Ordinary Differential Equations n

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1 Numerical Analsis MTH63 Ordinar Differential Equations Introduction Talor Series Euler Method Runge-Kutta Method Predictor Corrector Method Introduction Man problems in science and engineering when formulated mathematicall are readil expressed in terms of ordinar differential equations (ODE with initial and boundar condition. The trajector of a ballistic missile, the motion of an artificial satellite in its orbit, are governed b ordinar differential equations. Theories concerning electrical networks, bending of beams, stabilit of aircraft, etc., are modeled b differential equations. To be more precise, the rate of change of an quantit with respect to another can be modeled b an ODE Closed form solutions ma not be possible to obtain, for ever modeled problem, while numerical methods exist, to solve them using computers. In general, a linear or non-linear ordinar differential equation can be written as n n d d d = f t,,,, n n dt dt dt Here we shall focus on a sstem of first order d differential equations of the form f (, t dt = with the initial condition (t =, which is called an initial value problem (IVP. It is justified, in view of the fact that an higher order ODE can be reduced to a sstem of first order differential equations b substitution. For example, consider a second order differential equation of the form = f(, t, Introducing the substitution p =, the above equation reduces to a sstem of two first order differential equations, such as Theorem = p, p = f( t,, p Let f (t, be real and continuous in the strip R, defined b [ ] t t, T, Then for an t [ t, T] and for an,, there exists a constant L, satisfing the inequalit f (, t f(, t L so that f (, t L, for ever t, R Here, L is called Lipschitz constant. Copright Virtual Universit of Pakistan

2 Numerical Analsis MTH63 If the above conditions are satisfied, then for an, the IVP has a unique solution ( t, for t [ t T ], In fact, we assume the existence and uniqueness of the solution to the above IVP The function ma be linear or non-linear. We also assume that the function f (t, is sufficientl differentiable with respect to either t or. TAYLOR S SERIES METHOD Consider an initial value problem described b d f (, t, ( t dt = = Here, we assume that f (t, is sufficientl differentiable with respect to x and. If (t is the exact solution, we can expand (t b Talor s series about the point t = t and obtain 3 4 ( t t ( t t ( t t IV t = t ( + ( t t ( t + ( t + ( t + ( t +! 3! 4! Since, the solution is not known, the derivatives in the above expansion are not known explicitl. However, f is assumed to be sufficientl differentiable and therefore, the derivatives can be obtained directl from the given differential equation. Noting that f is an implicit function of, we have = f(, t f f d = + = fx + ff x dx Similarl = fxx + ffx + f ( fx + ff + f ( fx + ff = fxx + ffx + f f + f ( fx + ff IV = fxxx + 3ffxxx + 3f fx + f( fxx + ffx + f f + 3( fx + ff ( ffx + ff + f ( fx + ff Continuing in this manner, we can express an derivative of in terms of f (t, and its partial derivatives. Using Talor s series method, find the solution of the initial value problem d t, ( dt = + = at t =., with h =. and compare the result with the closed form solution Solution Let us compute the first few derivatives from the given differential equation as follows: Copright Virtual Universit of Pakistan

3 Numerical Analsis MTH63 = t+, = +, =, IV V IV =, = Prescribing the initial condition, that is, at t =, (t =, we have =, =, IV V = = = Now, using Talor s series method, we have 3 ( t t ( t t t ( = + ( t t ( t t IV ( t t V Substituting the above values of the derivatives, and the initial condition, we obtain... (. = + (.( + ( + ( + ( = = =.344 Therefore (. = = Taking =.3 at t =., the values of the derivatives are =.+.3 =.3 = +.3 =.3 IV V = = =.3 Substituting the value of and its derivatives into Talor s series expansion we get, after retaining terms up to fifth derivative onl ( t t (. = + ( t t ( t t ( t t IV ( t t V (. = = To obtain the closed form solution, we rewrite the given IVP as d t dt = or ( d e t = te t On integration, we get Copright Virtual Universit of Pakistan 3

4 Numerical Analsis MTH63 = e ( te + e + ce t t t t t = ce t Using the initial condition, we get Therefore, the closed form solution is = t + e t When t =., the closed form solution becomes (. =. + ( = Using Talor s Series method taking algorithm of order 3, solve the initial value problem = ; ( = with h=.5 at x=.5 Solution: Copright Virtual Universit of Pakistan 4

5 Numerical Analsis MTH63 Let us com pute the first three derivatives = = = The initial condition is t = =, we have = = = = = = = N ow, Talor ' s series m ethosd a lg rithm is ( t t ( t t 3 t = + t t + + 3! 3 OR ( t = + h + h + h h =.5 3! 3 (.5 (.5 (.5 = ! = = =.35 Taking =.35, Now = =.35 =.7786 = = = =.7786 =.7786 ( t t ( t t 3 t = + t t + + 3! 3 OR ( t = + h + h + h h =.5 6 = = (.5(.7786 (.5 (.7786 (.5 (.7786 = Copright Virtual Universit of Pakistan 5