The SIR Disease Model Trajectories and MatLab

Transcription

1 The SIR Disease Model Trajectories and MatLab James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University November 17, 2013 Outline Reviewing the SIR disease model The I vs. S curve Matlab!

2 Abstract This lecture discusses the SIR disease model trajectories and using MatLab to plot them. We will now build a simple model of an infectious disease. Assume the total population we are studying is fixed at N individuals. This population is then divided into three separate pieces: we have individuals that are susceptible to becoming infected are called Susceptible and are labeled by the variable S. Hence, S(t) is the number that are capable of becoming infected at time t. that can infect others. They are called Infectious and the number that are infectious at time t is given by I (t). that have been removed from the general population. These are called Removed and their number at time t is labeled by R(t).

3 Our complete Infectious Disease model is then I = r S I γ I S = r S I I (0) = I 0 S(0) =. where we can compute R(t) as N I (t) S(t). The Phase Plane in Quadrant I: S = 0 I axis S axis I = 0 S = 0 (, ) (, +) (S 1, I 1) S 1 S 0 (, I 0) S = γ/r I = 0 A plausible trajectory starting at the point > γ/r and I 0 > 0. Another trajectory starting at S 1 < γ/r and I 0 > 0 is also shown. In addition, the intersections with the S axis are labeled S0 and S1, respectively. Figure: The Disease Model in Quadrant One

4 We know that biologically reasonable solutions occur with initial conditions starting in Quadrant I and we know that our solutions satisfy S < 0 always with both S and I positive until we hit the S axis. Let the time where we hit the S axis be given by t. For any t < t, we can divide to obtain Thus, I (t) S (t) r S(t) I (t) γ I (t) = r S(t) I (t) = 1 + γ 1 r S(t). I (t) S (t) = 1 + γ r 1 S I (t) = S (t) + γ r S (t) S(t) Integrating, we find ) I (t) I 0 = (S(t) + γ ( ) S(t) r ln. simplify I (t) = I 0 + S(t) + γ ( ) S(t) r ln. Dropping the dependence on time t, we see the functional dependence of I on S. I = I 0 + S + γ r ln ( S ). It is clear that this curve has a maximum at γ/r. This value is very important in infectious disease modeling and we call it the infectious to susceptible rate ρ.

5 Now let s answer the question of whether or not the trajectory can hit the origin. If that happened, the terminal value would be S = 0 and I = 0. The I vs S equation would then say I = I 0 + S + γ ( ) S r ln or 0 = I γ r ln ( 0 ). Since ln(0) is undefined, this equation is impossible and so the trajectory can not hit the origin. Definition For the disease model I = r S I γ I, S = r S I I (0) = I 0, S(0) = the dependence of I on S is given by ( ) S I = I 0 + S + ρ ln. We say the infection becomes an epidemic if the initial value of susceptibles, exceeds the critical infectious to susceptible ratio ρ = γ r because the number of infections increases to its maximum before it begins to drop. This behavior is interpreted as an infection going out of control; i.e. it has entered an epidemic phase.

6 Homework 78 For the following disease models 1. Do the nullcline analysis for the first quadrant. 2. Explain why the trajectories must stay in Quadrant 1 if they start there. This means you find the trajectories on the positive I and positive S axis as part of answering the question. Draw a nice picture of this. 3. Derive the di ds equation and solve it. Explain why a trajectory can not hit the origin. 4. Answer the questions about whether or not the specific values given cause an epidemic. Draw the corresponding trajectory for each case. Homework S (t) = 150 S(t) I (t) I (t) = 150 S(t) I (t) 50 I (t) S(0) =, I (0) = 12.7 Is there an epidemic if is 1.8? (yes as γ/r = 1/3) Is there an epidemic if is 0.2? (no) S (t) = 5 S(t) I (t) I (t) = 5 S(t) I (t) 25 I (t) S(0) =, I (0) = 120 Is there an epidemic if is 4.9? ( no as γ/r = 5) Is there an epidemic if is 10.3? (yes )

7 Homework S (t) = 15 S(t) I (t) I (t) = 15 S(t) I (t) 40 I (t) S(0) =, I (0) = 12.7 Is there an epidemic if is 6? Is there an epidemic if is 2? S (t) = 18 S(t) I (t) I (t) = 18 S(t) I (t) 200 I (t) S(0) =, I (0) = 120 Is there an epidemic if is 20? Is there an epidemic if is 4? Here is a typical session to plot an SIR disease model trajectory for S = 5SI, I = 5SI 25I, S(0) = 10, I (0) = 5. T = 2. 0 ; h = ; N = c e i l (T/h ) ; x0 = [ 1 0 ; 5 ] ; [ ht, r k ] = FixedRK ( f, 0, x0, h, 4,N) ; X = r k ( 1, : ) ; Y = r k ( 2, : ) ; xmin = min (X) ; xmax = max (X) ; xtop = max ( abs ( xmin ), abs ( xmax ) ) ; ymin = min (Y) ; ymax = max (Y) ; ytop = max ( abs ( ymin ), abs ( ymax ) ) ; D = max ( xtop, y t o p ) x = l i n s p a c e (0,D,1 0 1 ) ; GoverR = 2 5 / 5 ; c l f h o l d on p l o t ( [ GoverR GoverR ], [ 0 D] ) ; p l o t (X, Y, k ) ; x l a b e l ( S a x i s ) ; y l a b e l ( I a x i s ) ; t i t l e ( Phase Plane for Disease Model S = 5 S I, I = 5 S I + 25 I, S (0) = 10, I ( 0 ) = 5 : E p i d e m i c! ) ; l e g e n d ( S = 2 5 / 5, S v s I, L o c a t i o n, Best ) ;

8 The resulting graph is shown here. Homework 79 For the following disease models, do the single plot corresponding to the an initial condition that gives an epidemic and also draw a phase plane plot using AutoPhasePlanePlot For the specific model 79.2 For the specific model S (t) = 150 S(t) I (t) I (t) = 150 S(t) I (t) 50 I (t) S(0) = S0, I (0) = I0 S (t) = 5 S(t) I (t) I (t) = 5 S(t) I (t) 25 I (t) S(0) = S0, I (0) = I0