1 / 57

Introduction to Differential Equations

CHAPTER 1. Introduction to Differential Equations. Chapter Contents. 1.1 Definitions and Terminology 1.2 Initial-Value Problems 1.3 Differential Equations as Mathematical Methods.

sandra-carver
Download Presentation

Introduction to Differential Equations

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CHAPTER 1 Introduction to Differential Equations

  2. Chapter Contents • 1.1 Definitions and Terminology • 1.2 Initial-Value Problems • 1.3 Differential Equations as Mathematical Methods

  3. An equation contains the derivates of one or more dependent variables with respect to one or more independent variables (DE). Definition 1.1.1 Differential equation 1.1 Definitions and Terminology • Introduction: differential equations means that equations contain derivatives, eg:dy/dx = 0.2xy (1) • Ordinary DE: An equation contains only ordinary derivates of one or more dependent variables of a single independent variable. eg: dy/dx + 5y = ex, (dx/dt) + (dy/dt) = 2x + y(2)

  4. Partial DE: An equation contains partial derivates of one or more dependent variables of tow or more independent variables.(3) • Notations: Leibniz notation dy/dx, d2y/ dx2 prime notation y’, y”, ….. subscript notation ux, uy, uxx, uyy, uxy , …. • Order: highest order of derivatives first order second order

  5. General form of n-th order ODE:(4) • Normal form of (4)(5)eg: normal form of 4xy’ + y = x, isy’ = (x – y)/4x • Linearity: An n-th order ODE is linear if F is linear in y, y’, y”, …, y(n). It means when (4) is linear, we have(6)

  6. The following cases are for n = 1 and n = 2 and (7) • Two properties of a linear ODE:(1) y, y’, y”, … are of the first degree.(2) Coefficients a0, a1, …, are at most on x • Nonlinear examples:

  7. Definition 1.1.2 Solution of an ODE Any function , defined on an interval I, possessing at least n derivatives that are continuous on I, when replaced into an n-th order ODE, reduces the equation into an identity, it said to be a solution of the equation on the interval. • That is, a solution of (4) is a function  possesses at least n derivatives and F(x, (x), ’(x),…, (n)(x)) = 0for all x in I, where I is the interval  is defined on.

  8. Example 1 Verification of a Solution Verify the indicated function is a solution of the given ODE on (－, ) (a) dy/dx = xy1/2; y = x4/16(b) Solution:(a) Left-hand side: Right-hand side: then left = right (b) Left-hand side: Right-hand side: 0then left = right

  9. Note: y = 0 is also the solution of example 1, called trivial solution

  10. Example 2 Function vs. Solution y = 1/x, is the solution of xy’ + y = 0, however, this function is not differentiable at x = 0. So, the interval of definition I is (－, 0), (0, ). Fig 1.1.1 Ex 2 illustrates the difference between the function y = 1/x and the solution y = 1/x

  11. Definition 1.3 Implicit solution of an ODE G(x, y) = 0 is said to be an implicit solution of (4) on I, provided there exists at least one function y = (x) satisfying the relationship as well as the DE on I. • Explicit solution: dependent variable is expressed solely in terms of independent variable and constants.Eg: solution is y = (x).

  12. Example 3 Verification of an Implicit Solution x2+ y2= 25 is an implicit solution of dy/dx = −x/y(8)on the interval -5 < x < 5. Since dx2/dx + dy2/dx = (d/dx)(25)then 2x + 2y(dy/dx) = 0 and dy/dx = -x/ysolution curve is shown in Fig 1.1.2

  13. Fig 1.1.2 An Implicit solution and two explicit solutions in Ex 3

  14. Families of solutions: A solution containing an arbitrary constant represents a set G(x, y) = 0 of solutions is called a one-parameter family of solutions. A set G(x, y, c1, c2, …, cn) = 0 of solutions is called a n-parameter family of solutions. • Particular solution: A solutionfree of arbitrary parameters. eg: y = cx – x cos x is a solution of xy’ – y = x2sin x on(-, ), y = x cos x is a particular solution according to c = 0. See Fig 1.1.3.

  15. Fig 1.1.3 Some solution of xy’-y=x2 sinx

  16. Example 4 Using Different Symbols x = c1cos 4t and x = c2 sin 4t are solutions ofx + 16x = 0we can easily verify that x = c1cos 4t + c2 sin 4t is also a solution.

  17. Example 5 A piecewise-Defined Solution We can verify y = cx4 is a solution of xy – 4y = 0 on (-, ). See Fig 1.1.4(a). The piecewise-defined function is a particular solution where we choose c = −1for x < 0and c = 1for x  0. See Fig 1.1.4(b).

  18. Fig 1.1.4 Some solution of xy’-4y=0 in Ex 5

  19. Singular solution: A solution can not be obtained by particularly setting any parameters. y = (x2/4 + c)2 is the family solution of dy/dx = xy1/2 , however, y = 0 is a solution of the above DE. • We cannot set any value of c to obtain the solution y = 0, so we call y = 0 is a singular solution.

  20. System of DEs: two or more equations involving of two or more unknown functions of a single independent variable. dx/dt = f(t, x, y) dy/dt = g(t, x, y)(9)

  21. 1.2 Initial-value Problems • IntroductionA solution y(x) of a DE satisfies an initial condition. • ExampleOn some interval I containing xo, solve:subject to:(1)This is called an Initial-Value Problem (IVP).y(xo) = yo , y(xo) = y1,are called initial conditions.

  22. First and Second IVPs(2)and(3)are first and second order initial-value problems, respectively. See Fig 1.2.1 and 1.2.2.

  23. Example 1 First-Order IVPs We know y = cex is the solutions of y’ = y on (-, ). If y(0) = 3, then 3 = ce0 = c. Thus y = 3ex is a solution of this initial-value problemy’ = y, y(0) = 3.If we want a solution pass through (1, -2), that is y(1) = -2, -2 = ce, or c = -2e-1. The function y = -2ex-1 is a solution of the initial-value problemy’ = y, y(1) = -2.

  24. Example 2 Interval / of Definition of a Solution In Problem 6 of Exercise 2.2, we have the solution of y’ + 2xy2 = 0 is y = 1/(x2 + c). If we impose y(0) = -1, it gives c = -1. Consider the following distinctions. 1) As a function, the domain of y = 1/(x2 - 1) is the set of all real numbers except x = -1 and 1. See Fig 1.2.4(a). 2) As a solution, the intervals of definition are (-, 1), (-1, 1), (1, ) 3) As a initial-value problem, y(0) = -1, the interval of definition is (-1, 1). See Fig 1.2.4(b).

  25. Fig 1.2.4 Graph of the function and solution of IVP in Ex 2

  26. Example 3 Second-Order IVP In Example 4 of Sec 1.1, we saw x = c1cos 4t + c2sin 4t is a solution ofx + 16x = 0Find a solution of the following IVP:x + 16x = 0, x(/2) = −2, x(/2) = 1 (4)Solution: Substitute x(/2) = − 2 into x = c1cos 4t + c2sin 4t, we find c1 = −2. In the same manner, from x(/2) = 1we have c2 = ¼.

  27. Existence and Uniqueness:Does a solution of the IVP exist? If a solution exists, is it unique?

  28. Example 4 An IVP Can Have Several Solution Since y= x4/16and y = 0 satisfy the DEdy/dx = xy1/2 , and also initial-value y(0) = 0, this DE has at least two solutions, See Fig 1.2.5.

  29. Theorem 1.2.1 Existence of a Unique Solution Let R be the region defined by a  x  b, c  y  d that contains the point (x0, y0) in its interior. If f(x, y) and f/y are continuous in R, then there exists some interval I0: (x0- h, x0 + h), h > 0,contained in [a, b]and a unique function y(x) defined on I0 that is a solution of the IVP (2).

  30. Fig 1.2.6 Rectangular region R • The geometry of Theorem 1.2.1 shows in Fig 1.2.6.

  31. Example 5 Example 3 Revisited For the DE: dy/dx = xy1/2 , inspection of the functionswe find they are continuous in y > 0. From Theorem 1.2.1, we conclude that through any point (x0, y0), y0> 0,there is some interval centered at x0 on which this DE has a unique solution.

  32. Interval of Existence / UniquenessSuppose y(x) is a solution of IVP (2), the following sets may not be the same: • the domain of y(x), the interval of definition of y(x)as a solution, the interval I0 of existence and uniqueness.

  33. 1.3 DEs as Mathematical Models • IntroductionMathematical models are mathematical descriptions of something. • Level of resolutionMake some reasonable assumptions about the system. • The steps of modeling process are as following.

  34. Assumptions Mathematics formulation Express assumptions in terms of differential equations If necessary, alter assumptions or increase resolution of the model Solve the DEs Obtain solution Check model Predictions with know facts Display model predictions, e.g., graphically

  35. Population DynamicsIf P(t) denotes the total population at time t, thendP/dt  P or dP/dt = kP(1)where k is a constant of proportionality, and k > 0. • Radioactive DecayIf A(t) denotes the substance remaining at time t, thendA/dt A or dA/dt = kA(2)where k is a constant of proportionality, and k < 0. • A single DE can serve as a mathematical model for many different phenomena.

  36. Newton’s Law of Cooling/WarmingIf T(t) denotes the temperature of a body at time t, Tm the temperature of surrounding medium, thendT/dt  T -Tm or dT/dt = k(T - Tm)(3)where k is a constant of proportionality.

  37. Spread of a DiseaseIf x(t) denotes the number of people who have got the disease and y(t) the number of people who have not yet, thendx/dt = kxy(4)where k is a constant of proportionality.From the above description, suppose a small community has a fixed population on n, If one inflected person is introduced into this community, we have x + y = n +1, anddx/dt = kx(n+1-x)(5)

  38. Chemical ReactionsInspect the following equation CH3Cl + NaOH  CH3OH + NaClAssume X is the amount of CH3OH,  and  are the amount of the first two chemicals, then the rate of formation isdx/dt = k( - x)( - x)(6)

  39. MixturesSee Fig 1.3.1. If A(t) denotes the amount of salt in the tank at time t, then dA/dt = (input rate) – (output rate) = Rin - Rout(7)We have Rin = 6lb/min, Rout = A(t)/100(lb/min),thendA/dt = 6 – A/100or dA/dt + A/100 = 6(8)

  40. Fig 1.3.1 Mixing tank

  41. Draining a TankReferring to Fig 1.3.2 and from Torricelli’s Law, if V(t) denotes the volume of water in the tank at time t, (9)From (9), since we have V(t) = Awh, then(10)

  42. Fig 1.3.2 Water draining from a tank

  43. Series CircuitsLook at Fig 1.3.3.From Kirchhoff’s second law, we have(11)where q(t) is the charge and dq(t)/dt = i(t), which is the current.

  44. Fig 1.3.3 Current i(t) and charge q(t) are measured in amperes (A) and coulumbs (C)

  45. Falling BodiesLook at Fig1.22.From Newton’s law, we have(12)Initial value problem(13)

  46. Fig 1.3.4 Position of rock measured from ground level

  47. Falling Bodies and Air ResistanceFrom Fig 1.3.5.We have the DE(14)and can be written as or (15)

  48. Fig 1.3.5 Falling body of mass m

  49. A Slipping ChainFrom Fig 1.3.6.We have(16)

More Related