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For a scalar, the Lie derivative of a function is its directional derivative:The Lie derivative of a vector field is their Lie bracket of vector fields:For a torsion-free connection,The last term vanishes because the connection coefficients are symmetric in $b,c$. This proves ${\mathcal{L}}_{\xi }X={\nabla }_{\xi }X-{\nabla }_X\xi$.

For the metric tensor statements use the Levi-Civita connection, which also has metric compatibility. The tensor derivation property of the Lie derivative of a tensor field givesExpanding the first term using $\nabla g=0$ and using the vector identity cancels the derivatives of $X$ and $Y$. What remains isSince $X,Y$ are arbitrary, the Killing equation isMetric compatibility is needed here, even though symmetry alone sufficed for the vector identity. In general the Lie derivative of a metric with nonmetricity contains a term ${\xi }^c{\nabla }_c{g}_{ab}$. For example, with the metric tensor $g=e^{2x}d{x}^2$, zero connection coefficients and $\xi ={\partial }_x$, the Lie derivative of a tensor field is $2e^{2x}$, whereas $2{\nabla }_x{\xi }_x=4e^{2x}$. The usual metric connection interpretation of the question supplies the required compatibility.

For the Riemann curvature tensor calculation, keep the original PDF's convention. With $R_{abcd}=g_{ae}{R}^e{}_{bcd}$ and $R_{abc}{}^d=g^{de}{R}_{abce}$, it meansThus the commutator on a covector has the opposite sign. Write $K_{abc}={\xi }_{a;bc}={\nabla }_c{\nabla }_b{\xi }_a$. Taking a covariant derivative of the Killing equation makes $K$ antisymmetric in its first two indices, while the Ricci identity givesCombining this with the first-two-index antisymmetry gives the three relationsAdd the first and third and subtract the second. Pair symmetry and the first Bianchi identity imply $R_{adbc}+R_{bdca}-R_{cdab}=-2R_{abcd}$. ConsequentlyThis is the second covariant derivative of a Killing vector in the paper's convention; its sign is tied to the declared curvature-sign convention in Killing derivative identities, not to raising a particular index.

Set $L_{ab}={\xi }_{a;b}$. Along any smooth curve with tangent vector $V$, the chain rule and the just-proved identity give the Killing transport equationsChoose a parallel-propagated orthonormal frame along the curve. These become a homogeneous linear system of ordinary differential equations for the components of $(\xi ,L)$. Uniqueness of its initial value problem implies that zero initial data remain zero. Any point in the same connected component as $P$ can be joined to $P$ by a piecewise smooth curve. Therefore zero value and zero first derivative at $P$ force the Killing field and its derivative to vanish throughout that component, with no completeness assumption. On a manifold with one connected component this is the requested global result. Connectedness cannot be dropped: on two disjoint flat spacetimes, take the zero field on one and a nonzero translation on the other.

On a pseudo-Riemannian manifold of dimension $n$ with one connected component, the map $\xi \mapsto ({\xi }_a(P),L_{ab}(P))$ is therefore injective. There are $n$ independent components of the value and $n(n-1)/2$ of its antisymmetric derivative. The dimension bound for Killing vector fields isIt is attained in flat space by ${\xi }_a=A_a+B_{ab}{x}^b$ with constant antisymmetric $B$, giving translations and infinitesimal rotations or Lorentz transformations. The four-dimensional maximum is 10. For a disconnected manifold the bound applies componentwise; there is no uniform global bound depending only on $n$ if the number of components is unrestricted.

The Gaussian integral over the full vertical velocity line is $\sqrt{2\pi }{\sigma }_z$. Thus the mass density of the self-gravitating isothermal slab isNeglecting radial derivatives makes the Poisson equation for Newtonian gravity ${\Phi }_{zz}=4\pi G\rho (z)$. Substituting $\Phi ={\sigma }_z^2\varphi$ and $z=z_0\zeta$, with $z_0^2={\sigma }_z^2/(8\pi G{\rho }_0)$, givesMultiply by ${\varphi }^{\prime }$ and integrate. The central value and symmetry under $z\mapsto -z$ fix the constant of integration:On the positive-$\zeta$ side, Newtonian gravity makes ${\varphi }^{\prime }\ge 0$. Put $y=e^{\varphi /2}$; then $4(y^{\prime }{)}^2=y^2-1$ with $y(0)=1$. Integration gives $2\mathrm{arcosh}y=\zeta$, and even reflection supplies the other side. ThereforeDirect differentiation verifies both the equation and central conditions. Uniqueness of the smooth initial value problem rules out other branches meeting those conditions. The surface density is $4{\rho }_0{z}_0$, and the conventional slab scale height is $2z_0$; the factor of two is only a definition of the length scale.

The Wronskian satisfies Abel identity. Differentiating and substituting the homogeneous equations givesso $W(x)=-\exp [-{\int }_0^{x}p(r)dr]\ne 0$. The two homogeneous solutions are therefore independent.

Define the variation of parameters kernelFor fixed $s$, it satisfies the homogeneous differential equation in $x$. Direct evaluation gives $K(s,s)=0$ and $K_x(s,s)=1$. For $y(x)={\int }_0^{x}K(x,s)f(s)ds$, differentiation under the integral therefore yieldsCombining these identities proves $y^{\prime \prime }+p{y}^{\prime }+qy=f$; both initial values vanish. Continuity of the coefficients gives uniqueness of this initial value problem, establishing the proposed formula rather than merely recognizing it as a standard method.

For the oscillator, $y_1=\sin x$, $y_2=\cos x$ and $W=-1$, so the kernel is $\sin (x-s)$. HenceUsing ${\int }_0^x\cos s\sin sds={\sin }^{2}x/2$ and ${\int }_0^x{\sin }^{2}sds=x/2-\sin (2x)/4$ givesThe factor $x\cos x$ reflects resonance with the homogeneous oscillator frequency.

The Euler-Lagrange equation is obtained by varying the scalar field, integrating the derivative terms by parts, and requiring the coefficient of an arbitrary compactly supported variation to vanish:The canonical momentum is $\pi ={\varphi }_t$, so the Hamiltonian density and conserved energy areIndeed,Thus $dE/dt=[{\varphi }_t{\varphi }_x{]}_{-\infty }^{+\infty }=0$ for the finite-energy boundary conditions with no flux at infinity. The scalar-field vacua are the constant fields $2\pi m$, $m\in \mathbb{Z}$.

For a static field, multiply ${\varphi }^{\prime \prime }=\sin \varphi$ by ${\varphi }^{\prime }$ to obtain $\frac{1}{2}{\varphi }^{\prime 2}+\cos \varphi =C$. The vacuum limits fix $C=1$, so on the increasing interval $0<\varphi <2\pi$,ConsequentlyIts derivative is $2\mathrm{sech}(x-a)$, and its energy is $\int 4{\mathrm{sech}}^2(x-a)dx=8$. The increasing sign cannot change inside $(0,2\pi )$ because the derivative does not vanish there. This proves uniqueness up to translation for the prescribed boundary conditions.

There are continuously many translated static solutions with those same endpoints, but only one modulo translations. In the full theory, the adjacent-vacuum kinks and antikinks areTheir endpoints are $2\pi m$ and $2\pi (m+\sigma )$. Thus there are countably many adjacent-vacuum boundary sectors, each with a continuous center parameter; integer field shifts and reflection relate their shapes. A Lorentz boost gives moving kinks, replacing $x-a$ by $(x-vt-a)/\sqrt{1-v^2}$, with energy $8/\sqrt{1-v^2}$. There is no finite static kink joining nonadjacent scalar-field vacua: reaching an intermediate vacuum at finite $x$ would give both the equilibrium value and zero derivative, forcing a constant solution by uniqueness of the second-order initial value problem. Dynamical multisolitons are a different matter.

The printed Sine-Gordon Bäcklund transformation has a normalization error. With precisely the printed coordinates $\tau =x+t$, $\rho =x-t$,is the field equation. Put $s=({\varphi }_1+{\varphi }_0)/2$, $d=({\varphi }_1-{\varphi }_0)/2$. The printed relations give $d_{\rho }=b\sin s$ and $s_{\tau }=b^{-1}\sin d$, henceAdding and subtracting yieldsThese are the equations with mass squared four, not the unit-mass equation derived from the Lagrangian. A concrete counterexample is the seed ${\varphi }_0=0$, parameter $b=1$, and transformed field ${\varphi }_1=4\arctan e^{2x}$. It obeys both printed first-order relations but satisfies ${\varphi }_{1,xx}=4\sin {\varphi }_1$, so its unit-mass field-equation residual is $-3\sin {\varphi }_1$, which is not identically zero. The requested preservation claim is false with the printed normalization.

There are two equivalent repairs. Keep the printed first-order coefficients but take $\tau =(x+t)/2$, $\rho =(x-t)/2$; then ${\partial }_{\tau }{\partial }_{\rho }={\partial }_x^2-{\partial }_t^2$ and the preceding calculation proves preservation. Alternatively keep the printed coordinates and replace the coefficients $2b,2b^{-1}$ by $b,b^{-1}$. Then $d_{\rho }=(b/2)\sin s$, $s_{\tau }=(b^{-1}/2)\sin d$, and the differentiated equations carry a factor $1/4$, giving exactly ${\varphi }_{j,\tau \rho }=\frac{1}{4}\sin {\varphi }_j$. This is the light-cone normalization of the sine-Gordon Bäcklund transformation. For a zero seed it gives ${\varphi }_1=4\arctan \exp [(b\rho +b^{-1}\tau )/2]$; $b=1$ recovers the unit-width static kink.

Use normal modes $e^{ikx+st}$, so $s=-i\omega$ in the convention $e^{ikx-i\omega t}$. For a translation-invariant, well-posed linear initial value problem on the whole line, localized data are propagated by a causal Green function. The Fourier transform in space and Laplace transform in time give an inverse integral whose denominator is the dispersion relation $D(k,s)$. The initial temporal contour is a Bromwich line to the right of every real-wave-number temporal singularity.

The Briggs-Bers technique lowers that temporal contour while deforming the spatial contour to avoid its poles. A spatial pinch point occurs when two spatial branches originating on opposite sides of the original Fourier contour collide and prevent further deformation. Algebraically a simple candidate satisfies $D=D_k=0$, with $D_s\ne 0$ and $D_{kk}\ne 0$; these equations alone do not establish the pinch topology. A same-side collision is a false spatial saddle in quartic dispersion. The opposite-side requirement is also explicit in the primary numerical study of pinch-point detection.

The rightmost contributing pinch determines the fixed-position exponential growth of the impulse response, ordinarily with a $t^{-1/2}$ saddle point prefactor. A positive rate is absolute wave-packet instability. If real-wave-number normal modes grow but the impulse response decays at each fixed position, it is convective wave-packet instability; growth can then occur along moving rays. If no temporal normal mode grows, the system is stable, with equality thresholds treated as marginal rather than exponential decay. Along a ray $x=vt$, apply the same argument to $s(k)+ikv$, whose saddle point condition is $s^{\prime }(k)+iv=0$. Thus the laboratory frame is part of the absolute/convective definition.

For the quartic problem the temporal dispersion relation isWrite $a=\mathrm{Re}{a}_1$, $b=\mathrm{Re}{a}_2$, $c=\mathrm{Re}{a}_3$. For $b>0$ the causal Fourier kernel is absolutely convergent for positive time and the finite maximum temporal growth rate isThe usual initial-value framework fails if $b<0$, or if $b=0$ and $a>0$, because arbitrarily large real wave numbers grow arbitrarily fast: no finite Bromwich line can start to their right. This is a high-frequency ill-posedness obstruction, not an absolute instability criterion. If $b=0$ and $a\le 0$, the growth bound is finite, but the dispersive limiting case needs oscillatory Fourier inversion or a limiting-absorption construction when there is no quadratic damping. It must not be justified by the absolutely convergent quartic-diffusion argument. The following explicit calculation first assumes $b>0$ and $a_2\ne 0$.

The double-root candidates areRather than assume all are physical, compute the causal resolvent at the impulse location. Put $S=s+a_3$, and for large $\mathrm{Re}s$ factorThen ${\kappa }_1{\kappa }_2=\sqrt{S/a_2}$ and $({\kappa }_1+{\kappa }_2{)}^2=-a_1/a_2+2\sqrt{S/a_2}$, with sheets selected from that large-$s$ region. Partial fraction decomposition and ${\int }_{\mathbb{R}}(k^2+{\kappa }^2{)}^{-1}dk=\pi /\kappa$ give the quartic impulse-response Laplace resolventAll roots in this expression are continued from the causal region, not selected afresh at a candidate collision. The two possible singularities are now transparent. $S=0$ is a square-root singularity when $a_1\ne 0$. The extra singularity requires $\sqrt{S}=a_1/(2\sqrt{a_2})$. Taking $\sqrt{a_2}$ with positive real part and putting $z=a_1/\sqrt{a_2}$, this lies on the physical $\sqrt{S}$ sheet precisely when $\mathrm{Re}z>0$. If that real part is zero the candidate is on a cut boundary and has nonpositive real $S$, so it cannot lie to the right of $S=0$. For $\mathrm{Re}z<0$, the apparently growing candidate may be on the wrong sheet altogether.

Therefore the physical-sheet growth rate of a quartic impulse response isThese are resolvent branch singularities on the causal continuation, so the formula implements the pinch test instead of merely listing formal saddle point rates. For $a_1=0$, the zero saddle point is fourth-order and the resolvent is proportional to $S^{-3/4}$; its impulse response is proportional to $t^{-1/4}{e}^{-a_{3}t}$ and the same exponential-rate formula applies. At nondegenerate candidates the prefactor is $t^{-1/2}$.

The classification in this diffusive case is now explicit: ${\sigma }_T<0$ gives stability; ${\sigma }_A>0$ gives absolute instability; ${\sigma }_T>0$ with ${\sigma }_A<0$ gives convective instability. When ${\sigma }_T>0$ and ${\sigma }_A=0$, the nondegenerate fixed-position response still decays algebraically, at the marginal absolute/convective threshold. Other equality cases require the stated saddle point prefactors. Temporal instability supplies a growing ray: at a real maximum of $\mathrm{Re}s(k)$, choose $v=-\mathrm{Im}{s}^{\prime }(k)$ so that the moving-ray saddle point is stationary there.

Three examples show why complex coefficients and sheets matter. For $(a_1,a_2,a_3)=(2,1,1/2)$, ${\sigma }_T={\sigma }_A=1/2$, so the instability is absolute. For $(1+2i,1,1/10)$, ${\sigma }_T=3/20>0$ but ${\sigma }_A=-1/10$, so it is convective even though there is no explicit first-derivative advection term: dispersion carries packets in opposite directions. For $(-2,1,1/2)$, both physical rates are $-1/2$. The formal candidates $k=\pm i$ have $s=1/2$ but fail the physical-sheet condition, so this temporally stable example is not absolutely unstable.

On the admissible boundary $b=0$, $a\le 0$, $a_2\ne 0$, the oscillatory or damped Fourier construction has ${\sigma }_T=-c$. The surviving zero-wave-number stationary response has the same exponential rate $-c$, with a quadratic or quartic algebraic factor; damping/dispersion therefore does not generate a separate convective-growth range there. If $a_2=0$, the problem is second order and its diffusion/dispersive kernel must be used instead of the quartic formula. Application to a finite interval or nonlocalized data also changes the spatial inversion problem, so the whole-line localized-data hypothesis is essential.

The space of test functions is $\mathcal{D}(\mathbb{R})=C_c^{\infty }(\mathbb{R})$: its elements are smooth functions with compact support. For a fixed compact set $K$, put ${\mathcal{D}}_K=\{\phi \in C^{\infty }(\mathbb{R}):\mathrm{supp}\phi \subset K\}$ and use the seminorms $p_m(\phi )={\max }_{0\le j\le m}\Vert {\phi }^{(j)}{\Vert }_{\infty }$. The space of test functions carries the usual test-function inductive limit topology of these spaces. In particular, a sequence converges precisely when its supports eventually lie in one compact set and every derivative converges uniformly. Thus ${\phi }_j\to \phi$ means that a common $K$ contains their supports and $p_m({\phi }_j-\phi )\to 0$ for every $m$.

A distribution is a continuous linear functional on this space of test functions, and ${\mathcal{D}}^{\prime }(\mathbb{R})$ denotes their space. We use complex-linear, bilinear pairings $⟨u,\phi ⟩$. Equivalently, for every compact set $K$ there are $C_K$ and a nonnegative integer $m_K$ such thatThe usual weak convergence of distributions is $u_j\to u$ if $⟨u_j,\phi ⟩\to ⟨u,\phi ⟩$ for every test function. This specifies the convergence used below; it does not require convergence in any norm.

The distributional derivative is defined byThe derivative map sends ${\mathcal{D}}_K$ continuously to itself, with $p_m({\phi }^{\prime })\le p_{m+1}(\phi )$. The preceding continuity estimate therefore proves that $u^{\prime }$ is again a distribution. This definition extends the ordinary derivative of a smooth function, by integration by parts.

Choose the translation of a distribution convention ${\tau }_{h}f(x)=f(x-h)$. Its action on a test function isFor fixed $h$, the translated test functions have translated compact support and unchanged derivative sup norms, so this defines a distribution. For the differentiability of distribution translations, apply the Taylor theorem in integral form:All supports lie in one slightly enlarged compact set for $|h|\le 1$, and the same identity for every derivative proves uniform convergence. The continuity of $u$ now givesThe minus sign in the translation parameter is necessary for this convention.

For (a), choose a test function $\eta$ with $\int \eta =1$. If $\int \psi =0$, then $F(x)={\int }_{-\infty }^x\psi (s)ds$ is a test function, since the zero integral makes it vanish beyond both ends of the compact support. If $u_1^{\prime }=0$, then $⟨u_1,\psi ⟩=⟨u_1,F^{\prime }⟩=0$. Decompose $\phi =(\int \phi )\eta +\psi$ to obtainConversely, these constant regular distributions have zero distributional derivative. This also proves the general fact that a distribution with zero derivative is constant.

For (b), choose a cutoff function $\rho$ equal to one near zero. Every test function decomposes asThe quotient extends as a smooth function at zero, and it has compact support. If $x{u}_2=0$, the definition of multiplication of a distribution by a smooth function gives $⟨u_2,\phi ⟩=⟨u_2,\rho ⟩\phi (0)$. Conversely, $x{\delta }_0=0$. Thus the kernel of multiplication by a coordinate is exactlyIn particular, derivatives of the Dirac delta distribution are not additional solutions: $x{\delta }_0^{\prime }=-{\delta }_0$.

Now write $D=d/dx$ and $P(z)=-z^n+{\sum }_{j=0}^{n-1}{a}_j{z}^j$, where $n\ge 1$. The characteristic roots of a constant-coefficient differential equation are the distinct roots of a polynomial ${\lambda }_1,\dots ,{\lambda }_r$ of $P$, with multiplicities $m_1,\dots ,m_r$. The distributional regularity of a constant-coefficient ordinary differential equation can be proved without assuming regularity in advance. Put $q_j(z)=(z-{\lambda }_j{)}^{m_j}$ and $Q={\prod }_j{q}_j=-P$. Since the $q_j$ are coprime polynomials, polynomial division and Bezout identity give polynomials $b_j$ such thatFor example, invert $Q/q_j$ modulo $q_j$, sum the resulting expressions, and absorb a remaining multiple of $Q$ into one coefficient. For $P(D)v=0$, the kernel decomposition for coprime polynomials therefore givesThe Leibniz rule for multiplication of a distribution by a smooth function implies $D^{m_j}(e^{-{\lambda }_{j}x}{v}_j)=0$. Repeatedly using a distribution with zero derivative is constant shows that a distribution with $m$th derivative zero is a polynomial of degree at most $m-1$: subtract the polynomial primitive of its constant $(m-1)$st derivative, and induct on $m$. Consequently the most general exponential polynomial solution of a constant-coefficient differential equation isEvery displayed term is annihilated by $P(D)$, so all coefficients are allowed. For the linear independence of these $n$ functions, on a relation, apply ${\prod }_{i\ne j}(D-{\lambda }_i{)}^{m_i}$; on $e^{{\lambda }_{j}x}$ times a polynomial of degree less than $m_j$, each remaining factor acts invertibly on that polynomial space. Hence the $j$th polynomial must vanish. Every distributional solution is an analytic function, and in particular a classical solution. For real coefficients and real-valued distributions, take complex conjugate coefficients at conjugate characteristic roots of a constant-coefficient differential equation, or equivalently use the corresponding real sine and cosine forms.

For the last equation, the operator is $xP(D)$: the coordinate multiplies the result of differentiation. The kernel of multiplication by a coordinate says precisely thatChoose the retarded fundamental solution of a constant-coefficient ordinary differential operator $E=Hw$, where $H$ is the Heaviside function and $w$ is the analytic solution of $P(D)w=0$ withSuch $w$ exists uniquely by the elementary initial value problem for a constant-coefficient ordinary differential equation; alternatively the residue construction in the next solution gives it explicitly. The distributional jump formula for a Heaviside product isIt follows by induction from $D(Hw)=H{w}^{\prime }+w(0){\delta }_0$, using integration by parts. With the chosen initial derivatives, every lower-order jump term vanishes and the leading coefficient $-1$ gives $P(D)E={\delta }_0$. Subtracting $CE$ reduces the last equation to the homogeneous one. Thus the coordinate-degenerate constant-coefficient differential equation has exactly the solutionsThere are $n+1$ independent constants. For $n\ge 2$, derivatives through order $n-2$ match at zero, while the derivative of order $n-1$ may jump; for $n=1$, the function itself may jump. An arbitrary distribution concentrated at zero cannot be added, because its image under the nonzero leading derivative would contain a nonvanishing highest derivative of the Dirac delta distribution.

The small root and the root near $-2$ are regular perturbation roots: they stay finite as the parameter vanishes. Substitution of power series givesFor the third root, dominant balance for algebraic roots requires $\lambda =\mu /\epsilon$. The largest terms give ${\mu }^2(\mu +1)=0$, and the missing branch is $\mu =-1$. The sum of the three roots then yieldsThus the leading roots are $-\epsilon /2$, $-2$, and $-1/\epsilon$; only the last is a divergent perturbation root. The first regular branch happens to have a zero limit, so retaining its first nonzero term is essential on the long spatial scale.

First take $\epsilon \to 0^{+}$, the decaying half-line regime. If the three exact roots are $r_j$, the exact initial value problem solution isThis follows either by solving the three initial-value equations or from the Laplace transform $\hat{y}(s)=\epsilon /(\epsilon s^3+s^2+2s+\epsilon )$. The coefficient formula has sums $\sum A_j=\sum r_j{A}_j=0$ and $\sum r_j^2{A}_j=1$.

Retain the three leading roots but compute their coefficients without expanding their denominators. Set $a=\epsilon /2$, $b=2$, $d=1/\epsilon$. A convenient composite asymptotic expansion isIt satisfies all three initial conditions exactly and retains the fast transient, the ordinary decay, and the slow decay. Its first two coefficients are $\epsilon /2+O({\epsilon }^2)$ and $-\epsilon /2+O({\epsilon }^2)$; the fast coefficient is ${\epsilon }^2+O({\epsilon }^3)$. Consequently the simpler bulk expression is $\frac{\epsilon }{2}(e^{-\epsilon x/2}-e^{-2x})$, but that expression alone does not reproduce the initial derivative layer.

The absolute error estimate isTo see its uniformity, the slow-root error is $O({\epsilon }^2)$, while its magnitude is $O(\epsilon )$ and its coefficient is $O(\epsilon )$. The uniform error bound for nearby decaying exponentials therefore gives an $O({\epsilon }^2)$ contribution even for $x=O({\epsilon }^{-1})$. The middle-root error is $O(\epsilon )$ with coefficient $O(\epsilon )$, again giving $O({\epsilon }^2)$. The fast-root error is $O(1)$ with magnitude $O({\epsilon }^{-1})$ and coefficient $O({\epsilon }^2)$, giving only $O({\epsilon }^3)$. Coefficient errors are $O({\epsilon }^2)$ or smaller. This estimate concerns $y$ itself, rather than asserting the same uniform order for all derivatives or a relative error at its zero.

For the sketch, on $x=\epsilon X$ one obtainsThus $y\sim x^2/2$ at the origin and rises smoothly after the fast layer. For $x=O(1)$ it rises towards a plateau of height approximately $\epsilon /2$, then decays on the much longer scale $x=O({\epsilon }^{-1})$. The bulk maximum lies near $x=\log (4/\epsilon )/(2-\epsilon /2)$ and has height asymptotic to $\epsilon /2$.

The sign of the parameter matters on an infinite interval. If $\epsilon \to 0^{-}$ is allowed, the singular mode grows rather than decays. For each fixed $x>0$, its dominant contribution is $y(x)\sim {\epsilon }^2\exp [(-1/\epsilon +2)x]$. The extra $2$ in that exponent is needed for relative leading accuracy at fixed $x$. The positive-parameter uniform absolute bound and decaying sketch do not extend to that regime.

Let ${\gamma }_X(s)$ be the affinely parametrized geodesic of the affine connection with ${\gamma }_X(0)=p$ and ${\dot{\gamma }}_X(0)=X$. For fixed $t$, the curve $\eta (u)={\gamma }_X(tu)$ has initial velocity $tX$. The geodesic equation is homogeneous of degree two in the velocity, so its left-hand side for $\eta$ is $t^2$ times the left-hand side for ${\gamma }_X$, and is zero. Uniqueness of the geodesic initial value problem gives $\eta ={\gamma }_{tX}$.

By the definition of the affine exponential map, rescaling the initial velocity rescales affine parameter:At $t=0$ both sides equal $p$. The statement is made where the geodesics exist; without geodesic completeness, the affine exponential map is defined only on a suitable neighbourhood of zero in $T_{p}M$, not necessarily all of $T_{p}M$. An affine parameter distance need not be a metric length.

Consider an initial value problem $y^{\prime }=f(t,y)$ on $[0,T]$, with a Lipschitz continuous vector field in the solution variable and enough smoothness for the stated error expansions. Convergence of a numerical method meanswith consistent starting values. An order of a numerical method $p$ quantifies this approximation: with sufficiently accurate starting data and the required stability, the global error is $O(h^p)$. Stability of a numerical method controls the amplification and accumulation of perturbations; it is the link between a small local defect and small global error.

For a one-step formula $Y_{n+1}={\Phi }_h(t_n,Y_n)$, define the unscaled local truncation errorThe method has local order $p$ if $\Vert d_{n+1}\Vert \le C{h}^{p+1}$ uniformly along sufficiently smooth solutions. If the defect is divided by $h$, its order is instead $O(h^p)$; the convention must be specified. Assume also the perturbation estimateThe global error $e_n=Y_n-y(t_n)$ then satisfiesIterating, or applying the discrete Gronwall inequality, givesThere are $O(1/h)$ steps on a fixed interval, so stable propagation turns $O(h^{p+1})$ local defects into $O(h^p)$ accumulated error. Starting error $O(h^p)$ is needed to retain order $p$. This explains why matching an exact Taylor expansion is insufficient unless errors are controlled during repeated steps.

For a fixed-coefficient linear multistep method, writeThe consistency of a numerical method conditions are $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)$. The higher order conditions for a linear multistep method arewith right side zero for $q=0$; exact order $p$ means first failure at $q=p+1$. They come from inserting the exact solution and comparing its Taylor expansion.

Errors at $h=0$ follow the homogeneous recurrence determined by $\rho$. Zero-stability is equivalent to the root condition for a multistep method: all roots of a polynomial have modulus at most one, and every unit-modulus root is simple. Roots outside the disk give exponential growth in the number of steps; repeated unit roots give polynomial growth in that number. Repeated interior roots are harmless because their polynomial factors are dominated by decay. Under the usual Lipschitz continuity and initialization assumptions, the Dahlquist equivalence theorem givesFor a zero-stable order-$p$ linear multistep method, local defects and starting errors of the stated orders yield global error $O(h^p)$ by a recurrence bound and the discrete Gronwall inequality. Every required starting value matters.

A concrete failure of “consistency implies convergence” isHere $\rho =(\zeta -1)(\zeta -2)$ and $\sigma =-1$, so $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)=-1$: the method is consistent, of order one. But its root $2$ violates zero-stability. For $y^{\prime }=0$ with exact solution zero and starting values $Y_0=0$, $Y_1=h^2$, the computed solution is $Y_n=h^2(2^n-1)$. These starting errors vanish, yet at $n\approx T/h$ the error diverges. Local accuracy cannot compensate for an unstable recurrence.

A different meaning of stability concerns a fixed step on decaying equations. The Dahlquist test equation $y^{\prime }=\lambda y$ introduces $z=h\lambda$. A one-step method gives $Y_{n+1}=R(z)Y_n$ and is absolutely stable when $|R(z)|\le 1$. For a linear multistep method, the corresponding amplification polynomial of a multistep method is $\rho (\zeta )-z\sigma (\zeta )$, and every root must satisfy the bounded root condition, not just the root approximating $e^z$. Strict asymptotic decay requires the roots to have modulus less than one for $\mathrm{Re}z<0$. This distinction matters at reducible methods with an undamped parasitic mode, as in question 1(c).

The Forward Euler method has order one and $R(z)=1+z$, so its linear stability domain is $|1+z|\le 1$. For $y^{\prime }=-\kappa y$, with $\kappa >0$, it requires $0\le h\kappa \le 2$. It converges as $h\to 0$, yet an excessively large step can produce growing oscillations on an exactly decaying problem. The Backward Euler method has the same order, but $R(z)=(1-z{)}^{-1}$ makes it A-stable and L-stable. For a stiff differential equation, fast decay can impose a severe explicit time-step restriction even when accuracy of the slow dynamics would permit a much larger step.

The trapezoidal rule has order two and is A-stable, but is not L-stable, since its stability function tends to $-1$ along the negative real axis. The Second Dahlquist barrier limits an irreducible A-stable linear multistep method to order at most two. This does not limit implicit Runge-Kutta methods: question 4's Lobatto IIIA method is fourth-order and A-stable. Conversely, question 1's fourth-order member is convergent but not A-stable. These examples show that order of a numerical method, zero-stability and A-stability answer different questions.

For nonlinear dissipative problems, B-stability controls distances between numerical solutions; algebraic stability of a Runge-Kutta method is a sufficient coefficient criterion. Question 4 shows that A-stability alone does not imply that coefficient criterion. Variable time steps require additional analysis of step ratios and error propagation, and stiff problems may require uniform-in-stiffness error estimates beyond the fixed-problem convergence result. The useful chain is local accuracy plus the appropriate perturbation bound, giving global accuracy; absolute and nonlinear stability then determine which practical time steps preserve the intended dynamics.

For a constant matrix, the matrix exponential series and its differentiated series converge uniformly on bounded time intervals, as follows by comparison with the scalar exponential in a operator norm. Termwise derivatives therefore giveHence $\mathbf{x}(t)=e^{tM}{\mathbf{x}}_0$ solves the initial value problem. For this particular matrix, direct multiplication gives $M^2=-I$. Splitting the matrix exponential into its even and odd powers provesThis is the matrix exponential when the square is minus the identity. If ${\mathbf{x}}_0=(x_0,y_0{)}^{\mathsf{T}}$, the general solution isThe constants in part (a) are $\alpha =x_0/5$ and $\beta =y_0+2x_0/5$, which makes the two forms identical.