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For $f(x,y)=g(x)+h(y)$, a degenerate Hessian matrix can be supplemented by the first nonzero term of each one-variable displacement. Two negative leading even powers give a strict local maximum, two positive ones a strict local minimum, and opposite signs give a saddle. Looking along the coordinate directions and bounding the separated terms avoids an inconclusive quadratic test.

Mathematical optimization studies extrema under constraints.

A stationary point is nonisolated if every neighbourhood contains other stationary points. For a twice differentiable function, a regular smooth curve of stationary points through a point forces its Hessian matrix to annihilate the curve's nonzero tangent: differentiate the vanishing gradient along the curve. Thus the Hessian matrix has a zero eigenvalue there, but that alone need not determine whether the point is a local minimum or local maximum.

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.

In the Landau theory approximation, choose the global minimum of the Landau free energy as a function of the magnetization. Since ${\mathcal{A}}_6>0$, the polynomial is a coercive function and a global minimum exists. The candidates are stationary points satisfyingCheck their values of the free-energy density; a stationary point or a metastable local minimum alone need not be the equilibrium state. Positive curvature ${\mathcal{A}}_2+3{\mathcal{A}}_4{m}^2+5{\mathcal{A}}_6{m}^4$ ensures local stability; when it vanishes, higher terms must be checked. The selected global minimum is the equilibrium magnetization within this approximation.

At zero magnetic field, equal opposite minima describe distinct phases exhibiting spontaneous symmetry breaking. A spontaneous magnetization is obtained by taking the thermodynamic limit before $h\to 0^{+}$ or $h\to 0^{-}$. A symmetric finite-volume average can instead have zero expectation even when the minima are nonzero. Thus one must specify the phase when interpreting the minimizer as the equilibrium expectation.