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Integrate the ordinary differential equation across one step, using the fundamental theorem of calculus:The midpoint quadrature rule approximates the integral by $hf(x_n+h/2,y(x_n+h/2))$. Approximate the unknown midpoint state by the midpoint value of linear interpolation between the endpoint states:Combining these approximations gives the implicit midpoint ruleIt is an implicit time-stepping method because its new state occurs inside the evaluation of $f$. An equivalent derivation uses a straight-line approximation on the step and imposes the ordinary differential equation at its midpoint, a one-stage collocation method.

The streamlines are contours of $\psi$. Since $A<0$, the exact streamfunction is negative inside the rectangle and zero on its perimeter. It has one interior minimum, with closed anticlockwise streamlines: a broad northward interior flow returns southward along the west wall; flow is westward near the north edge and eastward near the south edge. The western parts of the streamlines are closely packed because the western boundary current is much faster.

The figure uses the exact ordinary differential equation solution, so it also remains consistent with both wall conditions when the asymptotic scale separation is only moderate. It depicts the stipulated dynamical model, not observed winds or a reconstruction of the source illustration.

Seek $\psi =\Psi (x)\cos (my)$, with $m=\pi /b$. The ordinary differential equation isIts constant particular solution and characteristic exponents areBoth $\lambda ,\gamma$ are positive; the two homogeneous solutions are $e^{-\lambda x}$ and $e^{\gamma x}$. Their product satisfies $\lambda \gamma =m^2$. At small $rm/\beta$,For the rectangular basin, choose zero streamfunction on its connected impermeable boundary. The factor $\cos (my)$ already gives zero at $y=\pm b/2$. Enforcing $\Psi (0)=\Psi (L)=0$ yields the exact resultwhereThe narrow Stommel boundary layer approximation requiresThen $e^{-\lambda L}$ is negligible, $C_W\simeq \gamma L$ and $C_E\simeq 1$, givingIt is an asymptotic formula: at $x=0$ the remaining mismatch is of order $A(\gamma L{)}^2$. Expanding the eastern exponential as well gives the familiar leading Sverdrup balance interior plus western correction,Printed inequality. The original PDF explicitly has ${\lambda }^{-1}\gg L$ as well as ${\gamma }^{-1}\gg L$. The first sign is reversed for the requested small-friction limit. This can also be disproved directly: if $\lambda L=0.02$ and $\gamma L=0.01$, the printed approximate bracket at the western wall is $1-0.01-e^{-0.01}\simeq -4.98\times 10^{-5}$, while at the eastern wall it is $-0.01e^{-0.02}\simeq -0.00980$. The latter is of the same order as the retained basin-scale signal, violating the boundary condition at leading order. The exact expression above remains available without imposing the erroneous inequalities.

For the linear three-reservoir carbon model, with concentrations as a column vector, the linear ordinary differential equation isA homogeneous normal mode has $\mathbf{x}=\mathbf{v}{e}^{\lambda t}$, where $J\mathbf{v}=\lambda \mathbf{v}$. For negative real eigenvalues the e-folding relaxation time is $\tau =-1/\lambda$. The row sums vanish, so ${\mathbf{v}}_3=(1,1,1{)}^T$ has ${\lambda }_3=0$ andEqual effective concentrations generate no exchange. Summing the reservoir budgets givesThus injected carbon dioxide is conserved after emissions cease: the model redistributes it but contains no permanent removal. The zero normal mode records this conserved excess mass, rather than a long but finite decay.

To check the other normal modes are damped, set $D=\mathrm{diag}(R_a,R_b,R_d)$ and $L=-DJ$. ThenThe similarity transformation $D^{1/2}J{D}^{-1/2}=-D^{-1/2}L{D}^{-1/2}$ is a symmetric negative-semidefinite matrix with a one-dimensional kernel, so the remaining two eigenvalues are strictly negative.

Define the two concentration differences $u=C_a-C_b$ and $v=C_b-C_d$. For zero emissions and $R_a=R_b=R$, their ordinary differential equation isWriting the positive decay rates as ${\mu }_i=1/{\tau }_i$, their characteristic equation isConsequentlyFor $r\gg s$, the first normal mode rapidly equalizes the fast reservoirs, giving ${\mu }_1\simeq 2r/R$. The slower evolution treats them as a combined capacity $2R$ exchanging with $R_d$, giving ${\mu }_2\simeq s[1/(2R)+1/R_d]$. ThereforeThe timescale ratio is ${\tau }_2/{\tau }_1\simeq 4r/s\gg 1$ in the joint limit. Since the conserved normal mode has ${\tau }_3=\infty$, the hierarchy is ${\tau }_1\ll {\tau }_2\ll {\tau }_3$.

For a Hermitian operator $Q$, the operator exponential $C_{\lambda }=e^{i\lambda Q}$ is a unitary operator, with $C_{\lambda }^{†}=e^{-i\lambda Q}$. Differentiation and $[Q,C_{\lambda }]=0$ giveWhen $[Q,A]=cA$, this is the first-order ordinary differential equation $A^{\prime }(\lambda )=icA(\lambda )$ with $A(0)=A$. Multiplication by $e^{-ic\lambda }$ makes its derivative zero, provingEquivalently, the commutator expansion for exponential conjugation has successive nested commutators $c^{r}A$, whose exponential series sums to this expression.

Put $G(p)=\mathcal{L}[J_0](p)$, initially for real $p>0$. Besides the Laplace transform of a derivative, use multiplication by the original variable in the formTransforming the Bessel differential equation, with its given initial conditions, yieldsThe constant terms cancel, leaving $(1+p^2)G^{\prime }+pG=0$. This first-order ordinary differential equation integrates toThe initial value theorem for Laplace transforms gives ${\lim }_{p\to \infty }pG(p)=J_0(0)=1$, so $C=1$. Thus the Laplace transform of the Bessel function J0 isFor complex $p$ with $\mathrm{Re}p>0$, choose the analytic square-root branch agreeing with the positive square root on the positive real axis.

For the forcing $k=J_0$, part (b) now gives $F(p)=(1+p^2{)}^{-3/2}$. On the other hand, the same multiplication rule givesTaking an integral from zero divides the Laplace transform by $p$. By uniqueness of the Laplace transform,A direct check also explains the cancellation. This expression has $f^{\prime }=x{J}_0$ and $f^{\prime \prime }=J_0+x{J}_0^{\prime }$. The Bessel differential equation says $(x{J}_0^{\prime }{)}^{\prime }=-x{J}_0$, so integration from zero gives $x{J}_0^{\prime }=-{\int }_0^{x}t{J}_0(t)dt=-f$. Hence $f^{\prime \prime }+f=J_0$, and both initial conditions hold.

Write $f(N)=N[r/(1+b{N}^2)-\delta ]$. A steady state is a nonnegative zero of $f$. If $r<\delta$, the bracket is strictly negative for every $N\ge 0$, so only $N=0$ exists. If $r>\delta$, there are exactly two nonnegative steady states,For an autonomous scalar ordinary differential equation, a steady state with $f^{\prime }<0$ is locally asymptotically stable, while one with $f^{\prime }>0$ is unstable. HereAt the positive steady state, use $1+b(N^{\ast }{)}^2=r/\delta$ to obtainTherefore zero is stable for $r<\delta$ and unstable for $r>\delta$, whereas the positive steady state is stable whenever it exists. At the omitted equality $r=\delta$, $f(N)=-\delta b{N}^3/(1+b{N}^2)$; zero is nonhyperbolic but still attracts every nonnegative initial state.

Set $t=\sqrt{\lambda }x$ and $w(t)=y(t/\sqrt{\lambda })$. The chain rule turns the equation into the parameter-one Bessel differential equation. By the regular-at-zero solution property supplied in the paper,For a nonzero solution, $c\ne 0$, so $y(1)=0$ forces $J_k(\sqrt{\lambda })=0$. Conversely, every positive zero $z$ supplies the admissible eigenfunction $J_k(zx)$ with eigenvalue $z^2$.

Here is a direct justification of the infimum characterization that does not assume a minimizer exists. Let $u(x)=J_k(z_{k,1}x)$ and choose its sign so that $u>0$ on $(0,1)$. The first positive zero exists by the oscillatory large-argument behavior of the Bessel function of the first kind. At zero, its Frobenius series for Bessel functions gives $u(x)=C{x}^k(1+O(x^2))$. At one, its zero is simple by uniqueness for a regular second-order ordinary differential equation. Expanding $y=u(y/u)$ and using the equation for $u$ gives the Ground-state representation of a Sturm-Liouville energy:For completeness, apply the identity first on $[\epsilon ,1-\epsilon ]$. The discarded boundary term is $[x{u}^{\prime }{y}^2/u]$. It tends to zero at both ends: $x{u}^{\prime }/u\to k$ and $y=O(x)$ at zero, while $u^{\prime }/u=O((1-x{)}^{-1})$ and $y=O(1-x)$ at one. Taking $y=u$ achieves equality, soThe infimum is over the stated smooth admissible functions with finite energy; the equality also proves attainment by a multiple of $u$.

Put $y^{(0)}=y$ and $y^{(1)}=\dot{y}$. ThenThe second row is the original damped harmonic oscillator equation and the first row defines the auxiliary velocity. This converts a second-order ordinary differential equation into two first-order equations without changing the required two initial values.

Take the inner surface to be free of traction, as appropriate to an empty cavity with no surface tension. At the outer surface the imposed pressure is compressive when positive. The outward normal vectors are $-{\mathbf{e}}_r$ at $r=a$ and $+{\mathbf{e}}_r$ at $r=b$, givingIntegrating the ordinary differential equation from part (ii) between the two surfaces givesThe outer stretch follows from incompressibility:Hence an expression entirely in terms of the inner stretch and the reference radii isEquivalently $P=-{\int }_{{\lambda }_b}^{{\lambda }_a}{h}^{\prime }(\lambda )/({\lambda }^3-1)d\lambda$. For an expanding shell, $1<{\lambda }_b<{\lambda }_a$; when $h^{\prime }>0$ on this range, $P<0$, correctly describing an outward tensile loading. If an inner pressure $P_i$ were present, the same derivation would give $P-P_i$ on the left. Thus the empty-cavity condition specifies the pressure reference needed for the requested answer.

Let $W_r,W_{\theta },W_{\varphi }$ denote derivatives of the strain energy density with respect to its three principal stretches. The principal stress formula for incompressible hyperelasticity isIt follows by applying $\mathbf{T}=\mathbf{F}({\partial }_{\mathbf{F}}W{)}^T-p1$ in principal axes. Equal tangential principal stretches and material isotropy give $W_{\theta }=W_{\varphi }$ at $({\lambda }^{-2},\lambda ,\lambda )$. Differentiating the reduced strain energy density by the chain rule givesThe unknown pressure cancels in this difference. Substituting in part (i), and then using the supplied derivative of $\lambda$, givesConsequently the desired ordinary differential equation isFor completeness, the supplied derivative follows from $R^3=r^3-(a^3-A^3)$: $dR/dr=r^2/R^2={\lambda }^2$, and differentiating $\lambda =r/R$ gives $(1-{\lambda }^3)/R$. At the undeformed value $\lambda =1$ the division must be interpreted by a limit; the spherically symmetric deformation with $a=A$ has $\lambda \equiv 1$ and should be handled directly by the radial equilibrium equation.

The diffusion generator is $\mathcal{L}=\mu {\partial }_x+\frac{1}{2}{\sigma }^2{\partial }_{xx}$. By the Feynman-Kac formula, the required conditional characteristic function solvesUse separation of variables in the form $F(t,x)=e^{iux}h(t)$. Since $F_x=iuF$ and $F_{xx}=-u^{2}F$, the temporal factor satisfies the ordinary differential equationThus $h(t)=\exp [(iu\mu -\frac{1}{2}{\sigma }^2{u}^2)(T-t)]$, andThis is bounded in absolute value by one, so it lies in the uniqueness class of the Feynman-Kac formula. As an independent check, integrating the stochastic differential equation gives $X_T=x+\mu (T-t)+\sigma (B_T-B_t)$. The independent Brownian increments have the normal distribution with mean zero and variance $T-t$, yielding exactly the same characteristic function. For $\sigma =0$, the conditional distribution is a point mass at $x+\mu (T-t)$, and the same boxed expression still applies.

Here the diffusion generator is $\mathcal{L}=ax{\partial }_x+\frac{1}{2}{\sigma }^2{\partial }_{xx}$. The Feynman-Kac formula givesTry $F(t,x)=\exp [iu\alpha (t)x+\beta (t)]$, where $\alpha (T)=1$ and $\beta (T)=0$. Substitution and collection of the coefficients of $x$ and the constant term giveThe solutions of these ordinary differential equations, with the terminal conditions imposed, areFor $\Delta =T-t$, define the conditional varianceThen the required conditional characteristic function isIn particular, for $a\ne 0$ this exponent is $iu{e}^{a\Delta }x-\frac{{\sigma }^2{u}^2}{4a}(e^{2a\Delta }-1)$. The variance $v_{\Delta }$ is nonnegative for either sign of $a$, and its limit at $a=0$ is ${\sigma }^2\Delta$. Hence the solution is bounded in absolute value by one and the Feynman-Kac formula identifies it with the required expectation. The value $u=0$ gives one directly, including that case without dividing by $u$.

For an independent check, the integrating factor $e^{-as}$ and the Itô product rule giveThe stochastic integral has a normal distribution because its integrand is deterministic; the Itô isometry gives its variance as $v_{\Delta }$. For $a<0$ this is the Ornstein-Uhlenbeck transition distribution. The formula also covers the unstable case $a>0$, and degenerates to deterministic motion when $\sigma =0$.

For a constant $X$, set both ordinary differential equation derivatives to zero:These equilibrium concentrations are unique for each input. For a loss rate $\gamma$ other than one, divide the corresponding production term by $\gamma$.

Choose $a_Y=2$, $K_{XY}=1$, $n_{XY}=2$ and $a_Z=3$, $K_{XZ}=2$, $n_{XZ}=4$. The following Python code plots the requested equilibrium concentrations and defines the functions used in part (c):

import numpy as np
import matplotlib.pyplot as plt

def hill(u, K, n):
    u = np.asarray(u)
    return u**n / (K**n + u**n)

X_grid = np.linspace(0, 5, 501)
Y_eq = 2 * hill(X_grid, 1, 2)
Z_eq = 3 * hill(X_grid, 2, 4)
plt.figure(figsize=(6, 4))
plt.plot(X_grid, Y_eq, label="Y equilibrium: K=1, n=2")
plt.plot(X_grid, Z_eq, label="Z equilibrium: K=2, n=4")
plt.xlabel("X concentration")
plt.ylabel("Equilibrium concentration")
plt.legend()
plt.tight_layout()
plt.show()

The Hill constant sets the half-maximum input: $Y^{\ast }(1)=1$ and $Z_{\mathrm{direct}}^{\ast }(2)=1.5$. Increasing that constant shifts activation toward larger $X$. The Hill coefficient changes steepness rather than the half-maximum position; the normalized slope at $X=K$ is $n/(4K)$. Because the two examples also differ in $K$ and maximal production, compare normalized responses when isolating the effect of $n$ alone. Both curves vanish at zero and saturate toward their respective maxima.

For the Poisson process, $q_n=g$ and $r_n=0$, soThis is an ordinary differential equation in time for each fixed $z$, with solutionConsequentlyThe initial condition is interpreted as $P_0(0)=1$, with all other initial probabilities zero. Normalization follows from $F(1,t)=1$. The mean and variance are both $gt$, as expected for independent events arriving at constant rate.

Take profiles $b=b(y)$ and $g=g(y)$ with $y=x-ct$, and use a prime for differentiation with respect to $y$. The chain rule gives $b_t=-c{b}^{\prime }$, $b_{xx}=b^{\prime \prime }$ and $g_t=-c{g}^{\prime }$. ThereforeThese are coupled ordinary differential equations for a right-moving travelling wave with speed $c>0$. A first-order formulation, using $v=b^{\prime }$, is $b^{\prime }=v$, $v^{\prime }=-cv-gb$, $g^{\prime }=gb/c$. In particular a positive profile has $g^{\prime }>0$, consistent with depleted substrate behind the travelling front and fresh biological growth factor ahead.

Let $J_X={\lim }_{x\downarrow 0}X(x)X^{\prime }(x)>0$. The spatial ordinary differential equation gives $(X{X}^{\prime }{)}^{\prime }=-\lambda X$, so its leading outlet behavior isAgain this is a square-root drainage boundary. For the separable aquifer drawdown, the dimensional drainage flux isEquivalently, with ${\tau }_d=\varphi /(K\lambda )$, it is $Q_0(t^{\ast })[1+(t-t^{\ast })/{\tau }_d{]}^{-2}$ for an exact separated profile. Integrating the spatial ordinary differential equation gives $J_X=\lambda {\int }_0^{L}Xdx$. This verifies the global mass conservation checkThus the late drainage flux follows an inverse-square power law, while stored water decays inversely with time.

The aquifer drainage hydrograph starts at zero, rises linearly in the early rainfall similarity solution, and approaches the total recharge $RL$ if rainfall lasts long enough for a finite-length steady state. Indeed, steady mass conservation gives $q_x=R$ and $q(L)=0$, hence $Q_0=RL$. At $t^{\ast }$, the outlet flux stays continuous for a regular inherited flux profile, then recedes as water drains; the eventual decay is proportional to $(t-t^{\ast }+\text{constant}{)}^{-2}$. An arbitrary transition requires the full time-dependent partial differential equation, so matching an early rainfall formula directly to a separated drought formula is only a schematic approximation.

The curve illustrates these regimes and a possible $t^{\ast }$. Its numerical shape and the relative duration of the regimes are schematic; the early rainfall and late drought formulas apply in their respective asymptotic ranges.

With $R=0$, substitute $h=X(x)T(t)$ into the dimensional Boussinesq equation for an unconfined aquifer:For a positive draining profile, take $\lambda >0$. Normalize $T(t^{\ast })=1$, so $X$ has the dimensions of height. The temporal ordinary differential equation integrates toThe spatial ordinary differential equation and boundary conditions areHere $\lambda$ has units of inverse length with this normalization. The scale freedom between $X$ and $T$ is fixed by the chosen value $T(t^{\ast })=1$; the amplitude of $X$, and hence $\lambda$, must be matched to the available water.

This separable aquifer drawdown is exact if the inherited height profile has the separated spatial shape. For a general rainfall-generated initial condition, it describes the late drought regime; it cannot reproduce an arbitrary profile at the instant rainfall stops merely by choosing one temporal constant.

The two external stream velocities imposeLet $r=\Delta U/U_c\ll 1$ and seek $F=O(r)$ over $\xi =O(1)$. The nonlinear term $F{F}^{\prime \prime }$ is then $O(r^2)$, so the leading ordinary differential equation isFor $G=F^{\prime \prime }$, this is the first-order linear differential equation $G^{\prime }+\xi G=0$. Its integrating factor is $e^{{\xi }^2/2}$, giving $F^{\prime \prime }=M{e}^{-{\xi }^2/2}$. An integration using the error function givesThe two far-field boundary conditions require $E=0$ and $M=r/\sqrt{2\pi }$. A second integration usesConsequentlyThe relation between $A$ and $B$ cancels the extra $\xi e^{-{\xi }^2/2}$ term when differentiating. It gives the familiar error function mean velocity profileto first order in $r$. The additive constant $D=O(r)$ does not affect this velocity profile at the retained order; its interpretation comes from the transverse velocity.