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For Cartesian tensors, let $R$ be the orthogonal matrix taking old vector components to new ones, $v_i^{\prime }=R_{ij}{v}_j$. A rank-$n$ tensor transforms by one copy of $R$ on each index:Repeated indices are summed according to Einstein notation. In general coordinates a contravariant tensor index instead carries the corresponding Jacobian matrix $\partial x^{\prime i}/\partial x^j$, and a covariant tensor index carries its inverse; in orthonormal bases of Cartesian axes their component laws coincide with the displayed form.

An isotropic tensor has identical components after every proper orthogonal transformation, so $T^{\prime }=T$ for every $R$ with $R^{T}R=I$ and $\det R=1$. Requiring reflection invariance as well is a stronger convention for odd tensor ranks. The Kronecker delta is invariant under all orthogonal transformations, while the Levi-Civita symbol is invariant under proper rotations and acquires a determinant sign under reflections.

In Cartesian coordinates a rank-$r$ tensor is an array whose components transform under a change of orthonormal axes by one rotation matrix factor per index:An isotropic tensor has unchanged components under every proper rotation. The Levi-Civita symbol obeysfor $Q\in SO(3)$, proving rotation invariance of the Levi-Civita symbol. Under an orthogonal transformation of determinant $-1$ its sign reverses; including such transformations requires the pseudotensor distinction. Here isotropy is with respect to proper rotations.

Contract the determinant identity with ${ϵ}_{ijk}$. Since ${ϵ}_{ijk}{ϵ}_{ijk}=3!=6$, one obtainsEvery index on the right is contracted, so the tensor component transformation law leaves no free rotation factor: the determinant is a scalar. Equivalently, $A^{\prime }=QA{Q}^T$ gives $\det A^{\prime }=(\det Q{)}^2\det A=\det A$.

For the matrix inverse, defineThis is the cofactor matrix. To verify the inverse formula rather than just recognize it, contract with $A_{ib}$ and use the determinant identity again:Here the contraction of two Levi-Civita symbols gives ${ϵ}_{ajk}{ϵ}_{ijk}=2{\delta }_{ai}$. Therefore $A{C}^T=(\det A)I$ and, when $\det A\ne 0$,These are the Levi-Civita formulas for a three-dimensional determinant and inverse.

Now differentiate the cubic determinant formula with respect to the individual entry $A_{ab}$. The three differentiated factors give equal contributions: exchanging two paired indices changes the sign of both epsilon symbols, leaving their product unchanged. ThusSubstituting the inverse expression proves the entrywise derivative of the determinant:The cofactor version remains valid for singular matrices; only the inverse version requires invertibility.

For the time-dependent Jacobian matrix, the stated smooth expansion impliesApplying the chain rule and the entrywise determinant derivative gives Jacobi's formula near $t=0$:At zero, $C=I$, so ${{\partial }_t\det A|}_0={\sum }_a{\partial }_a{u}_a=\nabla \cdot u$. A Taylor expansion therefore yieldsThe Jacobian determinant measures local volume change, so this establishes volume preservation to first order. An arbitrary family specified only through its first-order expansion need not preserve volume exactly at higher orders.

The graph area functional is finite on $H^1$, since $\sqrt{1+|\nabla u{|}^2}\le 1+|\nabla u|$ and $\Omega$ has finite volume. For $p=\nabla u$, $q=\nabla v$, the difference quotient of $\sqrt{1+|p+tq{|}^2}$ is bounded in absolute value by $|q|$. This is integrable by the Cauchy-Schwarz inequality. The dominated convergence theorem therefore givesThis defines a bounded linear functional of $v$, since $|⟨A(u),v⟩|\le |\Omega {|}^{1/2}\Vert \nabla v{\Vert }_2$.

For $a(p)=p/\sqrt{1+|p{|}^2}$ the Jacobian matrix isThe inequality follows from the Cauchy-Schwarz inequality. Integrating this derivative along the segment between any two gradients shows $(a(p)-a(r))\cdot (p-r)\ge 0$. Integrating over $\Omega$ proves that $A$ is a monotone operator.

It is not coercive on $H^1(\Omega )$. The constant functions $u_k\equiv k$ satisfy $\Vert u_k{\Vert }_{H^1}=k|\Omega {|}^{1/2}\to \infty$ but $A(u_k)=0$. Thus $⟨A(u_k),u_k⟩/\Vert u_k{\Vert }_{H^1}=0$.

At a fixed point, the first equation gives $y=0$, the second then gives $x=0$, and the third gives $z=-\mu$. Thus there is exactly one equilibrium for every real $\mu$:The Jacobian matrix there isOne eigenvalue is $-1$. The other two have product one and sum $5ϵ\mu$. For $\mu <0$ their real parts are negative, so the equilibrium is asymptotically stable; for $\mu >0$ both real parts are positive, leaving two unstable directions and one stable direction. At $\mu =0$ the eigenvalues are $-1,i,-i$, withTherefore the only local bifurcation of this equilibrium is a Hopf bifurcation at $\mu =0$. A repeated real eigenvalue away from zero does not itself cause a bifurcation: the equilibrium remains hyperbolic there.

The cubic centre manifold calculation below gives the Hopf normal form radial coefficient $-9ϵ/8<0$, so this is a supercritical Hopf bifurcation. Small attracting periodic orbits occur on the $\mu >0$ side. In particular, at $\mu =0$ the leading cubic radial drift is inward, although linear stability alone cannot decide that boundary case.

The Jacobian matrix at the origin and its characteristic polynomial areThe fixed-point multipliers areBy the Jury stability criterion, both lie strictly inside the unit circle exactly whenThe last condition always holds, since ${\mu }^2+\mu +2=(\mu +1/2{)}^2+7/4$. Therefore the origin is asymptotically stable exactly for $-1<\mu <0$ among the hyperbolic parameter values.

For completeness, its linear stability throughout the parameter line is as follows. When $\mu <-1$, the complex-conjugate Floquet multipliers have modulus $|\mu |>1$, so the origin repels. When $-1<\mu <-1/3$, their modulus is $|\mu |<1$, so it attracts with oscillation. At $\mu =-1/3$ the repeated Floquet multiplier is $1/3$, still attracting. For $-1/3<\mu <0$, both Floquet multipliers are real and in $(0,1)$. For $0<\mu <1$, $p(1)<0$ and $p(0)>0$, so one Floquet multiplier lies in $(0,1)$ and the other exceeds one: the origin is a saddle fixed point of a map. For $\mu >1$, the complex-conjugate Floquet multipliers have modulus $\mu >1$, so it repels.

Possible changes at the unit circle are exhausted by $p(1)=0$, $p(-1)=0$, or a nonreal pair with product one. They giveAt $\mu =0$ the Floquet multipliers are $0,1$; the nonzero fixed point branches in part (a) meet the origin, and part (c) identifies a subcritical pitchfork bifurcation. Indeed at $\mu =0$ the exact restricted map is $y\mapsto y+y^3$, which makes the origin unstable despite its neutral linear direction.

At $\mu =-1$ the Floquet multipliers are $i,-i$: this is a 1:4 strong resonance of a planar map, coinciding with the change from repelling to attracting as $\mu$ increases. A generic nonresonant Neimark–Sacker bifurcation conclusion would be unjustified. At $\mu =1$ there is a double Floquet multiplier $+1$; the matrix has a nontrivial Jordan block, and the two further nonzero fixed point branches meet the origin. Linear stability alone is inconclusive at these unit-modulus boundary values. The repeated Floquet multiplier at $\mu =-1/3$ is strictly inside the unit circle and gives no bifurcation.

Adjoin ${\mu }_{n+1}={\mu }_n$. At the origin the stable direction is the $x$ axis, with Floquet multiplier zero, while the $y$ and parameter directions have Floquet multiplier one. By the centre manifold theorem for a discrete dynamical system, write the extended centre manifold asOn it the scalar map is $Y=(1+\mu )y-\mu h(y,\mu )+y^3$. Its discrete centre-manifold invariance equation isPut $h=h_2+O({\rho }^3)$ with $\rho =|y|+|\mu |$. Since $Y=y+O({\rho }^2)$, the quadratic equation is simply $h_2(y,\mu )=\mu y$. ThusIn particular, setting the stable variable identically to zero would lose the leading parameter correction. The reduced map through total degree three isIn the natural pitchfork bifurcation scaling $\mu =O(y^2)$ this has leading form $y_{n+1}=y_n+\mu y_n+y_n^3$. Its nonzero fixed points satisfy $y^2\sim -\mu$, and their reduced fixed-point multipliers satisfy $1-2\mu +O({\mu }^2)>1$ for small $\mu <0$. The transverse Floquet multiplier remains inside the unit circle. Therefore the bifurcation at $\mu =0$ is a subcritical pitchfork bifurcation: the origin attracts for small negative $\mu$, each nonzero branch is a saddle fixed point of a map, and the origin is a saddle fixed point of a map for small positive $\mu$.

An exact check uses the full Jacobian matrix on either nonzero branch. Its trace is $T_{\ast }=1-2\mu +3{\mu }^2$ and its determinant is $D_{\ast }={\mu }^2$. Thuswhenever a nonzero branch exists. Also $p_{\ast }(0)={\mu }^2>0$, so one real Floquet multiplier lies in $(0,1)$ and the other in $(1,\infty )$. This proves that each nonzero branch is a saddle fixed point of a map throughout $\mu <0$ and $\mu >1$, in particular near the requested bifurcation.

Work on a fixed finite interval $[a,a+T]$. Write $Y_n=y(x_n)$, $t=x_n+h/2$, and use the paper's normalized local truncation errorAssume $y$ has bounded continuous derivatives through order three on this interval, with $\Vert y^{\prime \prime }\Vert \le M_2$ and $\Vert y^{\prime \prime \prime }\Vert \le M_3$. Assume also that the Jacobian matrix $f_y$ has operator norm at most $L$ on the segments joining $y(t)$ to the endpoint averages. For a scalar equation this last hypothesis is simply $|\partial f/\partial y|\le L$ there.

Two symmetric estimates separate the errors. By Taylor's theorem applied to $y^{\prime }$ about $t$, the linear term integrates to zero on $[-h/2,h/2]$, givingLikewise the two linear terms in the endpoint expansions cancel, and their second-order remainders giveThe derivative bound on $f$ gives a Lipschitz condition in its second variable. Since $y^{\prime }(t)=f(t,y(t))$, the triangle inequality now yieldsThe constant is independent of $n$ and $h$ as long as the steps remain in the fixed interval. This implicit midpoint consistency bound proves numerical consistency of order two. The corresponding unnormalized residual is $h{\tau }_n=O(h^3)$; the two powers conventionally called local truncation error differ because the paper divides by the step size. If local error instead means the error of an exactly solved numerical step started at $Y_n$, the small-step Lipschitz condition gives the bound $h\Vert {\tau }_n\Vert /(1-hL/2)=O(h^3)$ as well.

Use the optimization Lagrangian $L(x,y)=f(x)-y^{T}c(x)$ and let $A=Dc$ be the $m\times n$ constraint Jacobian matrix. Under the standard regularity hypothesis $\mathrm{rank}A(x_{\ast })=m$, a local constrained minimizer satisfies, for some multiplier $y_{\ast }$,The second-order necessary condition isRegularity makes each such tangent vector the velocity of a feasible smooth curve. Twice differentiating the objective along that curve and using stationarity and the differentiated constraints gives the displayed nonnegative second derivative.

The regularity hypothesis is needed for these multiplier conditions to be necessary in this form. For example minimizing $f(x)=x$ subject to $c(x)=x^2=0$ has the unique feasible minimizer zero, but $f^{\prime }(0)-y{c}^{\prime }(0)=1$ for every $y$. The question does not explicitly state the qualification, so it must be supplied when using the usual KKT conditions.

With $y=-c/\mu$, differentiation of the quadratic penalty method objective givesThe primal Newton system is thereforeFor bounded multipliers and regular derivatives, $H$ remains of ordinary size while ${\mu }^{-1}{A}^{T}A$ becomes large in directions normal to the constraints. If the feasible tangent space is nontrivial and retains ordinary-size curvature, this produces a large condition number as $\mu \downarrow 0$.

Introduce $w=Ap/\mu$. The same step is obtained from the symmetric augmented systemEliminating $w$ recovers the primal system exactly. This matrix contains no explicit ${\mu }^{-1}$ entries and tends to a saddle-point matrix. Under full row rank of the limiting constraint Jacobian matrix and nonsingularity of the limiting reduced Hessian, that limiting matrix is nonsingular, so this formulation avoids the artificial penalty-induced separation of scales.

These conditioning statements require regularity beyond bounded $y$ alone. A rank-deficient constraint Jacobian matrix or singular reduced Hessian can still make the augmented system ill-conditioned. If $m=n$ and $A$ is nonsingular, all directions may scale together in the primal system, so a large condition number is not automatic there either.

The Jacobian matrix of $F_{\mu }$ in $(x,y)$ isThe top-left block is the Hessian matrix of $f-y^{T}c$ with $y$ held fixed. In the second block, differentiating $C(x)y$ in $x$ gives $YA$ and in $y$ gives $C$. The explicit derivative with respect to $\mu$ is $(0,-e{)}^T$. Differentiate $F_{\mu }(v(\mu ))=0$ by the chain rule:All blocks are evaluated on the current central path point.

The $j$th column of the matrix of $Df(a)$ in the standard bases is $Df(a)e_j$. Its $i$th component is $\partial f_i/\partial x_j(a)$, found by taking increments $t{e}_j$. Thus the matrix of the total derivative is the Jacobian matrix, with entry $(i,j)$ equal to $\partial f_i/\partial x_j(a)$.

The rows of the Jacobian matrix $Dg(a)$ are $\nabla g_1(a{)}^T,\dots ,\nabla g_k(a{)}^T$. Rank $k$ means precisely that these $k$ gradients are linearly independent. The Lagrange multiplier condition is that real numbers ${\lambda }_1,\dots ,{\lambda }_k$ exist withGeometrically, $\nabla f(a)$ is perpendicular to the tangent space of the constraint level set, whose tangent vectors form $\ker Dg(a)$. The full-rank hypothesis ensures this level set is a smooth local constraint manifold.

For $f=(f_1,\dots ,f_k)$, the Jacobian matrix is the matrix of its derivative in standard bases, with entries $(Df{)}_{ij}=\partial f_i/\partial x_j$. Here

For differentiable maps with compatible domains, the multivariable chain rule isIn Jacobian matrix notation this is matrix multiplication. In particular, if both maps are continuously differentiable, the composition is continuously differentiable. Componentwise,

The map $G$ is $C^1$ on the open ball $B$. Its Jacobian matrix isAt the origin $G(0)=0$ and $\boxed{\det DG(0)=1/3\ne 0}$. The inverse function theorem therefore supplies the required open sets $V_1\subset B$ and $V_2\subset {\mathbb{R}}^3$, both containing zero, with $G:V_1\to V_2$ a diffeomorphism. The explicit inverse formula on its image,also checks the local conclusion directly.

The inverse function theorem states: let $F:U\subseteq {\mathbb{R}}^n\to {\mathbb{R}}^n$ be $C^1$, with $U$ open, and suppose $DF(x_0)$ is invertible. Then there exist open neighbourhoods $U_0\subseteq U$ of $x_0$ and $V_0$ of $F(x_0)$ such that $F:U_0\to V_0$ is a $C^1$ diffeomorphism. Its inverse derivative satisfiesThe conclusion is local; invertibility of one Jacobian matrix does not assert global injectivity.

The map $G$ is $C^1$ on the open ball $B$. Its Jacobian matrix isAt the origin $G(0)=0$ and $\boxed{\det DG(0)=1/3\ne 0}$. The inverse function theorem therefore supplies the required open sets $V_1\subset B$ and $V_2\subset {\mathbb{R}}^3$, both containing zero, with $G:V_1\to V_2$ a diffeomorphism. The explicit inverse formula on its image,also checks the local conclusion directly.

The inverse function theorem states: let $F:U\subseteq {\mathbb{R}}^n\to {\mathbb{R}}^n$ be $C^1$, with $U$ open, and suppose $DF(x_0)$ is invertible. Then there exist open neighbourhoods $U_0\subseteq U$ of $x_0$ and $V_0$ of $F(x_0)$ such that $F:U_0\to V_0$ is a $C^1$ diffeomorphism. Its inverse derivative satisfiesThe conclusion is local; invertibility of one Jacobian matrix does not assert global injectivity.

The chain rule states that $f\circ g$ is differentiable at $x$ andIn terms of Jacobian matrices,The factors have dimensions $k\times m$ and $m\times n$, respectively, giving the required $k\times n$ Jacobian matrix.

Writing $f=(f_1,\dots ,f_m)$, the $j$th partial derivative at $x$ is the vector limitwhen it exists. The total derivative is a linear map $Df(x):{\mathbb{R}}^n\to {\mathbb{R}}^m$ satisfyingThe Jacobian matrix has entries $J_f(x{)}_{ij}={\partial }_j{f}_i(x)$. If the total derivative exists, its matrix in the standard bases is precisely $J_f(x)$, because substitution of $h=t{e}_j$ identifies every column. Existence of the partial derivatives alone does not in general imply existence of the total derivative.