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The Jacobian matrix at a diagonal steady state isIts eigenvectors $(1,1)$ and $(1,-1)$ represent common and opposing perturbations, with respective eigenvalues $(-1+g^{\prime })/\tau$ and $(-1-g^{\prime })/\tau$.

For a nonzero equilibrium, ${\sigma }^2+a^2{z}^2=m{a}^{2}z$ and ${\sigma }^2/a^2=z(m-z)$. Consequently $g^{\prime }(z)=2(m-z)/m$. The lower root has $g^{\prime }>1$ and is a saddle equilibrium; the upper root has $0<g^{\prime }<1$ and is a stable node. At the origin $g^{\prime }(0)=0$, so both eigenvalues are $-1/\tau$.

The supplied parameters give $\sigma /a=4$, and the nonzero roots solve $z^2-10z+16=0$, giving $2$ and $8$. The numerical stability results areBiologically, the origin is a quiescent state and the upper equilibrium a sustained mutually reinforced activity state. The saddle represents a collective activation threshold: a perturbation crossing its basin boundary can switch the network from low to high activity. This persistent-state mechanism can model a memory or decision variable, but the simplified rate equations do not establish that a particular biological circuit uses it.

For fixed weights during the input average,Thus $C$ is the input second-moment matrix. It is a positive semidefinite matrix because $v^{T}Cv=⟨(v^{T}x{)}^2⟩\ge 0$. If the inputs have been mean-centered, it is their covariance matrix; otherwise $C=\mathrm{Cov}(x)+⟨x⟩⟨x{⟩}^T$.

The stationary averaged learning equation isFor $w\ne 0$ this makes $w$ an eigenvector of $C$ with eigenvalue $\lambda =\alpha ⟨y^2⟩$. If $\lambda >0$, taking the inner product with $w$ givesand thereforeThe zero vector is also stationary but is not an eigenvector in the usual nonzero-vector definition. Likewise any vector in the nullspace of $C$ is stationary with zero output second moment and need not have the normalized length. These exceptional equilibria must be distinguished from a nonzero feature-learning solution.

Linearize the averaged rule about the genuine positive-eigenvalue equilibrium $w_i=v_i/\sqrt{\alpha }$, where $v_i$ is a unit eigenvector. The question's unit-length hint implicitly uses $\alpha =1$; for general $\alpha$ the scaling is required. Put $w=w_i+u$ and $u_i=v_i^{T}u$. To first order,Substituting into the mean learning equation and cancelling the equilibrium terms gives stability and normalization of averaged Oja learning:In the eigenbasis, the radial and transverse perturbations satisfyFor ${\lambda }_i>0$, the radial direction contracts. Any nonmaximal eigenvector has a transverse direction with ${\lambda }_j>{\lambda }_i$ and is unstable. If ${\lambda }_1>{\lambda }_2$ and ${\lambda }_1>0$, both $\pm v_1/\sqrt{\alpha }$ have strictly negative linearized eigenvalues and are locally asymptotically stable. Thus only the maximal-eigenvalue directions can be stable, with the usual simple-eigenvalue assumptions giving isolated attracting equilibria.

Convergence also needs an initial component along the top direction. For the full averaged equation, the coefficients $b_j=v_j^{T}w$ obeyFor $b_1(0)\ne 0$,With a strict top eigenvalue, all other ratios tend to zero. The norm equation from part a then drives the nonzero weight norm to $1/\sqrt{\alpha }$, so the weight converges to the top eigenvector with the sign of its initial top component. Random nonzero initialization has that component with probability one under a continuous distribution.

Several exceptions qualify the printed conclusion. A repeated largest eigenvalue gives neutral directions within the top eigenspace and convergence to a normalized vector in that space, not a unique eigenvector. An initial vector exactly orthogonal to the top eigenspace remains in that invariant subspace. The zero initialization stays zero. If $C=0$, every weight is stationary and no distinguished component can be learned. Nonzero nullspace equilibria are unstable when $C$ has a positive eigenvalue. Hence unconditional convergence from every initial weight, or uniqueness without a spectral gap, would be false.

Put $c=a^2-a_0^2$ and use reference cylindrical coordinates $(\rho ,\Theta ,Z)$ and current coordinates $(r,\theta ,z)$. The deformation map is $r=\sqrt{{\rho }^2+c}$, $\theta =\Theta +\alpha Z$, $z=Z$. In the associated orthonormal bases its deformation gradient isIts determinant is one, as required by incompressibility. The radial principal stretch is ${\lambda }_0=\rho /r$. The other two squared principal stretches are the eigenvalues of the tangential-axial block of $F{F}^T$:Writing $K=(r/\rho {)}^2+{\alpha }^2{r}^2+1$, they areThe positive square roots give ${\lambda }_{\pm }$, and ${\lambda }_0{\lambda }_{+}{\lambda }_{-}=1$.

The strain energy density is per reference volume, so the energy per height isThe work rates per height are $M\dot{\alpha }$ from the end torque and $2\pi ap\dot{a}$ from the inner pressure. The outer surface contributes no work because it is traction free, and the fixed height has zero axial work. Thus the virtual work identities for twist and inflation of an incompressible tube are

For a neo-Hookean solid, an irrelevant energy constant can be omitted:Differentiating under the finite integral givesAn antiderivative divided by $\mu$ is $\log \rho -\frac{1}{2}\log ({\rho }^2+c)-c/[2({\rho }^2+c)]+{\alpha }^2{\rho }^2/2$. With $b^2=b_0^2+c$, the neo-Hookean tube pressure under twist and inflation is thereforeIt vanishes for an untwisted undeformed tube, and the twist contributes the last term at fixed radii.

Write $c=\cos 2\varphi$, $s=\sin 2\varphi$ and $\sigma =({\sigma }_{11}+{\sigma }_{22})/2$. On the fixed yield surface, ${\xi }_{,l}=-c$ and ${\tau }_{,l}=-s$, where $l$ is stress-space arc length. Substituting ${\sigma }_{11}=\sigma +\xi$, ${\sigma }_{22}=\sigma -\xi$, ${\sigma }_{12}=\tau$ into planar force balance givesThus $\nabla \sigma =M\nabla l$, with $M=\left(\begin{array}{cc}c& s\\ s& -c\end{array}\right)$.

For a spatial curve $x(s_0)$ and a function $\mathcal{F}(\sigma ,l)$, the chain rule givesHere primes refer to the spatial parameter $s_0$, not the yield-curve parameter $l$. The displayed row-matrix product being zero is therefore sufficient for $\mathcal{F}$ to remain constant.

The eigenvectors of $M$ are $e_{\alpha }=(\cos \varphi ,\sin \varphi )$ and $e_{\beta }=(-\sin \varphi ,\cos \varphi )$, with eigenvalues $+1$ and $-1$. Along these slip lines, respectively, $d\sigma =dl$ and $d\sigma =-dl$. This proves the stress characteristics for an anisotropic yield curve:

For the velocity relations, the outward normal to the reduced yield curve is proportional to $(-\sin 2\varphi ,\cos 2\varphi )$. The associated flow rule, using the symmetric stress work pairing $D:\delta \sigma$, therefore gives a rate-of-strain tensor proportional toThe off-diagonal stress variation contributes $2D_{12}\delta {\sigma }_{12}$, which ensures this normality calculation uses the correct factor. Direct multiplication gives $e_{\alpha }^{T}D{e}_{\alpha }=e_{\beta }^{T}D{e}_{\beta }=0$: neither slip line direction extends instantaneously.

Resolve the velocity as $w=u{e}_{\alpha }+v{e}_{\beta }$. Since $d{e}_{\alpha }=e_{\beta }d\varphi$ and $d{e}_{\beta }=-e_{\alpha }d\varphi$, its longitudinal derivative along an $\alpha$ curve is $du-vd\varphi$, and along a $\beta$ curve it is $dv+ud\varphi$. Setting these extensions to zero gives the Geiringer velocity relations:

Put $p=\dot{u}$ and extend the system by $\dot{\lambda }=\dot{\kappa }=0$. At the zero equilibrium with zero parameters, the $(u,p)$ block is the nilpotent matrix $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, while the $v,w$ eigenvalues are $-1,-\tau$. Since $\tau >0$, these are stable. The extended centre manifold is a graph $v=h(u,p,\lambda ,\kappa )$, $w=k(u,p,\lambda ,\kappa )$ tangent to the four-dimensional centre space containing the two parameter directions. Its invariance equations areFor the quadratic jets in the ordinary total degree, the coefficient multiplying $h_p,k_p$ has no linear part. Hence $p{\partial }_u{h}_2=-h_2+u^2$ and $p{\partial }_u{k}_2=-\tau k_2+u^2/\tau$. With $h_2=a{u}^2+bup+c{p}^2$, comparison gives $a=1$, $b=-2$, $c=2$. The second graph is obtained in the same way. Terms involving parameters at quadratic order have zero forcing and vanish. Thus the quadratic centre manifold for a pair of linear filters isSubstitution into the centre equations gives the cubic termsThe last displayed cubic term is absent from the proposed second-order approximation, so its omission needs the weighted scaling rather than an ordinary-degree truncation.

Under the hinted scaling $u=\epsilon \tilde{u}$, $\kappa =\epsilon \tilde{\kappa }$, $\lambda ={\epsilon }^2\tilde{\lambda }$, $d/dt=\epsilon d/d\tilde{t}$, one has $p=O({\epsilon }^2)$ and $\dot{p}=O({\epsilon }^3)$. The invariance equations at weighted orders two and three give $v=u^2-2up+O({\epsilon }^4)$ and $w=u^2/{\tau }^2-2up/{\tau }^3+O({\epsilon }^4)$. In particular no additional parameter-dependent weighted-order-three graph terms appear. Therefore the $u{\dot{u}}^2$ term and the omitted manifold corrections affect the reduced acceleration only at order ${\epsilon }^5$, whereas $u^3$ has order ${\epsilon }^3$ and $u^2\dot{u}$ order ${\epsilon }^4$. After division by ${\epsilon }^3$, the slow equation isThis proves why the stated approximation retains the leading nonlinear restoring force and its first nonlinear damping correction. The centre manifold theorem removes the exponentially decaying transverse directions, while the retained generic coefficients determine the nearby reduced bifurcations. It is a local asymptotic description, not an exact global reduction. If either relevant coefficient vanishes, higher-order terms may be needed to resolve that degeneracy; the two extreme $s$ cases in part (b), with $\tau =1$, avoid it.

Set $A=s-1\ne 0$ and rescale $q=\sqrt{|A|}u$, $p=\dot{q}$. The reflection-symmetric cubic double-zero normal form becomesIt is invariant under $(q,p)\mapsto (-q,-p)$. Thus the large-$s$ case has $\sigma =1$ and the small-$s$ case $\sigma =-1$. At the origin the Jacobian matrix has determinant $\lambda$ and trace $\kappa$: it is a saddle for $\lambda <0$, stable for $\lambda >0,\kappa <0$, and unstable for $\lambda >0,\kappa >0$. In the latter two cases the discriminant ${\kappa }^2-4\lambda$ distinguishes nodes from foci; its zero is not an equilibrium-creation bifurcation.

The nonzero equilibria satisfy $p=0$, $q^2=\lambda /\sigma$. At either of them the determinant is $-2\lambda$ and the trace is $\kappa -2\lambda$. Therefore for $s\gg 1$, a pair of saddles exists when $\lambda >0$, while for $s\ll 1$, a pair of attracting equilibria exists when $\lambda <0,\kappa <2\lambda$, and a repelling pair when $\lambda <0,\kappa >2\lambda$. The repelling or attracting points may be nodes or foci. Their origin collision occurs on the pitchfork bifurcation line $\lambda =0$, away from $\kappa =0$. For fixed $\kappa \ne 0$, the scalar centre equation there beginsFor $\kappa <0$, the large-$s$ case is a subcritical pitchfork and the small-$s$ case a supercritical pitchfork. For $\kappa >0$, the scalar criticality reverses, but there is already an unstable transverse eigenvalue, so the full-plane branches have the stability just listed.

At the origin, $\kappa =0$, $\lambda >0$ is a Hopf bifurcation with frequency $\sqrt{\lambda }$. Averaging the leading weakly nonlinear oscillator gives the amplitude equationConsequentlyThese are respectively a supercritical Hopf bifurcation and a subcritical Hopf bifurcation, with $\lambda >0$ fixed and the cycle sufficiently small. The portraits change from a stable focus to an unstable focus surrounded by a stable cycle in the first case; in the second, a stable focus is surrounded by an unstable cycle before the focus becomes unstable and that cycle disappears.

The small-$s$ case has an additional pair of Hopf bifurcations at the nonzero equilibria:Their criticality includes quadratic terms and is not read off from the positive $q^{2}p$ term alone. Set $q=a+x$, $a^2=-\lambda$, and $\delta =\kappa -2\lambda$. ThenFor an oscillator $\ddot{x}=-{\omega }_H^{2}x+A_2{x}^2+B_{2}x\dot{x}+C_3{x}^3+D_3{x}^2\dot{x}$, the cubic radial coefficient is $D_3/8+A_2{B}_2/(8{\omega }_H^2)$. Here it is $2/8+(-3a)(4a)/(8\cdot 2a^2)=-1/2$. Thus both nonzero Hopf bifurcations are supercritical: for small $\delta >0$, each newly unstable well equilibrium is surrounded by a stable small cycle with $r^2\sim \delta$.

To constrain periodic orbits, use either the divergence or the energyFor $\sigma =1,\kappa \le 0$, the divergence has one strict sign except on a line, and the Bendixson-Dulac criterion excludes every nonconstant periodic orbit. For $\sigma =-1,\kappa \ge 0$, the same reasoning excludes every periodic orbit. When the divergence changes sign, this criterion makes no such exclusion; the Hopf cycles above are consistent with that restriction. In the large-$s$ case a periodic orbit also requires $\lambda >0$: if $\lambda \le 0$, a positive maximum of $q$ would have $\ddot{q}=-\lambda q+q^3>0$, or a negative minimum would have negative acceleration, both impossible. For $\lambda >0$ the same extremum argument confines a cycle to $|q|<\sqrt{\lambda }$, between the two saddle locations. The local bifurcation lines and representative portraits on both sides are summarized below; their global completion requires part (c).

Let a finite or compact group $G$ act linearly on a real space $V$, and let $\dot{x}=F(x,\mu )$ be a smooth equivariant dynamical system: $F(gx,\mu )=gF(x,\mu )$. Assume $F(0,\mu )=0$. The representation is an absolutely irreducible real representation when its complexification is irreducible; equivalently, every real linear operator commuting with $G$ is a scalar multiple of the identity. Consequently the linearization at the trivial branch is $D_{x}F(0,\mu )=\alpha (\mu )I$. A generic steady-state crossing has $\alpha (0)=0$ and ${\alpha }^{\prime }(0)\ne 0$, with stable noncritical directions removed by a centre manifold if this is a reduction of a larger system.

The isotropy group of a vector $x$ is ${\Sigma }_x=\{g\in G:gx=x\}$. For a subgroup $\Sigma$, its fixed-point subspace of a group action is $\mathrm{Fix}(\Sigma )=\{x:\sigma x=x\text{ for every }\sigma \in \Sigma \}$. Its normaliser isThe normaliser preserves $\mathrm{Fix}(\Sigma )$, and $N_G(\Sigma )/\Sigma$ describes its residual symmetry. A steady-state axial subgroup is an actual isotropy subgroup with ${\dim }_{\mathbb{R}}\mathrm{Fix}(\Sigma )=1$.

The equivariant branching lemma states that, under the smoothness, equivariance, absolute irreducibility, and transverse scalar-eigenvalue crossing hypotheses above, each conjugacy class of axial isotropy subgroup gives a branch of nonzero equilibria in its fixed-point line through $(0,0)$. The generic leading nonlinear coefficient determines which parameter side the branch occupies and whether it is transcritical-like or pitchfork-like; the theorem does not assert that these branches are stable or exhaust every possible branch. The usual nondegeneracy condition excludes a vanishing leading nonlinear coefficient. Conjugate subgroups produce symmetry-related copies of a branch, rather than new isotropy types.

To explain the dynamical role of the fixed-point subspaces of a group action, if $x$ is fixed by $\Sigma$, then $\sigma F(x,\mu )=F(\sigma x,\mu )=F(x,\mu )$ for every $\sigma \in \Sigma$. Thus the vector field is tangent to that subspace, and uniqueness makes it flow invariant. On a fixed-point line one solves a scalar equilibrium equation and determines radial stability. Stability in the full space additionally requires all transverse eigenvalues to be stable. Higher-dimensional fixed-point spaces support their own restricted dynamics and may contain symmetry-constrained connecting orbits.

Finally, the normalizer action determines axial branch parity: if $N_G(\Sigma )/\Sigma$ contains an element acting as $x\mapsto -x$ on the fixed line, its scalar vector field must be odd, so a generic cubic term gives a pitchfork bifurcation. If that residual sign reversal is absent, a quadratic term is permitted and can give a transcritical-like branch. This is why fixed-space dimension alone does not determine either the branch type or its full stability.

Use the natural dihedral group generators $\rho z=e^{2\pi i/n}z$ and $mz=\bar{z}$, and orient the unfolding parameter so the trivial equilibrium loses stability as $\mu$ increases through zero. Rotation equivariance requires a monomial $z^p{\bar{z}}^q$ in $\dot{z}$ to have $p-q\equiv 1(\mathrm{mod}n)$, while reflection makes its coefficient real. These constraints give the dihedral steady-state normal form. The reflection axes have angles $\theta =k\pi /n$ and one-dimensional fixed-point subspaces of a group action; their equilibria are the branches guaranteed by the equivariant branching lemma.

For $n=3$, a generic quadratic anisotropy dominates:In polar coordinates, $\dot{r}=\mu r+b{r}^2\cos 3\theta +O(r^3)$ and $\dot{\theta }=-br\sin 3\theta +O(r^2)$. At a nonzero equilibrium, $\chi =\cos 3\theta =\pm 1$ and $r=-\mu /(b\chi )+O({\mu }^2)>0$. There are three symmetry-related equilibria on each parameter side, on opposite half-axes as the sign of $\mu$ changes. The radial and angular eigenvalues are $-\mu +O({\mu }^2)$ and $3\mu +O({\mu }^2)$, respectively. Hence all three nonzero branches are saddles in the full plane, even though radial stability exchanges at the crossing. The origin is a sink for $\mu <0$ and a source for $\mu >0$. The reflection normaliser has no residual sign reversal in this odd dihedral case, so these are transcritical-like branches, not symmetry-forced pitchforks.

For $n=4$, both relevant nonlinearities are cubic:There are two conjugacy classes of reflection isotropy: the coordinate axes, with $\chi =\cos 4\theta =1$, and the diagonals, with $\chi =-1$. Generically $b\ne 0$ and $a\pm b\ne 0$. Each class has a pitchfork bifurcation, with four symmetry-related equilibria when its squared radius is positive:These formulas give all generic existence and stability cases. If $a<-|b|$, both classes emerge for $\mu >0$; the class with $b\chi >0$ consists of stable nodes and the other of saddles. If $a>|b|$, both classes exist for $\mu <0$; the class with $b\chi >0$ consists of saddles and the other of sources. If $|a|<|b|$, one class exists on each side and both consist of saddles. Equality $a=\pm b$ is nongeneric and requires higher-order terms. Unlike the odd dihedral cases, the residual normaliser quotient contains a sign reversal, enforcing exact paired pitchfork branches.

For $n=5$, cubic radial saturation occurs before the first angular anisotropy:For generic $a,b\ne 0$, ten equilibria lie on the reflection half-axes whenever $-\mu /a>0$, divided into two distinct five-point group orbits. Their radii and eigenvalues areIf $a<0$, they emerge for $\mu >0$: the five with $b\chi >0$ are stable nodes and the other five are saddles. If $a>0$, they exist for $\mu <0$: the five with $b\chi >0$ are saddles and the others are sources. The leading amplitude scaling is pitchfork-like, but the quartic term distinguishes opposite half-axes and generally gives different corrections to their radii; there is no exact sign-reversal symmetry on an individual fixed line. Thus $D_3$ has quadratic saddle branches, $D_4$ has two cubic axial branch types, and $D_5$ has cubic radial branches whose angular stability splits at quartic order.

Let a smooth $G$-equivariant vector field have a trivial equilibrium branch and a critical real eigenspace $V\oplus V$, where $V$ is an absolutely irreducible real representation of the finite group $G$. Assume the critical linearization is semisimple with eigenvalues $\pm i{\omega }_0$, ${\omega }_0\ne 0$, no other imaginary-axis eigenvalues are present, and the real part of the critical eigenvalue crosses zero transversely as the parameter varies. The equivariant Hopf theorem then gives, generically, a branch of small-amplitude periodic solutions for each conjugacy class of complex axial isotropy subgroup of $G\times S^1$ in the critical representation. The restricted Hopf bifurcation on its two-dimensional fixed space must be nondegenerate, with nonzero leading radial cubic coefficient. Each branch has the associated spatiotemporal symmetry of a periodic orbit and period tending to $2\pi /|{\omega }_0|$.

The spatial multiplicity means this is not the ordinary simple-eigenvalue Hopf hypothesis: the required critical representation is the whole two-copy isotypic component, and the semisimple pair has the corresponding symmetry multiplicity. The theorem guarantees primary branch existence and symmetry, but neither their stability nor the side of the parameter on which they exist. Those require the nonlinear coefficients and transverse perturbation analysis. A centre manifold and a symmetry-compatible normal form of a dynamical system isolate the critical dynamics and supply those coefficients.

Let $D=Q/V_0$ be the dilution rate. Population loss is $DN$, and nutrient loss by consumption is $NK(C)/V_0$, so mass conservation givesUse the dimensionless variables $\tau =Dt$, $c=C/K_0$, and $n=N{K}_{\max }/(Q{K}_0)$. Substitution into the Monod equation gives the stated system withFor a positive equilibrium, the first equation requires $\alpha c/(1+c)=1$. Hence $c_{\ast }=1/(\alpha -1)$ and $n_{\ast }=\alpha (\beta -c_{\ast })$, both positive exactly under the specified inequalities.

Writing $A=\alpha n_{\ast }/(1+c_{\ast }{)}^2>0$, the Jacobian matrix at this equilibrium isBoth eigenvalues are negative, so the state is locally asymptotically stable. The Monod chemostat equilibrium and relaxation also makes one decay rate transparent: $z=c+n/\alpha$ satisfies $z^{\prime }=\beta -z$, independently of the uptake kinetics.

Continuation around one positively oriented loop induces an invertible linear map $M$ on the two-dimensional solution space. Its inverse is continuation around the reversed loop. Over $\mathbb{C}$, choose an eigenvector with eigenvalue $\lambda \ne 0$, and let $w$ be the corresponding nonzero solution. After continuation it becomes $\lambda w$.

Choose $\sigma$ with $e^{2\pi i\sigma }=\lambda$. The product $h(z)=z^{-\sigma }w(z)$ has trivial monodromy, so it is single-valued and holomorphic on the punctured disc. Its Laurent series, convergent on compact subannuli, gives the Monodromy eigenfunction Laurent representationNo assumption of a regular singular point has been made, so the negative part need not terminate.

For a self-adjoint Hamiltonian operator bounded below, the Rayleigh-Ritz variational principle says $E[\psi ]=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩\ge E_0$. Expanding in energy eigenstates proves this, since the quotient is a weighted average of energies. Optimize a tractable trial family to obtain an upper bound on the ground energy.

If $H{\psi }_0=E_0{\psi }_0$, a regular trial error $\psi ={\psi }_0+ϵ\eta$ gives the exact cancellationThe linear terms vanish because $(H-E_0){\psi }_0=0$. Here $\eta$ is a fixed finite-energy direction, or is uniformly bounded in the quadratic-form norm. A Hilbert-space norm bound alone is insufficient for an unbounded Hamiltonian operator: a norm-$ϵ$ admixture of an eigenstate with energy of order ${ϵ}^{-2}$ can have order-one energy error. The usual quadratic-error claim therefore presupposes a controlled-energy perturbation.

For the first excited energy minimize the quotient on states orthogonal to the exact ground state. Alternatively diagonalize $H$ in a trial subspace and use the min-max principle: its second Ritz eigenvalue is an upper bound for the second exact eigenvalue. An arbitrary trial function without either orthogonality or a min-max construction need not bound the first excited level.

The one-dimensional even potential has even ground state and odd first excited state, by the Sturm oscillation ordering. Choose the automatically orthogonal odd trial function ${\psi }_a(x)=Nx{e}^{-a{x}^2/2}$, $a>0$. Integration by parts gives the kinetic numerator $\int |{\psi }_a^{\prime }{|}^2$, and the given Gaussian moments giveFor example, the kinetic factor is $(I_0-2a{I}_2+a^2{I}_4)/I_2=3a/2$, while the potential factors are $I_4/I_2$ and $I_6/I_2$. Differentiation gives the unique positive minimizer $a^3-a-5=0$. ThusThis is an upper estimate, not an exact eigenvalue. Improve it with odd polynomial-Gaussian trial functions such as $(x+b{x}^3)e^{-a{x}^2/2}$, or a larger odd Ritz basis, and optimize the width. The parity restriction keeps all such states orthogonal to the even ground state.