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Use a time Laplace transform and a spatial Green function. The first-order transport term must be retained: the Airy resolvent kernel for ${\partial }_x^3+p$ alone would solve a different equation. The governing linear partial differential equation is the linear dispersive Stokes equation, distinct from the viscous-fluid Stokes equation.

For the time interval of interest, extend the given boundary signal by zero after $T$ and putThe possible discontinuity at $T$ does not affect a solution at $t<T$. Equivalently, use any sufficiently regular continuation beyond $T$: the resulting causal Green function produces the same earlier solution. We seek the usual solution with spatial decay and a time transform, rather than an unrestricted spatially growing solution on the half-line.

Writing $Q(x,p)$ for the time Laplace transform, the Laplace transform of a derivative givesThe corresponding characteristic roots are the three roots of a polynomialFor $\mathrm{Re}p>0$, label the unique root of negative real part by $r_{-}(p)$, and the two roots of positive real part by $r_1(p),r_2(p)$. This spatial root splitting for the dispersive Stokes resolvent is easily checked: if $r=ik$, then $p=-i(k-k^3)$ is imaginary, so no root crosses the imaginary axis as $p$ varies in the right half-plane. For real $p>0$ the cubic is strictly increasing on the real line and has one negative real root, while its conjugate pair has positive real parts. The root count remains the same throughout that connected half-plane. A repeated root would satisfy $3r^2+1=0$ and again put $p$ on the imaginary axis. All roots are therefore simple there, and $r_{-}(p)$ is a nonzero holomorphic function.

Set $d_j=3r_j^2+1$, including $j=-$. Define the Stokes resolvent kernel with advectionThe sign on the negative half-line is essential. To verify the Green function, use the partial fraction decompositionIts expansion at infinity yields the reciprocal-polynomial root identitiesConsequently $R_p$ and $R_p^{\prime }$ are continuous at zero, and $R_p^{\prime \prime }$ has jump one. Each piece solves the homogeneous ordinary differential equation, so the jump conditions for a third-order Green function imply, in the sense of distributional derivatives,Both pieces decay in their respective spatial directions.

Now defineIt satisfies the transformed inhomogeneous equation. The difference $Q-V$ solves the homogeneous linear ordinary differential equation; spatial decay leaves only $e^{r_{-}x}$. Imposing the Neumann boundary condition fixes its coefficient, giving the half-line Stokes solution with a boundary derivativeNo division by a vanishing root is involved, since $r_{-}=0$ would require $p=0$. Moreover the remaining boundary trace is explicitly known from $q_0$:Thus introduce the completely explicit initial-data kernelFor any upward Bromwich contour $\mathrm{Re}p=\sigma >0$ to the right of any transform singularities, the requested integral representation isThe contour integral is understood in the standard Bromwich inversion formula sense, with truncation limits if it is not absolutely convergent. Every term is expressed in the supplied data and algebraic characteristic roots; there are no unknown Dirichlet or second-derivative boundary traces.

The transformed equation verifies the linear partial differential equation and the initial term, while taking the spatial derivative gives $Q_x(0,p)=G(p)$. Bromwich inversion formula therefore supplies the prescribed derivative for $0<t<T$ and the initial value for $x>0$. For zero initial and boundary data, the transformed solution is a decaying homogeneous mode with zero boundary derivative, hence vanishes; this proves uniqueness in the stated transform/decay class. A change of the boundary continuation after $T$ has transform $e^{-pT}$ times another causal transform and contributes only after $T$, by the Laplace transform time-shift rule. The printed dot on $q_0$ is interpreted as its derivative in its sole variable $x$, so the stated corner compatibility is $q_0^{\prime }(0)=g_1(0)$. Additional differentiability at the corner requires the corresponding higher-order compatibility, but is not needed for the representation at positive $x,t$.

For a variation $y+\epsilon \eta$ with $\eta ={\eta }^{\prime }=0$ at both endpoints, the first variation isTwo applications of integration by parts remove derivatives from $\eta$. The boundary terms vanish, and the fundamental lemma of the calculus of variations gives the higher-order Euler-Lagrange equation

For the stated integrand,so the equation reduces toThe general solution of this linear ordinary differential equation isDecay and finiteness of the integral force $A=B=0$; $y(0)=1$ gives $C=1$, and $y^{\prime }(0)=2$ gives $D=3$. Hence the unique stationary function is

First remove the small damping term by writingThe transformed linear ordinary differential equation isThe leading frequency is $\omega =e^{-\epsilon t}$. Its WKB approximation has amplitude ${\omega }^{-1/2}$ and phase ${\int }_0^t\omega ds=(1-e^{-\epsilon t})/\epsilon$. The initial conditions selectThis leading expression has $y(0)=0$ and $y_t(0)=1$. Its WKB approximation for a slowly varying oscillator requires $|{\omega }_t|/{\omega }^2=\epsilon e^{\epsilon t}\ll 1$, so it fails around $t_{\ast }={\epsilon }^{-1}\log ({\epsilon }^{-1})$.

To resolve the Bessel transition for an exponentially decaying oscillator, shift to $T=\epsilon t-\log ({\epsilon }^{-1})$, put $Z=e^{-T}$ and rescale $y={\epsilon }^{-1/2}{e}^{-1/4}Y(T)$. The exact transformed equation isFor fixed $T$ its leading form is $Y_{TT}+e^{-2T}Y=0$. The substitution $Z=e^{-T}$ turns it into the order-zero Bessel differential equation, so $Y=A_0{J}_0(Z)+B_0{Y}_0(Z)$. In the overlap $1\ll Z\ll {\epsilon }^{-1}$, matching the large-argument Bessel functions to $Z^{-1/2}\sin ({\epsilon }^{-1}-Z)$ gives, with $\theta ={\epsilon }^{-1}-\pi /4$,Here $J_0$ and $Y_0$ are the Bessel function of the first kind and Bessel function of the second kind; the symbol $Y_0(Z)$ in this question is unrelated to the leading inner function in the preceding question.

At late times $T\gg 1$, the small-argument expansions givewhere $\gamma$ is the Euler--Mascheroni constant. The solution becomes asymptotically linear rather than maintaining the exponentially growing WKB envelope. Its leading late-time slope is ${\epsilon }^{1/2}{e}^{-1/4}\sqrt{2/\pi }\cos \theta$. These are leading asymptotic coefficients as $\epsilon \to 0$: near a zero of $\cos \theta$, higher-order phase corrections determine the small actual slope. The formula is not an absolute-error estimate uniform to arbitrarily late times. The exact late slope of an exponentially damped oscillator, obtained from a Kummer function and a Wronskian, provides a separate check even near those exceptional phases.

Use the Minkowski metric $\eta =\mathrm{diag}(-1,1,1,1)$ and define the electromagnetic field tensor byAs a rank-two Lorentz tensor, it transforms under a Lorentz transformation asTwo independent quadratic electromagnetic field invariants are

With $A^{\mu }=(\varphi /c,\mathbf{A})$, $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$, and $\mathbf{B}=\nabla \times \mathbf{A}$, the components areDirect contraction giveswhere the sign of $I_2$ changes if the opposite spacetime orientation is chosen. Since both are complete tensor contractions, $S$ and $T$ are Lorentz scalars.

For a constant uniform field, the relativistic Lorentz force is a constant-coefficient linear ordinary differential equation. Its solution is the matrix exponential

Now $S=T=0$ implies that the field is a null electromagnetic field:The stated axes allow us to choosePut $\Omega =qB/m$. On the $(ct,x,y)$ coordinates, the mixed tensor divided by $B$ isThe particle starts with four-velocity $u(0)=(c,0,0,0)$, so the exponential terminates after its quadratic term and givesIntegrating with the particle initially at the origin gives the relativistic trajectory in a constant null crossed fieldEliminating $\tau$ yieldsIf the original positive $x$ direction is opposite to $\mathbf{E}$, both the sign of $x$ and the displayed signed coefficient reverse.

The characteristic polynomial of the homogeneous linear ordinary differential equation isso its complementary solution is $A{e}^{3x}+B{e}^{-x}$. Because the forcing contains the resonant factor $e^{-x}$, set $y=e^{-x}v$. Substitution givesWriting $w=v^{\prime }$, a polynomial particular solution of $w^{\prime }-4w=-16x$ is $w=4x+1$, and hence $v=2x^2+x$. ThusBoundedness as $x\to \infty$ forces $A=0$, while the initial condition $y(0)=1$ gives $B=1$. Therefore

For $p=0$ and $\lambda =1$, the linear ordinary differential equation is $y^{\prime \prime \prime \prime }=y$, whose general solution isBy the stated parity reduction, an even eigenfunction has the form $y=A\cos x+C\cosh x$. The conditions $y(c)=y^{\prime }(c)=0$ have a nonzero solution exactly whenthat is,An odd eigenfunction has the form $y=B\sin x+D\sinh x$, and the corresponding determinant givesThus $\lambda =1$ is an eigenvalue precisely under one of the two stated conditions.

For the minus sign, division by $\cos c\cosh c$ away from $c=\pi /2$ gives $\tan c=\tanh c$. On $(0,\pi /2)$, $\tan c>\tanh c$ for $c>0$, while on $(\pi /2,\pi )$ their signs differ, so there is no root. For the plus sign the equation isThere is no root in $(0,\pi /2)$. On $(\pi /2,\pi )$, the function $\tan c+\tanh c$ increases strictly from $-\infty$ to $\tanh \pi >0$, so it has exactly one root by the intermediate value theorem. Numerically,Hence the combined $\pm$ condition has exactly one solution in $0<c<\pi$.

Use the singular system of a compact operator convention $A{x}_j={\sigma }_j{y}_j$ and $A^{\ast }{y}_j={\sigma }_j{x}_j$. Starting from zero gives no component in $\ker A$. Write $x(t)={\sum }_j{c}_j(t)x_j$. The coefficient equations areSolving these scalar linear ordinary differential equations givesConsequently asymptotic regularization has the spectral regularization method representationThe component of $f$ orthogonal to the closure of the range of $A$ is annihilated by $A^{\ast }$ and contributes nothing. The scalar filter satisfies $(1-e^{-{\sigma }^2/\alpha })/\sigma \le \min (\sigma /\alpha ,1/\sigma )\le {\alpha }^{-1/2}$, so every fixed-$\alpha$ reconstruction is a bounded linear operator. Its long-time limit is the Moore–Penrose inverse of an operator on admissible data.

The normalized ansatz intended in the question isfor which $\varphi (1,t)=1$. Substitution into the equation from part (ii) and cancellation of $(s-1)\varphi$ gives the linear ordinary differential equationThusThis generating function is that of a Poisson distribution with parameter $f(t)$. Equivalently, differentiating at $s=1$ givesConsequently

Given a solution $u(t)$, construct $a_j(t)=\frac{1}{2}{e}^{u_j(t)/2}$ and hence the smooth matrices $L(t)$ and $B(t)$. Let $U(t)$ solve the linear ordinary differential equationBecause $B(t)$ is skew-symmetric,so $U(t)$ is an orthogonal matrix and is invertible. Moreover,It follows thatThus the Lax pair flow is an Isospectral Lax equation: every $L(t)$ is similar to $L(0)$ and has the same eigenvalues and characteristic polynomial.

Every symmetric function of these constant eigenvalues is a first integral. In particular, the trace invariants of a Lax equationare constant because the cyclic property of the trace givesFor example,is an explicit first integral of equation $(†)$.

For part (a), the adjoint operator isUsing the canonical commutation relation $[X,P]=i\hslash I$,Thus $A$ and $A^{†}$ are the annihilation and creation operators of the quantum harmonic oscillator.

For part (b), writeSince $[A,K]=A^{†}$ and $[A^{†},K]=A$, differentiation of the unitary conjugation givesThe initial conditions are $A(0)=A$ and $A^{†}(0)=A^{†}$. Solving this pair of linear ordinary differential equations gives the Bogoliubov transformationThis is the transformation implemented by the single-mode squeeze operator.

For part (c), invert the definition of $A$:The transformation above impliesThe ground state has zero position and momentum means and variances $\hslash /(2m\omega )$ and $m\hslash \omega /2$. Hence the squeezed vacuum state $|\gamma ⟩=S(\gamma )|0⟩$ hasThe reciprocal squeezing factors cancel, soThe state therefore continues to saturate the Heisenberg uncertainty relation.

Let the source's hydrogen-ionizing photon production rate bewhere $h{\nu }_0$ is the hydrogen ionization energy. The idealized Strömgren sphere has an almost fully ionized interior and a thin ionization front. In photoionization equilibrium, every ionization is balanced by a Case B recombination, so spherical symmetry givesFor pure hydrogen of constant number density $n$, the interior has $n_e=n_p=n$, and the Strömgren radius is therefore

Now let the effective number of dust grains per hydrogen nucleus be $f_d$, so the dust absorption coefficient is $a(r)=f_d{\sigma }_{d}n(r)$. If $Q(r)$ is the ionizing-photon rate crossing the sphere of radius $r$, recombinations and dust absorption give the linear ordinary differential equationIf $f_d$ denotes a dust mass fraction instead, the grain mass and gas mean particle mass are simply absorbed into the effective product $f_d{\sigma }_d$. Define the dust optical depthMultiplication by the integrating factor $e^{{\tau }_d(r)}$ and integration to the dusty front gives its governing equationThe exponential function weights recombinations at large optical depth by the extra source photons that dust must remove before those photons reach that radius.

For constant $n$, put $a=f_d{\sigma }_{d}n$ and $\tau =a{R}_d$. The elementary integralreduces the equation toEquivalently, with $y=R_d/R_S$ and ${\tau }_S=a{R}_S$,Its dust-free limit is $y\to 1$, while absorption makes $R_d<R_S$ for nonzero dust abundance.

Insert the planets' two secular eigenmodes and define their forcing strengthsSolving the first-order linear ordinary differential equation givesThe first term is the particle's freely precessing eccentricity. The remaining terms are its forced eccentricity, phase-locked to the planets' modes. In the complex plane, their vector sum makes the eccentricity and longitude of periapsis oscillate. A denominator becomes small at a secular resonance $A=g_k$; the nonresonant formula then ceases to be uniform.

At a regular point of a second-order linear ordinary differential equation, a nontrivial solution cannot have both $x$ and $x^{\prime }$ zero, by initial-value uniqueness. Its zeros are therefore simple. Under the linearization of a Riccati equation, such a zero $t_{\ast }$ produces $y=x^{\prime }/(cx)\sim 1/[c(t-t_{\ast })]$. Zeros separate the intervals on which the Riccati equation solution is finite.

For a radial function in three dimensions, $(\Delta -1)y=g(x)$ becomes $y^{\prime \prime }+2y^{\prime }/x-y=g(x)$. Setting $u=xy$ gives $u^{\prime \prime }-u=xg(x)$, reducing it to a constant-coefficient linear ordinary differential equation. If $g=e^{-3x}/x^2$, a decaying particular solution obtained by variation of parameters isThe small-argument expansion of the exponential integral givesThe undetermined decaying homogeneous coefficient $\alpha$ is fixed by a boundary condition or by a matched asymptotic expansion.

Suppose two complex amplitudes satisfy ${\partial }_t|A{|}^2\le 2a|A||B|-2d|A{|}^2$ and ${\partial }_t|B{|}^2\le 2c|A||B|-2d|B{|}^2$, with positive $a,c$ and $d\ge 0$. Then $N=c|A{|}^2+a|B{|}^2\ge 2\sqrt{ac}|A||B|$ gives $\dot{N}\le 2(\sqrt{ac}-d)N$. A positive weighted norm avoids treating two coupling rates of different size as a single arithmetic sum. This is useful for nonautonomous linear ordinary differential equations and the bounded-modulation alpha-Omega growth estimate.