Minimum energy paths string neb and saddle certificates — Examples

The gradient is $\mathrm{\nabla }V(x,y)=(x^3\mathrm{/}6,y)$. Thus $\mathrm{\nabla }V(ϵe_1)=({ϵ}^3\mathrm{/}6,0)$ and $\mathrm{\nabla }V(-ϵe_1)=(-{ϵ}^3\mathrm{/}6,0)$. Their central difference is $(({ϵ}^3\mathrm{/}3)\mathrm{/}(2ϵ),0)=({ϵ}^2\mathrm{/}6)e_1$. The Hessian at the origin is $\mathrm{diag}(0,1)$, so [F1] gives $H{e}_1=0$.

Adding the stated force errors changes the numerator by $2\eta e_1$; division by $2ϵ$ adds $\eta e_1\mathrm{/}ϵ$. Hence the computed estimate is exactly ${\widehat{D}}_{ϵ}=({ϵ}^2\mathrm{/}6+\eta \mathrm{/}ϵ)e_1$. This realizes both contributions in [F2] for this example.

For fixed $\eta >0$, the deterministic term ${ϵ}^2\mathrm{/}6$ tends to zero as $ϵ\to 0$, but the permitted error term $\eta \mathrm{/}ϵ$ grows without bound. Therefore reducing the dimer separation does not monotonically improve the Hessian-vector estimate when force evaluations are inexact.

For the corner, the one-sided derivatives at $0$ are $R_c^{\mathrm{\prime }}(0^{-})=(1,0)$ and $R_c^{\mathrm{\prime }}(0^{+})=(0,1)$. They are distinct, so there is no single tangent at the corner. The centered chord is $R_c(h)-R_c(-h)=(h,h)$, whose unit direction is $(1,1)\mathrm{/}\sqrt{2}$; it agrees with neither one-sided tangent.

For the cusp, $R_u(h)=R_u(-h)=(h^2,0)$, so its centered chord is exactly zero. Also $R_u^{\mathrm{\prime }}(0)=(0,0)$, so the path is not regular there.

The corner violates the differentiability/unique-tangent hypotheses of [F1], while the cusp violates regularity and makes the algorithmic centered direction undefined. By [F2], a finite string update encountering this zero chord must return geometry-failure; neither example can be covered by the second-order normalized-tangent estimate of [F1].

Write $T(s)=(-\sin s,\cos s)$ and $N(s)=(\cos s,\sin s)$. Then $X_s=(1+n)T(s)$ and $X_n=N(s)$ are orthogonal, with lengths $1+n>0$ and $1$. By [F3], $\mathrm{\nabla }V(X(s,n))\cdot T(s)=W^{\mathrm{\prime }}(s)\mathrm{/}(1+n)$ and $\mathrm{\nabla }V(X(s,n))\cdot N(s)=n$. At $n=0$, the path tangent is $T(s)$ and its unit normal is $N(s)$, so the normal gradient component vanishes. The path is regular because $\mathrm{\parallel }{R}^{\mathrm{\prime }}(t){\mathrm{\parallel }}_2=2a>0$; hence it is a regular MEP by [F1].

Since $W(s)\ge 0$ and $n^2\mathrm{/}2\ge 0$, the potential is nonnegative and equals zero exactly at $(s,n)=(-a,0)$ and $(a,0)$. Thus both specified endpoints are global minima on $U$.

At $(s,n)=(0,0)$, $W^{\mathrm{\prime }}(0)=0$ and $n=0$, so step 1.1 gives $\mathrm{\nabla }V(X(0,0))=0$. The coordinate Hessian of $V\circ X=W(s)+n^2\mathrm{/}2$ is $\mathrm{diag}(W^{\mathrm{\prime }\mathrm{\prime }}(0),1)=\mathrm{diag}(-4a^2,1)$. By [F4], the quadratic Taylor term of $V$ at $X(0,0)$ is $\frac{1}{2}{v}^T(D^{2}V)v+o(\mathrm{\parallel }v{\mathrm{\parallel }}_2^2)$. Substituting $v=X(s,n)-X(0,0)=DX(0,0)(s,n)+O(s^2+n^2)$ shows the coordinate quadratic form is $DX(0,0{)}^T(D^{2}V)DX(0,0)$; the linear term vanishes because the point is critical. Since $DX(0,0)=[T(0),N(0)]$ is orthogonal, this is an orthogonal representation of the ambient Hessian. By [F2] it has exactly one negative eigenvalue, making the point an index-one saddle.

The fixed-endpoint finite-image string of Finite-image projected-gradient string with arc-length redistribution can sample and redistribute this arc, but that finite procedure alone does not imply convergence to it from arbitrary initial images.

Along $C$, the first coordinate is $\cos a>0$, and $\mathrm{\mid }y\mathrm{\mid }\mathrm{/}\cos a\le \tan a<1$, so the polar angle lies in $[-a,a]\subset (-\pi \mathrm{/}4,\pi \mathrm{/}4)$. Its radius is between $\cos a=\sqrt{3}\mathrm{/}2>3\mathrm{/}4$ and $1<5\mathrm{/}4$. Thus the whole chord is in the declared sector.

The chord's tangent is $e_2=(0,1)$. At its midpoint $x_0=(\cos a,0)=X(0,n_0)$ with $n_0=\cos a-1\mathrm{\ne }0$. In the chart $X_s=(1+n)T$ and $X_n=N$, where $T(0)=e_2$ and $N(0)=e_1$. The coordinate derivatives of $V$ there are ${\mathrm{\partial }}_{s}V=W^{\mathrm{\prime }}(0)=0$ and ${\mathrm{\partial }}_{n}V=n_0$, so by [F3] $\mathrm{\nabla }V(x_0)\cdot e_1=n_0\mathrm{\ne }0$.

Since $e_1$ is perpendicular to the chord tangent $e_2$, step 1.2 gives a nonzero normal component of the gradient at the chord midpoint. This violates the MEP condition in [F1], proving the chord is not the circular MEP. The improved tangent rule in [F2] only changes the discrete force direction; it does not make this straight segment satisfy the continuous normal-gradient equation.

The gradient is $\mathrm{\nabla }V=(-2x,-2y)$, so it vanishes at $(0,0)$; the Hessian is $H=-2I$, with ordered eigenvalues ${\lambda }_1={\lambda }_2=-2$.

For any $\gamma >0$, an interval enclosing ${\lambda }_2=-2$ cannot have lower endpoint ${\mathrm{\ell }}_2\ge \gamma >0$. Thus no exact validated enclosure of this Hessian satisfies [F1].

The point is a maximum with Morse index two, not index one. Since the required enclosure premise in [F2] fails by step 2.1, the theorem makes no certification claim for it; this example illustrates why a small gradient or critical-point report alone does not establish the desired saddle type.

The eigenvalues of $H_{\delta }$ are $-\delta$ and $\delta$, with exact lowest eigenvector $e_1$ and spectral gap $2\delta$. Also $E_{\delta }^T{E}_{\delta }={\delta }^{2}I$, so for every $z$ we have $\mathrm{\parallel }{E}_{\delta }z{\mathrm{\parallel }}_2^2=z^T{E}_{\delta }^T{E}_{\delta }z={\delta }^2\mathrm{\parallel }z{\mathrm{\parallel }}_2^2$ and hence $\mathrm{\parallel }{E}_{\delta }z{\mathrm{\parallel }}_2=\delta \mathrm{\parallel }z{\mathrm{\parallel }}_2$.

The eigenvalues of ${\widehat{H}}_{\delta }$ are $\pm \sqrt{2}\delta$. For the negative eigenvalue, the first row equation gives $(-1+\sqrt{2})x+y=0$, so a lowest unit eigenvector is proportional to $(1,1-\sqrt{2})$. Its angle with $e_1$ is therefore $\arctan (\mathrm{\mid }1-\sqrt{2}\mathrm{\mid })=\arctan (\sqrt{2}-1)$, a positive constant independent of $\delta$.

As $\delta \downarrow 0$, the perturbation norm tends to zero while this angle stays fixed at $\arctan (\sqrt{2}-1)>0$. The spectral gap also tends to zero, so this does not contradict the separated-mode bound [F1]; it shows why that bound must include a gap scale.

For each $c\in \mathbb{R}$, $R_c(t)=(t,c\sin (\pi t))$ on $[0,1]$ is regular because its first coordinate derivative is $1$, and it has endpoints $A,B$. Since $\mathrm{\nabla }V=0$, every normal gradient component is zero, so [F1] makes every $R_c$ an MEP. Distinct $c$ give distinct paths.

For $M=2$, each listed interior point is equidistant from $A$ and $B$: the two distances are $1\mathrm{/}2$ when $R_1=(1\mathrm{/}2,0)$ and both $\sqrt{5}\mathrm{/}2$ when $R_1=(1\mathrm{/}2,1)$. Thus the adjacent lengths agree. Since the gradient is zero, the string projected force is zero by [F2], and the ordinary NEB spring term is also zero by [F3]. Both chains therefore have exactly zero force and spacing residuals.

The two chains differ in their interior image, and the continuum family in step 1.1 is nonunique. In this example the string residual map has no locally invertible Jacobian that can isolate a path: its force components vanish identically and its spacing equation alone cannot determine the two coordinates of the interior image. Thus the inverse/stability premise [F4] is absent. Independently of that missing premise, the two distinct MEPs and two zero-residual chains explicitly prove nonuniqueness. Each chain's midpoint lies on its corresponding exact MEP $R_c$, so this is not an assertion of inaccuracy relative to every MEP. Relative to the prescribed reference $R_0([0,1])$, however, the second chain's midpoint has Euclidean distance $1$ while its residual is zero. Therefore residual alone cannot select a unique continuum MEP or bound distance to that reference by an error bound that vanishes with the residual.