Welcome back. So, continue our discussion on analysis of dynamics of infectious diseases. The model that we had in front of us was a third order nonlinear autonomous dynamical system, where the dynamical variable was S, I, R corresponding to the population of susceptible, infectives and removed.
The three equations are in front of you. And with the help of these three equations and various assumptions that we made a mention of in the previous lecture, we were trying to answer several questions. Like if we know the parameters of the system, can we tell whether the infection would spread or not.
We came across the term reproduction number in the previous lecture, which I said that would give an idea that when the reproduction number R0 was greater than 1, then the infection would in fact spread. But now, we will also like to know the exact details of the dynamics and what would be the maximum number of infectives that one can expect during the pandemic and so on. And if these predictions can be made during the early stages of your infection, so that one can in principle, be prepared to face that the pandemic using various preventive measures.
So, we had various questions in front of us. Let us try to answer some of them using this dynamical analysis. So, we have d S by d t is equal to minus r S I, we have d I by d t is equal to r S I minus aI and we have d R by d t is equal to a I.
Let me assign these equations as equation 1, equation 2 and equation 3. So, let us try to get explicit expressions for S, I and R. So, from equation 1, I can write that since d S by d t is equal to minus r S I, I can write I is equal to 1 upon r minus 1 upon S d S by d t.
So, now this can be written as I is equal to minus 1 over r d by d t of ln S. So, therefore, if I have a mechanism of determination of S, then I will simply substitute, I will take the log of it, take the time derivative multiply it with minus 1 over r, I will get the number of infectives. Similarly, I have d R by d t is equal to aI and I comes from equation number 4.
So, I can write this as dR by d t is equal to minus a upon r d by d t of ln S, from where it is easy to see that R is simply minus a upon r ln S. So, how would you determine the number of removed persons from the population? If you can determine S, then you take the log of it multiply it with minus a upon r you will get the number of removed persons.
Now, the whole analysis basically is to determine the value of S or the expression for S, rather the time dependence of S. So, what I can now do is make use of a fact that S plus R plus I, number of susceptible, number of removed persons and number of infectives is equal to N, which is the total population at any instant of time. How do I know this?
This can be established very easily. I will add equations 1, 2 and 3. So, what is going to happen, d S by d t plus d I by d t plus d R by d t from equations 1, 2 and 3 you have minus r S I plus r S I 0 minus a I plus a I 0.
So, this is equal to 0 which means d by d t of S plus I plus R is equal to 0. In other words, S plus I plus R is a constant, constant c. And how do I determine this constant c, at t is equal to 0 S is S 0, I is I 0 and R 0 is simply 0.
So, therefore, the initial population consists of the number of infectives which you have introduced and the rest population becomes the susceptible population. So, therefore, infective I 0, susceptible S 0 in the beginning is the total population. So, therefore, I can write S plus I plus R is equal to N, the total population of your system.
So, if this is the case, then what can be done about my value functional form of S, I can write S plus I, what is I, from equation 4 minus 1 upon r d by d t of ln S plus r, r is minus a upon r ln S should be equal to capital N. So, if I assume let ln S be equal to some variable x, so I can write S is equal to e to the power x. So, let me call this equation 7, this as equation 6.
So, therefore, what I can do is I can use ln S is equal to x in equation number 7, so, I can write equation number 7 as e to the power x minus 1 upon r d x by d t minus a upon r x is equal to N. So, now I have one equation. So, I have here d x by d t, which is a function of x with parameters a, r, N.
So, what I will need to do is I will need to solve for x from this ordinary differential equation, first order, one variable with a, r, N and S parameters. And when I solve for x, I will then exponentiate it to get the population S. So, let me make it as equation 8.
So, I will solve for equation number 9. I will solve equation number 9 for x substituted in equation number 8 to get the value of S, which means, it would give me the time dependence of variation of susceptibles in the population. Then what I will do is I will then use that function S of t in equation number 4 to give me the time variation of infectives in my population and equation number 5 to give me the time variation of removed population from my system.
So, in principle, I can follow this entire method and get the time evolution of S, I and R in my system. This in principle is possible, but the mathematical form that we would get will not be very convenient. So, I encourage you to solve equation 9.
It is a nonlinear equation. So, it is a little tricky to solve it. But, if you manage to get an analytical expression, then also you will see that getting an explicit expression of x in terms of t and subsequently S in terms of t is going to be a little difficult.
So, therefore, can we do, can we take an alternative approach to do this analysis. So, let us see if we can do this. So, I have d S by d t is equal to minus r S I d I by d t is equal to r S I minus aI and dR by d t is equal to a I.
So, from equations 1 and 2, what I can do is I can write d I by d S as r S I minus a I divided by minus r S I. From where, I get d I by d t is equal to minus 1 plus a upon r multiplied by 1 upon S. So, what do I get this from here?
So, there is one correction it should be d I by dS, so, d I by d S. So, what does this equation tell me, this equation would tell me the variation in the number of infectives as a function of the variation in the number of susceptibles, which means, that for my third order system for which the phase portraits would be in three dimensions, I am now considering the S I projection of my system. So, let us see how I can understand the S I projection.
I can do this that I, rather d I is equal to minus 1 plus a upon r, 1 upon S times d S from where I can write I is equal to minus S plus a upon ln S plus a constant, integration constant c. I can determine the integration constant c from the initial conditions when the number of susceptibles in the population is S 0 at t is equal to 0. The corresponding number of infectives was I 0.
So, therefore, I can write I 0 is equal to minus of S 0 plus a upon r, ln S 0 plus c from where I can write c is equal to I 0 plus S 0 plus a upon r ln S 0. But what is I 0 plus S 0? You introduced I 0 number of infectives in the population and rest entire population became susceptible.
So, therefore, I 0 plus S 0 is the total population of your system. So, therefore, I can write this as c is equal to N the total population plus a upon r ln S 0. So equation 4, equation 5.
So, from 5 and 4, I can write I is equal to minus S plus a upon r ln S plus N plus a upon r ln S 0, equation 6. So, what do you see here? What do you see here is that the population of the infectives changes as the population of the susceptibles and the relationship is given like this, I is equal to minus S plus a upon r ln S plus N plus a upon r ln S 0.
So, now, I have a parameter here, a upon r. And let me call a upon r as c 1. I have another parameter here S 0, initial population, let me call this S 0 as another parameter c 2.
I have third parameter, N which is the total population and I am denoting N simply by N. So, I have I is equal to minus S plus c 1 ln S plus N plus c 1 ln c 2. This is what I want to get the nature of.
So, let me see if I can plot it using my online plotter. My equation is this, f of x is equal to minus x plus c 1 ln x plus N plus c 1 ln c 2. So, let me get rid of the text here.
Now, what did we learn from our analysis N, capital N is the population, let me consider a normalized population between 0 and 1. The population will always be positive, but I am considering a normalized population which means the population varies from 0 and 1. So, if I have to introduce I 0 it would be a fraction.
So, 0. 1 means 10 percent of the initial population is the infected persons. So, that is the meaning.
So, let me make it from 0 to 1. Then c 1 corresponded to the ratio a over R. a is positive, R is positive.
So, therefore, let me make this quantity say from 0 to 5, and c 2 is the initial population. So, c 2 is the initial population of the susceptible the range has to be between 0 and 1 because the maximum population is 1. So, the range will be between 0 and 1.
And therefore, now I will tweak around this to get the phase lines. So, let us see what happens. I changed the number of susceptibles.
And what I get is, I change this ratio, I change this ratio, I change this number of susceptible and this is what happens, I am going on changing here. So, let us qualitatively see the nature. I am going on changing this, these are the phase lines, which I am getting.
And similarly, I will change the initial population, these are the phase lands which I am getting. So, I got a general nature of variation of I with S, but is there any constraint in my system which I must take care off? Well, there is.
So, if S 0 is the initial susceptible population, I 0 is the initial infective population then I know that S 0 plus I 0 is equal to N, the total population. But as the infections grow the susceptible number of susceptibles change and some of them become recovered, S plus I should become less than N. Why?
Because some of them go to R. S plus I plus R is equal to N. Therefore, S plus I should be less than N because R is again, because it is a population, a positive quantity.
So, therefore, I must look at these solutions which are there I have changed c 1 you see here, I have changed c 2 you see here, I must look at only those solutions for which first I and S positive, negatives should not be considered because we are considering populations, but then I must consider the case where S plus I must be less than N. So, I must now here write g of x should be N minus x. So, now, I have a triangle so, let me zoom it in.
I have a triangle. So, this is the triangle. And any solution which lies within this triangle only is a feasible population.
So, for the current case, for example, you see this curve. So, therefore, all of these points on the curve are the feasible populations at any given instant of time. So, now, if I need to look at this particular solution, then, let me do one thing, let me try to draw different phase lines conforming to this constraint.
So, first of all what I will do is I will see I will fix a value for the parameter a upon r. So, let me fix it at 1. So, for c 2 is equal to 0.
73 and c 1 is equal to 1, which means a by r is equal to 1, you do not have any solution. Do I get any solution is something which I need to see. So, let me change the population and I do see a population which is here.
So therefore, it is, it, for c is equal a by r is equal to 1, which is just 1, ratio is 1 and S 0 is also 1. If you remember the reproduction ratio, this will result in reproduction ratio 1. So, we just have this point here.
So now, what I will do is I will, this basically means that I have the entire population which is susceptible, but for this particular point here, the point is 1, 0 which means that there is no infectives present in my system. Everyone is susceptible, but there is no infective. How would the infective grow if there is no initial number of infective.
Therefore, this particular observation is consistent with our physical intuition that if you do not reduce introduce any infected members in the population, the infection is not going to rise. So, what do I need to do? Let me do one thing, let me make the a by r ratio 1.
2, greater than 1, and let me reduce this number of infectives. Again, I do not see any solution. Physically, make sense.
And therefore, now, if I conform to this situation here, c 1 is a by r ratio, c 2, very large number of infective in a, small number of infective, large number of susceptible population, I see some solutions appearing here. What is the importance of several points here? c 2, which you can see here, c 2 is equal to S 0, initial number of susceptible population.
So, which means, I 0 will become n minus S 0 which is 1 minus 0. 846. And if you go here, you will see that you have 0.
86, if you see here, you have some number of initial population and then the population changes with time. So, now, what you have to see is that how does the pandemic evolve with time. So, when I set c 1 as 0.
44 and c 2 as 0. 4, 0. 846, you are evolving in time.
So, let me draw this. You are evolving in time and you are coming here. So, what is happening to I?
So, I have the axis I here, I have the axis S here. And remember, I am drawing I versus S. So, therefore, what happens is that I am going up, so my number of infectives go up.
And let me now tweak around with these parameters. I go up, you see here, let me tweak around with this parameter further, and so on. And I can tweak around with this parameter like this.
So, what is the general nature of the plot which you see? The general nature of the plot which you see here can be drawn here, let me draw that. So, the general nature is this.
This is I, this is S. I need to confine myself within this line. This is equal to, this is the line which is S plus I is equal to N.
This is the line which corresponds to S plus I is equal to N. So, any solution to my system, which is within this triangle is the feasible population. And what are the phase lines that we drew?
We saw the phase lines which looked like this. So, suppose this is one phase line as an example. So, what is going on?
I am starting with some initial population and therefore, my I as I go up, you will see here, I is increasing. I can have what is it dependent upon, its dependent upon S 0, it is dependent upon a upon r, which ultimately can be clubbed to the reproduction number R 0. So, therefore, if S 0, a and r during the beginning of the pandemic itself, you are in a good position to tell whether you are going.
So, if you look at this particular plot, can you say that the pandemic is going to happen? Yes, because you see that, you in fact, see an increase in the population of I. So, during the back calculation of a and r and S 0 in the beginning, you can simply say that your pandemic is going to happen or epidemic is going to happen and the number of infections are going to increase.
So, now, what happens further as you change these parameters, you see these kinds of plots and there would come a situation like this. So, what happens in this situation? You are only going down.
In the first situation, the infective increased, went till this point, and then they started decreasing. Do we see this these days for COVID-19 infection? Yes, every time we have a wave, number of infectives go on increasing, reach a peak, that is why we say that we have reached a peak, so that corresponds to this point.
So, once you reach the peak, then your infectives start coming down. But, will this always happen? In this last case, for example, you start with I and you always come down.
So, therefore, for such case, the severity of pandemic would not, the pandemic would not occur at all, because the number of infectives will always come down. What about the severity of pandemic? This is also one may comment upon.
So, for that, what we can do is we can redraw this plot. I have I, I have S. So, therefore, you would go through maxima and this can be one of the solutions.
To make it a little more elegant, this would be one of the solutions, for example. So, when you start with this at any point, you see here, you compare this situation versus this situation. Here, you have reached a maxima, where you have larger number of infectives I 1 versus I 2, I 2 less than I 1 and therefore, what you see is that although pandemic or epidemics is there in both the cases, the number of infectives increased with time in case of I 2, there did not occur much increase, or the peak value was smaller in case of I 1, the peak value was larger.
So, what we saw here today is that, you can in fact make certain predictions about the pandemic or epidemic right during the beginning of the spread itself. The parameters which would be associated for making these predictions would involve the infection rate, the removal rate as well as the initial susceptible population. Depending upon the combination of these three parameters, which would result into the parameter called reproduction number, what you will see is that you will sample different phase lines on the I-S projection of your phase portrait.
So, therefore, when you have this combination of a, r and S 0, such that your phase trajectory goes up along the I axis, you say that the pandemic has occurred. But then by looking at this itself, that why peaks during wave are observed, because once you reach the maximum, the number of infectives again start coming down. In certain cases for combination of S 0, a and r, you can simply have the decline in the value of I right from the beginning, in which case you do not observe a pandemic or epidemic at all.
We will further analyze the effect of these parameters in the next lecture. Till then, goodbye.
