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For an implicit Runge-Kutta method, algebraic stability means that $b_i\ge 0$ and the symmetric matrixis positive semidefinite. Here $b_1=b_2=1/2$ and $A+A^T=e{e}^T/2$, soThe method is algebraically stable. This also explains why the property matters for nonlinear problems. If a vector field satisfies $⟨f(v)-f(w),v-w⟩\le 0$, the Runge-Kutta contractivity identity for two solutions, with stage differences $D_i$ and vector-field differences $F_i$, isThe first sum is nonpositive and the second vanishes. Thus, whenever the implicit stage equations have the relevant solutions, their updates are contractive in the Hilbert space norm.

The zero Dirichlet boundary conditions permit an exact finite-interval calculation. Let $T=\mathrm{tridiag}(1,-2,1)$ on the $M$ interior grid points. The Dirichlet discrete Laplacian has an orthogonal eigenvector basisConsequently the update matrix $G=(I-rT{)}^{-1}$ has modal amplification factorsFor every physical $r\ge 0$, all denominators are at least one. Orthogonal diagonalization therefore provesuniformly in $M$, $r$ and $n$. The same bound controls initial perturbations; a forcing increment is propagated by contraction, so successive increments accumulate at most by their sum. The highest-order consistent method is unconditionally stable for all $r\ge 0$. This proof incorporates the boundary conditions, whereas a periodic Fourier mode calculation alone would not do so.

For completeness, if “range” is interpreted algebraically to include negative $r$ on a fixed grid, the exact power-stable range isIndeed, for $r<0$ every denominator is less than one, so $|g_j|\le 1$ requires $1+4r{s}_j\le -1$ for every $j$. The strongest restriction comes from $s_1$. Equality gives a simple amplification factor $-1$ and is allowed because the update is orthogonally diagonalizable. Other negative values either amplify a mode or make the update singular. This extra branch describes backward time stepping on a fixed spatial grid; it is not a positive-time diffusion discretization, and no fixed negative $r$ remains in that branch as $M\to \infty$.

For every test function $\varphi \in C_c^{\infty }({\mathbb{R}}_t\times {\mathbb{R}}_x^3\times {\mathbb{R}}_v^3)$, the distributional weak solution satisfiesThere is no initial-time boundary term because the test functions are on the whole time line. In divergence notation the flux in the seven coordinates is $(f,vf,0)$, whose divergence is ${\partial }_{t}f+v\cdot {\nabla }_{x}f$. For a general $N$-dimensional divergence the introductory flux must be vector-valued; the scalar codomain printed for $A$ is dimensionally inconsistent when $N>1$. The transport identity above has the correct flux dimensions.

A pure spin-one-half state on the equator of the Bloch sphere has $⟨Z_k⟩=0$, so $|{\alpha }_k{|}^2=|{\beta }_k{|}^2=1/2$. Its azimuthal phase does not enter the conditional environment overlap, because the interaction is diagonal in $Z_k$. Substitution in the decoherence factor givesThe Zurek spin-bath model now has a real coherence factor: positive and negative values correspond to opposite relative phases, while suppression of coherence depends on $|z|$.

If the gas can cool appreciably below the virial temperature, radiative cooling removes thermal energy and pressure support. In a dark-matter halo the gas then contracts, dissipating more energy as it falls. Efficient condensation requires the radiative gas cooling time to be short enough compared with the relevant dynamical or assembly time; merely having an available low-temperature transition does not guarantee that the gas reaches it quickly.

The collisionless dark matter cannot lose comparable energy through radiation and remains extended. Gas with appreciable conserved angular momentum stops radial collapse when rotation supports it, often forming a disk; lower-angular-momentum gas reaches a more compact central region. Cold dense gas can fragment into self-gravitating clouds and form stars if its gravitational instability overcomes remaining support. Stellar feedback subsequently reheats or expels gas and regulates the conversion. Efficient cooling enables central baryonic condensation and star formation; angular momentum and feedback determine the resulting galaxy.

The annihilation operators kill the vacuum, and their commutator givesTherefore the scale-invariant inflationary power spectrum has the superhorizon limitThe canonical mode grows as $1/|\tau |$ outside the Hubble radius, while the physical field perturbation $f/a$ freezes. A slowly varying $H$ produces an almost scale-invariant spectrum rather than exact scale invariance.

For a slowly rolling single field, fluctuations in the inflaton clock become the comoving curvature perturbation, $\mathcal{R}\simeq -H\delta \varphi /\dot{\bar{\varphi }}$ up to sign convention. Restoring the reduced Planck mass and using $ϵ={\dot{\bar{\varphi }}}^2/(2M_{\mathrm{pl}}^2{H}^2)$ givesThis conversion presumes nonzero background roll; a strictly constant test scalar in exact de Sitter spacetime does not by itself define a finite comoving curvature perturbation through this formula. The curvature perturbations seed the later density fluctuations.

When metric fluctuations and their Einstein-gravity normalization are restored, the two graviton polarizations have the same massless mode equation. The usual total primordial tensor power spectrum is ${\Delta }_t^2=2H^2/({\pi }^2{M}_{\mathrm{pl}}^2)$, so the single-field leading ratio is $r=16ϵ$. Tensor amplitudes therefore probe the inflationary energy scale, whereas scalar amplitudes also depend on the roll rate. These tensor statements explain the relevance of the mode result; they are not derived from a scalar action with metric perturbations omitted.

Promote the canonical field and momentum $\pi =f^{\prime }$ to operators satisfying the canonical commutation relation, $[\hat{f}(\tau ,\mathbf{x}),\hat{\pi }(\tau ,\mathbf{y})]=i{\delta }^{(3)}(\mathbf{x}-\mathbf{y})$. A real field has Fourier reality condition ${\hat{f}}_{\mathbf{k}}^{†}={\hat{f}}_{-\mathbf{k}}$. Its oscillator expansion isThe creation and annihilation operators add or remove excitations in the selected mode basis, withCanonical normalization requires the Wronskian $u_k{u}_k^{\ast \prime }-u_k^{\ast }{u}_k^{\prime }=i$.

For the Bunch-Davies vacuum, choose the positive-frequency short-wavelength behavior $u_k\sim e^{-ik\tau }/\sqrt{2k}$ as $k|\tau |\to \infty$. For the proposed solution, direct differentiation givesSubstitution cancels every term in $u_k^{\prime \prime }+(k^2-2/{\tau }^2)u_k$. Its Wronskian is $i$, and its large-$|k\tau |$ limit is the required flat-spacetime positive-frequency mode. ThusA term with the conjugate frequency is another allowed classical solution, but would describe a different quantum state, through a Bogoliubov transformation. Its coefficient is set to zero by the early-time vacuum condition, not by absence of a second solution to the differential equation.

Use primes for conformal-time derivatives in this question, reserving $V_{,\varphi \varphi }$ for a potential derivative. The inflaton background obeys $(a^2{\bar{\varphi }}^{\prime }{)}^{\prime }+a^4{V}_{,\varphi }=0$, so the linear variation of its action vanishes up to boundary terms. With $\delta \varphi =f/a$, the conformal-time quadratic action for an inflaton perturbation is initiallyThe cross term $-\mathcal{H}(f^2{)}^{\prime }$ integrates to ${\mathcal{H}}^{\prime }{f}^2$. Since ${\mathcal{H}}^{\prime }+{\mathcal{H}}^2=a^{\prime \prime }/a$, this becomesVarying $f$ gives $f^{\prime \prime }-{\nabla }^{2}f+(a^2{V}_{,\varphi \varphi }-a^{\prime \prime }/a)f=0$. The symmetric Fourier transform convention used in the question turns $-{\nabla }^2$ into $k^2$. Dropping the small mass term under the stated slow-roll inflation assumption yieldsFor de Sitter spacetime, $a=-1/(H\tau )$ with $\tau <0$, so $a^{\prime \prime }/a=2/{\tau }^2$. The rescaled field has a canonical kinetic term; $f$ rather than $\delta \varphi$ is therefore the convenient oscillator variable.

Take canonical charged chiral superfield kinetic terms, gauge coupling $e>0$, and no Fayet–Iliopoulos term, since none is specified. Elimination of the three complex auxiliary fields givesThe Abelian gauge auxiliary field obeys $D=-e(|{\phi }_{+}{|}^2-|{\phi }_{-}{|}^2)$ in this normalization. Thus the F-term scalar potential and the gauge D-term giveAll terms are nonnegative. The normalization of $e$ can be changed together with the vector-field normalization, but the relative charges and the zero-potential conditions cannot. A noncanonical Kähler potential would change the inverse-metric factors in the F-term potential; adding a Fayet–Iliopoulos term would shift $D$ and define a different model. Neither is silently introduced here.

Write $\sigma =\mathrm{sgn}(x_S)$ and $G=A_S^{\ast }{A}_S$. The injectivity of $A_S$ makes this Gram matrix a positive-definite matrix, since $u^{\ast }Gu=\Vert A_{S}u{\Vert }_2^2>0$ for $u\ne 0$. Thus its matrix inverse exists. Construct the least-norm dual certificateOn the active coordinates, $A_S^{\ast }h=G{G}^{-1}\sigma =\sigma$. For $l\notin S$, symmetry of the real Gram matrix and its matrix inverse givesCondition (iii) makes the absolute value of this coordinate strictly less than one. Consequently $h$ is a strict dual certificate for basis pursuit, and part (c) applies. If $S$ is empty, take $h=0$; the zero vector uniquely minimizes the L1 norm on its feasible set. Condition (iii) supplies an explicit certificate and therefore unique recovery. There is no claim that this particular least-norm dual certificate is necessary: other valid certificates may exist when this one fails.

Let $u=A^{\ast }h$ be the strict dual certificate for basis pursuit supplied by condition (ii). For $0\ne v\in \ker A$, we haveHere the adjoint operator is the real transpose. The injectivity of $A_S$ ensures $v_{S^c}\ne 0$: otherwise $A_S{v}_S=0$ would force $v=0$. Thus at least one nonzero term lies outside the support of a vector $x$. Since $|u_l|<1$ at every such index,This proves the fixed-sign null space condition, so part (a) gives uniqueness in basis pursuit. If $S^c$ is empty, the injectivity of $A_S=A$ instead means there is no nonzero null space vector, and the feasible set is a singleton. Injective active columns and a strict dual certificate for basis pursuit ensure unique recovery. Both ingredients matter: strictness outside $S$ cannot detect a nonzero null space direction supported entirely inside $S$.

Assume $x$ is the unique basis pursuit minimizer. Fix $0\ne v\in \ker A$ and setThere is $\epsilon >0$ such that the sign function of $x_j+t{v}_j$ equals that of $x_j$ at every $j\in S$ whenever $|t|<\epsilon$. To choose it, take less than the minimum of $|x_j|/|v_j|$ over the nonzero $v_j$ in $S$; if that set is empty, any positive $\epsilon$ works. On this interval the L1 norm has the exact expressionFor $0<t<\epsilon$, both $x+tv$ and $x-tv$ are distinct feasible vectors. Uniqueness forces their objective differences to be strictly positive, so $\alpha +\beta >0$ and $-\alpha +\beta >0$. ThereforeThe fixed-sign null space condition is necessary as well as sufficient. This argument also covers an empty support of a vector: then $\alpha =0$ and the nonzero null space vector has $\beta >0$. Strictness is indispensable: equality would make a sufficiently short feasible segment have the same objective as $x$.

Let $\sigma =\mathrm{sgn}(x_S)$, using the real sign function on the support of a vector $x$. The fixed-sign null space condition in the PDF uses $S^c$ on its right-hand side. This complement is essential. For any real coordinate $x_j\ne 0$, the supporting-line inequality for the absolute value isEvery distinct feasible vector is $x+v$ with $0\ne v\in \ker A$. Summing the coordinate inequalities on $S$ and adding the L1 norm on $S^c$ givesThe strict final inequality is precisely the fixed-sign null space condition. Hence $x$ is the unique basis pursuit minimizer. Unlike the null space property of Question 1, this condition concerns the particular sign function values of $x$, rather than every vector supported in $S$.

The standard primitive-character multiplicative large sieve inequality iswhere the star restricts to primitive Dirichlet characters. This is the form used in analytic arguments for Linnik's theorem. The prime-power Gauss identities extend to arbitrary primitive conductors by the Chinese remainder theorem. Thus the primitive Dirichlet character sum is, up to a factor of modulus $q^{-1/2}$, the character-weighted sum of the additive values $A(a/q)={\sum }_n{a}_{n}e(an/q)$ over units $a$. Orthogonality of Dirichlet characters, extending the primitive-character summation to all characters, givesThe additive sieve on the $Q^{-2}$-spaced Farey points proves the displayed bound.

For both prime-interval applications, use the following large sieve upper bound for sifted intervals. Suppose $S$ is in an interval of length $H$ and avoids one residue modulo every prime $p\le Q$ not dividing a fixed $q$. ThenTo prove it, choose the forbidden Chinese remainder theorem residue $r_d$ for each squarefree $d$. The Ramanujan sum $c_d(n-r_d)$ equals $\mu (d)$ on $S$, since $n-r_d$ is a unit modulo $d$. Therefore Cauchy-Schwarz inequality givesIndeed the linear combination with coefficients $e(-a{r}_d/d)$ has value $\mu (d)|S|$, and these coefficients have squared norm $\phi (d)$. Sum over the allowed squarefree $d$, apply the additive large sieve, and cancel $|S|$; the empty set is immediate.

Finally ${\mathcal{L}}_1(Q)\ge c\log (2Q)$. Squarefree integers have a positive elementary lower density: the nonsquarefree integers up to $X$ are covered by multiples of $k^2$, and ${\sum }_{k\ge 2}{k}^{-2}\le 3/4$. Partial summation turns this density into the harmonic lower bound. Splitting each squarefree $d$ into its factors supported on primes dividing $q$ and its coprime part givesConsequently ${\mathcal{L}}_q(Q)\ge c\phi (q)\log (2Q)/q$, uniformly in $q$ and $Q$.

Under the Riemann hypothesis, every $\beta =1/2$, so for $\sigma >1/2$ the absolutely convergent formula givesThe zero set is nonempty: otherwise Hadamard factorization would make $\xi$ an exponential of a linear polynomial, and its functional symmetry would force it to be constant, contrary to gamma growth on the positive real axis. Thus the inequality is strict in the open right half-plane. Continuity at $\sigma =1/2$, including at boundary zeros, proves the claimed increasing modulus on the closed half-line.

Conversely, suppose the modulus is nondecreasing for every fixed $t$. If a zero $\beta +i\gamma$ had $\beta >1/2$, then nonnegativity and monotonicity would force $|\xi (\sigma +i\gamma )|=0$ throughout $1/2\le \sigma \le \beta$. The identity theorem would make $\xi$ identically zero, a contradiction. A zero left of the line reflects to one right of the line by the functional equation and complex conjugation symmetry. Hence every zero lies on the critical line. This proves the xi modulus criterion for the Riemann hypothesis. The two following roman headers refer to supplied asymptotic assumptions, not further questions, and require no Solution sections.