Numerical Solution of Differential Equations

Transcription

1 Numerical Solution of Differential Equations Drs. Constance Schober & Alvaro Islas Tomasz H. Wlodarczyk Applications of Calculus I

2 Competition in Population Models The simplest population model assumes that dp/dt, the rate of growth, is proportional to the population P. This assumption is valid for small populations, when there is plenty of resources. But as populations increase and resources diminish, competition among individuals has to be taken into account. A simple model that includes competition is given by dp dt = kp(t) αp 2 (t) where α is a small interaction parameter.

3 One Week Project Assume, as before, that the initial population P 0 = 1, and a two percent growth, k = Set the interaction parameter α = Write Euler s method for this equation. 2 Use this method to advance the solution from t = 0 to t = 500 with the three different step sizes: h 1 = 100, h 2 = 50, and h 3 = 25. Call these numerical solutions P 1, P 2 and P 3, respectively, and make a table similar to the one given in class. 3 Graph all three solutions in a single graph. Use a dotted line for P 1, a dashed line for P 2 and a solid line for P 3. 4 What limit are these populations approaching? 5 Give an argument to explain the shape of the solutions and the fact that they approach a limit.

4 Solution: Problem #1 For h = 100 Euler s method looks like ( ) t 1 = t 0 + h = 100 P 1 = P 0 + h 0.02P P0 2 = 2.9 ( ) t 2 = t 1 + h = 200 P 2 = P 1 + h 0.02P P1 2 = 7.9 ( ) t 3 = t 1 + h = 300 P 3 = P 1 + h 0.02P P2 2 = 17.4 ( ) t 4 = t 1 + h = 400 P 4 = P 1 + h 0.02P P3 2 = 21.9 ( ) t 5 = t 1 + h = 500 P 5 = P 1 + h 0.02P P4 2 = 17.7

5 Solution: Problem #2 n x n P 1 n x n P 2 n x n P Table: Euler s method applied to P (t) = kp αp 2, P(0) = 1 with h = 100 (P 1 ), h = 50 (P 2 ) and h = 25 (P 3 ).

6 Solution: Problem #2 Continued n x n P 1 n x n P 2 n x n P Table: Euler s method applied to P (t) = kp αp 2, P(0) = 1 with h = 100 (P 1 ), h = 50 (P 2 ) and h = 25 (P 3 ).

7 Solution: Problem #2 Continued n x n P 1 n x n P 2 n x n P Table: Euler s method applied to P (t) = kp αp 2, P(0) = 1 with h = 100 (P 1 ), h = 50 (P 2 ) and h = 25 (P 3 ).

8 Solution: Problem # Population Time Figure: Numerical solutions P 1 (h = 100, dotted line), P 2 (h = 50, dashed line), and P 3 (h = 25, solid line).

9 Solution: Problem #4 The population P approaches 20 as t approaches 500.

10 Solution: Problem #5 Initially the population is small so it seems as if there are unlimited resources and the growth looks like before, without competition. As the population increases, competition slows the growth and an equilibrium is reached.

11 Time for Some Real Witchcraft

12 Solution: Problem #5 Extended Notice that if P = kp αp 2 we can also write P = P(k αp) which suggests two critical point of P. It is also not difficult to observe that for all 0 < P < k/α function P is increasing and for P > k/α function P is decreasing. This suggests that function P converges to k/α as t, a phenomenon which you observed in the project where k/α = 20.

13 P as a Function of P dp/dt P Figure: Monotonicity of P.

14 Solution: Problem #5 Extended We can learn more about the solution of this IVP by analyzing second derivative of P. By the chain rule we have that P = kp 2αPP = P(k αp)(k 2αP) suggesting that there is an inflection point for that value of t for which P = k/(2α). In our case k/(2α) = 10.

15 P as a Function of P 4 x d 2 P/dt P Figure: Concavity P.

16 Solution: Problem #5 Extended From the plot of the second derivative we see that our solution is concave upward whenever 0 < P < k/(2α) or P > k/α. It is concave downward whenever k/(2α) < P < k/α. This can be visualized by plotting solutions to our problem for different initial populations.

17 P as a Function of t Population Time Figure: P(t) for different initial data.

18 Euler s Method Error Let s go back to the original population problem dp dt = 0.02 P, P(0) = 1 The approximate solutions depend on the time step h. In fact, from the table, we see that the error is proportional to h, it doubles with h. h N P N Error Table: Error at t = 100 for Euler s method.

19 Euler s Method Error Continued This is not an attractive feature of Euler s method, if we are going to double our effort, we would like to get more out of it. Not only that, the table also shows that we need 160 steps to achieve an error less than 0.1. If we were looking for an error less than 0.001, we would need at least 100 times more steps, that is, over steps! Way too much.

20 Improving Euler s Method In class we discussed how given some fixed data we can choose different linear approximations to make an estimate and how sometimes taking the average of two of them works better. For example, consider the following population:

21 Improving Euler s Method Continued Population Time Figure: Population curve with two linear approximations

22 Improving Euler s Method Continued If we want to estimate the population at either 1984 or 1986, we could use either linear approximation or we could use their average, marked by the dashed straight line.

23 Improving Euler s Method Continued Population Time Figure: Population curve and the average linear approximation

24 Improving Euler s Method Continued Let s apply the average idea to Euler s method. Euler s method estimates the solution at x n+1 using the slope (derivative) at the previous point x n. Now we want to use the average of the slopes at the points x n and x n+1. That is, y n+1 = y n + h F(x n, y n ) + F(x n+1, y n+1 ) 2 This should provide a better approximation, except that this equation is much more difficult to use since it has the quantity we want to estimate, y n+1, on both sides of the equation, which in general, it cannot be solved for it and it has to be numerically estimated. This defeats the simplicity of linear approximations.

25 Improving Euler s Method Continued To resolve this difficulty, we are going to make use of the original Euler s method to get a first approximation of y n+1 and then use the average formula to get a second, more precise, approximation. That is, the new method requires two computations for every step: First step: k = y n + h F(x n, y n ) Second step: y n+1 = y n + h F(x n, y n ) + F (x n+1, k) 2 This method is known as the modified Euler s method.

26 The Modified Euler s Method Example 1 Let s apply the modified Euler s method to our old population problem. The table shows the errors for the same values of h use before. h N P N Error MP N Error Table: Error at t = 100 for Euler s and modified Euler s methods.

27 The Modified Euler s Method Example 1 Continued With N = 160 the error given by the modified Euler s method is less than To achieve the same accuracy using Euler s method, we would need N = Too many steps for our taste!

28 Predator Prey

29 Predator Pray Population Problem Let s apply these ideas to a more interesting problem, more subtable for Halloween. Suppose you have populations of rabbits and wolves sharing the same area and suppose the wolves survive only by eating rabbits. How do we describe their population growth?

30 Predator Pray Population Problem Continued In the absence of wolves, the rabbit population will increase proportionally to their population. But in the presence of wolves, it will decrease proportionally to the number of encounters between rabbits and wolves. On the other hand, in the absence of rabbits, the wolf population will decrease proportionally to their population. But in the presence of rabbits, it will increase proportionally to the number of encounters between rabbits and wolves.

31 Governing Equations To write these observations precisely, let R and W be the rabbit and wolf populations. Then the IVP associated with this problem can be written as a system of equations dw dt dr dt = k w W + e αrw W (t 0 ) = W 0 = k r R αrw R(t 0 ) = R 0 Here k w and k r are their rates of growth, α and e are measures of how often they cross their paths and how efficient are the wolves at catching rabbits.

32 Solution to the Predator Prey Problem To apply the modified Euler s method, it is useful to think of the wolf and rabbit population as a point P moving an the WR coordinate system as a function of t.

33 Solution to the Predator Prey Problem Continued Rabits Wolfes Figure: Wolf and Rabbit Populations

34 Solution to the Predator Prey Problem Continued So let P(t) = ( W (t) R(t) Then P satisfies the following IVP: ). where F(t, P) = dp dt = F(t, P), P(0) = P 0 ( kw W + eαrw k r R αrw ) and P 0 = ( W0 R 0 ).

35 Solution to the Predator Prey Problem Continued Applying the modified Euler s method we get after n steps ( ) ( ) ( ) K1 Wn kw W = + h n + eαr n W n K 2 R n k r R n αr n W n ( ) ( ) Wn+1 Wn = + h ( ) kw W n + eαr n W n R n+1 R n 2 k r R n αr n W n ( ) kw K 1 + eαk 2 K 1 + h 2 k r K 2 αk 2 K 1

36 Solution to the Predator Prey Problem Continued Let s look at the graph of the solution. Here we see that at first the rabbit population increases fast, while the wolf population increases slowly. At some point, the rabbit population ceases to growth and the wolf population begins to grow up to the point where they almost eat all the rabbits at which point the wolf population collapses. And the cycle repeats itself.

37 Solution to the Predator Prey Problem Continued Rabits Wolves Figure: Wolf and Rabbit Populations

38 Water Waves So far, we have seen examples where there is only one independent variable (t) and one or two dependent variables. But the same ideas apply if we have more independent variables, such as time and space. In that case we have surfaces representing the solutions. Consider a model equation that governs the motion of one directional water waves: ( ) u t = i u xx + 2 u 2 u, u(x, t = 0) = u 0 (x) where u(x, t) is the surface elevation.

39 Water Waves Continued To apply techniques like those we just described we need to discretized the second derivative with respect to x. One such discretization gives the following system of equations [ ] u2 2u 1 + u 0 u 1t = i x u 1 2 u 1 u nt. = [ un+1 2u n + u n 1 i x u n 2 u n ] u N t. = [ un+1 2u N + u N 1 i x u N 2 u N ]

40 Water Waves Continued For different initial conditions we can obtain a number of solutions. Here are just a couple examples Space Time Figure: Water Waves

41 Water Waves Continued Space 0 Time Figure: Water Waves

42 Movie

43 Trick or Treat!!!

44 Happy Halloween Enjoy your Halloween! Play it safe.