Determine the behavior of τ(t)=α(t)-β(t)>0 at large t in Malthus model (10), in order the keep t

Determine the behavior of τ(t)=α(t)-β(t)>0 at large t in...

Key Concepts

Malthusian Growth Model

The Malthusian growth model describes population dynamics through a simple differential equation where the rate of change of the population is proportional to the current population size. It provides a basic framework for understanding how populations can grow or decay exponentially over time depending on the net growth rate.

Net Growth Rate

The net growth rate, defined as the difference between the birth rate and death rate, is central to the model. In this context, it is represented by ?(t) = ?(t) - ?(t). Its sign and magnitude directly influence whether the population will increase, decrease, or potentially remain constant, and its time dependence plays a crucial role in long-term behavior.

Asymptotic Behavior

Analyzing the asymptotic behavior involves exploring how functions or solutions to differential equations behave as time approaches infinity. In population models, ensuring that the integrated effect of the net growth rate does not diverge is necessary for keeping the population finite.

Convergence of Integrals

For population size to remain finite over an infinite time horizon, the integral of the net growth rate must converge. This means that ?(t) must decay sufficiently fast as t grows large, such that the accumulated effect on population size does not result in unbounded growth.

Transcript

00:03 Okay, for part a, we're asked to show that 0 is less than y is 0, is less than a.

00:09 If we're given this, and why does not blow up as at any time tb.

00:16 All right, so it's like y is 0 equal y not.

00:21 And from our general solution, we have that y not is equal to a over 1 minus 1 over c.

00:28 So based on this, if we solve for c, we get c is equal to y0 over y not minus a.

00:35 Now if why not is less than a and c is less than 0, and we know that our denominator of our general solution is always positive.

01:00 And then let's see, then we have y is equal to a over 1 minus b kt over c.

01:24 So we know that this is always positive.

01:30 And we know that, what was it that? why not is less than a and c is less than 0? why not is less than a? this is a negative.

01:56 Okay, so it just turns into a positive, right? yeah, so if we have c is less, oh, not c is less than 0, why not is less than a.

02:05 So why not is smaller than a, so why not minus a is negative.

02:10 So we have that c is less than zero.

02:13 Based on this.

02:16 So if you plug in t here, you get that, this negative sign turns into a positive.

02:22 So this portion over here is always positive.

02:25 So based on this, whenever 0 is less than y is 0, is less than a, rt, or i mean ry does not blow up at any time.

02:39 Okay, now for part, we're asked to show that if y of 0 is greater than a, then y blows up at a time t, t, b, which is negative and hence does not correspond to a real time okay so we have 1 minus e k t over c is equal to 0 when c is equal to e negative k t correct right so if you plug in e negative k t we get 1 minus e k t sorry negative k t over e negative k t we get that equal to 1 minus 1 which is 0 okay so we have y of t is equal to a over 1 minus e negative k t over e negative k t so a over a sterile okay so if c is equal to e negative k t and we solve for t we get that it's c let's take all in on both sides of c is equal to e sorry negative k t is equal to rate of 1 over k lm of c and since we know that c is equal to y not over y not minus a and that y not is greater than a we can say that c is greater than a based on this assumption so our c value ln of c must also be greater than zero so based on this y blows up at a time which is negative so negative 1 over k ln of c this is the time which is the time where when it blows up...

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