Solve and interpret continuous logistic models

Use the separation and model foundations

Senior 6A establishes model formulation and verification; 6B supplies separation and connected solution domains. The earlier affine population recurrence is discrete and linear. It supplies modelling habits, but it is not the nonlinear continuous logistic equation

Define the parameter convention and units

Use P′=rP(1-P/K), with r>0 and K>0. P and K share population units; r has inverse-time units. The per-capita net rate is P′/P=r(1-P/K) for P>0. For populations much smaller than K, this is approximately r and the total rate approximately rP. This approximation is not exact for a positive population

Do not confuse two common coefficient conventions

If the equation is written P′=kP(K-P), its intrinsic rate is r=kK. The coefficient k has inverse population-times-time units. In expanded form P′=αP-βP² with α,β>0, r=α and K=α/β. State the convention before interpreting a number

For P′=P/2-P²/240, find K.

Separate total growth from growth per individual

For P′=(1/2)P(1-P/120), at P=30 the total rate is (1/2)(30)(3/4)=45/4 population units/day. The per-capita rate is 3/8 per day. At P=150 the rate is negative; the ideal model predicts decrease toward the carrying capacity

For P′=(1/2)P(1-P/120), give the per-capita rate at P=30.

Find and interpret both equilibria

Substitute P′=0: P=0 and P=K are constant solutions. For 0<P<K the rate is positive; above K it is negative. The zero equilibrium repels small positive populations; K attracts positive populations from both sides. These continuous solutions do not oscillate across K. A large-step numerical approximation may behave differently

What does the rate graph show?

Derive the non-equilibrium solution by partial fractions

For P≠0,K, separation gives dP/[P(1-P/K)]=r dt. Since 1/[P(1-P/K)]=1/P+1/(K-P), integration gives ln|P|-ln|K-P|=rt+C. Rearrangement yields P=K/(1+A e^(-rt)). Check P=0 and P=K separately because division excluded them

Apply a positive initial value

At t=0, P0=K/(1+A), so A=K/P0-1 for P0>0. The case P0=K gives A=0 and recovers that equilibrium. The case P0=0 remains the separate zero solution. For any positive P0, A>-1 and the denominator stays positive for every t≥0.

Work the logistic initial value

For r=1/2,K=120,P0=30, A=3 and P(t)=120/(1+3e^(-t/2)). At t=2ln2, the exponential is 1/2 and P=48. As t→∞ it tends to 120. It remains below 120 for every finite nonnegative time

For P(t)=120/(1+A e^(-t/2)) and P(0)=30, give A.

For P(t)=120/(1+3e^(-t/2)), give P at t=2ln2.

Find maximum total growth

Complete the square: P′=-(r/K)(P-K/2)²+rK/4. Thus the greatest positive rate occurs at P=K/2 and equals rK/4. For r=1/2,K=120, these are P=60 and rate 15. This differs from per-capita growth, which decreases as P increases

For P′=(1/2)P(1-P/120), give the maximum total rate

Relate the rate vertex to a trajectory inflection

Differentiating the rate along a solution gives P″=rP′(1-2P/K). For a nonconstant trajectory below K, P′>0: it is concave up below K/2 and concave down above K/2. A start below K/2 has a future inflection there. A start between K/2 and K has already passed that level in its mathematical past; a start above K decreases toward K and has no future crossing of K/2.

Which statement fits these three initial values?

Solve a reachable threshold before taking logarithms

For 0<P0<P*<K, rearrange the solution to e^(-rt)=(K-P*)/(A P*), so t=(1/r)ln[A P*/(K-P*)]. The ratio exceeds 1 and the time is positive. For P0=30,K=120,r=1/2 and target 90, the ratio is 9 and t=2ln9. Reaching K itself would require infinite time

For A=3, K=120 and target P*=90, give the argument A P*/(K-P*)

Infer parameters from instantaneous per-capita rates

Under the ideal model, q=P′/P=r-(r/K)P is a line in P. At populations 30 and 60, suppose exact instantaneous per-capita rates are 3/8 and 1/4 per day. Its slope is -1/240 and intercept 1/2, giving K=120. Finite-period percentage growth is not automatically an instantaneous derivative

In q=r-(r/K)P, the exact observations are (P,q)=(30,3/8) and (60,1/4). Give r.

Verify a proposed logistic expression

For candidate P=L/(1+A e^(-λt)), differentiate to obtain P′=λP-(λ/L)P². Compare that polynomial with rP-(r/K)P² on the candidate range, then check the initial value. A wrong A can preserve the rate equation while failing the initial condition; a wrong L or λ can fail the equation itself

For candidate 120/(1+4e^(-t/2)), give its initial-value error relative to P0=30.

Constant harvesting changes the equilibrium equation

For P′=rP(1-P/K)-H with H≥0, compare H with rK/4. Below that maximum there are two nonnegative equilibria: the lower is a survival threshold and the upper attracts nearby populations. At equality there is one double equilibrium, attracting only from above; above the maximum there is no equilibrium. Stop or change the physical model at extinction

Work the harvesting threshold

With r=1/2,K=120,H=45/4, factor the rate as -(P-30)(P-90)/240. The equilibria are 30 and 90. Rates are negative below 30, positive between them and negative above 90. A start below 30 declines; a start above 30 tends toward 90. Starting exactly at 30 remains there in the ideal equation

For P′=(1/2)P(1-P/120)-45/4, give the attracting equilibrium

Above-capacity starts have a different backward domain

For r=1/2,K=120,P0=240, A=-1/2 and P=120/(1-(1/2)e^(-t/2)). Its denominator vanishes at t=2ln(1/2)=-2ln2. The maximal connected interval through 0 is t>-2ln2. For physical time t≥0 it stays above 120 and decreases toward 120; its backward pole does not invalidate the forward model

For P(t)=120/(1-(1/2)e^(-t/2)), give B in the pole time t=2ln B.

Carry model assumptions into the interpretation

The logistic curve assumes a fixed resource setting and a linearly decreasing net per-capita growth rate. Continuous population is an approximation; real data can show changing capacity, migration, seasonality or delay. A fitted carrying capacity is a parameter of the model, not a guaranteed biological ceiling under every condition

Derive the family and classify initial values, ungraded

Derive the logistic solution for r,K>0 using separation. Recover both equilibria and classify every P0≥0 for forward time. State the maximal interval for P0>K and explain why crossing K is impossible for a classical solution

Model: preserve equilibria, signs and connected intervals

For P≠0,K integrate 1/P+1/(K-P) to obtain ln|P/(K-P)|=rt+C. Rearranging gives P=K/(1+A e^(-rt)). For P0>0,A=K/P0-1. If 0<P0<K,A>0, the denominator is greater than 1 and the solution increases toward K on all real times. P0=K is constant; P0=0 is the separate zero solution. If P0>K,-1<A<0, the pole is t=(1/r)ln(-A)<0 and the branch through 0 lies to its right; the solution decreases toward K for t≥0. The smooth polynomial right-hand side gives local uniqueness, preventing distinct finite solutions from crossing the equilibrium. Direct differentiation verifies the formula on each valid branch

Recover and assess a fitted model, ungraded

Use instantaneous observations P=30,P′=45/4 and P=60,P′=15 to infer r,K. Give the resulting model and explain why these two observations alone do not establish its validity for all time

Model: normalise the observations before fitting

For the instantaneous observations (P,P′)=(30,45/4) and (60,15), the per-capita values are (45/4)/30=3/8 and 15/60=1/4. The slope of q versus P is (1/4-3/8)/(60-30)=-1/240. Thus r=3/8+30/240=1/2 and K=-(1/2)/(-1/240)=120. The model is P′=(1/2)P(1-P/120). Check units and both supplied rates, then compare independent observations, residuals and resource conditions. Noise or finite-period differences require an estimation method and uncertainty; treating them as exact derivatives would add an assumption

Find threshold and inflection times, ungraded

For P′=(1/2)P(1-P/120),P(0)=30, find the time to reach 90 and the inflection time. Explain whether the population can reach 120 at finite time and whether a start at 100 has a future inflection

Model: check the target range and second derivative

For P′=(1/2)P(1-P/120) and P(0)=30, find the time to reach 90 and the inflection time. The solution is 120/(1+3e^(-t/2)). Setting P=90 gives e^(-t/2)=1/9 and t=2ln9. Setting P=60 gives e^(-t/2)=1/3 and t=2ln3. Since P″=(1/2)P′(1-P/60) and P′>0 below 120, its sign changes from positive to negative at 60. The denominator stays above 1 at every finite t, so 120 is only a limit. A start at 100 remains between100 and 120, hence has no future crossing of 60 or future inflection

Analyse a harvested population, ungraded

For r=1/2,K=120 compare harvesting H=45/4,H=15 and H=18. Find equilibria and sign patterns, explain the critical case, and state the physical stopping condition. Do not reuse the unharvested closed form

Model: compare harvesting with the rate maximum

For r=1/2 and K=120, the unharvested maximum is 15 at P=60. For H=45/4 the rate is -(P-30)(P-90)/240: 30 repels from both sides and 90 attracts. For H=15 the rate is -(P-60)²/240, nonpositive everywhere: 60 is constant, approached from above but left behind by starts below 60. For H=18 the rate is at most -3, so there is no equilibrium and every finite positive start reaches zero within at most P0/3 days if the ideal harvesting rule continues. A negative population has no physical meaning; stop or modify harvesting at extinction. These are qualitative classifications, not a claim that the unharvested solution formula still solves the changed equation

Check a model using technology, ungraded

For the start 240 solution with r=1/2,K=120, construct exact reference values at 0,2ln2 and4ln2. Describe reproducible checks of a numerical solution and identify which conclusions remain assumptions about the real population

Model: use exact reference points and separate validation

For P′=(1/2)P(1-P/120) and P(0)=240, the values at t=0, 2ln2 and 4ln2 are 240, 160 and 960/7. Check P(0)=240, evaluate the derivative residual and confirm monotone decrease above 120 over a stated interval. Record the numerical method, step size or tolerances and plotting axes; repeat with finer settings and compare with the exact reference values. Section 6D supplies explicit Euler steps. Numerical agreement tests implementation, while independent data and resource information test the fixed-r,fixed-K assumptions. Neither a smooth plot nor a fitted curve proves the model remains valid indefinitely

Assessment boundary

Practice grades bounded parameters, rates, initial-value constants,explicit time forms,verification residuals and equilibrium classifications. Full separation proofs, range arguments, model calibration and independent technology execution remain ungraded. The companion quadratic-equilibrium family is frozen 6A reuse, not a new logistic bank