= Solution
{c}
Here $R=7.8$, $R_d=78$, $r=0.4$, and $s=0.04$, in the stated reservoir units. The exact characteristic coefficients are $P=0.10820513\,\mathrm{yr}^{-1}$ and $Q=0.000315582\,\mathrm{yr}^{-2}$, giving
$$
\boxed{\tau_1=9.51\,\mathrm{yr},\qquad\tau_2=333.4\,\mathrm{yr},\qquad\tau_3=\infty.}
$$
The leading separation estimates give $9.75$ yr and $325$ yr if finite $R_d$ is retained, or $390$ yr if $R_d/R$ is also sent to infinity. Thus $\tau_1\ll100\,\mathrm{yr}<\tau_2$; a ratio of ten warrants using the exact values for the requested sketch.
Assume initially zero anomalies in all reservoirs. The atmospheric impulse response, obtained by decomposing $e^{Jt}$ into the three <normal modes>, is
$$
G(t)=w_0+w_1e^{-t/\tau_1}+w_2e^{-t/\tau_2},\qquad t\geq0,
$$
with $w_0=1/93.6=0.01068376$, $w_1=0.06087812$, and $w_2=0.05664324$, in ppmv/GtCO$_2$. The checks $w_0+w_1+w_2=1/R_a$ and $w_1/\tau_1+w_2/\tau_2=r/R_a^2$ fix the instantaneous injection response and its initial decay.
The concentration is the <convolution> $C_a(t)=\int_0^tG(t-t')E(t')\,dt'$. To evaluate it transparently, define $H_0(q)=q^2/2$ for $q>0$, and
$$
H_i(q)=\tau_iq-\tau_i^2(1-e^{-q/\tau_i}),\qquad q>0,\quad i=1,2,
$$
setting every $H_i(q)=0$ when $q\leq0$. Expressing the triangular emission history as $E(t)=0.5[t_+-2(t-100)_++(t-200)_+]$ gives
$$
C_a(t)=\frac12\sum_{i=0}^2w_i[H_i(t)-2H_i(t-100)+H_i(t-200)].
$$
Here times are in years and $0.5$ has the corresponding emission-ramp units. Differentiating or evaluating the <convolution> gives
$$
\boxed{\max\dot C_a\simeq3.27\,\mathrm{ppmv/yr}\ \text{at year }100,\qquad
C_a(200)\simeq267.6\,\mathrm{ppmv},\qquad C_a(400)\simeq169.4\,\mathrm{ppmv}.}
$$
The maximum concentration occurs earlier than the end of emissions, at approximately year $172$, with $C_a\simeq280.7$ ppmv. Initially there is a short adjustment over $\tau_1$; the curve then rises, turns over as emissions fall, and continues declining after year 200 as the deep <ocean> takes up <carbon dioxide>.
The total emitted <mass> is the area of the emission triangle, $\frac12(200)(50)=5000$ GtCO$_2$. All reservoirs eventually share one concentration, so <conservation of mass> gives
$$
\boxed{C_a(\infty)=\frac{5000}{7.8+7.8+78}=53.4\,\mathrm{ppmv}.}
$$
\Image[/past-exam-of-the-physics-of-atmospheres-and-oceans-course-of-the-university-of-oxford/2022-carbon-reservoir-response.png]
{title=Atmospheric concentration and its rate of change under a triangular emissions pulse}
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