Lectures in Numerical Analysis：在数值分析的讲座.doc

alwahip 2011Lectures in Numerical Analysis MethodsBushra H. AliwiDepartment of MathematicsBABYLON UNIVERSITYbushra_Course for year 2011Solution of Ordinary Differential Equations: Some equations has an analytical solution,while other has not such ;There are two types of conditions for the ordinary differential equations which are ; Initial Value Problem:Conditions are specified at only one value of the independent variable Analytically can solve ; with form; where and Both initial conditions have been specified at . An ordinary differential equation of order required conditions to be specified . Boundary Value Problem:Conditions are specified at two values of independent variable, such form ; where , Or where , , , and Methods of Solution: 1. Direct method using Taylar Series .2. One step method (self starting method):- solution is carried from to .3. Multistep method ;required information for . One Step Method (Self Starting Method):- I. Euler Method ; Consider the 1st order initial value problem; , replacing by forward difference; where then ; ,or can written as; we can use only two terms from Tayler Series II. Modified Euler Method ; The accuracy of method (I) can be implemented if better approximation is used for the derivative ; In each step we first calculate (by Euler Method); And then find new value as; ,or formed as ; This method is also known as Predictor_Correction Method ,because in each step we first calculate (by Euler Method) and then correct it .Same expression can obtained by using Tayler Series if 2nd derivative is approximated by forward differences ; where so ; III. Runge_Kutta Method ; The 4th order formula given as; Where ; Euler and Modified Euler are in reality 1st and 2nd order Runge_Kutta Method .The method requires four evaluation off to get more accurate results and the step size should be sufficiently small Example: Solve the differential equation ; , , by ; (1) Euler Method (2) Modified Euler Method (3) Ruge_Kutta Method ,also find on using Solution: (1) Euler Method; (2) Modified Euler Method ; (3) Ruge_Kutta Method; So the value of through the law calculated as; Then the table of calculating for in closed interval through this three methods as;byEulerby Modified Eulerby Runge_KuttaExact valuefor 011211.22211.22210.21.4421.49231.49771.49770.31.73841.82841.84321.8432Exercises:1. Solve to find