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 precalculus - logarithms integration basics of differential equations exponential growth and decay

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Single Variable Calculus

 Absolute Convergence Alternating Series Arc Length Area Under Curves Chain Rule Concavity Conics Conics in Polar Form Conditional Convergence Continuity & Discontinuities Convolution, Laplace Transforms Cosine/Sine Integration Critical Points Cylinder-Shell Method - Volume Integrals Definite Integrals Derivatives Differentials Direct Comparison Test Divergence (nth-Term) Test
 Ellipses (Rectangular Conics) Epsilon-Delta Limit Definition Exponential Derivatives Exponential Growth/Decay Finite Limits First Derivative First Derivative Test Formal Limit Definition Fourier Series Geometric Series Graphing Higher Order Derivatives Hyperbolas (Rectangular Conics) Hyperbolic Derivatives
 Implicit Differentiation Improper Integrals Indeterminate Forms Infinite Limits Infinite Series Infinite Series Table Infinite Series Study Techniques Infinite Series, Choosing a Test Infinite Series Exam Preparation Infinite Series Exam A Inflection Points Initial Value Problems, Laplace Transforms Integral Test Integrals Integration by Partial Fractions Integration By Parts Integration By Substitution Intermediate Value Theorem Interval of Convergence Inverse Function Derivatives Inverse Hyperbolic Derivatives Inverse Trig Derivatives
 Laplace Transforms L'Hôpital's Rule Limit Comparison Test Limits Linear Motion Logarithm Derivatives Logarithmic Differentiation Moments, Center of Mass Mean Value Theorem Normal Lines One-Sided Limits Optimization
 p-Series Parabolas (Rectangular Conics) Parabolas (Polar Conics) Parametric Equations Parametric Curves Parametric Surfaces Pinching Theorem Polar Coordinates Plane Regions, Describing Power Rule Power Series Product Rule
 Quotient Rule Radius of Convergence Ratio Test Related Rates Related Rates Areas Related Rates Distances Related Rates Volumes Remainder & Error Bounds Root Test Secant/Tangent Integration Second Derivative Second Derivative Test Shifting Theorems Sine/Cosine Integration Slope and Tangent Lines Square Wave Surface Area
 Tangent/Secant Integration Taylor/Maclaurin Series Telescoping Series Trig Derivatives Trig Integration Trig Limits Trig Substitution Unit Step Function Unit Impulse Function Volume Integrals Washer-Disc Method - Volume Integrals Work

Multi-Variable Calculus

 Acceleration Vector Arc Length (Vector Functions) Arc Length Function Arc Length Parameter Conservative Vector Fields Cross Product Curl Curvature Cylindrical Coordinates
 Directional Derivatives Divergence (Vector Fields) Divergence Theorem Dot Product Double Integrals - Area & Volume Double Integrals - Polar Coordinates Double Integrals - Rectangular Gradients Green's Theorem
 Lagrange Multipliers Line Integrals Partial Derivatives Partial Integrals Path Integrals Potential Functions Principal Unit Normal Vector
 Spherical Coordinates Stokes' Theorem Surface Integrals Tangent Planes Triple Integrals - Cylindrical Triple Integrals - Rectangular Triple Integrals - Spherical
 Unit Tangent Vector Unit Vectors Vector Fields Vectors Vector Functions Vector Functions Equations

Differential Equations

 Boundary Value Problems Bernoulli Equation Cauchy-Euler Equation Chebyshev's Equation Chemical Concentration Classify Differential Equations Differential Equations Euler's Method Exact Equations Existence and Uniqueness Exponential Growth/Decay
 First Order, Linear Fluids, Mixing Fourier Series Inhomogeneous ODE's Integrating Factors, Exact Integrating Factors, Linear Laplace Transforms, Solve Initial Value Problems Linear, First Order Linear, Second Order Linear Systems
 Partial Differential Equations Polynomial Coefficients Population Dynamics Projectile Motion Reduction of Order Resonance
 Second Order, Linear Separation of Variables Slope Fields Stability Substitution Undetermined Coefficients Variation of Parameters Vibration Wronskian

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17calculus > differential equations > population dynamics

This page builds on the discussion of exponential growth and decay and specifically applies it to the dynamics of population change, i.e. population growth and decline. In fact, we will touch on basic exponential growth and decay and then extend the discussion to more realistic models. Here are the four topics covered on this page. The initial condition for each equation is $$p(0)=p_0$$.

exponential growth

$$p' = rp$$

$$p(t) = p_0e^{rt}$$

exponential growth with critical threshold

$$\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)p }$$

$$\displaystyle{ p(t) = \frac{p_0T}{p_0+(T-p_0)e^{rt}} }$$

logistic growth

$$\displaystyle{ \frac{dp}{dt} = r \left( 1-\frac{p}{L} \right)p }$$

$$\displaystyle{ p(t) = \frac{p_0L}{p_0+(L-p_0)e^{-rt}} }$$

logistic growth with critical threshold

$$\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)\left( 1-\frac{p}{L} \right)p }$$

$$\displaystyle{ \frac{(L/p-1)^T}{(T/p-1)^L} = \frac{(L/p_0-1)^T)}{(T/p_0-1)^L}e^{-(L-T)rt)} }$$

Exponential Growth

plot 1 - exponential growth

$$p(t) = p_0e^{rt}$$

As discussed on the exponential growth and decay page, we start with the differential equation $$p' = rp$$, which solves to $$p(t) = p_0e^{rt}$$. In this equation, $$p(0) = p_0$$ is the initial population and $$r$$ is the growth rate. The graph of $$p(t)$$ with $$p_0 = 1$$ for various values of $$r > 0$$ is shown in plot 1.

There are two problems with this model when we are trying predict population growth.
1. Notice that as $$t$$ increases, $$p(t)$$ increases without bound. Although this model for population growth may be accurate over short time periods, it is not realistic as space, food supply and other resources would limit the growth in any reasonable situation.
2. Some populations cannot sustain itself if there are too few members. For example, our model shows that even when starting out with only one member, the population increases. Again, this is not realistic.

In order to handle the limitations of resources, we will use another model called logistic growth but first, let's look at how to handle the second issue, starting out with enough members to sustain a population.

autonomous equation

A special type of differential equation of the form $$y' = f(y)$$ where the independent variable does not explicitly appear in the equation.

Exponential Growth with Critical Threshold

plot 2 - exponential growth with threshold

$$\displaystyle{ p(t) = \frac{p_0T}{p_0+(T-p_0)e^{rt}} }$$

In order to handle the situation when a population needs more than one member to sustain itself or it will die out, we use another autonomous differential equation

 $$\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)p }$$

where $$r$$ is the growth rate and $$T$$ is called the critical threshold (or just threshold).

The solution to this differential equation is $$\displaystyle{ p(t) = \frac{p_0T}{p_0+(T-p_0)e^{rt}} }$$. (See the practice problems for the derivation of this equation.) Let's discuss what is going on with the equation. We have plotted $$p(t)$$ for various values of $$p(0)=p_0$$ in plot 2.

When $$p_0=T$$, i.e. the initial population is exactly the threshold value $$T$$, the coefficient of the exponential term is zero, so $$p(t)=T$$ and the population stays exactly at the threshold value for all time $$t>0$$. However, if the initial population is either above or below the threshold value, the exponential term $$e^{rt}$$ drives the plots away from the threshold. This seems logical since, if the initial population is higher than the threshold value, the population will increase and if the initial population is below the threshold value the population will decrease, eventually dying out. In the second case, there are not enough individuals to sustain the population.

Okay, so far, so good. We have dealt with the second problem of the basic exponential growth model. Now let's address the first problem dealing with limited resources.

Logistic Growth

plot 3 - logistic growth

$$\displaystyle{ p(t) = \frac{p_0L}{p_0+(L-p_0)e^{-rt}} }$$

In the real world, populations are often limited by things like resources and space. To handle this situation, we model the population with the autonomous differential equation

 $$\displaystyle{ \frac{dp}{dt} = r \left( 1-\frac{p}{T} \right)p }$$

where $$r$$ is the growth rate and $$L$$ is called the saturation level.

If you look carefully at this equation and compare it to the equation for critical threshold above, you will see that they are almost identical except for a minus sign. Even though they are similar, the behaviour of this equation is quite different from the previous one. If you followed the derivation of the threshold equation (in a practice problem), you can derive the solution here as $$\displaystyle{ p(t) = \frac{p_0L}{p_0+(L-p_0)e^{-rt}} }$$. Notice that the exponential here is $$e^{-rt}$$.

As we did before, plot 3 shows the solution for various values of $$p(0)=p_0$$. Notice that the exponential in this equation drives the solution toward the saturation level (as compared to away from the threshold in the previous section).

Logistic Growth with Critical Threshold

plot 4 - logistic growth with threshold $$p(t)$$

In the last two sections we have addressed both problems with the basic exponential individually. Let's combine the two solutions into one equation. Our differential equation is

 $$\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)\left( 1-\frac{p}{L} \right)p }$$

where $$r$$ is the growth rate, $$T$$ is the threshold and $$L$$ is the saturation level. In this discussion, we will assume that $$0 < T < L$$, i.e. the saturation level (limit on resources) is higher than the threshold.

The plot of $$p(t)$$ for various initial conditions is shown in plot 4. Notice how the plot tends to diverge away from the threshold but converge towards the saturation limit as we expected. The equation of the solution, in implicit form, is $$\displaystyle{ \frac{(L/p-1)^T}{(T/p-1)^L} = \frac{(L/p_0-1)^T)}{(T/p_0-1)^L}e^{-(L-T)rt)} }$$ (see the practice problems for a derivation of this equation)

Practice

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on.

GOT IT. THANKS!

Instructions - Solve the following differential equations using the initial conditions, if given.

Solve $$p' = rp$$, $$p(0)=p_0$$

Problem Statement

Solve $$p' = rp$$, $$p(0)=p_0$$

$$p(t) = p_0 e^{rt}$$

Problem Statement

Solve $$p' = rp$$, $$p(0)=p_0$$

Solution

$$\begin{array}{rcl} p' &=& rp \\ \displaystyle{ \frac{dp}{dt} } &=& rp \\ \displaystyle{ \frac{dp}{p} } &=& r~dt \\ \displaystyle{ \int{ \frac{dp}{p} } } &=& \displaystyle{ \int{ r~dt } } \\ \ln |p| &=& rt+C \\ e^{\ln |p|} &=& e^{rt+C} \\ p &=& ke^{rt} \\ \end{array}$$
general solution $$p=ke^{rt}$$
apply the initial condition $$p(0)=p_0$$
$$p_0=ke^0 \to k=p_0$$
particular solution $$p(t)=p_0e^{rt}$$

$$p(t) = p_0 e^{rt}$$

Solve $$\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)p }$$, $$p(0)=p_0$$

Problem Statement

Solve $$\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)p }$$, $$p(0)=p_0$$

$$\displaystyle{ p(t) = \frac{Tp_0}{p_0+(T-p_0)e^{rt} } }$$

Problem Statement

Solve $$\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)p }$$, $$p(0)=p_0$$

Solution

This differential equation is separable.
$$\displaystyle{ \frac{dp}{(1-p/T)p} = -r~dt }$$
On the left side of the equation, we need to use partial fractions.

 $$\displaystyle{ \frac{1}{(1-p/T)p} = \frac{A}{1-p/T} + \frac{B}{p} }$$ $$1 = Ap+B(1-p/T)$$ $$1 = Ap+B-Bp/T$$ $$1 = p(A-B/T) + B$$

Remember that the variable in the above partial fraction expansion is $$p$$. Now we equate coefficients to find the constants A and B.
$$B=1$$
$$A-B/T=0 \to A=B/T \to A=1/T$$
So now our integrals are

 $$\displaystyle{ \int{\frac{1/T}{1-p/T} + \frac{1}{p} dp} = \int{-r~dt} }$$ $$\displaystyle{ \int{\frac{1}{T-p} + \frac{1}{p} dp} } = -rt+C$$ $$\displaystyle{ \ln \left| \frac{p}{T-p} \right| } = -rt+C$$ $$-\ln |T-p| + \ln |p| = -rt+C$$ $$\displaystyle{ \frac{p}{T-p} = e^{-rt+C} }$$

Before we apply the initial condition to find C, let's solve the last equation for $$p$$.
$$\begin{array}{rcl} \displaystyle{ \frac{p}{T-p} } &=& e^{-rt+C} \\ \displaystyle{ \frac{1}{T/p-1} } &=& \\ \displaystyle{ \frac{T}{p} - 1 } &=& e^{rt-C} \\ \displaystyle{ \frac{T}{p} } &=& 1+e^{rt-C} \\ p &=& \displaystyle{ \frac{T}{1+ke^{rt}} } \end{array}$$
In the last line, we set $$k=e^{-C}$$ just for convenience. Now we will apply the initial condition $$p(0)=p_0$$.
$$\begin{array}{rcl} p_0 &=& \displaystyle{ \frac{T}{1+ke^0} } \\ 1+k &=& \displaystyle{ \frac{T}{p_0} } \\ k &=& \displaystyle{ \frac{T}{p_0}-1 } \\ k &=& \displaystyle{ \frac{T-p_0}{p_0} } \end{array}$$
Our particular solution is
$$\displaystyle{ p(t) = \frac{T}{1+e^{rt}(T-p_0)/p_0} }$$
$$\displaystyle{ p(t) = \frac{Tp_0}{p_0+(T-p_0)e^{rt} } }$$

$$\displaystyle{ p(t) = \frac{Tp_0}{p_0+(T-p_0)e^{rt} } }$$

Solve $$\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)\left( 1-\frac{p}{L} \right)p }$$, $$p(0)=p_0$$

Problem Statement

Solve $$\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)\left( 1-\frac{p}{L} \right)p }$$, $$p(0)=p_0$$

$$\displaystyle{ \frac{(L/p-1)^T}{(T/p-1)^L} = \frac{(L/p_0-1)^T)}{(T/p_0-1)^L}e^{-(L-T)rt)} }$$

Problem Statement

Solve $$\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)\left( 1-\frac{p}{L} \right)p }$$, $$p(0)=p_0$$

Solution

The differential equation is separable.
$$\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)\left( 1-\frac{p}{L} \right)p }$$
$$\displaystyle{ \frac{dp}{(1-p/T)(1-p/L)p} = -r~dt }$$
Using partial fraction expansion, we have
$$\displaystyle{ \int{ \frac{L}{T(L-T)}\frac{1}{1-p/T} + \frac{T}{L(T-L)}\frac{1}{1-p/L} + \frac{1}{p}dp } = \int{ -r~dt } }$$
$$\displaystyle{ \frac{-L}{L-T}\ln |T-p| - \frac{T}{T-L} \ln|L-p| + \ln|p| = -rt+C }$$
Multiply both sides of the equation by $$L-T$$ to get
$$\displaystyle{ -\ln|T-p|^L + \ln|L-p|^T + \ln|p|^{L-T} = (L-T)(-rt+C) }$$
Combining the natural log terms and doing some algebra, we get
$$\displaystyle{ \ln \left|\frac{p^{L-T}(L-p)^T}{(T-p)^L} \right| = (L-T)(-rt+C) }$$
$$\displaystyle{ \frac{p^{L-T}(L-p)^T}{(T-p)^L} = exp( (L-T)(-rt+C) ) }$$
$$\displaystyle{ \frac{(L/p-1)^T}{(T/p-1)^L} = k e^{-(L-T)rt} }$$

Now we apply the initial condition, $$p(0)=p_0$$ to get
$$\displaystyle{ k=\frac{(L/p_0-1)^T}{(T/p_0-1)^L} }$$
$$\displaystyle{ \frac{(L/p-1)^T}{(T/p-1)^L} = \frac{(L/p_0-1)^T)}{(T/p_0-1)^L}e^{-(L-T)rt)} }$$

We can leave this equation in implicit form, i.e. not explicitly solve for $$p(t)$$.

$$\displaystyle{ \frac{(L/p-1)^T}{(T/p-1)^L} = \frac{(L/p_0-1)^T)}{(T/p_0-1)^L}e^{-(L-T)rt)} }$$