Unit 8: PDE Modelling in Key AIMS Domains

Published

26/06/2026

This unit gives you just enough reef science to read the capstone in Unit 9 with confidence. You already know PDEs and ML; what you need is to recognise the capstone’s equation as something familiar — heat carried by a flow, smeared by turbulence, heated at the surface, and pinned to a cold reservoir below. That is the whole story. Everything else in this unit is detail you can take or leave.

TipHow to read this unit

The main text is the part you need. Wherever the genuine oceanography gets heavy — turbulence closures, air–sea flux formulae, the research literature — it is tucked into a collapsible “Deeper dive (optional)” box. Skip every one of those boxes and you will still understand the capstone completely. Open them only if you are curious.

Every term the project uses — upwelling, thermocline, eddy diffusivity, body source, and the rest — is defined in one place in the glossary at §8.9. Jump there any time a word is unfamiliar.

8.1 Why reef modelling? Why PDEs?

The Australian Institute of Marine Science studies the Great Barrier Reef — the largest coral ecosystem on Earth, and one of the most heat-sensitive. A reef typically sits in 20–100 m of water on a shallow shelf, stirred by tides, river plumes, storms, and the daily solar cycle. Most of the questions AIMS cares about are, underneath, the same PDE:

  • Coral bleaching — corals expel their algae when the water stays too warm for too long. Predicting which reefs heat up, and how deep the warm layer reaches, is a temperature model.
  • Water quality — nitrate, oxygen, and plankton are transported and react. Same transport equation, plus a reaction term.
  • Turbidity — river sediment blocks the light corals need. Same transport equation again.
  • Larval dispersal — where coral larvae drift and settle is a passive-tracer problem on the shelf currents.

All four are advection–diffusion problems: something at each point in space changes in response to its neighbours (diffusion) and to the flow carrying it (advection), plus sources and boundary exchanges. Switching from heat to nutrients to sediment is mostly a matter of which terms you keep. The capstone uses the thermal version — a single sensor chain near a reef, over a few weeks — but the machinery transfers directly.

The section claims sediment transport is “identical in form to the heat equation”. Earn that claim: write down the 1-D column PDE for suspended-sediment concentration C(z, t) near a reef after a resuspension event. Start from the temperature equation and decide, term by term, what changes: which transport terms survive unchanged, what plays the role of the body source, what new sinking term appears (sediment falls at a settling velocity w_s), and what the natural top and bottom boundary conditions become (no flux through the surface; erosion/deposition at the bed). Then say in one sentence how light attenuation for the corals would be computed from your C(z, t).

💡 Hint

Settling is just advection at velocity w_s relative to the water — so it enters exactly like w, with a sign. The top BC is the subtle part: ‘no flux’ means the total flux (diffusive + settling) vanishes, which makes it a Robin condition. For the light question, Beer–Lambert with a C-dependent attenuation coefficient.

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8.2 The column idealisation

Reef waters are fully three-dimensional. But many AIMS questions only need one slice through them:

What does temperature look like, as a function of depth, at a single mooring, over a few weeks?

That is the column idealisation. We follow one vertical pillar of water and track only T = T(z, t) — temperature as a function of depth z and time t — assuming nothing varies horizontally inside the column. Horizontal currents still exist; we just summarise their effect on the column through a single quantity, the vertical velocity w(z, t) they induce.

Figure 1: Schematic of the 1-D thermistor column. Five sensors at fixed depths z_1, \ldots, z_5 sample temperature T(z_k, t) while the column evolves under vertical advection w(z, t) (green), turbulent diffusion \kappa(z, t) (purple), the surface heat inputs (red — penetrating Q_{\text{SW}}, which enters as a body source, and non-penetrating Q_{\text{LW}} and Q_{\text{lat}} + Q_{\text{sens}}), penetrating shortwave decaying through the water (orange), and a fixed deep reservoir at z = -H. The right-hand panels collect the governing PDE, its boundary conditions, and the externally prescribed forcings.

The governing equation is the 1-D advection–diffusion equation:

\boxed{\;\; \frac{\partial T}{\partial t} \;=\; \underbrace{-\, w(z, t)\,\frac{\partial T}{\partial z}}_{\text{vertical advection}} \;+\; \underbrace{\frac{\partial}{\partial z}\!\left(\kappa(z, t)\,\frac{\partial T}{\partial z}\right)}_{\text{turbulent diffusion}} \;+\; \underbrace{\mathcal{S}(z, t)}_{\text{sunlight absorbed in the water}} \;\;} \tag{1}

with two boundary conditions, one at each end of the column:

\underbrace{\rho_0\, c_p\, \kappa(0, t)\,\frac{\partial T}{\partial z}\bigg|_{z = 0} \;=\; Q_{\text{np}}(t)}_{\text{surface: heat flux in}}, \qquad \underbrace{T(-H, t) \;=\; T_{\text{deep}}}_{\text{bottom: cold reservoir}}. \tag{2}

In words: temperature at a depth changes because the flow carries warmer/cooler water past it (advection), turbulence stirs out gradients (diffusion), and sunlight deposits heat in place (source). At the surface, the heat flux into the ocean is set by the atmosphere (Q_{\text{np}}); at the bottom, the temperature is pinned to a slowly-varying deep reservoir (T_{\text{deep}}). The constants \rho_0 \approx 1025\,\mathrm{kg/m^3} (seawater density) and c_p \approx 3990\,\mathrm{J\,kg^{-1}\,K^{-1}} (heat capacity) are the only “where do these numbers come from” constants — they convert a heat flux in W/m² into a temperature change.

Sections 8.3–8.5 take the three right-hand-side terms and the two boundary conditions one at a time. If you read nothing else, read those.

Equation 1 is what you get by collapsing the full 3-D advection–diffusion equation for a scalar T(\mathbf{x}, t),

\frac{\partial T}{\partial t} \;+\; \mathbf{u}\cdot\nabla T \;=\; \nabla\cdot\bigl(\kappa\,\nabla T\bigr) \;+\; \mathcal{S},

down to one dimension. Assume nothing varies horizontally inside the column (\partial_x T = \partial_y T = 0). The horizontal-advection terms then vanish, the velocity reduces to its vertical part w, and exactly Equation 1 survives.

The surface BC in Equation 2 is the air–sea heat-flux relation: with z measured upward, \rho_0 c_p\,\kappa\,\partial_z T is the downward heat flux entering the ocean (W/m²), set equal to the non-penetrating heat input Q_{\text{np}} (positive = heating).

The payoff is dimensionality: a 1-D problem in (z, t) instead of a 3-D one in (x, y, z, t). The price is that you must prescribe w(z, t) rather than solve for it — for the capstone, w is a prescribed linear-in-z upwelling profile that a cheaper 2-D shallow-water balance justifies (no shallow-water solve is actually run; see Unit 9 §9.2 and §8.3).

NoteWhen the column idealisation breaks

The column model assumes the interesting physics is vertical. If you care about river plumes meeting clear water, lagoon-vs-open-water exchange, or flow steered by the coastline, the horizontal physics matters and the column fails. The capstone is sited deliberately where the idealisation holds — a mooring on the open shelf, away from the reef break and away from a freshwater plume.

Check every term of Equation 1 and Equation 2 dimensionally — the fastest way to internalise what each symbol is:

  1. Confirm all four terms of Equation 1 have units of K/s, given [w] = \text{m/s}, [\kappa] = \text{m}^2/\text{s}.
  2. What must the units of \mathcal{S}(z, t) be, and how does dividing Q_{\text{SW}} (W/m²) by \rho_0 c_p deliver them? (Track the watts: \text{W} = \text{J/s}.)
  3. In the surface BC, show both sides have units W/m² (heat flux), then divide through by \rho_0 c_p and compute the equilibrium surface gradient \partial_z T implied by a net cooling Q_{\text{np}} = -200\,\text{W/m}^2 through a mixed-layer \kappa = 10^{-3}\,\text{m}^2/\text{s}. Is the answer (in K per metre) big or small compared to a typical thermocline gradient of ~0.3 K/m?

💡 Hint

Two facts do the whole exercise: (i) W = J/s, so a flux in W/m² is J s⁻¹ m⁻²; (ii) \rho_0 c_p \approx 1025\times3990 \approx 4.09\times10^6 J m⁻³ K⁻¹ is the single conversion factor between a heat flux and a “temperature flux” (K·m/s). Part 1: substitute [w]= m/s, [\kappa]= m²/s, [\partial_z T]= K/m into each term of Equation 1 and watch the units collapse to K/s. Part 2: divide Q_{SW} by \rho_0 c_p, then the Beer–Lambert factor k\,e^{kz} (units m⁻¹) finishes it. Part 3 is one division: from Equation 2, \partial_z T = Q_{np}/(\rho_0 c_p \kappa) = -200/(1025\cdot3990\cdot10^{-3}) — compute it and compare the magnitude against 0.3 K/m.

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8.3 Advection and diffusion

Two mechanisms move heat around the column. They are the first two terms of Equation 1.

Advection — heat riding along with the water

If a parcel of water is rising at speed w > 0, it carries its temperature with it. A fixed sensor at depth z then sees the temperature of the parcel that used to be at depth z - w\,\Delta t. That is advection:

\left.\frac{\partial T}{\partial t}\right|_{\text{advection}} = -w\,\frac{\partial T}{\partial z}.

The minus sign is the same one from the linear advection equation u_t + cu_x = 0 in Unit 6. In ocean terms: upwelling (w > 0) brings cold deep water up past your sensor; downwelling (w < 0) brings warm surface water down. The parcel doesn’t change temperature on the way — advection just relocates it.

NoteSign convention for w

w = \mathrm{d}z/\mathrm{d}t. With z measured upward from the surface (so z < 0 inside the ocean), w > 0 is upward motion of water. Upwelling — cold deep water rising — therefore corresponds to w > 0. Written as a contribution to \partial_t T the advection term is -w\,\partial_z T (as in Equation 1 above); moved to the other side of the equation it becomes +w\,\partial_z T, the form Unit 9 §9.3 uses. Either way the physical sign is the same: upwelling cools a stably stratified column. This same note appears in Unit 9 — keep the two conventions aligned.

For the capstone, w(z, t) is given — a prescribed linear-in-z upwelling profile, the form a shallow-water balance justifies (see Unit 9 §9.2; deriving w from an actual shallow-water solve is the optional scenario-5 design (Unit 9 §9.2 → Unit 10 §10.1), not something the capstone runs). What isn’t given — the one thing the capstone recovers — is the wind-stress envelope \tau(t): the strength over time of the storm’s wind drag on the sea surface. A real wind stress acts on the column several ways at once — it pumps water up (the upwelling w, via Ekman pumping), stirs the surface layer (raising \kappa), and brings cloud that dims the sunlight (Q_{\text{SW}}). The capstone keeps the inverse tractable by collapsing those into a single forcing envelope \tau(t) that it recovers as a heating source — a deliberate simplification, not the full three-way coupling (the framing is spelled out in Unit 9 §9.1). You don’t need to know where w comes from physically to solve the column — but if you’re curious:

The vertical velocity has three contributions, on three timescales:

  • Tides — a regular ~12.4-hour sloshing. The capstone folds the tidal part into the known prescribed w(z, t).
  • Storm-driven upwelling — the days-long pulse the storm’s wind stress \tau(t) produces; its time envelope is what the capstone recovers.
  • Mean shelf circulation — a slow near-constant background.

Real operational systems “de-tide” the data before inversion, so the recovery can focus on the storm-driven piece — exactly what the capstone does.

Diffusion — turbulence as effective stirring

Ocean turbulence is chaotic below ~1 m and we don’t resolve it. But averaged over a few minutes, turbulence mixes the column the way diffusion mixes a gas — down the gradient, with an effective diffusivity \kappa far larger than the molecular value:

\left.\frac{\partial T}{\partial t}\right|_{\text{diffusion}} = \frac{\partial}{\partial z}\!\left(\kappa(z, t)\,\frac{\partial T}{\partial z}\right).

The number to remember: \kappa ranges from about 10^{-5}\,\text{m}^2/\text{s} in quiet, deep, stratified water up to 10^{-3}\,\text{m}^2/\text{s} in the wind-stirred surface layer. That four-orders-of-magnitude span is itself a modelling choice — how you set \kappa(z, t) is the heart of any column model. The capstone keeps it simple by prescribing \kappa from a profile (large near the surface, small at depth); it does not run a live turbulence model.

Operational models compute \kappa from a turbulence closure instead of prescribing it. The common ones:

  • KPP (Large, McWilliams & Doney 1994) — the operational default: a smooth boundary-layer profile over a small interior background. The capstone’s prescribed \kappa is a frozen snapshot of what KPP would produce.
  • Mellor–Yamada 2.5 (Mellor & Yamada 1982) — carries turbulent energy as an extra prognostic variable; more honest about transients, heavier to run.
  • Pacanowski–Philander (Pacanowski & Philander 1981) — a cheap Richardson-number-based diffusivity used in many global models.

Inferring an effective \kappa from sparse mooring data — which is part of what the capstone inverse problem does — is the data-driven cousin of what these closures do prognostically.

Which one wins?

The two mechanisms compete, and one dimensionless number tells you who dominates: the Péclet number

\mathrm{Pe} = \frac{wH}{\kappa}.

\mathrm{Pe} \gg 1 is advection-dominated, \mathrm{Pe} \ll 1 is diffusion-dominated, and \mathrm{Pe} \sim 1 is the balanced regime the capstone column lives in (more in §8.6). The three capstone sites are deliberately chosen to span \mathrm{Pe} \ll 1, \sim 1, and \gg 1.

Everything in this section applies unchanged to any passive scalar: swap T for a nutrient or sediment concentration and only the source term and boundary conditions change.

A thought experiment with numbers. A sensor sits at z = -20 m in a 60 m column with \kappa = 10^{-4}\,\text{m}^2/\text{s} below the mixed layer and a thermocline gradient \partial_z T \approx 0.3 K/m at the sensor.

  1. Steady upwelling switches on at w = 2 \times 10^{-5} m/s (about 1.7 m/day). What cooling rate (K/day) does the advection term produce at the sensor?
  2. How large would the curvature \partial_{zz} T have to be for the diffusion term to match that cooling rate? Is that a plausible number for a smooth thermocline (where \partial_{zz}T \sim 0.3\,\text{K/m} / 10\,\text{m})?
  3. Conclude: at this depth and these parameters, which mechanism must a days-long cooling of ~0.5 K/day be attributed to — and what would have to change for diffusion to compete?

💡 Hint

Each rate is a single multiplication: w\cdot\partial_z T for advection, \kappa\cdot\partial_{zz}T for diffusion — convert to K/day with ×86 400 so the numbers mean something. For part 3, remember §8.5: the κ below the mixed layer is closer to 10^{-5} than the 10^{-4} given — does that strengthen or weaken the conclusion?

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8.4 Surface heating: the one distinction that matters

Heat enters and leaves through the sea surface. For the capstone, the only thing you must get right here is that surface heating splits into two kinds that enter the equation in two different places:

1. Sunlight that penetrates (Q_{\text{SW}}) — a body source. Sunlight passes through the surface and is absorbed inside the water, decaying with depth (red light in the top centimetres, blue–green reaching tens of metres). Because it deposits heat throughout the column, it is the source term \mathcal{S}(z, t) in Equation 1 — not a boundary condition. A simple model is a single exponential, \mathcal{S} \propto e^{z/\zeta} with penetration depth \zeta \approx 10 m; Unit 9 §9.5 gives the exact form the capstone uses.

2. Everything else (Q_{\text{np}}) — a surface flux. Longwave radiation, evaporation, and direct air–sea heat exchange all act right at the surface. Together they make up the non-penetrating flux Q_{\text{np}}(t), which is the surface boundary condition in Equation 2. Cooling usually wins at night and on cloudy days; heating wins on still, sunny afternoons.

For the capstone, both Q_{\text{SW}}(t) and Q_{\text{np}}(t) are prescribed time series — you don’t compute them from the atmosphere. The load-bearing point is just where each one enters: penetrating sunlight in the source, everything else in the surface BC. Getting this backwards is one of the two classic capstone bugs (§8.8).

The non-penetrating flux is itself a sum of three terms, each with a standard “bulk formula” tying it to the near-surface atmosphere (Fairall et al. 2003, COARE 3.0):

Q_{\text{np}} = \underbrace{Q_{\text{LW}}}_{\text{net longwave}} + \underbrace{Q_{\text{sens}}}_{\text{sensible}} + \underbrace{Q_{\text{lat}}}_{\text{latent (evaporation)}},

with, e.g., Q_{\text{sens}} = \rho_a c_{p,a} C_H U_{10}(T_a - T_s) and a similar form for the latent and longwave parts (U_{10} is the 10-m wind, C_H, C_E \sim 1.2\times10^{-3} exchange coefficients). The capstone never evaluates these — it takes Q_{\text{np}}(t) as given — but this is where the number comes from in a real model.

The penetrating sunlight, similarly, has a standard two-band form (Paulson & Simpson 1977): one band absorbed near the surface, one reaching the thermocline. The capstone collapses this to one band (\zeta \approx 10 m) for simplicity. (The k_1, k_2 in exercise 8.4 are rounded illustrative values for the two-band shape; the canonical Jerlov Type-I scales are 1/k_1 \approx 0.35 m and 1/k_2 \approx 23 m — the figures Unit 9 §9.5 quotes.)

TipA useful intuition

The ocean is a heat capacitor: it charges when the net surface flux is positive and discharges to the deep when it isn’t. The daily cycle wiggles the charge in a thin surface layer; a storm kicks it over a few days; the seasons shift it over months. Each timescale reaches a different depth, because diffusion penetrates a distance \sqrt{\kappa\,\tau} in time \tau.

Plot the Paulson–Simpson body source. With Q_{\text{SW}} = 800\,\text{W/m}^2 (noon sun), R = 0.58, k_1 = 0.5\,\text{m}^{-1}, k_2 = 1/14\,\text{m}^{-1}, plot \rho_0 c_p\,\mathcal{S}(z) over z \in [-50, 0] m on a log x-axis, where the two-band source is \mathcal{S}(z) = \frac{Q_{\text{SW}}}{\rho_0 c_p}\bigl[R\,k_1 e^{k_1 z} + (1-R)\,k_2 e^{k_2 z}\bigr]. Then integrate (analytically — the integrals are elementary exponentials) to answer:

  1. What fraction of the incoming shortwave is absorbed in the top 1 m? The top 10 m?
  2. What fraction passes below 30 m — the typical thermocline depth?
  3. The capstone’s single-band simplification uses \zeta = 10 m. Compare its below-30 m fraction with the two-band model’s. In which depth range do the two models disagree most, and would a thermistor at -40 m care?

💡 Hint

No numerical integration needed: the fraction of light still travelling downward past depth z is F(z) = R\,e^{k_1 z} + (1-R)\,e^{k_2 z}, so ‘absorbed above z’ is 1 - F(z). Compare the two models at z = -1, -10, -30 m and the answer to the design question falls out.

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8.5 Stratification and the deep boundary

Ocean temperature usually decreases with depth — warm sunlit water on top, cold water below, with a sharp transition called the thermocline in between. This is stable stratification, and it has two consequences the capstone relies on.

1. \kappa is not constant. In the wind-stirred mixed layer above the thermocline, \kappa is large (\sim 10^{-3}\,\text{m}^2/\text{s}). Below the thermocline, stratification suppresses mixing and \kappa collapses to a background \sim 10^{-5}\,\text{m}^2/\text{s}. So a surface heat anomaly mixes downward quickly through the mixed layer, then stalls at the thermocline — which is exactly why the deep sensors lag the surface ones in the capstone data.

2. The bottom needs a boundary condition. The capstone uses the simplest sensible one: a fixed deep reservoir, T(-H, t) = T_{\text{deep}}. Physically, below the modelled column lies a much larger volume of cold water whose temperature barely moves on our timescale. (A zero-flux “insulating floor” is the alternative; Unit 9 §9.4 discusses the trade-off.)

How sharply mixing shuts off is governed by the Richardson number \mathrm{Ri} = N^2/(\partial_z U)^2, comparing the stabilising buoyancy frequency N^2 = \alpha g\,\partial_z T to the destabilising shear \partial_z U. Mixing is suppressed where \mathrm{Ri} > 1/4. The capstone’s stratification-dependent closure (closure 3 in Unit 9 §9.5), \kappa = \kappa_b + \kappa_0/(1 + 5\,\mathrm{Ri})^2, encodes this: near the thermocline \mathrm{Ri} is large and \kappa drops toward its background floor.

Compute the stratification profile the capstone’s initial condition implies. Using T_0(z) from Unit 9 §9.4 (the \tanh thermocline: T_{\text{surface}} = 28, T_{\text{deep}} = 18, z_t = -30 m, \delta_t = 5 m) and the buoyancy frequency N^2 = \alpha g\,\partial_z T with \alpha = 2\times10^{-4}/K:

  1. Derive \partial_z T_0(z) in closed form and find where N^2 peaks and its peak value.
  2. With a background shear \partial_z U = 0.01\,\text{s}^{-1}, plot the Richardson number \mathrm{Ri}(z) and mark the region where \mathrm{Ri} > 1/4.
  3. Plot closure 3 from Unit 9 §9.5, \kappa = \kappa_b + \kappa_0/(1 + 5\mathrm{Ri})^2, through the column. By what factor is mixing suppressed at the thermocline relative to the surface — and what does that mean for how fast a surface heat anomaly can reach the deep sensors?

💡 Hint

Differentiate the tanh: \partial_z \tanh((z-z_t)/\delta_t) = \mathrm{sech}^2((z-z_t)/\delta_t)/\delta_t, so \partial_z T_0 peaks where the argument is 0, i.e. at z = z_t = -30 m (\mathrm{sech}^2(0)=1) — that’s where N^2 = \alpha g\,\partial_z T is largest. For the \mathrm{Ri} > 1/4 band, set \mathrm{Ri}(z) = N^2/(\partial_z U)^2 > 1/4 and solve the sech² threshold: |z-z_t| = \delta_t\,\operatorname{acosh}(1/\sqrt{\text{thr}}). For the consequence, compare diffusive crossing times L^2/\kappa through the suppressed band (\kappa \approx \kappa_b) against the mixed layer (\kappa \approx \kappa_0) — the ratio is the punchline.

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8.6 Dimensionless regimes — why the three sites differ

A few dimensionless groups tell you which terms dominate before you solve anything. They are the reason the capstone’s three sites behave so differently.

Group Definition What it tells you
Péclet \mathrm{Pe} wH/\kappa advection vs. diffusion across the column
Diffusive time T_\kappa H^2/\kappa how long diffusion takes to mix the column
Storm-to-diffusive \tau_{\text{storm}} / T_\kappa how deep a storm signal reaches

For a representative GBR column (H = 100 m, \kappa = 10^{-3} m²/s, w = 10^{-5} m/s):

  • \mathrm{Pe} \approx 1 — advection and diffusion compete on equal footing.
  • T_\kappa \approx 116 days — a 30-day run mixes only a fraction of the column.
  • The daily cycle is ~100× faster than bulk mixing, so it stays trapped in a thin surface layer.

Those three facts explain almost every behaviour you’ll see in the Unit 9 toy tasks: why the daily warm layer is thin, why a 3-day storm leaves a fingerprint deeper than naive cooling suggests, why upwelling cools the surface only slowly, and why true steady state takes months. The three sites (15 m, 60 m, 100 m deep) sit at different \mathrm{Pe} — set in the capstone by per-site advection strength (Site A is run diffusion-only) and the interior-vs-mixed-layer \kappa, not depth alone — which is why each records a different fingerprint of the same storm.

Do the §8.6 arithmetic for all three capstone sites (Unit 9 §9.1). Using each site’s depth (H_A = 15 m, H_B = 60 m, H_C = 100 m), with \kappa_m = 10^{-3}\,\text{m}^2/\text{s} and w_0 = 10^{-5}\,\text{m/s} for all three, compute per site: \mathrm{Pe} = w_0 H / \kappa_m, the diffusive time T_\kappa = H^2/\kappa_m in days, and the diffusive penetration depth \sqrt{\kappa_m \tau} for a 3-day storm. Check your numbers against the regime labels in the Unit 9 site table — do you reproduce “\mathrm{Pe} \ll 1 / \sim 1 / \gg 1”? One of the three labels deserves a caveat; identify which and explain (hint: how much of Site C’s column does a 30-day run actually mix?).

💡 Hint

Three formulas, three sites: \mathrm{Pe} = w_0H/\kappa, T_\kappa = H^2/\kappa, \sqrt{\kappa\tau} — a six-entry table you can fill in two minutes. The caveat hides in which κ you use: redo Site C’s Pe with the interior \kappa_b = 10^{-5} and watch the label change.

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8.7 Further reading

This column problem isn’t a workshop invention — it’s the standard way oceanographers reason about a single mooring. You do not need any of these to do the capstone; they’re here for when you want to go deeper. A few entry points, grouped by what they justify:

  • The column response to surface forcing — Price, Weller & Pinkel 1986 is the canonical observational + model account of the diurnal warm layer the toy tasks reproduce.
  • The mixing closure — Large, McWilliams & Doney 1994 (KPP) is the closure most operational column models use; the capstone’s prescribed \kappa profile is a frozen snapshot of it.
  • Why a column instead of just SST — Liu et al. 2014 and NOAA Coral Reef Watch use surface temperature and Degree-Heating-Weeks, which miss the vertical structure a column model resolves. That blind spot is the operational motivation for column modelling at AIMS.
  • Recovering a driver from one mooring — Kunze 1985 is the classic single-mooring inverse problem; modern variational data assimilation (ROMS 4D-Var) does the same thing at far greater cost than a CPU-scale PINN.
  • PINNs for ocean columns specifically — Bolton & Zanna 2019 is an early subgrid-closure precursor; the CSIRO AI4DA initiative is doing exactly this kind of PINN-flavoured column inverse problem today. This is the line of work the capstone is closest to.

The capstone deliberately simplifies relative to all of these: prescribed \kappa (not a live closure), prescribed w (linear-in-z, its shape justified by a shallow-water balance — no solve run), synthetic data with known truth, and a single scalar unknown \tau(t) instead of a full forcing field — so the focus stays on the inverse-problem mechanics.

A matching drill that doubles as a reading list. For each capstone design choice below, name the literature entry above that justifies it (paper or product), and the operational shortcoming the capstone’s approach addresses:

  1. The shape of the prescribed mixed-layer \kappa(z, t) profile.
  2. The claim that a diurnal warm layer forms and erodes daily in the top few metres.
  3. Using a vertical column of temperatures instead of satellite SST + Degree-Heating-Weeks for bleaching risk.
  4. The whole enterprise of recovering an unknown driver from a single mooring’s record.

Then the honest-framing question: which item in the §8.7 list is closest to the capstone’s actual method, and what does the capstone simplify away relative to it?

💡 Hint

Match by mechanism, not author: closure shape → the KPP paper; diurnal layer → the 1986 observational classic; bleaching proxies → the Coral Reef Watch line; mooring inversion → the 1985 ‘what does this mooring tell me’ paper. The last question wants the §8.7 bullet that mentions PINN-flavoured tooling.

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8.8 From concepts to the capstone column

That’s the whole model. The capstone in Unit 9 is just this column written down concretely:

  • One PDE — \partial_t T + w\,\partial_z T = \partial_z(\kappa\,\partial_z T) + \mathcal{S} (advection + diffusion + sunlight source).
  • Two boundary conditions — Neumann (heat flux) at the surface, Dirichlet (cold reservoir) at the bottom.
  • Four prescribed forcings — w(z, t), \kappa(z, t), Q_{\text{np}}(t), and Q_{\text{SW}}(t).
  • One initial condition — the \tanh thermocline profile.

Everything else — the toy tasks, the storm scenario, the inverse problem — is choosing which parts of that setup to perturb. The two mistakes worth inoculating against now, because a PINN will happily minimise the wrong physics without complaint:

  1. Sunlight is a body source, not a surface flux — it heats the interior where it’s absorbed (§8.4).
  2. The two boundary conditions are different types — a flux (Neumann) at the surface, a value (Dirichlet) at the bottom (§8.2, §8.5).

The science sits in §§8.2–8.6 above; the implementation sits in Unit 10.

Close this page (really). On a blank sheet, write down the complete capstone column problem: the PDE with all four terms, both boundary conditions, the initial condition, and the four prescribed forcings with their roles. Then reopen Unit 9 and grade yourself against §§9.3–9.5. The two items most commonly forgotten: where the penetrating shortwave enters (body source, not the surface BC), and which BC type sits at which end of the column. If you dropped either, reread §8.4 — that distinction is load-bearing for the whole inverse problem.

💡 Hint

Structure your sheet as: 4 PDE terms, 2 BCs (different types at the two ends), 1 IC, 4 forcings. The two classic mistakes are advertised in the exercise itself — so make sure your sheet takes an explicit position on where Q_{SW} enters and which BC type sits where.

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8.9 Glossary — the terms you need for the project

Every term Unit 9 assumes, defined once, in one place. If a symbol or word in the capstone spec is unfamiliar, it’s here.

Term What it means
Column / column idealisation Modelling one vertical pillar of water at a fixed location, tracking only T(z, t) — nothing varies horizontally inside it (§8.2).
Mooring / thermistor chain A cable of temperature sensors at fixed depths recording T over time — the data the capstone inverts (§8.2, Unit 9 §9.1).
z, H Depth coordinate, measured upward from the surface (z = 0), so z < 0 in the water; H is the column depth, with the bottom at z = -H.
Advection Heat carried bodily by the moving water: the -w\,\partial_z T term. The flow relocates a temperature, it doesn’t change it (§8.3).
Vertical velocity w(z, t) Up/down speed of the water. w > 0 is upward. Prescribed in the capstone (§8.3).
Upwelling / downwelling w > 0 (upwelling) lifts cold deep water up past a sensor → cooling; w < 0 (downwelling) pushes warm surface water down → warming (§8.3).
Diffusion Turbulent stirring that smooths gradients: the \partial_z(\kappa\,\partial_z T) term (§8.3).
Eddy diffusivity \kappa(z, t) The “strength” of turbulent mixing. Large (\sim 10^{-3} m²/s) near the surface, small (\sim 10^{-5} m²/s) at depth (§8.3, §8.5).
Mixed layer The well-stirred surface layer where \kappa is large and temperature is nearly uniform (§8.5).
Thermocline The sharp temperature drop between the warm mixed layer and the cold deep water (§8.5).
Stratification Warm-over-cold layering. Stable stratification resists mixing → suppresses \kappa (§8.5).
Body source \mathcal{S}(z, t) Heat deposited inside the water by penetrating sunlight — a source term in the PDE, not a boundary condition (§8.4).
Penetrating shortwave Q_{\text{SW}} Sunlight that passes through the surface and is absorbed with depth, feeding \mathcal{S} via Beer–Lambert decay (§8.4).
Non-penetrating flux Q_{\text{np}} Longwave + evaporation + air–sea exchange, all acting at the surface — this is the surface boundary condition (§8.4).
Beer–Lambert The exponential law \propto e^{z/\zeta} describing how light dims with depth; \zeta is the penetration scale (§8.4).
Neumann / Dirichlet BC Neumann = the flux (gradient) is set, used at the surface; Dirichlet = the value is set, used at the bottom (§8.2, §8.5).
Deep reservoir T_{\text{deep}} The fixed cold temperature the bottom of the column is pinned to (§8.5).
Péclet number \mathrm{Pe} = wH/\kappa Advection-vs-diffusion ratio. \gg 1 advective, \ll 1 diffusive, \sim 1 balanced — the three capstone sites span this range (§8.3, §8.6).
Diffusive time T_\kappa = H^2/\kappa How long diffusion takes to mix the whole column (§8.6).
Wind-stress envelope \tau(t) The storm’s strength over time — the unknown the capstone inverse problem recovers. Physically the storm’s wind stress drives upwelling, mixing, and cloud cover together; the capstone collapses these into one forcing it recovers as a heating source (§8.3, Unit 9 §9.1).
Ekman pumping Wind-driven vertical motion: how a storm’s wind stress \tau produces upwelling w (§8.3).
Shallow-water driver A cheap 2-D shallow-water balance that justifies the prescribed linear-in-z shape of w(z, t); it is not actually solved in the capstone — w is prescribed (§8.2, Unit 9 §9.2).
Forward vs inverse problem Forward: given the forcings, predict T(z, t). Inverse: given the measured T, recover an unknown forcing (\tau(t)) (Unit 9 §9.7–§9.9).