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With the dimensionless variables $x=r/r_s$ and $y=u/c_s$, mass conservation and part (c) giveSubstitution into the Bernoulli function, together with $GM/r=2c_s^2/x$, eliminates every dimensional parameter and gives the transcendental equationEquivalently,orThe transonic branch passes through $(x,y)=(1,1)$.

Source: released examination PDF, bundle pages 230–233.

Solutions to all seven questions, including every optional question and Section C. In Question 6(b), the printed transcendental equation excludes valid separated solutions at common poles of the tangents; the complete characteristic equation and an explicit counterexample are given below.

For $\omega \ne 0$, substitution of the separated solutions into the heat equations givesThe zero exterior boundary values are incorporated byTo obtain a eigenvalue equation for heat conduction in two joined slabs, write $u=\omega {\alpha }_{1}L$ and $v=\omega {\alpha }_{2}L$. Continuity of temperature and heat flux at zero gives the homogeneous linear systemA nonzero amplitude pair exists precisely when its determinant vanishes, equivalentlyIf $\cos u\cos v\ne 0$, division by $k_1{k}_2{\alpha }_1{\alpha }_2\cos u\cos v$ gives the printed transcendental equationThe printed necessity claim needs the qualification that both tangents are defined. If just one of $\cos u,\cos v$ vanishes, $D(\omega )\ne 0$. If both cosines vanish, however, $D(\omega )=0$ and legitimate normal modes remain. Their temperature at the thermal interface need not be zero; the temperature condition fixes the amplitude ratio, while each heat flux is zero there.

For a concrete counterexample, take identical slabs: ${\rho }_1={\rho }_2=\rho$, $c_1=c_2=c$, $k_1=k_2=k$, and $\alpha =\sqrt{\rho c/k}$. Thenwith each formula restricted to its own slab, satisfies both heat equations and all four boundary conditions. But $\omega \alpha L=\pi /2$, so neither tangent is defined. This directly disproves the unqualified printed assertion. The determinant condition retains these missing normal modes.

Finally, $\omega =0$ must be checked before dividing by it. Then $G_1=A(x+L)$, $G_2=B(L-x)$, and interface continuity gives $A=B$, while heat flux continuity gives $k_{1}A=-k_{2}B$. Positivity of the thermal conductivities forces $A=B=0$. Thus $\boxed{\omega =0\text{ gives only the trivial solution}.}$ In the determinant formula, its apparent zero at $\omega =0$ is spurious because the chosen sine basis degenerates there.

Introduce three positive functions of the unknown front parameter:Differentiating the error function profiles gives ${\theta }^{s\prime }=2P{\theta }_{\lambda }$, ${\theta }^{l\prime }=-2Q({\theta }_{\lambda }-1+{\mathcal{R}}_{\theta })$ and ${\mathcal{C}}^{l\prime }=-2R({\theta }_{\lambda }-1)$ at the interface. The three flux conditions are therefore exactlyThese undivided equations are the safest description of the constant-fraction mushy-layer similarity solution, including degenerate parameter choices. In the nondegenerate case the third equation first givesThe second then gives the solid fractionUse the third equation to replace the $Q$ term in the first, and eliminate $\varphi$. The front parameter is a positive root of the single transcendental equationThen substitute back for ${\theta }_{\lambda },\varphi$ and $h(t)=2\lambda \sqrt{\kappa t}$. Since the physical parameters have not been assigned numbers, an implicit scalar equation is the appropriate determination of $\lambda$.

The divided formulas require $R\ne Q$ and ${\theta }_{\lambda }-1+{\mathcal{R}}_C\ne 0$. If either vanishes, return to the boxed three-equation system. For example $Le=1$ gives $R=Q$, so the last equation demands ${\mathcal{R}}_{\theta }=0$; otherwise this proposed similarity solution cannot exist. A physical branch must also have $\lambda >0$ and $0\le \varphi \le 1$. The algebra alone does not establish existence or uniqueness of such a branch for arbitrary parameters.

Consider the closed SIR model in count variables, with $\dot{S}=-rSI$, $\dot{I}=rSI-aI$, transmission coefficient $r>0$ and recovery rate $a>0$. An ideal intervention stops all further transmission at the first upward crossing of $I=I_{\ast }$. This is a mathematical contact-removal rule; its instantaneous perfect effect is an assumption of the model.

Before intervention,so $I+S-(a/r)\log S$ is a first integral of an ordinary differential equation. At the trigger, the remaining susceptible individual count $S_{\ast }$ is a root of an equation satisfyingThis is a transcendental equation, with $S_{\ast }>0$, rather than a polynomial equation.

For an initially growing outbreak with $I_0>0$ and $S_0>a/r$, the unmodified SIR model reaches its infectious-count peak at $S=a/r$, withA trigger with $I_0<I_{\ast }\le I_{\mathrm{peak}}$ is therefore reachable on this growth branch. The function $S-(a/r)\log S$ is strictly increasing for $S>a/r$, so the trigger equation has exactly one solution in $a/r\le S_{\ast }<S_0$. A threshold above $I_{\mathrm{peak}}$ is never reached; choosing another mathematical root would not describe the first upward crossing.

Once the assumed intervention stops transmission, recovery can continue but $S$ no longer changes. The number of new infections among those initially susceptible individuals is thus $S_0-S_{\ast }$. The initial $I_0$ is excluded from that count. An intervention with residual transmission, or a model with unobserved exposed individuals, requires a different final-size calculation.