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The independent dimensionless variable is $S=s/s_{\infty }$. Since $g{Z}_i=s_{\infty }G{Z}_i$ and $dS/ds=1/s_{\infty }$, the chain rule gives $d(g{Z}_i)/ds=d(G{Z}_i)/dS$. The derivative in the normalized equation must consequently be with respect to $S$. Substituting the gas history in the retained-ejecta chemical evolution equation yieldsThis step allows a varying retention fraction $f(S)$; no derivative of $f$ is needed because $f$ enters the instantaneous metal source and loss terms.

Set $H=G{Z}_i$ and now take $f=S$. The coefficient in this first-order ordinary differential equation decomposes asHence an integrating factor is $\mu (S)=S^{1/(1-\beta )}/(1-S)$. With $q=1+1/(1-\beta )=(2-\beta )/(1-\beta )$, multiplication by this integrating factor givesThe general solution has an additional term $C/S^q$. Finite initial gas-phase metallicity rules it out. Consequently, for $0<S<1$,The upper limit is the dimensionless $S$, not the unnormalized stellar mass. On each compact subinterval of $0\le S<1$, the geometric series $1/(1-u)={\sum }_{k\ge 0}{u}^k$ converges uniformly, permitting integration term by term. ThusIn particular $Z_i\sim y_{i}S/(q+1)$ at small $S$. For positive stellar yield, the series is strictly increasing, so the inverse needed for the metallicity distribution function exists. Differentiating either representation gives $Z_i^{\prime }=y_i/(1-S)-q{Z}_i/S>0$. It diverges logarithmically as gas exhaustion approaches; the trace-metal approximation eventually becomes inadequate.

The quadratic gas-history enrichment model is not a closed gas reservoir: its gas mass initially increases. This can be made consistent with the metal equation by pristine galactic gas inflow. Indeed gas accounting gives $dg/ds=-(1-\beta f)/(1-\beta )+d{M}_{\mathrm{in}}/ds$. Since $dg/ds=1-2S$ and $f=S$, the required inflow is $d{M}_{\mathrm{in}}/ds=q(1-S)\ge 0$. It adds no metals and therefore leaves the preceding derivation intact.

For the specified affine linear PTT fluid, perturb about the relaxed state $A=0$, $\mathbf{v}=0$. Products of $A$ with the velocity gradient and the term $\alpha (\mathrm{tr}A)A$ are second order in the perturbation. The material-advection term is also second order. The linearized equation is thereforeWith vanishing stress in the remote past, an integrating factor givesSince ${\sigma }^{\prime }=G_{0}A$, comparison with the linear viscoelastic fluid convolution yields $G(s)=G_0{e}^{-s/\tau }$ and zero-shear viscosity $G_0\tau$. The nonlinear relaxation parameter does not appear at linear order; the response is that of a Maxwell fluid.

The basic quasi-geostrophic streamfunction $\bar{\psi }=-\Lambda yz$ gives $\bar{u}=\Lambda z$, $\bar{v}=0$ and constant quasi-geostrophic potential vorticity $\bar{Q}=f$. DefineBecause the background has no potential-vorticity gradient, the linear potential-vorticity equation is simplyIts characteristics have constant $y,z,x-\Lambda zt$, so the general solution is $q^{\prime }=\mathcal{F}(y,z,x-\Lambda zt)$. In particular, $q^{\prime }=F(y,z)G(x-\Lambda zt)$ is the requested arbitrary separated family; one product is not the most general initial field.

Linearizing the vertical velocity requires retaining advection of ${\bar{\psi }}_z=-\Lambda y$ by $v^{\prime }={\psi }_x^{\prime }$. ThereforeThe impermeable boundary at $z=0$ imposesThis is conservation of boundary buoyancy, not merely a condition that ${\psi }_z^{\prime }$ vanish.

For zero interior potential vorticity, write ${\psi }^{\prime }=\mathrm{Re}\{\varphi (z,t)e^{i(kx+ly)}\}$. With $K_h=\sqrt{k^2+l^2}>0$, stratified quasi-geostrophic inversion givesThe decaying solution isNo time dependence has been assumed to obtain this vertical structure. Applying the boundary condition gives $-\mu \dot{A}-i\Lambda kA=0$, henceThus the Eady edge wave travels eastward with horizontal phase velocity $\omega /k=\Lambda /\mu$ for $k\ne 0$.

The propagation mechanism can be made explicit using boundary parcels. Thermal wind gives the background buoyancy gradient ${\bar{b}}_y=f{\bar{\psi }}_{zy}=-f\Lambda$. A small meridional parcel displacement $d$ produces boundary buoyancy perturbation $b^{\prime }=-d{\bar{b}}_y=f\Lambda d$. Meanwhile the decaying potential-vorticity inversion gives $b^{\prime }=f{\psi }_z^{\prime }=-f\mu {\psi }^{\prime }$ at the boundary. Consequently ${\psi }^{\prime }=-\Lambda d/\mu$ there, and the parcel equation $d_t=v^{\prime }={\psi }_x^{\prime }$ becomesA displaced buoyancy contour therefore induces a velocity field that advances that displacement eastward. No interior potential-vorticity gradient is needed because the boundary buoyancy supplies the invertibility data.

Vortex stretching is what sets the induced flow's vertical reach. In the interior, the zero-potential vorticity condition requiresBoundary isopycnal displacement decreases with height as $e^{-\mu z}$, so fluid columns are stretched or compressed over depth ${\mu }^{-1}=f/(N{K}_h)$. Conservation of potential vorticity makes this stretching balance the relative vorticity. The resulting inversion factor $b^{\prime }=-f\mu {\psi }^{\prime }$ gives the propagation speed: stronger density stratification reduces penetration and speed, while larger $f$ increases them. Zero interior potential vorticity therefore does not imply zero interior flow.

For imposed boundary vertical velocity, retain the same decaying structure and writeThe boundary relation becomes the forced Eady edge wave equationThe integrating factor $e^{i\omega t}$ gives, for ${\omega }_0\ne \omega$,The arbitrary homogeneous coefficient $B$ sets the initial disturbance. With $A(0)=0$, this is $A=C(e^{-i{\omega }_{0}t}-e^{-i\omega t})/[i(\omega -{\omega }_0)]$.

At resonance, ${\omega }_0=\omega$, integration instead yieldsAn initially undisturbed fluid has ${\psi }^{\prime }=Ct{e}^{-\mu z}\cos (kx+ly-\omega t)$. The imposed pumping reinforces the same boundary normal mode every cycle, with no dissipative loss, so the linear amplitude grows secularly. The divergence describes resonance of the undamped linear model; sufficiently large wave amplitude invalidates the assumed small-disturbance approximation.

With its material state fixed by a prescribed deformation history, the differential pom-pom model is a simple-fluid constitutive law. The stress depends on the local history of a material element through its molecular stretch and orientation, without gradients of that history between neighbouring elements. A differential representation does not prevent a material from being a simple fluid.

To see the history dependence explicitly, use the component-first velocity gradient $K$, so the tensor evolution is $D_{t}A=KA+A{K}^T-(A-I)/{\tau }_1$. Let $F(t,s)$ be the relative deformation gradient, satisfying ${\partial }_{t}F=K(t)F$ and $F(s,s)=I$. Along a material trajectory, an integrating factor givesThus $A$ is a functional of the local deformation history and its prepared initial state. If the initial $A$ is a positive-definite matrix, it retains that property, so $B=A/\mathrm{tr}A$ is well defined. Given this $B(t)$, the scalar stretch solves $\dot{\lambda }=r(t)\lambda +1/{\tau }_2$, where $r=B:K-1/{\tau }_2$. Its variation of constants formula again depends only on that local history:The differential pom-pom model also satisfies material frame indifference. Under an observer rotation, $A$ and $B$ transform as objective tensors, while $\lambda$ is a scalar. The extra skew part of the transformed velocity gradient has zero contraction with symmetric $B$, so the stretch equation is unchanged. The tensor evolution uses an upper-convected derivative, and the final stress transforms objectively. Initial values cannot be independently forgotten when specifying a finite-past experiment: a fully specified prepared state, such as rest, is part of the history specification.

An integrating factor for a differential one-form makes the multiplied form exact. If $dF=\mu gdx+\mu fdy$, then $F_x=\mu g$ and $F_y=\mu f$. Equality of the mixed partial derivatives yieldsThis is a local compatibility condition under the usual smoothness assumptions.

For the trigonometric equation, where its coefficients are defined and division is legitimate, the slope is $y^{\prime }=\tan x\tan y$. In the first quadrant it is positive, small near either axis and large near the upper-right corner. The direction field below uses normalized line segments, so steep directions remain visible.

Multiplication by $\sin x\cos y$ givesThus the multiplying function is an integrating factor, and the implicit general solution isIn the larger square, $\cos x\ge 0$. The positive-$C$ curves have positive $y$ and are symmetric about the $y$-axis; the negative-$C$ curves are their reflections. For $0<|C|<1$, a graph representation iswith endpoints tending to $y=\mathrm{sgn}(C)\pi /2$. The $C=0$ interior curve is $y=0$. The domain boundaries include singular coefficients or zeros of the integrating factor, so these endpoint limits and smooth continuations through $x=0$ describe the regularized relation; they must not be mistaken for additional solutions of the literal original equation at undefined points.

For a linear viscoelastic fluid, causal superposition givesThe relaxation modulus $G(s)$ is the stress response to a unit small step of shear strain. The equation assumes an isotropic reference state and deformations and rotations small enough over the significant memory times that nonlinear transport and strain products can be neglected. Small-amplitude motion can have high frequency; linearity does not require low frequency by itself. The infinite-past integral encodes the preceding history, rather than assuming that stress vanishes whenever the present rate vanishes.

For a small constant shear rate maintained from the remote past, the rate can be taken outside the integral. Therefore the zero-shear viscosity isprovided the kernel is integrable, with any instantaneous contribution counted fully. A nondecaying solid-like modulus would not yield a finite fluid viscosity.

To linearize the stated Oldroyd-B model, put $A=I+C$ and absorb the isotropic $G_{0}I$ into pressure. The products of the small deformation tensor C and velocity gradient are second order. Using the paper's derivative-index-first gradient convention therefore givesAn integrating factor, with the prescribed preceding history, gives $C(t)=2{\int }_0^{\infty }{e}^{-s/\tau }E(t-s)ds$. Thus the linear nonpressure stress isThe regular relaxation tail is $G_0{e}^{-s/\tau }$, while the solvent is instantaneous. Equivalently, the instantaneous solvent term in a relaxation modulus can be represented byHere the causal endpoint delta is defined by ${\int }_0^{\infty }{\delta }_{+}(s)f(s)ds=f(0)$. If instead a symmetric Dirac delta is assigned half its mass at the endpoint, the same solvent term is written $2{\mu }_0\delta (s)$. Stating the convention prevents a factor-of-two error in the viscosity; the explicit stress formula above has no such ambiguity.

In a cone-and-plate rheometer, the small-angle gap is $\alpha r$ and the local wall speed is $\Omega r$, so the shear rate is independent of radius to leading order: $g(t)=\Omega (t)/\alpha$. Write $s(t)$ for shear stress and $P(t)$ for its polymer part. The linear constitutive response isThe uniform shear stress gives torqueFor the initial steady torque T, put $s_{+}=3T/(2\pi a^3)$ and ${\mu }_p=G_0\tau$, $\mu ={\mu }_0+{\mu }_p$. The steady polymer stress is $P_{+}={\mu }_p{g}_{+}$, with $g_{+}=s_{+}/\mu$. Hence the initial angular velocity is

For $t>0$, the total stress is $s=-s_{+}$, but the polymer stress is continuous at the torque reversal. Since ${\mu }_0>0$, a finite jump in shear rate is possible even though P cannot jump. Its immediate value isEliminating P from the constant-stress equations for $t>0$ givesThe decay time is the Retardation time of an Oldroyd fluid ${\lambda }_2={\mu }_0\tau /({\mu }_0+{\mu }_p)$. Solving this first-order equation with the actual initial value yields the complete torque reversal of a linear Oldroyd fluid:In particular, $\Omega (\infty )=-{\Omega }_{+}$ but $\Omega (0^{+})$ is more negative. The stored polymer stress initially retains its old positive sign, requiring a larger negative solvent shear stress to attain the new negative total torque. This produces the backward-speed overshoot and its subsequent relaxation. Apparatus inertia would smooth the instantaneous jump, but is neglected in the question. The torque must be sufficiently small for the entire transient, including its enhanced initial rate, to remain in the linear viscoelastic regime; small pre-reversal rate alone is not sufficient when $G_0\tau /{\mu }_0$ is large.

An integrating factor isMultiplication by this integrating factor turns the first-order linear differential equation into $(e^{x^3}y{)}^{\prime }=x^2{e}^{x^3}$. Integrating from zero to $x$ givesThus the initial value problem has the solutionIndeed, direct differentiation gives $y^{\prime }=-3x^2(a-1/3)e^{-x^3}$, which verifies both the differential equation and the initial condition.

On an interval where $p,q$ are continuous, define the WronskianUsing the two second-order linear differential equations to eliminate the second derivatives givesThe integrating factor yields the Abel identityThus the Wronskian either vanishes identically or is nonzero everywhere on the interval. If $u_1,u_2$ are linearly dependent, it is identically zero. Conversely, if $W(x_0)=0$, the two initial-data vectors are dependent, so some nontrivial linear combination has both value and first derivative zero at $x_0$. The uniqueness theorem for ordinary differential equations makes that combination identically zero. Hence the solutions are linearly independent exactly when their Wronskian is nonzero at one, and therefore every, point. Continuity and the common differential equation are essential to this converse.

For the particular equation, divide by $x^2$ on either interval $x>0$ or $x<0$. The coefficient of $y^{\prime }$ is zero, so its Wronskian is constant there. The proposed $y_1=x^2$ satisfies $x^2{y}_1^{\prime \prime }-2y_1=2x^2-2x^2=0$. To construct a second solution by reduction of order, imposeDividing by $x^4$ shows that $(y_2/x^2{)}^{\prime }=C/x^4$, givingTaking $C=-3,D=0$ producesThe nonzero Wronskian proves linear independence; hence the general solution on either nonsingular interval is $y=A{x}^2+B/x$. Zero is a singular point of a differential equation and is excluded from this construction.

A particle pathline follows the time-dependent velocity field, so its coordinates satisfy $\dot{x}-x=\sin t$ and $\dot{y}=-y$. The second equation and the initial data give $y=e^{-t}$. Using the integrating factor $e^{-t}$ in the first,Therefore the complete pathline isIf $x_0+\frac{1}{2}\ne 0$, its first coordinate grows without bound in magnitude. If $x_0=-\frac{1}{2}$, the first coordinate remains bounded and periodic while $y\to 0$. The unique release coordinate giving a bounded particle path is $x_0=-1/2$.

For a discrete-time martingale relative to the filtration $({\mathcal{F}}_n)$, each $M_n$ is ${\mathcal{F}}_n$-measurable, $\mathbb{E}|M_n|<\infty$, and $\mathbb{E}[M_n\mid {\mathcal{F}}_{n-1}]=M_{n-1}$ for $n\ge 1$. The tower property then gives the corresponding identity conditional on any earlier time.

A bounded previsible process has ${\varphi }_i$ measurable at time $i-1$ and $|{\varphi }_i|\le C$. Its martingale transform $V$ is adapted and integrable, since each increment has absolute expectation at most $C(\mathbb{E}|M_i|+\mathbb{E}|M_{i-1}|)$. Moreover,Thus $V$ is a martingale, with $V_0=0$.

For the Itô integral, begin with a bounded adapted step integrand $H_s=H_i$ on $(t_i,t_{i+1}]$. Its integral is $I={\sum }_i{H}_i(W_{t_{i+1}}-W_{t_i})$. Each summand has zero conditional mean. For $i<j$, conditioning the cross product on ${\mathcal{F}}_{t_j}$ makes its expectation zero. The conditional second moment of a Brownian increment is $t_{i+1}-t_i$, henceApproximate the predictable integrand $g(s,W_s)$ in $L^2(dsd\mathbb{P})$ by such step processes. The same identity makes their integrals Cauchy in $L^2$ and defines the limit; both its first and second moments converge. This gives the requested mean-zero result and the Itô isometry:

For the Ornstein-Uhlenbeck process, the deterministic integrating factor has no quadratic covariation with $X$. The Itô formula therefore gives $d(e^{\kappa t}{X}_t)=\kappa \theta e^{\kappa t}dt+\sigma e^{\kappa t}d{W}_t$. Integrating and dividing by $e^{\kappa t}$ yieldsThe mean-zero integral and Itô isometry giveAt $\kappa =0$ these formulas are read by continuity: $X_t=x+\sigma W_t$, with mean $x$ and variance ${\sigma }^{2}t$.

For the CIR model, let $C_0=c_0\ge 0$ and use the usual positive parameters. Before a possible visit to zero, the derivatives of $\sqrt{c}$ are $1/(2\sqrt{c})$ and $-1/(4c^{3/2})$. Applying the Itô formula gives the Square root of a CIR diffusion:The PDF's claim under $2ab={\sigma }^2$ is false: that choice leaves the nonzero drift ${\sigma }^2/(8Z_t)-a{Z}_t/2$, which is not affine in $Z_t$. The actual interior cancellation condition is $4ab={\sigma }^2$. Even then the principal square root is nonnegative and cannot be an unrestricted Gaussian OU process.

To give the intended corrected construction, let a signed process solve $d{Y}_t=-a{Y}_{t}dt/2+(\sigma /2)d{B}_t$, with $Y_0=\sqrt{c_0}$. The Itô formula givesWriting ${\tilde{W}}_t={\int }_0^t\mathrm{sgn}(Y_s)d{B}_s$, with either sign chosen at zero, produces a Brownian motion; the diffusion term is $\sigma |Y_t|d{\tilde{W}}_t$. Thus $Y^2$ is a CIR model with $ab={\sigma }^2/4$, while $Z=|Y|$ is reflected at zero. The reflection contributes local time of a semimartingale, so the interior SDE alone is not a global unrestricted OU equation. Its second moment is $c_0{e}^{-at}+{\sigma }^2(1-e^{-at})/(4a)$, agreeing with the CIR mean for that corrected parameter choice.

The requested mean under the parameters actually printed is still available without the false OU assertion. Taking expectation in the original CIR equation gives $m^{\prime }=a(b-m)$; the stochastic integral has zero mean after the usual moment/localization justification. Consequently, for the printed choice as well as general admissible parameters,

In single-photon emission computed tomography, the unknown is an emitting source, rather than the absorption coefficient reconstructed by transmission computed tomography. Photons emitted at different points on the same ray are weighted by their probability of surviving the path to the detector. If the known attenuation is $a$ and the detector is at the positive end of the ray, the data are an attenuated Radon transformThe directional transport equation is $\mathbf{d}\cdot \nabla u+au=f$ with zero incoming intensity. An integrating factor derives this attenuation weight. The same complexification, boundary-value and jump strategy gives spectral reconstruction of an attenuated Radon transform when attenuation is known. The unattenuated inverse from part (d) applies directly only when $a=0$; for nonzero attenuation the weighted jump and inverse must be used. This distinction allows emission imaging of the source without confusing it with reconstruction of the absorbing medium.

The spin tensor $W$ is skew, and $\dot{Q}{Q}^T=W$ means $\dot{Q}=WQ$ and ${\dot{Q}}^T=-Q^{T}W$. Differentiate all factors to obtain the corotating-frame representation of a Jaumann derivative:Thus the Jaumann derivative is the ordinary derivative of the tensor components in this rotating frame.

Put $S={\sigma }^d$, $\Delta \mu ={\mu }_1-{\mu }_0$, $A=S-2{\mu }_{0}D$ and $\tilde{A}=Q^{T}AQ$, $\tilde{D}=Q^{T}DQ$. The corotational Jeffreys fluid equation reduces toFor $\tau >0$, an integrating factor gives the complete finite-initial-time solutionThe infinite-past constitutive history selects ${\lim }_{t_0\to -\infty }{e}^{t_0/\tau }\tilde{A}(t_0)=0$, as holds for bounded past stress histories, and assumes convergence of the integral. Taking that limit and rotating back provesA general initial stress would retain the displayed homogeneous term. Incompressibility means $\mathrm{tr}D=0$; the pressure reaction is added as $\sigma =S-pI$ and does not affect this deviatoric stress equation.

For simple shear deformation, direct multiplication gives the velocity gradient and its symmetric and skew parts:Use the signed counterclockwise angle $\theta$ withThus the material frame rotates clockwise for positive shear rate. This angle convention is necessary for the signs of the sine terms. Set $\delta =\theta (t)-\theta (s)$. With $H=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ and $R_{\delta }=Q(t)Q^T(s)$, multiplication yieldsSubstitution in the history formula provesThe suppressed third row and column are zero for the selected infinite-past history.

For constant shear rate $g=\dot{\gamma }$, $2\delta =-g(t-s)$. Put $u=t-s\ge 0$. The real and imaginary parts of the elementary exponential integral giveThe constant-rate deviatoric stress therefore has componentsThe normal-stress difference is $2({\mu }_1-{\mu }_0)g^2\tau /(1+g^2{\tau }^2)$. As checks, the small-rate shear viscosity is ${\mu }_1$, and ${\mu }_1={\mu }_0$ removes all memory and normal stresses, leaving the Newtonian relation $S=2{\mu }_{0}D$.

Take the norm to be the Euclidean norm. The solution is $\mathbf{q}(t)=e^{tA}\mathbf{q}(0)$, so optimal growth is the operator norm $G(t)=\Vert e^{tA}{\Vert }_2$, equivalently the largest singular value of the matrix exponential. Let $s(A)$ be the spectral abscissa and let $w(A)={\lambda }_{\max }((A+A^{\ast })/2)$ be the numerical abscissa. The Euclidean semigroup growth bounds areFor the lower bound, the spectral radius of $e^{tA}$ is $e^{ts(A)}$ and is bounded by its operator norm. For the upper bound, differentiating $\Vert \mathbf{q}{\Vert }_2^2$ gives $2\mathrm{Re}({\mathbf{q}}^{\ast }A\mathbf{q})\le 2w(A)\Vert \mathbf{q}{\Vert }_2^2$, and integration gives the result. A non-normal matrix can have $w(A)>s(A)$, allowing transient amplification even when all its eigenvalues have negative real parts.

For the triangular matrix, $s(M)=-1/R$. The eigenvalues of its Hermitian part are $-3/(2R)\pm \frac{1}{2}\sqrt{1+R^{-2}}$, soThe upper exponent becomes positive when $R>\sqrt{8}$, even though the long-time spectrum is stable.

The two equations are $q_1^{\prime }=-q_1/R$ and $q_2^{\prime }=q_1-2q_2/R$. Solve the first, then use an integrating factor in the second. Writing $s=t/R$, $a=e^{-s}$, $b=e^{-2s}$ and $c=R(a-b)$ givesThe maximizing initial state is the unit eigenvector of $B^{\ast }B$ belonging to its largest eigenvalue. HereFor $t>0$, defineThe optimal initial state for triangular stable shear is thereforeMultiplication by any nonzero complex scalar gives the same growth ratio; at $t=0$, every nonzero initial state is optimal.

For $t\gg R$, $d\to \infty$ and $r\to 0$. Thus the optimal initial state tends to $(1,0{)}^T$, and $G(t)\sim e^{-t/R}\sqrt{1+R^2}$. The initial first component excites the slower eigenmode and its amplified second component.

For $t\ll R$, the exact expression gives $d=(R^{-1}+t/2)[1+O(t/R)]$. Accordingly the short-relative-time optimal state for triangular shear usesThis retains both the unequal damping rates and the shear. If $R\gg 1$, it reduces to $r\sim (\sqrt{t^2+4}-t)/2$, the maximizing direction for the shear matrix $\left(\begin{array}{cc}1& 0\\ t& 1\end{array}\right)$. If time also tends to zero at fixed $R$, it instead tends to $r=\sqrt{1+R^{-2}}-R^{-1}$, the largest-eigenvalue direction of the Hermitian part of $M$. Merely setting $t/R$ small does not justify setting $t$ small; the exact formula resolves both possibilities.

The Itô product rule with a deterministic integrating factor givesIntegrating from zero provesBoth coefficients are globally Lipschitz, so part (a) proves that this is the pathwise unique solution. Under a measure where $B$ is a Brownian motion, the solution is centered Gaussian with variance $e^{2t}{\int }_0^t{e}^{-2s}ds=(e^{2t}-1)/2$. The positive drift sign gives an unstable linear diffusion rather than a mean-reverting one.