Numerical Differential Equations: IVP
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1 Chapter 11 Numerical Differential Equations: IVP **** 4/16/13 EC (Incomplete) 11.1 Initial Value Problem for Ordinary Differential Equations We consider the problem of numerically solving a differential equation of the form dy dt = f(t,y), a t b, y(a) = α (given). Such a problem is called the Initial Value Problem or in short IVP, because the initial value of the solution y(a) = α is given. Since there are infinitely many values between a and b, we will only be concerned here to find approximations of the solution y(t) at several specified values of t in [a, b], rather than finding y(t) at every value between a and b. The following strategy will be used: Divide [a,b] into N equal subintervals each of length h: t 0 = a < t 1 < t < < t N = b. 491
2 49 CHAPTER 11. NUMERICAL DIFFERENTIAL EQUATIONS: IVP a = t 0 t 1 t t N = b Set h = b a N (step size) Compute approximations y i to y(t i ) at t = t i ; that is, given y(0) = y 0, compute, recursively, y 1,y,...,y n. y(t i ) = Exact value of y(t) at t = t i. y i = An approximation of y(t i ) at t = t i. Given The Initial Value Problem (i) y = f(y,t), a t b (ii) The initial value y(a) = y(t 0 ) = y 0 = α (iii) The step-size h. Find y i (approximate value of y(t i )), i = 1,...,N, where N = b a h. We will briefly describe here the following well-known numerical methods for solving the IVP: The Euler and Modified Euler Method (Taylor Method of order 1) The Higher-order Taylor Methods The Runge-Kutta Methods The Multistep Methods: The Adams-Bashforth and Adams-Moulton Method The Predictor-Corrector Methods Methods for Systems of IVP Differential Equations We will also discuss the error behavior and convergence of these methods. However, before doing so, we state a result without proof, in the following section on the existence and uniqueness of the solution for the IVP. The proof can be found in most books on ordinary differential equations.
3 11.1. INITIAL VALUE PROBLEM FOR ORDINARY DIFFERENTIAL EQUATIONS Existence and Uniqueness of the Solution for the IVP Theorem 11.1 (Existence and Uniqueness Theorem for the IVP). Suppose (i) f(t,y) is continuous on the domain defined by R = {a t b, < y < }, and (ii) the following inequality is satisfied: f(t,y) f(t,y ) L y y, whenever (t,y) and (t,y ) R, and L is constant. Then the IVP y = f(t,y),y(a)α has a unique solution. Definition 11.. The condition f(t,y) f(t,y ) L y y is called the Lipschitz Condition. The number L is called a Lipschitz Constant. Definition A set S is said to be convex if whenever (t 1,y 1 ) and (t,y ) belong to S, the point ((1 λ)t 1 +λt, (1 λ)y 1 +λy ) also belongs to S for each λ, 0 λ Simplification of the Lipschitz Condition for the Convex Domain Suppose that the domain R in Theorem 11.1 is convex set, then the The IVP has a unique solution if f y (t,y) L for all (t,y) R Liptischitz Condition and Well-Posedness Definition An IVP is said to be well-posed if a small perturbation in the data of the problem leads to only a small change in the solution. Since numerical computation may introduce some perturbations to the problem, it is important that the problem that is to be solved is well-posed. Fortunately, the Lipschitz condition is a sufficient condition for the IVP problem to be well-posed. Theorem 11.5 (Well-Posedness of the IVP problem). If f(t, y) satisfies the Lipschitz condition, then the IVP is well-posed.
4 494 CHAPTER 11. NUMERICAL DIFFERENTIAL EQUATIONS: IVP 11. The Euler Method One of the simplest methods for solving the IVP is the classical Euler method. The method is derived from the Taylor Series expansion of the function y(t). Recall that the function y(t) has the following Taylor series expansion of order n at t = t i+1 : y(t i+1 ) = y(t i )+(t i+1 t i )y (t i )+ (t i+1 t i ) y (t i )! + + (t n i+1 t i ) y (n) (t i )+ (t i+1 t i ) n! (n+1)! Substitute h = t i+1 t i. Then n+1 y n+1 (ξ i ), where ξ i is in (t i,t i+1 ). Taylor Series Expansion of y(t) of order n at t = t i+1 y(t i+1 ) = y(t i )+hy (t i )+ h! y (t i )+ + hn n! y(n) (t i )+ hn+1 (n+1)! y(n+1) (ξ i ). For n = 1, this formula reduces to y(t i+1 ) = y(t i )+hy (t i )+ h! y (ξ i ). The term h! y() (ξ i ) is called the remainder term. Neglecting the remainder term, we can write the above equation as: y(t i+1 ) y(t i )+hy (t i ) Using our notations, established earlier, we then have y i+1 = y i +hf(t i,y i ), since y i = f(t i,y i ) We can then resursively generate the successive approximations y i,y,...,y N to y(t 1 ), y(t ),... and y(t N ) as follows:
5 11.. THE EULER METHOD 495 y 0 = α (given) y 1 = y 0 +hf(t 0,y 0 ) y = y 1 +hf(t 1,y 1 ).. y N = y N 1 +hf(t N 1,y N 1 ) These iterations are known as Euler s iterations and the method is the historical Euler s method. Figure 11.1: Geometrical Interpretation y(t N ) = y(b) y(t ) y(t 1 ) y(t 0 ) = y(a) = α a t 1 t t N 1 b = t N
6 496 CHAPTER 11. NUMERICAL DIFFERENTIAL EQUATIONS: IVP Algorithm 11.6 (Euler s Method for IVP). Input: (i) The function f(t, y) (ii) The end points of the interval [a,b]: a and b (iii) The initial value: α = y(t 0 ) = y(a) (iv) The step size: h Output:Approximations y i+1 of y(t i +1), i = 0,1,...,N 1. Step 1.Initialization: Set t 0 = a, y 0 = y(t 0 ) = y(a) = α and N = b a h. Step.For i = 0,1,...,N 1 do Compute y i+1 = y i +hf(t i,y i ) End Example 11.7 Using Euler s method, solve numerically y = t +5, 0 t 1, with h = 0.5. Input Data: f(t,y) = t +5 t 0 = a = 0, t 1 = t 0 +h = 0.5, t = t 1 +h = 0.50 t 3 = t +h = 0.75, t 4 = b = t 3 +h = 1. y 0 = y(t 0 ) = y(0) = 0. Find: y 1 = approximation of y(t 1 ) = y(0.5) y = approximation of y(t ) = y(0.50) y 3 = approximation of y(t 3 ) = y(0.75) y 4 = approximation of y(t 4 ) = y(1) The exact solution obtained by direct integration: y(t) = t3 3 +5t. Solution. i = 0: Compute y 1 from y 0 (Set i = 0 in Step ) of Algorithm 11.6:
7 11.. THE EULER METHOD 497 y 1 = y 0 +hf(t 0,y 0 ) = 0+0.5(5) = 1.5 (Approximate value of y(0.5)) Actual value of y(0.5) = i = 1: Compute y from y 1 (set i = 1 in Step ). y = y 1 +hf(t 1,y 1 ) = (t 1 +5) = ((0.5) +5) =.5156 (Approximate value of y(0.50)) Note: of y(0.5) = (Four significant digits) i = : Compute y 3 from y (Set i = in Step ). y 3 = y +hf(t,y ) = ((0.5) +5) = (Approximate value of y(0.75)) Actual value of y(0.75) = Etc. Example 11.8 y = t +5, 0 t y(0) = 0, h = 0.5 Solution. i = 0 : Compute y 1 from y 0 (Set i = 0 in Step of Algorithm 11.6). y 1 = y 0 +hf(t 0,y 0 ) = y(0)+hf(0,0) = =.5 (Approximate value of y(0.50)) Note:y(0.50) =.5417 (four significant digists) i = 1 : Compute y from y 1 (Set i = 1 in Step in Algorithm 11.6).
8 498 CHAPTER 11. NUMERICAL DIFFERENTIAL EQUATIONS: IVP y = y 1 +hf(t 1,y 1 ) =.5+0.5((0.5) +5) = (Approximate value of y(1)) Note: y(1) = (four significant digits). i = 3 : Compute y 3 from y (Set i = in Step ). y 3 = y +hf(t,y ) = (t +5) = (1+5) = (Approximate value of y(1.5)) Note: of y(1.5) = (four significant digits). Etc The Errors in Euler s Method The approximations obtained by a numerical method to solve the IVP are usually subjected to three types of errors: Local Truncation Error Global Truncation Error Round-off Error We will not consider the round-off error for our discussions below. Definition The local error is the error made at a single step due to the truncation of the series used to solve the problem. Recall that the Euler Method was obtained by truncating the Taylor series y(t i+1 ) = y(t i )+hy (t i )+ h y (t i )+ after two terms. Thus, in obtaining Euler s method, the first term neglected is h y (t i ). The local error in Euler s method is:
9 11.. THE EULER METHOD 499 E L E = h y (ξ i ), where ξ i lies between t i and t i+1. In this case, we say that the local error is of order h, written as O(h ). Note that the local error EE L converges to zero as h 0. That means, the smaller h is, the better accuracy will be. Definition The global error is the difference between the true solution y(t i ) and the approximate solution y i at t = t i. Thus, Global error = y(t i ) y i. Denote this by E G E. The following theorem shows that the global error, EE G, is of order h. Theorem (Global Error Bound for the Euler Method). (i) Let y(t) be the unique solution of the IVP: y = f(t,y); y(a) = α, where a t b, < y <. (ii) Let L and M be two numbers such that f(t,y) y L, and y (t) M in [a,b]. Then the global error E G E at t = t i satisfies the following inequality: E G E = y(t i ) y i hm L (el(t i a) 1). Note: The global error bound for Euler s method depends upon h, whereas the local error depends upon h. Proof of the above theorem can be found in the book by G.W. Gear, Numerical Initial Value Problems in Ordinary Differential Equations, Prentice Hall, Inc. (1971). Remark: Since the exact solution y(t) of the IVP is not known, the above bound may not be of practical value as far as knowing how large the error can be a priori. However, from this error bound, we can say that the Euler method can be made to converge faster by decreasing the step-size. Furthermore, if the equalities, L and M of the above theorem can be found, then we can determine what step-size will be needed to achieve a certain accuracy, as the following example shows. Example 11.1 Consider the IVP: dy dt = t +y, y(0) = 0 0 t 1, 1 y(t) 1.
10 500 CHAPTER 11. NUMERICAL DIFFERENTIAL EQUATIONS: IVP Determine how small the step-size should be so that the error does not exceed ǫ = Input Data: f(t,y) = t +y y(0) = y 0 = 0 a = 0,b = 1,y(t) [ 1,1],t [0,1]. Compute L: Since we have Thus f(t,y) = t +y, f y = y. f y 1 for all y, giving L=1. Find M: To find M, we compute the second-derivative of y(t) as follows: y = dy dt By implicit differentiation, y = f t +f f y = f(t,y) (Given) ( t +y = t+ ) y = t+ y (t +y ) So, y (t) = t+ y (t +y ), for 1 y 1. Thus, M=, The, Global Error Bound: E G E = y(t i) y i h L (e(t i) 1) = h(e (t i) 1) Again, e (t i) e for 0 t 1 Thus, E G E h(e 1). Compute h: For the error not to exceed 10 4, we must have: h(e 1) < 10 4 or h < 10 4 e
11 11.3. HIGH-ORDER TAYLOR METHODS High-order Taylor Methods Euler s method was developed by truncating the Taylor series expansion of y(t) after just one term. Higher-order Taylor series can be developed by retaining more terms in the series. These higher-order methods will be more accurate than Euler s method; however, it will be computationally more demanding to implement these methods, because of the need for computing the higher-order derivatives, as shown below. Computations of these derivatives of y(t) = f(t, y) will require implicit differentiations. Recall that the Taylor s series expansion of y(t) of degree n is given by y(t i+1 ) = y(t i )+hy (t i )+ h y (t i )+ + hn n! y(n) (t i )+ hn+1 (n+1)! y(n+1) (ξ i ) In order to develop a computational method for IVP, based on the above series, we must write the various derivatives of y(t) in terms of the derivatives of f(t,y). Thus, we write (i) y (t) = f(t,y(t)) (Given). (ii) y (t) = f (t,y(t)). In general, (iii) y (i) (t) = f (i 1) (t,y(t)), i = 1,,...,n. With these notations, we can now write y(t i+1 ) = y(t i )+hf(t i,y(t i ))+ h f (t i,y(t i ))+ + hn 1 (n 1)! f(n ) (t i,y(t i )) + hn n! f(n 1) (t i,y(t i ))+ hn+1 (n+1)! f(n) (ξ i,y(ξ i )) (Remainder term) [ = y(t i )+h f(t i,y(t i ))+ h f (t i,y(t i ))+ ] + hn 1 f n 1 (t i,y(t i )) + Remainder Term n! Neglecting the remainder term the above formula can be written in compact form as follows: y i+1 = y i +ht k (t i,y i ), i = 0,1,...,N 1,
12 50 CHAPTER 11. NUMERICAL DIFFERENTIAL EQUATIONS: IVP where T k (t i,y i ) is defined by: T k (t i,y 1 ) = f(t i,y i )+ h f (t i,y i )+ + hk 1 f (k 1) (t i,y i ) k! (f (0) (t i,y i ) = f(t i,y i )). So, if we truncate the Taylor Series after k terms and use the truncated series to obtain the approximation y 1+1 of y(t i+1 ), we have the following of k-th order Taylor s algorithm for the IVP.
13 11.3. HIGH-ORDER TAYLOR METHODS 503 Algorithm (Taylor s Algorithm of order k for IVP). Input: (i) The function f(t, y) (ii) The end points: a and b (iii) The initial value: α = y(t 0 ) = y(a) (iv) The order of the algorithm: k (v) The step size: h Step 1.Initialization: Set t 0 = a, y 0 = α, N = b a h Step.For i = 0,...,N 1 do.1 Compute T k (t i,y i ) = f(t i,y i )+ h f (t i,y i )+ + hk 1 k! f (k 1) (t i,y i ). Compute y i+1 = y i +ht k (t i,y i ) End Remark: (i) Taylor s algorithm of order k reduces to Euler s method when k = 1. (ii) Higher order methods will involve implicit differentiation and thus, there will be an increasing complexity of computation as the order of the method increases. Example Using Taylor s algorithm of order, approximate y(0.) and y(0.4) of the following IVP: y = y t +1, 0 t, y(0) = 0.5, h = 0.. Input Data: f(t,y) = y t 1 t 0 = 0,t 1 = 0.,t = 0.4 Preparation: We compute f (t,y(t),f (t,y(t)), etc. which will be needed to compute y 1 and y.
14 504 CHAPTER 11. NUMERICAL DIFFERENTIAL EQUATIONS: IVP f(t,y(t)) = y t +1 (Given). f (t,y(t)) = d dt (y t +1) = y t = y t +1 t f (t,y(t)) = d dt (y t +1 t) = dy dt d dt (t 1+t)) = f(t,y) (t+) = y t +1 t = y t t 1. Solution. i = 0: Compute y 1 from y 0 (Set i = 0 in Step ): i = 1 : Compute y from y 1 (Set i = 1 in Step ). etc. y 1 = y 0 +hf(t 0,y(t 0 ))+ h f (t 0,y(t 0 )) = (0.) (0.5+1) = (approximate value of y(0.)). y = (approximate value of y(0.4)). Truncation Error for Taylor s Method of Order k Since the Taylor s method of order k was obtained by truncating the series after k terms, the remainder after k terms is R k = hk+1 y (k+1) (ξ), x k < ξ < x k+h k+1 and since y (k+1) (t) = f (k) (t,y(t)), the error term for Taylor s method of order k is: E Tk = hk+1 (k +1)! f(k) (ξ,y(ξ)) Thus, the truncation error (local error) of Taylor s algorithm of order k is O(h k+1 ).
15 11.4. RUNGE-KUTTA METHODS Runge-Kutta Methods The Euler s method is the simplest to implement; however, even for a reasonable accuracy the step-size h needs to be very small. The difficulties with higher order Taylor s series methods are that the derivatives of higher orders of f(t,y) need to be computed, which are very often difficult to compute; indeed, f(t,y) is not even explicitly known in many areas. The Runge-Kutta methods aim at achieving the accuracy of higher order Taylor series methods without computing the higher order derivatives. (At the cost of more function evaluations per step.) We first develop the simplest one: The Runge-Kutta Methods of order The Runge-Kutta Methods of order Suppose that we want an expression of the approximation y i+1 in the form: y i+1 = y i +α 1 k 1 +α k, (11.1) where and k 1 = hf(t i,y i ), (11.) k = hf(t i +αh, y i +βk 1 ). (11.3) The constants α 1 and α and α and β are to be chosen so that the formula is as accurate as the Taylor s Series Method of as high order as possible. To develop such a method we need an important result from Calculus: Taylor s series for a function to two variables.
16 506 CHAPTER 11. NUMERICAL DIFFERENTIAL EQUATIONS: IVP Taylor s Theorem for a Function of Two Variables Let f(t,y) and its partial derivatives of orders up to (n + 1) be continuous in the domain D = {(t,y) a t b, c y d}. Then [ f(t,y) = f(t 0,y 0 )+ (t t 0 ) f t (t 0,y 0 )+(y y 0 ) f ] y (t 0,y 0 ) + [ 1 n ( ) ] n + (t t 0 ) n i (y y 0 ) i n f n! i t n 1 y i(t 0,y 0 ) +R n (t,y), h=0 where R n (t,y) is the remainder after n terms and involves the partial derivative of order n+1. Substituting the value of k 1 and k, respectively from (11.) and (11.3) into (11.1), we obtain: y i+1 = y i +α 1 hf(t i,y i )+α hf(t i +αh,y i +βk 1 ) (11.4) Again, Taylor s Theorem of order n = 1 gives Thus, f(t i +αh, y i +βk 1 ) = f(t i,y i )+αh f t (t i,y i )+βk 1 f y (t i,y i ) (11.5) [ y i+1 = y i +α 1 hf(t i,y i )+α h f(t i,y i )+αh f t (t i,y i )+βhf(t i,y i ) f ] y (t i,y i ) [ = y i +(α 1 +α )hf(t i,y i )+α h α f t (t i,y i )+βf(t i,y i ) f ] (11.6) y (t i,y i ) Also, notethaty(t i+1 ) = y(t i )+hf(t i,y i )+ h So, neglecting the higher order terms, we can write ( ) f t (t i,y i )+f(t i,y i ) f y (t i,y i ) +higher order terms. y i+1 = y i +hf(t i,y i )+ h ( f t (t i,y i )+f f ) y (t i,y i ). (11.7)
17 11.4. RUNGE-KUTTA METHODS 507 If we want (11.6) and (11.7) to agree for numerical approximations, then we must have the corresponding coefficients be equal. This will give us α 1 +α = 1 (comparing the coefficients of hf(t i,y i )). α α = 1 (comparing the coefficients of h f t (t i,y i )). α β = 1 (comparing the coefficients of h f(t i,y i ) f y (t iy i )). Since the number of unknowns here exceeds the number of equations, there are infinitely many possible solutions. The simplest solution is: α 1 = α = 1, α = β = 1. With these choices we can generate y i+1 from y i as follows. The process is known as the Modified Euler s Method. Generating y i+1 from y i in Modified Euler s Method Given y = f(t,y), a t b, y 0 = y(t 0 ) = y(0) = α. h = t i+1 t i, i = 0,1,...,N 1. Compute k 1 = hf(t i,y i ). Compute k = hf(t i +h,y i +k 1 ). Compute y i + 1 [k 1 +k ], i = 0,1,...,N 1.
18 508 CHAPTER 11. NUMERICAL DIFFERENTIAL EQUATIONS: IVP Algorithm (The Modified Euler Method). Inputs:The given function: f(t,y) The end points of the interval: a and b The step-size: h The initial value: y(t 0 ) = y(a) = α Outputs:Approximations y i+1 of y(t i+1 ) = y(t 0 +ih), i = 0,1,,,N 1 Step 1. Initialization Set t 0 = a, y 0 = y(t 0 ) = y(a) = α N = b a h Step.For i = 0,1,,,N 1 do Compute k 1 = hf(t i,y i ) Compute k = hf(t i +h,y i +k 1 ) Compute y i+1 = y i + 1 (k 1 +k ). End Example Solve the Initial Value Problem: y = e t, y(0) = 1, and h = 0.5, 0 t 1. f(t,y) = e t, Input Data: y 0 = y(t 0 ) = y(0) = 1. h = 0.5 Step 1. t 0 = 0,y 0 = y(0) = 1 Step. i = 0 : Compute y 1 from y 0 : k 1 = hf(t 0,y 0 ) = 0.5e t 0 = 0.5 k = hf(t 0 +h,y 0 +k 1 ) = 0.5(e t 0+h ) = 0.5e 0.5 = y 1 = y (k 1 +k ) = 1+0.5( ) = 1.66 Note: y(0.5) = e 0.5 = (Four significant-digits). i = 1: Compute y from y 1 :
19 11.4. RUNGE-KUTTA METHODS 509 Example k 1 = hf(t 1,y 1 ) = 0.5e t 1 = 0.5e 0.5 = k = hf(t 1 +h,y 1 +k 1 ) = 0.5e = 0.5e = y = y (k 1 +k ) = ( ) =.7539 Note: y(1) =.7183 Given: y = t+y, y(0) = 1, compute y 1 (approximation to y(0.)) and y (approximation to y(0.0) by using Modified Euler Method. f(t,y) = t+y Input Data: t 0 = 0, y(0) = 1 h = 0. i = 0 : y 1 = y (k 1 +k ) k 1 = hf(t 0,y 0 ) = 0.(0+1) = 0. k = hf(t 0 +h, y 0 +k 1 ) = 0. f(0., 1+0.) = 0. (0.+1.) = = 0.0 Thus y 1 = 1+ 1 (0.+0.0) = 1. (Approximate value of y(0.)) i = 1 : y = y (k 1 +k ) k 1 = hf(t 1,y 1 ) = 0. f(0.,1.) = 0. (0.+1.) = 0.0 k = hf(t 1 +h, y 1 +k 1 ) = 0. f(0.0, ) = 0. ( ) = 0.04 y = ( ) = (Approximate value of y(0.0)). Local Error in the Modified Euler Method Since in deriving the modified Euler method, we neglected the terms involving h 3 and higher powers of h, the local error for this method is O(h 3 ). Thus with the modified Euler method, we will be able to use larger step-size h than the Euler method to obtain the same accuracy.
20 510 CHAPTER 11. NUMERICAL DIFFERENTIAL EQUATIONS: IVP The Midpoint and Heun s Methods In deriving the modified Euler s Method, we have considered only one set of possible values of α 1,α,α 1 and β. We will now consider two more sets of values. Choice 1: α = 0, α = 1, α = β = 1. This choice yields the Midpoint Method. Algorithm (The Midpoint Method). Inputs: (i) The given function: f(t, y) (ii) Step-size: h (iii) The initial value: y(t 0 ) = y(a) = α Output: Approximation y i+1 of y(t i+1 ) = y(t 0 +ih), i = 0,1,,N 1. For i = 0,1,,,N 1 do End Compute k 1 : k 1 = hf(t i,y i ) ( ) Compute k = hf t i + h, y i + k 1 Compute y i+1 from y i : y i+1 = y i +k Example For the IVP: y = e t, y(0) = 1, h = 0.5, 0 t 1, approximate y(0.5) and y(1). Solution Input Data:
21 11.4. RUNGE-KUTTA METHODS 511 f(t,y) = e t y 0 = y(0) = 1 h = 0.5 t 0 = 0, t 1 = 0.5 Compute y 1, an approximation to y(0.5) and y(1). i = 0 : k 1 = hf(t 0,y 0 ) = 0.5e t 0 = 0.5e 0 = 0.5 Note: y(0.5) = k = hf(t 0 + h,y 0 + k 1 ) = 0.5e 0.5 = y 1 = y 0 +k = = Compute y, an approximation of y(1): i = 1 : k 1 = hf(t 1,y 1 ) = 0.5e 0.5 = k = hf(t 1 + h,y 1 + k 1 ) = 0.5e.75 = y = y 1 +k = =.7005 (y(1) = e =.7183) Choice : α 1 = 1 4, α 1 = 3 4, α = β = 3. This choice gives us Heun s method. Heun s Method Generating y i+1 from y i by Heun s Method: Compute k 1 : k 1 = hf(t i,y i ) Compute k : k = hf ( t i + 3 h,y i + 3 k ) 1 Compute y i+1 from y i : y i+1 = y i k k, i = 0,1,...,N 1. Heun s Method and the Modified Euler s Method are classified as the Runge-Kutta methods of order. These methods have local errors of O(h 3 )
22 51 CHAPTER 11. NUMERICAL DIFFERENTIAL EQUATIONS: IVP 11.5 The Runge-Kutta Method of order 4 A method widely used in practice is the Runge-Kutta method of order 4. It s derivation is complicated. We will just state the method without proof. Algorithm 11.0 (The Runge-Kutta Method of Order 4). Inputs:f(t,y) : the given function a,b : the end points of the interval α : the initial value y(t 0 ) h : the step size = b a N Outputs:The approximations y i+1 of y(t i+1 ), i = 0,1,...,N 1 Step 1.(Initialization) Set t 0 = a, y 0 = y(t 0 ) = y(α)n = b a h. Step.For i = 0,1,,...,N 1 do Step.1. Compute the Runge-Kutta coefficients: k 1 = hf(t i,y i ) k = hf(t i + h, y i + 1 k 1) k 3 = hf(t i + h, y i + 1 k ) k 4 = hf(t i +h, y i +k 3 ) Step..Compute y i+1 from y i : y i+1 = y i (k 1 +k +k 3 +k 4 ) End The Local Truncation Error: The local truncation error of the Runge-Kutta Method of order 4 is O(h 5 ). Example 11.1 Apply Runge-Kulta Method of order 4 to the following IVP: y = t+y, 0 t 0.05 y(0) = 1 h = 0. Input Data:
23 11.6. MULTI-STEP METHODS (EXPLICIT-TYPE) 513 f(t,y) = t+y t 0 = a = 0, t 1 = t 0 +h = 0., t = t 1 +h = 0.0, t 3 = 0.03, t 4 = 0.04, t 5 = Step.1. Compute y 1 from y 0 (Set i = 0 in Step ): y 1 = y (k 1 +k +k 3 +k 4 ) Step.. Compute the Runge-Kutta coefficients. k 1 = hf(t 0,y 0 ) = 0.f(0,1) = 0. 1 = 0.. k = hf(t 0 + h, y 0 + k 1 ) = 0.f ( 0. 0.,1+ ( ) k 3 = hf t 0 + h, y 0 + k = h ) [ = ] = 0.. ( t 0 + h +y 0 + k ) = k 4 = hf(t 0 +h,y 0 +k 3 ) = h(t 0 +h+y 0 +k 3 ) = Step.3. Compute y 1 from y 0 : y 1 = y (k 1 +k +k 3 +k 4 ) = and so on Multi-Step Methods (Explicit-type) The methods we have discussed so far, Euler s method and Runge-Kutta methods of order and order 4, are single-step methods; because, given y(t 0 ) = f(t 0,y 0 ) = α, y i+1 is computed from y i. However, if f(t,y) is known at m + 1 points, say, at t i, t i 1, t i,..., t i m ; that is, if f(t k,y k ), k = i,i 1,...,i m are known, then we can develop higher-order methods to compute y i+1. One such class of methods can be developed based on numerical integration as follows: From we have, by integrating from t i to t i+1 : y = f(t,y) or ti+1 t i y dt = y i+1 y i = ti+1 t i ti+1 t i f(t,y)dt f(t,y)dt
24 514 CHAPTER 11. NUMERICAL DIFFERENTIAL EQUATIONS: IVP or ti+1 y i+1 = y i + f(t,y)dt t i Estimating the Integral ti+1 t i f(t,y)dt : Since the functional values f(t,y)are known at m points backward from t i ; namely at t = t i 1,...,t i m, the obvious thing to do is: Find the interpolating polynomial P m (t) that interpolates at t i,t i 1,...,t i m using Newton s backward difference formula: ( m P m (t) = ( 1) k k=0 s k ) k f i k, where s = t t i h, f t,k = f(t k,y(t k )) Integrate ti+1 t i P m (t)dt Substituting the expression of P m (t) and performing the integral, we obtain an explicit formula for computing y i+1 from y i that makes use of the (m + 1) values of f(t,y) at the points, t i, t i 1, t i,,t i+m,, yielding the (m+1) step Adams-Bashforth Formula.
25 11.6. MULTI-STEP METHODS (EXPLICIT-TYPE) 515 Given: f i,f i 1,...,f i m. The (m + 1) Step Adams-Bashforth Formula: Compute: y i +h[c 0 f i +c 1 f i 1 + +c m m f i m ] where ( 1 c k = ( 1) k 0 s k ) ds,k = 1,,,m f k = f(x k,y(x k )),k = i, i 1,,i m Special Cases: m = 0: Adams-Bashforth One-Step Formula Euler s Method. m = : Adams-Bashforth Three-Step Formula Given f i,f i 1,f i. Compute: y i+1 = y i + h 1 [3f i, 16f i 1 +5f i ] m = 3 : Adams-Bashforth Four-Step Formula Given: f i,f i 1,f i,f i 3 Compute: y i+1 = y i + h 4 [55f i 59f i 1 +37f i 9f i 3 ]. Other explicit higher multi-step methods can similarly be developed. Implicit Multistep Methods Consider now integrating ti+1 t i f(t,y)dt using an interpolating polynomial P m+1 (t) of degree m+1 that interpolates at (m+) points (rather than (m+1) points as in Adams-Bashforth formula): t i+1,t i,t i 1,...,t i m. Then using Newton s backward formula, we have ( m+1 P m+1 (t) = ( 1) k 1 s k k=0 ) k f i+1 k.
26 516 CHAPTER 11. NUMERICAL DIFFERENTIAL EQUATIONS: IVP For m =, We have Adams-Moulton Three-Step Formula: y i+1 = y i + h 4 (9f i+1 +19f i 5f i 1 +f i ). Since f k = f(t k,y(t k )), k = i,i 1,i, the above expression becomes: y i+1 = y i + h 4 (9f(t i+1,y(t i+1 ))+19f(t i,y(t i )) 5f(t i 1,y(t i 1 ))+f(t i,y(t i )). The formulas of the above type are implicit multi-step formulas; because computation of y i+1 requires y(t i+1 ). Error Term: The error term of three-step implicit Adams-Moulton formula is also of O(h 5 ), but with a smaller constant term. Specifically, it is E 3 AM = h5 y 5 (ξ), where ξ lies between t i and t i Predictor-Corrector Methods A class of methods called Predictor-Corrector methods, is based on the following principle: Predict the initial value of y (0) i+1 by using an explicit formula Correct the predicted value of y (0) i+1 iteratively to y(1) i+1, y() i+1,..., y(k) i+1 by using an implicit formula until two successive iterations agree to each other by a prescriben error tolerance. In the sequel, we will state two such methods Euler-Trapezoidal Predictor-Corrector Method This method uses Euler s method as the predictor and then a corrector formula is developed based on integrating t i+1 t i f(t,y)dt using trapezoidal rule. Euler s method gives y (0) i+1 = y i +hf(t i,y i ) (predicted value) Trapezoidal rule of integration applied to y i+1 y i y dt gives y i+1 = y i + h [f(t i,y i )+f(t i+1,y(t i+1 ))].
27 11.7. PREDICTOR-CORRECTOR METHODS 517 This is an implicit formula of compute y i+1 since the value of y(t i+1 ) is needed to compute y i+1. However, having found the initial gues of y (0) i+1 from the predictor, we can now correct this value iteratively using the above formula. Thus, compute y (1) i+1 = y i + h [ y () i+1 = y i + h ] f(t i,y i )+f(t i+1,y (0) i+1 [ ) ] f(t i,y i )+f(t i+1,y (1) i+1 ) In general, y (k) i+1 = y i + h [f(t i,y i )+f(t i+1,y (k 1) i+1 )], k = 1,,3,... We then have a second-order predictor-corrector method:
28 518 CHAPTER 11. NUMERICAL DIFFERENTIAL EQUATIONS: IVP Algorithm 11. (Euler-Trapezoidal Predictor-Corrector Method). Input: (i) y = f(t,y), y(t 0 ) = α (ii) Step-size h (iii) Error tolerance ǫ (iv) Points t i+1 = t i + h, i = 0,1,,N 1 at which approximations to y(t i ) are sought. Output:Approximations y (k) i+1, k = 1,, of y(t i+1) for a fixed i = 0,1,,,N 1. For i = 0,1,,,N 1 do Step 1.(Predict): Compute y (0) i+1 using y(0) i+1 = y i +hf(t i,y i ). Step.(Correct): For k = 1,, do End Step.1.Compute y (k) i+1 : y(k) i+1 = y i + h [f(t i,y i )+f(t i+1,y (k 1) i+1 )] Step..Stop when the relative change is less than ǫ: y (k) i+1 y(k 1) i+1 < ǫ. y (k) i+1 Step.3.Accept the current value of y (k) i+1 as y i+1. Example 11.3 Given the Initial Value Problem: y = t+y, y(0) = 1 h = 0. Compute an approximation of y(0.), using Euler-Trapezoidal Predictor-Corrector Method. f(t,y) = t+y y 0 = y(0) = 1 Input Data: h = 0. N = 1 Analytical Solution: y = t +t+1 Solution i = 0.
29 11.7. PREDICTOR-CORRECTOR METHODS 519 Step 1. Predict y 0 1 using formula in Step 1. y (0) 1 = y 0 +hf(t 0,y 0 ) = 1+0.(t 0,y 0 ) = 1+0.(1) = 1+0. = 1. Step. Correct y (0) 1 using formula in Step.1. k = 0: y (1) 1 = y 0 + h [f(t 0,y 0 )+f(t 1,y (0) 1 )] = [(t 0 +y 0 )+(t 1 +y (0) 1 )] = [1+(0.+1.)] = 1. k = 1: y () 1 = y 0 + h [f(t 0,y 0 )+f(t 1,y (1) 1 )] = [ ] = 1. y () 1 is accepted as y 1, the approximate value of y(0.). Error y(0.1) y 1 = = When does the iteration in.1 converge? It can be shown [Exercise] that if f(t,y) and f y will converge if h is chosen so that h <. f y are continuous on [a,b], then the iteration Convergence of the Iteration of Example 11.3: Note that for the above example, f y = 1, so the iteration will converge if h <. Since we had h = 0. <, the iteration converged after 1 iteration Higher-order Predictor-corrector Methods Higher-order predictor corrector methods can be developed by combining an explicit multistep method (corrector) with an implicit multistep method (predictor).
30 50 CHAPTER 11. NUMERICAL DIFFERENTIAL EQUATIONS: IVP For example, when a Four-Step explicit Adams-Bashforth formula is combined with a Three-Step implicit Adams-Moulton formula, the result is Adams-Bashforth-Moulton predictor-corrector method. Here: Predictor: Four-Step Adams-Bashforth Formula: y i+1 = y i + h 4 (55f i 59f i 1 +37f i 9f i 3 ) Corrector: Four-Step Adams-Moulton Formula: y i+1 = y i + h 4 (9f i+1 +19f i 5f i 1 + f i )
31 11.7. PREDICTOR-CORRECTOR METHODS 51 Algorithm 11.4 (Adams-Bashforth-Moulton Predictor-Corrector Method). Inputs: (i) f(t, y) - the given function (ii) h - the step size (iii) f(t 0,y(t 0 )), f(t 1,y(t 1 )), f(t,y(t )),f(t 3,y(t 3 )) the values of f(t,y) at four given points t 0, t 1, t, and t 3. (iv) ǫ - Error tolerance (v) points t i+1 = t i +h, i = 0,1,,...,N 1 at which approximations to y(t i+1 ) are sought. Outputs:Approximations of y (k) i+1, k = 1,,3,..., for fixed i = 0,1,,...,N 1. For i = 0,1,,...,N 1 do Step 1.Compute y (0) i+1, using explicit Four-Step Adams-Bashforth Formula: y (0) i+1 = y i + h 4 [55f(t i,y(t i )) 59f(t i 1,y(t i 1 )) +37f(t i,y(t i )) 9f(t i 3,y(t i 3 ))] Step.Predict y (1) i+1,y() i+1... using implicit Adams-Moulton formula. For k = 1,,..., do y (k) i+1 = y i + h 4 [9f(t i+1,y (k 1) i+1 +19f(t i,y(t i )) 5f(t i 1,y(t i 1 )+f(t i,y(t i )] Stop if y(k) i+1 y(k 1) i+1 y (k) i+1 < ǫ Accept the current value of y (k) i+1 as y i+1. End End Milne s Predictor-Corrector Method The well-known Milne s predictor-corrector method is obtained by using the corrector formula based on Simpson s rule of integration and the following formula as the predictor.
32 5 CHAPTER 11. NUMERICAL DIFFERENTIAL EQUATIONS: IVP Predictor: y (0) i+1 = y i 3 + 4h 3 (f i f i 1 +f i ) Corrector: y (1) i+1 = y i + h 3 (f(0) i+1 +4f i +f i 1 ) where f (0) i+1 = f(t i+1,y (0) i+1 ), and f k = f(t k,y k ), k = i+1, i, and i 1. Error Term: The error term for the predictor: 8 90 h5 y v (ξ) and that for the corrector: 1 90 h5 y v (η) where t i < ξ < t i, and t i+1 < η < t i 1. Both errors are of the same order, but the corrector has the lower constant term. EXAMPLES TO BE INSERTED 11.8 Systems of Differential Equations So far we have considered solution of a single first-order differential equation of the form: y f(t,y), given y(t 0 ) = α. However, many applications give rise to systems of differential equations. A system of n first order differential equations has the form: y 1 = f 1(t,y 1,y,,y n ) y = f (t,y 1,y,,y n ). y n = f n (t,y 1,y,,y n ) The numerical methods that we have discussed so far for a single equation can be applied to the system of equations as well. For the purpose of illustration, let s consider a system of two equations only, written as dy dt = f 1(t,y,z) dz dt = f (t,y,z), and suppose that Euler s method is applied to solve them. Then given the initial values y(t 0 ) = y(a) = α and z(t 0 ) = z(a) = β
33 11.8. SYSTEMS OF DIFFERENTIAL EQUATIONS 53 we can obtain successive approximations y 1,y,,y N to y(t 1 ), y(t ),, y(t N ) and those z 1, z,, z N to z(t 1 ), z(t ),, z(t N ), as follows: Euler s Method for a System of Two Equations for i = 0,1,,N 1. y i+1 = y i +hf 1 (t i,y i,z i ) z i+1 = z i +hf (t i,y i,z i ) Given Example 11.5 dy dt dz dt = y +3z, y(0) = 1 = y +z, z(0) = 1, h = 0.1 Find an approximation of (y(0.1), z(0.1)) uisng Euler s method. Exact Solution: y(t) = e t z(t) = e t Input Data: (i) f 1 (t,y,z) = y +3z (ii) f (t,y,z) = y +z (iii) Initial Values: y 0 = y(0) = 1; z 0 = z(0) = 1 (iv) Step size: h = 0.1
34 54 CHAPTER 11. NUMERICAL DIFFERENTIAL EQUATIONS: IVP i = 0 y 1 = y 0 +hf 1 (t 0,y 0,z 0 ) = 1+0.1(y 0 +3z 0 ) = 1+0.1( 3) = 1. = (Approximate value of y(0.1)) z 1 = z 0 +hf (t 0,y 0,z 0 ) = 1+0.1(y 0 +z 0 ) = 1+0.1( 1) = = (Approximate value of z(0.1)) Exact solution (correct up to four significant digits): ( y(0.1) = e 0.1 = z(0.1) = e 0.1 = ).
35 11.8. SYSTEMS OF DIFFERENTIAL EQUATIONS 55 Similarly, if Runge-Kutta method of order 4 is applied to the above two equations we have: Runge-Kutta Method of Order 4 for a System of Two Equations Input: (i) Two functions f 1 (t,y,z) and f (t,y,z). (ii) Step size h (iii) Initial values y(t 0 ) = α, and z(t 0 ) = β. (iv) points t 1, t,, t N 1, t N where the approximations are sought Output:Approximations to to y(t 1 ),, y(t N ) and z(t 1 ),, z(t N ). For i = 0,1,,N 1 do Step 1. Compute the following Runge-Kutta Coefficients: k 1 = hf 1 (t i,y i,z i ) l 1 = hf (t i,y i,z i ) k = hf 1 (t i + h,y i + k 1,z i + l 1 ) l = hf (t i + h,y i + k 1,z i + l 1 ) k 3 = hf 1 (t i + h,y i + k,z i + l ) l 3 = hf (t i + h,y i + k,z i + l ) k 4 = hf 1 (t i +h,y i +k 3,z i +l 3 ) l 4 = hf (t i +h,y i +k 3,z i +l 3 ). Step. Compute y i+1 = y i (k 1 +k +k 3 +k 4 ) z i+1 = z i (l 1 +l +l 3 +l 4 ). End Example will be inserted. More: Aspects of Stability, stiffness, etc. Much more to come
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