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For the three-colourability problem, represent a candidate by a complete colour assignment $c:V\to \{r,g,b\}$, initially chosen at random. Give it the conflict objective functionA neighbor changes the colour of one vertex to either other colour. For hill climbing, maximize $-H$, or equivalently repeatedly choose a recolouring that reduces $H$. One can inspect all $2|V|$ neighbors, or focus on conflicted vertices and evaluate their alternative colours using only incident edges. Stop successfully as soon as $H=0$, which is exactly a proper graph colouring with three colours.

If no improving move exists at positive $H$, the method is stuck in a local minimum. Random restarts, limited sideways moves or other exploratory local search choices can reduce this failure mode, but do not turn a finite unsuccessful run into a proof of impossibility. On a graph with no proper graph colouring using three colours, every complete colouring has $H>0$, so this procedure cannot return a valid colouring. Conversely, a positive-conflict stopping point also occurs on some graphs admitting a proper graph colouring with three colours. Failure of hill climbing alone does not certify non-colourability. For a concrete trap, use a bipartite graph with three vertices on each side and every cross-edge, and give each side one vertex of each colour. There are three conflicts. Recolouring any single vertex leaves it with exactly one same-colour neighbor, so no single change improves $H$, even though colouring each whole side differently solves the instance.

Use Model 3 and keep every prescribed row at its prescribed column. Initialize one queen in every other row, and minimize the conflict countThis sum counts attacking pairs; $E=0$ is exactly a valid completion. A neighbor changes the column of one unfixed row. Evaluate such neighbors and move to one with lower $E$, breaking equal choices arbitrarily. Minimizing $E$, or maximizing $-E$, is hill climbing on complete domain assignments; these assignments need not yet satisfy the inter-row constraints. Fixed rows are never moved.

The min-conflicts algorithm chooses a conflicted unfixed queen and places it in a column giving the smallest conflict count against the other queens. Recomputing only the affected pair contributions makes a move cheaper than reevaluating every pair. Equal-cost moves can help cross plateaus; bounded sideways moves or repeated random starts of only the unfixed queens can escape some traps. These modifications still preserve the prescribed positions.

Return a board only after checking $E=0$ and every prescribed position. A local minimum with positive $E$ is not proof of impossibility. A queens completion problem may be unsatisfiable even when its prescribed queens do not attack each other; unrestricted restart policies give no finite worst-case success bound. If every row is fixed, simply validate the one board. For a guaranteed decision procedure use finite backtracking, possibly with constraint propagation, instead of relying on hill climbing.

Taking the natural logarithm of the positive product constraint gives $h(x)={\sum }_i{a}_i\log x_i=0$. Its gradient has entries $a_i/x_i>0$, so the equality constraint qualification holds everywhere on the feasible set. Use the optimization LagrangianStationarity requires $a_i-\lambda a_i/x_i=0$, so every $x_i=\lambda$. Substitution into the constraint, using ${\sum }_i{a}_i=1$, gives $\lambda =1$. Thus the only candidate for a local minimum is $x=(1,\dots ,1)$.

To prove global minimality without assuming the inequality to be established, set $u_i=\log x_i$. The constraint becomes the affine hyperplane ${\sum }_i{a}_i{u}_i=0$, and the objective function becomes $F(u)={\sum }_i{a}_i{e}^{u_i}$. Its Hessian matrix $\mathrm{diag}(a_i{e}^{u_i})$ is positive-definite, so $F$ is strictly convex. The elementary tangent inequality $e^t\ge 1+t$, with equality only at zero, givesEquality forces every $u_i=0$. ThereforeThis point is the unique global minimizer and the only local minimizer. The logarithm change of variables also explains why a multiplicative constraint leads to a convex optimization argument.

Where $y\ne \beta$, taking the quotient of the two ordinary differential equations givesChoose the first integralThe chain rule checks conservation even at points excluded in taking that quotient:Thus each trajectory is contained in a connected component of a level set of $h$. At $(0,\beta )$ the gradient vanishes and the Hessian matrix is the identity, so $h$ has a strict local minimum there. To make the closed-trajectory argument explicit, put $z=y-\beta$ and $0<E<1/4$. The bounded component of $h=E$ isThe two arcs join at $(\pm r_E,0)$ to form a simple closed curve that is a regular level set. The gradient cannot vanish on it: the only critical point of a dynamical system in $|x|<1$ is the centre, whose level is $0$.

The vector field is tangent and nowhere zero on this compact set, so its speed has a positive minimum. It therefore traverses the curve in finite time and returns to its initial point, giving a periodic orbit. These ovals shrink to the centre as $E\downarrow 0$. This is the planar first integral criterion for a centre and proves the required nonlinear assertion; it also gives Lyapunov stability without asymptotic stability.

Linearization gives ${\theta }^{\prime \prime }+{\pi }^2\lambda \theta =0$. The two homogeneous Dirichlet conditions select $\boxed{\lambda =n^2,\theta =A\sin (n\pi \xi ),n=1,2,\dots }$. For any small trial amplitude in this mode, ${\int }_0^1{\sin }^2(n\pi \xi )={\int }_0^1{\cos }^2(n\pi \xi )=1/2$, soFor $\lambda <1$ every nonzero quadratic variation is positive by the sharp Poincaré inequality. For $\lambda >1$, the first mode gives negative energy at sufficiently small amplitude, so zero cannot be a local minimum.

The printed strict “only if” conclusion omits the critical case. At $\lambda =1$ the quadratic test is degenerate, not negative. In fact zero is still the unique minimum of the full displayed functional: use $1-\cos \theta \le {\theta }^2/2$ and $\int {\theta }^{\prime 2}\ge {\pi }^2\int {\theta }^2$ to get $U\ge 0$, with strict inequality for a nonzero continuous function because $1-\cos t<t^2/2$ when $t\ne 0$. In particular on the critical first mode,Thus strict positivity of the second variation holds for $\lambda <1$, while local minimality of the full energy holds also at $\lambda =1$. This is the distinction captured by critical-load minimality of the angular Euler strut.

The first instruction asks for maximization, whereas the later parts refer to minimization. For the literal maximization task, minimize the negative objective:

This gives a local maximum near $x=-0.927662$, $f(x)=1.839576$. If minimization was intended, use opt.minimize(lambda z: f(z[0]), x0=[-1.0]), which instead returns a local minimum near $x=-1.546182$, $f(x)=0.384751$. The optimizer's stopping criterion establishes local numerical convergence, not a certificate of a global optimum.

The product rule and chain rule giveThe exponential function is strictly positive, so there are exactly two stationary points, at $\boxed{x=\pm 1/\sqrt{2}}$. The derivative is negative outside these two values and positive between them. Thus the negative value is a strict local minimum and the positive value a strict local maximum.

At a critical point $x_0$ of a twice continuously differentiable real function, the Taylor expansion isThe Hessian matrix is real symmetric, so an orthogonal eigenbasis diagonalizes its quadratic form. If every eigenvalue is positive, that form is bounded below by a positive multiple of $|h{|}^2$, which dominates the remainder and gives a strict local minimum. If every eigenvalue is negative, the same argument gives a strict local maximum. If both signs occur, displacement along the corresponding eigenvectors produces values above and below $f(x_0)$, so the point is a saddle point of a scalar function. These exhaust a nonsingular Hessian matrix.

A singular Hessian matrix does not always make the classification impossible: nonzero eigenvalues of both signs still give a saddle. If the Hessian matrix is positive or negative semidefinite, higher-order terms in its null directions must be examined. For example, $x^2+y^4$ and $x^2-y^4$ have the same Hessian matrix $\mathrm{diag}(2,0)$ at zero, but the first has a strict minimum and the second a saddle. The function $x^2$ has a non-strict minimum along a whole line.

For the given polynomial,A stationary point has $y=3x^2/2$ and then $x(2\alpha -x^2)=0$. Hence the origin is always critical, and two additional points exist exactly for $\alpha >0$:The Hessian matrix isAt the origin it is $\mathrm{diag}(2\alpha ,2)$, giving a strict local minimum for $\alpha >0$ and a saddle for $\alpha <0$. At either extra critical point it has determinant $64\alpha -36(2\alpha )=-8\alpha <0$, so both extra points are saddles.

At $\alpha =0$, complete the square:Along $x=0$ it is positive away from zero, whereas along $y=3x^2/2$ it is negative away from zero. Thus the origin remains a degenerate saddle despite its nonnegative Hessian matrix. This is the parabolic quartic critical-point classification. The complete result isFor $\alpha >0$ the local minimum is not global, since along the same parabola $f=\alpha x^2-x^4/4\to -\infty$.

At a critical point $p$ of a twice continuously differentiable real function, $\nabla f(p)=0$ and the Taylor expansion isThe real symmetric Hessian matrix has an orthonormal eigenbasis, so writing $h={\sum }_i{a}_i{u}_i$ gives $h^{T}Hh={\sum }_i{\lambda }_i{a}_i^2$. If every eigenvalue is positive, this quadratic form is at least ${\lambda }_{\min }|h{|}^2$, and the remainder cannot change its sign for sufficiently small nonzero $h$. Thus $p$ is a strict local minimum. If every eigenvalue is negative, the analogous argument gives a strict local maximum. If positive and negative eigenvalues both occur, displacement along the corresponding eigenvectors gives values above and below $f(p)$, so $p$ is a saddle point of a scalar function.

The second-derivative test can be inconclusive when the Hessian matrix is singular and semidefinite: higher-order terms along directions in the kernel of a linear map matter. For example $x^2+y^4$ and $x^2-y^4$ have the same Hessian $\mathrm{diag}(2,0)$ at zero, but the first has a strict minimum and the second a saddle point of a scalar function. A zero eigenvalue does not prevent a saddle conclusion if other eigenvalues already have both signs.

For the given function write $r^2=x^2+y^2$. Its first derivatives areThe origin is a critical point for every real $\alpha$, withIts eigenvalues are $1,-1$, so it is always a saddle point of a scalar function. If one coordinate at a critical point is zero, the other must also be zero. If both are nonzero, the derivative equations force $x^2=y^2=1/(2\alpha )$, possible only for $\alpha >0$. Hence for $\alpha \le 0$ the origin is the only critical point.

For $\alpha >0$ there are four additional critical points,At such a point,Thus the two equal-sign points are strict maxima, and the two opposite-sign points are strict minima. They are global: $2|xy|\le r^2$ implieswhere the last maximum occurs at $r^2=1/\alpha$ and equality also requires $x^2=y^2$.

As $\alpha \downarrow 0$, the four nonzero critical points have distance ${\alpha }^{-1/2}$ from the origin and escape to infinity, rather than merging with the origin. Their values diverge as $\pm 1/(2e\alpha )$. As $\alpha \uparrow 0$ from below there are no extra critical points, while the origin stays a saddle throughout. This is critical points escaping to infinity in a Gaussian-weighted bilinear function.

For this quartic potential with fourfold symmetry, the first derivatives factor:The critical points are the origin, the four axis points $(\pm 1,0),(0,\pm 1)$, and four points with both coordinates nonzero. At the latter, $x^2+b{y}^2=y^2+b{x}^2=1$. Subtracting and using $b\ne 1$ gives $x^2=y^2=1/(1+b)$. Thus the complete list iswhere the two signs in the final group are chosen separately.

The Hessian isAt the origin it is $2I$, so there is a strict local minimum with value zero. At an axis point its eigenvalues are $-4$ and $2(1-b)$, giving a saddle point of a scalar function if $b<1$ and a strict local maximum if $b>1$. All four axis values are $1/2$.

At a diagonal point the Hessian has eigenvaluesHence these four points are strict local maxima for $b<1$ and saddle points of a scalar function for $b>1$. Their value is $1/(1+b)$. None of these critical points is degenerate when $b\ne 1$.

To describe the level curves precisely, use polar coordinates and writeFor a level $f=c$, the possible radii satisfyretaining only real, nonnegative values. This formula, together with reflection in each axis and interchange of the coordinates, determines the level curves. The radial maximum in a direction is $1/(2q)$, attained at $r^2=1/q$. It also proves that the local maxima identified above are global: the largest directional value is $1/(1+b)$ on the diagonals if $b<1$, and $1/2$ on the axes if $b>1$.

For negative $c$ only the plus sign gives a positive radius, producing one outer closed level curve. At $c=0$ there is the isolated origin and the outer curve $r^2=2/q$. For $0<c<c_s$, where $c_s$ is the saddle value, both radial branches exist in every direction, giving an inner loop around the origin and an outer loop. At $c=c_s$ they meet at the four saddle points of a scalar function. Above the saddle value, four separate loops surround the four maxima; they shrink to those points at the maximum value. There are no level curves above it.

For $0<b<1$, the saddle value is $c_s=1/2$ and the four high-level loops lie along the diagonals. For $b>1$, $c_s=1/(1+b)$ and the four high-level loops lie along the coordinate axes. The sketches show representative values in both regimes; the highlighted level curve is the saddle level.

Finally, at $b=1$ the expression depends only on radius:Consequently $\boxed{\max f=1/2\text{, attained exactly on }{x}^2+y^2=1.}$ The maximum set is the whole unit circle, not merely four isolated points.

For a single chiral superfield, the supergravity F-term potential, with $\kappa =m_p^{-1}$ restored, isThe canonical Kähler potential has $K_{z\bar{z}}=K^{z\bar{z}}=1$, so the Kähler covariant derivative of a superpotential isConsequently the Polonyi model has the scalar potentialThe negative term is essential; retaining only the square of the supergravity auxiliary field would give the global F-term scalar potential rather than the supergravity F-term potential.

Set $x=\kappa z=a+iy$ and $b=\kappa \beta =2-\sqrt{3}$. In this notation $V=|\mu {|}^2{m}_p^2{e}^{|x{|}^2}U$, whereAt $a_0=\sqrt{3}-1$, $y=0$, one has $a_0+b=1$ and $1+a_0(a_0+b)=\sqrt{3}$, so $U=0$. To prove that this is a local minimum in both real directions, rather than checking only the real axis, put $u=a-a_0$. Direct expansion gives the stable zero-energy Polonyi vacuum identityEvery coefficient outside the square is strictly positive, and the exponential factor is positive. Thus $V\ge 0$ everywhere and equality requires $u=y=0$, provided $\mu \ne 0$. This proves the stronger conclusiona unique global minimum of this Polonyi model. If $\mu =0$, the scalar potential vanishes identically and this uniqueness and supersymmetry breaking conclusion do not hold.

For an independent local stability check, the Hessian matrix of $U$ at this global minimum is diagonal:Since $U$ and its first derivatives vanish there, differentiating the exponential introduces no additional Hessian matrix terms at the global minimum. For canonically normalized real fluctuations $z=z_0+(s+it)/\sqrt{2}$, the squared scalar masses are thereforewhich are both positive.

Although the vacuum energy is zero, the supergravity auxiliary field is not. In the convention $F^z=-e^{{\kappa }^{2}K/2}\overline{D_{z}W}$, its vacuum expectation value obeysThe matter fermion transformation contains a term proportional to $F^zϵ$, so no nonzero constant supersymmetry parameter leaves this vacuum invariant. Hence supersymmetry is spontaneously broken even though the vacuum energy vanishes: the positive auxiliary field contribution cancels the negative gravitational term. The matter fermion supplies the goldstino, which is absorbed by the gravitino in the super-Higgs mechanism.

Using the gravitino mass from a superpotential, and $z_0+\beta =m_p$, givesAn auxiliary field $F^z$ has mass dimension two. Defining the supersymmetry breaking scale by ${\Lambda }_{\mathrm{SUSY}}=\sqrt{|F^z|}$, one obtainsThus the gravitino mass is small compared with the supersymmetry breaking scale when that scale is sub-Planckian, as in the usual hierarchy $|\mu |\ll m_p$. This smallness is gravitational suppression; it is not a conclusion valid for arbitrary choices of the mass parameter.

Use the scalar components of the three chiral superfields, and set $a=|{\varphi }_0{|}^2$, $b=|{\varphi }_{+}{|}^2$, $c=|{\varphi }_{-}{|}^2$ and $\Sigma =a+b+c$. We assume the physical sign $g>0$ and a nonzero coupling $\lambda$. The coefficient $g$ below is exactly the coefficient in the supplied D-term scalar potential; it should not be silently replaced by a differently normalized squared gauge coupling. The canonical Kähler potential has identity Kähler metric, and the three Kähler covariant derivatives of a superpotential areSubstitution in the supergravity F-term potential givesIn particular, the subtraction cancels only half of the six ${\kappa }^{2}abc$ terms contributed by the three squares. The full scalar potential isIts nonnegativity follows term by term from $a,b,c\ge 0$ and $g>0$. This is a special property of this cubic superpotential, not of a general supergravity F-term potential.

The D-term vacuum branches of a cubic superpotential must be described as branches: there are not two isolated field-space minima. First suppose $\zeta >0$. Vanishing of $V_F$ requires $ab=ac=bc=0$, and vanishing of the D-term scalar potential requires $b-c=\zeta$. Together these conditions give the vacuum expectation valuesThe phase of ${\varphi }_{+}$ lies on a gauge orbit, and one may choose it positive and real. At this global minimum, all $D_{i}W$ and the real auxiliary field of the vector multiplet vanish. It is a supersymmetric vacuum, even though the internal gauge symmetry is spontaneously broken. For $\zeta <0$, interchange ${\varphi }_{+}$ and ${\varphi }_{-}$: the nonzero vacuum expectation value has magnitude $\sqrt{-\zeta }$. For $\zeta =0$, the supersymmetric vacuum is the entire family ${\varphi }_{+}={\varphi }_{-}=0$, with ${\varphi }_0$ arbitrary.

The other branch has ${\varphi }_{+}={\varphi }_{-}=0$ and arbitrary ${\varphi }_0$. Here $W=D_{i}W=0$ but the real auxiliary field is nonzero for $\zeta \ne 0$, andThus this branch has pure D-term supersymmetry breaking and a flat direction of a scalar potential in the neutral scalar. It is not stable at every neutral-field value. At a fixed $a=|{\varphi }_0{|}^2$, expansion in the charged scalars yieldsHence the two charged squared masses and the stability condition areAbove this threshold, both charged directions have positive mass squared. Continuity makes them uniformly positive in a sufficiently small neighborhood of the chosen neutral-field value, so the stationary branch is a genuine, non-strict local minimum valley. Its neutral direction stays flat. Below the threshold, a charged mass squared is negative and the branch is a saddle point. At equality it is a boundary of the stable valley, not a local minimum in the full field space: changing $a$ slightly downward and switching on the unstable charged field lowers $V$. For example, for $\zeta >0$, put $a=a_c-ϵ$, $b={ϵ}^2$, $c=0$. Since $M(a)=|\lambda {|}^{2}a{e}^{{\kappa }^{2}a}$ has $M^{\prime }(a_c)>0$, one finds $V-g{\zeta }^2=-M^{\prime }(a_c){ϵ}^3+O({ϵ}^4)<0$.

There are no further local-minimum branches under these generic assumptions. If $b+c>0$, the expression for $V_F$ is strictly increasing in $a$, so a stationary configuration with a charged condensate must have $a=0$. At $a=0$, an interior critical point with $b,c>0$ is impossible, becauseOn the boundary $c=0$, the relevant scalar potential is $g(b-\zeta {)}^2$; on $b=0$ it is $g(c+\zeta {)}^2$. Their nonnegative-domain minima give exactly the supersymmetric vacuum described above, or the origin of the charged-field branch. Thus, for nonzero $\zeta$, the two types are the zero-energy supersymmetric gauge orbit and the positive-energy D-breaking minimum valley above the threshold. If $\zeta =0$, the two types coalesce into the flat supersymmetric vacuum; the literal two-minimum assertion needs this parameter qualification.

In the global supersymmetry limit $\kappa \to 0$, the F-term scalar potential becomes $|\lambda {|}^2(ab+ac+bc)$ and the D-term scalar potential retains its form. The same two types of branches occur, with the threshold $|\lambda {|}^{2}a>2g|\zeta |$. Supergravity instead introduces the exponential and additional positive terms. The goldstino of global supersymmetry breaking is a physical fermion; in local supersymmetry it participates in the super-Higgs mechanism. A positive vacuum energy also gravitates in supergravity. In particular, the positive-energy branch here has $W=0$ and hence vanishing gravitino mass from a superpotential; it is not the tuned zero-energy broken vacuum of Question 3.

There is a further gauge-consistency qualification. In conventional two-derivative matter-coupled supergravity, a constant Fayet–Iliopoulos term gauges an R-symmetry and requires a corresponding nontrivial transformation of the superpotential. The apparent charges $0,+1,-1$ make the supplied cubic superpotential neutral, so a nonzero independent constant $\zeta$ does not by itself specify a consistent conventional supergravity action. The computed branches are those of the supplied scalar potential; a full embedding needs compatible charges or additional fields. This restriction is explained in Van Proeyen's derivation of the superpotential gauge-covariance condition.

More generally, the supergravity moment-map constraint on D-term breaking follows directly from gauge covariance. If $k^i{\partial }_{i}W=-{\kappa }^{2}rW$ and the moment map is $\mathcal{P}=i(k^i{K}_i-r)$, then at a point with $W\ne 0$,Consequently all chiral auxiliary fields vanishing implies the D-term also vanishes there. This relation has no analogue forcing that implication in an arbitrary global supersymmetry model. Its $W\ne 0$ hypothesis matters: the formal D-breaking branch above has $W=0$, so it does not contradict the identity. These conventional gauge-completion restrictions are additional to the elementary minimization calculation.

At a critical point $p$ of a twice continuously differentiable function, the linear term in the Taylor expansion vanishes:The real symmetric Hessian matrix has an orthonormal basis of eigenvectors. If all its eigenvalues are positive, its quadratic term is bounded below by a positive multiple of $\Vert h{\Vert }^2$, giving a strict local minimum. If all are negative, it gives a strict local maximum. If positive and negative eigenvalues both occur, directions along their eigenvectors give values on both sides of $f(p)$, so the point is a saddle point of a scalar function. The non-singularity assumption excludes zero eigenvalues; with them the quadratic test may be inconclusive and higher-order terms are needed.

Put $\xi =x$ and $\eta =y-1$. The polynomial becomesThere are no terms of degree one or two, so the gradient and the entire Hessian matrix vanish at $(0,1)$. It is a critical point with singular Hessian matrix. The lowest nontrivial approximation isunderstood as the fourth-degree leading term rather than a uniform ratio asymptotic along its zero directions.

On the path $\eta =0$ the exact value is $-{\xi }^4<0$ for $\xi \ne 0$. On the path $\eta =-2\xi$ it is ${\xi }^4(1-2\xi {)}^2>0$ for sufficiently small nonzero $\xi$. Both paths approach the point, where $g=0$. It is neither a local maximum nor a local minimum; it is a degenerate saddle point of a scalar function. The higher-order saddle test supplies the conclusion the singular Hessian matrix cannot supply.

A stationary point requires $x=z=0$ and $y(a-y^2)=0$. The Hessian matrix is diagonal:For $a>0$, the origin has positive definite Hessian matrix, hence is a strict local minimum, with value zero. The two additional stationary points $(0,\pm \sqrt{a},0)$ have value $a^2$ and Hessian matrix $\mathrm{diag}(2,-8a,2)$; both are saddle points.

For $a<0$, the origin is the only stationary point and has an indefinite Hessian matrix, so it is a saddle point. For $a=0$, the Hessian matrix test is inconclusive, but $f(x,y,z)=x^2+z^2-y^4$ takes positive values along the $x$-axis and negative values along the $y$-axis arbitrarily near zero. Thus the origin is a degenerate saddle point. There are no local maxima in any case: varying $x$ or $z$ away from a stationary point increases $f$.

In the $(x,y)$-plane take $z=0$; holding another fixed $z$ merely shifts all contour labels by $z^2$. The level sets satisfyFor $a=1$, the origin is a local minimum and $(0,\pm 1)$ are saddle points of a scalar function at contour value $1$. For $0<c<1$, a closed contour around the origin is accompanied by unbounded outer components. At $c=1$ the contour satisfies $x=\pm (y^2-1)$ and crosses at the two saddle points of a scalar function. For $c>1$, two unbounded components remain; negative contours have only outer components.

For $a=-1$, the origin is a saddle point of a scalar function, and the zero contour is $x=\pm y\sqrt{y^2+2}$, with tangent lines $x=\pm \sqrt{2}y$. Positive contours form two components extending on opposite sides of the $y$-axis; negative contours form upper and lower components. These level sets are illustrated below.

A first-order phase transition exchanges two distinct global minima of the Landau free energy. The equilibrium free energy remains continuous, but a first derivative such as the magnetization or entropy jumps. A temperature-driven transition generically has latent heat; crossing $h=0$ between symmetry-related ordered phases instead has a magnetization jump without requiring latent heat. A continuous phase transition has no such order-parameter jump and usually has a divergent correlation length and nonanalytic higher responses.

For $u>0$, $h=0$ and $r$ crossing zero, minimizing $V=r{M}^2/2+u{M}^4/4$ gives $M=0$ for $r>0$ and $M=\pm \sqrt{-r/u}$ for $r<0$. The minima merge continuously at zero. The curvature vanishes there, so the long-distance response becomes singular.

For $u<0<v$, put $z=M^2$. A nonzero stationary minimum satisfies $r+uz+v{z}^2=0$. Equality with the $M=0$ free energy givesThe finite separation demonstrates the first-order phase transition. Metastability occurs where a nonglobal local minimum persists; its disappearance at a spinodal point is not the equilibrium coexistence criterion.

For the stable branch found above, $t=x+\beta =1$ and $s=1$. Hence the vacuum expectation value isin the stated Planck units. To verify that it is a vacuum rather than merely a zero-energy stationary point, evaluate the Hessian matrix. At either zero-energy stationary branch,Because $H$ and its first derivatives vanish there, the Hessian matrix of $V$ is just $|m{|}^4{e}^{x^2}$ times this Hessian matrix. For $s=1$, both eigenvalues are positive. This proves a strict local minimum in both real scalar directions. For $s=-1$, $H_{xx}<0$, so the alternative $\beta =2+\sqrt{3}$, $⟨z⟩=-\sqrt{3}-1$ is a saddle point and is excluded from the stable zero-energy Polonyi vacuum.

There is also a useful global check. Set $u=x-(\sqrt{3}-1)$ on the stable branch. Directly completing squares givesBoth remaining coefficients are positive. Thus $H\ge 0$ everywhere, with equality only at $u=y=0$. The positive exponential prefactor proves that this is the unique global minimum, not just a metastable vacuum.

Finally, at the stable vacuum $z+\beta =1$ and $D_{z}W=\sqrt{3}{m}^2$. Its supergravity auxiliary field hasThus the Minkowski vacuum breaks supersymmetry, even though its cosmological constant vanishes. If $m=0$, the potential is flat and the displayed tuned parameter and scalar value are not selected.

The function $f$ is continuous at $y_0$ when for every $\epsilon >0$ there is a $\delta >0$ such thatSuppose $f$ has a local minimum at $c$ and is differentiable there. For sufficiently small $h>0$,Both one-sided limits equal $f^{\prime }(c)$, so it is both nonnegative and nonpositive. Therefore

The first partial derivatives areBoth vanish at $(\lambda ,\lambda )$, so every positive diagonal point is a stationary point. The Hessian matrix there isIts eigenvectors $(1,1)$ and $(1,-1)$ have respective eigenvaluesThe zero eigenvalue reflects the entire line of nonisolated stationary points: $f(\lambda ,\lambda )=2$. In fact $f-2=(x-y{)}^2[1/(xy)-1/a^2]$, showing a non-strict local minimum when $\lambda <|a|$, a non-strict local maximum when $\lambda >|a|$, and values on both sides of two near $\lambda =|a|$. At that last point the whole Hessian matrix vanishes, so its signs alone cannot classify the point.