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One form of the Householder-John theorem is as follows. Suppose $A=A^{\ast }$ and $A=M-N$, with $M$ invertible and $W=M+M^{\ast }-A$ positive definite. Then the stationary iteration $M{x}^{(k+1)}=N{x}^{(k)}+b$ converges for every initial vector exactly when $A$ is positive definite; its iteration matrix is $B=M^{-1}N$ and convergence means spectral radius below one.

For a known positive definite system, design a splitting with $M$ easy to solve, such as diagonal or triangular, and arrange $M+M^{\ast }-A>0$. The theorem then guarantees convergence. For example $M=sI$ has $W=2sI-A>0$ when $2s$ exceeds the largest eigenvalue of $A$. In the converse direction, the discrete energy identity $A-B^{\ast }AB=(I-B{)}^{\ast }W(I-B)$ explains the criterion: stable $B$ gives $A$ as a convergent sum of positive matrices, while positive $A,W$ force every iteration eigenvalue inside the unit disk.

The simultaneous zeros are the origin and the four corners $(\pm 1,\pm 1)$. The Jacobian matrix isAt the origin its eigenvalues are $1,-1$, with eigenvectors $(1,1),(1,-1)$, so the origin is a saddle equilibrium of Poincaré index $-1$. At a corner $J=-2xyI$. Thus $(1,1)$ and $(-1,-1)$ are stable nodes; $(1,-1)$ and $(-1,1)$ are unstable nodes, all of index $+1$. Although the node eigenvalues coincide, the equilibria are hyperbolic and their stability follows from their strictly signed real eigenvalues.

The lines $x=\pm 1$ and $y=\pm 1$ are invariant and divide the plane into nine convex regions. A nonconstant periodic trajectory cannot cross any of these lines by uniqueness, and cannot be contained in one of them because one-dimensional autonomous flows have no nonconstant periodic solutions. It must therefore lie inside one region. The central region contains only the origin, whose index is $-1$; all other regions contain no equilibria. But a simple periodic orbit has index $+1$, equal to the sum of the enclosed equilibrium indices. Its interior stays in the same convex region, giving a contradiction in either case. There are no periodic orbits.

Away from the invariant lines, trajectory curves have the first integralsince both logarithmic derivatives equal $-2xy$. The diagonal $y=x$ is the unstable separatrix of the origin, leading toward the two sinks; $y=-x$ is its stable separatrix, coming from the two sources. The diagram shows these separatrices, the invariant lines and the direction field.

For a Laplacian mode with ${\nabla }^2=-k^2$, the linearized chemotaxis system isIts trace is negative for all $k$. Writing $s=k^2$, its determinant isInstability occurs when this determinant is negative, because then one real eigenvalue is positive. A quadratic with positive leading and constant coefficients is negative for some $s>0$ precisely when its middle coefficient is negative and its discriminant positive. These conditions combine toThis proves the printed threshold. In a bounded spatial domain, only its permitted Laplacian eigenmodes are available; the continuous-$k$ threshold describes the onset when wavenumbers can vary freely.

The positive homogeneous equilibrium is $n=n_0$, $C=\alpha n_0/\beta$. For spatially constant perturbations $u,v$, linearization gives $\dot{u}=-\gamma n_{0}u$ and $\dot{v}=\alpha u-\beta v$. The triangular matrix has eigenvalues $-\gamma n_0$ and $-\beta$, both negative, proving asymptotic stability to homogeneous perturbations. The other homogeneous equilibrium, zero population and zero chemical, has positive population eigenvalue $\gamma n_0$ and is not the desired stable state.

The centre manifold theorem says that near an equilibrium of a sufficiently smooth finite-dimensional vector field, there is a locally invariant centre manifold tangent to the generalized eigenspace of eigenvalues with zero real part. It can be represented as a graph over that space, with graph value and derivative zero at the equilibrium; the dynamics on it determine the local centre-direction behaviour. For parameter-dependent systems the parameters may be adjoined as variables with zero derivative, producing an extended centre manifold.

At $r=1$, the linearization has the $x,y$ block $\left(\begin{array}{cc}-\sigma & \sigma \\ 1& -1\end{array}\right)$ and the independent eigenvalue $-b$. Its other eigenvalues are zero and $-(\sigma +1)$. Hence the equilibrium is nonhyperbolic, the stable subspace has dimension two, and the non-extended centre subspace has dimension one. A centre eigenvector is $(1,1,0)$.

The permutation representation on the complex vector space with basis $X$ has permutation character ${\pi }_X(g)=|{\mathrm{Fix}}_X(g)|$. The multiplicity of the trivial character is $⟨{\pi }_X,1_G⟩=|G{|}^{-1}{\sum }_g|\mathrm{Fix}(g)|$. Count pairs $(g,x)$ with $gx=x$: each orbit contributes $|G|$, by the orbit-stabilizer theorem. Hence this multiplicity equals the number of orbits.

For a transitive action the average number of fixed points is one. The identity fixes $|X|>1$ points, so not every other element can fix at least one; otherwise the average would exceed one. Thus there is a derangement $g$ with ${\pi }_X(g)=0$.

If ${\pi }_X=1_G+m\chi$, evaluating at that derangement gives $\chi (g)=-1/m$. Character values are algebraic integers, since a finite-order representation matrix has roots of unity as eigenvalues, and the sum of algebraic integers is integral. A rational algebraic integer is an integer. Therefore $-1/m$ is an integer, forcing $m=1$.

The diagonal action on $X^2$ is $g(x,y)=(gx,gy)$. Its permutation character is ${\pi }_X(g{)}^2$, since both coordinates must be fixed. The orbit-count result and the real-valued character ${\pi }_X$ giveIf any $m_j\ge 2$ and another nontrivial constituent occurs, this is at least $1+4+1=6$. If just one nontrivial constituent occurs, the previous result forces its multiplicity to one. Thus $r\le 5$ implies every multiplicity is one, and the formula then gives $k=r$.

Write $Q=1r^T-RI$, with $r=R\pi$. Then ${\pi }^{T}1=1$ andThus $\pi$ is a stationary probability vector, a left eigenvector for eigenvalue zero. If any probability vector $\nu$ is stationary, the same calculation gives ${\nu }^{T}Q=r^T-R{\nu }^T=0$, forcing $\nu =\pi$. Hence the equilibrium is unique, even if some $r_j$ vanish and the chain is not irreducible; those zero-rate states are transient or inaccessible rather than extra stationary classes.

Use the Fourier convention ${\hat{q}}_0(k)={\int }_{\mathbb{R}}{e}^{-ikx}{q}_0(x)dx$ and sufficiently rapidly decaying smooth initial data. A scalar Lax pair for the linear equation isDifferentiate the first equation in time and use the second and its spatial derivative. The terms involving $m$ and $k$ cancel, leaving $q_t=i{q}_{xx}$, exactly the linear equation.

The spatially normalized solutions are $m^{+}={\int }_{-\infty }^x{e}^{ik(x-y)}q(y,t)dy$ for $\mathrm{Im}k>0$, and $m^{-}=-{\int }_x^{\infty }{e}^{ik(x-y)}q(y,t)dy$ for $\mathrm{Im}k<0$. Integration by parts gives $m^{\pm }=O(k^{-1})$. Their difference is $e^{ikx}\hat{q}(k,t)$; Fourier transformation of the evolution gives $\hat{q}(k,t)=e^{-i{k}^{2}t}{\hat{q}}_0(k)$. Set $M^{\pm }=\left(\begin{array}{cc}1& m^{\pm }\\ 0& 1\end{array}\right)$. ThenThe Cauchy integral solution is $m(k)=(2\pi i{)}^{-1}\int e^{isx-i{s}^{2}t}{\hat{q}}_0(s)/(s-k)ds$, and reconstruction $q=-i{\lim }_{k\to \infty }km(k)$ gives the usual inverse Fourier solution.

For the nonlinear deformation, put $D=\mathrm{diag}(1,-1)$ and $Q=\left(\begin{array}{cc}0& q\\ \vartheta & 0\end{array}\right)$. The matrix Lax pair isA direct multiplication gives the off-diagonal entries of $U_t-V_x+[U,V]$ as $q_t-i{q}_{xx}+2i{q}^2\vartheta$ and ${\vartheta }_t+i{\vartheta }_{xx}-2i{\vartheta }^{2}q$, with zero diagonal. Hence compatibility is precisely the two requested nonlinear equations.

Here is a normalized Riemann-Hilbert problem for this pair, in the case without discrete scattering poles. Let $a(k),b(k)$ be its two reflection data determined from the initial spatial scattering problem, and set $\theta =kx-k^{2}t$. Deform the triangular jump toThe data $a,b$ are time-independent in this convention; their spatial and temporal oscillations are displayed explicitly. Setting $b=0$ and $a={\hat{q}}_0$ recovers the linear problem. Thus the diagonal correction $-ab$ is necessary when the second off-diagonal jump is introduced; omitting it would destroy determinant one.

To verify the deformation rather than merely name it, note $J_x=(ik/2)[D,J]$ and $J_t=-(i{k}^2/2)[D,J]$. Therefore $(M_x-(ik/2)[D,M])M^{-1}$ and $(M_t+(i{k}^2/2)[D,M])M^{-1}$ have no jump. Analyticity and the expansion at infinity make them, by Liouville's theorem, respectively $Q$ and $-kQ+V_0$. The first expansion gives $Q=-(i/2)[D,M_1]$, soThe constant term of the second expression is $V_0=(M_1{)}_x$. In the first equation's next coefficient, its off-diagonal part equals $iD{Q}_x$ and its diagonal part equals $-iD{Q}^2$. Hence $V_0=iD(Q_x-Q^2)$, exactly the displayed nonlinear Lax pair. This proves that the reconstructed functions obey the required PDEs.

Finally let $C$ be the Cauchy integral operator on the real axis, $C_{-}$ its lower boundary value, and $\mu =M^{-}$. The jump relation and normalization giveThis is the requested linear singular integral equation for the inverse problem: it is linear in the unknown matrix $\mu$ once the scattering data are given. Reconstruction is $q=(2\pi {)}^{-1}\int [\mu (J-I){]}_{12}ds$ and $\vartheta =-(2\pi {)}^{-1}\int [\mu (J-I){]}_{21}ds$.

For completeness, generic simple discrete eigenvalues add the following residue data. At an upper-half-plane pole $k_j$, use $C_j=\left(\begin{array}{cc}0& c_j{e}^{i\theta (k_j)}\\ 0& 0\end{array}\right)$; at a lower-half-plane pole use $C_j=\left(\begin{array}{cc}0& 0\\ d_j{e}^{-i\theta (k_j)}& 0\end{array}\right)$, with the norming constants determined by the initial spatial scattering problem. Impose ${\mathrm{Res}}_{k_j}M={\lim }_{k\to k_j}M{C}_j$. The nilpotent matrices make this limit well-defined. If $A_j$ denotes the residue, the full linear inverse system isThe reconstructed field formulas acquire $-i{\sum }_j(A_j{)}_{12}$ and $i{\sum }_j(A_j{)}_{21}$ respectively. This includes the soliton terms when they occur; the preceding pure-jump formula is their pole-free special case. These standard scattering formulations assume no spectral singularity on the real contour and simple discrete eigenvalues; repeated poles require the corresponding higher-order pole conditions.

The oscillator operators satisfy $[a,a^{†}]=1$ and $[a,a]=[a^{†},a^{†}]=0$. Substitution into the kinetic and potential energies givesThe number operator $N=a^{†}a$ is nonnegative because $⟨\psi ,N\psi ⟩=\Vert a\psi {\Vert }^2$. The commutators $[N,a]=-a$ and $[N,a^{†}]=a^{†}$ lower and raise an eigenvalue by one. If an eigenvalue were a noninteger $\nu \ge 0$, repeated lowering would produce a nonzero eigenvector with eigenvalue between zero and one, and one further lowering would have negative eigenvalue; its norm is nonzero because $\Vert a\psi {\Vert }^2=\nu \Vert \psi {\Vert }^2$. This contradicts nonnegativity. Thus only nonnegative integers are allowed. The normalized Gaussian solves $a|0⟩=0$, and raising it supplies every level, withFor the three-dimensional unperturbed oscillator the product states are $|n_1,n_2,n_3⟩=|n_1⟩\otimes |n_2⟩\otimes |n_3⟩$, with energy $\hslash \omega (n_1+n_2+n_3+3/2)$.

Write the perturbation as $\lambda W$, where $W=(\hslash \omega /2){\sum }_{i<j}(a_i+a_i^{†})(a_j+a_j^{†})$. Its ground-state expectation is zero. Acting on the ground state, it produces $|110⟩,|101⟩,|011⟩$, each with coefficient $\hslash \omega /2$. They lie $2\hslash \omega$ above the ground state, so second-order perturbation theory givesIn the first excited subspace with basis $|100⟩,|010⟩,|001⟩$, the first-order perturbation matrix is $(\lambda \hslash \omega /2)\left(\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right)$. Its symmetric eigenvector has eigenvalue $\lambda \hslash \omega$, while the perpendicular two-dimensional space has eigenvalue $-\lambda \hslash \omega /2$. Thus the lowest excited triplet splits intoAs an independent check, diagonalizing the quadratic potential gives normal frequencies $\omega \sqrt{1+2\lambda }$ and $\omega \sqrt{1-\lambda }$ twice; their small-$\lambda$ expansions reproduce both results.