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For a fully discrete linear partial differential equation evolution, write $U^{n+1}=C_{h,k}{U}^n$ in a specified discrete norm. Finite-time stability of a numerical method meanswith $C_T$ independent of the allowed meshes. For a multistep method, augment the state with its time-history values and include a stable starter. Bounds may grow as $e^{\omega T}$ when the continuous problem has growth; demanding decay for all time would be a stronger assertion. The treatment of initial data, forcing, Dirichlet boundary conditions, and the norm is part of the hypothesis, not a detail supplied by an interior calculation.

For constant coefficients on an infinite or periodic uniform grid, Von Neumann stability analysis substitutes the Fourier mode $U_j^n=G(\theta {)}^n{e}^{ij\theta }$. The discrete Fourier transform and the Parseval identity turn a uniform multiplier estimate into a discrete L2 norm estimate. In a one-step scalar scheme, $|G(\theta )|\le 1$ proves contractivity, while $|G(\theta )|\le 1+\omega k$ gives finite-time stability. In a multilevel scheme, every root of a polynomial in its amplification polynomial of a multilevel finite difference scheme matters, as does uniform control of the associated companion matrix.

For the Forward Euler diffusion scheme with $\mu =k/h^2$,Thus $|G|\le 1$ for all frequencies exactly whenSufficiency follows since $G\in [-1,1]$; necessity follows from the highest-frequency mode $\theta =\pi$ on even periodic grids. This gives the usual parabolic mesh restriction $k\le h^2/2$. For the Backward Euler diffusion scheme,so every $\mu \ge 0$ is stable. The Crank--Nicolson method hasagain of modulus at most one for all $\mu \ge 0$, but stiff modes approach $-1$ rather than zero. These are PDE manifestations of the A-stability and L-stability distinctions for time integration.

For advection $u_t+a{u}_x=0$, assume $a>0$ and set $\nu =ak/h$. The upwind finite difference scheme is $U_j^{n+1}=(1-\nu )U_j^n+\nu U_{j-1}^n$, withIt is contractive when $0\le \nu \le 1$. This also has a direct maximum-norm proof: each newest value is a convex combination of two old values. The resulting Courant–Friedrichs–Lewy condition expresses that the numerical domain of dependence covers the physical one. For negative $a$, the upwind direction must be reversed. By contrast, Forward Euler method time stepping with the centered first finite difference has $G=1-i\nu \sin \theta$ and $|G|>1$ for nonzero $\nu \sin \theta$. Under fixed nonzero $\nu$ refinement it is unstable, since a fixed nontrivial mode grows geometrically over $T/k$ steps. This is not a claim of instability under every imaginable mesh coupling: $k=O(h^2)$ instead bounds its spurious finite-time growth, since $\log |G|\le a^2{k}^2/(2h^2)$.

The leapfrog advection scheme illustrates a multilevel subtlety. Its amplification polynomial of a multilevel finite difference scheme isFor $|\nu |<1$, both roots of a polynomial have unit modulus and separation at least $2\sqrt{1-{\nu }^2}$. A uniformly conditioned eigenbasis of the companion matrix then proves power boundedness of a two-level Fourier scheme for arbitrary history data. At $|\nu |=1$ and a grid admitting $\theta =\pi /2$, the roots coincide on the unit circle, and a Jordan block produces $n{\xi }^n$ growth. Thus the endpoint fails the ordinary arbitrary-history root condition for a multistep method; checking only that both moduli equal one misses the instability. Restricting the starter or filtering the parasitic mode is an additional hypothesis.

The energy method handles variable coefficients and finite boundaries for which Fourier stability analysis may be unavailable. For the Backward Euler diffusion scheme with homogeneous Dirichlet boundary conditions, define $\Vert U{\Vert }_h^2=h{\sum }_j|U_j{|}^2$. The discrete summation by parts identity isTake the inner product of $U^{n+1}-U^n=k{L}_h{U}^{n+1}$ with $2U^{n+1}$. The elementary identity $2\mathrm{Re}⟨x-y,x⟩=\Vert x{\Vert }^2-\Vert y{\Vert }^2+\Vert x-y{\Vert }^2$ givesThis proves unconditional contractivity including the actual boundary treatment. More generally, a dissipative operator $L_h$ gives $(I-k{L}_h{)}^{-1}$ of operator norm at most one: with $V=(I-k{L}_h)U$, dissipativity implies $\mathrm{Re}⟨V,U⟩\ge \Vert U{\Vert }^2$, and the Cauchy-Schwarz inequality gives $\Vert U\Vert \le \Vert V\Vert$. In finite dimensions this also proves invertibility. For the second-order backward differentiation formula, the BDF2 discrete energy identity used in question 4 controls both time levels and proves unconditional diffusion stability. A repeated root strictly inside the disk is harmless here; the energy argument supplies the uniform bound without a singular eigenvector formula.

A second technique is eigenvalue stability analysis of a finite difference method. Under the method of lines, a time integrator advances $U^{\prime }=L_{h}U$ by $R(k{L}_h)$. If $L_h$ is a normal matrix, the scalar linear stability domain criterion on all $k{\lambda }_j(L_h)$ controls its powers exactly in the corresponding L2 norm. For the centered Dirichlet discrete Laplacian, its eigenvalues lie between $-4/h^2$ and zero; intersecting this interval with a time integrator's stability interval determines its mesh restriction. A uniform bound on the diagonalization of a matrix is needed for nonnormal diagonalizable systems. Eigenvalues alone can be misleading. For example,has both eigenvalues inside the disk for $0<k<1$, but the upper-right entry of $C_{h,k}^n$ is $nk(1-k{)}^{n-1}/h$. Taking $k=h$, $nk\to T>0$, this grows like $T{e}^{-T}/h$. Hence stable scalar eigenvalues do not give mesh-uniform stability. Boundary closures can introduce precisely the extra growth that an interior Fourier symbol overlooks, so boundary stability of a finite-difference method must be checked separately.

For nonlinear spatial discretizations, a useful replacement for Fourier stability analysis is a convexity argument. If the Forward Euler method map $E_k(U)=U+kF(U)$ is nonexpansive in a chosen norm for $k\le k_{\mathrm{FE}}$, the second-order strong stability preserving Runge-Kutta methodis also nonexpansive under the same restriction. Indeed, for two inputs, the triangle inequality gives $\Vert U_{\mathrm{new}}-{\tilde{U}}_{\mathrm{new}}\Vert \le \frac{1}{2}\Vert U-\tilde{U}\Vert +\frac{1}{2}\Vert E_k(Y)-E_k(\tilde{Y})\Vert \le \Vert U-\tilde{U}\Vert$. Its Taylor expansion is $U+kF(U)+k^2{F}^{\prime }(U)F(U)/2+O(k^3)$, proving order two. Such strong stability preserving Runge-Kutta methods transfer suitable forward-step bounds without relying on a linear spectral calculation.

Finally, consistency of a numerical method explains what stability buys. For a linear Hadamard well-posed problem, the Lax equivalence theorem equates convergence of a consistent discretization with its stability, under the stated approximation-space and norm hypotheses. Directly, if the error satisfies $e^{n+1}=C_{h,k}{e}^n+k{\tau }^n$, iteration gives the discrete Duhamel principleThus vanishing normalized local truncation error and initial error yield convergence. With a nonlinear Lipschitz step estimate, the discrete Gronwall inequality plays the same role. A stable but inconsistent stencil, such as the literal defective formulas in questions 3 and 4 under their usual refinement, is not rescued by any amplification bound. The decisive checks are a mesh-uniform evolution bound, a valid boundary closure, and consistency with the actual PDE.

Let $A,B$ be the stated diagonal matrices and $C$ the interchange matrix. Since ${\epsilon }^3={\eta }^3=1$, direct matrix multiplication givesThe complete presentation established by the preceding normal-form argument therefore defines a group homomorphism $\rho :E\to {\mathrm{GL}}_2(\mathbb{C})$. More explicitly, setMultiplication of normal forms uses $c{a}^i{b}^j=a^{-i}{b}^{-j}c$, and the matrices obey exactly the same rule, so this is well-defined and multiplicative. It is the required complex group representation.

Choose an integral basis $u_1,\dots ,u_d$ of the nonzero ideal $\mathfrak{a}$. Its rational span is $L$, since it contains a nonzero integer multiple of ${\mathcal{O}}_L$. The condition $x\mathfrak{a}\subseteq \mathfrak{a}$ gives integers $m_{ij}$ with $x{u}_j={\sum }_i{m}_{ij}{u}_i$. Thus multiplication by $x$ on $L$ has an integer matrix $M$ in this basis.

Its characteristic polynomial $P(t)=\det (tI-M)$ is monic with integer coefficients. By the Cayley-Hamilton theorem, $P(M)=0$. On the field $L$, this is multiplication by $P(x)$, so $P(x)=0$. Hence $x$ is an algebraic integer, andThis is the determinant trick for an ideal preserved by multiplication.

An embedded smooth $d$-dimensional manifold $X\subset {\mathbb{R}}^N$ is locally carried by a smooth ambient coordinate change to ${\mathbb{R}}^d\times \{0\}$. A map between such manifolds is a smooth map between manifolds if its coordinate representatives are smooth. Its differential $d{f}_x:T_{x}X\to T_{f(x)}Y$ sends the tangent vector represented by a curve $\gamma$ to $(f\circ \gamma {)}^{\prime }(0)$; equivalently it is the derivative in charts.

A regular value $y$ is one for which $d{f}_x$ is surjective at every $x\in f^{-1}(y)$, including vacuous regularity when the fibre is empty. For example zero is not a regular value of $f:\mathbb{R}\to \mathbb{R}$, $f(x)=x^2$, because $d{f}_0=0$.

For an invertible matrix $A$, the derivative of the determinant isAt a determinant-one matrix it is nonzero, since taking $H=A/n$ gives value one. The regular level set theorem therefore makes $S{L}_n(\mathbb{R})={\det }^{-1}(1)$ an embedded manifold of dimension $n^2-1$, also for $n=1$ when it is a single point. Matrix multiplication has polynomial ambient coordinates and restricts to the group, so it is smooth. At the identity,Finally suppose the singular-matrix set were an embedded smooth manifold near zero for $n\ge 2$. Every curve $t\mapsto t{E}_{ij}$ lies in it, since its rank is at most one. Thus every matrix unit belongs to its tangent space at zero. These span ${\mathbb{R}}^{n^2}$, forcing its local dimension to be $n^2$. An embedded manifold of full ambient dimension is locally open, but arbitrarily small scalar multiples of the identity are invertible. This contradiction proves the singular-matrix set is not an embedded smooth manifold. The argument addresses the subset-manifold convention specified at the start.

A Lie point symmetry is a local one-parameter family of diffeomorphisms of $(t,\mathbf{x})$ that takes solution graphs of the differential equation to solution graphs wherever the transformed graph remains a graph over time. Its infinitesimal generator is $Y=\tau (t,\mathbf{x}){\partial }_t+{\sum }_i{\eta }_i(t,\mathbf{x}){\partial }_{x_i}$. Its first prolongation isFor a regular first-order equation the infinitesimal criterion is ${\mathrm{pr}}^{(1)}Y(\Delta )=0$ on $\Delta =0$; it says that the prolonged vector field is tangent to the equation submanifold.

With the constant nondegenerate canonical matrix $J$, define the Poisson bracket by $\{f,g\}=(\nabla f{)}^{T}J\nabla g$ and the Hamiltonian vector field by $V_f=J\nabla f$. Our convention gives $V_f[h]=\{h,f\}$. Applying the Jacobi identity to a test function $h$,soA first integral $F$ obeys $dF/dt=\{F,H\}=0$. Consequently $[V_F,V_H]=0$, and the local flows commute. The time-independent transformation generated by $V_F$, with $t$ unchanged, therefore maps integral curves of a vector field for $V_H$ to integral curves of a vector field: it is a Lie point symmetry.

Conversely, such a time-independent symmetry gives $[V_F,V_H]=0$, hence $V_{\{F,H\}}=0$. Nondegeneracy of $J$ makes $\{F,H\}$ constant on each connected component. At a fixed point ${\mathbf{x}}_0$, $J\nabla H({\mathbf{x}}_0)=0$ implies $\nabla H({\mathbf{x}}_0)=0$, so that constant is zero on its component. Thus the converse holds on a connected phase space containing a fixed point, or on each component that contains one.

The printed global converse needs this connectedness convention: a fixed point on one component does not control the constants on other components. For example take one component with $H=(q^2+p^2)/2$, $F=0$, and another with $H=p$, $F=q$. The Hamiltonian vector fields commute everywhere and there is a fixed point on the first component, but $F$ is not a first integral on the second. Even on a connected space the fixed-point condition matters: $H=p$, $F=q$ commute but $\dot{F}=1$.

Invert the operator definitions to obtain $a=\sqrt{m\omega /(2\hslash )}\hat{x}+i\hat{p}/\sqrt{2\hslash m\omega }$ and the adjoint formula for $a^{†}$. The canonical commutation relation yields $[a,a^{†}]=1$, with $[a,a]=[a^{†},a^{†}]=0$. Substitution givesWrite $N=a^{†}a\ge 0$. Since $[N,a]=-a$ and $[N,a^{†}]=a^{†}$, these are lowering and raising operators. Applying $a$ to a ground state would lower its energy unless it vanishes, so $a|0⟩=0$. From the normalized unique ground state constructThe commutator gives the factorial normalization and $a|n⟩=\sqrt{n}|n-1⟩$. For any energy eigenstate, repeated lowering cannot pass below $N=0$; positivity of each lowered norm forces its $N$ eigenvalue to be a nonnegative integer and lowering eventually reaches the ground state. Its uniqueness then yields the entire eigenstate ladder. The usual oscillator on the real line has discrete complete spectrum because its potential is confining, so these states span its Hilbert space.

For the perturbation $\lambda W$ with $W=\hslash \omega (a^2+a^{†2})$, nondegenerate time-independent perturbation theory gives $E_n^{\prime }=E_n+\lambda W_{nn}+{\lambda }^2{\sum }_{m\ne n}|W_{mn}{|}^2/(E_n-E_m)+O({\lambda }^3)$. The first-order term vanishes. Only $m=n-2,n+2$ contribute, with squared matrix elements $(\hslash \omega {)}^{2}n(n-1)$ and $(\hslash \omega {)}^2(n+1)(n+2)$ respectively. ThusThe absent downward transition for $n=0,1$ is automatically represented by $n(n-1)=0$. Direct substitution also givesFor $|\lambda |<1/2$ this is another oscillator with mass $m^{\prime }=m/(1-2\lambda )$ and frequency ${\omega }^{\prime }=\omega \sqrt{1-4{\lambda }^2}$. ThereforeIts expansion is $E_n[1-2{\lambda }^2+O({\lambda }^4)]$, agreeing with the perturbative result and sharpening its remainder. At the excluded endpoints confinement is lost; beyond them the quadratic Hamiltonian is not bounded below.

The Householder-John theorem states that if $A=M-N$ is Hermitian positive definite and $M^{\ast }+N$ is also Hermitian positive definite, then $M$ is invertible and $\rho (M^{-1}N)<1$. The associated stationary iteration $x^{(k+1)}=M^{-1}N{x}^{(k)}+M^{-1}b$ therefore converges from every starting vector. For designing a method, choose a splitting with easily solved $M$ and verify the positive-definiteness condition on $M^{\ast }+N$; it is a sufficient certificate and does not require computing every iteration-matrix eigenvalue.

For completeness, put $C=M^{\ast }+N=M^{\ast }+M-A$. If $Mv=0$ for $v\ne 0$, then $v^{\ast }Cv=-v^{\ast }Av<0$, a contradiction. If $M^{-1}Nv=zv$, then $Av=(1-z)Mv$. The case $z=1$ contradicts invertibility of $A$. Taking scalar products and using Hermitian symmetry giveshence $|z|<1$. The usual spectral-radius convergence criterion now proves the stated conclusion.

For the given scheme take the matrix splittingSince $A$ is real symmetric, $U=L^T$, and since it is positive definite, every diagonal entry in $D$ is positive. Thusis positive definite exactly for $0<\omega <2$. The theorem applies, and the iteration matrix has spectral radius below one. The unique solution $x_{\ast }=A^{-1}b$ is its fixed point, so $x^{(k)}-x_{\ast }=(M^{-1}N{)}^k(x^{(0)}-x_{\ast })\to 0$. ThereforeThe scheme is successive over-relaxation; at $\omega =1$ it becomes the Gauss-Seidel method.

Let $g$ have order $n$ in a finite group. In a complex group representation, its matrix is diagonalizable with eigenvalues ${\zeta }_j$ satisfying ${\zeta }_j^n=1$. If each $g^i$ with $\gcd (i,n)=1$ is conjugate to $g$, the character of a representation is constant on these powers. Henceusing the stated integrality of each Ramanujan sum. Therefore $\chi (g)$ is rational. It is also an algebraic integer, being a sum of roots of unity; a rational algebraic integer is an integer. Thus $\boxed{\chi (g)\in \mathbb{Z}}$.

In a symmetric group, raising a permutation to a power coprime to its order preserves each cycle length and therefore its entire cycle type. Since cycle type determines the conjugacy class, the hypothesis holds, and all character values of symmetric groups are integers.

Suppose $\chi$ is an irreducible character of a group of odd order and $\chi =\overline{\chi }$. Write $d=\chi (1)$. Every nonidentity element pairs with its distinct inverse, since there is no element of order $2$. Also $\chi (g^{-1})=\overline{\chi (g)}=\chi (g)$. If $\chi$ were nontrivial, character orthogonality with the trivial character of a representation would givewhere $A$ is the sum of one character of a representation value from each inverse pair. This $A$ is an algebraic integer, so $A=-d/2\in \mathbb{Q}$ is an integer. Hence $d$ is even.

But an irreducible character degree divides the group order, so $d$ must be odd. One quick justification of that divisibility is that the central class sum of a conjugacy class $C$ acts by the algebraic integer ${\lambda }_C=|C|\chi (g_C)/d$. Its integrality follows from its integer matrix on the regular group representation. Character orthogonality then givesa rational algebraic integer, hence an integer. The parity contradiction proves $\boxed{\chi =1_G}$: the trivial character is the only self-conjugate irreducible character of an odd-order group.

A point $P$ belongs to the smooth locus of a variety if its local ring ${\mathcal{O}}_{X,P}$ is a regular local ring, meaningEquivalently, the Zariski tangent space has the expected local dimension. Over the algebraically closed field, this agrees with smoothness for a reduced variety. In affine coordinates, the Jacobian criterion expresses it by the rank of the matrix of first derivatives of defining equations.

For a regular parameter $t$, the curvature and torsion of a space curve areThe torsion of a space curve requires $\kappa \ne 0$. Reparameterize by arc length $s$ and set $T={\alpha }_s$, $N=T_s/\kappa$, $B=T\times N$. Differentiating the pairwise inner products of this orthonormal Frenet frame shows that its derivative matrix is skew symmetric. Since $T_s=\kappa N$, defining $\tau =N_s\cdot B$ gives the Frenet-Serret formulasThe fundamental theorem of regular space curves says that prescribed smooth $\kappa >0$ and smooth $\tau$ on an interval determine a unit-speed curve uniquely up to an orientation-preserving Euclidean rigid motion. Solve this frame's linear differential equation with an initial oriented orthonormal frame, then integrate ${\alpha }_s=T$; uniqueness proves the assertion.

Under the intended convention that the shared parameter $s$ is arc length, equality on an open interval gives equal position and equal Frenet frames at one point. Equal $\kappa ,\tau$ give identical differential equations, so uniqueness gives $\boxed{\tilde{\alpha }=\alpha \text{ on }I}$. The printed “regular smooth” hypothesis alone does not establish this convention. Without common arc length, an orientation-preserving smooth change of speed that is the identity on $J$ but not elsewhere reparameterizes a circle into a counterexample: both curves have $\kappa =1$, $\tau =0$ and agree on $J$, but differ elsewhere as parameterized curves.

For an oriented surface with unit normal $n$, define normal curvature ${\kappa }_n=T_s\cdot n$ and geodesic curvature ${\kappa }_g=T_s\cdot (n\times T)$. ThusA unit-speed curve is a geodesic exactly when ${\kappa }_g=0$. If also ${\kappa }_n\equiv 0$, then $T_s=0$, so it is a straight line. A vertical ruling of a circular cylinder is such a geodesic, although the cylinder contains no open piece of a plane. A nonconstant regular straight line cannot be a simple closed curve. Hence no planar patch is required, and such a geodesic cannot be simple and closed.

Finally let the principal curvatures be ${\kappa }_1,{\kappa }_2$. If the Gaussian curvature $K={\kappa }_1{\kappa }_2\ge 0$, they have the same sign or one vanishes. Euler formula for normal curvature makes ${\kappa }_n$ a weighted average of them. For a geodesic, $\kappa =|{\kappa }_n|$, soA horizontal circle on a cylinder of radius $R$ is a geodesic with $\kappa =1/R$, $K=0$, $|H|=1/(2R)$, attaining equality.

Compatibility of the auxiliary system, computed on a fundamental matrix of solutions, gives the zero-curvature conditionIndeed equality of the mixed derivatives gives $(U_t+UV-V_x-VU)\Psi =0$. A single possibly zero vector solution would not justify the matrix identity; compatibility means a full fundamental family.

Put $H=W_t-VW$. Using the zero-curvature condition gives $H_x=UH$. At $x=0$, $W_t=0$ and hence $H(0)=-V(0)$. Uniqueness of the initial-value problem gives $H=-WV(0)$, or $W_t=VW-WV(0)$. Periodicity implies $V(2\pi )=V(0)=v$, so the monodromy matrix $w=W(2\pi )$ obeys the Lax equationCyclicity of the trace yields $\frac{d}{dt}\mathrm{tr}(w^k)=k\mathrm{tr}(w^{k-1}[v,w])=0$ for every $k\ge 1$.

There is a genuine factor-of-two error in the printed sine-Gordon pair. Assign the second printed matrix to $U$ and the first to $V_{\mathrm{print}}$:Direct multiplication givesThus the printed matrices encode $u_{xt}=2\sin u$.

For the stated Sine-Gordon equation, explicitly repair the first prefactor by taking $V=V_{\mathrm{print}}/2$, that is $1/(4i\lambda )$. Its zero-curvature condition becomes $u_{xt}=\sin u$. For this corrected pair, $\mathrm{tr}(w(t,\lambda {)}^k)$ is conserved for every $\lambda \ne 0$. Since $U$ is entire in $\lambda$, so are $W$ and these traces. Their Taylor coefficients in $\lambda$ supply an infinite family of first integrals, by the identity theorem extending conservation through $\lambda =0$. At a fixed spectral parameter the higher traces of a $2\times 2$ matrix are related by the Cayley-Hamilton theorem; no independence of the resulting first integrals is claimed.

For a normalized trial state $|\psi ⟩={\sum }_n{c}_n|n⟩$ in the form domain of a bounded-below Hamiltonian, the quantum variational principle gives $⟨H⟩={\sum }_n|c_n{|}^2{E}_n\ge E_0$. Thus each trial expectation is an upper bound for the ground-state energy. The same spectral argument covers a continuum: for a decaying potential whose essential spectrum begins at $0$, a negative trial expectation guarantees a negative-energy bound state.

For the stated exponential trial function, normalization follows from $4{\alpha }^3{\int }_0^{\infty }{r}^2{e}^{-2\alpha r}dr=1$. The kinetic expectation, using $\int |\nabla \psi {|}^2$, is ${\hslash }^2{\alpha }^2/(2m)$. The potential expectation is $-4A{\alpha }^3{\int }_0^{\infty }r{e}^{-(\mu +2\alpha )r}dr$. ThereforeThe exponential variational bound for the Yukawa potential at $\alpha =\mu /2$ gives $E=(\mu /8)({\hslash }^2\mu /m-A)<0$ if $\mu <Am/{\hslash }^2$. Thus a bound state necessarily exists in the stated range.

For $\mu \ge Am/{\hslash }^2$, this single-exponential family cannot prove binding: the maximum of $4A\alpha /(\mu +2\alpha {)}^2$ is $A/(2\mu )$, so every $E(\alpha )\ge 0$. This does not prove absence of bound states.

The equality case can in fact be strengthened. At $\mu =Am/{\hslash }^2$, use normalized exponentials ${\psi }_{\mu /2},{\psi }_{\mu }$ and write $Q={\hslash }^2{\mu }^2/m$. Their matrix elements areThe Rayleigh numerator for ${\psi }_{\mu /2}+ϵ{\psi }_{\mu }$ is negative for sufficiently small $ϵ>0$, and its norm is nonzero. Hence binding also holds at equality, and by continuity in a small interval beyond it. For arbitrary larger $\mu$, these trials alone do not settle existence; a nonnegative variational upper bound cannot establish a nonnegative ground-state energy.