How do I fit time series data to a logistic differential equ

How do I fit time series data to a logistic differential equation?

I have the following data and I am supposed to should fit my data to the logistic differential equation dP/dt = P(a?bP), P(0) = P0. using regression method brute force method and Discuss how good the fit is.

Unfortunately I am very confused on how to do this and really really need some help.

I know that if the model is a good fit to the data, the residuals should be normally distributed about the mean but I don\'t really know how to get there.

I do get to use python to do this problem as I am assuming there are a lot of tedious calculations that would have to be done by hand. If any one could please please help me that would be super duper greatly appreciated!!

TABLE 3 Time-Specific Hazard of Lung Cancer Death for White Males Given the Presence of a Tumor Time of Tumor Growth (in years) Hazard 4.03 x 10-\" 3.96 x 10 7.87 x 10 7.11 x 10 4.04 × 10-6 1.69 x 10 5.71 x 10s 1.64 x 10 4.18 x 10* 9.66 x 10 2.06 x 10 4.13 x 10 7.81 × 10-s 1.41 x 10 2.45 x 10* 4.09 x 102 664 × 10-7 1.05 x 101 1.61 × 10-1 2.43 x 10-1 3.59 x 10- 5.21 × 10-1 7.43 × 10-1 1.04 10 12 14 15 16 18 19 20 21 23 24

Solution

Dr. Cavallini uses the logistic equation in the form dm dt = rm 1 m K , (1) where t is time, m = m(t) is the biomass, and r and K are positive parameters. Using separation of variables, it can be shown that the solution of the logistic equation is m(t) = K 1 + Cert , (2) where C is an arbitrary constant. It is interesting to allow Matlab to provide a solution. 2 syms m r k t m=dsolve(’Dm=r*m*(1-m/K)’,’t’) Matlab responds with the following solution. m = K/(1+exp(-r*t)*C1*K) Letting C = KC1 in this result provides equation (2). A somewhat tricky calculation provides the second derivative of m with respect to t. m00(t) = CKr 2 e rt (C e rt ) (C + e rt ) 3 . (3) Of course, you can also allow Matlab to do this calculation for you. m2=diff(m,t,2) m2 = 2*K^3/(1+exp(-r*t)*C1*K)^3*r^2*exp(-r*t)^2*C1^2-K^2/(1+exp(-r*t)*C1*K)^2*r^2*exp(-r*t)*C1 Of course, you will want to simplify this result. m2=simple(m2) m2 = K^2*r^2*exp(-r*t)*C1*(exp(-r*t)*C1*K-1)/(1+exp(-r*t)*C1*K)^3 A little algebraic manipulation (let C = KC1 and multiply numerator and denominator by e 3rt ) will reveal that this is identical to equation (3). The graph of m = m(t) has a point of inflection when the second derivative in equation (3) is zero. The second derivative provided by equation (3) is zero when its numerator is zero; that is, when C = e rt . So, if we put C = e rt0 , where t0 is the time when the point of inflection occurs, then the solution (2) become m(t) = K 1 + er(tt0) . (4) Note that the biomass m approaches the carrying capacity K as t . 4 The Curve Fitting Algorithm Dr. Cavallini now states: “We now show a procedure for fitting, in the least squares sense, a logistic curve (4) to a given data set (ti , mi) for i = 1, 2, . . . , n. In symbols, the problem is to minimize, for K, r and t0 varying in the real line, the error e = Xn i=1 m(ti) mi 2 . (5) 3 0 20 40 60 80 0 1 2 3 4 5 6 Time evolution of algal sample. Time (days) Biomass (mm2 ) (ti ,mi ) (ti ,m(t1 )) Figure 2: Finding the square of the error. This warrants some explanation. In Figure 2, we’ve plotted the data set as discrete points and overlayed a potential solution to our curve fitting goal. Note that the coordinates of the given data point are (ti , mi), while the coordinates of the point with the same abscissa on the fitted curve are (ti , m(ti)). The “error” is m(ti) mi . Because some of the data points lie below the fitted curve while others lie above, we square to insure that our error is positive. That is, the squared error made at the time value ti is m(ti) mi 2 . (6) Thus, the total squared error in fitting the curve to our n data points is given by equation (5). Clearly, the object of the game is to minimize the total least squared error presented by equation (5). This is why we say that we are fitting a logistic to the data set in a “least squares sense