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Apply the time Laplace transform to the Airy equation. Extend $g$ past $T$ in any suitable way and writeFor example, setting $g=0$ after $T$ makes $G(p)={\int }_0^T{e}^{-pt}g(t)dt$; the final solution for $t<T$ is independent of this extension. The Laplace transform of a derivative givesTake the principal cube root $\lambda =p^{1/3}$ for $\mathrm{Re}p>0$. The three characteristic roots of the spatial ordinary differential equation areSince $|\arg \lambda |<\pi /6$, only $-\lambda$ has negative real part. Thus spatial decay leaves just one homogeneous exponential, which the single prescribed Neumann boundary condition determines.

The Airy resolvent kernel on the whole real line isIt decays at both ends, is continuous together with its first derivative, and satisfies $R_p^{\prime \prime }(0+)-R_p^{\prime \prime }(0-)=1$. Consequently $({\partial }_s^3+p)R_p={\delta }_0$ as a distribution. A particular solution is the Green-function representationAdding the decaying homogeneous mode to impose the boundary derivative givesEvery term is known. If ${\tilde{u}}_0(r)={\int }_0^{\infty }{e}^{-ry}{u}_0(y)dy$ denotes the spatial Laplace transform, thenThe Bromwich inversion formula now gives the required integral representation:Here $\gamma >0$ is to the right of any singularities required by the growth of the data. The usual decay or growth hypotheses are understood for this Laplace transform construction; when absolute inversion is unavailable, the vertical integral is interpreted as the limit of truncated Bromwich contours. The transformed ordinary differential equation verifies the partial differential equation and initial condition, while differentiating at $x=0$ gives exactly $G(p)$ and hence $u_x(0,t)=g(t)$. The derivative compatibility $u_0^{\prime }(0)=g(0)$ makes the two data agree at the corner.

Choose $a=1$. The power-transformed Taylor–Goldstein energy identity reduces toThe right side is real. For an unstable mode, $c_i>0$, so taking the imaginary part givesConsequently the phase velocity is a weighted average:The weights are nonnegative and not identically zero. For a nonconstant shear profile, equality at an extremum would force $q$ to vanish on an interval where $U$ differs from that extremum; uniqueness for the regular ordinary differential equation would then give the zero mode. This is the phase-speed bound for unstable stratified shear modes.

The shallow-water approximation requires a small depth-to-horizontal-scale ratio, slow streamwise variation of the channel and interface, small vertical acceleration, and rapid enough cross-sectional mixing for $u$ and $\varphi$ to be representative averages. Near the channel apex, or during the initial collapse at a sharp front, these conditions need not hold. The high Reynolds number permits neglect of leading viscous resistance; weak stirring mixes the particles without appreciable ambient fluid entrainment.

Put $\lambda =\beta /H$. At elevation $z$, the channel width is $2\lambda xz$, so the occupied area and interface width areVolume conservation and suspended-particle conservation giveThe excess hydrostatic pressure is ${\rho }_{0}d\varphi (h-z)$. Its streamwise gradient averaged over the triangular section gives acceleration $-d\varphi h_x-(dh/3){\varphi }_x$: the weighted mean of $h-z$ is $h/3$. Thus the triangular-channel shallow water equations areThe quasilinear coefficient matrix in $(h,u,\varphi )$ isIts three distinct real eigenvalues, for $h,d\varphi >0$, establish strict hyperbolicity:On the material characteristic curve, $d\varphi /dt=-2W_s\varphi /h$. On each of the other two characteristic curves, multiplying by the left eigenvector $(\pm 2c/h,1,\pm dh/(3c))$ gives the compatibility ordinary differential equationThe derivatives are along the indicated characteristic, with $g^{\prime }=d\varphi$. At vanishing concentration, the gravity-wave speeds coalesce and strict hyperbolicity is lost.

Put $y(x)=w(z)$ with $z={\sin }^{2}x$. The chain rule givesand henceAfter substitution and division by $4z(1-z)$, the ordinary differential equation becomesComparison with the Gauss hypergeometric equation gives, up to interchanging $A$ and $B$,

At $z=0$ and $z=1$, the coefficient of $w^{\prime }$ has at most a simple pole and that of $w$ has at most a double pole, so both are regular singular points. The regular singular point at infinity criterion gives the same conclusion at infinity. The corresponding Papperitz symbol isIndeed, the exponent pairs are $(0,1-C)=(0,1/2)$ at zero, $(0,C-A-B)=(0,1/2)$ at one, and $(A,B)=(k/2,-k/2)$ at infinity; their sum also satisfies the Fuchs relation.

For ${\lambda }_n=\alpha n+\beta$,so part (b) ensures nonexplosion. The Markov jump-process generator acts on the identity function $f(n)=n$ asConsequently $m(t)=\mathbb{E}N(t)$ satisfiesSolving this linear ordinary differential equation gives the mean of the linear birth process with immigration:

The per-capita net growth rate is the birth rate minus the death rate,Multiplying by the population size gives the ordinary differential equationThis is the exponential density-dependent birth model.

Substitution of $y=e^{-x}$ givesFor a second solution $y=ax+b$, the ordinary differential equation reduces to $a\lambda -b=0$. Taking $a=1$ gives the linearly independent solution

The ordinary differential equations areThe conservation of population identity $S+I=N$ follows by adding the equations.

Use a well-mixed gravity-current box model of length $L(t)$ and depth $h(t)=V_0/L(t)$. Its fixed volume requires $w_e=w_d$. The integrated chemical balance and the standard gravity-current front condition areThese two ordinary differential equations are the required integral model.

First suppose $R_1>0$ and putEliminating time givesWithintegration from $(L,\varphi )=(L_0,1)$ yieldsAs $t\to \infty$, the concentration tends to zero and the runout length of a gravity current isThe formula has a regular $a\to 0$ limit. If $R_1=0<R_2$, direct integration instead giveswhose value at $\varphi =0$ gives $L_{\infty }$.

Assume the condition in part (ii), so $m_0<0$. Substituting the wavevector solution into the vertical ray equation gives the linear ordinary differential equationwhereThe integrating factor $e^{\gamma t}$ givesThe bracket is strictly positive, so $z(t)>0$ at every finite time. Moreover,so the improper integral converges to a finite positive value. It follows thatfor some $C>0$. The packet therefore approaches $z=0$ asymptotically and never reaches it at finite time. This is the vertical trapping of an internal gravity wave by planar strain.

Integrating the ordinary differential equation four times gives a quartic polynomial. It is useful to use the linearity of the equation and split the solution into a gravity part and a force part:wheresatisfies the clamped and torque-free boundary conditions with $F=0$, whilesatisfies the homogeneous equation and contributes the endpoint force. Direct differentiation verifiesThusthe clamped-free beam under uniform load and endpoint force.

HereSubstitution into the higher-order Euler-Lagrange equation gives the fourth-order ordinary differential equation

Apply separation of variables to the Laplace equation in polar coordinates by writing $\varphi (r,\theta )=R(r)\Theta (\theta )$. Division by $R\Theta$ givesThe requirement that $\Theta$ be a real $2\pi$-periodic function restricts the separation constants to $n^2$, with angular factors $\cos (n\theta )$ and $\sin (n\theta )$. For $n\ge 1$, the radial ordinary differential equation is an Euler equation with solutions $r^n$ and $r^{-n}$. For the zero mode, $\Theta$ is constant andso $R=a_0+c_0\log r$. By linearity, superposition givesThis is the separated expansion of a harmonic function in a circular region.

Apply a one-step numerical method to the scalar test ordinary differential equationIf one step has the formthen $R$ is the stability function and the linear stability domain isThe method is A-stable whenso it does not amplify any mode that the exact differential equation does not amplify.

Take the Fourier transform in $x$. The wave equation becomes the ordinary differential equationwith initial data $\tilde{u}(k,0)=\tilde{f}(k)$ and ${\partial }_t\tilde{u}(k,0)=\tilde{g}(k)$. Therefore

Part (c) and the Translation property of the Fourier transform invert the first term toFor the second term choose an antiderivative $p$ of $g$. By part (b), $\tilde{g}=ik\tilde{p}$, and henceIts inverse transform isCombining the terms gives the D'Alembert formula

In free space $Q=0$, and the fourth moment obeysFor the two-dimensional Fourier transformthe Fourier transform of a derivative turns this equation into the ordinary differential equationThusUsing the screen-exit value from part (a), the inverse transform givesthrough first order in $\Delta x$. Equivalently, the free-space fourth-moment propagator has kernel

Put $w=g_T^{-1}(U_T+z)$ and defineThen $f_0(z)=z$. Differentiating with the Chordal Loewner equation givesUniqueness for this ordinary differential equation shows that $f_s=h_s$. At $s=T$,which is the endpoint identity for the Reverse Loewner flow.

For $0<s<T$, the pathwise identity $h_s(z)=g_s^{-1}(U_s+z)-U_s$ is generally false. The left side is built from the reversed final driver segment $(U_{T-r}-U_T{)}_{0\le r\le s}$, whereas the right side is built from the initial segment $(U_r{)}_{0\le r\le s}$. By time reversal and symmetry of Brownian motion they have the same probability distribution, but they are not equal for the given Brownian path.

PutFor an axisymmetric vector field depending only on the cylindrical radius $R$, the solenoidal vector field condition isHence $R{B}_R$ is constant. The hypothesis $B_R\to 0$ on the axis forcesThe azimuthal and axial components of $\nabla \times \mathbf{B}=q\mathbf{B}$ are thenEliminating $B_{\varphi }=-B_z^{\prime }/q$ gives the closed ordinary differential equationOnce $B_z$ is known, $B_{\varphi }=-B_z^{\prime }/q$ and $B_R=0$ determine the full cylindrical force-free magnetic field.

For $\alpha =2$, part iii gives $a=1$, $b=2$, and $\beta =1$. PutSubstitution into the nonlinear partial differential equation gives the ordinary differential equationwith boundary conditions

To find the leading edge, set $y={\eta }_N-\eta$ and suppose $F\sim C{y}^p$. The two singular terms in the ordinary differential equation have orders $y^{4p-2}$ and $y^{p-1}$. Their exponents agree only when $p=1/3$. Their leading coefficients then satisfywhereas $F$ and the constant drainage $D$ are lower-order terms. HenceThe draining gravity current therefore has a one-third-power leading edge and its flux vanishes there.