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Fin500J Topic 6Fall 2010 Olin Business School 1 Fin500J: Mathematical Foundations in Finance Topic 6: Ordinary Differential Equations Philip H. Dybvig.

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Presentation on theme: "Fin500J Topic 6Fall 2010 Olin Business School 1 Fin500J: Mathematical Foundations in Finance Topic 6: Ordinary Differential Equations Philip H. Dybvig."— Presentation transcript:

1 Fin500J Topic 6Fall 2010 Olin Business School 1 Fin500J: Mathematical Foundations in Finance Topic 6: Ordinary Differential Equations Philip H. Dybvig Reference: Lecture Notes by Paul Dawkins, 2007, page 20-33, page 102-121, page 137-155 and page 340-344 http://tutorial.math.lamar.edu/terms.aspx Slides designed by Yajun Wang

2 Introduction to Ordinary Differential Equations (ODE) Recall basic definitions of ODE, order linearity initial conditions solution Classify ODE based on( order, linearity, conditions) Classify the solution methods Fin500J Topic 6Fall 2010 Olin Business School 2

3 Derivatives Fin500J Topic 6Fall 2010 Olin Business School Partial Derivatives u is a function of more than one independent variable Ordinary Derivatives y is a function of one independent variable 3

4 Differential Equations Differential Equations Fin500J Topic 6Fall 2010 Olin Business School involve one or more partial derivatives of unknown functions Ordinary Differential Equations involve one or more Ordinary derivatives of unknown functions 4 Partial Differential Equations

5 Ordinary Differential Equations Fin500J Topic 6Fall 2010 Olin Business School Ordinary Differential Equations (ODE) involve one or more ordinary derivatives of unknown functions with respect to one independent variable y(x): unknown function x: independent variable 5

6 Order of a differential equation Fin500J Topic 6Fall 2010 Olin Business School The order of an ordinary differential equations is the order of the highest order derivative Second order ODE First order ODE Second order ODE 6

7 Solution of a differential equation Fin500J Topic 6Fall 2010 Olin Business School A solution to a differential equation is a function that satisfies the equation. 7

8 Linear ODE Fin500J Topic 6Fall 2010 Olin Business School An ODE is linear if the unknown function and its derivatives appear to power one. No product of the unknown function and/or its derivatives Linear ODE Non-linear ODE 8

9 Boundary-Value and Initial value Problems Boundary-Value Problems The auxiliary conditions are not at one point of the independent variable More difficult to solve than initial value problem Fin500J Topic 6Fall 2010 Olin Business School Initial-Value Problems  The auxiliary conditions are at one point of the independent variable same different 9

10 Classification of ODE ODE can be classified in different ways Order First order ODE Second order ODE N th order ODE Linearity Linear ODE Nonlinear ODE Auxiliary conditions Initial value problems Boundary value problems Fin500J Topic 6Fall 2010 Olin Business School 10

11 Solutions Analytical Solutions to ODE are available for linear ODE and special classes of nonlinear differential equations. Numerical method are used to obtain a graph or a table of the unknown function We focus on solving first order linear ODE and second order linear ODE and Euler equation Fin500J Topic 6Fall 2010 Olin Business School 11

12 First Order Linear Differential Equations Def: A first order differential equation is said to be linear if it can be written Fin500J Topic 6Fall 2010 Olin Business School 12

13 First Order Linear Differential Equations How to solve first-order linear ODE ? Sol: Fin500J Topic 6Fall 2010 Olin Business School 13 Multiplying both sides by, called an integrating factor, gives assuming we get

14 First Order Linear Differential Equations Fin500J Topic 6Fall 2010 Olin Business School 14 By product rule, (4) becomes Now, we need to solve from (3)

15 First Order Linear Differential Equations Fin500J Topic 6Fall 2010 Olin Business School 15 to get rid of one constant, we can use

16 Summary of the Solution Process Fin500J Topic 6Fall 2010 Olin Business School 16 Put the differential equation in the form (1) Find the integrating factor, using (8) Multiply both sides of (1) by and write the left side of (1) as Integrate both sides Solve for the solution

17 Example 1 Sol: Fin500J Topic 6Fall 2010 Olin Business School 17

18 Example 2 Sol: Fin500J Topic 6Fall 2010 Olin Business School 18

19 Second Order Linear Differential Equations Homogeneous Second Order Linear Differential Equations o real roots, complex roots and repeated roots Non-homogeneous Second Order Linear Differential Equations o Undetermined Coefficients Method Euler Equations Fin500J Topic 6Fall 2010 Olin Business School 19

20 where a, b and c are constant coefficients Let the dependent variable y be replaced by the sum of the two new variables: y = u + v Therefore If v is a particular solution of the original differential equation The general solution of the linear differential equation will be the sum of a “complementary function” and a “particular solution”. purpose Second Order Linear Differential Equations The general equation can be expressed in the form Fin500J Topic 6Fall 2010 Olin Business School 20

21 Let the solution assumed to be : characteristic equation Real, distinct roots Double roots Complex roots The Complementary Function (solution of the homogeneous equation) Fin500J Topic 6Fall 2010 Olin Business School 21

22 Real, Distinct Roots to Characteristic Equation Let the roots of the characteristic equation be real, distinct and of values r 1 and r 2. Therefore, the solutions of the characteristic equation are: The general solution will be Example Fin500J Topic 6Fall 2010 Olin Business School 22

23 Let where V is a function of x Equal Roots to Characteristic Equation Let the roots of the characteristic equation equal and of value r 1 = r 2 = r. Therefore, the solution of the characteristic equation is: Fin500J Topic 6Fall 2010 Olin Business School 23

24 Complex Roots to Characteristic Equation Let the roots of the characteristic equation be complex in the form r 1,2 =λ±µi. Therefore, the solution of the characteristic equation is: Fin500J Topic 6Fall 2010 Olin Business School 24

25 (I) Solve characteristic equation Examples (II) Solve characteristic equation Fin500J Topic 6Fall 2010 Olin Business School 25

26 When g(x) is a polynomial of the form where all the coefficients are constants. The form of a particular solution is Non-homogeneous Differential Equations (Method of Undetermined Coefficients) When g(x) is constant, say k, a particular solution of equation is Fin500J Topic 6Fall 2010 Olin Business School 26

27 Example Solve equating coefficients of equal powers of x characteristic equation complementary function Fin500J Topic 6Fall 2010 Olin Business School 27

28 Non-homogeneous Differential Equations (Method of Undetermined Coefficients) When g(x) is of the form Te rx, where T and r are constants. The form of a particular solution is When g(x) is of the form Csinnx + Dcosnx, where C and D are constants, the form of a particular solution is Fin500J Topic 6Fall 2010 Olin Business School 28

29 Example Solve characteristic equation complementary function Fin500J Topic 6Fall 2010 Olin Business School 29

30 Example Solve characteristic equation complementary function Fin500J Topic 6Fall 2010 Olin Business School 30

31 Example Solve characteristic equation complementary function Fin500J Topic 6Fall 2010 Olin Business School 31

32 Euler Equations Def: Euler equations Assuming x>0 and all solutions are of the form y(x) = x r Plug into the differential equation to get the characteristic equation Fin500J Topic 6Fall 2010 Olin Business School 32

33 Solving Euler Equations: (Case I) Fin500J Topic 6Fall 2010 Olin Business School 33 The characteristic equation has two different real solutions r 1 and r 2. In this case the functions y = x r 1 and y = x r 2 are both solutions to the original equation. The general solution is: Example:

34 Solving Euler Equations: (Case II) Fin500J Topic 6Fall 2010 Olin Business School 34 The characteristic equation has two equal roots r 1 = r 2 =r. In this case the functions y = x r and y = x r lnx are both solutions to the original equation. The general solution is: Example:

35 Solving Euler Equations: (Case III) Fin500J Topic 6Fall 2010 Olin Business School 35 The characteristic equation has two complex roots r 1,2 = λ±µi. Example:


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