3. 3E: Runge-Kutta Usuli (Mashqlar) Matematika

Exercise 8
Consider the solution to the differential equation (dot = -100y) with (y(0) = 1) .
Use explicit (forward, usual) Euler’s method with a step size of (0.2) to estimate the value of (y(1)) .
Repeat part (a) with the same step size but use implicit (backwards) Euler instead.
What function are we approximating here, and which method got closer to the correct value?
Does increasing the number of steps help with any shortcomings you saw with either of the methods? Nima uchun yoki nima uchun bunday emas?
Qaror
We have (f(t, y) = -100y) and (h = frac<1><5>) , so the next estimate comes from the equation (y_= -19y_n) . That gives the following chart: [egin t & 0 & 0.2 & 0.4 & 0.6 & 0.8 & 1 hline y & 1 & -19 & 361 & -6859 & 130321 & -2476099 end] so the estimate is (y(1) approx -2476099) . 1
With backwards Euler we wind up with (21y_ = y_n) , so (y_ = frac<1><21>,y_). We divide each number by (21) as we go. [oshlanishi t & 0 & 0.2 & 0.4 & 0.6 & 0.8 & 1 hline y & 1 & 0.0476 & 0.0023 & 0.0001 & 5.14 imes 10^ <-6>& 2.45 imes 10^ <-7>end] The estimate we get is (y(1) approx 21^<-5>) , which is a very small number. Notice that this differs from the explicit Euler.
The solution is (y(t) = e^<-100t>) , so we were trying to estimate (frac<1> approx 3.72 imes 10^<-44>) . Implicit Euler wins, by virtue of being less than two million.The explosion in the forward Euler approximation could make us skeptical that the prospects are good, but the problem came from the fact that the step size (frac<1><5>) wasn’t big enough to subdue the coefficient (100) . If we use (h) steps instead of five steps, then the equation governing the next estimate is (y_ = (1 - frac<100>)y_n) . Using induction, since (y(0) = 1) , we see that (y_ = (1 - frac<100>)^). When we chose (h = 5) , we got ((-19)^n) which went nuts. But if (h > 100) , then (1 - frac<100>) is a number smaller than one, so its powers are going to shrink. As we continue doing this we will get a number closer and closer to the correct value. If we let the number of steps go to infinity, then we wind up calculating the limit [lim_ left( 1 - frac<100> ight)^n = e^ <-100>,.] This kind of limit may look familiar to you from calculus classes.