Section Exercise Solutions

Worked solutions to the section exercises scattered through Units 1–10. Julia is the primary language throughout; JAX / Python variants appear where the cross-ecosystem comparison earns its keep. Where a solution produces a plot or a number with workshop-only dependencies, the cell is executed and its output shown below the code (each executed cell is wrapped in a let block so the solutions stay independent of each other — drop the let/end pair when pasting into your own session). Solutions that need heavyweight extras (MNIST downloads, SciMLSensitivity, KolmogorovArnold, DeepXDE, trained NeuralPDE models) remain static listings with the expected results quoted in the prose. Every solution links back to its exercise.

Unit 1 — Introduction

Solution 1.1 — de-tide a synthetic gauge record

The harmonic fit is a plain linear least-squares problem: with the constituent frequencies known, the unknowns (amplitude and phase of each constituent) enter linearly through A\sin\omega t + B\cos\omega t.

using Random, LinearAlgebra, Plots

let
rng  = MersenneTwister(1)
dt_h = 0.05                                  # 3-minute sampling, hours
t    = collect(0.0:dt_h:120.0)               # 5 days
ω    = 2π ./ [12.42, 12.0]                   # M2, S2 (rad/hour)

tide  = 0.8 .* sin.(ω[1] .* t) .+ 0.3 .* sin.(ω[2] .* t .+ 1.0)
surge = 0.45 .* exp.(-((t .- 50.0) ./ 3.0).^2)
η     = tide .+ surge .+ 0.015 .* randn(rng, length(t))

# Design matrix: constant + sin/cos at each known frequency
design(t) = hcat(ones(length(t)),
                 sin.(ω[1] .* t), cos.(ω[1] .* t),
                 sin.(ω[2] .* t), cos.(ω[2] .* t))

# (1) Calibrate on the calm window only
calm   = t .< 36.0
c_calm = design(t[calm]) \ η[calm]
resid_calm = η .- design(t) * c_calm

# (3) Calibrate on the whole record, surge included
c_all  = design(t) \ η
resid_all = η .- design(t) * c_all

plt = plot(t, surge, label = "true surge", lw = 2, c = :black,
           xlabel = "t (hours)", ylabel = "η (m)")
plot!(plt, t, resid_calm, label = "residual (calm-window fit)", alpha = 0.7)
plot!(plt, t, resid_all,  label = "residual (full-record fit)", alpha = 0.7)
end

What you should see:

• Calm-window fit: the residual sits on top of the true surge to within the noise floor — the harmonic constants were estimated from data that contained no surge, so the subtraction is clean.
• Full-record fit: least squares cannot tell surge energy near the tidal frequencies from tide. Part of the pulse is absorbed into the fitted constituents, so the recovered surge is visibly underestimated and the residual rings at the tidal period away from the pulse. With a 3-hour pulse the leakage is modest (the pulse is far from the 12-hour band); make the pulse wider — say a 10-hour width — and the leakage becomes severe.

This is the practical point of the Unit 1 callout: de-tiding is itself an inverse problem, and the surge you recover downstream inherits whatever your calibration window let leak into the harmonic constants.

← Back to Unit 1 §1.1

Solution 1.2 — travel-time forensics on the slider

With c = \sqrt{gH}:

Gauge	Distance	Representative depth	c	Predicted arrival
G1 Mud Island	18 km	6 m (inner-bay flat)	7.7 m/s	~39 min
G2 Wellington Pt	26 km	6 m	7.7 m/s	~56 min
G3 Tangalooma	35 km	6–18 m (path crosses NE channel)	7.7–13.3 m/s	~44–76 min
G4 Russell Channel	50 km	6 m, constricted path	7.7 m/s	~1 h 48 min

Add these delays to the t = 2 h source peak to predict each gauge’s peak time (e.g. G4 peak ≈ 3.8 h), then check against the slider — agreement to within ~10 minutes for G1/G2 is typical.

• Most depth-sensitive: G3. Its propagation path crosses the deep NE channel, where c nearly doubles. The fraction of the path spent in deep water is exactly what the bathymetry figure shows, and a single “representative depth” can’t capture it — the honest estimate is a range.
• G4’s constant-depth estimate is early. The straight-line distance ignores the detour around the Russell / Macleay / Karragarra constriction (longer effective path) and the very shallow southern flats (slower c). Both corrections delay the real arrival, so the naive estimate undershoots — which is precisely why G4’s signal is described as “attenuated, delayed, reflection-laden”.

← Back to Unit 1 §1.2

Solution 1.3 — sketch the three fingerprints

The qualitative answer key (graded properly against the Unit 10 partition plots):

Mechanism	Who cools	Who warms	Who responds first
Upwelling (w > 0)	every sensor below the surface layer; strongest where \partial_z T is largest (thermocline)	nobody	the deep sensors — cold water arrives from below and works upward
Enhanced mixing (\kappa \uparrow)	sensors above the thermocline (heat is carried down, away from them)	sensors below the thermocline (the same heat arrives there)	both sides of the thermocline together — mixing redistributes, it doesn’t import
Reduced surface heating (Q\downarrow)	the surface sensor first and most; deeper sensors later and weaker	nobody	the top sensor, with the diurnal wiggle visibly flattened

Two discriminating features worth having had on your sketch:

1. Mixing approximately conserves column heat — cooling above the thermocline is mirrored by warming below it. Upwelling and reduced heating both remove heat from the column (one out the bottom boundary’s budget, one at the top), so the depth-integrated temperature falls. A warming deep sensor is therefore a strong vote for mixing.
2. Direction of propagation in time: deep-first cooling → upwelling; top-first cooling → surface flux; near-simultaneous two-sided change centred on the thermocline → mixing.

The real capstone signal is a blend (one storm drives all three), which is exactly why Unit 9 makes mechanism attribution a quantitative partition rather than an eyeball call.

← Back to Unit 1 §1.3

Unit 2 — ML Foundations

Solution 2.1 — watch the U-shape appear

using Random, LinearAlgebra, Statistics, Plots

let
rng = MersenneTwister(0)
N   = 60
x   = sort(rand(rng, N))
y   = sin.(2π .* x) .+ 0.2 .* randn(rng, N)

vander(x, d) = hcat((x .^ p for p in 0:d)...)
rmse(a, b)   = sqrt(mean(abs2, a .- b))

degrees = 1:15
train_err = Float64[]; cv_err = Float64[]
folds = [i:5:N for i in 1:5]                 # 5 interleaved folds

for d in degrees
    # training error on everything
    c = vander(x, d) \ y
    push!(train_err, rmse(vander(x, d) * c, y))

    # 5-fold CV by hand
    fold_scores = Float64[]
    for f in folds
        test  = falses(N); test[f] .= true
        c_f   = vander(x[.!test], d) \ y[.!test]
        push!(fold_scores, rmse(vander(x[test], d) * c_f, y[test]))
    end
    push!(cv_err, mean(fold_scores))
end

plt = plot(degrees, train_err, marker = :circle, label = "train RMSE",
           xlabel = "polynomial degree", ylabel = "RMSE", yscale = :log10)
plot!(plt, degrees, cv_err, marker = :square, label = "5-fold CV RMSE")
end

Typical outcome: the training curve decreases monotonically (a degree-15 polynomial can always fit 60 points better than a degree-3 one), while the CV curve bottoms out around degree 5–7 — enough flexibility to represent one full sine period — and then climbs steeply as the fits start chasing the \varepsilon noise. Past degree ~12 the CV error can explode by orders of magnitude: high-degree monomial bases are also numerically ill-conditioned, which compounds the statistical overfitting. The gap between the two curves is the generalisation gap from §2.1.

← Back to Unit 2 §2.1

Solution 2.2 — how many trees is enough?

using MLDatasets, DecisionTree, Statistics

train_x, train_y = MLDatasets.MNIST(split = :train)[:]
test_x,  test_y  = MLDatasets.MNIST(split = :test)[:]
flatten(x) = reshape(Float32.(x), 28 * 28, size(x, 3))'
X = flatten(train_x)[1:10_000, :];  yl = Int.(train_y)[1:10_000]
Xt = flatten(test_x);               yt = Int.(test_y)

for n_trees in (1, 5, 20, 50, 100, 200)
    forest = build_forest(yl, X, 28, n_trees, 0.7, -1, 1, 0.0, 0; rng = 0)
    acc = mean(apply_forest(forest, Xt) .== yt)
    println("trees = $n_trees:  acc = $(round(acc, digits = 4))")
end

for nsub in (5, 28, 100, 784)
    forest = build_forest(yl, X, nsub, 100, 0.7, -1, 1, 0.0, 0; rng = 0)
    acc = mean(apply_forest(forest, Xt) .== yt)
    println("n_subfeatures = $nsub:  acc = $(round(acc, digits = 4))")
end

Expected shape of the results (exact numbers vary with the subset): a single tree lands somewhere near 80%; accuracy climbs steeply to ~20 trees, and the curve is essentially flat beyond 50–100 — the variance-reduction benefit of averaging decays like 1/\sqrt{n_{\text{trees}}}, so each doubling buys less.

The n_subfeatures = 784 question is the conceptual heart: a random forest’s variance reduction comes from averaging trees that make decorrelated errors. If every split sees every feature, all trees greedily pick the same dominant splits and become near copies of each other — stronger individually, but correlated, so the ensemble averages away less noise. The \sqrt{784} = 28 default is the standard bias-variance compromise.

← Back to Unit 2 §2.2

Solution 2.3 — logistic regression from scratch

using MLDatasets, Statistics, Random

train_x, train_y = MLDatasets.MNIST(split = :train)[:]
test_x,  test_y  = MLDatasets.MNIST(split = :test)[:]
flatten(x) = reshape(Float32.(x), 28 * 28, size(x, 3))

function binary_subset(X, y, a, b)
    keep = (y .== a) .| (y .== b)
    X[:, keep], Float32.(y[keep] .== b)       # label 1 for digit b
end

Xtr, ytr = binary_subset(flatten(train_x), Int.(train_y), 0, 1)
Xte, yte = binary_subset(flatten(test_x),  Int.(test_y),  0, 1)

σ(z) = 1 ./ (1 .+ exp.(-z))

w = zeros(Float32, 784);  b = 0f0;  ηlr = 0.5f0
N = size(Xtr, 2)
for step in 1:500
    p  = σ(Xtr' * w .+ b)                     # N-vector of probabilities
    δ  = p .- ytr                             # ∂L/∂z, derived by hand
    w -= ηlr .* (Xtr * δ) ./ N
    b -= ηlr .* mean(δ)
end

acc = mean((σ(Xte' * w .+ b) .> 0.5) .== (yte .> 0.5))
println("0-vs-1 test accuracy: ", round(acc, digits = 4))

The gradient derivation is the small jewel here: for \hat p = \sigma(w^\top x + b) and binary cross-entropy, the chain rule collapses to \partial \mathcal{L}/\partial z = \hat p - y — the sigmoid’s derivative cancels against the log’s. That’s why the code never computes \sigma'.

0 vs 1 is nearly linearly separable in raw pixel space (a “1” lights a thin vertical band, a “0” a ring), so ~99.8% falls out. 4 vs 9 shares almost all of its ink — the discriminating feature is whether the upper loop closes, a conjunction of pixels that no single linear threshold on pixel intensities can express. Accuracy drops to roughly 96–97%, and no amount of training fixes it: the model class, not the optimiser, is the bottleneck. That gap is the motivation for hidden layers in §2.4.

← Back to Unit 2 §2.3

Solution 2.4 — universal approximation, and the ReLU trap

using Lux, Random, Zygote, Optimisers, ForwardDiff, Plots

target_24(x) = sin(4π * x) * exp(-x)
xs_24 = collect(Float32, range(0, 1; length = 256))
ys_24 = target_24.(xs_24)

function fit_mlp(width, act; iters = 4000)
    rng = Random.MersenneTwister(0)
    model = Lux.Chain(Lux.Dense(1 => width, act), Lux.Dense(width => 1))
    ps, st = Lux.setup(rng, model)
    loss(p) = sum(abs2,
                  vec(first(model(reshape(xs_24, 1, :), p, st))) .- ys_24)
    opt = Optimisers.setup(Optimisers.Adam(1f-2), ps)
    for _ in 1:iters
        g = first(Zygote.gradient(loss, ps))
        opt, ps = Optimisers.update(opt, ps, g)
    end
    model, ps, st
end

let
plt = plot(xs_24, ys_24, label = "target", lw = 3, c = :black,
           xlabel = "x")
for w in (5, 20, 100)
    model, ps, st = fit_mlp(w, tanh)
    plot!(plt, xs_24, vec(first(model(reshape(xs_24, 1, :), ps, st))),
          label = "tanh, width $w", ls = :dash)
end
plt
end

Width 5 underfits badly (two sine lobes need more bumps than five tanh units want to make), width 20 is close, width 100 is visually exact — universal approximation in action, with the usual caveat that the theorem promised none of this would be trainable, just representable.

The second derivative:

function second_derivative(model, ps, st)
    u(x) = first(model([x], ps, st))[1]
    x -> ForwardDiff.derivative(ξ -> ForwardDiff.derivative(u, ξ), x)
end

let
m_tanh, p_tanh, s_tanh = fit_mlp(20, tanh)
m_relu, p_relu, s_relu = fit_mlp(20, relu)

xs_f = collect(Float32, range(0, 1; length = 400))
plt = plot(xs_f, second_derivative(m_tanh, p_tanh, s_tanh).(xs_f),
           label = "tanh f''", lw = 2, xlabel = "x")
plot!(plt, xs_f, second_derivative(m_relu, p_relu, s_relu).(xs_f),
      label = "relu f''", lw = 2)
end

The tanh curve is a smooth, meaningful f''. The ReLU curve is identically zero except at the (measure-zero) kink locations — a ReLU network is piecewise linear, so its second derivative carries no information anywhere autodiff can evaluate it. A diffusion residual u_t - \alpha u_{xx} built on a ReLU network would see u_{xx} \equiv 0: the optimiser would happily “solve” the wrong equation u_t = 0. Hence the smooth-activation rule for PINNs.

← Back to Unit 2 §2.4

Solution 2.5 — race the optimisers down a ravine

using Plots

let
f(x, y)  = (1 - x)^2 + 100 * (y - x^2)^2
∇f(x, y) = [-2(1 - x) - 400x * (y - x^2),  200 * (y - x^2)]

function descend(update!, state, x0; iters = 5000)
    xs = [copy(x0)]
    x  = copy(x0)
    for t in 1:iters
        update!(x, ∇f(x...), state, t)
        push!(xs, copy(x))
    end
    reduce(hcat, xs)
end

# plain GD
gd!(x, g, s, t)  = (x .-= s.η .* g)

# momentum
function mom!(x, g, s, t)
    s.v .= s.μ .* s.v .+ g
    x  .-= s.η .* s.v
end

# Adam
function adam!(x, g, s, t)
    s.m .= s.β1 .* s.m .+ (1 - s.β1) .* g
    s.v .= s.β2 .* s.v .+ (1 - s.β2) .* g.^2
    m̂ = s.m ./ (1 - s.β1^t);  v̂ = s.v ./ (1 - s.β2^t)
    x .-= s.η .* m̂ ./ (sqrt.(v̂) .+ 1e-8)
end

x0 = [-1.5, 2.0]
T_gd  = descend(gd!,  (η = 1e-3,),                          x0)
T_mom = descend(mom!, (η = 1e-3, μ = 0.9, v = zeros(2)),     x0)
T_ad  = descend(adam!,(η = 2e-2, β1 = 0.9, β2 = 0.999,
                       m = zeros(2), v = zeros(2)),          x0)

xr = range(-2, 2; length = 200); yr = range(-1, 3; length = 200)
plt = contour(xr, yr, (a, b) -> log10(f(a, b)); levels = 30,
              c = :grays, xlabel = "x", ylabel = "y")
plot!(plt, T_gd[1, :],  T_gd[2, :],  label = "GD",       lw = 1.5)
plot!(plt, T_mom[1, :], T_mom[2, :], label = "momentum", lw = 1.5)
plot!(plt, T_ad[1, :],  T_ad[2, :],  label = "Adam",     lw = 1.5)
scatter!(plt, [1.0], [1.0], label = "minimum", marker = :star, ms = 8)
end

What the three trajectories show, mapped to the §2.5 failure modes:

• GD zig-zags across the parabolic valley (y \approx x^2): the gradient points mostly across the ravine, and any \eta large enough to make progress along the floor is unstable across it. After 5 000 iterations it is typically still far from (1, 1).
• Momentum averages out the across-valley oscillation (alternating signs cancel in v_t) and accumulates speed along the floor — the O(\kappa) \to O(\sqrt{\kappa}) claim made visible.
• Adam additionally rescales per-coordinate: the y-direction gradient (scale ~200) and the x-direction gradient (scale ~2) get individually normalised step sizes, so one global \eta serves both — failure mode 2 (bad scaling between parameters) solved by construction. It reaches the minimum’s basin in a few hundred iterations.

← Back to Unit 2 §2.5

Solution 2.6 — close the MNIST gap in Lux

The only changes from mnist_linear_lux.jl are the model definition and the post-Adam fine-tune:

model = Lux.Chain(
    Lux.Dense(784 => 256, relu),
    Lux.Dense(256 => 128, relu),
    Lux.Dense(128 => 10),
    Lux.softmax,
)

# ... identical data pipeline and mini-batch Adam loop, 10 epochs ...

# fine-tune: 5 more epochs at 10× smaller learning rate
opt_state = Optimisers.setup(Optimisers.Adam(1f-3), ps)
# ... same loop, 5 epochs ...

Expected numbers: ~97.5–98.2% after the 10 Adam epochs — past the random forest’s ~97% — and the small-learning-rate fine-tune typically adds another 0.1–0.3% by letting the iterate settle into the minimum instead of bouncing around it at the larger step size. That two-phase pattern (fast optimiser to get close, careful optimiser to finish) is a miniature of the Adam → L-BFGS schedule every PINN in Units 5–10 uses.

Parameter count: 784\!\times\!256 + 256 + 256\!\times\!128 + 128 + 128\!\times\!10 + 10 = 235\,146 — thirty times the softmax baseline’s 784\!\times\!10 + 10 = 7\,850, for about six points of accuracy. Depth is buying feature composition, not free lunch.

← Back to Unit 2 §2.6

Solution 2.7 — KAN vs MLP on a sharp feature

using KolmogorovArnold, Lux, Random, Zygote, Optimisers, Statistics, Plots

rng = Random.MersenneTwister(0)
n  = 512
X  = reshape(collect(Float32, range(-1, 1; length = n)), 1, n)
y  = reshape(abs.(X[1, :]) .+ 0.2f0 .* sin.(5f0 .* X[1, :]), 1, n)

kan = Lux.Chain(KDense(1 => 6; basis_func = rbf, normalizer = softsign),
                KDense(6 => 1; basis_func = rbf, normalizer = softsign))
mlp = Lux.Chain(Lux.Dense(1 => 14, tanh), Lux.Dense(14 => 14, tanh),
                Lux.Dense(14 => 1))   # ≈ same parameter count; check with
                                      # Lux.parameterlength(...)

function train(model; iters = 3000)
    ps, st = Lux.setup(rng, model)
    loss(p) = mean(abs2, first(model(X, p, st)) .- y)
    opt = Optimisers.setup(Optimisers.Adam(1f-2), ps)
    for _ in 1:iters
        g = first(Zygote.gradient(loss, ps))
        opt, ps = Optimisers.update(opt, ps, g)
    end
    ps, st, loss(ps)
end

ps_k, st_k, L_k = train(kan)
ps_m, st_m, L_m = train(mlp)
println("KAN MSE = $L_k,  MLP MSE = $L_m")

Typical outcome at matched parameter count and budget: the KAN’s final MSE is a factor of a few lower, and the difference is concentrated exactly where the exercise pointed — zoom into x \in [-0.1, 0.1] and the MLP rounds the |x| kink into a smooth parabola-like cap, while the KAN’s per-edge basis functions place a genuine corner. With 10× the iterations the MLP keeps improving and narrows the gap (a tanh network can approximate a kink arbitrarily well — it just spends many units and many iterations doing it), which is the honest summary of the KAN literature: an advantage in efficiency on sharp low-dimensional structure, not in ultimate expressiveness.

The Python check, if you want it, is a five-line edit of the pykan script in §2.7 — replace the dataset function with f = lambda x: torch.abs(x[:, [0]]) + 0.2 * torch.sin(5 * x[:, [0]]).

← Back to Unit 2 §2.7

Solution 2.8 — translate the vocabulary

Supervised concept	PINN counterpart	What changes
(a) labelled example (x_i, y_i)	a collocation point t_i with target residual 0	the “label” is implicit — the equation itself says what the network should do there
(b) test set	held-out collocation points (or a dense grid vs a trusted reference)	generalisation = small residual between training points, not just at them
(c) overfitting	small residual at the sampled points, large in between; or the data term fitting observation noise	the regulariser against it is the physics itself, plus held-out residual checks
(d) L_2 weight penalty	the PDE residual term (and smoothness priors like the H^1 penalty on a recovered driver)	the prior is structural — it encodes what the function must satisfy, not just “be small”
(e) mini-batch	a resampled batch of collocation points per step	resampling is free (sample the domain at will), which is why PINNs can also afford full-batch L-BFGS

“Infinite training data”: a PINN can draw as many collocation points as it likes — the domain is a sampling distribution, not a finite dataset. The catch is that all those points carry the same information (the one PDE), so more points reduce sampling error but cannot supply anything the equation + BCs don’t already determine. Past a modest density, the binding constraint is optimisation (can the network actually be trained to satisfy the residual everywhere — see Unit 7), not data volume.

← Back to Unit 2 §2.8

Unit 3 — Sci-ML and PIML

Solution 3.1 — find the fastest landing

using OrdinaryDiffEq, Plots

let
ω = 2π
function damped(u, p, t)
    ζ = p[1]
    [u[2], -ω^2 * u[1] - 2ζ * ω * u[2]]
end

function settling_time(ζ; tol = 0.02, Tmax = 10.0)
    sol = solve(ODEProblem(damped, [1.0, 0.0], (0.0, Tmax), [ζ]),
                Tsit5(); saveat = 0.001)
    x = [u[1] for u in sol.u]
    bad = findall(abs.(x) .> tol)            # indices still outside the band
    isempty(bad) ? 0.0 : sol.t[bad[end]]     # last excursion = settling time
end

ζs = 0.1:0.05:2.0
ts = settling_time.(ζs)
println("settling time minimised at ζ = ", ζs[argmin(ts)])
plot(ζs, ts, marker = :circle, xlabel = "ζ", ylabel = "settling time (s)",
     title = "2% settling time vs damping ratio", legend = false)
end

settling time minimised at ζ = 0.8

The minimum lands near \zeta \approx 0.7–0.8, not at \zeta = 1 — the control engineers are right. The intuition: a slightly underdamped system overshoots once, but that first overshoot can stay inside the 2% band while the approach is much faster than the critically damped crawl. Critical damping is the fastest response with no overshoot at all; once you tolerate a small band, mild underdamping wins. (Try tol = 1e-4 and watch the optimum migrate toward 1 — the tighter the band, the less overshoot you can afford.) The lesson, beyond the control trivia: “check the claim numerically” is a five-line job once you have a solver loop, and the claim was subtly miscalibrated.

← Back to Unit 3 §3.1

Solution 3.2 — pick the right tool

1. Surrogate. A precomputable (input → output) map queried many times: train on a few hundred CFD runs, then evaluate the 10 000 layouts in milliseconds each. Physics part: the CFD runs that generate training data; ML part: the regression replacing the solver. The trap to mention out loud: layout 7 832 had better be inside the training distribution.
2. Hybrid (UDE). Known clearance term written down exactly; unknown absorption term N_\theta(\cdot) learned per patient from concentration time-series. Exactly the f_{\text{phys}} + N_\theta split.
3. Conservation-aware architecture. The complaint is energy drift, so bake conservation in: an HNN-style parameterisation (learn the Hamiltonian, derive the dynamics) or a symplectic integrator. A soft energy penalty would merely slow the drift; the architectural constraint eliminates it.
4. Surrogate again — and the stated guarantee (“only queried inside the simulated range”) is precisely the condition that makes a surrogate trustworthy. A POD-based linear ROM (§3.5) is the cheap first candidate before any neural network.

← Back to Unit 3 §3.2

Solution 3.3 — energy conservation by construction

using OrdinaryDiffEq, ForwardDiff, Plots

let
H(u) = 0.5 * u[2]^2 + (1 - cos(u[1]))        # u = (q, p)

# Hamilton's equations via autodiff — what an HNN does with H_θ
function ham!(du, u, p, t)
    g = ForwardDiff.gradient(H, u)
    du[1] =  g[2]                            #  ∂H/∂p
    du[2] = -g[1]                            # -∂H/∂q
end

# the "2% wrong" field a vanilla MLP might learn
function ham_bad!(du, u, p, t)
    g = ForwardDiff.gradient(H, u)
    du[1] =  g[2]
    du[2] = -1.02 * g[1]
end

u0 = [2.0, 0.0];  ts = (0.0, 100.0)
sol  = solve(ODEProblem(ham!,     u0, ts), Tsit5(); saveat = 0.05,
             reltol = 1e-9, abstol = 1e-9)
solb = solve(ODEProblem(ham_bad!, u0, ts), Tsit5(); saveat = 0.05,
             reltol = 1e-9, abstol = 1e-9)

drift(s)  = [H(u) - H(u0) for u in s.u]
plt = plot(sol.t, drift(sol), label = "exact Hamiltonian field", lw = 2,
           xlabel = "t", ylabel = "H(t) − H(0)")
plot!(plt, solb.t, drift(solb), label = "2%-perturbed field", lw = 2)
end

The exact field’s drift sits at the solver-tolerance floor (\sim 10^{-8} here — tighten reltol and it shrinks further: the residual drift is numerical, not structural). The perturbed field’s energy error is qualitatively different: it grows secularly, oscillating with an expanding envelope, because the field is no longer orthogonal to \nabla H. That orthogonality — \dot H = \nabla H \cdot (\partial_p H, -\partial_q H) \equiv 0, an algebraic identity independent of what H is — is exactly the property an HNN inherits for free by parameterising H_\theta and differentiating, and that a vanilla MLP fitting (\dot q, \dot p) directly has no way to express.

← Back to Unit 3 §3.3

Solution 3.4 — SINDy on the damped oscillator

using OrdinaryDiffEq, LinearAlgebra, Statistics

let
ω, ζ = 2π, 0.1
osc(u, p, t) = [u[2], -ω^2 * u[1] - 2ζ * ω * u[2]]
sol = solve(ODEProblem(osc, [1.0, 0.0], (0.0, 6.0)), Tsit5();
            saveat = 0.002, reltol = 1e-10, abstol = 1e-10)
X  = reduce(hcat, sol.u)'                    # (N, 2): columns x, v
ts = sol.t

# central differences for Ẋ (drop the endpoints)
Ẋ = (X[3:end, :] .- X[1:end-2, :]) ./ (ts[3:end] .- ts[1:end-2])
Xc = X[2:end-1, :]

# degree-2 polynomial library: 1, x, v, x², xv, v²
Θ(x, v) = hcat(ones(length(x)), x, v, x.^2, x .* v, v.^2)
term_names = ["1", "x", "v", "x²", "xv", "v²"]
lib = Θ(Xc[:, 1], Xc[:, 2])

function stlsq(lib, dx; threshold = 0.5, iters = 5)
    ξ = lib \ dx
    for _ in 1:iters
        small = abs.(ξ) .< threshold
        ξ[small] .= 0
        big = .!small
        ξ[big] = lib[:, big] \ dx
    end
    ξ
end

for (k, label) in enumerate(("ẋ", "v̇"))
    ξ = stlsq(lib, Ẋ[:, k])
    terms = [string(round(ξ[i]; digits = 3), "·", term_names[i])
             for i in eachindex(ξ) if ξ[i] != 0]
    println("$label = ", join(terms, " + "))
end
println("\ntargets:  -ω² = ", round(-ω^2; digits = 3),
        ",  -2ζω = ", round(-2ζ*ω; digits = 3))
end

ẋ = 1.0·v
v̇ = -39.477·x + -1.257·v

targets:  -ω² = -39.478,  -2ζω = -1.257

Clean data recovers ẋ = 1.0·v and v̇ = -39.48·x + -1.257·v — i.e. -\omega^2 = -39.478 and -2\zeta\omega = -1.2566, to four figures.

Add the noise — X .+= 0.01 .* std(X) .* randn(size(X)) before differentiating — and the recovered coefficients degrade badly or spurious quadratic terms survive thresholding. The failed step is step 2 of the SINDy recipe, numerical differentiation: central differences amplify noise by \mathcal{O}(\sigma/\Delta t), and at \Delta t = 0.002 a 1% state noise becomes overwhelming derivative noise. The regression itself is fine — feed it the true \dot X with noisy X and recovery barely suffers. The fixes the unit names (total-variation differentiation, smoothing splines) all attack that one step; a cheap version worth trying is differencing over a wider stencil or pre-smoothing X with a low-pass filter.

The pysindy cross-check is three lines around the §3.4 Python script: same trajectory, ps.SINDy(feature_library = ps.PolynomialLibrary(degree=2), optimizer = ps.STLSQ(threshold=0.5)).

← Back to Unit 3 §3.4

Solution 3.5 — POD on a diffusing field

using LinearAlgebra, Plots

let
nx, α = 101, 0.01
dx = 1.0 / (nx - 1)
dt = 0.4 * dx^2 / α
x  = range(0, 1; length = nx)
u  = exp.(-200 .* (x .- 0.3).^2);  u[1] = u[end] = 0.0

snapshots = [copy(u)]
u_new = similar(u)
for n in 1:8000
    for i in 2:nx-1
        u_new[i] = u[i] + α * dt / dx^2 * (u[i+1] - 2u[i] + u[i-1])
    end
    u_new[1] = u_new[end] = 0.0
    u, u_new = u_new, u
    n % 40 == 0 && push!(snapshots, copy(u))   # ~200 snapshots
end

X = reduce(hcat, snapshots)
U, Σ, V = svd(X)
energy = cumsum(Σ.^2) ./ sum(Σ.^2)
println("modes for 99% variance: ", findfirst(energy .≥ 0.99))

plot(x, U[:, 1:3], label = ["mode 1" "mode 2" "mode 3"],
     xlabel = "x", title = "first three POD modes")
end

modes for 99% variance: 3

Typically 2–4 modes carry 99% of the variance — versus ~6 for the 50-mass oscillator chain, and the tail decays much faster here. Diffusion actively destroys high-frequency content (each Fourier mode dies like e^{-\alpha (n\pi)^2 t}), so the snapshot set is dominated by a handful of slowly-decaying smooth shapes; an oscillatory system keeps recycling its energy through many modes instead.

The first POD modes look like slightly distorted \sin(n\pi x) — distorted because the off-centre initial bump weights the snapshots asymmetrically. That was predictable from separation of variables: the heat equation’s evolution is diagonal in the sine eigenbasis (Unit 6 §6.2), so the snapshot covariance is nearly diagonal there too, and POD — which diagonalises the snapshot covariance — has little choice but to recover (approximately) the same basis.

← Back to Unit 3 §3.5

Solution 3.6 — evaluate a residual by hand

using ForwardDiff, Statistics

let
ω = 2π
candidates = (
    a = t -> cos(ω * t),
    b = t -> cos(0.9 * ω * t),
    c = t -> exp(-t / 2) * cos(ω * t),
)

d2(f, t) = ForwardDiff.derivative(τ -> ForwardDiff.derivative(f, τ), t)
ts = range(0, 2; length = 100)
for (name, f) in pairs(candidates)
    L = mean(t -> (d2(f, t) + ω^2 * f(t))^2, ts)
    println("candidate $name:  L_PDE = ", round(L; sigdigits = 3))
end
end

candidate a:  L_PDE = 4.65e-30
candidate b:  L_PDE = 27.4
candidate c:  L_PDE = 8.46

The ranking (printed above): (a) \approx 10^{-30} — zero up to rounding through the nested duals — then (c) \approx 8.5, and (b) \approx 27 worst of all.

Working the algebra for (c), u = e^{-t/2}\cos\omega t gives \ddot u + \omega^2 u = e^{-t/2}\bigl(\tfrac{1}{4}\cos\omega t + \omega \sin\omega t\bigr) — a residual of amplitude \approx \omega e^{-t/2}, largest near t \approx 0.12 (the first place \sin\omega t = 1, before the envelope decays). So the impostor that perfectly matches the initial condition is loudly rejected by the residual, and rejected earliest in time — the residual sees the spurious damping immediately, no data required. Note also the mildly surprising ordering: a 10% frequency error (b) costs more residual than visible damping (c), because the residual of (b) is proportional to \omega^2 \,(1 - 0.81), and \omega^2 is large. Residual loss punishes frequency errors harshly — a foreshadowing of why spectral content dominates so much of PINN behaviour in Units 5–7.

← Back to Unit 3 §3.6

Unit 4 — Neural DEs

Solution 4.1 — test the invariant

using OrdinaryDiffEq, Plots

let
α, β, γ, δ = 1.5, 1.0, 3.0, 1.0
function lv!(du, u, p, t)
    du[1] = α*u[1] - β*u[1]*u[2]
    du[2] = δ*u[1]*u[2] - γ*u[2]
end

V(x, y) = δ*x - γ*log(x) + β*y - α*log(y)
u0 = [1.0, 1.0];  V0 = V(u0...)

plt = plot(xlabel = "t", ylabel = "V(t) − V(0)")
for rtol in (1e-3, 1e-10)
    sol = solve(ODEProblem(lv!, u0, (0.0, 50.0)), Tsit5();
                saveat = 0.05, reltol = rtol, abstol = rtol)
    plot!(plt, sol.t, [V(u...) - V0 for u in sol.u],
          label = "reltol = $rtol")
end
plt
end

1. The drift is numerics. At reltol = 1e-3 the invariant drifts visibly (typically 10^{-3}–10^{-2} over 50 time units, growing with t); at reltol = 1e-10 it collapses by many orders of magnitude. If the drift were physics — a genuinely non-conserved V — tightening the solver tolerance would not reduce it. That two-tolerance comparison is the standard cheap test for “is my conservation violation real?”

2. A neural network fit to one trajectory has no reason to keep \nabla V \cdot f_\theta = 0; its ~percent-level vector-field errors act like the perturbed field in Solution 3.3, so over 500 time units V drifts secularly and the orbit spirals inward or outward — qualitatively wrong long-run behaviour even when the short-run fit looks perfect. Encoding the invariant structurally (HNN/LNN for mechanical systems, or learning V itself) is the fix, which is why §3.3 exists.

← Back to Unit 4 §4.1

Solution 4.2 — train a tiny Neural ODE on a spiral

using Lux, OrdinaryDiffEq, SciMLSensitivity, Zygote, Optimisers,
      ComponentArrays, Random, Plots

# ground truth: decaying linear spiral
A = [-0.1 2.0; -2.0 -0.1]
true_rhs!(du, u, p, t) = (du .= A * u)
ts   = range(0.0f0, 6.0f0; length = 30)
data = Array(solve(ODEProblem(true_rhs!, Float32[2, 0], (0f0, 6f0)),
                   Tsit5(); saveat = ts))

rng    = Random.MersenneTwister(0)
net    = Lux.Chain(Lux.Dense(2 => 16, tanh), Lux.Dense(16 => 2))
ps, st = Lux.setup(rng, net)
ps     = ComponentArray(ps)

function predict(p, tspan, saveat)
    rhs!(du, u, p_, t) = (du .= first(net(u, p_, st)))
    Array(solve(ODEProblem(rhs!, Float32[2, 0], tspan, p), Tsit5();
                saveat, sensealg = InterpolatingAdjoint()))
end

loss(p) = sum(abs2, predict(p, (0f0, 6f0), ts) .- data)

opt = Optimisers.setup(Optimisers.Adam(1f-2), ps)
for i in 1:1000
    g = first(Zygote.gradient(loss, ps))
    opt, ps = Optimisers.update(opt, ps, g)
    i % 100 == 0 && println("iter $i  loss = $(loss(ps))")
end

# extrapolate to t = 12 against the truth
t_ext   = range(0f0, 12f0; length = 200)
pred    = predict(ps, (0f0, 12f0), t_ext)
truth   = Array(solve(ODEProblem(true_rhs!, Float32[2, 0], (0f0, 12f0)),
                      Tsit5(); saveat = t_ext))
plot(truth[1, :], truth[2, :], label = "truth", lw = 2, c = :black)
plot!(pred[1, :], pred[2, :], label = "Neural ODE", ls = :dash, lw = 2)
vline!([0], alpha = 0); # cosmetic

Expected behaviour: inside the training window the fit is excellent (final loss \sim 10^{-2} on this scale). Extrapolating to t = 12, the spiral continues roughly correctly for another turn or two — the trajectory keeps visiting states near ones the network saw — then degrades as the decaying radius enters state-space the network was never supervised on; the learned field there is unconstrained and the decay rate drifts. Training on only the first half of the spiral makes this much worse: the network never sees small-radius states at all, and the extrapolated orbit can stall, spiral at the wrong rate, or invent a limit cycle.

The moral, stated in UDE terms: a black-box f_\theta is only as good as its state-space coverage. If you know the dynamics are linear (or know any structural piece), encoding it shrinks what the data must cover — which is §4.3’s argument.

JAX users: the same experiment is ~30 lines with diffrax.diffeqsolve + equinox.nn.MLP + optax.adam, with diffrax’s default backsolve adjoint playing the role of InterpolatingAdjoint.

← Back to Unit 4 §4.2

Solution 4.3 — the functional inverse, in miniature

using Lux, OrdinaryDiffEq, SciMLSensitivity, Zygote, Optimisers,
      ComponentArrays, Random, Plots

α, β, γ, δ = 1.5, 1.0, 3.0, 1.0
lv!(du, u, p, t) = (du[1] = α*u[1] - β*u[1]*u[2];
                    du[2] = δ*u[1]*u[2] - γ*u[2])

ts    = range(0.0f0, 8.0f0; length = 60)
rng   = Random.MersenneTwister(0)
truth = Array(solve(ODEProblem(lv!, Float32[1, 1], (0f0, 8f0)),
                    Tsit5(); saveat = ts))
data  = truth .* (1 .+ 0.02f0 .* randn(rng, Float32, size(truth)))

# UDE: keep everything except the predator-growth term, learn that
nn      = Lux.Chain(Lux.Dense(2 => 8, tanh), Lux.Dense(8 => 1))
ps, st  = Lux.setup(rng, nn)
ps      = ComponentArray(ps)

function ude!(du, u, p, t)
    N = first(nn(u, p, st))[1]
    du[1] = α*u[1] - β*u[1]*u[2]
    du[2] = N - γ*u[2]                      # δxy replaced by N_θ(x, y)
end

predict(p) = Array(solve(ODEProblem(ude!, Float32[1, 1], (0f0, 8f0), p),
                         Tsit5(); saveat = ts,
                         sensealg = InterpolatingAdjoint()))
loss(p) = sum(abs2, predict(p) .- data)

opt = Optimisers.setup(Optimisers.Adam(5f-3), ps)
for i in 1:600
    g = first(Zygote.gradient(loss, ps))
    opt, ps = Optimisers.update(opt, ps, g)
end

# compare N_θ(x, y) with δ·x·y over a grid
xs = range(0.2f0, 4f0; length = 60); ys = range(0.2f0, 2.5f0; length = 60)
Nθ = [first(nn(Float32[x, y], ps, st))[1] for y in ys, x in xs]
Ntrue = [δ * x * y for y in ys, x in xs]
p1 = heatmap(xs, ys, Nθ,    title = "learned N_θ(x, y)")
p2 = heatmap(xs, ys, Ntrue, title = "true δ·x·y")
p3 = heatmap(xs, ys, Nθ .- Ntrue, title = "difference", c = :balance)
plot(p1, p2, p3, layout = (1, 3), size = (1100, 320))
# overlay the training orbit on the difference panel:
plot!(p3, truth[1, :], truth[2, :], c = :black, lw = 1.5, label = "orbit")

Inside the band of phase space the training orbit actually visits, the difference panel is near zero — N_\theta recovers \delta x y to a few percent, without ever being told the term was bilinear. Outside the orbit (corners of the grid the trajectory never enters), the difference grows arbitrarily: the loss never constrained those inputs, so the network freelances. Plot the orbit over the difference heatmap and the correspondence is unmistakable — the trustworthy region of a learned closure is the data’s footprint in state space.

That picture is what you should reproduce, in grown-up form, for the capstone’s learned wind-stress envelope: report the recovered function together with the region where the data actually constrains it.

← Back to Unit 4 §4.3

Unit 5 — First PINN

Solution 5.1 — count the solutions

1. Any x(t) = C e^{-t} has zero residual — e.g. C = 0 (the trivial solution x \equiv 0, a favourite of lazy optimisers), C = 1, and C = 5. The zero-residual set is the full one-parameter family \{C e^{-t} : C \in \mathbb{R}\}, and the IC term selects C = 1. This is why §5.1 says “initial conditions are what pin it down” — a residual-only loss has a continuum of global minima, most of them useless.
2. With \lambda_{\text{IC}} = 10^{-6} and total loss 10^{-8}, the IC term alone satisfies \lambda_{\text{IC}} |x_\theta(0) - 1|^2 \leq 10^{-8}, i.e. |x_\theta(0) - 1| \leq \sqrt{10^{-8}/10^{-6}} = 0.1. The network can start 10% off while the headline loss looks superb — and since the residual then locks the trajectory to the (wrong) member 0.9\,e^{-t}, the whole solution is 10% off, uniformly. The takeaway generalises: never report or monitor only the total PINN loss; track each term in its own units, because the weights make raw magnitudes incomparable.

← Back to Unit 5 §5.1

Solution 5.2 — trust, then verify, the autodiff stack

using ForwardDiff

let
u(x, t) = sin(3x) * exp(-2t)

∂x  = ForwardDiff.derivative(ξ -> u(ξ, 0.4), 0.7)
∂xx = ForwardDiff.derivative(
          ξ -> ForwardDiff.derivative(η -> u(η, 0.4), ξ), 0.7)
∂t  = ForwardDiff.derivative(τ -> u(0.7, τ), 0.4)

println("∂x  = ", ∂x,  "   vs closed form ", 3cos(2.1) * exp(-0.8))
println("∂xx = ", ∂xx, "  vs closed form ", -9sin(2.1) * exp(-0.8))
println("∂t  = ", ∂t,  "  vs closed form ", -2sin(2.1) * exp(-0.8))

α = 2 / 9
println("heat-equation residual ∂t − α∂xx = ", ∂t - α * ∂xx)
end

∂x  = -0.6805259316554049   vs closed form -0.6805259316554054
∂xx = -3.490784734793593  vs closed form -3.490784734793592
∂t  = -0.7757299410652428  vs closed form -0.7757299410652426
heat-equation residual ∂t − α∂xx = 0.0

Numbers: \partial_x u = -0.68047\ldots, \partial_{xx} u = -3.4902\ldots, \partial_t u = -0.77560\ldots, and \partial_t u - \tfrac{2}{9}\partial_{xx} u \approx 10^{-16}. The why: u = \sin(3x)e^{-2t} is a single separated heat-equation mode with decay rate \alpha k^2 = \alpha \cdot 9 = 2, so \alpha = 2/9 — the same eigenmode arithmetic as Unit 6 §6.2.

The JAX twin:

import jax, jax.numpy as jnp

u = lambda x, t: jnp.sin(3 * x) * jnp.exp(-2 * t)
du_dx   = jax.jacfwd(u, argnums=0)
d2u_dx2 = jax.jacfwd(jax.jacfwd(u, argnums=0), argnums=0)
du_dt   = jax.jacfwd(u, argnums=1)

x, t = 0.7, 0.4
print(du_dx(x, t), d2u_dx2(x, t), du_dt(x, t))
print(du_dt(x, t) - (2 / 9) * d2u_dx2(x, t))   # ~1e-8 in float32

Same numbers (to float32 precision unless you enable jax.config.update("jax_enable_x64", True)). The habit this builds is the important part: before trusting any PINN residual, run its derivative pipeline on a function with known derivatives. Five minutes, and it catches a remarkable fraction of sign, argument-order, and nesting bugs.

← Back to Unit 5 §5.2

Solution 5.3 — your second PINN: the oscillator

The three changes, grafted onto the §5.3 script (everything not shown is unchanged):

ω = 2.0f0
model = Lux.Chain(Lux.Dense(1 => 32, tanh), Lux.Dense(32 => 32, tanh),
                  Lux.Dense(32 => 1))
ps, st = Lux.setup(rng, model)

x_θ(t, ps, st) = first(model([t], ps, st))[1]

# first and second time derivatives through the network, by central
# differences — the same robust choice as §5.3's dxdt, one order up.
# Zygote differentiates exactly through the three network evaluations.
h = 1f-2
d1(t, ps) = (x_θ(t + h, ps, st) - x_θ(t - h, ps, st)) / (2h)
d2(t, ps) = (x_θ(t + h, ps, st) - 2x_θ(t, ps, st) + x_θ(t - h, ps, st)) / h^2

ts = collect(range(0f0, 4f0; length = 128))

function loss(ps)
    L_res = sum(t -> (d2(t, ps) + ω^2 * x_θ(t, ps, st))^2, ts) / length(ts)
    L_ic1 = (x_θ(0f0, ps, st) - 1f0)^2       # x(0) = 1
    L_ic2 = (d1(0f0, ps))^2                  # ẋ(0) = 0  ← the new IC term
    L_res + 100f0 * (L_ic1 + L_ic2)
end

opt_state = Optimisers.setup(Optimisers.Adam(1f-2), ps)
for epoch in 1:4000
    g = first(Zygote.gradient(loss, ps))
    opt_state, ps = Optimisers.update(opt_state, ps, g)
end

(In JAX the pure-AD version is clean and safe: jax.jacfwd(jax.jacfwd(x_theta)) for \ddot x_\theta with jax.grad outside — no finite differences needed. The Julia stack’s AD-inside-AD caveats are §5.3’s listing comment; the central difference keeps the training gradient exact for the discretised residual.)

Expected results: on [0, 4] the PINN overlays \cos(2t) to plotting accuracy after ~4 000 Adam iterations. Why a second IC? Same counting as Solution 5.1: a second-order ODE has a two-parameter solution family (A\cos\omega t + B\sin\omega t), so two conditions are needed to pin one member.

On [0, 12] — about 3.8 periods — the vanilla setup degrades in a characteristic way: the early-time fit stays good, but the amplitude at late times shrinks toward zero, sometimes with a phase drift. Nothing in the loss tells late times to be a consequence of early times (causal violation), and the oscillatory target sits exactly where spectral bias makes optimisation slow; the easiest way for the optimiser to cut residual at late t is the flat function. Both Unit 7 fixes (causal weighting; Fourier features) directly target this experiment — worth rerunning after reading §7.3.

← Back to Unit 5 §5.3

Solution 5.4 — change the equation, not the plumbing

The whole edit, in ModelingToolkit terms:

@parameters t
@variables x(..)
Dₜ = Differential(t)

eq  = Dₜ(x(t)) ~ x(t) * (1 - x(t))           # was: ~ -x(t)
bcs = [x(0.0) ~ 0.1]                          # was: x(0) ~ 1.0
domains = [t ∈ ClosedInterval(0.0, 8.0)]      # was: 0..4

Everything else — chain, PhysicsInformedNN, discretize, Optimization.solve — is untouched. That’s the declarative payoff: the symbolic layer derives the new residual; you never wrote a dxdt.

Against the exact x(t) = 1/(1 + 9e^{-t}), expect agreement to \sim 10^{-3} with the default settings. The thing you did have to reconsider: the solution spends t \gtrsim 5 glued to the plateau x \approx 1, and a uniform GridTraining lavishes half its collocation points on that trivially-satisfied region while the interesting sigmoid transition around t = \ln 9 \approx 2.2 gets the same density as everywhere else. The fit is still fine here (the problem is easy), but the observation scales: collocation density should follow solution activity, which is the motivation for the adaptive sampling strategies mentioned in §5.1 — and trainable-quadrature strategies like QuasiRandomTraining in NeuralPDE.jl are the one-line upgrade.

← Back to Unit 5 §5.4

Solution 5.5 — sharpen the bump until it breaks

Concretely, the IC line in the script becomes u(x, 0) ~ exp(-S * (x - 0.5)^2) with S ∈ (200, 800, 3200) — bump full-widths of roughly 0.17, 0.083, 0.042.

Indicative outcomes (your numbers will differ, the trend won’t):

Sharpness S	final loss	PINN at t = 0
200	\sim 10^{-4}	clean fit
800	\sim 10^{-3}	peak clipped by ~10–20%
3200	\sim 10^{-2}	bump smeared to a fraction of its true height

And the punchline experiment: at S = 3200, quadrupling the collocation density (GridTraining([0.025, 0.025]) or finer) barely moves the result. The failure is not a sampling problem — the residual is already being measured near the bump — it’s an optimisation problem: the loss-landscape directions that encode high-frequency structure are poorly conditioned, so Adam and even L-BFGS crawl along them (spectral bias, §5.6). More points sharpen the measurement of an error the optimiser already can’t fix. What does fix it: give the network high-frequency inputs to work with — Fourier feature embeddings, Unit 7 §7.3 — at which point S = 3200 becomes as easy as S = 200 was.

← Back to Unit 5 §5.5

Solution 5.6 — diagnose before you medicate

1. Loss imbalance. The BC term is numerically tiny but the solution is wrong at the boundary: the optimiser spent its budget on the dominant residual term and the BC term’s gradient never competed at \lambda_b = 1. First fix: hard BC enforcement (remove the term from the contest entirely); otherwise adaptive loss weighting.
2. Causal violation. Early times fine, late times a frozen smear that doesn’t follow from them — the textbook symptom. Fix: causal training (weight late-time residuals by early-time convergence).
3. Spectral bias. Same network, 100× more iterations purely because the target frequency went from 2\pi to 16\pi. Fix: Fourier feature embeddings with B covering \sim 16\pi.
4. Loss imbalance again — the see-saw form. Cranking one weight starves the other terms; manual tuning is a zero-sum game across 10^2-plus dynamic range. Fix: adaptive loss weighting (gradient-balancing or NTK), or restructure so the conflict disappears (hard IC enforcement removes \lambda_{\text{IC}} from the game).

The meta-skill: each disease has a cheap diagnostic (per-term loss tracking for imbalance; residual-vs-t histogram for causality; iteration-count-vs-frequency for spectral bias). Run the diagnostics before reaching for any fix — the fixes are not interchangeable, and applying the wrong one wastes a tuning cycle.

← Back to Unit 5 §5.6

Unit 6 — PDE Bootcamp

Solution 6.1 — classify, then prescribe

#	(a, 2b, c)	\Delta = b^2 - ac	Type	Needs
1	(1, 0, 4)	-4	elliptic	BCs on a closed boundary, no IC
2	(1, 0, -4)	4	hyperbolic	ICs (u and u_t) + BCs
3	heat form	0	parabolic	one IC + BCs
4	(1, 2, 1)	1 - 1 = 0	parabolic (degenerate)	like heat after a change of variables
5	(y, 0, 1)	-y	mixed: elliptic for y > 0, hyperbolic for y < 0, parabolic on y = 0	depends on region — the equation models transonic flow for exactly this reason

The traps:

• Both u and \partial_n u on an elliptic boundary is Cauchy data on Laplace — Hadamard’s original example. A solution exists only for very special (analytic, compatible) data, and even then it doesn’t depend continuously on it: perturb the Neumann data at spatial frequency k by \varepsilon and the interior solution changes by \varepsilon e^{k d} at distance d from the boundary. Existence and stability both fail → ill-posed.
• Backward heat: write the solution in the sine basis — each mode evolves like e^{+\alpha k^2 t} when you run diffusion in reverse. Any measurement noise at high k, however small, explodes faster than any polynomial; continuous dependence fails catastrophically. (This is also why §6.3 says diffusion “loses information” — the forward map is smoothing, so its inverse is unbounded. Every regularised deconvolution method is a response to exactly this.)

PINN corollary worth repeating: a PINN will happily minimise the residual of an ill-posed problem — it just converges to something that isn’t a solution, because no solution exists or none is stable. Well-posedness is a property of the problem, not the method.

← Back to Unit 6 §6.1

Solution 6.2 — two modes, two decay rates

1. The IC is already a finite sine series with b_1 = 1, b_3 = 0.5, all other b_n = 0. So u(x, t) = e^{-\alpha \pi^2 t}\sin(\pi x) + 0.5\, e^{-9 \alpha \pi^2 t}\sin(3\pi x). No integrals — orthogonality has done the work.
2. Mode n hits 1% of its initial amplitude when e^{-\alpha (n\pi)^2 t} = 0.01, i.e. t = \ln(100) / (\alpha n^2 \pi^2). With \alpha = 0.01: t_3 = 4.605/(0.01 \cdot 9\pi^2) \approx 5.2 and t_1 \approx 46.7. The ratio is exactly n^2 = 9 — the “high frequencies die first” slogan with a number on it. (At t = 5, the profile is already visually a pure \sin \pi x.)
3. FTCS check: swap the IC line for u = sin.(π .* x) .+ 0.5 .* sin.(3π .* x) and the boundary pinning for u_new[1] = u_new[nx] = 0.0, run to T = 5, and compare with the closed form — the whole experiment:

let
nx, α, T = 101, 0.01, 5.0
dx = 1.0 / (nx - 1)
dt = 0.4 * dx^2 / α
nt = ceil(Int, T / dt)
x  = range(0, 1; length = nx)

u = sin.(π .* x) .+ 0.5 .* sin.(3π .* x)
u[1] = u[nx] = 0.0
u_new = similar(u)
for n in 1:nt
    for i in 2:nx-1
        u_new[i] = u[i] + α * dt / dx^2 * (u[i+1] - 2u[i] + u[i-1])
    end
    u_new[1] = u_new[nx] = 0.0
    u, u_new = u_new, u
end

exact = exp(-0.01 * π^2 * T) .* sin.(π .* x) .+
        0.5 * exp(-0.01 * 9π^2 * T) .* sin.(3π .* x)
println("max |FTCS − exact| at T = 5:  ", maximum(abs.(u .- exact)))
end

max |FTCS − exact| at T = 5:  4.616195025480829e-5

← Back to Unit 6 §6.2

Solution 6.3 — match the equation to the movie

Movie	Equation	Mechanism
(1) bump splits into two frozen half-copies	wave	propagation (d’Alembert: \tfrac12 u_0(x - ct) + \tfrac12 u_0(x + ct))
(2) bump slides rigidly	advection	transport
(3) bump spreads, centre fixed	heat	smoothing
(4) front steepens from smooth data, then creeps	Burgers	shock formation (then viscous propagation)

The erfc check, by substitution: let \eta = x / (2\sqrt{\alpha t}), so u = u_0\operatorname{erfc}(\eta) and \operatorname{erfc}'(\eta) = -\tfrac{2}{\sqrt{\pi}}e^{-\eta^2}. Then u_t = u_0\operatorname{erfc}'(\eta)\cdot\Bigl(-\frac{\eta}{2t}\Bigr), \qquad u_x = u_0\operatorname{erfc}'(\eta)\cdot\frac{1}{2\sqrt{\alpha t}}, \qquad u_{xx} = u_0\operatorname{erfc}''(\eta)\cdot\frac{1}{4\alpha t}, and since \operatorname{erfc}''(\eta) = -2\eta\operatorname{erfc}'(\eta), \alpha u_{xx} = u_0 \operatorname{erfc}'(\eta) \cdot \Bigl(-\frac{\eta}{2t}\Bigr) = u_t.\;\checkmark

Or let the machine do it at a few points:

using ForwardDiff, SpecialFunctions
u(x, t) = erfc(x / (2sqrt(0.01t)))
r(x, t) = ForwardDiff.derivative(τ -> u(x, τ), t) -
          0.01 * ForwardDiff.derivative(
              ξ -> ForwardDiff.derivative(η -> u(η, t), ξ), x)
maximum(abs(r(x, t)) for x in 0.1:0.2:1.5, t in 0.5:0.5:3.0)  # ~1e-16

← Back to Unit 6 §6.3

Solution 6.4 — cross the CFL line on purpose

using Plots

let
nx, α = 101, 0.01
dx = 1.0 / (nx - 1)
x  = range(0, 1; length = nx)

plt = plot(xlabel = "step", ylabel = "max |u|", yscale = :log10,
           title = "FTCS at four Courant numbers", legend = :topleft)
for r in (0.49, 0.50, 0.51, 0.60)
    dt = r * dx^2 / α
    u  = exp.(-200 .* (x .- 0.75).^2);  u[1] = u[nx] = 0.0
    u_new = similar(u)
    steps = Int[];  maxima = Float64[]
    for n in 1:4000
        for i in 2:nx-1
            u_new[i] = u[i] + r * (u[i+1] - 2u[i] + u[i-1])
        end
        u_new[1] = u_new[nx] = 0.0
        u, u_new = u_new, u
        if n % 100 == 0
            push!(steps, n); push!(maxima, maximum(abs.(u)))
            maxima[end] > 1e6 && break
        end
    end
    plot!(plt, steps, maxima, marker = :circle, ms = 2, label = "r = $r")
    r > 0.5 && println("r = $r exceeds 1e6 by step ",
                       isempty(steps) ? "?" : steps[end])
end
plt
end

r = 0.51 exceeds 1e6 by step 1000
r = 0.6 exceeds 1e6 by step 100

What you’ll observe, and why von Neumann predicts it exactly: the FTCS amplification factor at wavenumber k is G(k) = 1 - 4r\sin^2(k\Delta x/2), worst at the grid-scale sawtooth k\Delta x = \pi where G = 1 - 4r.

• r = 0.49: G = -0.96 — stable, with a decaying odd-even flicker if you look closely.
• r = 0.50: G = -1 exactly — marginal; the sawtooth neither grows nor decays, it flips sign each step. Roundoff-level sawtooth persists indefinitely.
• r = 0.51: |G| = 1.04. Starting from roundoff-scale sawtooth content (\sim 10^{-16} relative), reaching 10^6 takes \ln(10^{22})/\ln(1.04) \approx 1\,300 steps; starting from the \sim 10^{-2}-scale sawtooth content of the actual profile near its kinks, more like \ln(10^8)/\ln(1.04) \approx 470 steps. Either way: the blow-up appears first as a two-cell-period zigzag superimposed on the smooth profile — measure the spatial period of the garbage and it’s exactly 2\Delta x, as predicted.
• r = 0.60: |G| = 1.4; NaN within ~100 steps.

The deeper point for this course: instability is selective in frequency. The smooth part of the solution remains perfectly well-behaved while one Fourier mode silently exponentiates — which is why “it looks fine for the first 400 steps” is no defence, and why the same Fourier-mode lens reappears for PINN spectral bias.

← Back to Unit 6 §6.4

Solution 6.5 — watch upwind diffuse a step

using Plots

let
nx = 100;  dx = 1.0 / nx;  c = 1.0
x  = (0.5:nx) .* dx
u0 = Float64.((x .> 0.1) .& (x .< 0.3))

function advect(u0, cfl; T = 1.0)
    dt = cfl * dx / c
    u  = copy(u0)
    for _ in 1:round(Int, T / dt)
        u .= u .- cfl .* (u .- circshift(u, 1))    # upwind, periodic
    end
    u
end

plt = plot(x, u0, label = "exact (initial step)", lw = 2, c = :black,
           xlabel = "x", ylabel = "u")
for cfl in (0.3, 0.8, 0.99)
    plot!(plt, x, advect(u0, cfl), label = "upwind, CFL = $cfl", lw = 1.5)
end
plt
end

At CFL 0.99 the step comes back nearly intact; at 0.8 the face has smeared over ~5–8 cells; at 0.3 it’s a sad mound. The resolution of the paradox is the modified equation: Taylor-expanding the upwind scheme shows it solves, to leading error, u_t + c\,u_x = \underbrace{\frac{c\,\Delta x}{2}\bigl(1 - \mathrm{CFL}\bigr)}_{\nu_{\text{num}}}\,u_{xx}, an advection–diffusion equation with an artificial viscosity that vanishes as CFL → 1. At CFL exactly 1, upwind shifts the solution one whole cell per step — the discrete characteristics coincide with the physical ones and the scheme is exact. The diffusion comes from the interpolation implied by fractional-cell shifts, and a smaller time step means more steps, hence more interpolation events, hence more smearing. (“Run at the largest stable step” is, for upwind advection, also an accuracy prescription — a genuine numerics curiosity.)

← Back to Unit 6 §6.5

Solution 6.6 — assemble a stiffness matrix by hand

1. Each hat \phi_i has slope \pm 1/\Delta x on its two supporting elements. So K_{ii} = \int (\phi_i')^2 = 2\Delta x \cdot (1/\Delta x)^2 = 2/\Delta x, and for neighbours the slopes are +1/\Delta x and -1/\Delta x on the one shared element: K_{i,i\pm1} = -1/\Delta x. Non-neighbours don’t overlap → 0. Load: \int \phi_i = \Delta x (area of a unit-height triangle of base 2\Delta x… careful: \tfrac12 \cdot 2\Delta x \cdot 1 = \Delta x ✓).
2. With \Delta x = 1/4: K = 4\,\mathrm{tridiag}(-1, 2, -1) (3×3) and b = (1/4)(1, 1, 1)^\top. Solving 4(2u_1 - u_2) = 1/4, etc., gives u = (3/32,\ 1/8,\ 3/32) = (0.09375, 0.125, 0.09375).
3. The exact solution at the nodes: u(0.25) = 0.25 \cdot 0.75 / 2 = 0.09375, u(0.5) = 0.125 — exact, as promised. (Why: for -u'' = f in 1-D with linear elements, the FE solution is the interpolant of the exact solution at the nodes — Galerkin orthogonality plus the fact that the Green’s function of the 1-D Laplacian is piecewise linear. This nodal-exactness is special to 1-D second-order problems; don’t expect it in 2-D.)

using LinearAlgebra

let
N  = 8;  h = 1 / N
K  = (1/h) * SymTridiagonal(fill(2.0, N - 1), fill(-1.0, N - 2))
b  = fill(h, N - 1)
u  = K \ b
nodes = h .* (1:N-1)
println("FE nodal values:    ", round.(u; digits = 6))
println("exact x(1-x)/2:     ", round.(nodes .* (1 .- nodes) ./ 2; digits = 6))
println("interior maximum:   ", maximum(u), "  (= 1/8 at the midpoint)")
end

FE nodal values:    [0.054688, 0.09375, 0.117188, 0.125, 0.117188, 0.09375, 0.054688]
exact x(1-x)/2:     [0.054688, 0.09375, 0.117188, 0.125, 0.117188, 0.09375, 0.054688]
interior maximum:   0.12500000000000003  (= 1/8 at the midpoint)

← Back to Unit 6 §6.6

Solution 6.7 — four problems, four methods

1. FE. Patient-specific curved geometry from segmentation is the canonical unstructured-mesh problem; FE triangulates it natively and its error theory supports refining near the wall. (FV is defensible for the flow itself; FD is not.)
2. FV. A dam-break is a hyperbolic conservation law with a moving shock — exactly what FV-with-limiters was built for, and “volume conserved to machine precision” is FV’s by-construction guarantee that no other method offers.
3. FD (via MethodOfLines.jl, as the capstone does). Smooth solution, 1-D interval, no geometry: the structured-grid method is the simplest trustworthy thing, and “reference fast” rules out anything with meshing overhead.
4. PINN — workflow W2/W3. The unknown is a function (spatially-varying diffusivity), the data are scattered and noisy, and the optimiser needs gradients of the fit with respect to the unknown — the differentiable-forward-map property is the deciding feature. (A classical adjoint FD inversion is the honourable runner-up if the geometry is simple and an adjoint code exists — Unit 1’s lesson.)

← Back to Unit 6 §6.7

Solution 6.8 — audit the Unit 1 solver

1. c_{\max} = \sqrt{9.81 \times 50} = 22.1 m/s → \Delta t \leq 500/22.1 = 22.6 s ✓. The production run’s Courant number is c\,\Delta t/\Delta x = 22.1 \times 12 / 500 = 0.53 — a safety factor of ~1.9. Comfortable: deep enough into the stable region that bathymetry tweaks (a deeper hand-edited channel) won’t silently cross the line.
2. Pulse width in the channel: 0.55\,\text{h} \times 3600 \times 15\,\text{m/s} \approx 30 km \approx 60 cells. A feature resolved by 60 cells suffers negligible dispersion error in a second-order scheme — supporting the claim. (The flip side, also claimed in §6.8: a sharp impulse a few cells wide would disperse visibly.)
3. c_{\max} = \sqrt{9.81 \times 200} = 44.3 m/s → \Delta t \leq 11.3 s. The production 12 s step now violates CFL and the run would blow up from the deep-shelf corner outward. Fix: drop to \Delta t = 10 s (cost: +20% runtime) — or coarsen \Delta x over the deep shelf if runtime mattered, since deep-water waves are long and smooth anyway. The general habit: every time the bathymetry changes, re-derive the CFL bound; it is not a constant of the code.

← Back to Unit 6 §6.8

Unit 7 — PINNs Meet PDEs

Solution 7.1 — grade the disk PINN against the exact harmonic

# after `res = Optimization.solve(...)` in the §7.1 disk script:
using Statistics

phi   = discretization.phi
ps_tr = res.u

rs = range(r_min, R; length = 50)
θs = range(0.0, 2π;  length = 100)
u_pinn  = [first(phi([r, θ], ps_tr))    for r in rs, θ in θs]
u_exact = [r^3 * sin(3θ)                for r in rs, θ in θs]

rel_L2 = sqrt(sum(abs2, u_pinn .- u_exact) / sum(abs2, u_exact))
worst  = findmax(abs.(u_pinn .- u_exact))
println("relative L² error = $rel_L2")
println("worst error $(worst[1]) at (r, θ) = ",
        (rs[worst[2][1]], θs[worst[2][2]]))

Typical findings:

1. Relative L^2 error of 10^{-3}–10^{-2}, with the worst pointwise error near the outer boundary, often close to the \theta = 0 / 2\pi seam. Two collocation facts explain it: the solution r^3\sin 3\theta has all its amplitude and gradient near r = R, while uniform-in-(r,\theta) sampling spends most of its points at small r where the solution is nearly zero (and, bonus subtlety, uniform (r, \theta) sampling over-densifies small radii in area terms); and the boundary itself is only enforced at ~36 soft points. The r_{\min} cutoff is usually not the worst spot — the solution there is \mathcal{O}(r_{\min}^3) \approx 10^{-4}, so even large relative errors are absolutely tiny.

2. With \sin(6\theta) (exact solution r^6\sin 6\theta), the error typically grows several-fold at the same budget: six lobes double the angular frequency the network must represent, and r^6 concentrates the action even harder against the rim. Same architecture, same optimiser, harder spectrum — the §7.2 spectral bias story arriving on schedule.

← Back to Unit 7 §7.1

Solution 7.2 — reproduce the failure, then prove it

The d’Alembert reference is the cheapest honest check: for \eta_t + H u_x = 0, u_t + g\eta_x = 0 with u(x,0) = 0, the elevation splits exactly into \eta(x, t) = \tfrac12\eta_0(x - ct) + \tfrac12\eta_0(x + ct) with c = \sqrt{gH} \approx 9.9 m/s. At the half-time snapshot t = 0.025 the two half-bumps are centred at 1.0 \pm 0.248.

# residual-vs-time histogram for the trained two-network system
phi = discretization.phi          # vector of two networks (η, u)
p_η, p_u = res.u.depvar.η, res.u.depvar.u_vel   # per-network params

using ForwardDiff, Statistics
ηf(x, t) = first(phi[1]([x, t], p_η))
uf(x, t) = first(phi[2]([x, t], p_u))
r1(x, t) = ForwardDiff.derivative(τ -> ηf(x, τ), t) +
           10.0 * ForwardDiff.derivative(ξ -> uf(ξ, t), x)
r2(x, t) = ForwardDiff.derivative(τ -> uf(x, τ), t) +
           9.81 * ForwardDiff.derivative(ξ -> ηf(ξ, t), x)

xs = range(0, 2; length = 200)
tbins = range(0, 0.05; length = 11)
means = [mean(abs2(r1(x, t)) + abs2(r2(x, t))
              for x in xs, t in range(tbins[i], tbins[i+1]; length = 20))
         for i in 1:10]
bar(midpoints(collect(tbins)), means, xlabel = "t bin",
    ylabel = "mean residual", legend = false)

What each diagnostic exposes:

• Snapshot vs d’Alembert → spectral bias (one over-smoothed lump instead of two distinct bumps) and loss imbalance (the visible drift at the x = 0, 2 boundaries).
• Residual histogram → causal violation: the late bins sit at or above the early ones; a causally-trained run would show a clean decreasing staircase.

The BC×100 experiment: the boundary drift improves (that was loss imbalance), but the single-lump oversmoothing and the non-decreasing histogram persist — they were never about the BC weight. One knob, one disease: empirical proof that the three failure modes are separate, each needing its own §7.3 fix.

← Back to Unit 7 §7.2

Solution 7.3 — Fourier features in twenty lines

using Lux, Random, Zygote, Optimisers, Statistics, Plots

let
rng = Random.MersenneTwister(0)
xs  = rand(rng, Float32, 1, 200)
ys  = sin.(25f0 .* xs)

# (a) plain MLP
mlp = Lux.Chain(Lux.Dense(1 => 64, tanh), Lux.Dense(64 => 64, tanh),
                Lux.Dense(64 => 1))

# (b) Fourier-feature front end: x ↦ [sin(Bx); cos(Bx)], 16 frequencies
function ff_model(σB)
    B = σB .* randn(rng, Float32, 16, 1)
    embed = x -> vcat(sin.(B * x), cos.(B * x))
    Lux.Chain(Lux.WrappedFunction(embed),
              Lux.Dense(32 => 64, tanh), Lux.Dense(64 => 64, tanh),
              Lux.Dense(64 => 1))
end

function fit(model; iters = 2000)
    ps, st = Lux.setup(rng, model)
    L(p) = mean(abs2, first(model(xs, p, st)) .- ys)
    opt = Optimisers.setup(Optimisers.Adam(1f-3), ps)
    hist = Float32[]
    for _ in 1:iters
        g = first(Zygote.gradient(L, ps))
        opt, ps = Optimisers.update(opt, ps, g)
        push!(hist, L(ps))
    end
    ps, st, hist
end

_, _, h_plain = fit(mlp)
_, _, h_ff10  = fit(ff_model(10f0))
plt = plot(h_plain, yscale = :log10, label = "plain MLP",
           xlabel = "Adam iteration", ylabel = "MSE")
plot!(plt, h_ff10, label = "Fourier features σ = 10")
end

Expected: the plain MLP’s loss plateaus high for most of the budget — at 2 000 iterations it has often barely begun to carve out the 25 rad/unit oscillation — while the \sigma = 10 FF model drops orders of magnitude within a few hundred iterations.

The two failure directions: \sigma = 1 gives embeddings whose frequencies are all far below 25 — the bias is not broken and the curve looks like the plain MLP’s. \sigma = 100 scatters most frequencies far above the target: the model can represent wild oscillations between the 200 samples, so the loss gets noisy, optimisation jitters, and the fit between samples is garbage — spectral bias replaced by spectral overshoot/overfitting. The one-sentence takeaway: the embedding’s frequency distribution must cover (roughly bracket) the target solution’s spectrum — \sigma is a prior on content, not a free lunch.

The hard-BC factor: x(1-x) vanishes at x = 0, 1 identically, so u_\theta = x(1-x)N_\theta satisfies the homogeneous Dirichlet pair for any N_\theta — the BC is removed from the optimisation entirely. The IC analogue from the exercise: u_\theta(x, t) = g(x) + t\,N_\theta(x, t), which equals g(x) at t = 0 regardless of the network (this is exactly the “hard IC” used in Task B, Unit 10 §10.3). Composing both: u_\theta = g(x) + t\,x(1-x)\,N_\theta — at the price that the prefactors also rescale the network’s gradients, which is why hard enforcement of everything sometimes trains worse than hard BC + soft flux.

← Back to Unit 7 §7.3

Solution 7.4 — file these five papers

1. W3 (with a defensible W1 reading — this is the arguable one, see below).
2. W1 — forward solve, no data term, benchmark against a classical solver.
3. W2 — the van-Genuchten parameters are a handful of scalars optimised jointly with the field network. (That the head field is also recovered doesn’t change the classification: the field network exists in every workflow; the column that matters is what’s unknown in the PDE.)
4. W4 — known dynamical core + learned closure from high-resolution output: the textbook hybrid-simulator setup.
5. W3 — the course’s running example: unknown boundary-condition function ψ(t), data from gauges.

The arguable case is (1) Hidden Fluid Mechanics. Reading A (W3): the unknowns are functions — entire velocity and pressure fields never measured anywhere — recovered from auxiliary observations; “recover an unknown function from indirect data” is W3’s definition. Reading B (W1-plus-data): there is no unknown in the PDE — Navier–Stokes is fully specified — so this is “just” a forward solve where unusual data (concentration) replaces unknown BCs/ICs. The honest verdict: HFM shows the taxonomy’s edges are about what plays the role of the missing information; the concentration data substitutes for unknown boundary and initial conditions, which are functions — so W3 by spirit, W1 by letter.

← Back to Unit 7 §7.4

Solution 7.5 — recover a diffusivity (Workflow 2, end-to-end)

using Lux, Random, Zygote, Optimisers, ComponentArrays, Statistics

# --- synthetic truth and data --------------------------------------
α_star = 0.07
u_exact(x, t) = exp(-α_star * π^2 * t) * sin(π * x)

rng = Random.MersenneTwister(0)
Nd  = 50
xd, td = rand(rng, Float32, Nd), rand(rng, Float32, Nd)
ud  = Float32.(u_exact.(xd, td)) .* (1 .+ 0.01f0 .* randn(rng, Float32, Nd))

# --- PINN with a trainable scalar α --------------------------------
net = Lux.Chain(Lux.Dense(2 => 32, tanh), Lux.Dense(32 => 32, tanh),
                Lux.Dense(32 => 1))
ps0, st = Lux.setup(rng, net)
θ = ComponentArray(net = ComponentArray(ps0), logα = Float32[log(0.02)])
# (log-parameterise α so it stays positive; init deliberately wrong)

# hard BC/IC ansatz: u = sin(πx)·(1 + t·N(x,t)) — satisfies the IC
# u(x,0)=sin(πx) and the boundaries u(0,t)=u(1,t)=0 by construction.
u_θ(x, t, θ) = sin(π * x) * (1 + t * first(net([x, t], θ.net, st))[1])

# input derivatives by central differences (§5.3's robust pattern) so
# Zygote's training gradient flows exactly through every term
hfd = 1f-2
function residual(x, t, θ)
    α   = exp(θ.logα[1])
    ut  = (u_θ(x, t + hfd, θ) - u_θ(x, t - hfd, θ)) / (2hfd)
    uxx = (u_θ(x + hfd, t, θ) - 2u_θ(x, t, θ) + u_θ(x - hfd, t, θ)) / hfd^2
    ut - α * uxx
end

xr, tr = rand(rng, Float32, 400), rand(rng, Float32, 400)
function loss(θ)
    Lr = mean(abs2, residual.(xr, tr, Ref(θ)))
    Ld = mean(abs2, u_θ.(xd, td, Ref(θ)) .- ud)
    Lr + 100f0 * Ld
end

opt = Optimisers.setup(Optimisers.Adam(1f-2), θ)
for i in 1:3000
    g = first(Zygote.gradient(loss, θ))
    opt, θ = Optimisers.update(opt, θ, g)
    i % 500 == 0 && println("iter $i  α̂ = $(exp(θ.logα[1]))")
end

(Why central differences rather than ForwardDiff inside the training loss: Zygote cannot track parameter gradients through ForwardDiff.derivative of a parameter-closure — it warns and silently drops that contribution, which would leave the residual term untrainable. See the §5.3 listing comment; the JAX equivalent has no such trap.)

Indicative results:

Data regime	recovered \hat\alpha
50 points, 1% noise	0.070 \pm 0.001 — clean
50 points, 5% noise	\sim 0.066–0.074, run-to-run scatter, mild bias
10 points, 1% noise	usually still within ~5% — the physics is doing the interpolation

The first “when it doesn’t work” item you hit is very noisy data: with 5% noise and no noise model, the data term drags u_\theta toward the noise realisations and \alpha absorbs part of the error as a biased decay rate. Note what you don’t hit: ill-posedness. A single global decay rate observed through 50 points is massively over-determined — which is why W2 problems are the gentle on-ramp and the capstone’s W3 (a whole unknown function) needs the extra regularisation machinery.

← Back to Unit 7 §7.5

Solution 7.6 — pick the package

Job	First pick	Cross-ecosystem runner-up
1. FD reference from a symbolic PDESystem	MethodOfLines.jl	no direct Python analogue in the table — closest is hand-rolling in numpy/scipy (or PhysicsNeMo-Sym’s FD utilities)
2. Symbolic-PDE PINN in Julia	NeuralPDE.jl	DeepXDE (or PhysicsNeMo-Sym for the most symbolic Python workflow)
3. Backprop through an ODE solve in JAX	Diffrax	SciMLSensitivity.jl
4. Sparse symbolic equation discovery	PySINDy	the Unit 3 STLSQ script / DataDrivenDiffEq.jl in Julia
5. PyTorch team, industrial PINN + operator tooling	NVIDIA PhysicsNeMo (+-Sym)	NeuralPDE.jl + NeuralOperators.jl
6. KAN inside a Lux PINN	KolmogorovArnold.jl	pykan

Worth noticing while filling this in: rows 1–2 are where the Julia side’s symbolic layer (ModelingToolkit) has no full Python counterpart, and row 5 is where the Python side’s industrial weight has no Julia counterpart. The ecosystems are genuinely asymmetric, which is why the workshop teaches both.

← Back to Unit 7 §7.6

Solution 7.7 — read one keystone paper properly

A filled-in example for Hidden Fluid Mechanics (Raissi, Yazdani & Karniadakis, Science 2020):

1. PDE enforced: incompressible Navier–Stokes (momentum + continuity) plus the passive-scalar transport equation for the observed concentration.
2. Measured vs reconstructed: measured — spatio-temporal concentration fields (dye/smoke/contrast agent) and geometry; reconstructed — full velocity field and pressure (pressure is never measured anywhere, and is recovered only up to the constant the equations leave free).
3. Workflow: W3 in spirit (unknown fields recovered from indirect data) — see Solution 7.4 for the W1-by-letter argument.
4. Acknowledged limitation: long training times; degradation at high Reynolds number; fine-scale turbulence out of reach.
5. Pre-dated fixes: causal training (2022) and modern adaptive weighting both post-date it; the multi-scale/turbulence limitation is exactly where Fourier features and causal weighting would plausibly help — and where follow-up literature went.

Calibration verdict: the headline (“velocity and pressure from concentration visualisations alone”) survives within its stated regime — laminar-to-moderate-Re, well-resolved scalar fields. The over-claim to watch for is in citations of the paper rather than the paper itself: it is routinely cited as if turbulent flows were demonstrated. Reading the limitations section of the original is the antidote — which was the point of this exercise.

← Back to Unit 7 §7.7

Solution 7.8 — the marketing audit

Claim	Bucket	The confirming question
1. trained on 40 000 Fluent runs	surrogate	“Is any PDE residual evaluated during training, or only data misfit?”
2. minimises the NS residual at collocation points	residual-loss PINN	“At collocation points or only at sensor locations?” (the former is the PINN signature)
3. GNN on ERA5, 10-day forecast	neural operator (GraphCast-class)	“Does the loss contain the primitive equations, or only reanalysis targets?”
4. “respects conservation because the training data did”	surrogate (the trap)	“What enforces conservation at inference time on inputs unlike the training set?”
5. governing equations as soft constraints, 100× less data	PINN (residual-loss hybrid, W4-flavoured)	“Soft constraints on what — the full PDE, or a low-order proxy?”

Why claim 4 is hollow: a network trained on energy-conserving trajectories learns to imitate conservation on the training distribution. Conservation is a property of the data, not of the hypothesis class — nothing in the architecture or loss penalises a violating output, so the first out-of-distribution input (longer horizon, unseen forcing, extrapolated parameter) is free to leak energy, and typically does, compounding over rollout steps. The fixes are exactly the Unit 3 §3.2 inductive-bias menu: conservation as an architectural constraint (predict a flux and take its divergence; HNN-style potentials), or at minimum as a residual penalty evaluated on the model’s own outputs — which is the line separating buckets 1 and 2 in the first place.

← Back to Unit 7 §7.8

Unit 8 — PDE Modelling in AIMS Domains

Solution 8.1 — write the turbidity model

Term by term from the temperature equation:

\frac{\partial C}{\partial t} \;+\; w(z, t)\,\frac{\partial C}{\partial z} \;=\; \frac{\partial}{\partial z}\!\left(\kappa(z, t)\,\frac{\partial C}{\partial z}\right) \;+\; \underbrace{\frac{\partial}{\partial z}\bigl(w_s\, C\bigr)}_{\text{settling}},

• Unchanged: vertical advection by w and turbulent diffusion by \kappa — turbulence stirs sediment exactly as it stirs heat (same Reynolds-closure argument, same eddy diffusivity to leading order).
• The body source is gone — sunlight heats water but doesn’t create sediment. Its place is taken by the settling term: particles fall at w_s > 0 relative to the water, a downward flux -w_s C whose divergence adds +\partial_z(w_s C) to the right-hand side (with z up). Equivalently: sediment is advected at the composite velocity w - w_s.
• Top BC (z = 0): no sediment crosses the sea surface — total flux zero: \kappa\,\partial_z C + w_s C = 0. Note this is a Robin condition, not Neumann: the settling flux must be cancelled by a diffusive flux at the surface.
• Bottom BC (z = -H): the bed exchanges in both directions — deposition w_s C|_{-H} out, erosion E(\tau_b) in, with E an increasing function of bed shear stress (this is where a resuspension event enters the model). A minimal version: \kappa\,\partial_z C|_{-H} = E(t) - w_s C(-H, t).

Light for the corals: Beer–Lambert again, but with an attenuation coefficient that depends on your solution — I(z, t) = I_0(t)\exp\bigl(\int_z^0 -k_d(C(z', t))\,dz'\bigr) with k_d increasing in C. (Note the pleasant symmetry: in the thermal problem light was a source in the PDE; in the turbidity problem the PDE’s solution feeds back into the light. Couple the two and you have a genuinely interesting — and AIMS-relevant — multiphysics column.)

← Back to Unit 8 §8.1

Solution 8.2 — units police

1. [\partial_t T] = \text{K/s}. [w\,\partial_z T] = (\text{m/s})(\text{K/m}) = \text{K/s} ✓. [\partial_z(\kappa\,\partial_z T)] = \frac{1}{\text{m}}\cdot (\text{m}^2/\text{s})(\text{K/m}) = \text{K/s} ✓. So [\mathcal{S}] must also be K/s ✓.
2. Q_{\text{SW}} is W/m² = \text{J s}^{-1}\text{m}^{-2}. Dividing by \rho c_p (\text{kg m}^{-3} \cdot \text{J kg}^{-1}\text{K}^{-1} = \text{J m}^{-3}\text{K}^{-1}) gives \text{K·m/s} — a flux of temperature. The Beer–Lambert factor k\,e^{kz} carries units \text{m}^{-1}, turning the flux into the volumetric heating rate K/s. Watts in, kelvin per second out, with \rho c_p \approx 4.09 \times 10^6\, \text{J m}^{-3}\text{K}^{-1} as the exchange rate.
3. Left side: (\text{m}^2/\text{s})(\text{K/m}) = \text{K·m/s}; right side: (\text{W/m}^2)/(\text{J m}^{-3}\text{K}^{-1}) = \text{K·m/s} ✓. Equilibrium gradient: \partial_z T = \frac{Q_{\text{np}}}{\rho c_p \kappa} = \frac{-200}{1025 \times 3990 \times 10^{-3}} \approx -0.049\ \text{K/m}. Small — about a sixth of a typical thermocline gradient. That’s worth internalising: even a hefty 200 W/m² surface flux sustains only a gentle gradient through a well-mixed layer, because mixed-layer turbulence is so efficient. Sharp gradients in the ocean live where \kappa is small, not where Q is large.

← Back to Unit 8 §8.2

Solution 8.3 — which mechanism does your sensor see?

1. Advective cooling rate: w\,\partial_z T = 2\times10^{-5} \times 0.3 = 6\times10^{-6}\ \text{K/s} \approx 0.52\ \text{K/day}.
2. To match it, diffusion needs \kappa\,\partial_{zz}T = 6\times10^{-6}, i.e. \partial_{zz}T = 0.06\ \text{K/m}^2 at \kappa = 10^{-4}. A smooth thermocline’s curvature is more like 0.3/10 = 0.03\ \text{K/m}^2 — a factor of two short even with this generous \kappa; with the background interior value \kappa_b = 10^{-5} (§8.5) diffusion delivers only ~0.026 K/day, twenty times short. And there’s a sign subtlety the careful reader catches: near the top of the thermocline the curvature is negative, so diffusion there actually warms the sensor’s depth.
3. So a sustained ~0.5 K/day cooling at -20 m must be advective under quiet-ocean parameters — unless the storm has boosted \kappa by an order of magnitude or more, which is precisely hypothesis 2 (storm-enhanced mixing). The competition between “upwelling at quiet-\kappa” and “quiet-w with storm-\kappa” can’t be settled by one sensor’s cooling rate alone — it needs the vertical pattern across sensors (warming below the thermocline indicts mixing; uniform deep cooling indicts upwelling). Which is the capstone’s whole premise.

← Back to Unit 8 §8.3

Solution 8.4 — where does the sunlight actually go?

The fraction still travelling downward past depth z is F(z) = R\,e^{k_1 z} + (1 - R)\,e^{k_2 z} (with z \le 0), so the fraction absorbed above z is 1 - F(z).

Depth	two-band absorbed above	single-band (\zeta = 10 m)
1 m	1 - (0.58\,e^{-0.5} + 0.42\,e^{-1/14}) \approx 26\%	1 - e^{-0.1} \approx 9.5\%
10 m	1 - (0.58\,e^{-5} + 0.42\,e^{-10/14}) \approx 79\%	1 - e^{-1} \approx 63\%
past 30 m	0.42\,e^{-30/14} \approx 4.9\% reaches	e^{-3} \approx 5.0\% reaches

using Plots

let
R, k1, k2 = 0.58, 0.5, 1/14
F2(z) = R * exp(k1 * z) + (1 - R) * exp(k2 * z)
F1(z) = exp(z / 10)
z = range(-50, 0; length = 500)
plt = plot(F2.(z), z, xscale = :log10, label = "two-band", lw = 2,
           xlabel = "fraction of Q_SW remaining", ylabel = "z (m)")
plot!(plt, F1.(z), z, label = "single band ζ = 10 m", lw = 2)
end

The striking coincidence: below 30 m the two models agree almost exactly (4.9% vs 5.0%) — the capstone’s single-band \zeta = 10 m was evidently chosen to get the thermocline-depth energy right. Where they part company is the top metre or two: the two-band model dumps a quarter of the noon flux into the top metre (the red band), the single-band model spreads it over ~10 m. So the diurnal warm layer is the casualty of the simplification — too thick and too cool in the single-band model — while a thermistor at -40 m (receiving ~2% of surface flux either way) genuinely doesn’t care. Right model complexity depends on which sensor you’re trying to explain; for the storm-at-depth question, single-band is fine.

← Back to Unit 8 §8.4

Solution 8.5 — find the mixing barrier

1. \partial_z T_0(z) = \frac{T_{\text{surface}} - T_{\text{deep}}}{2\,\delta_t}\, \mathrm{sech}^2\!\bigl((z - z_t)/\delta_t\bigr) = \frac{10}{10}\,\mathrm{sech}^2\!\bigl((z + 30)/5\bigr) — peaking at 1 K/m at z = -30 m. So N^2_{\max} = \alpha g\,\partial_z T = 2\times10^{-4} \times 9.81 \times 1 \approx 2.0\times10^{-3}\ \text{s}^{-2} (buoyancy period 2\pi/N \approx 2.3 min — a briskly stratified thermocline).
2. \mathrm{Ri}(z) = N^2 / (0.01)^2 peaks at \mathrm{Ri}_{\max} \approx 20. The \mathrm{Ri} > 1/4 band: solve \mathrm{sech}^2((z+30)/5) > 0.25/19.6, giving |z + 30| \lesssim 14 m — stability from roughly -16 m down to -44 m.
3. With \kappa = \kappa_b + \kappa_0/(1 + 5\mathrm{Ri})^2, \kappa_0 = 10^{-3}, \kappa_b = 10^{-5}: near the surface \mathrm{Ri} \approx 0 so \kappa \approx 1.0\times10^{-3}; at the thermocline \kappa \approx 10^{-5} + 10^{-3}/(1+98)^2 \approx 1.01\times10^{-5} — a suppression factor of ~100, with the closure pinned essentially at its background floor.

using Plots

let
dTdz(z) = (10 / (2 * 5)) * sech((z + 30) / 5)^2
Ri(z)   = 2e-4 * 9.81 * dTdz(z) / 0.01^2
κ(z)    = 1e-5 + 1e-3 / (1 + 5 * Ri(z))^2
z = range(-60, 0; length = 600)
plot(κ.(z), z, xscale = :log10, lw = 2, legend = false,
     xlabel = "κ (m²/s)", ylabel = "z (m)",
     title = "Ri-dependent closure through the initial thermocline")
end

The consequence: the thermocline is a mixing barrier. A surface heat anomaly diffuses through the upper mixed layer in hours (T \sim h_m^2/\kappa_m \approx 20^2/10^{-3}\,\text{s} \approx 4.6 days for the full 20 m, much less for the top few metres) but crossing the ~30 m-wide suppressed band at \kappa \approx 10^{-5} takes \sim 30^2/10^{-5}\,\text{s} \approx 1\,000 days. Deep sensors simply cannot be reached diffusively on storm timescales unless the storm smashes the barrier — i.e. raises \mathrm{Ri}’s denominator via shear until the closure reopens. That’s hypothesis 2 in dimensionless clothing.

← Back to Unit 8 §8.5

Solution 8.6 — three sites, three regimes, by the numbers

With \kappa_m = 10^{-3} m²/s and w_0 = 10^{-5} m/s:

Site	H	\mathrm{Pe} = w_0H/\kappa_m	T_\kappa = H^2/\kappa_m	\sqrt{\kappa_m \cdot 3\,\text{d}}
A — Cleveland Bay	15 m	0.15	2.6 days	16 m (> H!)
B — Davies Reef	60 m	0.6	42 days	16 m
C — Myrmidon	100 m	1.0	116 days	16 m

Site A: \mathrm{Pe} \ll 1 ✓ and $T_< $ the 30-day window — diffusion wins and has time to act on the whole column; a 3-day storm’s diffusive reach exceeds the full depth. Site B: \mathrm{Pe} \sim 1 ✓. Site C: the computed Pe is 1.0, not \gg 1 — the label fails if you use the mixed-layer \kappa_m for the whole column. The label is rescued by the §8.5 structure: below the mixed layer the relevant diffusivity is \kappa_b = 10^{-5}, giving an interior \mathrm{Pe} = 10^{-5} \times 100 / 10^{-5} = 100 \gg 1. So the honest statement is “advection-dominated in the interior, where four of the five sensors sit” — the regime label depends on which \kappa governs the depths you care about. And the hinted caveat doubles it: T_\kappa = 116 days \gg 30 days means a 30-day run mixes only the top fraction of Myrmidon’s column, so whatever happens at its deep sensors over the storm window is advection’s doing almost by default.

← Back to Unit 8 §8.6

Solution 8.7 — place the capstone in the literature

1. Mixed-layer \kappa profile shape → Large, McWilliams & Doney 1994 (KPP): the capstone’s “large in a surface layer, decaying with depth” profile is a frozen snapshot of what KPP computes prognostically. Shortcoming addressed: running a full closure adds state, cost, and tuning that would bury the inverse-problem pedagogy.
2. Diurnal warm layer forming and eroding → Price, Weller & Pinkel 1986 — the canonical observational + model account of the diurnal cycle the toy tasks reproduce.
3. Column instead of SST + DHW → Liu et al. 2014 / NOAA Coral Reef Watch: DHW is the operational standard, and its acknowledged blind spot — no vertical structure — is exactly what a thermistor-chain column model fixes. This is the capstone’s “why does AIMS care” row.
4. Recovering an unknown driver from one mooring → Kunze 1985 (the classic single-mooring inverse) and, at modern scale, coastal 4D-Var. Shortcoming addressed: variational assimilation at GCM cost vs a laptop-scale PINN.
5. Closest method → the PINN-flavoured mixed-layer line (Bolton & Zanna 2019 onward, and the CSIRO AI4DA work). What the capstone simplifies away relative to it: prescribed (not learned) \kappa closure, prescribed w from an offline SWE driver, synthetic data with known truth, and a single scalar driver \tau(t) instead of a full forcing field.

← Back to Unit 8 §8.7

Solution 8.8 — reconstruct the spec from memory

The answer key, compactly (grade yourself against Unit 9 §§9.3–9.5 for the full forms):

• PDE: \partial_t T + w\,\partial_z T = \partial_z(\kappa\,\partial_z T) + \mathcal{S} on z \in [-H, 0] — (a) storage, (b) vertical advection,
  3. turbulent diffusion, (d) penetrating-shortwave body source.
• BCs: Neumann at the top — \kappa\,\partial_z T|_{0} = Q_{\text{np}}/(\rho_0 c_p), the non-penetrative flux only; Dirichlet at the bottom — T(-H) = T_{\text{deep}}.
• IC: the \tanh thermocline profile (T_{\text{surface}} = 28, T_{\text{deep}} = 18, z_t = -30 m, \delta_t = 5 m).
• Forcings (all prescribed): w(z, t) from the offline SWE driver via w = -(z+H)\nabla_h\!\cdot\!\mathbf{u}_h; \kappa(z, t) from one of the three closures; Q_{\text{np}}(t) (longwave + sensible + latent, the BC); Q_{\text{SW}}(t) (diurnal half-cosine, entering through \mathcal{S} via Beer–Lambert — not through the BC).

The two load-bearing distinctions, as advertised: shortwave is a body source (it heats the interior where it is absorbed), and the BC types differ top vs bottom (flux known at the surface, value known at the deep reservoir). Get either wrong in a PINN and the loss will happily minimise the wrong physics — there is no error message for a mis-specified problem.

← Back to Unit 8 §8.8

Unit 9 — Project Specification

Solution 9.1 — read the mooring CSVs before modelling them

using CSV, DataFrames, Statistics, Plots

let
# paths are relative to this page; from the repo root use
# units/unit_10/data/mooring_$site.csv
panels = []
for site in ("A", "B", "C")
    df = CSV.read("../unit_10/data/mooring_$site.csv", DataFrame)
    tday = df.time_hours ./ 24
    cols = names(df)[2:end]                       # the five depth columns
    p = plot(title = "Site $site", xlabel = "day", ylabel = "T (°C)",
             legend = :outerright)
    for c in cols
        plot!(p, tday, df[!, c], label = c)
    end
    vline!(p, [10], c = :gray, ls = :dash, label = "")
    push!(panels, p)
end
plot(panels..., layout = (3, 1), size = (800, 900))
end

2. What the raw traces should tell you (the spec’s regime table, recovered by eye): Site A — all five traces move nearly together (well-mixed 15 m column), modest storm dip, fast recovery within days. Site B — clear thermocline separation; the 25–45 m sensors show the largest storm cooling with the deep sensor lagging the surface by several hours to a day; partial recovery by day 30. Site C — surface nearly untouched; the deep sensors (40–95 m) cool first and most, with the slowest recovery (its T_\kappa \gg window — Solution 8.6’s arithmetic in the flesh).
3. One-hour increments of a slowly-varying signal are dominated by noise: \mathrm{Var}(\Delta X) = 2\sigma^2 + (\text{signal increment})^2, so \hat\sigma \approx \mathrm{SD}(\Delta X)/\sqrt{2}, minus whatever real signal moves within an hour. In the calm window the deep sensors give \hat\sigma \approx 0.05\,°C cleanly; the top sensor over-estimates because the diurnal cycle contributes genuine hourly increments — divide by \sqrt 2 and you’ll still sit slightly high there. (This “difference twice, divide by \sqrt2” trick is worth keeping: it estimates sensor noise without de-trending.)

← Back to Unit 9 §9.1

Solution 9.2 — build w(z,t) from a divergence trace

using Plots

let
H = 100.0
div_h(t_d) = -2e-7 * exp(-((t_d - 10) / 1)^2)     # s⁻¹, t in days
w(z, t_d)  = -(z + H) * div_h(t_d)

td = range(0, 30; length = 601)
plt = plot(td, [w(-10, t) * 86400 for t in td], label = "z = -10 m",
           xlabel = "day", ylabel = "w (m/day)")
plot!(plt, td, [w(-50, t) * 86400 for t in td], label = "z = -50 m")
plot!(plt, td, [w(-90, t) * 86400 for t in td], label = "z = -90 m")
end

• w(-H, t) = -((-H) + H)\cdot(\ldots) = 0 ✓ for all t — the rigid bottom can’t pump water through itself.
• Sign: the trace is a negative divergence — convergence. Continuity says \partial_z w = -\nabla_h\!\cdot\!\mathbf{u}_h > 0; integrating up from w(-H) = 0 makes w > 0 everywhere above the bottom: horizontally converging water has nowhere to go but up. So this storm trace drives upwelling — peak w(-10) = 90 \times 2\times10^{-7} = 1.8\times10^{-5} m/s ≈ 1.6 m/day near the surface, ~0.9 m/day at mid-column — right on the reference w_0 \sim 1 m/day scale.
• What linear-in-z assumes: that \nabla_h\!\cdot\!\mathbf{u}_h is independent of depth, i.e. the horizontal flow is barotropic (depth-uniform) — which is what a shallow-water model produces by construction. A real stratified shelf has baroclinic structure and w(z) bends accordingly; that’s a known modelling debt the capstone carries knowingly.

← Back to Unit 9 §9.2

Solution 9.3 — get the signs right, once and forever

1. With \partial_z T > 0 (warm above) and w > 0 (upwelling), the advective tendency is \partial_t T = -w\,\partial_z T < 0: every fixed depth cools, because the water now occupying it came from below, where it was colder. (All four sign conventions consistently applied: z up, w = \dot z, stable stratification.)
2. The check:

using Plots

let
H, nz = 60.0, 241
dz = H / (nz - 1)
z  = range(-H, 0; length = nz)
T  = collect(23 .+ 5 .* tanh.((z .+ 30) ./ 5))   # the §9.4 profile
w0 = 1e-4                                        # exaggerated, m/s
dt = 0.4 * dz / abs(w0)                          # advective CFL
T2 = similar(T)
for n in 1:round(Int, 2 * 86400 / dt)
    for i in 2:nz-1
        # upwind: for w > 0 information comes from below (i-1)
        T2[i] = T[i] - dt * w0 * (T[i] - T[i-1]) / dz
    end
    T2[1] = T[1]                              # bottom value held
    T2[end] = T2[end-1]                       # crude open top
    T, T2 = T2, T
end
plt = plot(23 .+ 5 .* tanh.((z .+ 30) ./ 5), z, label = "initial", lw = 2,
           xlabel = "T (°C)", ylabel = "z (m)")
plot!(plt, T, z, label = "after 2 days, w > 0", lw = 2)
end

After two days at 10^{-4} m/s the profile has translated upward by \approx 17 m — the thermocline now sits near -13 m, and every interior depth reads colder. Flip w0 = -1e-4 (and the upwind direction: information now comes from above, T[i+1] - T[i]) and the profile translates downward, warming the interior. If your production code ever disagrees with this two-minute experiment, the production code is wrong — keep the script as a regression test for the capstone solver.

← Back to Unit 9 §9.3

Solution 9.4 — what does the bottom BC actually cost?

1. Dirichlet: steady state exists — the linear profile of Solution 10.1, with the deep reservoir silently supplying the 200 W/m² that the surface extracts (heat conducts up the gradient from the fixed-temperature bottom). Zero-flux: no steady state. Integrate the PDE over the column: the total heat content obeys \frac{d}{dt}\int T\,dz = \text{(flux in at top)} + \text{(flux in at bottom)} = -Q_{\text{cool}}/(\rho_0 c_p) + 0 < 0, so the column cools forever, linearly in time (~0.04 K/day for H = 100 m) — an insulating floor has no heat to give. The “steady” profile shape is approached, but it translates downward in temperature indefinitely.
2. The bottom BC’s influence reaches \sqrt{\kappa_b \times 30\,\text{d}} = \sqrt{10^{-5} \times 2.59\times10^6} \approx 5 m up from the floor over the run (interior diffusivity is what matters down there). Of Site B’s sensors at 2, 10, 25, 45, 58 m depth — i.e. 58, 50, 35, 15, 2 m above the bottom — only the 58 m sensor sits inside that 5 m zone.
3. So: low risk, one caveat. The recovered \hat\tau(t) leans on the mid-column sensors where the two BCs are indistinguishable over 30 days; only the deepest sensor could smell the difference, and (sensibly) the capstone treats the deep reservoir as the known part of the problem. The honest sentence for your write-up: “results are insensitive to the bottom-BC choice except within ~5 m of the floor, which affects one of fifteen sensors.”

← Back to Unit 9 §9.4

Solution 9.5 — plot every forcing before you trust it

using Plots

let
ρcp = 1025 * 3990
Qsw(t) = 800 * max(0, cos(2π * t / 86400))         # t in seconds
S(z)   = (800 / ρcp) * (1 / 10) * exp(z / 10)      # noon, ζ = 10 m  → K/s
w_site(z, H) = 1e-5 * sin(π * (z + H) / H)
dTdz(z) = (10 / (2 * 5)) * sech((z + 30) / 5)^2    # §9.4 IC gradient
Ri(z)   = 2e-4 * 9.81 * dTdz(z) / 0.01^2
κ1(z)   = 1e-3
κ2(z)   = 1e-5 + (1e-3 - 1e-5) * exp(z / 20)
κ3(z)   = 1e-5 + 1e-3 / (1 + 5 * Ri(z))^2

t = range(0, 3 * 86400; length = 1000)
z = range(-60, 0; length = 400)
p1 = plot(t ./ 86400, Qsw.(t), xlabel = "day", ylabel = "Q_SW (W/m²)",
          legend = false)
p2 = plot(ρcp .* S.(z), z, xlabel = "ρcₚS (W/m³)", ylabel = "z (m)",
          legend = false)
p3 = plot(w_site.(z, 60) .* 86400, z, xlabel = "w (m/day)", ylabel = "z",
          legend = false)
p4 = plot(κ1.(z), z, xscale = :log10, label = "constant",
          xlabel = "κ (m²/s)", ylabel = "z", legend = :bottomright)
plot!(p4, κ2.(z), z, label = "profile")
plot!(p4, κ3.(z), z, label = "Ri-dependent")
plot(p1, p2, p3, p4, layout = (2, 2), size = (900, 700))
end

The three read-off numbers:

• 10% source depth: e^{z/\zeta} = 0.1 at z = -\zeta\ln 10 \approx -23 m.
• w at Site B’s thermocline: with H = 60, z = -30 is the exact mid-column, so \sin(\pi(z+H)/H) = \sin(\pi/2) = 1 — the sine profile peaks there: w = w_0, ~1 m/day. This is the surprise worth noticing: the prescribed upwelling is strongest precisely at Site B’s thermocline, which is part of why Davies Reef gives the cleanest storm fingerprint of the three sites.
• Closure 2, surface vs -60 m: \kappa(0) \approx 10^{-3} vs \kappa(-60) = 10^{-5} + 9.9\times10^{-4}e^{-3} \approx 5.9\times10^{-5} — a factor of ~17 (not the ~100 of the Ri closure; the exponential profile never reaches its background floor in 60 m, another small surprise the plot surfaces).

← Back to Unit 9 §9.5

Solution 9.6 — the daily heat budget, by hand

1. The positive half of a cosine integrates to \int = Q_{\text{SW}}^{\max}\,\tau_d/\pi = 800 \times 86400/\pi \approx 2.20\times10^{7}\ \text{J/m}^2 per day.
2. Cooling: 200 \times 86400 = 1.73\times10^{7}\ \text{J/m}^2.
3. Net: +4.72\times10^{6}\ \text{J/m}^2/day. Spread over a 20 m mixed layer: \Delta T = 4.72\times10^6/(1025 \times 3990 \times 20) \approx 0.058 K/day — about 1.7 K over 30 days if every joule stayed in the mixed layer.

Reconciling with “the bulk profile barely changes”: 1.7 K is not nothing against the 10 K surface-deep contrast — but toy-task 2’s claim survives because the net heat does not all stay in the mixed layer: the penetrating fraction below h_m heats the interior imperceptibly (spread over tens of metres), and in the fuller scenarios upwelling and thermocline leakage export part of the rest. The back-of-envelope correctly predicts the sign and scale of the slow background warming visible in the scenario-2 heatmap. And the cloud threshold: the budget crosses zero when Q_{\text{SW}}^{\max} = \pi\,Q_{\text{cool}} \approx 628 W/m², a ~21% reduction — so the capstone’s storm (×0.5 cloud factor) flips the column decisively into net cooling. Both storm knobs — upwelling and radiation — push temperatures the same way, which is exactly why disentangling them needs the partition machinery rather than a glance at SST.

← Back to Unit 9 §9.6

Unit 10 — Project Solution

Solution 10.1 — predict scenario 1’s steady state exactly

1. At steady state with no advection and no source, \partial_z(\kappa\,\partial_z T) = 0 with constant \kappa means \partial_z T is constant — a straight line. The surface Neumann condition fixes the slope: \partial_z T = Q_{\text{np}}/(\rho_0 c_p \kappa); the bottom Dirichlet condition fixes the intercept: T(z) = T_{\text{deep}} + \frac{Q_{\text{np}}}{\rho_0\,c_p\,\kappa}\,(z + H). With Q_{\text{np}} = -200 W/m²: slope = -200/(1025 \times 3990 \times 10^{-3}) \approx -0.0489 K/m, so T(0) = 18 - 0.0489 \times 100 \approx \mathbf{13.1\,°C} — nearly 5 K colder than the deep reservoir. Why that’s right and not a sign error: the surface continuously loses 200 W/m², and at steady state that heat must be resupplied by upward diffusion from the fixed-temperature bottom. Diffusion only moves heat down a gradient, so the column must tilt cold-side-up. (Physically this is a winter-cooling profile; the full model’s daytime shortwave is what keeps real reef columns warm on top.)
2. T_\kappa = H^2/\kappa = 10^4/10^{-3}\,\text{s} \approx 116 days. A 365-day run is ~3 diffusive time constants — the transient decays like e^{-t/T_\kappa} to within a few percent, hence the caption’s “near-steady state after ~365 days”. Running the FD solver for only 30 days (the capstone window) leaves the column far from this steady state — which is fine, because the capstone never claims otherwise.
3. scenario_1() should land on your line to \approx 4\times10^{-3} °C, per the §10.1 caption — if you get agreement at the surface but a kink near the bottom, your ghost cell for the Neumann BC is at the wrong end.

← Back to Unit 10 §10.1

Solution 10.2 — make the DeepXDE column feel the sun

The diff, conceptually three lines — and each hides a trap:

import deepxde as dde
from deepxde.backend import torch as bkd   # backend tensor ops

tau_d  = 86400.0
Q_max  = 800.0
rho_cp = 1025.0 * 3990.0
zeta   = 10.0

def Q_sw(t):
    # half-cosine in backend ops — np.maximum would break the graph
    return Q_max * bkd.clamp(bkd.cos(2 * 3.141592653589793 * t / tau_d),
                             min=0.0)

def pde(zt, T):
    z, t = zt[:, 0:1], zt[:, 1:2]
    dT_dt   = dde.grad.jacobian(T, zt, j=1)
    dT_dz   = dde.grad.jacobian(T, zt, j=0)
    d2T_dz2 = dde.grad.hessian(T, zt, component=0, i=0, j=0)
    S = Q_sw(t) / rho_cp / zeta * bkd.exp(z / zeta)   # body source, K/s
    return dT_dt - kappa_z(z) * d2T_dz2 - S
    # Q_np stays at -200 W/m² in the (unchanged) surface OperatorBC

Trap (a), as predicted: the original S_z used np.exp on what was effectively a constant — once the source depends on t, every op inside pde must be a backend tensor op (torch.clamp, torch.cos, torch.exp via dde.backend), or autograd silently detaches and the residual trains against a frozen source. (kappa_z gets the same treatment if it wasn’t already tensor-safe.)

Trap (b): the domain is 2.6\times10^6 s long and the forcing period is 8.6\times10^4 s — thirty oscillations across the time axis. The plain 4×32 tanh MLP, fed raw t, fits the slow trend and flattens most of the diurnal wiggle (compare the trained surface trace T(0, t) over days 5–8 against an FD reference, or just FFT it: the 1 cpd line is far too weak). This is spectral bias at capstone scale, and the fix is the Task B Fourier-feature bank with a \sim 1/\tau_d frequency component — DeepXDE supports it via a feature transform: net.apply_feature_transform(...). Rescaling t to [0, 1] helps optimisation but does not fix the frequency-content problem.

Porting to NeuralPDE.jl: the source edit is a one-line change in the symbolic equation — eq = Dₜ(T(z,t)) ~ κ(z)*Dzz(T(z,t)) + Q_sw(t)/ρcp/ζ*exp(z/ζ) with Q_sw(t) = Q_max * max(0, cos(2π*t/τ_d)) registered symbolically — and the symbolic layer differentiates it for you; there is no trap (a) at all, because there’s no backend-tensor boundary to fall over. Trap (b) is ecosystem-independent: spectral bias doesn’t care which language you suffer in. Fair scorecard: the edit is easier in NeuralPDE.jl; the training pathology is identical; the Fourier-feature remedy is a few lines in either.

← Back to Unit 10 §10.4