# Math Help - Diff. EQ. Help - model for disease

1. ## Diff. EQ. Help - model for disease

The following problem is an SIS disease problem:

Calling: I(t) = number of infectives at time t
N = the total population (assumed constant)
b = infection rate (here, a positive constant)
v = recovery rate (also, a positive constant)

a model for this disease is given bu the following:

dI/dt = bI(N-I) - vI

And since the population is assumed constant, we can just take S(t) to be N -I(t). Derive a condition for when the number of infectives goes to zero.

Is there anyone out there than can help me, even if it's just a little bit?

2. Hi. That's a Bernoulli equation which can be solved exactly but wouldn't the recovery rate have to be greater than the infection rate to drop the number of infected down to zero? Not sure. May need to just solve the equation with test values of the constants to see what's happening.

3. Originally Posted by shawsend
Hi. That's a Bernoulli equation which can be solved exactly but wouldn't the recovery rate have to be greater than the infection rate to drop the number of infected down to zero? Not sure. May need to just solve the equation with test values of the constants to see what's happening.
My teacher gave us a hint that said solving this equation is definitely not the best way, and for us to think of a way to model long-term equations.

By this I think he wants us to use the phase diagram.

So i found that I is zero when I*= 0 and I*= (bN-v)/B, but this is where I get stuck. I choose values less than 0, but that doesn't tell me if the function is increasing of decreasing.

Is there anything that I am missing?

4. "Differential Equations" by Blanchard, Devaney, and Hall is what I use. It goes over this really well. So you figured out the equilibrium points of zero and $N-v/b$. That's when the derivative is zero and since this is an autonumous equation, the value of I completely determines the value of the derivative so once I reaches one of the equilibrium points, then the infections stay at that value (although below it's negative which is not applicable in real life). So we have:

$\frac{dI}{dt}=bI(N-I)-vI$

and I want to study just one particular case:

$\frac{dI}{dt}=0.005 I(100-I)-0.55I$

That means of course the recovery rate is much higher than the infection rate. Suppose I then plot $y=0.005x(100-x)-0.55x$. That's the first plot and it represents the derivative as a function of the number of infections (I).. Note the derivative is positive between -10 and zero. That means if we start I anywhere in that range (not realistic in actual practice), the derivative is positive and I will increase until it reaches the nearest equilibrium point of zero. If I<-10, then the derivative is negative and the value of I will continue to decrease. And if I is greater than zero, then again the derivative is negative and so I will decrease again towards it's nearest equilibrium point. I've shown all these cases in the second plot which is I(t) with the red lines as the equilibrium points. Note that if under these conditions, the initial value of I>0, since it's derivative is negative, the value of I will continue to drop until it reaches the nearest equilibrium point which is zero (the number of infections is then zero and stays there right?). So what happens if I then use different values of b and v (just leave N=100)? How will this change the first plot and by the shape of that plot, since it's the value of the derivative, how then will the solutions behave for various starting values of I based on the value of the derivative there? Just try a few different values and see what happens

5. Originally Posted by shawsend
"Differential Equations" by Blanchard, Devaney, and Hall is what I use. It goes over this really well. So you figured out the equilibrium points of zero and $N-v/b$. That's when the derivative is zero and since this is an autonumous equation, the value of I completely determines the value of the derivative so once I reaches one of the equilibrium points, then the infections stay at that value (although below it's negative which is not applicable in real life). So we have:

$\frac{dI}{dt}=bI(N-I)-vI$

and I want to study just one particular case:

$\frac{dI}{dt}=0.005 I(100-I)-0.55I$

That means of course the recovery rate is much higher than the infection rate. Suppose I then plot $y=0.005x(100-x)-0.55x$. That's the first plot and it represents the derivative as a function of the number of infections (I).. Note the derivative is positive between -10 and zero. That means if we start I anywhere in that range (not realistic in actual practice), the derivative is positive and I will increase until it reaches the nearest equilibrium point of zero. If I<-10, then the derivative is negative and the value of I will continue to decrease. And if I is greater than zero, then again the derivative is negative and so I will decrease again towards it's nearest equilibrium point. I've shown all these cases in the second plot which is I(t) with the red lines as the equilibrium points. Note that if under these conditions, the initial value of I>0, since it's derivative is negative, the value of I will continue to drop until it reaches the nearest equilibrium point which is zero (the number of infections is then zero and stays there right?). So what happens if I then use different values of b and v (just leave N=100)? How will this change the first plot and by the shape of that plot, since it's the value of the derivative, how then will the solutions behave for various starting values of I based on the value of the derivative there? Just try a few different values and see what happens
can you please explain where you got .005, 100 and .55 from?

6. That's just any old numbers. Just a set of numbers to illustrate a particular case of the differential equation. Your mission, should you choose to accept it, is to try other values of the infection and recovery rate and substitute it into the equation $y=bI(100-I)-vI$, generate the parabola, and since that parabola represents the value of the derivative, use that information like I did above to explain the long-term behavior of the infection function I(t). Alright then, here's some to try:

$\begin{array}{cc} \text{b} & \text{v} \\
.1 & .2 \\
.2 & .1 \\
.01 & 0.01 \\
.2 & .2 \\
.79 & .8
\end{array}
$

So I plot $y=.1 I(100-I)-0.2 I$ then I say, ok, it's negative here. That means if I start the initial conditions with this value, the derivative is negative so the infections will continue to drop. Ok, here, it's positive. So if I start the initial conditions with this value, since the derivative is positive, the infections will increase. And just do that like I did above and from studying those examples, make some conclusion about the relationship between infection and recovery rate that is necessary for the infection to drop to zero based on the values of b and v. For example, if the recovery rate is ANY value above the infection rate like b=0.1 and v=0.10001 would that still result in the infections eventually dropping to zero? I don't know. It depends entirely on what the parabola $0.1 I(100-I)-0.10001 I$ tells you. So just try doing that for the first set of values for b and v and risk doing it wrong cus' look, I'll tell you a secret: sometimes the wrong ones are on the road to the right ones.

7. Originally Posted by tactical
The following problem is an SIS disease problem:

Calling: I(t) = number of infectives at time t
N = the total population (assumed constant)
b = infection rate (here, a positive constant)
v = recovery rate (also, a positive constant)

a model for this disease is given bu the following:

dI/dt = bI(N-I) - vIAnd since the population is assumed constant, we can just take S(t) to be N -I(t). Derive a condition for when the number of infectives goes to zero.

Is there anyone out there than can help me, even if it's just a little bit?
In order that I go to 0, it must decrease. And for that to be true, the derivative must be negative. Under what conditions is
$\frac{dI}{dt}= bI(N-I)- vI< 0$? That's the same as $I(bN- v- I)< 0$. The only way a product can be negative is if one factor is positive and the other is negative. Since I, the number of infected persons cannot be negative, we must have bN- v-I< 0. That is the same as I> bN- v. Of course, if I< bN- v it will start going up again. So a condition that I go to 0 is that bN- v be negative; that is, that v> bN.