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Take upward velocity as positive. The linear drag force is $-mkv$, so it opposes the motion both during ascent and descent. Newton's second law gives $\dot{v}+kv=-g$. With $v(0)=u$, the integrating factor $e^{kt}$ yieldsThe terminal velocity is $-g/k$. The limit $k\to 0$ at fixed $t$ recovers $u-gt$; expanding the exponential is necessary to see the cancellation of the apparently singular terms.

Take $x$ horizontal, $y$ upward and $g>0$. Resolve Newton's second law into components for projectile motion with linear drag:For $\alpha >0$, let ${\tau }_d=m/\alpha$. An integrating factor $e^{t/{\tau }_d}$ givesThe relaxation time is ${\tau }_d$, and the terminal velocity is $(0,-g{\tau }_d)$. For fixed finite $t$, expanding the exponential function as $\alpha \to 0$ gives the continuous limit $v_x=v_0\cos \theta$ and $v_y=v_0\sin \theta -gt$; the two terms proportional to ${\tau }_d$ cancel before taking the limit.

Put $s=e^{-R\Delta t/L}$ and $J=V/R$, with $0<s<1$. The first interval leaves electric current $I_1=s{I}_0$. Position B reconnects the battery, so the driven RL circuit obeys $L\dot{I}+RI=V$. Multiplication by the integrating factor $e^{Rt/L}$, or subtraction of the constant solution $J$, givesConsequentlyThe same two intervals give the recurrence relation $I_{2n+2}=s^2{I}_{2n}+J(1-s)$. Its fixed point is $I_{\ast }=J/(1+s)$. Subtracting this fixed point makes the recurrence relation homogeneous:This is the upper endpoint of the eventual periodically switched RL circuit response, immediately before switching from B to A. The lower endpoint is $sJ/(1+s)$. Their sum is $J$, consistent with equal charging and discharging intervals.

Set $s=r_0+\mu t$. The proposed velocity and its derivative areThe remaining term in the linear ordinary differential equation isso the sum is exactly $g$. Also $v(0)=g(r_0-r_0)/(4\mu )=0$, establishing both the equation and its initial condition:The integrating factor $(r_0+\mu t{)}^3$ explains where the expression comes from:The result has units of velocity. Its continuous $\mu \to 0$ limit is $gt$, despite the apparent $1/\mu$ in the written expression.

For a small additional carbon mass, logarithmic forcing has derivative inversely proportional to the background atmospheric carbon inventory. With other conversion factors fixed, $b_0$ therefore decreases as background concentration rises. Solve the temperature equation using an integrating factor and the pulse burden:At $b_1=2k$ the second term is its continuous limit $t{e}^{-b_{1}t}$. For $0<k\ll b_1$, temperature initially rises over $b_1^{-1}$ toward roughly $b_0/b_1$, then relaxes over $(2k{)}^{-1}$ toward half that value. This produces an overshoot, rather than permanent growth. The two numerical limits requested separately below also follow directly from the differential equation.

In the steady state, conservation of probability requires $J_x=0$, so $J=j_0$ is constant. The sign is positive because increasing gap reaches the incorporation boundary, whereas the deterministic drift points towards smaller gap. ThusAn integrating factor $e^{\beta Fx}$ givesSubstitution makes the check particularly transparent:so ${\partial }_{t}p=-J_x=0$. The constant offset is necessary for nonzero transport; a pure Boltzmann distribution would have $J=0$ and could not describe a steady succession of addition events. The $F=0$ limit is taken after applying the boundary conditions, avoiding the apparent singularity of the displayed parametrization.

Each addition starts a new statistically identical waiting episode at zero gap. Thus the addition rate is the reciprocal of a mean first-passage time, not the reciprocal of the mean gap:This can also be derived without the steady density. If $T(x)$ is the mean time to reach $\delta$ starting at gap $x$, its backward Smoluchowski equation isThe first boundary is reflecting and the second absorbing. An integrating factor gives $T^{\prime }(x)=[1-e^{\beta Fx}]/(D\beta F)$, whenceAt weak force, $\tau \simeq {\delta }^2/(2D)$ is the time for a freely diffusing wall to move a monomer length, with reflection at the tip. At strong force, opening the gap requires work $F\delta$, with Boltzmann distribution suppression $e^{-\beta F\delta }$. The drift establishes a near-tip thermal length ${\ell }_F=k_{B}T/F$ and a diffusion time ${\ell }_F^2/D$; consequentlyThe exponentially long wait explains why the growth slows. There is no additional chemical waiting time in the imposed absorbing-threshold model; such a time would be needed to describe the separate reaction-limited polymerization Brownian ratchet regime.

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 $\sigma =0$.

Inside the fan, $u+c=(4c_0+x/t)/3$. Hence a right-running characteristic satisfiesMultiplication by the integrating factor $t^{-1/3}$ givesIntegration yields $x_{+}=2c_{0}t+C{t}^{1/3}$. Requiring $x_{+}(\tau )=-c_0\tau$ determines $C=-3c_0{\tau }^{2/3}$, soThe invariant $Q_{+}=2c_0$ stays constant along it until it enters a region affected by new boundary data.

Set $C(\beta )=[A,e^{-\beta B}]$. The product rule givesMultiplying by the integrating factor $e^{\beta B}$ and integrating yields the commutator of an operator with an exponential:Equivalently, differentiate $e^{\beta B}A{e}^{-\beta B}-A$ directly; its derivative is the integrand and it vanishes at zero. The argument is exact for finite matrices, and for bounded operators with their norm-convergent exponentials; unbounded linear operators require an appropriate common domain.

The first integral and $b_0=1$ yield the linear identity $b^{\prime }+cb=c(1-g)$. Multiply by the integrating factor $e^{cy}$ and integrate from $-\infty$. Since the bounded limit $b(-\infty )=1$ makes $e^{cy}b(y)\to 0$ there,Positivity of $g$ makes the last integral nonnegative, and strictly positive at any finite point of a strictly positive travelling wave. Thuswith equality approached only at the far-behind boundary. This bound uses the first integral rather than an unsupported assumption that the bacterial profile is monotone.

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.

Set $g=f_1^{\prime }$. The boundary value problem becomesFor $y_1=1+{\zeta }^2$, direct substitution gives $y_1^{\prime \prime }+\zeta y_1^{\prime }-2y_1=2+2{\zeta }^2-2(1+{\zeta }^2)=0$. This growing polynomial solution is useful for reduction of order.

Write $g=y_{1}v$. Substitution into the linear ordinary differential equation cancels the terms proportional to $v$ and givesAn integrating factor for this first-order equation for $v^{\prime }$ yieldsThe far-field boundary condition selects the convergent integral from $-\infty$:Any additional multiple of $1+{\zeta }^2$ would grow at $-\infty$. The chosen expression tends to zero: for $r=-\zeta \ge 1$,Here $(1+t^2{)}^{-2}\le (1+r^2{)}^{-2}$ and ${\int }_r^{\infty }{e}^{-t^2/2}dt\le r^{-1}{\int }_r^{\infty }t{e}^{-t^2/2}dt$. Thus the far-field boundary condition is verified without a cancellation of large terms.

At $\zeta =0$, differentiating gives $g^{\prime }(0)=C$, so $C=1$. Using the antiderivative supplied in the paper gives the explicit answerHere $\mathrm{erf}$ is the error function. Finally,recovers the unique correction satisfying $f_1(0)=0$. The decay bound also shows that $f_1$ approaches a finite constant at $-\infty$, as allowed by the normal velocity matching. This is the weak surface-shear perturbation of stagnation flow.

If the printed $+\infty$ condition is instead imposed literally on a positive half-line, reduction of order givesIt satisfies $g_{+}^{\prime }(0)=1$ but grows as $\zeta \to -\infty$, so it cannot describe the fluid domain specified in this paper. This makes the consequence of the printed sign error explicit.

Divide by $x^2$, which is allowed because $x>0$:The integrating factor for this linear ordinary differential equation is $e^{\int dx/x}=x$. Multiplying through makes the left side a product rule derivative:Part (b) now gives $xy=2e^{x^2}+C$. Therefore

After division by $x>0$, the equation is linear with coefficient $-1/x$ on $y$. Its integrating factor is $1/x$, and the product rule givesUsing the antiderivative from the preceding part,The arbitrary constant multiplies the homogeneous solution $x$.

For $x\ne \pm 1$, write the linear ordinary differential equation asAn integrating factor on each interval avoiding $\pm 1$ is $\mu (x)=1/(1-x^2)$; multiplication by a nonzero constant, including a sign, is immaterial. ThereforeThe partial fraction decomposition isIntegrating gives the general solutionThe constant is independent on each connected interval of definition. At the singular points the undivided differential equation forces $y(1)=0$ and $y(-1)=1$. The displayed family tends to these values, but its derivative has a logarithmic divergence there, so no member extends as a differentiable function through either singular point.

Since $x>-1$, divide by $x+1$ to obtain the first-order linear ordinary differential equationAn integrating factor is $\mu (x)=e^{-3\log (x+1)}=(x+1{)}^{-3}$. Multiplication by this factor givesAfter integration, $(x+1{)}^{-3}y=\frac{1}{2}(x+1{)}^2+C$, so

On an interval avoiding $x=0$, divide by $x$ to obtain the first-order linear ordinary differential equationAn integrating factor is $\mu (x)=\exp \left(\int -2/xdx\right)=x^{-2}$. Multiplying through and using the product rule givesBy integration by parts, $\int x\cos xdx=x\sin x+\cos x+C$. ConsequentlyThe formula differentiates back to the original equation. Although division excluded $x=0$, this formula has a smooth extension there with $y(0)=0$, satisfying the original equation at zero as well.

Factor the right side as $(x-1{)}^2(x-3)$. On $-1<x<1$, division by $x^2-1$ gives a first-order linear differential equation with integrating factorMultiplication makes the equation $(\mu y{)}^{\prime }=3-x$, so $\mu y=3x-x^2/2+C$. At zero, $\mu (0)y(0)=1$, hence $C=1$. Simplifying givesIts derivative isThe derivation solves the initial-value problem on $(-1,1)$. The simplified expression also extends smoothly through $x=1$ and satisfies the original undivided equation there, giving a solution on all $x>-1$; its pole prevents continuation through $x=-1$.

Here the integrating factor is $e^{2\sin x}$, since its logarithmic derivative is $2\cos x$. Multiplying the differential equation by it cancels the right-hand exponential and yields $(e^{2\sin x}y{)}^{\prime }=1$. HenceThis formula is actually defined for every real $x$.