Numerical Methods for Ordinary Dierential Equations Episode 6 ppsx

RUNGE–KUTTA METHODS 159 The resulting sum is the value of Φ(t). A similar formula for Φ i (t),wherei is not a member of V , is found by replacing (312e) by a ij  (k,l)∈E a kl (312f) and summing this as for Φ(t). Note that, although c does explicitly appear in Definition 312A or Lemma 312B, it is usually convenient to carry out the summations  s l=1 a kl to yield aresultc k if l denotes a leaf (terminal vertex) of V . This is possible because l occurs only once in (312e) and (312f). We illustrate the relationship between the trees and the corresponding elementary weights in Table 312(I). For each of the four trees, we write Φ(t) in the form given directly by Lemma 312B, and also with the summation over leaves explicitly carried out. Finally, we present in Table 312(II) the elementary weights up to order 5. 313 The Taylor expansion of the approximate solution We show that the result output by a Runge–Kutta methods is exactly the same as (311d), except that the factor γ(t) −1 is replaced by Φ(t). We first establish a preliminary result. Lemma 313A Let k =1, 2, ,.If Y i = y 0 +  r(t)≤k−1 1 σ(t) Φ i (t)h r(t) F (t)(y 0 )+O(h k ), (313a) then hf(Y i )=  r(t)≤k 1 σ(t) (Φ i D)(t)h r(t) F (t)(y 0 )+O(h k+1 ). (313b) Proof. Use Lemma 310B. The coefficient of σ(t) −1 F (t)(y 0 )h r(t) in hf(Y i )is  n j=1 Φ i (t j ), where t =[t 1 t 2 ···t k ].  We are now in a position to derive the formal Taylor expansion for the computed solution. The proof we give for this result is for a general Runge– Kutta method that may be implicit. In the case of an explicit method, the iterations used in the proof can be replaced by a sequence of expansions for Y 1 ,forhf (Y 1 ), for Y 2 ,forhf (Y 2 ),andsoonuntilwereachY s , hf(Y s )and finally y 1 . 160 NUMERICAL METHODS FOR ORDINARY DIFFERENTIAL EQUATIONS Theorem 313B The Taylor expansions for the stages, stage derivatives and output result for a Runge–Kutta method are Y i = y 0 +  r(t)≤n 1 σ(t) Φ i (t)h r(t) F (t)(y 0 )+O(h n+1 ),i=1, 2, ,s, (313c) hf(Y i )=  r(t)≤n 1 σ(t) (Φ i D)(t)h r(t) F (t)(y 0 )+O(h n+1 ),i=1, 2, ,s,(313d) y 1 = y 0 +  r(t)≤n 1 σ(t) Φ(t)h r(t) F (t)(y 0 )+O(h n+1 ). (313e) Proof. In a preliminary part of the proof, we consider the sequence of approximations to Y i given by Y [0] i = y 0 ,i=1, 2, ,s, (313f) Y [k] i = y 0 + h s  j=1 a ij f  Y [k−1] j  ,i=1, 2, ,s. (313g) We prove by induction that Y [n] i agrees with the expression given for Y i to within O(h n+1 ). For n = 0 this is clear. For n>0, suppose it has been proved for n replaced by n − 1. From Lemma 313A with k = n − 1andY i replaced by Y [n−1] i ,weseethat hf(Y [n−1] i )=  r(t)≤n 1 σ(t) (Φ i D)(t)h r(t) F (t)(y 0 )+O(h n+1 ),i=1, 2, ,s. Calculate Y [n] i using (313c) and the preliminary result follows. Assume that h is sufficiently small to guarantee convergence of the sequence (Y [0] i ,Y [1] i ,Y [2] i , )toY i and (313c) follows. Finally, (313d) follows from Lemma 313A and (313e) from (312d).  314 Independence of the elementary differentials Our aim of comparing the Taylor expansions of the exact and computed solutions to an initial value problem will give an inconclusive answer unless the terms involving the various elementary differentials can be regarded as independent. We introduce a special type of differential equation for which any finite number of elementary differentials evaluate to independent vectors. Let U denote any finite subset of T , such that if t i =[t m 1 1 ,t m 2 2 , ,t m k k ] ∈ U, (314a) RUNGE–KUTTA METHODS 161 Table 314(I) Trees to order 4 with corresponding differential equations it i y  i = f i 1 [] y  1 =1, 2 [t 1 ] y  2 = y 1 , 3 [t 2 1 ] y  3 = 1 2 y 2 1 , 4 [t 2 ] y  4 = y 2 , 5 [t 3 1 ] y  5 = 1 6 y 3 1 , 6 [t 1 t 2 ] y  6 = y 1 y 2 , 7 [t 3 ] y  7 = y 3 , 8 [t 4 ] y  8 = y 4 . then each of t 1 , t 2 , , t k is also a member of U. For example, U might consist of all trees with orders up to some specified integer. Assume that when we write a tree in this way, the t i , i =1, 2, ,k, are all distinct. Suppose that N is the number of members of U , and consider the m-dimensional differential equation system in which y  i = k  j=1 y m j j m j ! , (314b) corresponding to tree number i defined in (314a). The initial values are supposed to be y i (0) = y i (x 0 )=0,fori =1, 2, ,N. The interesting property of this initial value problem is encapsulated in the following result: Theorem 314A The values of the elementary differentials for the differential equation (314b), evaluated at the initial value, are given by F (t i )(y(x 0 )) = e i ,i=1, 2, ,N. Because the natural basis vectors e 1 ,e 2 , ,e N are independent, there cannot be any linear relation between the elementary differentials for an arbitrary differential equation system. We illustrate this theorem in the case where U consists of the eight trees with up to four vertices. Table 314(I) shows the trees numbered from i =1 to i = 8, together with their recursive definitions in the form (314a) and the corresponding differential equations. Note that the construction given here is given as an exercise in Hairer, Nørsett and Wanner (1993) . 162 NUMERICAL METHODS FOR ORDINARY DIFFERENTIAL EQUATIONS 315 Conditions for order Now that we have expressions for the Taylor expansions of the exact solution, and also of the computed solution, we have all we need to find conditions for order. If the exact solution has Taylor series given by (311d) and the approximate solution has Taylor series given by (313e), then we need only compare these term by term to arrive at the principal result on the order of Runge–Kutta methods. Theorem 315A A Runge–Kutta method with elementary weights Φ:T → R, has order p if and only if Φ(t)= 1 γ(t) , for all t ∈ T such that r(t) ≤ p. (315a) Proof. The coefficient of F(t)(y 0 )h r(t) in (313e) is 1 σ(t) Φ(t), compared with the coefficient in (311d), which is 1 σ(t)γ(t) . Equate these coefficients and we obtain (315a).  316 Order conditions for scalar problems Early studies of Runge–Kutta methods were built around the single scalar equation y  (x)=f(x, y(x)). (316a) Even though it was always intended that methods derived for (316a) should be interpreted, where appropriate, in a vector setting, a subtle difficulty arises for orders greater than 4. We adopt the notation f x , f y for partial derivatives of f with respect to the first and second arguments, with similar notations for higher derivatives. Also, for simplicity, we omit the arguments in expressions like f x (x, y). By straightforward differentiation of (316a), we have y  = f x + f y y  = f x + f y f, where the two terms together correspond to the elementary differential associated with t = . Similarly, for the third derivative we have y  =  f xx +2f xy f + f yy f 2  +  f y (f x + f y f)  , where the grouped terms correspond to t = and t = , respectively. RUNGE–KUTTA METHODS 163 The expressions that arise here, and for the fourth derivative, are more complicated, because of the presence of derivatives with respect to x. However, the terms can be grouped together according to the elementary differentials to which they correspond. Furthermore, the order conditions are identical to those found in the general vector case. When similar expressions are worked out for the 17 elementary differentials of order 5, we find a confusion between the results for two particular trees. In fact for each of t 1 = and t 2 = , F (t) reduces to F (t)=f y (f yy f + f xy )(f y f + f x ), and instead of two order conditions Φ(t 1 )=  b i c i a ij a jk c k = 1 γ(t 1 ) = 1 30 and Φ(t 2 )=  b i a ij c j a jk c k = 1 γ(t 2 ) = 1 40 , (316b) we have the single condition  b i c i a ij a jk c k +  b i a ij c j a jk c k = 7 120 . We discuss in Subsection 325 the construction of fifth order methods. These usually satisfy the so-called D(1) condition, which we introduce in Subsection 321. This simplifying assumption has, as one of its consequences, the dependence of (316b) on other conditions, for which there is no confusion. Hence, for methods satisfying D(1), scalar and vector order 5 conditions are equivalent. For orders 6 and higher, the confusion between the order conditions for the scalar case becomes more pronounced. The first published methods of this order(Hu ˇ ta, 1956, 1957) were derived for scalar problems but, nevertheless, have order 6 for the general vector case (Butcher, 1963a). 317 Independence of elementary weights We show in Subsection 324 that, given a positive integer p,thereexistsan integer s such that there is a Runge–Kutta method with s stages with order p. We now present a more general result on the independence of the elementary weights but without a specific value of s given. Theorem 317A Given a finite subset T 0 ,ofT and a mapping φ : T 0 → R, there exists a Runge–Kutta method such that the elementary weights satisfy Φ(t)=φ(t), for all t ∈ T 0 . 164 NUMERICAL METHODS FOR ORDINARY DIFFERENTIAL EQUATIONS Proof. Let #T 0 = n. The set of possible values that can be taken by the vector of Φ(t)values,forallt ∈ T 0 , is a vector space. To see why this is the case, consider Runge–Kutta methods given by the tableaux c A b and c A b (317a) with s and s stages, respectively. If the elementary weight functions for these two Runge–Kutta methods are Φ and Φ, then the method given by the tableau c A 0 c 0 A θb θb has elementary weight function θΦ+θΦ. Let V ⊂ R n denote this vector space. We complete the proof by showing that V = R n . If this were not the case, there would exist a non-zero function ψ : T 0 → R such that  t∈T 0 ψ(t)Φ(t)=0, for all Runge–Kutta methods. Because every coefficient in a Runge–Kutta tableau can be multiplied by an arbitrary scalar θ to give a new method for which Φ(t) is replaced by θ r(t) Φ(t), we may assume that every non-zero value of ψ corresponds to trees with the same order k.Thisisimpossiblefork =1, because in this case there is only a single tree τ. Suppose the impossibility of this has been proved for all orders less than k, but that there exist trees t 1 , t 2 , , t m ,eachoforderk, such that  m i=1 ψ(t i )Φ(t i ) = 0, for all Runge– Kutta methods with ψ(t i ) =0,fori =1, 2, ,m.Writet i =[t l i1 i1 t l i2 i2 ···], for i =1, 2, ,m.Let ˆ t denote a tree appearing amongst the t ij which does not occur with the same exponent in each of the t i . Construct an s-stage Runge–Kutta method c A b for which each of Φ(t ij ) = 1, except for Φ( ˆ t)=θ. Define second Runge–Kutta tableau with s + 1 stages of the form c A 0 1 b 0 01 . If q i is the exponent of ˆ t in t i , then it follows that m  i=1 ψ(t i )θ q i =0. Since θ cantakeanyvalueandsinceq i is not constant, it is not possible that ψ is never zero.  RUNGE–KUTTA METHODS 165 318 Local truncation error The conditions for order give guarantees that the Taylor expansions of the exact and computed solutions agree up to terms in h p . Obtaining an understanding of the respective terms in h p+1 is regarded as a key to deriving methods that not only have a specific order, but also have a small truncation error. Because the number of terms of this order rises rapidly as p increases, it is extremely difficult to know how this sort of optimality should be arrived at. Picking out just the terms of order p+1, we can write the local truncation error in a single step as h p+1  r(t)=p+1 1 σ(t)  1 γ(t) −Φ(t)  F (t)(y 0 )+O(h p+2 ). (318a) Since we are interested in asymptotic behaviour, that is, limiting behaviour for h small, we do not devote much attention to the term O(h p+2 ). The coefficient of h p+1 in (318a) is bounded in magnitude by  r(t)=p+1 1 σ(t)     Φ(t) − 1 γ(t)     ·F (t)(y 0 ), (318b) and this should somehow be made small. There is simply no general rule interrelating the magnitudes of the various elementary differentials, and some assumptions need to be made. The first approach that can be considered is to compare, term by term, the expression for 1 (p+1)! y (p+1) (x 0 ), which is proportional to the local truncation error coefficient for linear multistep methods or for implicit Runge–Kutta methods of collocation type. The coefficient in this expression, corresponding to t,is 1 σ(t)γ(t) , so that the corresponding multiplier to yield the corresponding term in (318b) is |γ(t)Φ(t) − 1|. Hence, we can bound (318b) by max r(t)=p+1 |γ(t)Φ(t) − 1|  r(t)=p+1 1 σ(t)γ(t) ·F (t)(y 0 ) and hence, it might be desirable to minimize max r(t)=p+1 |γ(t)Φ(t) − 1| in seeking an efficient method. 166 NUMERICAL METHODS FOR ORDINARY DIFFERENTIAL EQUATIONS Another approach would be to assume a bound M on f, a bound L on the linear operator f  , and further bounds to make up the sequence f≤M, f  ≤L, f  ≤ L 2 M , f  ≤ L 3 M 2 , . . . . . . f (p) ≤ L p M p−1 . This will mean that for any tree of order p +1,F (t)(y 0 )≤L p M and that  r(t)=p+1 1 σ(t)     Φ(t) − 1 γ(t)     ·F (t)(y 0 )≤  r(t)=p+1 1 σ(t)     Φ(t) − 1 γ(t)     · L p M. In studying the behaviour of a particular method of order p when used to solve a particular initial value problem, we wish to assume that the local truncation error is bounded asymptotically by some constant multiplied by h p+1 . This assumption will hinge on smoothness of the solution and the differentiability, sufficiently many times, of f. 319 Global truncation error We consider the cumulative effect of errors in many steps leading to an error in a final output point. Suppose that n steps are performed to carry the solution from an initial point x 0 to a final point x. If a constant stepsize is used, this would need to be equal to ( x −x 0 )/n to exactly reach the final point. Denote the approximations computed by a Runge–Kutta method by y 1 , y 2 , , y n , with y 0 = y(x 0 ). If the error committed in each of the n steps is bounded by Ch p+1 then the total contribution to the error would seem to be nCh p+1 = C(x − x 0 )h p . We attempt to make this argument more precise by noting that an error in the initial value input to a step will lead to an error in the output value consisting of two terms. The first of these is the perturbation to the output due to the error in the input, and the second is the truncation error due to the method itself. In the statement of a preliminary lemma that we need, |A| and |b | will denote the matrix A and the vector b , respectively, with every term replaced by its magnitude. RUNGE–KUTTA METHODS 167 δ 1 δ 2 δ 3 δ n−2 δ n−1 δ n ∆ 1 ∆ 2 ∆ 3 ∆ n−2 ∆ n−1 x 0 x 1 x 2 x 3 x n−2 x n−1 x n y 0 y 1 y 2 y 3 y n−2 y n−1 y n y(x 0 ) y(x 1 ) y(x 2 ) y(x 3 ) y(x n−2 ) y(x n−1 ) y(x n ) Figure 319(i) Growth of global errors from local errors referred to the computed solution Lemma 319A Let f denote a function R m → R m ,assumedtosatisfya Lipschitz condition with constant L.Lety 0 ∈ R m and z 0 ∈ R m be two input values to a step with the Runge–Kutta method (A, b ,c), using stepsize h ≤ h 0 , where h 0 Lρ(|A|) < 1,andlety 1 and z 1 be the corresponding output values. Then y 1 − z 1 ≤(1 + hL  )y 0 − z 0 , where L  = L|b |(I −h 0 L|A|) −1 1. Proof. Denote the stage values by Y i and Z i , i =1, 2, ,s, respectively. From the equation Y i −Z i =(y 0 −z 0 )+h  s j=1 a ij (f(Y j ) −f(Z j )), we deduce that Y i − Z i ≤y 0 − z 0  + h 0 L s  j=1 |a ij |Y j − Z j , so that, substituting into y 1 − z 1 ≤y 0 − z 0  + hL s  j=1 |b j |Y j − Z j , we obtain the result.  168 NUMERICAL METHODS FOR ORDINARY DIFFERENTIAL EQUATIONS δ 1 δ 2 δ 3 δ n−2 δ n−1 δ n ∆ 1 ∆ 2 ∆ 3 ∆ n−2 ∆ n−1 x 0 x 1 x 2 x 3 x n−2 x n−1 x n y 0 y 1 y 2 y 3 y n−2 y n−1 y n y(x 0 ) y(x 1 ) y(x 2 ) y(x 3 ) y(x n−2 ) y(x n−1 ) y(x n ) Figure 319(ii) Growth of global errors from local errors referred to the exact solution To see how to use this result, consider Figures 319(i) and 319(ii). Each of these shows the development of global errors generated by local truncation errors in individual steps. In Figure 319(i), the local truncation errors are referred to the computed solution. That is, in this figure, δ k is the difference between the exact solution defined by an initial value at the start of step k and the numerical solution computed in this step. Furthermore, ∆ k is the contribution to the global error resulting from the error δ k in step k.An alternative view of the growth of errors is seen from Figure 319(ii), where δ k is now the difference between the exact solution at x k and the computed solution found by using an input value y k−1 at the start of this step exactly equal to y(x k−1 ). As in the previous figure, ∆ k is the contribution to the global error resulting from the local error δ k . To obtain a bound on the global truncation error we first need an estimate on δ 1 , δ 2 , , δ n using these bounds. We then estimate by how much δ k can grow to ∆ k , k =1, 2, ,n. The global error is then bounded in norm by  n k=1 ∆ k . We have a bound already from (110c) on how much a perturbation in the exact solution can grow. If we were basing our global error bound on Figure 319(i) then this would be exactly what we need. However, we use Figure 319(ii), and in this case we obtain the same growth factor but with L replaced by L  . The advantage of using an argument based on this figure, rather than on Figure 319(i), is that we can then use local truncation error defined in the standard way, by comparing the exact solution at step value x n with the numerically computed result over a single step with initial value y(x n−1 ). [...]... (326h)–(326k) below, where we have listed all the conditions we need to specify a method: b2 = 0, (326a) 7 i=1 bi (1 − ci )(ci − c6 )(ci − c3 )(ci − c4 )ci = 1 c3 + c4 + c6 c3 c4 + c3 c6 + c4 c6 c3 c4 c6 − + − , 30 20 12 6 7 c4 + c6 c4 c6 1 − + , bi (1 − ci )(ci − c6 )(ci − c4 )ci = 20 12 6 i=1 7 bi (1 − ci )(ci − c6 )ci = i=1 c6 1 − , 12 6 (326b) (326c) (326d) RUNGE–KUTTA METHODS 193 7 1 , 6 (326e)... c2 1 2 1 c2 1 − 8c2 1 2c2 − 1 1 2 1 6 1 8c2 1 − 2c2 0 (322f) 2 2 3 ; 1 6 for case III, 0 1 2 0 1 1 2 1 − 12b3 1 − 2 − 6b3 1 6 − b3 1 12b3 3 2 2 3 (322g) 6b3 b3 1 6 ; RUNGE–KUTTA METHODS 179 for case IV, 0 1 1 1 2 3 8 1− 1 1 8 1 − 12b4 1 6 − b4 1 4b4 1 6 (322h) 1 3b4 2 3 ; b4 and for case V, 0 1 2 1 2 1 2 1 2 1 1 − 6b3 0 1 6 1 6b3 (322i) 1 − 3b3 2 3 − b3 3b3 b3 1 6 Some interesting special choices within... (326c) (326d) RUNGE–KUTTA METHODS 193 7 1 , 6 (326e) 1 , 2 (326f) bi = 1, (326g) bi (1 − ci )ci = i=1 7 bi ci = i=1 7 i=1 7 1 c3 − , 60 24 (326h) bi (1 − ci )(ci − c6 )aij (cj − c3 )cj = c3 c6 c3 c6 1 − − + , 90 40 60 24 (326i) bi (1 − ci )aij (cj − c4 )(cj − c3 )cj = c3 + c4 c3 c4 1 − + , 120 60 24 (326j) c3 1 − , 360 120 (326k) 1 2 c , 2 i (326l) bi (1 − ci )aij (cj − c3 )cj = i,j=1 7 i,j=1 7 i,j=1 7... 1, 2, , 7, (326m) j=1 7 bi aij = bi (1 − cj ), j = 1, 2, , 7, (326n) i=1 7 bi (1 − ci )ai2 = 0, (326o) bi (1 − ci )ci ai2 = 0, (326p) bi (1 − ci )aij aj2 = 0 (326q) i=1 7 i=1 7 i,j=1 This rather formidable set of equations can be solved in a systematic and straightforward manner except for one detail: there are three equations, (326i), (326j) and (326k), each involving a54 and a65 and no other... ± 63 , 1}, c3 = 1 − c2 , 2 2 = 0, c2 = 0, c3 = 1 , 2 = 0, c2 = 1 , c3 = 0, 2 = 0, c2 = 1, c3 = 1 , 2 = 0, c2 = c3 = 1 2 The coefficient tableaux are for case I, 0 1−c3 c3 1 1 − c3 c3 (1−2c3 ) 2(1−c3 ) 12c3−24c2+17c3−4 3 3 2(1−c3 )(6c3−1−6c2 ) 3 6c3 −1−6c2 3 12c3 (1−c3 ) c3 2(1−c3 ) c3 (1−2c3 ) 2(1−c3 )(6c3−1−6c2 ) 3 1 12c3 (1−c3 ) (322e) 1−c3 6c3−1−6c2 3 1 12c3 (1−c3 ) 6c3−1−6c2 3 12c3 (1−c3 ) ; for. .. A For i corresponding to a member of row k for k = 1, 2, , m, the only non-zero 190 NUMERICAL METHODS FOR ORDINARY DIFFERENTIAL EQUATIONS aij are for j = 1 and for j corresponding to a member of row k − 1 Thus, the quadrature formula associated with this row has the form ci k−1 φ(x)dx ≈ w0 φ(0) + 0 wj φ(ξj ) j=1 and the coefficients are chosen to make this exact for φ a polynomial of degree k − 1 For. .. 8 0 0 3 16 −3 7 7 90 −3 8 8 7 0 1 2 3 8 6 7 16 45 9 16 − 12 7 2 15 8 7 16 45 7 90 192 NUMERICAL METHODS FOR ORDINARY DIFFERENTIAL EQUATIONS The first methods of this order, derived by Kutta (1901), have a different structure One of these, as corrected by Nystr¨m (1925), is o 0 1 3 2 5 1 2 3 4 5 1 3 4 25 1 4 2 27 2 25 23 192 6 25 −3 10 9 12 25 0 15 4 − 50 81 2 15 125 192 8 81 8 75 0 0 − 27 64 125 192... computed 180 NUMERICAL METHODS FOR ORDINARY DIFFERENTIAL EQUATIONS We also present the choices b3 = 1 12 in case III, 0 1 2 1 2 −1 −1 0 1 1 3 2 2 3 1 12 and b4 = 1 6 1 2 1 12 , 1 6 in case IV, 0 1 1 2 1 1 3 8 −1 2 1 6 1 8 1 −2 2 2 3 0 1 6 Amongst the methods in case V, the ‘classical Runge–Kutta method’ is especially notable The tableau is 0 1 2 1 2 1 2 1 2 0 0 0 1 1 6 1 1 3 1 3 1 6 Also in case... )2  Find formulae for the elementary differentials F (t), for t = [τ ], [τ 2 ] and [τ [τ ]] 31.2 For the Runge–Kutta method 1 3 5 12 3 4 3 4 1 1 − 12 1 4 1 4 find the elementary weights for the eight trees up to order 4 What is the order of this method? 31.3 For an arbitrary Runge–Kutta method, find the order condition corresponding to the tree 170 NUMERICAL METHODS FOR ORDINARY DIFFERENTIAL EQUATIONS. .. Butcher (1995) 3 26 Methods of order 6 The first methods of order 6 were derived by Huˇa (19 56, 1957) Although his t methods used s = 8 stages, it is possible to find methods of this order with s = 7 Just as for order 5, we assume the modified C(2) condition and the D(1) condition We also assume the quadrature conditions so that the only order conditions that remain are Φ(t) = 1/γ(t) for the trees t= , . expansions for Y 1 ,forhf (Y 1 ), for Y 2 ,forhf (Y 2 ),andsoonuntilwereachY s , hf(Y s )and finally y 1 . 160 NUMERICAL METHODS FOR ORDINARY DIFFERENTIAL EQUATIONS Theorem 313B The Taylor expansions for. Φ(t 3 )=1/γ(t 3 ). For explicit methods, D(2) cannot hold, for similar reasons to the impossibility of C(2). For implicit methods D(s) is possible, as we shall see in Section 342. 174 NUMERICAL METHODS FOR ORDINARY. method? 31.3 For an arbitrary Runge–Kutta method, find the order condition corresponding to the tree . 170 NUMERICAL METHODS FOR ORDINARY DIFFERENTIAL EQUATIONS 32 Low Order Explicit Methods 320 Methods