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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.

The first equation determines $u_m$ linearly from ${\theta }_m$:Substitution into the second equation gives an ordinary differential equation on the finite-dimensional space ${\tilde{H}}_m$. Its right-hand side is polynomial, and therefore locally Lipschitz continuous. The Picard-Lindelof theorem supplies a unique local solution. The energy estimate in part (iii) bounds ${\theta }_m$ on every finite time interval, so the finite-dimensional continuation criterion rules out finite-time escape. The solution is consequently unique on every interval $[0,T]$.

For an unstable Taylor–Goldstein equation mode in a finite channel, take $a=1$ in the power-transformed Taylor–Goldstein energy identity. Its imaginary part givesThus the real phase velocity lies strictly between the extremes of a nonconstant smooth shear profile. Equality would force the regular eigenfunction to vanish on an interval, hence everywhere by uniqueness for its ordinary differential equation.

The maximal time on which the coefficient solution of the exponential-quadratic transform stays finite. For $R^{\prime }=2R^2-2R+1$, $R(0)=0$, the solution is $R=(1+\tan (\tau -\pi /4))/2$ and explodes at $3\pi /4$. Thus local solvability of the coefficient ordinary differential equations does not imply an unrestricted global moment formula. For payoff exponent $0\le \theta \le 1$, the Riccati forcing is nonpositive and a negative equilibrium bounds the solution.

For the massless Klein-Gordon equation on a Kerr black hole, the covariant wave operator separates into radial and angular ordinary differential equations. With $K=(r^2+a^2)\omega -am$ and angular separation constant $\Lambda$, the radial equation is $(\Delta R^{\prime }{)}^{\prime }+[K^2/\Delta -a^2{\omega }^2+2am\omega -\Lambda ]R=0$. The angular equation is $(\sin \theta {\Theta }^{\prime }{)}^{\prime }/\sin \theta +[a^2{\omega }^2{\cos }^2\theta -m^2/{\sin }^2\theta +\Lambda ]\Theta =0$. No extra radial factor is included in the displayed ansatz.

A classical solution is a sufficiently differentiable function that satisfies a differential equation and any prescribed initial or boundary conditions. A weak solution instead satisfies an appropriate integral or distributional identity in a specified function space. For a smooth locally Lipschitz continuous vector field, an ordinary differential equation has a unique local solution for each initial state; this does not by itself ensure global existence or boundedness.