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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.

Assume an inviscid homogeneous ocean, no horizontal pressure gradient, and no wind stress. Horizontal uniformity then removes advective acceleration, and the parcel equations on a beta plane arewith $u=\dot{x}$ and $v=\dot{y}$.

Since $\dot{u}=f(y)\dot{y}$, integration from the stated initial data givesThis is conservation of the parcel's zonal canonical momentum: its zonal speed records the meridionally accumulated Coriolis acceleration. Multiplying the two momentum equations by $u$ and $v$ and adding givesThus kinetic energy and speed are constant:Eliminating $\dot{x}$ gives the required ordinary differential equation

For $f_0,\beta >0$, the equator is at $y_e=-f_0/\beta$. ThereThe parcel must encounter a meridional turning point before reaching the equator if it is to remain strictly in the Northern Hemisphere. The energy relation therefore requiresor $\beta V/f_0^2<1/2$. Equality is the limiting trajectory that reaches the equator with zero meridional speed.

Setwhere $s>0$. Applying first-order perturbation of a simple eigenvalue to the flux Jacobian, or expanding its quadratic formula directly, separates the characteristic that changes total concentration from the characteristic that changes composition:At $ϵ=0$, both species move with the common hindered velocity $\hat{W}(1-s)$. Summing their conservation laws givesso the total concentration is a nonlinear kinematic wave:Taking the ratio $c={\varphi }_1/s$ instead gives the composition wave in a bidisperse suspensionand henceThese are the requested leading-order ordinary differential equations along the two characteristic families.

The adsorption mechanism is uniform along the two vertical sides, and $X$ evolves independently of $Y$. Consequently the mean first-passage time depends only on the initial $x$-coordinate. Its Kolmogorov backward equation for $a_1=1$ is

The forward partially absorbing boundary condition for a diffusion is $\mathbf{J}\cdot \mathbf{n}=\kappa p$. The boundary term in the adjoint relation isbecause the $X$ diffusion coefficient is one. It vanishes for every admissible $p$ precisely whenFor $\kappa =1$, the two Robin boundary conditions are therefore

The general solution of the ordinary differential equation isThe right condition gives $C=2$, and the left gives $D=-2/e$. At the prescribed initial position,

The characteristic curves are $x=\xi +at$. Along one of them, $U(t)=u(t,\xi +at)$ obeys the separable ordinary differential equationThereforeFor $0\le t<1$, the denominator is positive everywhere. At $t=1$, it first vanishes where $\cos \xi =1$, namely at $\xi =2\pi k$. The solution consequently has finite-time blowup at timealong the points $x=a+2\pi k$. No finite classical solution can continue through that time.

The characteristic flow map solves the ordinary differential equationand henceAlong this characteristic curve, the chain rule givesThe value is therefore constant, and tracing $(t,x)$ back to time zero gives the classical solutionDirect differentiation verifies both the linear transport equation and its initial value.

Regard $\lambda$ as a running coupling $\lambda (M)$. At fixed $s,t,u$ and to order ${\lambda }^2$, differentiating the one-loop answer with respect to $\log M$ givesbecause $m\ll M$ implies $F(M^2)\simeq -x(1-x)M^2$. Thus the beta function isSolving this ordinary differential equation with $\lambda (M_0)={\lambda }_0$ givesto leading-logarithmic order.

WriteThe quasi-steady model is the coupled pairFor a thin shell, $T_s$ is close to $T_m$. A first-order asymptotic expansion givesand thereforeTo leading order $F_c\sim F_{\ast }$, so an initially freezing shell grows linearly:

For a thick shell the conductive correction in the surface balance is small. The radiative equilibrium temperature and its first correction areThus, with $A=a\log (T_m/T_e)$,When $F_w$ is negligible over an intermediate range, this ordinary differential equation gives the square-root growth lawRetaining $F_w$, its implicit solution isConsequently a sketch of $h(t)$ starts approximately linearly, crosses to square-root growth, and approaches $h_{\infty }$ with exponential decay of $h_{\infty }-h$. A shell placed above $h_{\infty }$ instead thins because the basal oceanic heat flux exceeds the conductive loss.