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Take the current to be dilute, vertically well mixed and Boussinesq approximation, with constant channel width $w$, a horizontal absorbing bed, negligible ambient fluid entrainment, negligible bed friction, and no resuspension. The shallow-water approximation neglects vertical inertia and gives hydrostatic pressure. Particle loss alters reduced gravity but does not alter the leading-order carrier-fluid volume or inertia; otherwise additional mass and momentum terms would be necessary. The current occupies depth $h(x,t)$, has depth-uniform speed $u(x,t)$, and has particle volume fraction $c(x,t)$, with $g^{\prime }=g_{o}c$ and constant $g_o>0$.

Volume conservation of a short channel segment gives $h_t+(uh{)}_x=0$. Its suspended particle amount per unit bed area is $hc$, and the downward particle deposition flux is $v_{s}c$. Thus $(hc{)}_t+(uhc{)}_x=-v_{s}c$, which becomes $(h{g}^{\prime }{)}_t+(uh{g}^{\prime }{)}_x=-v_s{g}^{\prime }$ on multiplying by $g_o$.

At vertical coordinate $y$ above the bed, excess hydrostatic pressure divided by reference mass density is $g^{\prime }(h-y)$. Integrating across the depth gives the depth-integrated force flux $g^{\prime }{h}^2/2$. The horizontal momentum per unit bed area is $hu$ and the advected kinematic momentum flux is $h{u}^2$. The resulting variable-buoyancy shallow water equations areThe derivative of the pressure term acts on both $h$ and $g^{\prime }$; treating reduced gravity as horizontally constant while particles settle would lose part of the force.

For a steady interior, let $q=Q/w=uh>0$ denote volume flux per unit width. Then the particle balance is a first-order linear differential equation,Here $g_0^{\prime }=g^{\prime }(0)>0$, $v_s\ge 0$ is constant, and deposited particles are not returned to the current. In particular the decay length is $q/v_s=Q/(w{v}_s)$ for $v_s>0$, and the cancellation of depth follows from $uh=q$: dilution and transit time compensate in the deposition balance.

The steady momentum equation also givesThe inlet depth selects $K=q^2/h_0+g_0^{\prime }{h}_0^2/2$. Depth and speed at every downstream position are determined by the continuous positive root of a polynomial inThis implicit expression also makes clear why the inlet condition is needed: there are generally two possible positive depths.

The depth branches of a steady depositing gravity current are distinguished by the Froude number ${\mathrm{Fr}}^2=q^2/(g^{\prime }{h}^3)$. Differentiating the momentum invariant, and using the exponential decay of reduced gravity, givesFor $v_s>0$, supercritical flow has $\mathrm{Fr}>1$ and decreasing depth, with $h\to q^2/K$ and $u\to K/q$ as $x\to \infty$. Subcritical flow has $\mathrm{Fr}<1$ and increasing depth, with $h\sim \sqrt{2K/g^{\prime }(x)}\propto e^{v_{s}x/(2q)}$. This indefinitely deepening branch eventually violates the assumed shallow or confined-channel geometry. Hence the depth does not universally decrease: that conclusion needs a supercritical inlet.

For fixed $g^{\prime }>0$, $q^2/h+g^{\prime }{h}^2/2$ has its unique minimum at $h_c=(q^2/g^{\prime }{)}^{1/3}$, with $K_c=3q^{4/3}(g^{\prime }{)}^{1/3}/2$. The two positive roots when $K>K_c$ lie below and above $h_c$. Since $K_c$ decreases downstream while $K$ is constant, an initially strict branch remains separate from the other branch and cannot cross a critical point. A critical inlet with $v_s>0$ has no finite depth derivative: at $\mathrm{Fr}=1$ the first differentiated equation would equate zero to a positive value. It requires a singular entrance adjustment or physics beyond this steady model. If $v_s=0$, both reduced gravity and the continuous depth branch are constant. These steady solutions describe the supplied current interior, not a moving finite-release front.

For two particle populations, write $g^{\prime }=g_1^{\prime }+g_2^{\prime }$ and $g_i^{\prime }=g_{0i}{c}_i$, allowing different constant particle-density conversion factors. The bidisperse particle-laden gravity current equations areFor the steady case,Thus the depth is obtained from the same cubic polynomial invariant with the sum of the two exponentially decreasing reduced gravity contributions.

The local deposit consists of the two particle deposition fluxes $v_1{c}_1$ and $v_2{c}_2$. For a positive total deposition rate, the bidisperse gravity-current deposition fraction by particle volume isIf the particles have a common material density, this is also their local mass fraction. If they have common $g_o$, it can be written in the source's buoyancy notation asFor unequal $g_{0i}$ replace each $g_{i0}^{\prime }$ by $g_{i0}^{\prime }/g_{0i}$. The exponent sign matters: with both populations present and $v_1>v_2>0$, the faster-settling population has decreasing local deposit fraction downstream, since $f_1^{\prime }=(v_2-v_1)f_1(1-f_1)/q<0$. Equal positive settling speeds give a constant fraction. A nonsettling population contributes no deposit; if the entire local deposition rate is zero the deposit fraction is undefined.

If “fraction of particles” means number fraction rather than particle-volume or mass fraction, let $a_i$ be a single particle's volume in population $i$. The correct weights are then $v_i{c}_i/a_i$, soSettling speeds alone do not determine $a_i$ without particle properties and a drag law. Finally, if the intended deposit is cumulative over the bed from $0$ to $x$, rather than local at $x$, integrating particle deposition flux gives the species volumes per unit channel width $D_i(x)=q{c}_{i0}(1-e^{-v_{i}x/q})$. The cumulative particle-volume fraction isFor both $v_i>0$, its limit at infinity is the incoming particle-volume fraction, whereas its limit as $x\downarrow 0$ is the local deposition fraction at the inlet. Local sorting and the cumulative inventory are different observables.

For $t>2$, the Heaviside step function contributes a constant unit forcing. Solving the resulting first-order linear differential equation from the value at $t=2$ gives

For $V(\varphi )=\lambda {\varphi }^4$, the slow-roll approximation givesTaking the expanding branch $H>0$ and using $H=\sqrt{8\pi G\lambda /3}{\varphi }^2$ gives the first-order linear differential equationThus the quartic-potential slow-roll solution with the normalization used in this question isIf ${\varphi }_0<0$, the field remains negative and $\dot{\varphi }>0$, so it rolls toward the potential minimum at zero.

Invert the equation and regard $x$ as a function of $y$:This is a first-order linear differential equation. Multiplication by the integrating factor $e^{-y}$ givesso $x=e^{2y}+C{e}^y$. The initial condition $x=1$ at $y=0$ gives $C=0$. Hence $x=e^{2y}$ and

Three distinguished limits are needed.

In the origin layer, put $x=\epsilon X$ and $y(x;\epsilon )=Y(X)$. Since $\mathrm{sech}(\epsilon X)=1+O({\epsilon }^2)$, the leading equation isIt has the first integralSolving this first-order linear differential equation givesRegularity at $X=0$ requires $D=-C$, and $Y(0)=1$ then fixes $C=1$. Thus Region I, $x=O(\epsilon )$, hasIts matching limit is $Y\sim X^{-1}$ as $X\to \infty$.

In Region II, $x=O(1)$ with $x\gg \epsilon$, setting $\epsilon =0$ reduces the second-order equation toHence $y_0=C{e}^{-x}/x$. Matching $C/x$ with the inner limit $\epsilon /x$ fixes $C=\epsilon$, so

The correction generated by $\epsilon x^{3}y$ changes the local decay rate when $x=O({\epsilon }^{-1/2})$. For Region III setThe terms of order ${\epsilon }^{-1/2}$ give the eikonal equationand the next balance gives $Z{G}^{\prime }+G=0$. Matching to Region II selectsThereforeThis solution is already beyond all algebraic orders in $\epsilon$ in the distant region. A leading multiplicative composite expansion that contains all three balances and satisfies the boundary value exactly iswith its value at $x=0$ understood by continuity.

Finally, the coefficient of $y^{\prime \prime }$ vanishes at the origin. Taking the regular limit of the original equation there givesso the prescribed value also fixes $y^{\prime }(0)=-1/(2\epsilon )$. Equivalently, the second inner solution behaves as $1/X$ and is excluded by regularity. This regular singular point is why one boundary condition determines the unique regular solution.

On an interval excluding $t=0$, an integrating factor for this first-order linear differential equation is $t^4$. Multiplying through givesAn antiderivative then gives $t^{4}y=3t^5/5+C$, soThe constant is arbitrary on each connected interval of the domain; the original differential equation is undefined at zero.

The dependent variable itself does not occur, so reduce order by putting $p=y^{\prime }$. Then $p^{\prime }+2p=3x$, a first-order linear differential equation. Its integrating factor is $e^{2x}$:Integrating once more and renaming the constants givesThe two homogeneous terms are the constant solution and the exponential associated with the roots $0,-2$ of the characteristic equation.

For this first-order linear differential equation, the integrating factor is $e^x$. Thusand the general solution is

Because the original differential equation divides by $xy$, its domain requires $x\ne 0$ and $y\ne 0$. Set $u=y^2$, so $u^{\prime }=2y{y}^{\prime }$. The equation becomes the first-order linear differential equationAn integrating factor is $x^2$, since $(x^{2}u{)}^{\prime }=x^2{u}^{\prime }+2xu$. ThusReturning to $y$ gives the explicit general real solutionsEach sign is fixed on a connected interval where $x\ne 0$ and the radicand is strictly positive. The strict condition excludes $y=0$, where the original differential equation is undefined. There is no additional zero solution.

First suppose $\alpha >0$. Differentiation under the integral sign, justified on compact positive parameter intervals by an integrable bound, givesThe first integral is $f(\alpha )/\alpha$. For $\alpha \ne 1$, use the partial fraction decompositionThe standard inverse tangent antiderivatives then giveThe same value holds at $\alpha =1$ by continuity, or by direct integration. Hence the proposed first-order linear differential equation is valid on the positive parameter range:Divide by $\alpha$ to recognise a quotient rule derivative:To fix $C$, the value $f(0)=0$ alone is insufficient, since every displayed solution tends to zero there. Instead considerThis limit follows from the dominated convergence theorem: for $0<\alpha \le 1$ and $0<x<1$, the logarithm's absolute value is bounded by $2|\ln x|+\ln 2$; for $x\ge 1$, it is at most $\ln (1+x^2)$. After division by $1+x^2$, these are integrable. The limiting improper integral is absolutely convergent, and the substitution rule $x=1/t$ changes it into its own negative. Its value is therefore zero, giving $C=0$. ThusThe question calls $\alpha$ a real parameter, but this last formula and the proposed differential equation do not hold on the entire real axis. The defining integral is odd in $\alpha$, because its logarithm depends on ${\alpha }^2$. Applying the positive-parameter result to $|\alpha |$ therefore gives the correct extensionwith $f(0)=0$. In particular, for $\alpha <0$, the right-hand side of the differential equation must be $-\pi \alpha /(1-\alpha )$; for every $\alpha \ne 0$ the unified expression is $\pi |\alpha |/(1+|\alpha |)$. The logarithmic quadratic parameter integral therefore provides the requested identity on its valid range and the corrected answer elsewhere.

For this first-order linear differential equation, an integrating factor on an interval avoiding zeros of $\cos x$ is $\cos x$. The product rule then givesThe initial condition gives $3/\sqrt{2}=1/\sqrt{2}+C$, hence $C=\sqrt{2}$. ThusThe displayed interval is the maximal one containing the specified initial point, since the coefficient $\tan x$ is undefined at its endpoints.

At $x=0$, the integrand vanishes for $0<t<1$, so $I(0)=0$ andSolving the first-order linear differential equation from part (i) on $(-1,\infty )$ gives $g=C(x+1)$. The initial condition fixes $C=1$, soEquivalently, the defining exponent is the logarithmic integral representation of the natural logarithm, $I(x)=\ln (x+1)$.

Write $I(x)=\ln g(x)$. For $x>-1$, differentiation under the integral sign givesIndeed, on any compact parameter interval $c\le x\le d$ with $c>-1$, the differentiated integrand is bounded by the integrable function $t^c$. The dominated convergence theorem therefore justifies the operation despite the improper lower endpoint. The apparent singularity at $t=1$ is removable.

Since $g=e^I$, the chain rule gives the first-order linear differential equation

Writing $v=y-1$ turns the first-order linear differential equation into $v^{\prime }=v$. Its general solution is $v=C{e}^x$, so $y=1+C{e}^x$. The initial condition gives $C=-1$, hence

The general solution of the first-order linear differential equation is $y=1+C{e}^x$. The initial condition $y(0)=1$ gives $C=0$, so the solution of this initial value problem is the constant function $y(x)=1$. Thus $\boxed{y(1)=1}$.

Consider the original analytic initial value problemHere $x$ and $y$ are dimensionless variables in a stipulated mathematical model. The trajectories below are exact solutions of a differential equation, with no measured data or empirical parameter fit. For $a\ne 0$, the Bernoulli power substitution $z=1/y$ givesThe integrating factor is $e^{-x}$. Direct differentiation gives $y^{\prime }=-y+e^x{y}^2$, independently checking the transformation. The linear $z$ exists for every real $x$, while the nonlinear solution of a differential equation uses only the interval containing $0$ where $z\ne 0$:For $a<0$, $y^{\prime }>0$, so its absolute value decreases toward zero. The lost zero solution of a differential equation is recovered by substitution into the original equation. Since $f(x,y)=-y+e^x{y}^2$ is locally Lipschitz in $y$ everywhere, the Picard-Lindelöf theorem ensures uniqueness through every finite point, including $y=0$. Thus a nonzero trajectory never meets or crosses zero at a finite $x$.

For $a=1/2$, the exact solution of a differential equation is $y=e^{-x}/(2-x)$ on $(-\infty ,2)$. It first decreases, reaches a local minimum $y(1)=e^{-1}$, then increases without bound. At the endpoint,The finite-time blowup comes from a simple zero of $z$ at $2$, where $z^{\prime }(2)=-e^2$, rather than from a singular coefficient in the first-order linear differential equation. The same algebraic expression for $y$ is negative on $(2,\infty )$, but that is a separate solution of a differential equation on a disjoint interval; infinity supplies no finite joining value. For $a=-1$, $y=-e^{-x}/(1+x)$ on $(-1,\infty )$, giving a distinct decaying branch whose finite blowup is instead in the past.

The direction field also explains the positive turning point. Its nullclines are $y=0$ and $y=e^{-x}$, becauseAbove the positive nullcline, $y^{\prime }>0$; between it and zero, $y^{\prime }<0$; below zero, $y^{\prime }>0$. The moving nullcline $y=e^{-x}$ is not a solution of a differential equation: its derivative is $-e^{-x}$, whereas the prescribed slope there is zero. The $a=1/2$ trajectory crosses it at $x=1$, exactly its local minimum.

Here is an explicitly stipulated elementary kinetic model, not experimental data. A well-mixed ideal closed system has constant temperature and volume. Alongside an uncatalysed elementary reaction, allow an illustrative catalytic encounter that returns C immediately:Both channels obey the law of mass action. In the second, the forward reaction rate is ${\alpha }_f[\mathrm{A}][\mathrm{C}]$ and the reverse reaction rate is ${\alpha }_r[\mathrm{B}][\mathrm{C}]$. C cancels in each encounter, so its free concentration $c_C$ stays constant without an external supply. This toy elementary reaction encodes regeneration without binding; it is not a claim that real catalytic cycles universally consist of one encounter. A binding cycle can accumulate a reaction intermediate, in which case constant total C does not imply constant free $c_C$.

Adding the parallel reaction rates gives effective first-order reaction rate constantsBecause $k_f^u/k_r^u={\alpha }_f/{\alpha }_r=3$, each channel balances at the same $[\mathrm{B}]/[\mathrm{A}]=3$, and so does their sum. At $c_C=0$ the pair is $(0.030,0.010){\mathrm{s}}^{-1}$; at $c_C=0.001\mathrm{mol}{\mathrm{L}}^{-1}$ it is $(0.300,0.100){\mathrm{s}}^{-1}$. Both give $K=3$ with ideal thermodynamic activities and a common concentration standard state. At $T=300\mathrm{K}$ their compatible standard reaction change in Gibbs free energy is ${\Delta }_r{G}^{\circ }=-RT\ln 3\approx -2.7403\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Choosing an unrelated reverse rate constant would break this consistency.

Let $S=[\mathrm{A}]+[\mathrm{B}]$ be the conserved concentration, and define the dimensionless quantity $x=[\mathrm{B}]/S$. ThenSolving this first-order linear differential equation givesThe chemical relaxation time is $25\mathrm{s}$ without C and $2.5\mathrm{s}$ with the chosen C. From either $x_0=0$ or $x_0=1$, the exponential decay of the deviation leads to the same chemical equilibrium $x=3/4$. Reducing that deviation to one percent takes $\ln (100)\tau$: about $115.13\mathrm{s}$ versus $11.51\mathrm{s}$.

At this chemical equilibrium, conversion continues in both directions. For $S=1\mathrm{mol}{\mathrm{L}}^{-1}$, the uncatalysed channel has equal forward and reverse reaction rates $0.0075\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$. Adding C gives total directional reaction rates $0.075\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$ each, yet zero net reaction rate. Faster molecular exchange and unchanged composition coexist. This exact calculation explains how catalysis can matter enormously within an observation window while leaving the final equilibrium constant intact.

An RL circuit combines electrical resistance $R$ and self-inductance $L$. In the series model, the same electric current $I$ flows through both components. An ideal battery maintains terminal electric potential difference $V$; the drops in the direction of positive electric current obey Ohm's law and the inductor relation:This is Kirchhoff's voltage law with the inductor's terminal drop, rather than its oppositely signed induced electromotive force. For constant $R>0$, $L>0$, and $V$, the first-order linear differential equation has solutionThe time constant $\tau$ measures exponential decay of the difference from the final electric current. ^[ref].

For a stipulated example, take electrical resistance $R=2\Omega$, self-inductance $L=0.5\mathrm{H}$, terminal electric potential difference $V=6\mathrm{V}$, and initial electric current $I(0)=0$. ThenWith $t$ measured numerically in seconds, this is $I(t)=3(1-e^{-4t})$ in amperes. At one time constant,and the terminal electric potential difference drops areTheir sum is exactly $6\mathrm{V}$. At $t=0^{+}$, the inductor takes the entire terminal electric potential difference while the electric current is still zero. At steady electric current, the inductor's terminal drop is zero and the entire drop is across the electrical resistance.

The final magnetic energy stored by the inductor is $U_{\infty }=L{I}_{\infty }^2/2=2.25\mathrm{J}$. At one time constant, the electric current has reached about $63.21\%$ of its final value, but its magnetic energy has reached only about $39.96\%$:This difference follows from the square of electric current in the inductor's magnetic energy.

For release, take $t=0$ at the start of a separate experiment with initial electric current $3\mathrm{A}$. Remove the battery from the model while maintaining a closed path through the same electrical resistance and inductor. ThenNow $v_L=-v_R$: the induced electromotive force supports the continuing positive electric current as its magnitude decreases. This closed-loop model describes exponential decay, not opening the sole path and imposing an instantaneous jump to zero electric current. The inductor's magnetic energy and the power of Joule heating areTheir time constant is $\tau /2=0.125\mathrm{s}$. The total energy dissipated by Joule heating is exactly the initial magnetic energy:The second diagram in inductor shows this energy transfer. A finite closed path lets us predict both how long the electric current persists and how much energy becomes heat. ^[ref].

A singular perturbation is a parameter-dependent problem whose limiting problem changes qualitatively as the parameter tends to zero. In an ordinary differential equation, a small coefficient of the highest derivative can reduce the order of the limiting differential equation, preventing it from satisfying all the original boundary conditions. Different length or time scales often require different asymptotic expansions.^[ref]

Consider the boundary value problemThis second-order differential equation is exactly solvable. Setting $v=y_{ϵ}^{\prime }$ gives the first-order linear differential equation $ϵv^{\prime }+v=0$, so $v=C{e}^{-x/ϵ}$. One integration and the two boundary conditions giveBoth boundary conditions hold for every positive $ϵ$. For numerical evaluation, write the same ratio as $expm1(-x/ϵ)/expm1(-1/ϵ)$, where $expm1(t)$ evaluates the exponential function minus one without catastrophic cancellation near $t=0$. Keep the denominator for finite $ϵ$: at $ϵ=0.2$ it is approximately $0.993262$, rather than exactly one.

Setting $ϵ=0$ gives $y_0^{\prime }=0$. The outer solution selected by the right boundary condition is therefore $y_0=1$, which fails the left boundary condition. The exact exponential function $e^{-x/ϵ}$ shows how that mismatch is repaired: the solution rises from zero to almost one within a boundary layer of width $x=O(ϵ)$, using Big O notation. For each fixed $x>0$, $y_{ϵ}(x)\to 1$, while $y_{ϵ}(0)=0$ always. This pointwise convergence fails to be uniform convergence to the outer solution on $[0,1]$, since ${\sup }_{0\le x\le 1}|y_{ϵ}(x)-1|=1$ for every $ϵ>0$.

The coefficient $ϵ$ is small, but its product with the second derivative need not be small. Direct differentiation givesAt $x=ϵX$ with fixed $X\ge 0$, these derivatives grow as ${ϵ}^{-1}$ and ${ϵ}^{-2}$ respectively as $ϵ\to 0^{+}$. Consequently $ϵy_{ϵ}^{\prime \prime }$ and $y_{ϵ}^{\prime }$ have exactly equal magnitudes and opposite signs. Even away from the boundary layer, their magnitudes remain equal in this example; both become exponentially small at each fixed positive $x$. The outer expansion neglects an exponentially small transient there, rather than establishing that one derivative term is smaller than the other everywhere.

To predict the width without knowing the exact solution, suppose an order-one change occurs over distance $\delta$. The derivative scales are $y^{\prime }=O({\delta }^{-1})$ and $y^{\prime \prime }=O({\delta }^{-2})$. The dominant balance $ϵ/{\delta }^2\sim 1/\delta$ gives $\delta \sim ϵ$. The resulting inner variable $X=x/ϵ$ makes the boundary layer visible at a fixed scale. This is the useful feature of singular perturbation: an apparently disappearing term identifies the missing structure when one measures its derivatives on the appropriate scale.