalid curvature")
next_x <- x - g/h
if (!is.finite(next_x)) stop("Nonfinite Newton iterate")
trace <- c(trace, next_x)
n <- length(trace)
if (n >= 3L && abs(trace[n] - trace[n - 2L]) <= tol*(1 + abs(next_x)))
return(list(x = next_x, converged = FALSE,
reason = "two-cycle", trace = trace))
x <- next_x
}
list(x = x, converged = FALSE, reason = "iteration limit", trace = trace)
}A sufficiently close start near converges to the minimum. The derivative cubic has exactly one real root: after it is proportional to , whose local maximum and minimum are both positive, so it crosses zero only once, on the left. Since the quartic tends to in both directions and its curvature at that stationary point is positive, this is the unique global minimum. The function reports convergence rather than assuming that every starting point must reach it.
Let , with , . Taylor expansion givesDivision in the Newton update yieldsThus convergence is generically quadratic when that coefficient is nonzero. With a locally Lipschitz second derivative and nonzero , the weaker estimate also proves local quadratic convergence.
“Quadratic at best” is false as a universal upper bound. If the displayed coefficient vanishes, the order can be higher. For example has a minimum at zero, and its Newton optimization update is , which has cubic convergence. For a nondegenerate quadratic objective the method reaches the stationary point in one step. The expansion proves the standard generic quadratic statement while retaining and correcting the source's stronger wording explicitly.
For scalar minimization, apply Newton's root-finding method to the stationarity equation . Linearize at the current point to obtainThe iteration requires . Near a nondegenerate local minimum, where , , suitable smoothness and a sufficiently close starting value give local convergence. A stationary limit must still be checked to be a minimum. This undamped method is not globally convergent and need not decrease at every step.
At , the derivative and curvature are , , so the next iterate is . At , they are , , so the next iterate is . HenceThe undamped algorithm does not converge to a minimum; it has an exact two-cycle. The implementation in (iii) detects this after returning to one and reports failure. This example illustrates why the local convergence result in (ii) does not justify global success.
At , the two spin occupations per site are and . With , the trial energy per site isIts derivative is . Since is a boundary point, its local minimum test is one-sided: . At , where this derivative vanishes, its Taylor series begins for small positive . HenceThe same applies to by spin interchange.
The symmetry makes a stationary point for every . Here , so the second-derivative test settles all cases except equality. Writing , at the Taylor series is . ThusThe minimizing branches interpolate continuously. To find all of them, put on . An interior stationary point obeysThe right side decreases strictly from to , sinceThe positivity follows from . There is therefore exactly one such stationary point on each side of when . The signs of show that both are global minima, whereas is then a local maximum. The complete answer isThe pair merges through a supercritical pitchfork bifurcation at . Comparing only and would miss the intervening ground states with partial spin polarization.
Assume and temperature-independent for the local model. The zero-field stationary points obeyFor , the favoured ordered sign is opposite to . The nonzero extrema first appear at ; their existence is not yet a coexistence condition. To find the equilibrium transition, require both and . Substituting into the free energy givesHenceAt coexistence the entire potential factorizes:It has equal minima at zero and separated by a barrier. Every fixed nonzero cubic coefficient produces a first-order transition in this quartic model, with a finite magnetization jump before the quadratic coefficient reaches zero.
For , the ordered minimum is metastable and zero is globally stable. For , the ordered state is globally stable and zero is metastable. At the zero minimum loses local stability. Thus the spinodals are and , distinct from . For , the opposite-sign nonzero state can also be a local minimum, but the cubic term favours the state of sign .
If , the transition is shifted to . With constant and , the entropy jump from the disordered to ordered phase is , giving a latent heat of magnitude . As , the jump and shift vanish and the original continuous transition is recovered.
A genuine zero-field uniaxial ferromagnet with exact spin-reversal Z2 symmetry forbids this cubic term. The modified model therefore presumes that symmetry has been changed; it must not be presented as a symmetry-allowed correction for the original ferromagnet.
Choose a basis whose first axis is . The tensor with uniaxial nematic order then has eigenvalues . Hence , and . Substitution givesFor , remains a local minimum with curvature . Its nonzero stationary points satisfy , soThey first appear at as a stationary inflection point; for more negative , is a local maximum and a local minimum. The latter is initially metastable, becomes degenerate with at , and then becomes the global minimum as 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- scan.
The illustrative plots set and show . A cubic invariant is permitted because head-tail reversal leaves unchanged; it does not implement . The physical range of a uniaxial nematic order parameter imposes ; 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]
Lectures on Landau Theory of Phase Transitions by Peter D. Olmsted. §4.3, printed pp.11–13, tensor invariants and cubic-term-driven coexistence; coefficient normalization here is derived from this paper's definition of
Past exam of the theoretical physics I course of the University of Cambridge 2022 3 g Solution 2026-10-18
For a homogeneous saddle-point approximation define . With the cubic source coupling, minimizing the Landau free energy givesIts curvature is positive, so there is a unique local minimum and global minimum. The solution near hasbecause substituting this leading term makes the cubic self-interaction of order . Therefore the cubic source coupling and third-order susceptibility giveThus the first nonzero response diverges like from above. The factor is essential: the coefficient of is not itself the third derivative. This is susceptibility to the specified cubic source variable, rather than the usual third-order response to a linearly coupled field.
Fix . Stationary points satisfy . With , the nonzero candidates areKeep only real positive roots. At such a point,so gives local minima and gives local maxima when both exist. The solution is a local minimum for . To choose the phase, compare free energies: using the stationary point relation givesFor , this expression is negative whenever a nonzero stationary minimum exists. The paramagnetism phase is stable for , and the ferromagnetism phase with is stable for . The phase transition at , , is continuous and second order.
For , equality with requires . Substitution into the stationary equation gives the first-order coexistence curveThe disordered phase is the global minimum above this curve, and the ordered phase is the global minimum below it. At coexistence the factorizationexhibits three equally low minima, and , proving the discontinuous jump.
The ordered spinodal is , , where its local minimum meets the intervening local maximum. The disordered spinodal is , . Thus the ordered phase is metastable for ; the disordered phase is metastable for . These spinodals are limits of metastability, not equilibrium transition curves.
The first-order curve meets the continuous transition line at the tricritical point . On the specially tuned path , the equilibrium ordered solution is for and tends continuously to zero at the origin.
Phase diagram of the sextic Landau free energy, including coexistence, spinodals and the tricritical point
. Details. Past exam of the thermal and statistical physics course of the University of Cambridge 2015 Paper 1 4 Solution Created 2026-10-20 Updated 2026-10-20
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.
Local behavior at the origin. To determine constrained order-parameter boundary stability, note that and , the origin is a local minimum for and a local maximum for . At , the quadratic term vanishes, but for small positive . ThereforeThe equality case uses the stipulated domain . On an unrestricted real line, the nonzero cubic term would instead make it a stationary inflection point.
Global stability in both inline cases. For , complete the square in :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, , has a nonzero globally stable phase with for every physical temperature. Case b, , has such a phase exactly forAt , the square's centre is at the boundary, but its negative constant still gives for small positive , so this endpoint is included.
At ,The two global minima have equal free energy, and . The coexistence barrier of a cubic-quartic Landau transition has its maximum at , with barrier height .
For locating the extrema away from coexistence,Retain only real positive roots; when both are positive, is a local maximum and a local minimum. For , only is positive. If , the pair first appears on cooling at ; if , 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 also favors a nonzero minimum at every finite temperature, but has no finite .
Cooling and transition temperature. For , equilibrium starts at the zero minimum at high temperature and switches to the nonzero minimum at . The jump is finite, , 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 is an energy density rather than total energy, is likewise a density. An actual system can supercool below before nucleation, so is the equilibrium transition temperature.
For thermal symmetry nonrestoration in a cubic-quartic model, with , 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 ; that minimum loses local stability at . This is the sense in which a change is possible rather than inevitable. There is no equilibrium coexistence temperature for in this model. A kinetic transition can depend on preparation and nucleation dynamics; is the limiting loss of local stability of the zero state, not a second coexistence temperature. Continuing 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 . The supplied ratios implyThe two strict conditions therefore requireSince is increasing for , the finite window isUsing the specified and givesAt u, , but . 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 . The result is a bound within the specified simplified model, not a model-independent Higgs-mass bound.
Write . Its three positive zeros of a function are , and expansion and differentiation giveThus its local minimum and maximum occur respectively atwith valuesThe function graph approaches the value as , decreases through its first zero to the minimum, increases through its second zero to the maximum, and then decreases through to . This fixes the shape and all axis crossings of the cubic polynomial.
Changing translates the function graph vertically without moving either stationary point. For both extrema are above the horizontal axis: their heights are approximately and . For both are below it, at approximately and . The function graph for therefore has only its final positive-axis crossing, whereas the function graph for has only its first positive-axis crossing. The original plots mark the extrema and also show the equilibrium point branches used below.
Cubic graphs and equilibrium branches
. Details. Solving gives . Substitution into gives . Therefore is always a critical point. When as well as the printed restriction , there is a second critical pointAt , the Hessian matrix is with determinant . The second-derivative test gives a strict local minimum when and a saddle point of a scalar function when .
The allowed value requires higher-order terms. With , completing the square gives the exact identityThe coordinate change is invertible, so the Hessian at a critical point under a change of coordinates preserves the classification. At , this is a cubic degenerate saddle: along , the difference from the critical value is and takes both signs. Thus, throughout the printed parameter range,There is no local maximum. When the second point is distinct, it is a local minimum for and a saddle point of a scalar function for : in the coordinates its Hessian matrix is . The printed instruction to split only at misses the negative-parameter cases.
Near zero, has the sign of , so just to the left and just to the right. Thus the critical point is neither a local minimum nor a local maximum. Moreover the Taylor series gives , which changes sign at zero. ThereforeThis is the point called a “saddle point” in the question; stationary inflection point specifies the one-variable meaning.
The critical points satisfy both first partial derivatives equal to zero:Substitute into the second equation to get . Thus the candidates are and .
The Hessian matrix and its determinant areAt the determinant is , so the Hessian matrix is indefinite and the point is a saddle point. At the determinant is and , so the Hessian matrix is positive definite and the point is a strict local minimum, with valueThere is no local maximum. Along , is unbounded above and below, so there are no global extrema either.
Separate inactive and active optimization constraints. At an interior feasible point with , a differentiable local minimum must satisfy . Retain only solutions lying in the interior.
On the boundary , if , apply the Lagrange multiplier condition. Using the optimization Lagrangian , the Karush-Kuhn-Tucker conditions areThe sign follows from feasible motion into : the directional derivative along at a minimum cannot be negative. Thus a boundary candidate has and a nonnegative Lagrange multiplier. Boundary points with must be examined separately, because the usual constraint qualification fails there.
Solve the interior and boundary systems, check membership in the feasible set, and establish which candidates are minima using second-order tests or a direct global comparison. The Karush-Kuhn-Tucker conditions alone are not sufficient for general nonconvex functions or feasible sets. On an unbounded feasible set, also check whether an infimum is attained and whether escaping sequences can improve the objective.
The stationary point equations give and , hence . At the Hessian matrix has negative determinant , so is a saddle point. At each of and , the Hessian matrix has positive leading minors and , proving a strict local minimum. More strongly,Thus and are exactly the global minima, with value . There are no local maxima, since any differentiable interior local maximum would be a stationary point, and all three have been classified.
The eigenvector identity gives . Since 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 . Thus is the unique global minimum, with value .
At , 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 , these values have opposite signs. Thus 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.
Substitution givesIn particular, is a stationary point. For sufficiently close to , Taylor's theorem with fifth-order remainder giveswhere lies between and . The fifth derivative is bounded near , so the remainder is and is smaller in magnitude than half the positive quartic term for sufficiently small nonzero . Therefore is a strict local minimum, despite the vanishing second derivative.
An exact algebraic check gives the stronger global conclusion:Equality holds only at , so it is the unique global minimum on the actual domain.