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Simulation

Module by: C. Sidney Burrus. E-mail the author

Dynamic Simulation on a Digital Computer

If a system is modeled by a differential equation, and if the equations are numerically solved on a digital computer or calculator, the system is said to be simulated on the computer. If the model is valid and the numerical methods accurate, experiments can be performed on the computer simulation that might be impossible to conduct otherwise.

Consider several examples that use the models already discussed. If a population is governed by a linear first-order equation

dp d t = r p dp d t = r p
(1)

one would not be able to "solve" this equation on a computer. If, however, we use Euler's method as was done in (14) by approximating the derivative as

d p d t = p ( n T + T ) - p ( n T ) T d p d t = p ( n T + T ) - p ( n T ) T
(2)

where time is considered at intervals of TT,

t = n T n = 0 , 1 , 2 , . . . t = n T n = 0 , 1 , 2 , . . .
(3)

This gives for (1) .EQ (34)

p ( n T + T ) = p ( n T ) + T r p ( n T ) p ( n T + T ) = ( 1 + r T ) p ( n T ) p ( n T + T ) = p ( n T ) + T r p ( n T ) p ( n T + T ) = ( 1 + r T ) p ( n T )
(4)

If we include the time interval TT in the functional notation by

p ( n T ) = x ( n ) p ( n T ) = x ( n )
(5)

then (3) becomes

x ( n + 1 ) = ( 1 + r T ) x ( n ) x ( n + 1 ) = ( 1 + r T ) x ( n )
(6)

which is now in a form that one can easily calculate successive values of x(n)x(n) given any initial value. This can be programmed on a computer or simply done on a hand calculator.

Next, consider the nonlinear equation that models a population with a simple limit given by (23).

d p d t = 1 - p k r o p d p d t = 1 - p k r o p
(7)

Using Euler's method again gives

x ( n + 1 ) = x ( n ) + T r o x ( n ) - T r o k x ( n ) 2 x ( n + 1 ) = x ( n ) + T r o x ( n ) - T r o k x ( n ) 2
(8)

This equation is complex enough to illustrate several points; therefore, we will examine several numerical solutions. Equation 8 was programmed on a Tektronix 31 programmable calculator with a plotter automatically plotting the solutions by drawing straight lines between successive x(n)x(n).

First, consider a low-density growth rate of ro=0.1ro=0.1 or 10% per year for an initial population of po=100po=100 over a time period of 100 years. We will use for the reducing effect on the growth rate in (2), a value of β=0.0001β=0.0001, which implies a carrying capacity for the system of k=10,000k=10,000. For the Euler method, a time interval of T=2T=2years is used, which means 50 calculations of (6) will be necessary for the 100-year period.

The curves in Figure A are the output of the simulation for the above parameters and also for other growth rates of 5% and 20%. Note the solution always approaching the same limit but requiring different amounts of time.

In Figure B, the model is run assuming several different initial populations. Again, the solutions always approach the limit of kk, even if the initial population is greater than kk.

Figure C shows the effects of various amounts of limiting by considering various values for the factor ββ, and therefore kk, the carrying capacity. When the limit is removed (k=k=), the growth is exponential.

These examples illustrate the kinds of questions that can be pursued by running experiments on the computer simulation. There is one more point that should be considered. It has nothing to do with the differential equation model (23) but with the numerical procedure, Euler's method. Consider the effects of using various time intervals TT while holding everything else constant. Figure D shows the results of this experiment. The curve resulting from using a time interval of T=0.2T=0.2years looks essentially the same as the exact solution of the differential equation. The numerical solution deviates more as TT is increased until, for T=20T=20years, it has lost the character of the exact solution. A method for checking to see if TT is sufficiently small is to try halving it until the change is small.

One last point should be made concerning this numerical simulation. Euler's method is the only approach to numerically solve (23) that has been discussed. That is not because it is the best – there are far more efficient and sophisticated methods – but that is not our subject here, so we will continue with the straightforward algorithm of Euler.

The super-exponential logistic equation of (31) was simulated on the calculator and run with a low population growth rate of ro=0.1ro=0.1, a maximum rate of M=.2M=.2, and a carrying capacity of k=10,000k=10,000. This solution is shown in Figure E and compared with an exponential of the same roro, and a simple logistic of the same roro and kk. The model was run again with a maximum growth rate of M=.5M=.5, and the results are shown in Figure F. Note the initial exponential growth which becomes super-exponential, growing extremely rapidly, then abruptly leveling off to an equilibrium.

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