[MUSIC] In today's video, we discuss the logistic function used to model population dynamics where an inhibition factor is included in the expression describing the rate of growth. The logistic model includes a ceiling on the size of the population. The resulting curve is increasing, has a sigmoid shape, and sits between two horizontal asymptotes. There's a point of inflection halfway between, about which the curve has 180-degree rotational symmetry. We begin with a simpler exponential growth model. Suppose we have a population of size x(t), which is a function of time t and want to understand its behavior. This model assumes that the derivative of x is some multiple k of x, with k is a positive constant. One can imagine without any limit of resources that the more there is of the colony, the faster it should grow. So the growth rate should be proportional to the size of the population. This is an example of differential equation. Note that though x appears on the right-hand side, x is being used as a dependent variable. The independent variable is t, not x, and anti-differentiating the right-hand side with respect to x will not make any progress. It needs some careful preparation to unravel this equation. You first multiply both sides by the differential to t, and dividing through by x gives dx on x = k dt. We now apply integral symbols to both sides to get equality of antiderivatives of the respective integrands. The differential on the left-hand side involves x, whilst the differential on the right-hand side involves t. An antiderivative 1 on x with respect to x is ln of x, and an antiderivative of a constant k with respect to t is just kt. We amalgamate constants of integration together with a single + C on the right. We can raise Euler's number e to the power of each side. On the left we get x back again, and on the right we get e to the kt + c, which we can expand. Thus x = A times e to the kt for some constant A by putting A = e to the c. Notice that we've reproduced the familiar exponential growth function whose graph is the exponential curve. Note that x labels the vertical axis and t the horizontal axis. Exponential growth is unlimited. The x values get arbitrarily large as t increases. Of course, this isn't a realistic model when resources are limited. We now modify the previous model to take into account constraints placed on populations by the environment and limited resources. We adjust the right hand side of the equation dx, dt equals kx by introducing a factor 1 minus x on M, for some new constant capital M. Think of M for the largest, or maximal sustainable population. The factor kx remains untouched, and contributes towards the growth of the population. The factor 1 minus x on M is carefully designed to model inhibition. If x is small, then this factor is close to 1, and the left-hand side is almost exactly kx, as in the original exponential model. However, if x is large, getting close to M, the factor 1- x on M gets close to 0, causing the overall growth rate, dx dt, to slow down to almost nothing. This modification is called the logistic growth model widely used in population dynamics. We get what's called a logistic equation whose solutions are called logistic functions. Let's now try to unravel the logistic equation. We first multiply through both sides through the differential, dt. Keep the constant, k, on the right-hand, next to dt, but move the factors x, and 1 minus x on M to the other side, by division. We separate the variables by putting x's to the left and t's to the right, including their differentials. We assume implicitly throughout that the population x always lies between 0 and M. We add integral signs to both sides to create functions by anti-differentiation. On the right hand side a general antiderivative to the constant k with respect to t is kt plus C. The left-hand side, however, is a mystery. Here's the left-hand side again, rewritten to express the integrand as a rational function. To motivate the next step, we first play with some ordinary fractions involving numbers. For example, one-sixth is 1 divided by 2 times 3. Which a half minus a third, which we can write as 1 over 2 plus negative 1 over 3. Similarly, one-thirty-third = 1 over 3 x 11. And you can check quickly this becomes two lots of one-third + negative 7 lots of one-eleventh. Both one-sixth and one-thirty-third decompose into combinations involving simpler fractions, a half, a third, and one-eleventh. It's like recognizing that water is H2O or common salt is sodium chloride. Here's a tougher fraction, one-ninety-ninth = 1 over 9 x 11. How can we decompose that? Let's postulate that it becomes A lots of the 9th, plus B lots of 11th. We put the right-hand side together as a single fraction, where the numerator becomes 11A plus 9B. We want this to equal 1. And it helps to find multiples of 11 and 9 that differ by 1. Say 44 which is 4 times 11, and 45 which is 5 times 9. By using negative 44 and positive 45, we get a total of 1. Thus, we should take A to be negative 4 and B to be 5. And then one ninety ninth decomposes as -4 to the ninth plus 5 lots of one eleventh. Let's try something similar with the integrand above, which is a fraction involving polynomials in x, rather than just numbers. The numerator is 1. The denominator is complicated, with factors x and 1-x on M. Let's postulate a decomposition using A lots of 1 on x and B lots of 1 over 1 minus x on M. Rewrite the right hand side with a common denominator, obtaining a numerator that has to become 1 to agree with the numerator on the left-hand side. If we take A equals 1, then we get 1 minus x on M, and just need to choose B to make the minus x on M go away, and B equal to 1 on M does the trick. Thus the rational function decomposes as this sum. And the right-hand side can be written more simply as 1 on x + 1 over M- x. What we've done in an ad hoc way is a special case of the method of partial fractions, a topic you'll meet in more advanced courses in calculus. To perform this integration, it suffices to integrate 1 on x and 1 on M-x separately, and add the answers together. The population x lies between 0 and M, so that both x and M-x are positive. An antiderivative of 1 on x is ln of x and of 1 over M-x is negative ln of M- x. For the second one you can use the guess and check method, or integration by substitution. We combine the constants of integration together with a plus C dash at the right. The difference of logarithms is the logarithm of the quotient, and we rewrite the expression inside the logarithm to isolate x, in preparation for the next step. Our aim, remember, is to unravel the logistic equation and we produced this difficult integral which we've just managed to solve. We can put everything together and amalgamate the constants on the right using, again, just a single plus C. We want explicit information about x and it's buried in the expression on the left hand side. We first raise Euler's number e to the power of each side of the equation. So the ln is stripped away from the left hand side, and the right hand side becomes e to the kt plus c, which can be expanded. We reciprocate everything and rewrite the right-hand side. We add 1 to both sides, and now rewrite the right hand side as 1 plus capital K e to the negative little kt, with capital K is the positive constant, e to the negative C. We rearrange this to obtain an explicit form of x, finally solving the logistic equation and its solution is known as the logistic function. In writing down the logistic equation, there were two positive constants involved, little k and capital M. In developing the general solution of the logistic function, we introduced a new positive constant capital K. This function has several important features. Firstly, the population x always lies between 0 and M where M represents a limiting upper bound or maximum. Indeed, the limit of x of t as t gets arbitrary large and positive to M because the expression e to the negative kt, which is one of e to the kt, goes to 0. You can think of M as the largest sustainable population in an environment with only limited resources. On the other hand, the limit of x(t), as t gets arbitrarily large and negative, will be 0, because now, the expression e to the negative kt will become arbitrary large and positive, and it appears in the denominator. The variable t represents time, so the first limit represents limiting behavior in the future. The second limit represents limiting behavior going backwards in time to the distance past. In the original differential equation we can rewrite the right hand side as k on M times x times M-x, which is a quadratic in x. Let's graph this quadratic regarding dx, dt as a function of x. You can see that it must be an upside down parabola which passes through the x axis when x equals 0 and when x equals M, and has an apex when x is exactly halfway between, that is when x equals M on 2. At this apex, the derivative dx/dt is maximised. This tells us that the growth rate is maximized when x happens to be halfway between 0 and M. We can now sketch the graph of x as a function of t. From features 2 and 3, we can see that x equals M and the the t axis form two horizontal asymptotes, and the curve has to fit in between. From feature 4, the growth rate is maximized when x has M on 2 and the corresponding point on the curve will be a point of inflection. Moving down to the t axis will reveal the time on which the growth rate is maximized. You can see a direct connection with the usual exponential function. If you look at the rule for x of t when the denominator is large, the +1 becomes negligible. By removing the +1, you get a simplified expression, which becomes a constant times e to the kt, which is the usual exponential function. This approximation effect occurs towards the left of the graph at time t before the curve gets near the point of inflection. The model indicates that the population is growing approximately exponentially until factors kick in to halt the growth. The population is growing as fast as possible at the moment, it reaches M on 2, and after that slows down and continues to slow down as it gets closer and closer to the maximum carrying capacity of M. [MUSIC]
