Definition 51Example 151 The Elementary Theory of Initial-Value Problems 261 A function f (t y) is said to satisfy a Lipschitz condition in the variable y on a set D sub R2if a constant L gt 0 exists with|f(ty1)minusf(ty2)|leL|y1 minusy2|whenever (t y1) and (t y2) are in D The constant L is called a Lipschitz constant for f 52 A set D sub R2 is said to be convex if whenever (t1y1) and (t2y2) belong to D then((1minusλ)t1 +λt2(1minusλ)y1 +λy2)also belongs to D for every λ in m[01]53 Suppose f (t y) is defined on a convex set D sub R2 If a constant L gt 0 exists with |part f party (ty)| le L for all (ty) isin D (51) then f satisfies a Lipschitz condition on D in the variable y with Lipschitz constant LDEFINITION 54 Suppose that D=(ty)|aletleb andminusinfinltyltinfinand that f(ty) is continuouson D If f satisfies a Lipschitz condition on D in the variable y then the initial-value problemyprime(t)=f(ty) aletleb y(a)=α has a unique solution y(t) for a le t le b

DEFINITION 55 The initial-value problemdy

=f(ty) aletleb y(a)=α (52)dtis said to be a well-posed problem ifbull A unique solution y(t) to the problem exists andbullThere exist constants ε0 gt0 and kgt0 such that for any εwith ε0 gtε gt0 whenever δ(t) is continuous with |δ(t)| lt ε for all t in [a b] and when |δ0| lt ε the initial-value problemdz

=f(tz)+δ(t) aletleb z(a)=α+δ0 (53) dthas a unique solution z(t) that satisfies|z(t)minusy(t)|ltkε foralltin[ab]Theorem 56 SupposeD=(ty)|aletlebandminusinfinltyltinfinIf f is continuous and satisfies a Lipschitz condition in the variable y on the set D then the initial-value problemDydt =f(ty) alt=tlt=b y(a) =α is well-posed52 Eulerrsquos Methodti =a+ih foreachi=012Nh = (b minus a)Ndydt = f(ty) aletleb y(a)=α (56)y(ti+1) = y(ti) + hf (ti y(ti)) +h22

yrsquorsquo (ξi) (57)

w0 = αwi+1 =wi +hf(tiwi) for eachi=01Nminus1 (58)Lemma 57 For all xgeminus1 and any positive mwe have 0le(1+x)m leemx

Lemma 58 If s and t are positive real numbers aiki=0 is a sequence satisfying a0 ge minusts and ai+1 le(1+s)ai +t for eachi=012kminus1ai+1lt=e(i+1)s (a0+ts) ndashtsTheorem 59 Suppose f is continuous and satisfies a Lipschitz condition with constant L on D=(ty)|aletleband minusinfinltyltinfinand that a constant M exists with|yprimeprime(t)| le M for all t isin [ab]where y(t) denotes the unique solution to the initial-value problem yprime =f(ty) aletleb y(a)=αLetw0w1wN be the approximations generated by Eulerrsquos method for some positive integer N Then for each i = 012N|y(ti)-wi|lthM2L [e^(L(ti-a) -1]

y (t)= dyrsquo

dt (t)= df

dt (ty(t))= df

part t(ty(t)) + partfparty(ty(t))middotf(ty(t))

53 Higher-Order Taylor MethodsDefinition 511 The difference method has local truncation error

τi+1(h) = (yi+1 minus (yi + hφ(ti yi)) )h

= (yi+1 minus yi )h

minus φ(ti yi)Taylor method of order nw0 = αwi+1 =wi +hT(n)(tiwi) foreach i=01Nminus1T(n)(tiwi) = f(tiwi) +h2frsquo(tiwi) +hellip+hn-1n f(n-1)(tiwi)Theorem 512If Taylorrsquos method of order n is used to approximate the solution to yprime(t)=f(ty(t)) aletleb y(a)=αwith step size h and if y isin Cn+1[a b] then the local truncation error is O(hn)The Taylor methods outlined in the previous section have the desirable property of high- order local truncation error but the disadvantage of requiring the computation and evaluation of the derivatives of f (t y) This is a complicated and time-consuming procedure for most problems so the Taylor methods are seldom used in practiceRunge-Kutta MethodsRunge-Kutta methods have the high-order local truncation error of the Taylor methodsbut eliminate the need to compute and evaluate the derivatives of f (t y) Before presenting the ideas behind their derivation we need to consider Taylorrsquos Theorem in two variables The proof of this result can be found in any standard book on advanced calculus (see for exa

mple Midpoint Methodw0 = αwi+1=wi+hf( ti+h2wi+h2f(tiwi) fori=01Nminus1

Modified Euler Methodw0 = αwi+1=wi+h2 x [f(tiwi)+f(ti+1wi+hf(tiwi))] for i=01Nminus1Runge-Kutta Order FourThe most common O(h3) is Heunrsquos method given byw0 = α

w0 = αk1 =hf(tiwi) k2=hf( ti+h2wi+12k1) k3=hf( ti+h2wi+12k2) k4 =hf(ti+1wi +k3)wi+1 = wi +16(k1+ 2k2 + 2k3 + k4)for each i = 0 1 N minus 1 This method has local truncation error O(h4 ) provided the solution y(t) has five continuous derivatives We introduce the notation k1 k2 k3 k4 into the method is to eliminate the need for successive nesting in the second variable of f (t y) Exercise 32 shows how complicated this nesting becomesTheRunge-Kutta method of order four requires four evaluations per stepwhereas Eulerrsquos method requires only one evaluation Hence if the Runge-Kutta method of order four is to be superior it should give more accurate answers than Eulerrsquos method with one-fourth the step size Similarly if the Runge-Kutta method of order four is to be superior to the second-order Runge-Kutta methods which require two evaluations per step it should give more accuracy with step size h than a second-order method with step size h255 Error Control and the Runge-Kutta-Fehlberg MethodThey incorporate in the step-size procedure an estimate of the truncation error that does not require the approximation of the higher derivatives of the function These methods are called adaptive because they adapt the number and position of the nodes used in the approximation to ensure that the truncation error is kept within a specified bound

Runge-Kutta-Fehlberg Method p314An advantage to this method is that only six evaluations of f are required per step Arbitrary Runge-Kutta methods of orders four and five used together (see Table 59 on page 290) require at least four evaluations of f for the fourth-order method and an additional six for the fifth-order method for a total of at least ten function evaluations So the Runge-Kutta- Fehlberg method has at least a 40 decrease in the number of function evaluations over the use of a pair of arbitrary fourth- and fifth-order methodsMultistep Methods p320

Predictor-Corrector Methods p310Consider the following fourth-order method for solving an initial-value problem Thefirst step is to calculate the starting values w0 w1 w2 and w3 for the four-step explicit Adams-Bashforth method To do this we use a fourth-order one-step method the Runge- Kutta method of order four The next step is to calculate an approximation w4p to y(t4) using the explicit Adams-Bashforth method as predictor57 Variable Step-Size Multistep Methods p31558 Extrapolation Methods p321wi+1 = wiminus1 + 2hf (ti wi) for i ge 1 (543)One starting value is the initial condition for w0 = y(a) = α To determine the second starting value w1 we apply Eulerrsquos method59 Higher-Order Equations and Systems of Differential Equations p352 A one-step difference-equation method is said to be convergent with respect to

the differ- ential equation it approximates if

A one-step difference-equation method is said to be convergent with

respect to the differ- ential equation it approximates if

Let p be a solution of the equation x = g(x) Suppose that gprime( p) = 0 and gprimeprime is continuous with |gprimeprime(x)| lt M on an open interval I containing p Then there exists a δ gt 0 such that forp0 isin[pminusδp+δ]thesequencedefinedbypn =g(pnminus1)whennge1convergesat least quadratically to p Moreover for sufficiently large values of n|pn+1 minusp|ltM2x|pn-p|2Definition 210 Asolutionpoff(x)=0isazeroofmultiplicitymoffifforx=pwecanwrite f (x) = (x minus p)mq(x) where limxrarrp q(x) = 0The function f isin C1[ab] has a simple zero at p in (ab) if and only if f(p) = 0 butfprime(p) = 0 P100The function f isin Cm[a b] has a zero of multiplicity m at p in (a b) if and only if0 = f(p) = fprime(p) = fprimeprime(p) = middotmiddotmiddot = f(mminus1)(p) but f(m)(p) = 0

g(x) = x minusf(x)fprime(x)

[fprime(x)]2 minusf(x)fprimeprime(x)Aitkenrsquos 11130892 methodpˆ n = p n minus(pn+1 minuspn)2(pn+2-2pn+1 +pn)Nevillersquos MethodA practical difficulty with Lagrange interpolation is that the error term is difficult to apply so the degree of the polynomial needed for the desired accuracy is generally not known until computations have been performed A common practice is to compute the results given from various polynomials until appropriate agreement is obtained as was done in the previous Illustration However the work done in calculating the approximation by the second polynomial does not lessen the work needed to calculate the third approximation nor is the fourth approximation easier to obtain once the third approximation is known and so on We will now derive these approximating polynomials in a manner that uses the previous calculations to greater advantage

Predictor-Corrector Methods p310Consider the following fourth-order method for solving an initial-value problem Thefirst step is to calculate the starting values w0 w1 w2 and w3 for the four-step explicit Adams-Bashforth method To do this we use a fourth-order one-step method the Runge- Kutta method of order four The next step is to calculate an approximation w4p to y(t4) using the explicit Adams-Bashforth method as predictor57 Variable Step-Size Multistep Methods p31558 Extrapolation Methods p321wi+1 = wiminus1 + 2hf (ti wi) for i ge 1 (543)One starting value is the initial condition for w0 = y(a) = α To determine the second starting value w1 we apply Eulerrsquos method59 Higher-Order Equations and Systems of Differential Equations p352 A one-step difference-equation method is said to be convergent with respect to

the differ- ential equation it approximates if

A one-step difference-equation method is said to be convergent with

respect to the differ- ential equation it approximates if

Let p be a solution of the equation x = g(x) Suppose that gprime( p) = 0 and gprimeprime is continuous with |gprimeprime(x)| lt M on an open interval I containing p Then there exists a δ gt 0 such that forp0 isin[pminusδp+δ]thesequencedefinedbypn =g(pnminus1)whennge1convergesat least quadratically to p Moreover for sufficiently large values of n|pn+1 minusp|ltM2x|pn-p|2Definition 210 Asolutionpoff(x)=0isazeroofmultiplicitymoffifforx=pwecanwrite f (x) = (x minus p)mq(x) where limxrarrp q(x) = 0The function f isin C1[ab] has a simple zero at p in (ab) if and only if f(p) = 0 butfprime(p) = 0 P100The function f isin Cm[a b] has a zero of multiplicity m at p in (a b) if and only if0 = f(p) = fprime(p) = fprimeprime(p) = middotmiddotmiddot = f(mminus1)(p) but f(m)(p) = 0

g(x) = x minusf(x)fprime(x)

[fprime(x)]2 minusf(x)fprimeprime(x)Aitkenrsquos 11130892 methodpˆ n = p n minus(pn+1 minuspn)2(pn+2-2pn+1 +pn)Nevillersquos MethodA practical difficulty with Lagrange interpolation is that the error term is difficult to apply so the degree of the polynomial needed for the desired accuracy is generally not known until computations have been performed A common practice is to compute the results given from various polynomials until appropriate agreement is obtained as was done in the previous Illustration However the work done in calculating the approximation by the second polynomial does not lessen the work needed to calculate the third approximation nor is the fourth approximation easier to obtain once the third approximation is known and so on We will now derive these approximating polynomials in a manner that uses the previous calculations to greater advantage