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Use the printed coordinate derivative $u_x(0,t)$, not an outward normal derivative; at the left endpoint these have opposite signs. Assume the usual bounded or admissibly growing solution at infinity, with an initial trace $b_0=u_0(0)$ and forcing possessing a Laplace transform. Set $a=\alpha /2$, so $\beta =a^2$, and use $z=\sqrt{p}$ on the principal square root branch. DefineThe half-line Dirichlet Green function for $p-{\partial }_x^2$ givesWrite the transformed solution as $U=U_D+e^{-zx}B(p)$. The Laplace transform of a derivative in the dynamic boundary condition is $pB-b_0$, soThe special choice of $\beta$ gives the perfect squareIn particular, the initial boundary trace cannot be omitted. Withan integral representation isChoose the Bromwich contour to the right of the forcing's growth abscissa and every pole; $\gamma >\max (0,a^2)$ and sufficiently large for the data is a safe choice for real $a$. No sign of $\alpha$ is explicitly imposed in this question.

The inverse transform can also be performed explicitly. For the repeated-root dynamic-boundary heat kernel, setFirst integrate the Heat Poisson kernel against $e^{ar}$, or differentiate the result with respect to $a$:Since differentiation of $(\sqrt{p}-a{)}^{-1}$ with respect to $a$ produces $(\sqrt{p}-a{)}^{-2}$, and ${\partial }_a{E}_a=(-x+2at)E_a+2\sqrt{t/\pi }{e}^{-x^2/(4t)}$, the desired kernel isHere $\mathrm{erfc}$ is the complementary error function. A direct integral characterization, also proving the sign and the transform, isThe convolution theorem for Laplace transforms now gives the fully real time-domain representationAll quantities here are known from the prescribed data. For $x>0$ it tends to $u_0(x)$ as $t\downarrow 0$. At $x=0$, $J_a(0,t)\to 1$ and the spatial correction tends to zero, recovering $b_0$. A solution classical through the initial corner additionally needs $u_0^{\prime \prime }(0)+\alpha u_0^{\prime }(0)+a^2{b}_0=f(0)$; weaker corner regularity does not invalidate the formula for positive times.

The repeated-root unstable heat boundary mode gives a useful sign check on the dynamic boundary condition for the heat equation. If $a>0$, $(\sqrt{p}-a{)}^2$ has a genuine double pole at $p=a^2$ on the physical branch, and the exact homogeneous mode $e^{-ax+a^{2}t}$ satisfies both the heat equation and the printed boundary condition. Generic data can also excite a $t{e}^{a^{2}t}$ contribution. A contour deformation must retain this repeated-pole contribution. If $a<0$, the putative root $\sqrt{p}=a$ is outside the chosen branch and is not a physical pole. At $a=0$, $J_0=\mathrm{erfc}(x/(2\sqrt{t}))$, as expected when the boundary trace satisfies $b^{\prime }(t)=f(t)$.

Let $Y(s)$ be the Laplace transform of $y$. The initial values give $\mathcal{L}(y^{\prime })=sY$ and $\mathcal{L}(y^{\prime \prime })=s^{2}Y-1$, while $\mathcal{L}(t{e}^{-3t})=(s+3{)}^{-2}$. Therefore, for $\mathrm{Re}s$ sufficiently large,The partial fraction decomposition isInverting the elementary Laplace transforms givesThese coefficients give $y(0)=0$ and $y^{\prime }(0)=1$; the transformed equation, or direct differentiation, verifies the forcing. Uniqueness follows from uniqueness for the corresponding linear initial-value ordinary differential equation.

A frailty model represents unobserved heterogeneity using a nonnegative frailty random variable $U$ that multiplies an individual's baseline hazard. With $H_0(t)={\int }_0^t{h}_0(s)ds$, the individual and population survivor functions areThus the population survivor function is the Laplace transform of the frailty distribution at $H_0(t)$; at points where differentiation is justified, $\overline{h}(t)=h_0(t)\mathbb{E}[U\mid T>t]$. The last identity follows by differentiating the integral and dividing $-{\overline{S}}^{\prime }$ by $\overline{S}$.

Here $g(u)=e^{-u}$ for $u\ge 0$, $\theta >0$, and $H_0(t)=\theta t^2/2$. Direct integration givesThis is population hazard under exponential frailty. For every fixed $u>0$ the individual hazard function $u\theta t$ increases linearly. In contrast,so the population hazard increases up to $t=\sqrt{2/\theta }$, then decreases, tending to zero like $2/t$. More vulnerable individuals leave the risk set earlier, leaving survivors with smaller frailties. This population decline is selection, not a decline in any fixed individual's hazard. If $\theta =0$, survival is identically one and both hazards vanish; the positive-rate comparison is then inapplicable.

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 $x,t$.

The drift makes the natural measure weighted. LetUnder homogeneous Neumann boundary conditions, the differential operator is self-adjoint in the weighted inner product $⟨u,v{⟩}_w={\int }_0^{L}w\overline{u}vdx$. Indeed integration by parts giveswhen the derivative boundary terms vanish. In particular its eigenvalues are nonpositive and it has the constant stationary eigenfunction. We first construct the weighted Neumann heat kernel with constant drift, then incorporate the two prescribed coordinate derivatives.

Put $b=\beta /2$. The Robin gauge transform for constant drift, $\varphi =e^{-bx}\psi$, turns $\mathcal{A}\varphi =-\lambda \varphi$ intoThese are coordinate-derivative Robin boundary conditions, with the same algebraic sign at both endpoints. The left condition for an oscillatory solution of wavenumber $k$ gives $\psi =\cos (kx)+(b/k)\sin (kx)$. The right condition then reduces to $(k+b^2/k)\sin (kL)=0$, giving $k_n=n\pi /L$. There is also a separate mode $\lambda =0$, $\psi =e^{bx}$, which must not be discarded by restricting to oscillatory $k$. For $0\le \lambda <b^2$, write $k=i\eta$: the endpoint determinant reduces to $\lambda \sinh (\eta L)=0$, leaving only that zero mode. At $\lambda =b^2$, the linear solution $\psi =A+Bx$ satisfies both endpoint conditions only if it vanishes, because $b>0$. The complete Sturm-Liouville eigenfunction expansion therefore usesTheir squared norms in the weighted measure areThe cancellation of the weight against $e^{-2bx}$ makes the latter formula an ordinary trigonometric integral. DefineFor $t>0$ this kernel is smooth in the interior. It integrates initial values against $w(y)dy$, not against unweighted Lebesgue measure. It is symmetric in $x,y$ as a weighted kernel, even though the drift operator does not look symmetric in an unweighted inner product.

To determine the boundary forcing, set $a_n(t)=⟨{\varphi }_n,q(\cdot ,t){⟩}_w/N_n$. Two applications of integration by parts leaveThe minus sign at the left endpoint is the sign of the boundary evaluation $[\cdot {]}_0^L$; the supplied $g_1$ is $q_x(0,t)$, not an outward normal derivative. Solving this linear differential equation by an integrating factor and recombining the modes gives the requested space-time integral representation:This is the Neumann boundary-forcing formula. The zero mode also gives the useful consistency checkOrdinary unweighted mass instead obeys $d{\int }_0^{L}qdx/dt=h_1-g_1+\beta [q(L,t)-q(0,t)]$. Replacing the weighted measure by $dy$ or deleting the constant mode would therefore give an incorrect solution.

There is a subtle point when checking nonzero derivative data: every homogeneous eigenfunction has zero endpoint derivative, but differentiating the boundary-forcing sum and its time integral term by term at an endpoint is not valid. The kernel is singular as $t-s\downarrow 0$, and the prescribed derivatives are interior limits of the full solution. An explicit Laplace transform form verifies those limits without this interchange and gives an alternative purely contour-integral version.

For $\mathrm{Re}p>0$, set $\kappa =\sqrt{p+b^2}$ using the principal square root, and defineBoth solve $(p-\mathcal{A})f=0$, with $u_p(0)=v_p(L)=1$, $u_p^{\prime }(0)=v_p^{\prime }(L)=0$. Their weighted Wronskian is the nonzero constantHence the resolvent kernel for Neumann advection-diffusion on an interval isThe first derivative in $x$ has jump $-e^{-\beta y}$, giving $(p-{\mathcal{A}}_x){\mathcal{R}}_p=e^{-\beta y}{\delta }_y$ in the sense of distributional derivatives. If $G(p)={\int }_0^T{e}^{-ps}{g}_1(s)ds$ and $H(p)={\int }_0^T{e}^{-ps}{h}_1(s)ds$, an equivalent transformed solution isIn fact $v_p^{\prime }(0)=-D_p$ and $u_p^{\prime }(L)=e^{-\beta L}{D}_p$, so $Q_x(0,p)=G(p)$ and $Q_x(L,p)=H(p)$ exactly. The inverse Bromwich contour integral of $e^{pt}Q(x,p)/(2\pi i)$ is an alternative final answer. Its poles at $p=0$ and $p=-{\lambda }_n$ reproduce the constant and decaying modes of $K_{\beta }$, respectively; apparent square-root singularities in the normalized hyperbolic factors are removable. This also independently checks the kernel normalization and boundary signs.

As in question 1, the printed dot on $q_0$ denotes its spatial derivative: the two corner conditions are $q_0^{\prime }(0)=g_1(0)$ and $q_0^{\prime }(L)=h_1(0)$. The initial eigenfunction expansion converges to $q_0$ in the weighted space and, for the stated smooth compatible data, has the appropriate classical initial and boundary limits. Differences of solutions with zero data satisfy $\frac{1}{2}d\Vert q{\Vert }_w^2/dt=-{\int }_0^{L}w|q_x{|}^{2}dx$, proving uniqueness in the usual regular solution class. In the limit $\beta \to 0$, $N_0\to L$, ${\varphi }_n\to \cos (n\pi x/L)$, and the ordinary Neumann heat kernel on an interval and its boundary signs are recovered.

Let $r_i$ be the probability of returning to $i$ after its first departure. Transience means $r_i<1$; in particular $q_i>0$. By the Strong Markov property, the number $N$ of visits, including the initial visit, has the geometric distribution $\mathbb{P}(N=n)=(1-r_i)r_i^{n-1}$, $n\ge 1$. The holding times of these visits are independent $\mathrm{Exp}(q_i)$ variables and are independent of the jump-chain return decisions.

For the total occupation time of a continuous-time Markov chain $T_i$ and $s\ge 0$, summing over $N$ givesUniqueness of the Laplace transform identifiesStarting elsewhere adds an atom at zero if the state might never be hit; the purely exponential conclusion uses the specified initial state $i$.

An M-M-1 queue has a Poisson arrival process of rate $\lambda$, independent exponential service times of rate $\mu$, and one server, with an unlimited waiting room and the usual first-come-first-served discipline. Its number-in-system process is a continuous-time birth-death chain with $q_{j,j+1}=\lambda$ and $q_{j,j-1}=\mu$ for $j\ge 1$. For $\rho =\lambda /\mu <1$, detailed balance givesThis sums to one and satisfies ${\pi }_j\lambda ={\pi }_{j+1}\mu$. The nonexplosive irreducible chain therefore has an invariant probability and is positive recurrent. Equivalently its negative drift above zero gives finite mean return cycles. Here queue length includes the customer in service.

In equilibrium, PASTA gives an arriving customer's number of predecessors the distribution $\pi$. Conditional on $j$ predecessors, memoryless property makes its total waiting-plus-service time a sum of $j+1$ independent $\mathrm{Exp}(\mu )$ variables. Its Laplace transform isThereforeThis is the sojourn time including service, not just the time until service begins.

The Laplace transform isUsing the supplied Gaussian integral after $t=x^2$ gives

Because $\mathrm{Re}a>0$, the endpoint factor $(1+t^2{)}^a$ vanishes at the branch points $t=\pm i$. A finite contour may therefore run from $-i$ to $i$ on either side of a chosen branch cut; the diagram shows contours bowing through the left and right half-planes.

For $a=1$, $f(t)=1$ is entire, so path independence makes either contour give, up to orientation,Hence a nonzero constant multiple gives the solution

Taking the Laplace transform in $t$ and using the initial condition $u(r,0)=u_0$, the radial heat equation in three dimensions becomesPut $W=rU$ and $q=\sqrt{s}$. ThensoFiniteness at the centre forces $B(s)=0$.

The transformed boundary condition isSincesubstitution and cancellation of the two $u_0/s$ terms givesHenceand the required explicit transform is

A frailty random variable is an unobserved positive multiplicative risk factor. A proportional frailty model hasThe scale of $U$ is not separately identifiable from $h_0$: multiplying $U$ by a constant and dividing $h_0$ by it leaves the model unchanged. We may therefore normalize $\mathbb{E}U=1$, which lets $h_0$ represent the mean initial hazard multiplier and makes relative frailty interpretable.

If $U\sim \mathrm{Exponential}(1)$ and $h_0(t)=\lambda$, then the Laplace transform of $U$ givesThus $h(0)=\lambda$, while $h(t)\to 0$ as $t\to \infty$. Survivors become increasingly enriched for low-frailty individuals.

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.

IfthenA nonnegative random variable with finite expectation is finite almost surely, so the process explodes.

Conversely, suppose the series diverges. For $s>0$, independence and the Laplace transform of an exponential variable giveThe sum in the exponent diverges. If infinitely many ${\lambda }_n\le s$, infinitely many terms are at least $\log 2$; otherwise, eventually $s/{\lambda }_n\le 1$ andThus $\mathbb{E}{e}^{-s{T}_{\infty }}=0$. Since $e^{-s{T}_{\infty }}$ is strictly positive on $\{T_{\infty }<\infty \}$, this forces $T_{\infty }=\infty$ almost surely. Hence the nonexplosion criterion for a pure birth process is

The string is initially at rest, so $y(x,0)=y_t(x,0)=0$. Taking the Laplace transform in $t$ converts the wave equation intoFor $\mathrm{Re}p>0$, the solution that decays as $x\to \infty$ is $\hat{y}=A(p)e^{-px/c}$. The transformed boundary condition $\hat{y}(0,p)=\hat{f}(p)$ fixes $A=\hat{f}$, and therefore

Let $N$ be the number of visits to the server made by one customer. Then $N$ has the geometric distribution with success parameter $\delta$, and its service periods $E_1,E_2,\dots$ are independent exponential random variables of rate $\mu$. For $s\ge 0$, the Laplace transform of the total service time $T=E_1+\cdots +E_N$ isThus

At each service completion, a successful departure reduces the queue length by one and occurs at rate $\delta \mu$ while the queue is nonempty; feedback completions leave its length unchanged. The queue-length process is therefore the birth-death process of an ordinary $M/M/1$ queue with service rate $\delta \mu$. It is positive recurrent exactly when $\lambda <\delta \mu$, and then, with $\rho =\lambda /(\delta \mu )$,