Ordinary differential equations solving methods

Transcription

1 Radim Hošek, A07237 Ordinary differential equations solving methods Problem: y = y2 (1) y = x y (2) y = sin ( + y 2 ) (3) Where it is possible we try to solve the equations analytically, then we use both implicit and explicit Euler methods to get approximate numeric solutions. To prove the qualities of both methods, we compare them to analytic solution (if available) or to Ode23, which is more accurate (second order) method then Euler methods. 1 Analytic solution 1.1 y = y2 For solving equation (1) we can use a basic method called separation of variables, because the algebra allows us to rewrite the equation in such a way that each of the variables occurs at opposite side of the equation. First we try to get a constant solution derivative of a constant is zero: y = K 0 = K which equals only for K = 0. Now we rewrite the derivative of y as a fraction: dy dx = y2 In the next step we can separate each variable to one side of the equation: Now let s integrate the equation. dy y 2 = dx 1

2 After simple integration we obtain: 1 y dy dx y 2 = = 1 x + C, C R where C stands for both integration constants. The solution, explicitly: x y = 1 Cx, C R Figure 1: Visualisation of the solution depending on C: (C = 4 blue; C = 2 violet; C = 0 linear; C = 2 dark red; C = 4 red). To prove that the function found really solves the equation (1), we can substitute y. Just briefly: 1 Cx + Cx (1 Cx) 2 = 1 (1 Cx) 2 = 0 = 0 (1 Cx) 2 (1 Cx) 2 2

3 We have proved the correctness of our calculation. 1.2 y = x y More, the equation (2) can be solved analytically using the separation of variables. The procedure is the same as in 1.1. y = x y ydy = xdx ydy = xdx y 2 = + C, C R where C stands for a real constant. Remark that not every C gives us a solution. Obviously, after rewriting into the explicit form: y = + x2 + C Figure 2: Visualisation of the solution of (2) depending on C (see the legend). The choice of the positive (or negative) solution depends on initial condition. 3

4 2 Numerical solution - Euler methods Only a small part of all the differential equations is possible to be solved analytically with some of the basic methods. There s no analytic solution for most of first order ordinary differential equations (ODEs), just like for (3), for example. The only way to handle them is to use numerical methods for their solving. 2.1 Explicit Euler method Often called shortly just Euler 1 method, is the simpliest one step method for solving first order ODEs with initial condition. Like other numerical methods it is based on computing the function value in a set of discrete figures of the variable (in our case equidistant). Euler method requires sufficient smoothness of the function which solves the equation. Let s have a sequence of values x, in which we ll calculate the function values, we can mark them x 0, x 1,... x k. For each x n+1 let f(x n+1 ) = f(x n )+hf (x n ). Tildes indicate the calculation with approximate values since the first step. We can see that it s nothing else but the approximation given by the first order Taylor polynomial of the function - here we can see the connection with our request of sufficient smoothness. Let s explain the algorithm using the equation (2) with initial condition y(0) = 1 and the step of the method (the difference between discrete values of x) h = y = x y x 0 = 0 as the problem is set. We can calculate f(x 1 ) as an approximation of f(x 1 ). f(x 1 ) = f(x 0 ) + hf (x 0 ) f(x 1 ) = f(x 1 ) = The derivative of f in x 0 was evaluated by using the initial condition of the equation (2), thus f(x 1 ) = 1. Thanks to having the precise analytic solution, we can compare the error of the method in the first step, which is expressed by the difference f(x 1 ) f(x 1 ), which is approx The value of f(x 1 ) is now our approximation of f(x 1 ) that we use for calculating f( ): f (x 1 ) = x 1 f(x 1 ) = = Leonhard Euler ( ), Swiss mathematician and physicist 4

5 Figure 3: Explicit Euler method in comparison with analytic solution. f( ) = f(x 1 ) + hf (x 1 ) f( ) = f( ) = and so on through the recursion. We want to know whether the method gives us acceptable values it means, that the method converges, which is fulfilled for Euler method. Intuitively it seems that better approximation can be reached by decreasing the step of the method as the picture (Fig.4) shows. It s true with expectation of no rounding errors which is unreal. Hence this proposition is false. We always have to be careful about sufficient number of decimal places to ensure that the rounding error doesn t exceed the error of the method. 5

6 Figure 4: Explicit Euler method using basic step (0.10) and half step (0.05). Can we get more information from our approximation? Consider following problem. y = y (4) y(0) = 1. Solution of this problem is simple (y = e x ), so there s no need to use any numerical method. But its simplicity allows us to demonstrate some more interesting things. We know that the solution is a convex function in R. Can we recognize it at the beginning? Sure we can. After calculating the derivative of (4) we obtain y = y, which, after substitution from itself, gives y = y. (5) Now let s use the initial condition. It says that y is positive at x 0. But also y is positive, which gives us: n 0 : y(x n ) 1 > 0. 6

7 so (according to (5)) n 0, y(x n ) is also positive, the solution of our problem must be convex. Let s see what happens in a single step during the algorithm of Euler method: We approximate the function f at I n = [x n ; x n+1 ] with its tangent in x n. We know that f is strictly increasing and convex at I n, which gives us that x (x n ; x n+1 ] : y (x) > y (x n ) Which means that the Explicit Euler method approximates f at I n with piecewise linear function f in such a way, that f < f. As it holds for all n 0, when considering the very first interval [x 0 ; x 1 ], we know that f(x 1 ) f(x 1 ), so in each next step we start the approximation even lower and x (x 0 ; + ) : f(x) < f(x) Thus, explicit Euler method gives us lower approximation of the solution. Figure 5: Demonstration of output given by explicit Euler method for a convex solution - problem (4). 2.2 Implicit Euler method Implicit Euler method has almost the same algorithm but in the approximation at I n = [x n ; x n+1 ] we don t use the derivative at x n, but at x n+1. This slight change causes a big problem let s see the step of recursion in solving the equation (2) with initial condition y(0) = 1. f (x n+1 ) = x n+1 f(x n+1 ) f(x n+1 ) = f(x n ) + hf (x n+1 ) 7

8 Figure 6: Implicit Euler method in comparison with analytic solution. After substitution from the first equation to the second one: f(x n+1 ) = f(x n ) + h x n+1 f(x n+1 ) Where h is the given step of the method, f(xn ) is the value (known from the previous step). But there s a problem left: we have to calculate f(x n+1 ) as a root of an implicit function. This is made in the appendant programme by using original MATLAB function fzero which looks for a root close to previous value f(x n ). This is what makes this method inexpedient looking for roots of implicit function in each step rapidly elongates the time necessary for the calculation. 2.3 Euler methods in use On previous pages we showed the algorithm of both methods in analytically solvable problems, which has given us sufficient knowledge to handle the equation (3): y = sin ( + y 2 ) The two following figures show this ODE with following attributes: step of the method h = 0.005, beginning of the interval x 0 = 10 and initial conditions varying from -1 to 1. Axes: horizontal x, vertical y. 8

9 Figures 7 & 8: Both Euler methods (explicit in green, implicit in blue) in use for solving equation (3) 9

10 Even if the two previous pictures look pretty the same, it gives us no information about the precisity. As (3) can t be solved analytically, we ll use for our comparison the second order ode23 method, which we ll consider being a precise solution to establish the absolute value of the error of each method. Figure 9: Comparison of both methods solving (3) with initial condition y( 10) = 0. We can see that, in this case, implicit method gives us more accurate values especially in a greater distance from the initial value. However, we often need the approximation in a very close neighbourhood of the intial value, where both these methods are comparable. 10

11 3 Conclusion Euler methods are suitable for solving ODEs with sufficiently smooth solution. It is impossible to say in general which of both methods is better for solving ODEs. The simplicity of the explicit EM predetermines it for wider use, because where eem collapses (due to unfulfilling the condition of smoothness, for example), iem doesn t work either. For numerical solving of ODEs especially MATLAB contains a lot of much better algorithms than EMs. As for its help ode45 is recommended to be the first try for solving most of the problems. ode23 and ode45 are both one step methods based on Runge-Kutta formulas. The undiscutable advantage of EMs is their simplicity, which allows us to observe what s happening during the algorithm. EMs are great when we need to know just rough shape of the solution and don t insist on high accuracy. 11