Multiplying the inflaton equation by $\dot{\varphi }$ givesComparison with $\dot{\rho }+3H(\rho +P)=0$ yields $\rho +P={\dot{\varphi }}^2$, and therefore

InsertMatching powers of $t$ requires $\lambda \alpha =2$, so $\alpha =2/\lambda$. Writing $\overline{V}=V_0{e}^{-\lambda {\varphi }_0}$, the scalar equation and Friedmann equation giveThusExistence with positive potential requires $\gamma >1/3$, or ${\lambda }^2<48\pi G$. Accelerated expansion requires $\gamma >1$, hence

Usingwe see that $z=-n$ is generally a simple pole, inherited from $\Gamma (z)$. A cancellation occurs exactly when $p-n$ is a nonpositive integer. Thus $z=0$ is always a simple pole; for $n\ge 1$, $z=-n$ is removable when $p$ is one of $1,2,\dots ,n$, and otherwise is a simple pole.

For any $z\notin \{0,-1,-2,\dots \}$, choose $N$ with $\mathrm{Re}(z+N)>0$ and defineThe recurrence makes this independent of $N$ and agrees with the original integral where $\mathrm{Re}z>0$. It is the desired analytic continuation away from the listed points.

Integration by parts, or the identities obtained by splitting $1=t+(1-t)$, givesTherefore

The substitution $t\mapsto 1-t$ gives $B(p,q)=B(q,p)$. Direct integration gives

For $b(a)=\beta e^{-\lambda a}$ and constant death rate $d$,ThusA separable solution exists with this exponent, and it grows precisely when

Substitution into the renewal boundary condition and cancellation of $A_{0}T(t)$ gives the Euler-Lotka equationThis is the necessary condition selecting the growth exponent $r$.

Put $n(a,t)=A(a)T(t)$. Separation givesfor a constant $r$. Hence

The boundary condition says that all newborn individuals enter at age zero, and that their influx equals the total birth rate obtained by summing the age-specific births $b(a)n(a,t)$ over the population.

Let $A_h$ be the symmetric Dirichlet second-difference matrix. Its eigenvalues are nonpositive, so every eigenvalue ofhas modulus at most one. Thus $\Vert R_h{\Vert }_{2,h}\le 1$.

Insert the exact smooth solution into the scheme. The local truncation error per step has discrete norm $O(k^3+k{h}^2)$, from the order-three trapezoidal time residual and the order-two centred spatial residual. If $e^n$ is the grid error, stability and iteration giveThere are at most $1/k$ steps and $e^0=0$, henceThis proves convergence for every $\mu =k/h^2>0$ without invoking the Lax equivalence theorem.

For the discrete Fourier mode $u_m^n=G^n{e}^{im\theta }$, the centred second difference has multiplier $-4{\sin }^2(\theta /2)$. Hence the amplification factor isFor every $\mu >0$, $|G(\theta )|\le 1$. The discrete Parseval identity then shows directly that the discrete $L^2$ norm cannot increase at any time step, proving unconditional stability for the Cauchy problem.

Taking the squared modulus and using ${\cosh }^{2}u=1+{\sinh }^{2}u$ givesThe incident wavevector makes an angle $\psi$ with the downward vertical, so $\sin 2\psi =2km/(k^2+m^2)$. Substitution gives the required formula.

Matching at $z=L$ first expresses the middle solution in terms of $A_T$. Propagating its value and derivative across the layer of thickness $2L$ uses the transfer matrixMatching this result to $A_I+A_R$ and $-im{A}_I+im{A}_R$ at $z=-L$, then eliminating $A_R$, gives

Continuity of normal velocity and pressure at each density interface requires continuity of $\hat{w}$ and $d\hat{w}/dz$ at $z=\pm L$. These four conditions determine the four unknown amplitudes.

Substitution of the monochromatic ansatz givesThus the outer solutions oscillate with $m=k\sqrt{N_0^2/{\omega }^2-1}$ and the middle solutions are combinations of $\cosh kz$ and $\sinh kz$. The incident wave has phase $kx-mz-\omega t$; its group velocity is upward, whereas the reflected sign has downward group velocity. The transmitted wave must again use the upward-energy sign. This gives exactly the three forms stated.

The similarity form givesUsing the outer pressure gradient ${\rho }^{-1}dP/dr=2A^2/r^5$ in the correctly written radial boundary-layer equation $u{u}_r+w{u}_z=-{\rho }^{-1}{P}_r+\nu u_{zz}$ givesNo slip and matching to the sink require $F(0)=F^{\prime }(0)=0$ and $F^{\prime }(\infty )=1$. The boundary-layer flux is of order $rU\delta$, relative to the sink flux $A$ it isMass conservation therefore requires an outer correction of this relative size, which is small in the assumed region.

The definitions through the Stokes stream function giveso incompressibility is automatic.

Balancing inertia $A^2/r^5$ with viscosity $\nu U/{\delta }^2$ givesThe streamfunction scale is $rU\delta \sim (A\nu r{)}^{1/2}$, motivating

The sink speed scales as $U\sim A/R^2$, so the Reynolds number is $Re\sim UR/\nu =A/(\nu R)$. The approximation applies in the high-Reynolds-number region $R\ll A/\nu$. Radial inertia scales as $UdU/dR\sim A^2/R^5$, so the radial pressure gradient has the same $R^{-5}$ dependence.