Models & Qualitative Behavior

Study Sheet

Models & Qualitative Behavior

Growth, cooling, mixing, and equilibria

The Classic Models

Tip
Logistic growth

dPdt=rP(1−PK)\tfrac{dP}{dt} = rP\left(1 - \tfrac{P}{K}\right): growth is proportional to BOTH the population and the remaining room. The rate is a downward parabola in PP, fastest exactly at half capacity P=K/2P = K/2, and every positive solution flows to the carrying capacity KK.

Tip
Newton cooling & mixing tanks

Cooling: the EXCESS over ambient decays exponentially — if the excess halves in 10 minutes it halves again in the next 10. Mixing: dSdt=(rate in)−SV(flow out)\tfrac{dS}{dt} = (\text{rate in}) - \tfrac{S}{V}(\text{flow out}); the equilibrium is where inflow balances outflow, always equal to incoming concentration times tank volume.

Example
Euler’s method, honestly

One step of size hh: yn+1=yn+h f(yn)y_{n+1} = y_n + h\,f(y_n) — follow the tangent line briefly, then re-aim. On y′=yy' = y from y(0)=4y(0)=4 with h=0.5h = 0.5: estimate 66 vs. the true 4e0.5≈6.594e^{0.5} \approx 6.59; Euler undershoots convex solutions, and halving hh halves the error.

Side note
Significance: the same equation, twelve costumes

Newton cooling = drug elimination = RC discharge = mixing = depreciation: all u′=−kuu' = -ku about the right variable. Recognizing the ARCHETYPE is the modeling skill — once you see "rate proportional to amount (or excess)," you already know the solution, the half-life, and the long-run behavior.

Try it
Try it: build then solve

Coffee at 90∘90^\circ in a 20∘20^\circ room cools to 70∘70^\circ in 5 minutes. When does it hit 45∘45^\circ? Work: excess halves from 7070 to 5050... compute: u=70ektu = 70e^{kt} with u(5)=50u(5) = 50: k=15ln⁡57k = \tfrac15\ln\tfrac57. Want u=25u = 25: t=ln⁡(25/70)k≈−1.0296−0.0673≈15.3t = \tfrac{\ln(25/70)}{k} \approx \tfrac{-1.0296}{-0.0673} \approx 15.3 minutes. Always solve in the EXCESS variable.

Proofs & Why It Matters

Tip
Proof: solving the logistic equation

Separate dPP(1−P/K)=r dt\tfrac{dP}{P(1 - P/K)} = r\,dt and split by partial fractions: 1P(1−P/K)=1P+1/K1−P/K\tfrac{1}{P(1-P/K)} = \tfrac1P + \tfrac{1/K}{1 - P/K}. Integrating gives ln⁡PK−P=rt+C\ln\tfrac{P}{K-P} = rt + C, and solving for PP: P(t)=K1+Ae−rtP(t) = \dfrac{K}{1 + Ae^{-rt}} — an S-curve rising to KK. Every positive start flows to the carrying capacity because e−rt→0e^{-rt} \to 0. ■\blacksquare

Tip
Proof: Newton cooling is exponential in the excess

The law says T′=−k(T−Tamb)T' = -k(T - T_{\text{amb}}). Let u=T−Tambu = T - T_{\text{amb}} (the excess); since TambT_{\text{amb}} is constant, u′=T′=−kuu' = T' = -ku — the pure exponential equation, so u=u0e−ktu = u_0e^{-kt}.

Halving in equal time intervals is exactly the property of exponentials, which is why the "excess halves every 10 minutes" phrasing works. ■\blacksquare

Tip
Proof sketch: Euler's error halves with h

One Euler step commits error ≈h22y′′\approx \tfrac{h^2}{2}y'' (the first Taylor term the tangent line misses). Reaching a fixed time TT takes T/hT/h steps, so the accumulated error is on the order of Th⋅h22max⁡∣y′′∣=O(h)\tfrac{T}{h}\cdot\tfrac{h^2}{2}\max|y''| = O(h) — proportional to hh. Halve the step, halve the error: Euler is a first-order method. ■\blacksquare

Going Deeper: Explanations & Worked Problems

Concept
Equilibria and stability: the phase-line shortcut

For an autonomous equation y′=f(y)y' = f(y), enormous insight costs one sketch. Find the zeros of ff — the EQUILIBRIA, where the system sits still. Between consecutive zeros, ff keeps one sign: yy moves right (up) where f>0f > 0, left (down) where f<0f < 0.

Mark arrows on a number line and stability is visible: arrows converging on an equilibrium make it STABLE (perturb, return), diverging make it UNSTABLE (perturb, flee). Logistic growth y′=ry(1−y/K)y' = ry(1 - y/K): zeros at 00 and KK; f>0f > 0 between them, so arrows point from 00 toward KK — the origin is unstable, the carrying capacity stable, and every positive solution S-curves to KK, all WITHOUT solving the equation. The phase line is the qualitative theory in miniature: solve only when you need numbers; sketch when you need understanding.

Example
Worked: half-life to dosage, a complete exponential-decay problem

A drug has half-life 66 hours; a dose raises blood concentration by 1010 units. When does a single dose drop below 11 unit?

Step 1 — model: y′=kyy' = ky with y=10ekty = 10e^{kt}.

Step 2 — pin kk from the half-life: y(6)=5y(6) = 5 gives e6k=12e^{6k} = \tfrac12, so k=−ln⁡26≈−0.1155k = -\tfrac{\ln2}{6} \approx -0.1155.

Step 3 — solve 10ekt=110e^{kt} = 1: ekt=110e^{kt} = \tfrac1{10}, so t=ln⁡10ln⁡2⋅6=6log⁡210≈19.9t = \dfrac{\ln 10}{\ln 2}\cdot6 = 6\log_2 10 \approx 19.9 hours.

Step 4 — sanity check in halvings: 10→5→2.5→1.2510 \to 5 \to 2.5 \to 1.25 after three half-lives (1818 h), and 1.25→11.25 \to 1 needs about a third more of one half-life — 18+2≈2018 + 2 \approx 20 h ✓.

Step 5 — the modeling lesson: the answer depended only on the RATIO 10:110:1, so every decade of decay costs the same log⁡210≈3.32\log_2 10 \approx 3.32 half-lives, whatever the starting dose.

Example
Worked: a mixing tank, from setup to equilibrium

A 100100 L tank starts with pure water. Brine at 22 g/L flows in at 55 L/min; the well-mixed tank drains at 55 L/min. Find the salt S(t)S(t).

Step 1 — build the equation from bookkeeping: rate in =2⋅5=10= 2 \cdot 5 = 10 g/min; rate out == (concentration)(flow) =S100⋅5=S20= \tfrac{S}{100}\cdot5 = \tfrac{S}{20}; so S′=10−S20S' = 10 - \tfrac{S}{20}.

Step 2 — equilibrium first: S′=0S' = 0 at S=200S = 200 g — which is just (incoming concentration)(volume) =2⋅100= 2\cdot100, as it must be.

Step 3 — solve: linear with μ=et/20\mu = e^{t/20}: (et/20S)′=10et/20(e^{t/20}S)' = 10e^{t/20}, so et/20S=200et/20+Ce^{t/20}S = 200e^{t/20} + C, and S(0)=0S(0) = 0 gives C=−200C = -200: S(t)=200(1−e−t/20)S(t) = 200\left(1 - e^{-t/20}\right).

Step 4 — read the time scale: the "time constant" is 2020 min; after 2020 min the tank is at 200(1−e−1)≈126200(1 - e^{-1}) \approx 126 g, after an hour ≈190\approx 190 g — the exponential approach to equilibrium that governs tanks, cooling coffee, and charging capacitors alike.