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Sliding and cavities. Regelation is melting under increased pressure and refreezing when the pressure is released. For ice near its melting point, the Clapeyron equation gives a lower melting temperature at greater pressure. Against an obstacle, ice melts on the high-pressure upstream face, the water passes around it, and it refreezes on the lower-pressure downstream face. Transport of the released latent heat back towards the melting face sustains the process. Together with viscous deformation around obstacles, this permits basal sliding; regelation is especially effective for small obstacles, for which heat conduction distances are short.

Subglacial cavitation occurs where the contact pressure predicted behind an obstacle falls below the water pressure: the ice separates from the bed and a water-filled cavity opens. The effective pressure in subglacial drainage isthe ice overburden pressure minus water pressure. Smaller $N$ facilitates larger cavities and reduces the resisting ice–bed contact. The exponent $n$ comes from Glen's flow law, with strain rate proportional to the $n$th power of deviatoric stress, commonly $n\simeq 3$. Its homogeneity makes $N^n$ the natural velocity scale, yielding $\tau /N=f(u/N^n)$ after including the fixed bed and ice-rheology scales.

Without cavities, material creep gives a drag law with $f(x)\propto x^{1/n}$. Cavities lower the resistance relative to that extrapolation. On a bounded-slope bed the drag is bounded in proportion to $N$; depending on bed geometry, the curve may level off or pass through a maximum and weaken as contact diminishes. The illustrated weakening curve is a qualitative possibility, not a universal formula for all beds.

Channels versus distributed drainage. In Röthlisberger channel drainage, frictional dissipation of flowing water supplies heat to melt the channel walls, while material creep of ice closes them under effective pressure. An area balance has the schematic formwhere $\Psi$ is hydraulic-potential drop per length, $L_f$ is latent heat per mass, and the positive constant $\mathcal{A}$ incorporates ice material creep and geometry. Water-flow resistance relates the flux to channel size and potential gradient. Balancing melting against closure produces the stated increasing relation $N_R(Q)$.

A channel gaining flux can maintain lower water pressure, since larger $N$ means smaller $p_w$. It draws still more water from neighboring routes, melts wider, and causes competing small conduits to lose flux and close. This positive feedback concentrates flow into an arterial network.

A linked cavity drainage system consists of water-filled lee-side cavities connected by narrow passages. Sliding over bed obstacles opens cavities; ice material creep tends to close them. The given decreasing relation $N_K(Q)$ means that carrying more water requires higher water pressure. A route gaining flux then has a higher pressure and diverts water towards neighboring routes. This negative redistribution feedback supports many connected routes and a distributed system.

Coexistence instability. At a coexistence point let $N_R(Q_R)=N_K(Q_K)$ and hold $Q=Q_R+Q_K$ fixed. A loss $\Delta Q_R=-\epsilon$ from the channel gives $\Delta Q_K=\epsilon$. The changes in the two equilibrium effective pressures areConsequentlyIf the bracket is positive, the cavity system's common effective pressure exceeds what the reduced-flux channel can support against material creep closure. The channel shrinks, loses further flux, and closes down. Thus the channel-cavity drainage instability condition is

Balances and meanings. The first equation is conservation of energy per unit area. Here $c$ is effective areal heat capacity, $Q$ is the solar constant, and $Q/4$ accounts for the ratio of the Earth's intercepting disc area to its total surface area. The albedo $a$ reflects a fraction of that input. The outgoing thermal radiation is the Stefan–Boltzmann law modified by $\gamma (p)$, an effective transmission factor for the greenhouse effect; it normally decreases as atmospheric carbon dioxide increases.

The albedo relaxes towards its temperature-dependent equilibrium on the ice-adjustment time $t_i$. The atmospheric mass inventory per unit pressure is $M_{{\mathrm{CO}}_2}A/(M_{a}g)$, using the molar mass ratio of carbon dioxide to air and hydrostatic atmospheric mass. Its sources and sinks are volcanic input $v$, removal by chemical weathering over land area $A_L$ at rate $W$, and the atmosphere-to-ocean exchange $h(p-p_s)$. The ocean equation is a molar mass conservation balance: ${\rho }_{{\mathrm{H}}_2\mathrm{O}}{V}_{\mathrm{oc}}$ is ocean water mass, division by $M_{{\mathrm{CO}}_2}$ converts the incoming carbon-dioxide mass to moles, and $bC$ removes dissolved carbon by burial/export. With $C$ in moles per kilogram, $b$ has units of kilograms of water-equivalent per year.

Times and reduction. Linearize the temperature equation at fixed $a,p$ about radiative equilibrium. Its exponential relaxation time isThe equality uses $\sigma \gamma T^4=Q(1-a)/4$. A stock-over-flux thermal time, sometimes used as a scale, is four times this, $0.381\mathrm{y}$; both describe a rapid thermal response relative to the other processes.

For pressure in atmospheres, write $I=M_{{\mathrm{CO}}_2}A10^5/(M_{a}g)$ for the atmospheric carbon-dioxide mass per atmosphere. The factor $10^5$ is essential because $h$ is given per atmosphere. The pressure and slow ocean exponential relaxation times areThus $t_T\ll t_p\ll t_i=10^4\mathrm{y}<t_C$. The leading quasi-steady approximation givesSubstitution in the two remaining equations yieldsThe coefficient of $\dot{a}$ remains $t_i$; the printed $t_T$ in the reduced equation is a typographical error. Eliminating a fast temperature variable does not replace the ice-response time. The ocean time here is the slow time after atmospheric balance cancels the exchange flux; holding pressure fixed after imposing a carbon-dependent $p_s$ would describe a different conditional relaxation.

The supplied volcanic input givesabout $41.1\%$ of the stated present value $2.0\times 10^{-3}$; that value is about $2.43C_0$.

Carbon-dioxide levels. Since $A_{L}W=3.0\times 10^{11}\mathrm{kg}{\mathrm{y}}^{-1}=v$, the atmospheric balance with $p_s=C/K$ givesThis uses the specified present ocean concentration, as requested, rather than replacing it by the model's smaller volcanic-only equilibrium.

After sustained additional input $v_i$, $C$ barely changes during the atmospheric adjustment. At this stageThis is an intermediate balance, approached over several centuries. On the slower ocean time,with pressure following $p\simeq C/K+v_i/h$. Since $C(0)<C_{\infty }$, pressure continues to rise slowly. Its eventual level in this stipulated constant-parameter model isThe two-reservoir carbon adjustment therefore has a rapid rise towards the frozen-ocean balance followed by a much longer increase. The figure solves the coupled linear pressure–ocean equations, retaining their small inventory correction to the reduced time scales.

The official Oxford course archive advertises an answer for this entire question. Its introductory ask and all seven leaves are covered externally and are not re-solved here.