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At a differentiable fixed point, linear perturbations advance by the Jacobian matrix. A hyperbolic source fixed point of a map has both eigenvalues of modulus greater than one, so nearby perturbations expand under forward iteration of a map. Real signs can cause alternating directions and a complex pair can cause spiral expansion. Eigenvalues on the unit circle are excluded from this linear classification.

A fixed point $X_{\ast }$ satisfies $F(X_{\ast })=X_{\ast }$. The Jacobian matrix is $DF=\left(\begin{array}{cc}{f}_x& f_y\\ g_x& g_y\end{array}\right)$; its eigenvalues are the linear fixed-point multipliers used in (i)–(iii).

For the Lozi map, $y=bx$ and $x=1+bx-a|x|$. At a positive point this gives $(1+a-b)x=1$, and at a negative point $(1-a-b)x=1$. ThereforeThere is no fixed point at $x=0$, because then $y=0$ and the first component would be one. On an equality boundary the corresponding scalar equation is $0=1$, so that branch is absent, not a line of fixed points.

For $x\ne 0$, $DF=\left(\begin{array}{cc}-a\mathrm{sgn}x& 1\\ b& 0\end{array}\right)$. If $a=0$, there is exactly one fixed point $(1,b)/(1-b)$ unless $b=1$, when there is none. Its eigenvalues satisfy ${\lambda }^2=b$. It is a sink fixed point of a map for $|b|<1$ and a source fixed point of a map for $|b|>1$. At $b=-1$, the affine map about its fixed point is a quarter turn, $D{F}^2=-I$ and $F^4=I$, so it is neutral, not attracting. At $b=1$ no fixed point exists. The positive/negative branch labels follow the sign of $1-b$.

If $b=0$, $X_{+}=(1/(1+a),0)$ exists for $a>-1$ with fixed-point multipliers $-a,0$. It is a sink fixed point of a map for $-1<a<1$ and a forward saddle fixed point of a map for $a>1$. At $a=1$ the fixed-point multiplier $-1$ is neutral; on the invariant axis the interval $0\le x\le 1$ is mapped by $x\mapsto 1-x$, giving two-cycles and the fixed midpoint. The negative point $X_{-}=(1/(1-a),0)$ exists only for $a>1$ and has fixed-point multipliers $a,0$, again a forward saddle fixed point of a map. For $a\le -1$ there is no fixed point. At $a=b=0$, both fixed-point multipliers vanish and two iterates send all initial points to the fixed point.

For a genuine two-cycle let the first coordinates be $x_1,x_2$ and put $h=1-b$. The two equations are $h{x}_2=1-a|x_1|$, $h{x}_1=1-a|x_2|$. If both first coordinates had the same sign $s$, subtracting would give $(h-as)(x_2-x_1)=0$. Since distinct cycle points must have different first coordinates, this contradicts $a^2\ne h^2$. If one coordinate were zero, the remaining equations would instead force $a=\pm h$, also excluded. Hence $\boxed{x_1{x}_2<0}$.

Choose $x_1>0>x_2$. Solving $a{x}_1+h{x}_2=1$, $h{x}_1-a{x}_2=1$ givesThe required signs hold exactly when $\boxed{a>|1-b|}$. This is the unique two-cycle up to interchanging its two points, under the stipulated nondegeneracy. Borderline parameter lines can carry different degenerate families and were explicitly excluded. These are Lozi fixed points and the alternating-sign two-cycle.

For the cosine mode in the question, the Laplacian acts by multiplication by $-s$, whereThese modes obey the Neumann boundary conditions. In the chemotactic term, put $q=1+u$, $c=\eta +v$. Then $\nabla c=\nabla v$, and $q\nabla c=\nabla v+u\nabla v$; the second product is a quadratic polynomial in perturbation amplitude. Thus the linearized equations areSubstituting the modal time dependence givesSeparating this matrix into its transport part and the reaction Jacobian matrix gives the required form, with the same $J$ as above.

Let $S(x,y,z)=(-x,-y,z)$. In the first two components of the vector field, replacing $x,y$ by their negatives changes the sign of the whole component; in the third, the product $xy$ is unchanged. Thus $F(Sv)=SF(v)$, establishing the symmetry, with solutions mapped to solutions by $S$.

At an equilibrium point, positivity of $a$ makes $y=x$. The remaining equations give $x(2c-a-z)=0$ and $z=x^2/b$. Hencewhere the two nonzero equilibrium points exist when $0<a<2c$. They merge with the origin at $a_p=2c$, in a pitchfork bifurcation.

At the origin the Jacobian matrix and its eigenvalues areAt $a=2c$ these are $0,-c,-b$. To see the local pitchfork bifurcation dynamics more explicitly put $\delta =2c-a$, $y=x+v$, and use $z=x^2/b+\cdots$ on the center manifold. The equations becomeAt $\delta =0$, invariance at cubic order gives $v=-x^3/(bc)+\cdots$, hence the reduced equation isIt has the supercritical pitchfork bifurcation form as $\delta$ increases, with the two branches $x\simeq \pm \sqrt{b\delta }$ attracting near onset while the origin becomes unstable.

For $0<a<2c$, the determinant of the planar block is $a(a-2c)<0$, so the origin has one positive and two negative eigenvalues and is a saddle equilibrium. At $a=c$, the planar eigenvalues are explicitly $\pm c$. For $a>2c$, the planar trace $c-a$ is negative and its determinant positive, so the origin is asymptotically stable. To distinguish its type putFor $2c<a<a_{\ast }$ both planar eigenvalues are real negative, giving a stable node. At $a=a_{\ast }$ they coincide at $-c/\sqrt{3}$; the planar block is not a scalar matrix, so it is a defective stable node. For $a>a_{\ast }$ they are nonreal with negative real part, giving a stable spiral in the planar block together with the stable third direction. This refines the requested node-versus-saddle classification.

The origin cannot have a Hopf bifurcation: a planar imaginary pair would require trace zero, namely $a=c$, but the determinant there is $-c^2<0$, so the pair is real and of opposite signs.

For either nonzero equilibrium point put $s=\pm \sqrt{b(2c-a)}$. The Jacobian matrix isIts characteristic polynomial isA nonzero imaginary root of an equation $i\omega$ requires, by separating real and imaginary parts, ${\omega }^2=B$ and $AB=C$. Equivalentlyso the candidate Hopf bifurcation values areAt such an admissible value the factorization is $p(\lambda )=(\lambda +A)({\lambda }^2+bc)$, giving the pair $\pm i\sqrt{bc}$ and a negative third eigenvalue $-A$, since $A=C/B>0$ on the nonzero branches.

The mathematical model parameter restrictions matter. A real candidate requires $b/c\le 17/8$, and the nonzero equilibrium points also require $0<a_h<2c$. The upper-sign value is admissible for $0<b/c<17/8$; the lower-sign value additionally requires $b/c>1$. For fixed $b,c$, differentiation of the characteristic equation at the imaginary root of an equation gives the eigenvalue crossing speedThus the transverse eigenvalue crossing condition holds except at the coincident-root case $b/c=17/8$, $a_h=3c/4$. At a transverse eigenvalue crossing, a generic nonzero first Lyapunov coefficient gives a Hopf bifurcation. The displayed radical is the spectral threshold; it does not justify an unconditional generic-Hopf claim for all positive mathematical model parameters, nor does it by itself exclude a degenerate Hopf bifurcation point. For example $b>17c/8$ yields no real candidate at all. The cubic Routh-Hurwitz stability criterion gives local Lyapunov stability on the side with $A>0$ and $AB-C>0$, where

For the modified equation, the planar block at the origin is $\left(\begin{array}{cc}-a& a\\ c& c\end{array}\right)$, with determinant $-2ac<0$ for the stipulated positive mathematical model parameters. Thus the origin is always a saddle equilibrium for nonzero $a,c$ in that mathematical model parameter range. The nonzero equilibrium points now satisfy $y=x$, $z=2c$ and $x^2=2bc$, so they areFor $b,c>0$ these branches stay a positive distance from the origin and do not merge with it as $a$ varies. The origin also has no zero eigenvalue for $a,c>0$. Therefore the pitchfork bifurcation is absent in the modified system.

In coordinate order $(n,m)$ the Jacobian matrix isAt the origin it is $\mathrm{diag}(1,q)$, giving a repelling node equilibrium; when $q=1$ the repeated positive eigenvalue is semisimple. At coexistence it isIts trace is $-2(1+q)/3$, determinant $q/3$, and characteristic discriminant $4(q^2-q+1)/9>0$. Hence both eigenvalues are real and negative, proving a stable node equilibrium for every $q>0$.

The other equilibria are $(1,0)$ and $(0,1)$. Their triangular Jacobians have eigenvalues $(-1,q/2)$ and $(1/2,-q)$ respectively. Thus both single-species equilibria are saddles, with their attracting directions on the corresponding coordinate axes and unstable directions allowing invasion by the other species. There are no further nonnegative equilibria because the two non-axis nullclines intersect only at $(2/3,2/3)$.

The coexistence Jacobian matrix calculated in part (a) has negative diagonal entries and positive determinant. For any spatial mode with squared wave number $k^2\ge 0$, the linearized matrix is $J_{\ast }-k^2\mathrm{diag}(1,\eta )$. Its trace is more negative than before, and its determinant isHence both eigenvalues retain negative real parts for every mode and every $\eta >0$. This is the mechanism by which negative self-response excludes two-species Turing instability, excluding diffusion-driven instability using only the reaction Jacobian matrix already computed, together with the diagonal diffusive damping. In particular unequal diffusion is not enough to produce a Turing pattern when both local self-responses are negative in this two-species competitive model.

For $\nu =1$, the equilibria are $(0,0)$ and $(\pm 1,0)$. At the origin the eigenvalues are $1,-1$, so it is a saddle for every $\kappa$.

At either nonzero point the Jacobian matrix is $\left(\begin{array}{cc}0& 1\\ -2& \kappa \end{array}\right)$, with eigenvaluesBoth points are attracting for $\kappa <0$ and repelling for $\kappa >0$. For $-2\sqrt{2}<\kappa <0$ they are stable foci; for $\kappa <-2\sqrt{2}$ stable nodes. For $0<\kappa <2\sqrt{2}$ they are unstable foci; for $\kappa >2\sqrt{2}$ unstable nodes. At the two repeated-root boundaries they are degenerate nodes. At $\kappa =0$, linearization has purely imaginary eigenvalues; the conserved energy in the next part shows that they are actual nonlinear centres. Thus the zero-real-part case is resolved by an additional first integral, rather than by the linear test alone.

An equilibrium is asymptotically stable when it is Lyapunov stable and attracts every initial state in some neighbourhood. Lyapunov stability means that sufficiently small initial displacement stays small for all future times; attraction additionally requires convergence to the equilibrium. A Lyapunov function is a differentiable positive-definite scalar function near the equilibrium whose derivative along solutions is nonpositive. Strict negativity, or an appropriate invariant-set argument for a non-strict derivative, gives asymptotic stability.

For $\nu =-1$, equilibria satisfy $y=0$ and $x+x^3=0$, so only $(0,0)$ is real. Its Jacobian matrix is $\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, with eigenvalues $\pm i$ for every $\kappa$. It is a linear centre equilibrium, so linearization alone cannot decide nonlinear attraction.

Use the positive proper energyFor $\kappa <0$, energy sublevel sets are compact and invariant. The zero-derivative set is the union of the axes, but its only invariant subset is the origin: on $x=0,y\ne 0$ one immediately leaves that axis, and on $y=0,x\ne 0$ one has $\dot{y}=-x-x^3\ne 0$. LaSalle's invariance principle therefore proves global asymptotic stability. At $\kappa =0$ the conserved energy levels are closed periodic orbits, giving a nonlinear centre equilibrium. For $\kappa >0$, time reversal together with $y\mapsto -y$ converts the system to the attracting negative-$\kappa$ case, so the origin is a nonlinear repeller and is unstable. Nonzero forward trajectories cannot remain bounded; this does not assert existence for all time if finite-time escape occurs.

A fixed point $z_{\ast }$ of a map $F:{\mathbb{R}}^2\to {\mathbb{R}}^2$ satisfies $F(z_{\ast })=z_{\ast }$. Its linear perturbations evolve by the Jacobian matrix $DF(z_{\ast })$: both eigenvalues must lie strictly inside the unit disk for linear asymptotic stability. One inside and one outside gives a saddle; both outside give a repeller. Unit-modulus eigenvalues require nonlinear analysis, unlike the continuous-time test using signs of real parts.

Here a fixed point has $x=y=p$ and $2p=a(p-p^3/3)$. ThereforeAt zero the characteristic equation is ${\lambda }^2+\lambda -a=0$. Its negative root is less than $-1$ for every $a>0$, while its positive root crosses $+1$ at $a=2$. Thus zero is a saddle for $0<a<2$ and a real repeller for $a>2$, with a pitchfork of fixed-point branches at $a=2$.

At either nonzero point the characteristic equation is ${\lambda }^2+\lambda +2a-6=0$, whose discriminant is $25-8a$. For $2<a<3$ the fixed points are saddles. At $a=3$ their periodic-orbit multipliers are $0,-1$, giving a flip threshold. For $3<a<25/8$ they are attracting real nodes; at $25/8$ repeated-eigenvalue degenerate nodes; and for $25/8<a<7/2$ attracting foci. The Jury stability criterion conditions reduce to $a>3$ and $a<7/2$, confirming the full stable interval.

At $a=7/2$ the periodic-orbit multipliers are $e^{\pm 2\pi i/3}$. For larger $a$ they form a repelling focus equilibrium. This is a unit-circle crossing at a strong resonance of a planar map with ratio $1:3$; linearization alone does not justify asserting a generic nonresonant Neimark-Sacker bifurcation. The period-two branch at the flip threshold is derived next, including an additional family absent from the source's proposed quadratic.

A genuine two-cycle must have the form $(x,y)\mapsto (y,x)$ with $x\ne y$. Hence $a(x-x^3/3)=x+y=a(y-y^3/3)$. Subtracting givesLet $S=x+y$ and $P=xy$. Using $x^3=3x-PS$ in either original equation gives $S(aP/3-1)=0$. The cubic swap-map two-cycle factorization exhibits two cases, so cancellation of $S$ would lose an actual branch.

If $S\ne 0$, then $P=3/a$ and $S^2=3+3/a$. Thus the source's quadratic holds for this family, withThere are two reflected genuine cycles for $a>3$, coalescing with the fixed points $p=\pm 1$ at $a=3$. For their two-step Jacobian matrix, $\mathrm{tr}D{F}^2=4-a$ and $\det D{F}^2=-2a^2+3a+9$. The unit-disk necessary inequality $1-\mathrm{tr}D{F}^2+\det D{F}^2>0$ fails, since it is $-2(a-3)(a+1)<0$. Therefore these quadratic-family cycles are unstable for every $a>3$. For these unstable asymmetric cycles of a cubic swap map, their periodic-orbit multipliers are real: one exceeds $1$, while the other crosses $-1$ at $a=(1+\sqrt{29})/2$. They are saddles before that secondary threshold and repelling beyond it. Their presence on the stable side of the fixed-point flip identifies the unstable branch of a subcritical flip.

If $S=0$, then $y=-x$ and the preceding quadratic identity gives $x^2=3$. Hence there is also the exact cycleThis antipodal cycle of a cubic swap map does not satisfy the printed $P=3/a$ condition: its product is $-3$. Along it the one-step Jacobian matrix is the same matrix $B=\left(\begin{array}{cc}0& 1\\ -2a& -1\end{array}\right)$ at both points, so the cycle periodic-orbit multipliers are the squares of the roots of ${\lambda }^2+\lambda +2a=0$. The Jury stability criterion conditions giveAt $a=1/2$ its periodic-orbit multipliers are on the unit circle and linearization is inconclusive. Its real-to-complex transition occurs at $a=1/8$. Thus the printed assertion about all period-two components is incomplete: it describes only $S\ne 0$, whereas the omitted antipodal cycle supplies the actual stable period-two interval.

For an equilibrium at zero, the local stable manifold consists of nearby points whose forward trajectories remain nearby and tend to zero; the local unstable manifold is defined by backward trajectories with the analogous property. For a smooth system $\dot{z}=Az+f(z)$ with $f(0)=0$, $Df(0)=0$ and no zero-real-part eigenvalues, the stable-manifold theorem gives manifolds tangent to the negative- and positive-real-part eigenspaces. Their dimensions are the corresponding eigenvalue counts. Real distinct eigenvalues alone are insufficient if one is zero; then a centre equilibrium direction requires additional analysis. In the saddle case each manifold is a curve, found by imposing invariance on a power-series graph through the relevant eigenvector.

In this system, the $x$ nullcline is $y=x^2+2x$ and the $y$ nullclines are $x=0$ and $y=x-3$. Substituting the first into the second gives $x(x^2+x+3)=0$. Since the quadratic has discriminant $-11$, the origin is the only real equilibrium. Its Jacobian matrix $\left(\begin{array}{cc}-2& 1\\ 3& 0\end{array}\right)$ has eigenvalues $-3,1$, with eigenvectors $(1,-1)$ and $(1,3)$. It is therefore a hyperbolic saddle.

Set $y=h(x)$ for either manifold. Invariance requiresFor $h=sx+b{x}^2+c{x}^3+O(x^4)$, matching the linear coefficient gives $s=-1$ or $s=3$. The quadratic and cubic coefficients respectively satisfyFor the stable slope $s=-1$, this yields $b=3/7$, $c=-9/98$; for the unstable slope $s=3$, it gives $b=1$, $c=1/6$. ThusSubstituting back into $\dot{x}$ gives the reduced dynamicsThe signs of the linear terms confirm attraction and repulsion along the respective curves.

In the phase portrait, $\dot{x}$ is positive above its parabolic nullcline and negative below it; $\dot{y}$ has sign $x(3+y-x)$. The stable branches approach the saddle and the unstable branches leave it in opposite directions. There can be no periodic orbit, since such an orbit would require total enclosed equilibrium index $+1$, while the only equilibrium has index $-1$. There is no homoclinic excursion either: eliminating $y$ gives $\ddot{x}+(2+x)\dot{x}=x(x^2+x+3)$, so a positive turning point is a strict local minimum and a negative one a strict local maximum. A loop returning to zero would require the opposite extremum somewhere. Hence bounded forward trajectories lie on the stable manifold and approach zero; other trajectories eventually leave compact regions, without a claimed universal rate or infinite-time existence at infinity. The figure shows the global nullclines and the local cubic manifold approximations; the latter are used only near zero.

The equilibrium points areThe third is biologically interior only for $\beta >1$, and coincides with $(1,0)$ at $\beta =1$. The Jacobian matrix isAt $(0,0)$ the eigenvalues are $r,1$, so it is a repelling node. At $(1,0)$ they are $-r,1-\beta$, giving a stable node for $\beta >1$ and a saddle for $\beta <1$.

At the interior equilibrium,so coexistence is a saddle. The formal equilibrium with negative $v$ for $\beta <1$ is not a population state; there its trace is negative and determinant positive.

At $\beta =1$ the boundary equilibrium is nonhyperbolic. To determine its behavior into the positive quadrant, put $x=u-1$. A local center manifold has $x=-v+v^2/r+O(v^3)$, so $\dot{v}=v^2+O(v^3)>0$ for small positive $v$. Thus it is unstable to introduction of species $v$, though attracting along the $v=0$ axis. The boundary/interior branches undergo a transcritical bifurcation at this parameter.

The nonzero homogeneous equilibrium is $\boxed{n_{\ast }=1,a_{\ast }=\alpha /\beta }$. Without transport its Jacobian matrix iswith negative eigenvalues $-1,-\beta$.

Zero chemical flux gives $a_x=0$ at the endpoints; zero bacterial flux then gives $n_x=0$. Linear perturbations therefore have Neumann boundary conditions and modes $\cos (k_{m}x)$, $k_m=m\pi$. For each mode, the amplitude matrix isIts trace is always negative. Its determinant isA mode grows exactly when $\Delta (k_m)<0$. In particular, any nonzero mode is destabilized by sufficiently large $\chi$. For continuously variable $k$, a nonempty unstable band exists precisely whenWriting $C=\alpha \chi -D_a-\beta D_n$, its squared-wavenumber endpoints are $[C\pm \sqrt{C^2-4\beta D_n{D}_a}]/(2D_n{D}_a)$. On the finite interval, at least one allowed $k_m^2$ must lie inside this band. The uniform $m=0$ mode stays stable. This is the chemotactic pattern-instability criterion.

A positive equilibrium must have $v=u=s$. LetIt is the unique positive root of $(1-s)(s+d)=as$. At this point the Jacobian matrix isConsequentlyIt is linearly asymptotically stable when $b>L$, and unstable when $0<b<L$. In either region it is a node or focus according as $(L-b{)}^2-4bsD/(s+d)$ is nonnegative or negative. At $b=L>0$ the linear eigenvalues are purely imaginary. If $L\le 0$, it is stable for every $b>0$.

The requested boundary follows from $1-d-2s=a-D$ and $s/(s+d)=(1-s)/a$:

To prove a limit cycle exists in the unstable case, a compact trapping set must stay away from the singular boundary $u=0$. The following construction supplies it. Choose $M>1$ large enough to contain the positive equilibrium and the orbit being considered. On $u=M$, $\dot{u}<0$, and on $v=M$ with $u\le M$, $\dot{v}\le 0$. Put $P=u{e}^{-v}$. Using $u+d\ge d$ and completing the square in $v$ givesChoose $\epsilon >0$ so small that $\epsilon e^M$ times the square bracket is less than $1/2$, and then choose $0<\eta <\epsilon$. The setis compact in the positive quadrant and forward invariant. On the curved boundary $P=\epsilon$, $u\le \epsilon e^M$ and $\dot{P}>0$. On $v=\eta$, $u\ge \epsilon >\eta$ gives $\dot{v}>0$. Its other boundaries were checked above. The constants can be chosen so the equilibrium is interior.

When $b<L$, the equilibrium is a hyperbolic repeller. Remove a sufficiently small outward-flow neighborhood of it from $\mathcal{K}$. This leaves a compact trapping region containing no equilibria. The Poincaré-Bendixson theorem gives a periodic omega-limit orbit of a nonequilibrium trajectory, hence a limit cycle. The analytic vector field excludes an accumulating family of periodic return-map zeros unless the whole local return map is the identity, which is incompatible with the repelling equilibrium and the nonperiodic approaching trajectory. This is the positive trapping construction for the Leslie-Gower model.

There is always the fixed point $(0,0)$. For $P>0$, the first fixed-point equation gives $e^{-P}=1/r$, and the second gives $P=N(1-1/r)$. Thus for $r>1$ there is exactly one positive fixed point,For $r=1$, every point $(N,0)$, $N\ge 0$, is fixed. There are no other nonnegative fixed points.

At the origin the Jacobian matrix has eigenvalues $r,0$. It is asymptotically stable for $0<r<1$ and a saddle for $r>1$. For $r<1$, the bounds $N_{t+1}\le r{N}_t$ and $P_{t+1}\le N_t$ also prove global extinction in the nonnegative quadrant.

At the positive fixed point,Here $\log r<r-1$ and $\log r>1-1/r$ for $r>1$. The discriminant is negative because the positive trace is less than two and determinant exceeds one. The conjugate multipliers have modulus $\sqrt{N_{\ast }}>1$, so the positive fixed point is always unstable. This is the instability of the Nicholson-Bailey equilibrium.

For completeness, on the $r=1$ fixed line,For $0\le N<1$, linear perturbations are bounded but have a neutral component along the fixed line; there is no asymptotic stability of an individual point. For $N>1$ there is transverse instability. At $N=1$, the repeated unit eigenvalue has a nontrivial Jordan block, so powers have an unbounded off-diagonal entry and the linearized system is unstable. This degenerate case is not covered by a distinct-eigenvalue assumption.

On a branch with fixed signs of $u,v$ and $uv\ne 0$, use $u^2=x+y$ and $v^2=y-x$. Differentiating these identities, or inverting the Jacobian matrix, givesThe inverse Jacobian matrix is thereforeThese are local partial derivatives. Choosing the signs specifies an inverse function on $y>|x|$; the original map has four preimages there and has no single global inverse. At $u=0$ or $v=0$, the displayed inverse derivatives are not finite.

Differentiating the polynomial coordinate map givesIts Jacobian matrix isThe determinant identifies where the inverse function theorem applies.

The determinant of the Jacobian matrix is $s^2+t^2$. Wherever $(s,t)\ne (0,0)$, the inverse function theorem permits a local inverse of the parabolic coordinates transformation, andReading off its entries givesThese are local formulas on a chosen branch: $(s,t)$ and $(-s,-t)$ have the same image, and at the origin the Jacobian matrix is singular, so this inverse-derivative formula does not apply there.

Taking the partial derivatives of the parabolic coordinates transformation, holding the other coordinate fixed each time, givesThus its Jacobian matrix is $J=\left(\begin{array}{cc}t& s\\ -s& t\end{array}\right)$.

For the vector equation, the Jacobian matrix isWrite the updated coordinates as $(X,Y)$. The second equation of the multivariable Newton iteration gives $X-Y=0$. Substituting $X=Y$ into its first linearized equation givesHence $h(x,y)=6x^5-20y^3$, and, whenever $h\ne 0$,

Starting from $(0,1)$ gives $(x_1,y_1)=(21/20,21/20)$. Every subsequent iterate remains on the diagonal, where this is exactly the scalar Newton root-finding iteration from (b). The value $f(21/20)>0$, so $21/20<r$.

For $1\le x<r$, concavity again puts the tangent's zero at or above $r$. Alsoso $r\le N(21/20)\le N(1)=8/7<6/5$. After one scalar step the sequence therefore enters the invariant interval proved in (b). It converges to $(r,r)$ at least quadratically. If the vector error is measured in the Euclidean norm, then $\Vert (x_k,y_k)-(r,r)\Vert =\sqrt{2}|x_k-r|$, so the scalar quadratic estimate also gives a quadratic vector estimate. The initial off-diagonal step causes no difficulty because its Jacobian matrix is invertible and lands exactly on the diagonal.