Upcoming SlideShare
×

# 258 lecnot2

1,772 views

Published on

Published in: Technology, Education
1 Comment
2 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• best

Are you sure you want to  Yes  No
Views
Total views
1,772
On SlideShare
0
From Embeds
0
Number of Embeds
5
Actions
Shares
0
76
1
Likes
2
Embeds 0
No embeds

No notes for slide

### 258 lecnot2

1. 1. ISBN: Copyright Notice: Lecture Notes on Diﬀerential Equations Emre Sermutlu
2. 2. ˙ To my wife Nurten and my daughters Ilayda and Alara
3. 3. vi CONTENTS 5.2 Diﬀerential Operators . . . . . . . . . . . . . . . . . . . . . . . 34 5.3 Homogeneous Equations . . . . . . . . . . . . . . . . . . . . . 35 5.4 Nonhomogeneous Equations . . . . . . . . . . . . . . . . . . . 37 6 Series Solutions Contents 41 6.1 6.2 1 First Order ODE 1.1 Deﬁnitions . . . . . . . . 1.2 Mathematical Modeling 1.3 Separable Equations . . 1.4 Transformations . . . . . ix . . . . . . . . . . . . 2 Exact Equations 2.1 Exact Equations . . . . . . . 2.2 Integrating Factors . . . . . . 2.3 Linear First Order Equations 2.4 Bernoulli Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1 3 3 5 . . . . 9 9 11 13 14 Classiﬁcation of Points . . . . . . . . . . . . . . . . . . . . . . 43 6.3 Preface Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 Power Series Method . . . . . . . . . . . . . . . . . . . . . . . 43 7 Frobenius’ Method 49 7.1 An Extension of Power Series Method . . . . . . . . . . . . . . 49 7.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 8 Laplace Transform I 57 8.1 Deﬁnition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 8.2 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 8.3 Initial Value Problems . . . . . . . . . . . . . . . . . . . . . . 61 9 Laplace Transform II 69 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 17 19 20 22 4 Nonhomogeneous Equations 25 4.1 General and Particular Solutions . . . . . . . . . . . . . . . . 25 4.2 Method of Undetermined Coeﬃcients . . . . . . . . . . . . . . 27 4.3 Method of Variation of Parameters . . . . . . . . . . . . . . . 29 5 Higher Order Equations 33 5.1 Linear Equations of Order n . . . . . . . . . . . . . . . . . . . 33 v Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 9.2 Unit Step Function . . . . . . . . . . . . . . . . . . . . . . . . 72 9.3 3 Second Order Equations 3.1 Linear Diﬀerential Equations . 3.2 Reduction of Order . . . . . . 3.3 Constant Coeﬃcients . . . . . 3.4 Cauchy-Euler Equation . . . . 9.1 Diﬀerentiation of Transforms . . . . . . . . . . . . . . . . . . . 73 9.4 Partial Fractions Expansion . . . . . . . . . . . . . . . . . . . 74 9.5 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 10 Fourier Analysis I 81 10.1 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 10.2 Convergence of Fourier Series . . . . . . . . . . . . . . . . . . 84 10.3 Parseval’s Identity . . . . . . . . . . . . . . . . . . . . . . . . 85 11 Fourier Analysis II 91 11.1 Fourier Cosine and Sine Series . . . . . . . . . . . . . . . . . . 91 11.2 Complex Fourier Series . . . . . . . . . . . . . . . . . . . . . . 94 11.3 Fourier Integral Representation . . . . . . . . . . . . . . . . . 96
4. 4. CONTENTS vii 12 Partial Diﬀerential Equations 12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2 Modeling a Vibrating String . . . . . . . . . . . . . . . . . . 12.3 Method of Separation of Variables . . . . . . . . . . . . . . . 101 . 101 . 103 . 104 13 Heat Equation 13.1 Modeling Heat Flow . . . . . . . . 13.2 Homogeneous Boundary Conditions 13.3 Nonzero Boundary Conditions . . . 13.4 Two Dimensional Problems . . . . 111 . 111 . 113 . 115 . 117 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Laplace Equation 121 14.1 Rectangular Coordinates . . . . . . . . . . . . . . . . . . . . . 121 14.2 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . 126 To the Student 133 References 135 Index 137
5. 5. Preface This set of lecture notes for ordinary and partial diﬀerential equations grew out of the course Engineering Mathematics I have taught at Cankaya Univer¸ sity since 1999. It is a one-semester course for second year students. The main audience for this text, of course, is students. Presentation is user-friendly. There are more examples and fewer theorems than usual. The material is based on a solid background in calculus. The student is assumed to be familiar with algebra, trigonometry, functions and graphs, series, diﬀerentiation, and most importantly, integration techniques of various kinds. It is my (and my students’) sad experience that if you cannot diﬀerentiate and integrate, you cannot solve diﬀerential equations. Knowledge of Linear Algebra, except for the determinants of a simple nature, is not assumed. There are 14 chapters. Each chapter can be covered in one week. After a summary of methods and solved exercises, there are a number of end of chapter exercises with answers. The exercises that take exceptionally longer times are marked with a star. ( ) Nobody can learn how to solve problems by watching someone else solve problems. So I advise the students to try each problem on their own. I would like to thank all my students who helped me write this book by the valuable feedback they provided. In particular, special thanks are for ˙ Nuh Co¸kun, Nevrez Imamo˜lu, Nilg¨n Din¸arslan and I¸ıl Lelo˜lu who have s g u c s g made a very extensive and meticulous check of the whole manuscript. You may send all kinds of comments, suggestions and error reports to sermutlu@cankaya.edu.tr. Assist. Prof. Dr. Emre Sermutlu ix
6. 6. 2 CHAPTER 1. FIRST ORDER ODE are partial diﬀerential equations. (Partial Diﬀerential Equations are usually much more diﬃcult) Order: The order of a diﬀerential equation is the order of the highest derivative that occurs in the equation. A ﬁrst order diﬀerential equation contains y , y and x so it is of the form F (x, y, y ) = 0 or y = f (x, y). For example, the following diﬀerential equations are ﬁrst order: Chapter 1 First Order Diﬀerential Equations y + x2 y = ex xy = (1 + y 2 ) y 2 = 4xy While these are second order: The subject of diﬀerential equations is an important part of applied mathematics. Many real life problems can be formulated as diﬀerential equations. In this chapter we will ﬁrst learn the basic concepts and classiﬁcation of diﬀerential equations, then we will see where they come from and how the simplest ones are solved. The concepts and techniques of calculus, especially integration, will be necessary to understand diﬀerential equations. 1.1 Deﬁnitions Ordinary Diﬀerential Equation: An ordinary diﬀerential equation is an equation that contains derivatives of an unknown function y(x). Partial Diﬀerential Equation: A partial diﬀerential equation is a diﬀerential equation involving an unknown function of two or more variables, like u(x, y). For example, y − 4y + y = 0 y 2 + 1 = x2 y + sin x are ordinary diﬀerential equations. uxx + uyy = 0 u2 + u2 = ln u x y 1 y − x2 y + y = 1 + sin x y + 6yy = x3 General and Particular Solutions: A general solution of a diﬀerential equation involves arbitrary constants. In a particular solution, these constants are determined using initial values. As an example, consider the diﬀerential equation y = 2x. y = x2 + c is a general solution , y = x2 + 4 is a particular solution . Example 1.1 Find the general solution of the diﬀerential equation y = 0. Then ﬁnd the particular solution that satisﬁes y(0) = 5, y (0) = 3. y = 0 ⇒ y = c ⇒ y = cx + d. This is the general solution. y (0) = 3 ⇒ c = 3, y(0) = 5 ⇒ d = 5 Therefore y = 3x + 5. This is the particular solution. Explicit and Implicit Solutions: y = f (x) is an explicit solution, F (x, y) = 0 is an implicit solution. We have to solve equations to obtain y for a given x in implicit solutions, whereas it is straightforward for explicit solutions. For example, y = e4x is an explicit solution of the equation y = 4y. x3 + y 3 = 1 is an implicit solution of the equation y 2 y + x2 = 0
7. 7. 1.2. MATHEMATICAL MODELING 1.2 3 Mathematical Modeling 4 CHAPTER 1. FIRST ORDER ODE Example 1.4 Solve the initial value problem y + y 2 xex = 0, y(0) = 2 Diﬀerential equations are the natural tools to formulate, solve and understand many engineering and scientiﬁc systems. The mathematical models of most of the simple systems are diﬀerential equations. y = −y 2 xex dP = αP dt ex dx Example 1.3 The downward acceleration of an object in free fall is g. Find the height as a function of time if the initial height is y0 and initial speed is v0 . 2= d2 y = −g dt2 1 = xex − ex + c y ⇒ c= 1 xex − ex + 3 2 3 2 Example 1.5 Find the general solution of the diﬀerential equation y + y 2 = 1. dy + y2 = 1 dx 1 y = − gt2 + v0 t + y0 2 ⇒ dy = 1 − y2 dx dy = 1 − y2 Separable Equations If we can separate x and y in a ﬁrst order diﬀerential equation and put them to diﬀerent sides as g(y)dy = f (x)dx, it is called a separable equation. We can ﬁnd the solution by integrating both sides. (Don’t forget the integration constant!) f (x)dx + c ⇒ xex 1 −1 + c y= dy = −gt + v0 dt g(y)dy = xex dx 1 − ex + c This is the general solution. Now we will use the condition y(0) = 2 to determine the constant c. y= where P0 = P (0) 1.3 dy = xex dx y2 Using integration by parts, we have u = x, dv = ex dx, du = dx, v = ex therefore 1 = xex − y P = P0 eαt − dy = y2 − Example 1.2 The rate of growth of a population is proportional to itself. Find the population as a function of time. ⇒ (1.1) 1 2 1 1 + 1−y 1+y ⇒ dy = dx 1 − y2 dx dy = 1 1+y ln =x+c 2 1−y 1+y = e2x+2c 1−y dx
8. 8. 1.4. TRANSFORMATIONS 5 6 CHAPTER 1. FIRST ORDER ODE After some algebra, we obtain y= u3/2 = 3 ln x + c 3/2 ke2x − 1 ke2x + 1 where k = e2c u= Example 1.6 Solve the initial value problem y = x3 e−y , y(1) = 0. ey dy = ey = 9 ln x + c1 2 e0 = Let’s use the substitution u = x + y. Then, 1 +c 4 y = u − x, ⇒ 1.4 x4 3 + 4 4 (u + 6)dx = (−u − 3)(du − dx) 3dx = (−u − 3)du Sometimes a change of variables simpliﬁes a diﬀerential equation just as y substitutions simplify integrals. For example if y = f , the substitution x y u = will make the new equation separable. x y +3 x x . y √ u du = 3dx = 3x = − 3x = − If y = ux, then y = u x + u and u x + u = u + 3 ux=3 dy = du − dx and the equation can be expressed in terms of u and x. Transformations Example 1.7 Solve y = 2/3 Example 1.8 Solve the diﬀerential equation (x + y + 6)dx = (−x − y − 3)dy. 3 c= 4 y = ln 2/3 x3 dx x4 +c 4 ⇒ y(1) = 0 y=x 9 ln x + c1 2 1 u 3dx x 1 u This is an implicit solution. (−u − 3) du u2 − 3u + c 2 (x + y)2 − 3(x + y) + c 2
9. 9. EXERCISES 7 Exercises CHAPTER 1. FIRST ORDER ODE Answers Solve the following diﬀerential equations. 1) y 3 y + x3 = 0 2) y + 4x3 y 2 = 0 y 3) xy = x + y Hint: y = f x 4) (x2 + y 2 ) dx + xydy = 0 Hint: y = f 8 y x 1) x4 + y 4 = c 1 2) y = 4 x +c 3) y = x(ln |x| + c) c x2 4) y 2 = 2 − x 2 5) y = − ln c + 2 5) y = xey−x 1 + ln x 6) y = 4y 3 7) y = 3x2 sec2 y 8) y = y(y + 1) 9) y + 2y = y 2 + 1 10) (1 + y 2 )dx + x2 dy = 0 y x 12) y = eax+by 13) y = x2 y 2 − 2y 2 + x2 − 2 14) y = − 2x + y x Solve the following initial value problems: 15) (y 2 + 5xy + 9x2 )dx + x2 dy = 0, y(1) = −4 16) y 3 y + x3 = 0, y(0) = 1 17) y = −2xy, y(0) = 3 18) y = 1 + 4y 2 , y(0) = 0 19) (x2 + 1)1/2 y = xy 3 , y(0) = 2 20) dx x x2 = − , x(0) = 1 dt 5 25 2 6) y 4 = x ln x + c 7) 2y + sin 2y = 4x3 + c ex 8) y = c − ex 9) y = 1 − 1 x+c 10) y = tan c + 11) y = a e−x 2 1 x 11) y = cxa 12) eax e−by + =c a b 13) y = tan x3 − 2x + c 3 14) y = −x + c x 15) y = 16) 17) 18) 19) x − 3x ln x − 1 x4 + y 4 = 1 2 y = 3e−x 1 y = 2 tan 2x √ y = ( 9 − 2 x2 + 1)−1/2 4 20) x = 5et/5 4 + et/5
10. 10. 10 CHAPTER 2. EXACT EQUATIONS So, the solution of this equation is very simple, if du is zero, u must be a constant, therefore x4 + x2 y 2 + y 4 = c ∂N ∂M = is necessary and suﬃcient for the ∂y ∂x equation M (x, y)dx + N (x, y)dy = 0 to be exact. Method of Solution: To solve M dx + N dy = 0, Theorem 2.1: The condition Chapter 2 • Check for Exactness Exact and Linear Diﬀerential Equations • If the equation is exact, ﬁnd u by integrating either M or N . u= M dx + k(y) or u = N dy + l(x) Note that we have arbitrary functions as integration constants. In this chapter, we will learn how to recognize and solve three diﬀerent types of equations: Exact, linear, and Bernoulli. All of them are ﬁrst order equations, therefore we expect a single integration constant in the solution. At this stage it seems like there’s a special trick for every diﬀerent kind of question. You will gain familiarity with exercise and experience. • Determine the arbitrary functions using the original equation. The solution is u(x, y) = c Example 2.1 Solve the equation 3y 2 dx + (3y 2 + 6xy)dy = 0. Let’s check for exactness ﬁrst. 2.1 ∂(3y 2 ) = 6y, ∂y Exact Equations A ﬁrst order diﬀerential equation of the form M (x, y)dx + N (x, y)dy = 0 The equation is exact. (2.1) u(x, y) = is called an exact diﬀerential equation if there exists a function u(x, y) such that ∂u ∂u = M, =N (2.2) ∂x ∂y In other words, du = M dx + N dy, so M dx + N dy is a total diﬀerential. For example, the equation (4x3 + 2xy 2 )dx + (4y 3 + 2x2 y)dy = 0 is exact, and u = x4 + x2 y 2 + y 4 9 ∂(3y 2 + 6xy) = 6y ∂x 3y 2 dx + k(y) = 3y 2 x + k(y) ∂u = 6yx + k (y) = 3y 2 + 6xy ∂y k (y) = 3y 2 ⇒ k(y) = y 3 We do not need an integration constant here because u(x, y) = c already contains one u(x, y) = 3y 2 x + y 3 = c
11. 11. 2.2. INTEGRATING FACTORS 2.2 11 Integrating Factors 12 CHAPTER 2. EXACT EQUATIONS But this equation is more diﬃcult than the one we started with. If we make a simplifying assumption that F is a function of one variable only, we can solve for F and obtain the following theorem: Consider the equation Theorem 2.2: Consider the equation P dx + Qdy = 0. Deﬁne P dx + Qdy = 0 (2.3) that is not exact. If it becomes exact after multiplying by F , i.e. if F P dx + F Qdy = 0 R= (2.4) is exact, then F is called an integrating factor. (Note that P, Q and F are functions of x and y) 1 For example, ydx − xdy = 0 is not exact, but F = 2 is an integrating x factor. Example 2.2 Solve (2xex − y 2 )dx + 2ydy = 0. Use F = e−x . a) If R depends only on x, then F (x) = exp factor. ˜ b) If R depends only on y, then F (y) = exp factor. ∂Q ∂P − ∂x ∂y R= ˜ R(y)dy is an integrating 8x2 y + 2 − 6x2 y − 1 2x2 y + 1 1 = 3 = 3y + x 2x 2x y + x x F (x) = e R(x)dx = eln x = x Multiply the equation by x to obtain the exact equation ∂(2x − y 2 e−x ) ∂(2ye−x ) = −2ye−x , = −2ye−x ∂y ∂x (4x3 y 2 + 2yx)dx + (2x4 y + x2 )dy = 0 Now the equation is exact. We can solve it as we did the previous example and obtain the result u(x, y) = x2 + y 2 e−x = c How To Find an Integrating Factor: Let P dx+Qdy = 0 be a diﬀerential equation that is not exact, and let F = F (x, y) be an integrating factor. By deﬁnition, (2.5) (2.6) R(x) dx is an integrating Example 2.3 Solve (4x2 y 2 + 2y)dx + (2x3 y + x)dy = 0 (2x − y 2 e−x )dx + 2ye−x dy = 0 Fy P + F P y = Fx Q + F Qx 1 ˜ and R = P The equation is not exact. The equation is not exact. Let’s multiply both sides by e−x . The new equation is: ⇒ ∂P ∂Q − ∂y ∂x ∂(2x3 y + x) ∂(4x2 y 2 + 2y) = 8x2 y + 2, = 6x2 y + 1 ∂y ∂x ∂(2y) ∂(2xex − y 2 ) = −2y, =0 ∂y ∂x (F P )y = (F Q)x 1 Q (4x3 y 2 + 2yx) dx + k(y) = x4 y 2 + yx2 + k(y) ∂u = 2x4 y + x2 + k (y) = 2x4 y + x2 ∂y u(x, y) = x4 y 2 + x2 y = c ⇒ k(y) = 0
12. 12. 2.3. LINEAR FIRST ORDER EQUATIONS 2.3 13 Linear First Order Equations 2.4 If a ﬁrst order diﬀerential equation can be written in the form y + p(x)y = r(x) CHAPTER 2. EXACT EQUATIONS Bernoulli Equation The equation y + p(x)y = g(x)y a (2.7) it is called a linear diﬀerential equation. If r(x) = 0, the equation is homogeneous, otherwise it is nonhomogeneous. We can express the equation (2.7) as [p(x)y − r(x)]dx + dy = 0. This is not exact but it has an integrating factor: R = p(x), F = e 14 p dx is called Bernoulli equation. It is nonlinear. Nonlinear equations are usually much more diﬃcult than linear ones, but Bernoulli equation is an exception. It can be linearized by the substitution u(x) = [y(x)]1−a (2.14) (2.8) Then, we can solve it as other linear equations. Method of Solution: • Given a ﬁrst order linear equation, express it in the following form: y + p(x)y = r(x) p dx y +e p dx py = re 2 p dx y − ex 2x y= 3 3xy 2 p(x) dx to obtain e Example 2.5 Solve the equation (2.9) • Multiply both sides by the integrating factor F (x) = exp (2.10) Here a = −2 therefore u = y 1−(−2) = y 3 ⇒ u = 3y 2 y Multiplying both sides of the equation by 3y 2 we obtain 2 • Express the left hand side as a single parenthesis. e p dx y = re p dx 3y 2 y − 2xy 3 = (2.11) y(x) = e−h eh r dx + c (2.12) 2 ex x ⇒ u − 2xu = e −2x dx = e−x 2 2 Multiplying both sides by e−x , we get 2 p dx. 2 e−x u − 2xe−x u = Example 2.4 Solve y + 4y = 1 2 The integrating factor is F = e equation by e4x to obtain 4 dx (e−x u) = 4x = e . Multiply both sides of the 2 e4x y + 4e4x y =e4x e−x u = ln x + c ⇒ ⇒ 1 y = + ce−4x 4 1 x 1 x u = (ln x + c)ex (e4x y) =e4x e4x e4x y = +c 4 ex x This equation is linear. Its integrating factor is • Integrate both sides. Don’t forget the integration constant. The solution is: where h = (2.13) y = (ln x + c)ex 2 1/3 2
13. 13. EXERCISES 15 16 Exercises Answers Solve the following diﬀerential equations. (Find an integrating factor if necessary) 1) (yex + xyex + 1)dx + xex dy = 0 2) (2r + 2 cos θ)dr − 2r sin θdθ = 0 3) (sin xy + xy cos xy)dx + (x2 cos xy)dy = 0 4) 2 cos ydx = sin ydy 5) 5dx − ey−x dy = 0 6) (2xy + 3x2 y 6 ) dx + (4x2 + 9x3 y 5 ) dy = 0 7) (3xey + 2y) dx + (x2 ey + x) dy = 0 1 5 8) y + y = x x 9) y + 1 1 y= x ln x ln x c − 1 e−x x r2 + 2r cos θ = c x sin xy = c F = e2x , e2x cos y = c F = ex , 5ex − ey = c F = y 3 , x 2 y 4 + x3 y 9 = c F = x, x3 ey + x2 y = c c 1 y= + 5 5 x 1) y = 2) 3) 4) 5) 6) 7) 8) 9) y = x+c ln x 10) y = −1 + x4 12) y = 4 − 5e− 4 Reduce to linear form and solve the following equations: 2 sin x 1/2 13) y − 4y tan x = y cos3 x x y 5 ln x 4/5 25 y= y 15) y + x x5 13) y = 14) y = c − ln cos x cos2 x 1 2 2 − x + ce−2x 15) y = y 1 =− 9 3 x xy x ln x − x + c x5 16) y = 14) y + y = − 1 c + 4 8 x x 19) x = y −2 2 1/4 1 3 cosh 3y + c Hint: x ↔ y 20) y = 20) 2xyy + (x − 1)y 2 = x2 ex 5 17) y = arcsin[c(x − 1)] 1 c + 3 18) F = y, x = 2y y tan y 17) y = x−1 18) y 2 dx + (3xy − 1)dy = 0 19) y (sinh 3y − 2xy) = y c cos x 11) y = x4 cos x + c cos x 10) y − y tan x = tan x 11) y + y tan x = 4x3 cos x 12) y + x3 y = 4x3 , y(0) = −1 16) y + CHAPTER 2. EXACT EQUATIONS Hint: z = y 2 cxe−x + 1 xex 2
14. 14. 18 Chapter 3 Second Order Homogeneous Diﬀerential Equations For ﬁrst order equations, concepts from calculus and some extensions were suﬃcient. Now we are starting second order equations and we will learn many new ideas, like reduction of order, linear independence and superposition of solutions. Many diﬀerential equations in applied science and engineering are second order and linear. If in addition they have constant coeﬃcients, we can solve them easily, as explained in this chapter and the next. For nonconstant coeﬃcients, we will have limited success. 3.1 Linear Diﬀerential Equations If we can express a second order diﬀerential equation in the form y + p(x)y + q(x)y = r(x) (3.1) it is called linear. Otherwise, it is nonlinear. Consider a linear diﬀerential equation. If r(x) = 0 it is called homogeneous, otherwise it is called nonhomogeneous. Some examples are: y + y 2 = x2 y Nonlinear sin xy + cos xy = 4 tan x Linear Nonhomogeneous x2 y + y = 0 Linear Homogeneous 17 CHAPTER 3. SECOND ORDER EQUATIONS Linear Combination: A linear combination of y1 , y2 is y = c1 y1 + c2 y2 . Theorem 3.1: For a homogeneous linear diﬀerential equation any linear combination of solutions is again a solution. The above result does NOT hold for nonhomogeneous equations. For example, both y = sin x and y = cos x are solutions to y + y = 0, so is y = 2 sin x + 5 cos x. Both y = sin x + x and y = cos x + x are solutions to y + y = x, but y = sin x + cos x + 2x is not. This is a very important property of linear homogeneous equations, called superposition. It means we can multiply a solution by any number, or add two solutions, and obtain a new solution. Linear Independence: Two functions y1 , y2 are linearly independent if c1 y1 + c2 y2 = 0 ⇒ c1 = 0, c2 = 0. Otherwise they are linearly dependent. (One is a multiple of the other). For example, ex and e2x are linearly independent. ex and 2ex are linearly dependent. General Solution and Basis: Given a second order, linear, homogeneous diﬀerential equation, the general solution is: y = c1 y1 + c2 y2 (3.2) where y1 , y2 are linearly independent. The set {y1 , y2 } is called a basis, or a fundamental set of the diﬀerential equation. As an illustration, consider the equation x2 y − 5xy + 8y = 0. You can easily check that y = x2 is a solution. (We will see how to ﬁnd it in the last section) Therefore 2x2 , 7x2 or −x2 are also solutions. But all these are linearly dependent. We expect a second, linearly independent solution, and this is y = x4 . A combination of solutions is also a solution, so y = x2 + x4 or y = 10x2 − 5x4 are also solutions. Therefore the general solution is y = c1 x 2 + c2 x 4 and the basis of solutions is {x2 , x4 }. (3.3)
15. 15. 3.2. REDUCTION OF ORDER 3.2 19 Reduction of Order 3.3 If we know one solution of a second order homogeneous diﬀerential equation, we can ﬁnd the second solution by the method of reduction of order. Consider the diﬀerential equation y + py + qy = 0 (3.4) Suppose one solution y1 is known, then set y2 = uy1 and insert in the equation. The result will be y1 u + (2y1 + py1 )u + (y1 + py1 + qy1 )u = 0 (3.5) (3.6) This is still second order, but if we set w = u , we will obtain a ﬁrst order equation: y1 w + (2y1 + py1 )w = 0 (3.7) Solving this, we can ﬁnd w, then u and then y2 . Example 3.1 Given that y1 = x2 is a solution of CHAPTER 3. SECOND ORDER EQUATIONS Homogeneous Equations with Constant Coeﬃcients Up to now we have studied the theoretical aspects of the solution of linear homogeneous diﬀerential equations. Now we will see how to solve the constant coeﬃcient equation y + ay + by = 0 in practice. We have the sum of a function and its derivatives equal to zero, so the derivatives must have the same form as the function. Therefore we expect the function to be eλx . If we insert this in the equation, we obtain: λ2 + aλ + b = 0 But y1 is a solution, so the last term is canceled. So we have y1 u + (2y1 + py1 )u = 0 20 (3.8) This is called the characteristic equation of the homogeneous diﬀerential equation y + ay + by = 0. If we solve the characteristic equation, we will see three diﬀerent possibilities: Two real roots, double real root and complex conjugate roots. Two Real Roots: The general solution is y = c1 eλ1 x + c2 eλ2 x (3.9) 2 x y − 3xy + 4y = 0 Example 3.2 Solve y − 3y − 10y = 0 ﬁnd a second linearly independent solution. Let y2 = ux2 . Then y2 = u x2 + 2xu and 2 Try y = eλx . The characteristic equation is λ2 − 3λ − 10 = 0 with solution λ1 = 5, λ = −2, so the general solution is y = c1 e5x + c2 e−2x y2 = u x + 4xu + 2u Inserting these in the equation, we obtain x4 u + x3 u = 0 If w = u then 1 x4 w + x3 w = 0 or w + w = 0 x 1 This linear ﬁrst order equation gives w = , therefore u = ln x and x 2 y2 = x ln x Example 3.3 Solve the initial value problem y −y = 0, y(0) = 2, y (0) = 4 We start with y = eλx as usual. The characteristic equation is λ2 − 1 = 0. Therefore λ = ±1. The general solution is: y = c1 ex + c2 e−x Now, we have to use the initial values to determine the constants. y(0) = 2 ⇒ c1 + c2 = 2 and y (0) = 4 ⇒ c1 − c2 = 4. By solving this system, we obtain c1 = 3, c2 = −1 so the particular solution is: y = 3ex − e−x
16. 16. 3.3. CONSTANT COEFFICIENTS 21 Double Real Root: One solution is eλx but we know that a second order equation must have two independent solutions. Let’s use the method of reduction of order to ﬁnd the second solution. y − 2ay + a2 y = 0 ⇒ y1 = eax (3.10) 22 CHAPTER 3. SECOND ORDER EQUATIONS 3.4 Cauchy-Euler Equation The equation x2 y + axy + by = 0 is called the Cauchy-Euler equation. By inspection, we can easily see that the solution must be a power of x. Let’s substitute y = xr in the equation and try to determine r. We will obtain Let’s insert y2 = ueax in the equation. r(r − 1)xr + arxr + bxr = 0 ax ax e u + (2a − 2a)e u = 0 Obviously, u = 0 therefore u = c1 + c2 x. The general solution is y = c1 eλx + c2 xeλx (3.12) Example 3.4 Solve y + 2y + y = 0 λx r2 + (a − 1)r + b = 0 (3.11) (3.17) (3.18) This is called the auxiliary equation. Once again, we have three diﬀerent cases according to the types of roots. The general solution is given as follows: • Two real roots 2 y = e . The characteristic equation is λ + 2λ + 1 = 0. Its solution is the double root λ = −1, therefore the general solution is y = c1 e−x + c2 xe−x (3.13) This can be proved using Taylor series expansions. If the solution of the characteristic equation is λ1 = α + iβ, λ2 = α − iβ y = c1 e (cos βx + i sin βx) + c2 e αx (cos βx − i sin βx) (3.14) (3.15) By choosing new constants A, B, we can express this as y=e αx (A cos βx + B sin βx) y = c1 xr + c2 xr ln x (3.20) • Complex conjugate roots where r1 , r2 = r ± si y = xr [c1 cos(s ln x) + c2 sin(s ln x)] then the general solution of the diﬀerential equation will be αx (3.19) • Double real root Complex Conjugate Roots: We need the complex exponentials for this case. Euler’s formula is eix = cos x + i sin x y = c1 xr1 + c2 xr2 (3.21) Example 3.6 Solve x2 y + 2xy − 6y = 0 Insert y = xr . Auxiliary equation is r2 + r − 6 = 0. The roots are r = 2, r = −3 therefore y = c1 x2 + c2 x−3 (3.16) Example 3.5 Solve y − 4y + 29y = 0. Example 3.7 Solve x2 y − 9xy + 25y = 0 y = eλx . The characteristic equation is λ2 −4λ+29 = 0. Therefore λ = 2±5i. The general solution is Insert y = xr . Auxiliary equation is r2 − 10r + 25 = 0. Auxiliary equation has the double root r = 5 therefore the general solution is y = e2x (A cos 5x + B sin 5x) y = c1 x5 + c2 x5 ln x
17. 17. EXERCISES 23 24 Exercises CHAPTER 3. SECOND ORDER EQUATIONS Answers 1) 2) 3) 4) 5) Are the following sets linearly independent? 1) {x4 , x8 } 2) {sin x, sin2 x} 3) {ln(x5 ), ln x} Use reduction of order to ﬁnd a second linearly independent solution: 4) x2 (ln x − 1) y − xy + y = 0, y1 = x 1 5) x2 ln x y + (2x ln x − x)y − y = 0, y1 = x 6) y + 3 tan x y + (3 tan2 x + 1)y = 0, y1 = cos x Yes Yes No y2 = ln x y2 = ln x − 1 6) y2 = sin x cos x 7) y = (1 + x)e−x 1 8) y = c1 e−2x + c2 e− 2 x Solve the following equations: 7) y + 2y + y = 0, y(0) = 1, y (0) = 0 9) y = e8x 5 8) y + y + y = 0 2 10) y = c1 e−12x + c2 xe−12x 9) y − 64y = 0, y(0) = 1, 10) y + 24y + 144y = 0 y (0) = 8 11) y = 4e−x + 3xe−x 7 11) y + 2y + y = 0, y(−1) = e, y(1) = e 12) 5y − 8y + 5y = 0 π2 13) y + 2y + 1 + y = 0, y(0) = 1, y (0) = −1 4 14) y − 2y + 2y = 0, y(π) = 0, y(−π) = 0 15) xy + y = 0 16) x2 y − 3xy + 5y = 0 17) x2 y − 10xy + 18y = 0 18) x2 y − 13xy + 49y = 0 19) Show that y1 = u and y2 = u y − v u +2 v u y + 12) y = e0.8x [A cos(0.6x) + B sin(0.6x)] 13) y = e−x cos 14) y = ex sin x 15) y = c1 + c2 ln x 16) y = x2 [c1 cos(ln x) + c2 sin(ln x)] vdx are solutions of the equation vu u2 u +2 2 − vu u u π x 2 y=0 20) Show that y1 = u and y2 = v are solutions of the equation (uv − vu )y + (vu − uv )y + (u v − v u )y = 0 17) y = c1 x2 + c2 x9 18) y = c1 x7 + c2 x7 ln x
18. 18. 26 CHAPTER 4. NONHOMOGENEOUS EQUATIONS the nonhomogeneous one. The general solution is of the form y = yh + yp (4.3) Example 4.1 Find the general solution of y − 3y + 2y = 2x − 3 using yp = x. Chapter 4 Let’s solve y − 3y + 2y = 0 ﬁrst. Let yh = eλx . Then λ2 − 3λ + 2 = 0 Second Order Nonhomogeneous Equations which means λ = 2 or λ = 1. The homogenous solution is yh = c1 ex + c2 e2x therefore the general solution is: y = x + c1 ex + c2 e2x In this chapter we will start to solve the nonhomogeneous equations, and see that we will need the homogeneous solutions we found in the previous chapter. Of the two methods we will learn, undetermined coeﬃcients is simpler, but it can be applied to a restricted class of problems. Variation of parameters is more general but involves more calculations. 4.1 Consider the nonhomogeneous equation (4.1) Let yp be a solution of this equation. Now consider the corresponding homogeneous equation y + p(x)y + q(x)y = 0 (4.2) Let yh be the general solution of this one. If we add yh and yp , the result will still be a solution for the nonhomogeneous equation, and it must be the general solution because yh contains two arbitrary constants. This interesting property means that we need the homogeneous equation when we are solving 25 The solution of y = 0 is simply yh = c1 x + c2 , therefore the general solution must be y = − cos x + c1 x + c2 As you can see, once we have a particular solution, the rest is straightforward, but how can we ﬁnd yp for a given equation? Example 4.3 Find a particular solution of the following diﬀerential equations. Try the suggested functions. (Success not guaranteed!) General and Particular Solutions y + p(x)y + q(x)y = r(x) Example 4.2 Find the general solution of y = cos x using yp = − cos x. a) y + y = ex , Try yp = Aex b) y − y = ex , Try yp = Aex c) y + 2y + y = x Try yp = Ax + B d) y + 2y = x Try yp = Ax + B e) y + 2y + y = 2 cos x Try yp = A cos x and yp = A cos x + B sin x As you can see, some of the suggestions work and some do not. yp is usually similar to r(x). We can summarize our ﬁndings as: • Start with a set of functions that contains not only r(x), but also all derivatives of r(x). • If one of the terms of yp candidate occurs in yh , there is a problem.
19. 19. 4.2. METHOD OF UNDETERMINED COEFFICIENTS 4.2 27 Method of Undetermined Coeﬃcients 28 CHAPTER 4. NONHOMOGENEOUS EQUATIONS The homogeneous equation is 3y + 10y + 3y = 0 To solve the constant coeﬃcient equation d2 y dy + a + by = r(x) 2 dx dx (4.4) • Solve the corresponding homogeneous equation, ﬁnd yh . • Find a candidate for yp using the following table: Term in r(x) Choice for yp xn eax cos bx or sin bx xn eax xn cos bx or xn sin bx An xn + · · · + A1 x + A0 Aeax A cos bx + B sin bx (An xn + · · · + A1 x + A0 )eax (An xn + · · · + A0 ) cos bx +(Bn xn + · · · + B0 ) sin bx eax cos bx or eax sin bx Aeax cos bx + Beax sin bx xn eax cos bx or xn eax sin bx (An xn + · · · + A0 )eax cos bx +(Bn xn + · · · + B0 )eax sin bx (You don’t have to memorize the table. Just note that the choice consists of r(x) and all its derivatives) • If your choice for yp occurs in yh , you have to change it. Multiply it by x if the solution corresponds to a single root, by x2 if it is a double root. • Find the constants in yp by inserting it in the equation. • The general solution is y = yp + yh Note that this method works only for constant coeﬃcient equations, and only when r(x) is relatively simple. Its solution is yh = c1 e−3x + c2 e−x/3 To ﬁnd a particular solution, let’s try yp = Ax + B. Inserting this in the equation, we obtain: 10A + 3Ax + 3B = 9x Therefore, A = 3, B = −10. The particular solution is: yp = 3x − 10 The general solution is: y = c1 e−3x + c2 e−x/3 + 3x − 10 Example 4.5 Find the general solution of y − 4y + 4y = e2x The solution of the associated homogeneous equation y − 4y + 4y = 0 is yh = c1 e2x + c2 xe2x Our candidate for yp is yp = Ae2x . But this is already in the yh so we have to change it. If we multiply by x, we will obtain Axe2x but this is also in yh . Therefore we have to multiply by x2 . So our choice for yp is yp = Ax2 e2x . Now we have to determine A by inserting in the equation. yp = 2Ax2 e2x + 2Axe2x yp = 4Ax2 e2x + 8Axe2x + 2Ae2x Example 4.4 Find the general solution of the equation 3y + 10y + 3y = 9x 4Ax2 e2x + 8Axe2x + 2Ae2x − 4(2Ax2 e2x + 2Axe2x ) + 4Ax2 e2x = e2x
20. 20. 4.3. METHOD OF VARIATION OF PARAMETERS 2Ae 2x =e 2x ⇒ 29 CHAPTER 4. NONHOMOGENEOUS EQUATIONS Therefore the particular solution is 1 1 A = , yp = x2 e2x 2 2 yp (x) = −y1 1 y = yh + yp = c1 e2x + c2 xe2x + x2 e2x 2 4.3 30 y1 r dx aW (4.11) e−x Example 4.6 Find the general solution of y + 2y + y = √ x yh = c1 e−x + c2 xe−x Method of Variation of Parameters Consider the linear second order nonhomogeneous diﬀerential equation a(x)y + b(x)y + c(x)y = r(x) y2 r dx + y2 aW W = (4.5) e−x xe−x −x −x −e e − xe−x yp = −e−x If a(x), b(x) and c(x) are not constants, or if r(x) is not among the functions given in the table, we can not use the method of undetermined coeﬃcients. In this case, the variation of parameters can be used if we know the homogeneous solution. Let yh = c1 y1 + c2 y2 be the solution of the associated homogeneous equation a(x)y + b(x)y + c(x)y = 0 (4.6) Let us express the particular solution as: xe−x e−x √ dx + xe−x e−2x x yp = −e−x √ x dx + xe−x = e−2x e−x e−x √ dx e−2x x 1 √ dx x 4 x3/2 x1/2 + xe−x = e−x x3/2 3/2 1/2 3 4 y = yh + yp = c1 e−x + c2 xe−x + e−x x3/2 3 2 Example 4.7 Find the general solution of x y − 5xy + 8y = x5 yp = −e−x We can ﬁnd the homogeneous solution of the Cauchy-Euler equation as: yp = v1 (x)y1 + v2 (x)y2 (4.7) yh = c1 x4 + c2 x2 There are two unknowns, so we may impose an extra condition. Let’s choose v1 y1 + v2 y2 = 0 for simplicity. Inserting yp in the equation, we obtain r a = 0 v1 y1 + v2 y2 = v1 y1 + v2 y2 −y2 r , aW v2 = y1 r aW (4.8) (4.9) where W is the Wronskian W = y1 y2 y1 y2 = y1 y2 − y2 y1 x4 x 2 4x3 2x = −2x5 Therefore the particular solution is The solution to this linear system is v1 = W = (4.10) yp (x) = −x4 1 yp (x) = x4 2 1 5 yp (x) = x 3 The general solution is x2 x5 dx + x2 x2 (−2x5 ) 1 dx − x2 x2 dx 2 1 y = c1 x 4 + c2 x 2 + x 5 3 x4 x5 dx x2 (−2x5 )
21. 21. EXERCISES 31 Exercises 1) 2) 3) 4) 5) 6) 32 CHAPTER 4. NONHOMOGENEOUS EQUATIONS Answers Find the general solution of the following diﬀerential equations y + 4y = x cos x y − 18y + 81y = e9x y = −4x cos 2x − 4 cos 2x − 8x sin 2x − 8 sin 2x y + 3y − 18y = 9 sinh 3x y + 16y = x2 + 2x y − 2y + y = x2 ex 7) 2x2 y − xy + y = 1 x 1) y = c1 sin 2x + c2 cos 2x + 1 x cos x + 2 sin x 3 9 1 2) y = c1 e9x + c2 xe9x + x2 e9x 2 3) y = c1 + c2 x + x cos 2x + 3 cos 2x + 2x sin 2x + sin 2x 1 1 4) y = c1 e3x + c2 e−6x + e−3x + xe3x 4 2 1 2 1 1 5) y = c1 sin 4x + c2 cos 4x + x + x − 16 8 128 1 6) y = c1 ex + c2 xex + x4 ex 12 √ 1 7) y = c1 x + c2 x + 6x 8) x y + xy − 4y = x ln x 9) y − 8y + 16y = 16x 10) y = x3 11) y + 7y + 12y = e2x + x 1 1 x2 8) y = c1 x2 + c2 x−2 + x2 ln2 x − x2 ln x + 8 16 64 1 9) y = c1 e4x + c2 xe4x + x + 2 x5 10) y = + c1 + c2 x 20 12) y + 12y + 36y = 100 cos 2x 11) y = c1 e−3x + c2 e−4x + 2 2 13) y + 9y = ex + cos 3x + 2 sin 3x 1 7 1 2x e + x− 30 12 144 12) y = c1 e−6x + c2 xe−6x + 2 cos 2x + 3 sin 2x 2 1 x 1 1 e − x cos 3x + x sin 3x 10 3 6 14) y + 10y + 16y = e−2x 13) y = c1 cos 3x + c2 sin 3x + 15) y − 4y + 53y = (53x)2 1 14) y = c1 e−2x + c2 e−8x + xe−2x 6 16) y + y = (x2 + 1)e3x 15) y = e2x (c1 cos 7x + c2 sin 7x) + 53x2 + 8x − 17) y + y = csc x 16) y = e3x (0.1x2 − 0.12x + 0.152) + c1 sin x + c2 cos x 17) y = c1 sin x + c2 cos x − x cos x + sin x ln | sin x| 18) y = c1 sin x + c2 cos x − cos x ln | sec x + tan x| − sin x ln | csc x + cot x| 18) y + y = csc x sec x 19) y − 4y + 4y = e2x ln x x 20) y − 2y + y = e2x (ex + 1)2 19) y = c1 e2x + c2 xe2x + xe2x 74 53 (ln x)2 − ln x + 1 2 20) y = c1 ex + c2 xex + ex ln(1 + ex )
22. 22. 34 CHAPTER 5. HIGHER ORDER EQUATIONS means that all the constants c1 , c2 , . . . , cn are zero, then this set of functions is linearly independent. Otherwise, they are dependent. For example, the functions x, x2 , x3 are linearly independent. The functions cos2 x, sin2 x, cos 2x are not. Given n functions, we can check their linear dependence by calculating the Wronskian. The Wronskian is deﬁned as Chapter 5 Higher Order Equations W (y1 , y2 , . . . , yn ) = y1 y1 . . . (n−1) y1 In this chapter, we will generalize our results about second order equations to higher orders. The basic ideas are the same. We still need the homogeneous solution to ﬁnd the general nonhomogeneous solution. We will extend the two methods, undetermined coeﬃcients and variation of parameters, to higher dimensions and this will naturally involve many more terms and constants in the solution. We also need some new notation to express nth derivatives easily. ... ... yn yn . . . (5.4) (n−1) . . . yn and the functions are linearly dependent if and only if W = 0 at some point. 5.2 Diﬀerential Operators We can denote diﬀerentiation with respect to x by the symbol D Dy = dy =y, dx D2 y = d2 y =y dx2 (5.5) etc. A diﬀerential operator is 5.1 An n form Linear Equations of Order n th L = a0 Dn + a1 Dn−1 + · · · + an−1 D + an order diﬀerential equation is called linear if it can be written in the a0 (x) dn y dn−1 y dy + a1 (x) n−1 + · · · + an−1 (x) + an (x)y = r(x) n dx dx dx (5.1) and nonlinear if it is not linear.(Note that a0 = 0) If the coeﬃcients a0 (x), a1 (x), . . . an (x) are continuous, then the equation has exactly n linearly independent solutions. The general solution is y = c1 y1 + c2 y2 + · · · + cn yn (5.2) We will only work with operators where coeﬃcients are constant. We can add, multiply, expand and factor constant coeﬃcient diﬀerential operators using common rules of algebra. In this respect, they are like polynomials. So, the following expressions are all equivalent: (D − 2)(D − 3)y = (D − 3)(D − 2)y = (D2 − 5D + 6)y = y − 5y + 6y Let’s apply some simple operators to selected functions: (D − 2)ex = Dex − 2ex = ex − 2ex = −ex Linear Independence: If c1 y1 + c2 y2 + · · · + cn yn = 0 33 (5.3) (5.6)
23. 23. 5.3. HOMOGENEOUS EQUATIONS 35 36 CHAPTER 5. HIGHER ORDER EQUATIONS (D − 2)e2x = De2x − 2e2x = 2e2x − 2e2x = 0 Now we are in a position to solve very complicated-looking homogeneous equations. (D − 2)2 xe2x = (D − 2)(D − 2)xe2x = (D − 2)(e2x + 2xe2x − 2xe2x ) = (D − 2)e2x = 0 Method of Solution: • Express the given equation using operator notation (D notation). (D2 − 4) sin(2x) = (D − 2)(D + 2) sin(2x) = (D − 2)(2 cos(2x) + 2 sin(2x)) = −4 sin(2x) + 4 cos(2x) − 4 cos(2x) − 4 sin(2x) = −8 sin 2x • Factor the polynomial. • Find the solution for each component. • Add the components to obtain the general solution. 5.3 Homogeneous Equations Example 5.1 Find the general solution of y (4) − 7y + y − 7y = 0. Based on the examples in the previous section, we can easily see that: The general solution of the equation (D − a)n y = 0 is y = eax (c0 + c1 x + . . . + cn−1 xn−1 ) In operator notation, we have (D4 − 7D3 + D2 − 7D)y = 0 (5.7) Factoring this, we obtain if a is real. Some special cases are: D(D − 7)(D2 + 1)y = 0 Dn y = 0 ⇒ y = c0 + c1 x + . . . + cn−1 xn−1 (D − a)y = 0 ⇒ y = eax (D − a)2 y = 0 ⇒ y = c1 eax + c2 xeax We know that (5.8) We can extend these results to the case of complex roots. If z = a + ib is a root of the characteristic polynomial, then so is z = a − ib. (Why?) Consider the equation (D − a − ib)n (D − a + ib)n y = (D2 − 2aD + a2 + b2 )n y = 0 (5.9) The solution is y = eax cos bx(c0 + c1 x + . . . + cn−1 xn−1 ) +eax sin bx(k0 + k1 x + . . . + kn−1 xn−1 ) (5.10) 2 (D + b )y = 0 ⇒ Therefore the general solution is y = c1 + c2 e7x + c3 sin x + c4 cos x Note that the equation is fourth order and the solution has four arbitrary constants. Example 5.2 Solve D3 (D − 2)(D − 3)2 (D2 + 4)y = 0. Using the same method, we ﬁnd: A special case is obtained if a = 0. 2 Dy = 0 ⇒ y = c (D − 7)y = 0 ⇒ y = ce7x (D2 + 1)y = 0 ⇒ y = c1 sin x + c2 cos x y = c1 cos bx + c2 sin bx (5.11) y = c1 + c2 x + c3 x2 + c4 e2x + c5 e3x + c6 e3x x + c7 cos 2x + c8 sin 2x
24. 24. 5.4. NONHOMOGENEOUS EQUATIONS 5.4 37 Nonhomogeneous Equations 38 CHAPTER 5. HIGHER ORDER EQUATIONS Then we will proceed similarly to simplify the steps. Eventually, we will obtain the system In this section, we will generalize the methods of undetermined coeﬃcients and variation of parameters to nth order equations. Undetermined Coeﬃcients: Method of undetermined coeﬃcients is the same as given on page 27. We will use the same table, but this time the modiﬁcation rule is more general. It should be: • In case one of the terms of yp occurs in yh , multiply it by xk where k is the smallest integer which will eliminate any duplication between yp and yh . v1 y1 v1 y1 . . . (n−1) v1 y1 (n) v1 y1 yp = y1 The homogeneous solution is yh = (c0 + c1 x + c2 x2 + c3 x3 )ex . According to the table, we should choose yp as Aex + Bxex , but this already occurs in the homogeneous solution. Multiplying by x, x2 , x3 are not enough, so, we should multiply by x4 . yp = Ax4 ex + Bx5 ex Inserting this in the equation, we obtain: 24Aex + 120Bxex = xex 1 5 x xe 120 n−1 d y d y dy + a1 (x) n−1 + · · · + an−1 (x) + an (x)y = r(x) n dx dx dx (5.12) Let the homogeneous solution be yh = c1 y1 + · · · + cn yn Then the particular solution is yp = v1 y1 + · · · + vn yn Here, vi are functions of x. Since we have n functions, we can impose n − 1 conditions on them. The ﬁrst condition will be v1 y1 + · · · + vn yn = 0 = = + + (n−1) vn y n (n) vn yn = = 0 0 . . . (5.14) 0 r(x) a0 (x) v1 dx + · · · + yn vn dx (5.15) x3 y − 6x2 y + 15xy − 15y = 8x6 We can ﬁnd the homogeneous solution yh = c1 x + c2 x3 + c3 x5 using our method for Cauchy-Euler equations. Then, the particular solution will be yp = xv1 + x3 v2 + x5 v3 . Using the above equations, we obtain the system xv1 + x3 v2 + x5 v3 = 0 v1 + 3x2 v2 + 5x4 v3 = 0 6xv2 + 20x3 v3 = 8x3 Variation of Parameters: The idea is the same as in second order equations, but there are more unknowns to ﬁnd and more integrals to evaluate. Consider a0 (x) vn yn vn yn . . . Example 5.4 Find the general solution of Therefore A = 0, B = 1/120 and the general solution is n + ··· + ··· + + Then, we will solve this linear system to ﬁnd vi , and integrate them to obtain yp . Example 5.3 Solve the equation (D − 1)4 y = xex . y = (c0 + c1 x + c2 x2 + c3 x3 )ex + + ··· + ··· (5.13) The solution of this system is v1 = x4 , v2 = −2x2 , v3 = 1 therefore the particular solution is yp = x x4 dx + x3 (−2x2 ) dx + x5 and the general solution is y = c1 x + c2 x 3 + c3 x 5 + 8 6 x 15 dx = 8 6 x 15
25. 25. EXERCISES 39 Exercises CHAPTER 5. HIGHER ORDER EQUATIONS Answers 1) y = c0 + c1 x + c2 x2 + c3 x3 + c4 x4 1) D5 y = 0 2) y = c1 ex + c2 xex + c3 x2 ex 2) (D − 1)3 y = 0 3) y − 4y + 13y = 0 4) (D − 2)2 (D + 3)3 y = 0 5) (D2 + 2)3 y = 0 d4 y d2 y + 5 2 + 4y = 0 dx4 dx 7) (D2 + 9)2 (D2 − 9)2 y = 0 6) 4 40 3 2 dy dy dy −2 3 +2 2 =0 4 dx dx dx 9) y − 3y + 12y − 10y = 0 8) 3) y = c1 e2x cos 3x + c2 e2x sin 3x + c3 4) y = c1 e2x + c2 xe2x + c3 e−3x + c4 xe−3x + c5 x2 e−3x √ √ √ √ 5) y = c1 cos 2x + c2 sin 2x + c3 x cos 2x + c4 x sin 2x √ √ + c5 x2 cos 2x + c6 x2 sin 2x 6) y = c1 cos 2x + c2 sin 2x + c3 cos x + c4 sin x 7) y = c1 e3x + c2 xe3x + c3 e−3x + c4 xe−3x + c5 cos 3x + c6 sin 3x + c7 x cos 3x + c8 x sin 3x 8) y = c1 + c2 x + c3 ex cos x + c4 ex sin x 10) (D2 + 2D + 17)2 y = 0 9) y = c1 ex + c2 ex cos 3x + c3 ex sin 3x 11) (D4 + 2D2 + 1)y = x2 10) y = c1 e−x sin 4x + c2 e−x cos 4x + c3 xe−x sin 4x + c4 xe−x cos 4x 12) (D3 + 2D2 − D − 2)y = 1 − 4x3 11) y = c1 cos x + c2 sin x + c3 x cos x + c4 x sin x + x2 − 4 √ √ √ 13) (2D4 + 4D3 + 8D2 )y = 40e−x [ 3 sin( 3x) + 3 cos( 3x)] 14) (D3 − 4D2 + 5D − 2)y = 4 cos x + sin x 15) (D3 − 9D)y = 8xex 12) y = c1 ex + c2 e−x + c3 e−2x + 2x3 − 3x2 + 15x − 8 √ √ √ 13) y = c1 + c2 x + c3 e−x cos 3x + c4 e−x sin 3x + 5xe−x cos 3x 14) y = c1 ex + c2 xex + c3 e2x + 0.2 cos x + 0.9 sin x 3 15) y = c1 + c2 e3x + c3 e−3x + ex − xex 4
26. 26. 42 CHAPTER 6. SERIES SOLUTIONS term wise, i.e. ∞ (an ± bn )(x − x0 )n f (x) ± g(x) = n=0 • In the interval of convergence, the series can be multiplied or divided to give another power series. Chapter 6 ∞ cn (x − x0 )n f (x)g(x) = Series Solutions n=0 where cn = a0 bn + a1 bn−1 + · · · + an b0 If none of the methods we have studied up to now works for a diﬀerential equation, we may use power series. This is usually the only choice if the solution cannot be expressed in terms of the elementary functions. (That is, exponential, logarithmic, trigonometric and polynomial functions). If the solution can be expressed as a power series, in other words, if it is analytic, this method will work. But it takes time and patience to reach the solution. Remember, we are dealing with inﬁnitely many coeﬃcients! 6.1 Power Series • In the interval of convergence, derivatives and integrals of f (x) can be found by term wise diﬀerentiation and integration, for example ∞ n an (x − x0 )n−1 f (x) = a1 + 2a2 (x − x0 ) + · · · = n=1 (n) (x • The series ∞ f n! 0 ) (x − x0 )n is called the Taylor Series of the funcn=0 tion f (x). The function f (x) is called analytic if its Taylor series converges. Examples of some common power series are: ∞ Let’s remember some facts about the series ex = ∞ an (x − x0 )n = a0 + a1 (x − x0 ) + a2 (x − x0 )2 + · · · n=0 (6.1) n=0 ∞ cos x = n=0 from calculus. ∞ sin x = • There is a nonnegative number ρ, called the radius of convergence, such that the series converges absolutely for |x − x0 | < ρ and diverges for |x − x0 | > ρ . The series deﬁnes a function f (x) = ∞ an (x − x0 )n n=0 in its interval of convergence. n=0 1 = 1−x 41 =1+x+ (−1)n x2n 2n! =1− x2 + ··· 2! x2 x4 + − ··· 2! 4! (−1)n x2n+1 x3 x5 =x− + − ··· (2n + 1)! 3! 5! ∞ xn = 1 + x + x2 + · · · (−1)n+1 xn n =x− n=0 ∞ ln(1 + x) = • In the interval of convergence, the series can be added or subtracted xn n! n=1 x2 x3 + − ··· 2 3
27. 27. 6.2. CLASSIFICATION OF POINTS 6.2 43 Classiﬁcation of Points 44 CHAPTER 6. SERIES SOLUTIONS Example 6.1 Solve y + 2xy + 2y = 0 around x0 = 0. First we should classify the point. Obviously, x = 0 is an ordinary point, so we can use power series method. Consider the equation R(x)y + P (x)y + Q(x)y = 0 (6.2) ∞ ∞ n y= If both of the functions P (x) , R(x) Q(x) R(x) an x , y = n=0 (6.3) are analytic at x = x0 , then the point x0 is an ordinary point. Otherwise, x0 is a singular point. Suppose that x0 is a singular point of the above equation. If both of the functions Q(x) P (x) , (x − x0 )2 (6.4) (x − x0 ) R(x) R(x) are analytic at x = x0 , then the point x0 is called a regular singular point. Otherwise, x0 is an irregular singular point. For example, the functions 1+x+x2 , sin x, ex (1+x4 ) cos x are all analytic cos x 1 ex 1 + x2 at x = 0. But, the functions , , , are not. x x x x3 We will use power series method around ordinary points and Frobenius’ method around regular singular points. We will not consider irregular singular points. ∞ nan x n−1 n(n − 1)an xn−2 , y = n=1 n=2 Inserting these in the equation, we obtain ∞ ∞ n(n − 1)an x n−2 ∞ + 2x n=2 nan x n−1 an x n = 0 +2 n=1 ∞ n=0 ∞ ∞ n(n − 1)an xn−2 + n=2 2nan xn + n=1 2an xn = 0 n=0 To equate the powers of x, let us replace n by n + 2 in the ﬁrst sigma. (n → n + 2) ∞ ∞ ∞ n=1 n=0 2an xn = 0 2nan xn + (n + 2)(n + 1)an+2 xn + n=0 Now we can express the equation using a single sigma, but we should start the index from n = 1. Therefore we have to write n = 0 terms separately. ∞ [(n + 2)(n + 1)an+2 + (2n + 2)an ] xn = 0 2a2 + 2a0 + n=1 6.3 Power Series Method If x0 is an ordinary point of the equation R(x)y + P (x)y + Q(x)y = 0, then the general solution is ∞ an (x − x0 )n y= −2(n + 1) −2 an = an (n + 2)(n + 1) (n + 2) This is called the recursion relation. Using it, we can ﬁnd all the constants in terms of a0 and a1 . a2 = −a0 , an+2 = (6.5) 2 1 a4 = − a2 = a0 4 2 2 1 a6 = − a4 = − a0 6 6 2 a3 = − a1 , 3 2 4 a 5 = − a3 = a1 5 15 n=0 The coeﬃcients an can be found by inserting y in the equation and setting the coeﬃcients of all powers to zero. Two coeﬃcients (Usually a0 and a1 ) must be arbitrary, others must be deﬁned in terms of them. We expect two linearly independent solutions because the equation is second order linear. We can ﬁnd as many coeﬃcients as we want in this way. Collecting them together, the solution is : 1 1 y = a0 1 − x 2 + x 4 − x 6 + · · · 2 6 2 4 + a1 x − x 3 + x 5 + · · · 3 15
28. 28. 6.3. POWER SERIES METHOD 45 In most applications, we want a solution close to 0, therefore we can neglect the higher order terms of the series. Remark: Sometimes we can express the solution in closed form (in terms of elementary functions rather than an inﬁnite summation) as in the next example: Example 6.2 Solve (x − 1)y + 2y = 0 around x0 = 0. Once again, ﬁrst we should classify the given point. The function analytic at x = 0, therefore x = 0 is an ordinary point. ∞ ∞ ∞ an x n , y = y= n=0 2 is x−1 nan xn−1 , y = n=1 n(n − 1)an xn−2 n=2 Inserting these in the equation, we obtain ∞ (x − 1) ∞ n(n − 1)an x n−2 nan xn−1 = 0 +2 n=2 n=1 ∞ ∞ ∞ n(n − 1)an xn−2 + n(n − 1)an xn−1 − n=1 n=2 n=2 2nan xn−1 = 0 46 CHAPTER 6. SERIES SOLUTIONS Exercises Find the general solution of the following diﬀerential equations in the form of series. Find solutions around the origin (use x0 = 0). Write the solution in closed form if possible. 1) (1 − x2 )y − 2xy = 0 2) y + x4 y + 4x3 y = 0 3) (2 + x3 )y + 6x2 y + 6xy = 0 4) (1 + x2 )y − xy − 3y = 0 5) (1 + 2x2 )y + xy + 2y = 0 6) y − xy + ky = 0 7) (1 + x2 )y − 4xy + 6y = 0 8) (1 − 2x2 )y + (2x + 4x3 )y − (2 + 4x2 )y = 0 9) (1 + 8x2 )y − 16y = 0 10) y + x2 y = 0 The following equations give certain special functions that are very important in applications. Solve them for n = 1, 2, 3 around origin. Find polynomial solutions only. To equate the powers of x, let us replace n by n+1 in the second summation. ∞ ∞ n(n − 1)an x n−1 − n=2 ∞ (n + 1)nan+1 x n−1 + n=1 2nan x n−1 =0 n=1 Now we can express the equation using a single sigma. 11) 12) 13) 14) (1 − x2 )y − 2xy + n(n + 1)y = 0 y − 2xy + 2ny = 0 xy + (1 − x)y + ny = 0 (1 − x2 )y − xy + n2 y = 0 (Legendre’s Equation) (Hermite’s Equation) (Laguerre’s Equation) (Chebyshev’s Equation) ∞ [(n(n − 1) + 2n)an − n(n + 1)an+1 ] xn−1 = 0 (−2a2 + 2a1 ) + n=2 a2 = a1 , an+1 = n2 − n + 2n an for n n(n + 1) 2 So the recursion relation is: an+1 = an All the coeﬃcients are equal to a1 , except a0 . We have no information about it, so it must be arbitrary. Therefore, the solution is: y = a0 + a1 x + x 2 + x 3 + · · · x y = a0 + a1 1−x Solve the following initial value problems. Find the solution around the point where initial conditions are given. 15) 16) 17) 18) xy + (x + 1)y − 2y = 0, y + 2xy − 4y = 0, 4y + 3xy − 6y = 0, (x2 − 4x + 7)y + y = 0, x0 x0 x0 x0 = −1, = 0, =0 =2 y(−1) = 1, y(0) = 1, y(0) = 4, y(2) = 4, y (−1) = 0 y (0) = 0 y (0) = 0 y (2) = 10 19) Find the recursion relation for (p + x2 )y + (1 − q − r)xy + qry = 0 around x = 0. (Here p, q, r are real numbers, p = 0) 20) Solve (1 + ax2 )y + bxy + cy = 0 around x0 = 0
29. 29. EXERCISES 47 Answers x3 x5 + + ··· 3 5 n=3 x3 x6 x9 + − + ··· 2 4 8 a1 x + 3 1+ x 2 3) y = a0 1 − a0 3 1+ x 2 3 1 3 4) y = a0 1 + x2 + x4 − x6 + · · · 2 8 16 2 2 5) y = a0 1 − x2 + x4 − x6 + · · · 3 3 6) y = a0 1 − +a1 x − 2 + a1 x + x3 3 17 1 + a1 x − x 3 + x 5 + · · · 2 40 k − 1 3 (k − 1)(k − 3) 5 (k − 1)(k − 3)(k − 5) 7 x + x − x + ··· 3! 5! 7! x3 3 x4 x6 1+x + + + ··· 2 6 2 10) y = a0 1 − ⇒ ⇒ n=3 ⇒ x8 x4 + + ··· 12 672 + a1 x y = a1 x y = a0 (1 − 3x2 ) 5 y = a1 (x − x3 ) 3 + a1 x − ⇒ 14) n = 1 n=2 ⇒ ⇒ ⇒ x5 x9 + + ··· 20 1440 y = a1 x y = a0 (1 − 2x2 ) 2 y = a1 (x − x3 ) 3 y = a0 (1 − x) 1 y = a0 (1 − 2x + x2 ) 2 3 1 y = a0 (1 − 3x + x2 − x3 ) 2 6 y = a1 x y = a0 (1 − 2x2 ) 4 y = a1 (x − x3 ) 3 1 1 15) y = 1 − (x + 1)2 − (x + 1)3 − (x + 1)4 − · · · 3 6 16) y = 1 + 2x2 17) y = 4 + 3x2 1 1 18) y = 4 1 − (x − 2)2 + (x − 2)4 + · · · 6 72 1 7 + 10 (x − 2) − (x − 2)3 + (x − 2)5 + · · · 18 1080 19) an+2 = − 8 64 9) y = a0 (1 + 8x2 ) + a1 x + x3 − x5 + · · · 3 15 11) n = 1 n=2 OR k 2 k(k − 2) 4 k(k − 2)(k − 4) 6 x + x − x + ··· 2! 4! 6! 7) y = a0 (1 − 3x2 ) + a1 x − 8) y = a0 x4 x7 x10 + − + ··· 2 4 8 ⇒ n=3 + a1 x − ⇒ n=3 x5 x10 x15 + − + ··· 5 5 · 10 5 · 10 · 15 x11 x16 x6 + − + ··· x− 6 6 · 11 6 · 11 · 16 ⇒ n=2 1 1+x ln 2 1−x ⇒ ⇒ 13) n = 1 OR y = a0 + a1 2) y = a0 1 − y= CHAPTER 6. SERIES SOLUTIONS 12) n = 1 n=2 1) y = a0 + a1 x + +a1 48 (n − q)(n − r) an p(n + 2)(n + 1) x4 x2 + c(2a + 2b + c) 2 4! x6 + ··· − c(2a + 2b + c)(12a + 4b + c) 6! x3 x5 + a1 x − (b + c) + (b + c)(6a + 3b + c) 3! 5! x7 −(b + c)(6a + 3b + c)(20a + 5b + c) + ··· 7! 20) y = a0 1 − c
30. 30. 50 CHAPTER 7. FROBENIUS’ METHOD Case 2 - Equal roots: A basis of solutions is ∞ ∞ an x n , y 1 = xr y2 = y1 ln x + xr bn x n (7.3) n=1 n=0 Case 3 - Roots diﬀering by an integer: A basis of solutions is Chapter 7 ∞ ∞ an xn , y 1 = xr 1 bn x n y2 = ky1 ln x + xr2 n=1 (7.4) n=0 In this chapter, we will extend the methods of the previous chapter to regular singular points. The calculations will be considerably longer, but the basic ideas are the same. The classiﬁcation of the given point is necessary to make a choice of methods. where r1 − r2 = N > 0 (r1 is the greater root) and k may or may not be zero. In all three cases, there is at least one relatively simple solution of the form y = xr ∞ an xn . The equation is second order, so there must be a n=0 second linearly independent solution. In Cases 2 and 3, it may be diﬃcult to ﬁnd the second solution. You may use the method of reduction of order. This is convenient especially if y1 is simple enough. Alternatively, you may use the above formulas directly, and determine bn one by one using the an and the equation. 7.1 7.2 Frobenius’ Method An Extension of Power Series Method Suppose x0 is a regular singular point. For simplicity, assume x0 = 0. Then p(x) q(x) the diﬀerential equation can be written as y + y + 2 y = 0 where x x p(x) and q(x) are analytic. We can try a solution of the form ∞ y = xr an x n (7.1) n=0 The equation corresponding to the lowest power xr−2 , in other words r(r − 1) + p0 r + q0 = 0 is called the indicial equation, where p0 = p(0), and q0 = q(0). Now we can ﬁnd r, insert it in the series formula, and proceed as we did in the previous chapter. We can classify the solutions according to the roots of the indicial equation. Case 1 - Distinct roots not diﬀering by an integer: A basis of solutions is ∞ ∞ y 1 = xr 1 an x n , y2 = xr2 bn x n n=0 n=0 49 (7.2) Examples Example 7.1 Solve 4xy + 2y + y = 0 around x0 = 0. 2 First we should classify the given point. The function 4x is not analytic at x = 0 therefore x = 0 is a singular point. We should make a further test to determine whether it is regular or not. x2 The functions 2x and 4x are analytic therefore x = 0 is a R.S.P., we can 4x use the method of Frobenius. ∞ ∞ an xn+r , y = y= n=0 ∞ (n + r)an xn+r−1 , y = n=0 (n + r)(n + r − 1)an xn+r−2 n=0 Note that the summation for the derivatives still starts from 0, because r does not have to be an integer. This is an important diﬀerence between methods of power series and Frobenius. Inserting these in the equation, we obtain ∞ ∞ (n + r)(n + r − 1)an xn+r−2 + 2 4x n=0 ∞ (n + r)an xn+r−1 + n=0 an xn+r = 0 n=0
31. 31. 7.2. EXAMPLES 51 ∞ ∞ 4(n + r)(n + r − 1)an x n+r−1 ∞ + n=0 2(n + r)an x n=0 n+r−1 an xn+r = 0 + 52 CHAPTER 7. FROBENIUS’ METHOD For simplicity, we may choose a0 = 1. Then an = We want to equate the powers of x, so n → n + 1 in the ﬁrst two terms. ∞ ∞ ∞ 4(n + r + 1)(n + r)an+1 xn+r + n=−1 2(n + r + 1)an+1 xn+r + n=−1 an xn+r = 0 Therefore the second solution is : ∞ n=0 Now we can express the equation using a single sigma, but the index of the common sigma must start from n = 0. Therefore we have to write n = −1 terms separately. [4r(r−1)+2r]a0 xr−1 + {[4(n + r + 1)(n + r) + 2(n + r + 1)]an+1 + an } xn+r = 0 n=0 We know that a0 = 0, therefore 4r2 − 2r = 0. This is the indicial equation. Its solutions are r = 0, r = 1 . Therefore this is Case 1. 2 If r = 0, the recursion relation is n=0 The general solution is y = c1 y1 + c2 y2 2 First we should classify the given point. The function x x−x is not analytic 2 at x = 0 therefore x = 0 is a singular point. The functions x − 1 and 1 + x are analytic at x = 0 therefore x = 0 is a R.S.P., we can use the method of Frobenius. Evaluating the derivatives of y and inserting them in the equation, we obtain −1 1 an 4(n + 1)(n + 2 ) an+1 = ∞ For simplicity, we may choose a0 = 1. Then n=0 ∞ − (n + r)an x (−1) 2n! y1 = n=0 n=0 + an x n+r ∞ an xn+r+1 = 0 + n=0 n=0 ∞ ∞ (n + r)(n + r − 1)an x n=0 ∞ n n (n + r)an xn+r+1 + Let’s replace n by n − 1 in the second and ﬁfth terms. Therefore the ﬁrst solution is: ∞ n+r ∞ n+r n=0 n an = ∞ (n + r)(n + r − 1)an x a0 a1 a0 a2 a0 a1 = − , a 2 = − , a3 = − ,... 3 = 5 = − 2 4! 6! 4.2. 2 4.3. 2 − √ (−1) x = cos x 2n! 1 If r = , the recursion relation is 2 a1 = − √ (−1)n xn = sin x (2n + 1)! y2 = x1/2 Example 7.2 Solve x2 y + (x2 − x)y + (1 + x)y = 0 around x0 = 0. ∞ an+1 = (−1)n (2n + 1)! n=0 −1 −an an = (2n + 3)(2n + 2) 4(n + 3 )(n + 1) 2 a0 a1 a0 a2 a0 , a2 = − = , a3 = − = − ,... 3.2 5.4 5! 7.6 7! n+r n=1 ∞ (n + r)an x n=0 n+r an x + n=0 (n + r − 1)an−1 xn+r + n+r ∞ an−1 xn+r = 0 + n=1 [r2 − 2r + 1]a0 xr + ∞ {[(n + r)(n + r − 1) − (n + r) + 1]an + [(n + r − 1) + 1]an−1 } xn+r = 0 n=1 The indicial equation is r2 − 2r + 1 = 0 ⇒ r = 1 (double root). Therefore this is Case 2. The recursion relation is an = − n+1 an−1 n2
32. 32. 7.2. EXAMPLES 53 54 CHAPTER 7. FROBENIUS’ METHOD Exercises For simplicity, let a0 = 1. Then 3 3 4 2 a1 = −2, a2 = − a1 = , a3 = − a2 = − 4 2 9 3 Find two linearly independent solutions of the following diﬀerential equations in the form of series. Find solutions around the origin (use x0 = 0). Write the solution in closed form if possible. 1) 2x2 y − xy + (1 + x)y = 0 Therefore the ﬁrst solution is : 3 2 y1 = x 1 − 2x + x2 − x3 + · · · 2 3 2) 2xy + (1 + x)y − 2y = 0 To ﬁnd the second solution, we will use reduction of order. Let y2 = uy1 . Inserting y2 in the equation, we obtain 3) (x2 + 2x)y + (3x + 1)y + y = 0 4) xy − y − 4x3 y = 0 2 2 2 x y1 u + (2x y1 − xy1 + x y1 )u = 0 Let w = u then −2 To evaluate the integral u = 1−x+ w= u= 1 x ⇒ w= 8) (2x2 + 2x)y − y − 4y = 0 xe−x 2 y1 9) 2x2 y + (2x2 − x)y + y = 0 1 w dx we need to ﬁnd 2 . This is also a series. y1 3 2 1 − 2x + x2 − x3 + · · · 2 3 xe−x =x 2 y1 7) x2 y + (x2 − x)y + y = 0 y1 1 + − 1 dx y1 x ln w = −2 ln y1 + ln x − x w= 6) 3x2 y + (−10x − 3x2 )y + (14 + 4x)y = 0 1 y w + 2 1 − +1 w =0 y1 x dw = w 1 1 = 2 2 y1 x 5) xy + y − xy = 0 −2 = x2 x3 − + ··· 2! 3! 1 + 3x + 1 x2 1 x2 1 + 4x + 9x2 + 1 + 4x + 9x2 + 11 2 13 3 x + x + ··· 2 6 w dx = ln x + 3x + 11 2 13 3 x + x + ··· 4 18 13 3 y2 = uy1 = y1 ln x + x 3x − x2 + x3 + · · · 4 2 46 3 x + ··· 3 46 3 x + ··· 3 10) 4x2 y + (2x2 − 10x)y + (12 − x)y = 0 11) (x2 + 2x)y + (4x + 1)y + 2y = 0 Use Frobenius’ method to solve the following diﬀerential equations around origin. Find the roots of the indicial equation, ﬁnd the recursion relation, and two linearly independent solutions. 12) (x2 + cx)y + [(2 + b)x + c(1 − d)]y + by = 0 (b = 0, c = 0, d is not an integer). 13) x2 y + [(1 − b − d)x + cx2 ]y + [bd + (1 − b)cx]y = 0 (c = 0, b − d is not an integer). 14) x2 y + [(1 − 2d)x + cx2 ]y + (d2 + (1 − d)cx)y = 0 (c = 0) 15) xy + [1 − d + cx2 ]y + 2cxy = 0 (c = 0, d is not an integer).
33. 33. EXERCISES 55 ∞ n=1 (−1)n xn n! · 3 · 5 · 7 · · · (2n + 1) 1 1+ n=1 2) y = c1 (−1)n xn n! · 1 · 3 · 5 · · · (2n − 1) 1 1 + 2x + x2 3 ∞ + c2 x 1 2 4) y = a0 n=0 1+ n=1 6 2 3) y1 = 1−x+ x2 − x3 +· · · , 3 15 ∞ n+b an c (n + 1 − d) b b(b + 1) y1 = 1 − x+ 2 x2 − · · · c(1 − d) c (1 − d)(2 − d) 12) r = 0 ∞ + c2 x 2 (−1)n 3xn 2n n!(2n − 3)(2n − 1)(2n + 1) 3 15 35 3 y2 = x1/2 1 − x + x2 − x + ··· 4 32 128 2 x4 x6 x2 + + + ··· 22 (2 · 4)2 (2 · 4 · 6)2 x2 3x4 11x6 y2 = y1 ln x − − − − ··· 4 8 · 16 64 · 6 · 36 9 27 3 3 6) y1 = x7/3 1 + x + x2 + x + ··· 4 28 280 x2 x3 y 2 = x2 1 + x + + + · · · = x2 e x 2! 3! x x − + · · · = xe−x 2! 3! x2 x3 y2 = xe−x ln x + xe−x x + + + ··· 2 · 2! 3 · 3! 1 1 1 y2 = x3/2 1 + x − x2 + x3 − · · · 2 8 16 ∞ y2 = x 1 + n=1 (−1)n (2x)n 1 · 3 · 5 · · · (2n + 1) ∞ 10) y1 = x2 e−x/2 , ⇒ an+1 = − y2 = x3/2 1 + n=1 (−1)n xn 1 · 3 · 5 · · · (2n − 1) an+1 = − ⇒ an = − n+b+d an c (n + 1) d+b (d + b)(d + b + 1) 2 1− x+ x − ··· c 2! c2 y 2 = xd c an−1 n+b−d c2 c x+ x2 − · · · 1+b−d (1 + b − d)(2 + b − d) c r = d ⇒ an = − an−1 n c2 2 c3 3 y2 = xd 1 − c x + x − x + · · · = xd e−cx 2! 3! y 1 = xb 1 − c an−1 n c3 c2 1 − c x + x2 − x3 + · · · 2! 3! 14) r = d (double root) an = − y 1 = xd 3 7) y1 = x 1 − x + 8) y1 = 1 − 4x − 8x2 , r=d 2 OR y = c1 ex + c2 e−x 5) y1 = 1 + 2 ⇒ 13) r = b ∞ x4n x4n+2 + a2 , (2n)! (2n + 1)! n=0 9) y1 = x1/2 e−x , CHAPTER 7. FROBENIUS’ METHOD 8 11) y1 = 1 − 2x + 2x2 − x3 + · · · 5 5 35 2 105 3 1/2 1− x+ y2 = x x − x + ··· 4 32 128 Answers 1) y = c1 x 1 + 56 y2 = xd e−cx ecx dx x y2 = xd e−cx ln x + xd e−cx cx + 15) r = 0 ⇒ = xd e−cx an+2 = − c2 2 c3 3 x + x + ··· 2 · 2! 3 · 3! c an (n + 2 − d) c c2 c3 x2 + x4 − x6 + · · · 2−d (2 − d)(4 − d) (2 − d)(4 − d)(6 − d) c r = d ⇒ an+2 = − an n+2 c c2 4 c3 y 2 = xd 1 − x2 + x − x6 + · · · 2 2·4 2·4·6 y1 = 1 −
34. 34. 58 CHAPTER 8. LAPLACE TRANSFORM I Example 8.1 Evaluate the Laplace transform of the following functions: a) f (t) = 1 ∞ 0 Chapter 8 0 1 = , s s>0 b) f (t) = eat ∞ Laplace Transform I 0 c) f (t) = Laplace transform provides an alternative method for many equations. We ﬁrst transform the diﬀerential equation to an algebraic equation, then solve it, and then make an inverse transform. Laplace transform has a lot of interesting properties that make these operations easy. In this chapter, we will see the deﬁnition and the basic properties. We will also compare this method to the method of undetermined coeﬃcients, and see in what ways Laplace transform is more convenient. 0 1 if if ∞ 1 ∞ e−st f (t)dt (8.1) 0 then, the inverse transform will be f (t) = L−1 {F (s)} (8.2) Note that we use lowercase letters for functions and capital letters for their transforms. 57 = 0 1 , s>a s−a ∞ = 1 e−s , s s>0 d) f (t) = t ∞ te−st dt L {t} = 0 Using integration by parts, we obtain L {t} = − The Laplace transform of a function f (t) is deﬁned as: e−st −s e−st dt = L {f } = Deﬁnition, Existence and Inverse of Laplace Transform ∞ 0<t<1 1 t L {t} = −t F (s) = L {f (t)} = e(a−s)t a−s eat e−st dt = L eat = 8.1 ∞ e−st −s e−st dt = L {1} = e−st s ∞ e−st s2 ∞ ∞ + 0 0 = 0 e−st dt s 1 , s>0 s2 The integral that deﬁnes the Laplace transform is an improper integral, it may or may not converge. In the above examples, the transform is deﬁned for a certain range of s. In practice, we can use Laplace transform on most of the functions we encounter in diﬀerential equations. The following deﬁnitions and the theorem answer the question Which functions have a Laplace transform? Piecewise Continuous Functions: A function f (t) is piecewise continuous on [a, b] if the interval can be subdivided into subintervals [ti , tj ], a = t0 < t1 < t2 · · · < tn = b such that f (t) is continuous on each interval and has ﬁnite one-sided limits at the endpoints (from the interior). An example can be seen on Figure 8.1.
35. 35. 8.2. PROPERTIES 59 60 CHAPTER 8. LAPLACE TRANSFORM I • Transform of Derivatives L {f } L {f } L f (n) = sL {f } − f (0) = s2 L {f } − sf (0) − f (0) = sn L {f } − sn−1 f (0) − sn−2 f (0) − · · · − f (n−1) (0) • Transform of Integrals t L f (x) dx = 0 F (s) s Example 8.2 Find the Laplace transform of sin at and cos at. Hint: Use Euler’s formula eix = cos x + i sin x and linearity. Figure 8.1: A piecewise continuous function Exponential Order: f (t) is of exponential order as t → ∞ if there exist real constants M, c, T such that |f (t)| M ect for all t T . In other words, a function is of exponential order if it does not grow faster than ect . Theorem 8.1: If f (t) is of exponential order and piecewise continuous on [0, k] for all k > 0, then its Laplace transform exists for all s > c. For example, all the polynomials have a Laplace transform. The function t2 e does NOT have a Laplace transform. 8.2 Basic Properties of Laplace Transforms It is diﬃcult to evaluate the Laplace transform of each function by performing an integration. Instead of this, we use various properties of Laplace transform. Let L {f (t)} = F (s), then, some basic properties are: (assuming these transforms exists) • Linearity L {af + bg} = aL {f } + bL {g} • Shifting L {eiat } − L {e−iat } 2i 1 1 1 a L {sin at} = − = 2 2i s − ia s + ia s + a2 Similarly, we can show that the transform of f (t) = cos at is s F (s) = 2 s + a2 1 Example 8.3 Find the inverse Laplace transform of F (s) = . (s + 5)2 Hint: Use shifting. sin at = eiat − e−iat 2i We know that L−1 1 s2 L−1 {F (s − a)} = eat f (t) L {sin at} = = t. Therefore 1 (s + 5)2 L−1 = te−5t Example 8.4 Find the Laplace transform of f (t) = t2 . Hint: Use Derivatives. Using L {f } = sL {f } − f (0), we obtain L {2t} 2 = 3 s s Example 8.5 Find the Laplace transform of f (t) = t3 . Hint: Use Integrals. L {2t} = sL t2 − 0 ⇒ L t2 = Using the integral rule, we see that L L eat f (t) = F (s − a) ⇒ L {t2 } 2 = 4 s s 6 L t3 = 4 s t3 3 =
36. 36. 8.3. INITIAL VALUE PROBLEMS 8.3 61 Initial Value Problems Consider the constant-coeﬃcient equation y + ay + by = r(t) (8.3) y(0) = p, y (0) = q 62 CHAPTER 8. LAPLACE TRANSFORM I The only disadvantage is that, sometimes ﬁnding the inverse Laplace transform is too diﬃcult. We have to ﬁnd roots of the polynomial s2 + as + b, which is the same as the characteristic polynomial we would encounter if we were using method of undetermined coeﬃcients. (8.4) with initial values Here y is a function of t (y = y(t)). We can solve it by the method of undetermined coeﬃcients. The method of Laplace transform will be an alternative that is more eﬃcient in certain cases. It also works for discontinuous r(t). Let us evaluate the Laplace transform of both sides. Example 8.6 Solve the initial value problem y + 4y = 0, y(0) = 5, y (0) = 3. Let’s start by ﬁnding the transform of the equation. L {y } + 4L {y} = 0 L {y } + aL {y } + bL {y} = L {r(t)} (8.5) Using L {y} = Y (s) and L {r(t)} = R(s) s2 Y − sp − q + a(sY − p) + bY = R (8.6) (s2 + as + b)Y = R + (s + a)p + q s2 Y − 5s − 3 + 4Y = 0 ⇒ (s2 + 4)Y = 5s + 3 5s + 3 Y = 2 s +4 Now, we have to ﬁnd the inverse transform of Y to obtain y(t). (8.7) Y = y = L−1 R + (s + a)p + q s2 + as + b R + sp + ap + q s2 + as + b (8.8) (8.9) Y = 5s 3 2 + + 4 2 s2 + 4 s2 3 sin 2t 2 Note that we did not ﬁrst ﬁnd the general solution containing arbitrary constants. We directly found the result. y(t) = L−1 {Y } = 5 cos 2t + Note that this method can be generalized to higher order equations. The advantages compared to the method of undetermined coeﬃcients are: Example 8.7 Solve the initial value problem • The initial conditions are built in the solution, we don’t need to determine constants after obtaining the general solution. • There is no distinction between homogeneous and nonhomogeneous equations, or single and multiple roots. The same method works in all cases the same way. y − 4y + 3y = 1, y(0) = 0, y (0) = − Transform both sides: L {y − 4y + 3y} = L {1} Use the derivative rule • The function on the right hand side r(t) belongs to a wider class. For example, it can be discontinuous. s2 Y − s.0 + 1 1 − 4(sY − 0) + 3Y = 3 s 1 3
37. 37. 8.3. INITIAL VALUE PROBLEMS Isolate Y 63 3−s 1 1 − = s 3 3s s−3 (s − 1)(s − 3)Y = − 3s 1 1 1 1 = − Y =− 3s(s − 1) 3 s s−1 64 CHAPTER 8. LAPLACE TRANSFORM I (s2 − 4s + 3)Y = f (t) F (s) f (t) F (s) 1 1 s eat − ebt a−b 1 (s − a)(s − b) t 1 s2 aeat − bebt a−b s (s − a)(s − b) n! eat sin bt b (s − a)2 + b2 s−a (s − a)2 + b2 Find the inverse transform y(t) = L−1 {Y } = 1 1 t − e 3 3 As you can see, there’s no diﬀerence between homogeneous and nonhomogeneous equations. Laplace transform works for both types in the same way. tn sn+1 eat Example 8.8 Solve the initial value problem y + 4y + 4y = 42te−2t , y(0) = 0, y (0) = 0 1 s−a eat cos bt teat 1 (s − a)2 a s 2 + a2 s s 2 + a2 tn eat −2t L {y } + 4L {y } + 4L {y} = 42L te 1 s2 Y + 4sY + 4Y = 42 · (s + 2)2 42 (s2 + 4s + 4)Y = (s + 2)2 42 Y = (s + 2)4 sin at cos at sinh at cosh at n! (s − a)n+1 a s 2 − a2 s s 2 − a2 42 3 −2t te 3! 2as (s2 + a2 )2 t sinh at 2as (s2 − a2 )2 t cos at s 2 − a2 (s2 + a2 )2 t cosh at s 2 + a2 (s2 − a2 )2 sin at − at cos at y(t) = L−1 {Y (s)} = t sin at 2a3 (s2 + a2 )2 sin at + at cos at 2as2 (s2 + a2 )2 y(t) = 7t3 e−2t If you try the method of undetermined coeﬃcients on this problem, you will appreciate the eﬃciency of Laplace transforms better. Table 8.1: A Table of Laplace Transforms
38. 38. EXERCISES 65 Exercises 3) f (t) = 2e−t cos2 t 4) f (t) = (t + 1)2 et 5) f (t) = t3 e3t 6) f (t) = 7) f (t) = CHAPTER 8. LAPLACE TRANSFORM I Answers Find the Laplace transform of the following functions: t 2) f (t) = et sin 3t 1) f (t) = cos2 2 t 0<t<a 0 a<t 66 1 0<t<a 0 a<t   t 0<t<a  8) f (t) = 1 a<t<b   0 b<t 1) F (s) = 2) F (s) = 3) F (s) = 4) F (s) = 5) F (s) = 1 s + 2 2s 2s + 2 3 (s − 1)2 + 9 1 s+1 + s + 1 s2 + 2s + 5 2 2 1 + + (s − 1)3 (s − 1)2 s − 1 6 (s − 3)4 1 − e−as s 1 ae−as e−as 7) F (s) = 2 − − 2 s s s −as −as 1−e e − ae−as − e−bs 8) F (s) = + s2 s 9) f (t) = cosh 2t − 2 sinh 2t 6) F (s) = Find the inverse Laplace transform of the following functions: s−4 9) F (s) = 2 s −4 3 10) F (s) = (s − 2)2 6 11) F (s) = s(s + 4) 1 12) F (s) = s(s2 + 9) 1 13) F (s) = 2 s (s + 1) 5s + 1 14) F (s) = 2 s +4 15) F (s) = 1 s+8 1 16) F (s) = (s − a)n Solve the following initial value problems using Laplace transform: 17) y − 2y + y = 0, y(0) = 4, y (0) = −3 18) y − 2y + 2y = 0, y(0) = 0, y (0) = 1 19) y + 2y = 4t2 + 12, y(0) = 4, y (0) = 0 20) y + 6y + 9y = e−3t , y(0) = 0, y (0) = 0 10) f (t) = 3te2t 11) f (t) = (3 − 3e−4t )/2 12) f (t) = (1 − cos 3t)/9 13) f (t) = e−t + t − 1 1 14) f (t) = 5 cos 2t + sin 2t 2 −8t 15) f (t) = e 16) f (t) = tn−1 eat (n − 1)! 17) y(t) = 4et − 7tet 18) y(t) = et sin t 19) y(t) = 4 + 2t2 1 20) y(t) = e−3t t2 2
39. 39. 68 CHAPTER 9. LAPLACE TRANSFORM II Reversing the order of integration, we obtain: ∞ ∞ f (x) g(t − x)e−st dt dx = 0 x Making the substitution u = t − x, we obtain: ∞ ∞ f (x) g(u)e−su−sx dudx L {f ∗ g} = Chapter 9 0 ∞ 0 ∞ 0 0 Laplace Transform II g(u)e−su du f (x)e−sx dx = =F (s) G(s) Example 9.1 Find the inverse Laplace transform of F (s) = In this chapter, we will study more advanced properties of Laplace transform. At the end, we will be able to ﬁnd transform and inverse transform of a wider range of functions. This will enable us to solve almost any linear constant coeﬃcient equation, including discontinuous inputs. 9.1 L−1 1 s2 1 s+4 = t, L−1 t (9.1) 0 The convolution operation is commutative, in other words f ∗ g = g ∗ f Theorem 9.1: The transform of convolution of two functions is equal to the product of their transforms, i.e. L {f ∗ g} = F (s) · G(s) −1 L {F (s) · G(s)} = f ∗ g where L {f } = F (s) and L {g} = G(s). Proof: Using the deﬁnitions of convolution and Laplace transform, t L {f ∗ g} =L f (x) g(t − x) dx 0 ∞ t f (x) g(t − x)e−st dx dt = 0 0 67 L−1 1 1 · 2 s+4 s (9.2) (9.3) = t ∗ e−4t xe−4(t−x) dx 0 xe4x e4x − 4 16 −4t t 1 e = − + 4 16 16 t = e−4t 0 Example 9.2 Find the inverse Laplace transform of F (s) = The convolution of two functions f and g is deﬁned as f (x)g(t − x) dx ⇒ 1 . + 4s2 t f (t) = t ∗ e−4t = Convolution h(t) = (f ∗ g)(t) = = e−4t s3 s . (s2 + 1)2 s 1 · 2 = L {cos t} · L {sin t}, + 1) (s + 1) we will see that f (t) = L−1 {F } = cos t ∗ sin t. If we express F as F (s) = (s2 t cos(x) sin(t − x) dx f (t) = 0 t = 0 1 = 2 = 1 [sin(t − x + x) + sin(t − x − x)] dx 2 t [sin(t) + sin(t − 2x)] dx 0 1 cos(t − 2x) x sin t + 2 2 t 0 1 1 = t sin t + (cos t − cos t) 2 2 1 = t sin t 2
40. 40. 9.2. UNIT STEP FUNCTION 9.2 69 70 CHAPTER 9. LAPLACE TRANSFORM II Unit Step Function ∞ e−st f (t)dt F (s) = The Heaviside step function (or unit step function) is deﬁned as 0 F (s) = ua (t) = u(t − a) = 0 1 if if t<a t a (9.4) This is a simple on oﬀ function. It is especially useful to express discontinuous inputs. Theorem 9.2: [t−shifting] Let L {f (t)} = F (s), then (9.5) Proof: Using the deﬁnition, ∞ e−st f (t − a) u(t − a) dt ∞ e−st f (t − a) dt a ∞ e−sa−sx f (x) dx = 0 −as =e ( where x = t − a) F (s) Example 9.3 Find the Laplace transform of g(t) = 0 t if if t<5 t 5 We can express g(t) as g(t) = u(t − 5)f (t − 5) where f (t) = (t + 5). Then F (s) = L {f (t)} = 9.3 In other words L {tf (t)} = −F (s) 5 1 + 2 s s ⇒ (9.7) Repeating this procedure n times, we obtain: dn F (s) dsn Using the derivative formula, we ﬁnd L {t sin t} = − = f (t)dt 0 Example 9.4 Find the Laplace transform of f (t) = t sin t. L {f (t − a) u(t − a)} = e−as F (s) 0 (−t)e −st L {tn f (t)} = (−1)n Figure 9.1: u(t − a) and its eﬀect on f (t) L {f (t − a) u(t − a)} = (9.6) ∞ L {g(t)} = e−5s 1 5 + 2 s s Diﬀerentiation of Transforms If f (t) is piecewise continuous and of exponential order, then we can diﬀerentiate its Laplace transform integral. d ds 1 1 + s2 = 2s (1 + s2 )2 (9.8)
41. 41. 9.4. PARTIAL FRACTIONS EXPANSION 9.4 71 Partial Fractions Expansion CHAPTER 9. LAPLACE TRANSFORM II 9.5 In many applications of Laplace transform, we need to expand a rational function in partial fractions. Here, we will review this technique by examples. 2x + 1 A B C = + + (x − 2)(x + 3)(x − 1) x−2 x+3 x−1 2 x + 4x − 5 B C A D + + = + (x − 2)(x − 1)3 x − 2 x − 1 (x − 1)2 (x − 1)3 x3 + 1 Dx + E A Bx + C + 2 + 2 = 2 + 4)2 x(x x x +4 (x + 4)2 3 A B x3 − 4x2 + x + 9 = x+1+ 2 =x+1+ + x2 − 5x + 6 x − 5x + 6 x−2 x−3 • We can express any polynomial as a product of ﬁrst and second order polynomials. • For second order polynomials in the expansion, we have to use Ax + B (not simply a constant) in the numerator. • If numerator’s degree is greater or equal to the denominator, we should ﬁrst divide them using polynomial division. 2 Example 9.5 Find the inverse Laplace transform of F (s) = 72 −s + 7s − 1 . (s − 2)(s − 5)2 Applications Now we are in a position to solve a wider class of diﬀerential equations using Laplace transform. Example 9.6 Solve the initial value problem y − 6y + 8y = 2e2t , y(0) = 11, y (0) = 37 We will ﬁrst ﬁnd the Laplace transform of both sides, then ﬁnd Y (s) L {y } − 6L {y } + 8L {y} = L 2e2t s2 Y − 11s − 37 − 6(sY − 11) + 8Y = (s2 − 6s + 8)Y = −s2 + 7s − 1 = A(s − 5)2 + B(s − 2)(s − 5) + C(s − 2) Inserting s = 2, we see that 9 = 9A ⇒ A = 1. Inserting s = 5, we see that 9 = 3C ⇒ C = 3. The coeﬃcient of s2 : A + B = −1 therefore B = −2. So 1 −s2 + 7s − 1 2 3 = − + 2 (s − 2)(s − 5) s − 2 s − 5 (s − 5)2 Now we can easily ﬁnd the inverse Laplace transform: L−1 {F (s)} = e2t − 2e5t + 3te5t 2 + 11s − 29 s−2 The factors of s2 − 6s + 8 are (s − 2) and (s − 4), so Y = 2 11s − 29 + (s − 2)(s − 2)(s − 4) (s − 2)(s − 4) Y = First, we have to express F (s) in terms of simpler fractions: −s2 + 7s − 1 A B C = + + (s − 2)(s − 5)2 s − 2 s − 5 (s − 5)2 2 s−2 11s2 − 51s + 60 (s − 2)2 (s − 4) Now we need to ﬁnd the inverse Laplace transform. Using partial fractions expansion Y = A C B + + 2 s − 2 (s − 2) s−4 After some algebra we ﬁnd that A = 3, B = −1, C = 8 so Y (s) = 8 1 3 − + 2 s − 2 (s − 2) s−4 y(t) = L−1 {Y (s)} = 3e2t − te2t + 8e4t
42. 42. 9.5. APPLICATIONS 73 Example 9.7 Solve the initial value problem 74 CHAPTER 9. LAPLACE TRANSFORM II Example 9.8 Solve the initial value problem y + y = f (t), y(0) = 0, y (0) = 3 where f (t) = 0 2 cos t if if 0 < t < 5π 5π < t y + 2y + y = r(t), y(0) = 0, y (0) = 0 where r(t) = As you can see, the input function is discontinuous, but this makes no diﬀerence for Laplace transform. L {y } + L {y} = L {f } t 0 if if 0<t<1 1<t Once again we have a discontinuous input. This time we will use unit step function. First, we have to express r(t) with a single formula. r(t) = t − u(t − 1)t = t − u(t − 1)(t − 1) − u(t − 1) 2 s Y −3+Y =F F +3 Y = 2 s +1 1 Using the fact that L {sin t} = 2 , we can obtain y(t) by convolution: s +1 y(t) = L−1 {Y } = f (t) ∗ sin t + 3 sin t Its Laplace transform is R(s) = L {r(t)} = Finding the Laplace transform of the equation, we obtain (s2 + 2s + 1)Y = R Using the deﬁnition of convolution, t f ∗ sin t = Y = f (x) sin(t − x) dx 0 If t < 5π, f = 0 therefore this integral is also zero. If t > 5π we have Y = t f ∗ sin t = 1 e−s e−s − 2 − 2 s s s 2 cos x sin(t − x) dx s2 (s R (s + 1)2 1 e−s − 2 2 + 1) s (s + 1) Using partial fractions expansion 5π Using the trigonometric identity 2 sin A cos B = sin(A + B) + sin(A − B) we obtain t f ∗ sin t = sin t + sin(t − 2x) dx 2 1 2 1 1 1 1 Y =− + 2 + + − e−s − + 2 + s s s + 1 (s + 1)2 s s s+1 Using the fact that L−1 {e−as F (s)} = f (t − a)u(t − a), we obtain 5π = x sin t + cos(t − 2x) 2 t 5π = (t − 5π) sin t Therefore y(t) = −2 + t + 2e−t + te−t − u(t − 1) −1 + (t − 1) + e−(t−1) We know that u(t − 1) = 0 for t > 1 and u(t − 1) = 1 for t > 1 so y(t) = y(t) = 3 sin t (t − 5π + 3) sin t if if 0 < t < 5π 5π < t −2 + t + 2e−t + te−t (2 − e)e−t + te−t if if 0<t<1 1<t
43. 43. EXERCISES 75 Exercises 1) F (s) = 3 s+3 (s2 + 4)2 8) F (s) = 2 2) F (s) = s s4 + 4a4 10) F (s) = sin 4t − 4t cos 4t 128 4t sin 2t + 3 sin 2t − 6t cos 2t 7) f (t) = 16 1 t 9) f (t) = sin 2t + cos 2t 4 2 5) f (t) = 6) f (t) = 10) f (t) = e2t + 2 cos 3t + 12) y = 3 cos t + (4 + t) sin t 13) y = −25 + 8t2 Solve the following initial value problems : (where y = y(t)) 1 −t e − et cos t + 7et sin t 5 15) y = t − sin t 14) y = 11) y − y − 2y = 0, y(0) = 8, y (0) = 7 12) y + y = 2 cos t, y(0) = 3, y (0) = 4 13) y + 0.64y = 5.12t2 , y(0) = −25, y (0) = 0 1 0 < t < 2π cos t 2π < t 16) y = 14) y − 2y + 2y = e−t , y(0) = 0, y (0) = 1 15) y + y = t, y(0) = 0, y (0) = 0 16) y + y = r(t), y(0) = 1, y (0) = 0 where r(t) = 1 0 if if 0 < t < 2π 2π < t 5 0 if if 17) y + y = e−2t sin t, y(0) = 0, y (0) = 0 18) y +2y +5y = r(t), y(0) = 0, y (0) = 0 where r(t) = 0<t<π π<t 19) 4y + 4y + 17y = g(t), y(0) = 0, y (0) = 0 sin t 0 if if 0 < t < 3π 1 7 , y(0) = , y (0) = − 50 50 3π < t 1 1 17) y = (sin t − cos t) + e−2t (sin t + cos t) 8 8    1 − e−t cos 2t + sin 2t , 0 < t < π  2 18) y =  −t π  e (e − 1) cos 2t + sin 2t π<t  2 19) y = 1 8    t 1 e− 2 (t−x) sin 2(t − x)g(x) dx 0 1 (cos t − 7 sin t) 50 20) y = 1 −9π 3t 2   e e − e6π e−2t 50 50 1 e−t e−3t − + 3 2 6 8) f (t) = cosh at cos at 11) y = 3e−t + 5e2t 3s − 2s + 5 (s − 2)(s2 + 9) 12s2 − 16 (s2 + 4)3 4) f (t) = u(t − 1) cos(2t − 2) 2 s (s2 + 4)2 20) y − y − 6y = s2 + 2s (s2 + 2s + 2)2 3) f (t) = u(t − 3) sin(t − 3) Find the inverse Laplace transform transform of the following functions: se−s e−3s 4) F (s) = 2 3) F (s) = 2 s +1 s +4 1 1 5) F (s) = 2 6) F (s) = 3 2 (s + 16) s + 4s2 + 3s 9) F (s) = CHAPTER 9. LAPLACE TRANSFORM II Answers Find the Laplace transform transform of the following functions: 1) f (t) = te−t cos t 2) f (t) = t2 sin 2t 7) F (s) = 76 if 0 < t < 3π if 3π < t 2 sin 3t 3
44. 44. 78 CHAPTER 10. FOURIER ANALYSIS I L cos nπx mπx cos dx = 0 (m = n) L L (10.3) sin nπx mπx sin dx = 0 (m = n) L L (10.4) −L L Chapter 10 −L L cos2 Fourier Analysis I −L −L nπx mπx sin dx = 0 (for all m, n) L L 77 kπx dx = L L a0 cos −L ∞ + + (10.1) (10.2) kπx dx L L cos nπx kπx cos dx L L sin an nπx kπx cos dx L L −L n=1 ∞ ∞ If possible, this expansion would be very useful in all kinds of applications. Once we solve a question for sine and cosine functions, we will be able to solve it for any periodic f . Here, an and bn are the coordinates of f in the space of sine and cosine functions. But then how can we ﬁnd an and bn ? The following identities will help us: (10.6) Now, suppose the expansion (10.1) exists. To ﬁnd ak , we will multiply both sides by cos kπx and then integrate from −L to L. L L nπx nπx f (x) = a0 + an cos + bn sin L L n=1 n=1 −L (10.5) 2 cos A sin B = sin(A + B) − sin(A − B) Let f (x) be a periodic function with period 2L. It is suﬃcient that f be deﬁned on [−L, L]. Is it possible to express f as a linear combination of sine and cosine functions? cos −L nπx dx = L L 2 sin A sin B = cos(A − B) − cos(A + B) f (x) cos L sin2 2 cos A cos B = cos(A − B) + cos(A + B) Fourier Series ∞ L In the terminology of linear algebra, the trigonometric functions form an orthogonal coordinate basis. We can easily prove these formulas if we remember the following trigonometric identities: The trigonometric functions sine and cosine are the simplest periodic functions. If we can express an arbitrary periodic function in terms of these, many problems would be simpliﬁed. In this chapter, we will see how to ﬁnd the Fourier series of a periodic function. Fourier series is important in many applications. We will also need them when we solve partial diﬀerential equations. 10.1 nπx dx = L L bn n=1 −L (10.7) Using the property of orthogonality, we can see that all those integrals are zero, except the kth one. Therefore L f (x) cos −L kπx dx = ak L L ⇒ ak = 1 L L f (x) cos −L kπx dx L (10.8) We can apply the same procedure to ﬁnd a0 and bn . In the end, we will obtain the following formulas for a function f deﬁned on [−L, L].
45. 45. 10.1. FOURIER SERIES 79 a0 = Fourier coeﬃcients: an = bn = 1 2L 1 L 1 L CHAPTER 10. FOURIER ANALYSIS I L f (x) dx −L L f (x) cos nπx dx L f (x) sin nπx dx L −L L −L (10.9) ∞ ∞ Fourier series: f (x) = a0 + 80 an cos n=1 nπx nπx + bn sin L L n=1 (10.10) Example 10.1 Find the Fourier series of the periodic function f (x) = x2 , −L x L having period= 2L. a0 = = 1 2L L x2 dx −L 3 L 1 x 2L 3 = −L L2 3 Using integration by parts two times we ﬁnd: an = = bn = 1 L L x2 cos −L nπx dx L 4L2 cos nπ n2 π 2 1 L L x2 sin −L nπx dx = 0 L Therefore the Fourier series is: ∞ x2 = 4L2 nπx L2 + (−1)n 2 2 cos 3 nπ L n=1 The plot of the Fourier series up to n = 1, 2 and 3 is given in Figure 10.1. Figure 10.1: Fourier Series of f = x2 for n = 1, 2, 3
46. 46. 10.2. CONVERGENCE OF FOURIER SERIES 10.2 81 82 CHAPTER 10. FOURIER ANALYSIS I Convergence of Fourier Series Like any inﬁnite series, Fourier series is of no use if it is divergent. But most functions that we are interested in have Fourier series that converge and converge to the function. Theorem 10.1: Let f be periodic with period 2L and let f and f be piecewise continuous on the interval [−L, L]. Then the Fourier expansion of f converges to: • f (x) if f is continuous at x. f (x+ ) + f (x− ) if f is discontinuous at x. 2 Example 10.2 Find the Fourier series of the periodic function • a b f (x) = Figure 10.2: Convergence at a discontinuity 10.3 Theorem 10.2: Let f be continuous on [−L, L], f (L) = f (−L) and let f be piecewise continuous. Then the Fourier coeﬃcients of f satisfy: ∞ 2a2 + 0 having period= 2L. Then evaluate the series at x = L. a0 = 1 2L a dx + −L 1 2L L b dx = 0 a+b 2 L 1 L f (x)2 dx ∞ f 2 (x) = a0 f (x) + ∞ L f 2 (x) dx = a0 f (x) dx + −L ∞ n=1 an cos nπx + L bn = 1 L 0 nπx 1 a cos dx + L L −L 0 a sin −L nπx 1 dx + L L a L nπx =− cos L nπ L 0 L 0 n=1 ∞ L an f (x) cos −L nπx bn dx + L n=1 b sin 0 L 0 b−a = (1 − (−1)n ) nπ Therefore the Fourier series is: ∞ a+b b−a nπx + [1 − (−1)n ] sin f (x) = 2 nπ L n=1 a + b 2(b − a) + 2 π πx 1 3πx 1 5πx + sin + sin + ··· L 3 L 5 L a+b If we insert x = L in that series, we obtain f (L) = . Thus the value at 2 discontinuity is the average of left and right limits. The summation of the series up to n = 1, 5 and 9 is plotted on Figure 10.2. = f (x) sin −L 1 1 1 = 1 + 4 + 4 + ··· 4 n 2 3 n=1 2 (Hint: Use the Fourier series of f (x) = x on the interval −π < x < π) Example 10.3 Find the sum of the series S = nπx dx L b L nπx − cos L nπ L −L L Using equation (10.9) to evaluate these integrals, we can obtain the result. nπx b cos dx = 0 L L sin nπx . L nπx nπx + bn f (x) sin L L n=1 ∞ 1 an = L ∞ n=1 bn ∞ an f (x) cos n=1 L (10.11) −L Proof: We can express f (x) as f (x) = a0 + Now multiply both sides by f and integrate −L 0 (a2 + b2 ) = n n n=1 −L < x < 0 0<x<L if if Parseval’s Identity Evaluating the integrals in (10.9) for f (x) = x2 we obtain π2 4(−1)n and bn = 0 so a0 = , a n = 3 n2 f (x) = 1 π2 1 − 4 cos x − cos 2x + cos 3x − · · · 3 4 9 Using Parseval’s theorem, we have sin 1 2π 4 1 + 16 1 + 4 + 4 + · · · 9 2 3 Therefore 1 π 4 x dx π −π 2 = π4 5 = nπx d L
47. 47. 10.3. PARSEVAL’S IDENTITY 1 1 2 2 − 16 1 + 4 + 4 + · · · = π 4 2 3 5 9 1 1 π4 S = 1 + 4 + 4 + ··· = 2 3 90 83 84 CHAPTER 10. FOURIER ANALYSIS I Exercises Find the Fourier series of the periodic function f (x) deﬁned on the given interval 1) f (x) = x, −π < x < π 3) f (x) = 0 1 2) f (x) = x, 0 < x < 2π −π < x < 0 0<x<π if if 5) f (x) = sin2 x, −π < x < π 7) f (x) = −π/4 π/4 if if −1 < x < 0 0<x<1 9) f (x) = |x|, −2 < x < 2 11) f (x) = x 1−x if if 4) f (x) = x2 , 0 < x < 2π 6) f (x) = x + |x|, −π < x < π 8) f (x) = π x if if 10) f (x) = | sin x|, −π < x < π 0<x<1 1<x<2 13) f (x) = ax + b, −L < x < L 15) f (x) = x3 , −π < x < π 12) f (x) = −a a 1 π2 1 + + ··· = . 9 25 8 −L < x < 0 0<x<L 16) f (x) = ex , −π < x < π x sin ax cos ax + a a2 x cos ax sin ax + x sin ax dx = − a a2 2 x sin ax 2x cos ax 2 sin ax x2 cos ax dx = + − a a2 a3 2 x cos ax 2x sin ax 2 cos ax x2 sin ax dx = − + + a a2 a3 x cos ax dx = if if 14) f (x) = 1 − x2 , −1 < x < 1 17) Using integration by parts, show that: 18) Show that 1 + −π < x < 0 0<x<π
48. 48. EXERCISES 85 Answers 1 1 sin 2x + sin 3x − · · · 2 3 2) f (x) = π − 2 sin x + 1 1 sin 2x + sin 3x + · · · 2 3 1 1 sin x + sin 3x + sin 5x + · · · 3 5 4π 2 1 1 + 4 cos x + cos 2x + cos 3x + · · · 3 4 9 −4π sin x + 5) f (x) = 6) f (x) = π 4 − 2 π 1 1 sin 2x + sin 3x + · · · 2 3 1 1 − cos 2x 2 2 cos x + +2 sin x − 7) f (x) = sin πx + 1 1 cos 3x + cos 5x + · · · 9 25 1 1 1 sin 2x + sin 3x − sin 4x + · · · 2 3 4 1 1 sin 3πx + sin 5πx + · · · 3 5 ∞ 3π (−1)n − 1 1 + cos nx − sin nx 2 4 πn n n=1 9) f (x) = 1 − 10) f (x) = 8 π2 cos 2 4 − π π 11) f (x) = − + 4 π2 2 π sin 13) f (x) = b + 2aL π 14) f (x) = 2 4 + 2 3 π πx 1 3πx 1 5πx + sin + sin + ··· L 3 L 5 L sin ∞ n=1 3πx 1 5πx πx 1 + cos + cos + ··· 2 9 2 25 2 cos 2nx 4n2 − 1 cos πx + sin πx + 1 1 cos 3πx + cos 5πx + · · · 9 25 1 sin 3πx + · · · 3 (−1)n+1 15) f (x) = 2 n=1 πx 1 2πx 1 3πx − sin + sin − ··· L 2 L 3 L cos πx − ∞ ∞ 1 − (−1)n 1 2 1 sin nx = + 3) f (x) = + 2 n=1 nπ 2 π 8) f (x) = CHAPTER 10. FOURIER ANALYSIS I 4a π 12) f (x) = 1) f (x) = 2π sin x − 4) f (x) = 86 1 1 cos 2πx + cos 3πx + · · · 4 9 (nπ)2 − 6 sin nx n3 ∞ 16) f (x) = 2 sinh π 1 (−1)n + (cos nx − n sin nx) π 2 n=1 1 + n2 18) Use the function in exercise 12 in Parseval’s identity
49. 49. 88 CHAPTER 11. FOURIER ANALYSIS II Figure 11.1: Plots of Some Even and Odd Functions Chapter 11 Fourier Analysis II In this chapter, we will study more advanced properties of Fourier series. We will ﬁnd the even and odd periodic extensions of a given function, we will express the series using complex notation and ﬁnally, we will extend the idea of Fourier series to nonperiodic functions in the form of a Fourier integral. As you can see in Figure 11.1, an even function is symmetric with respect to y−axis, an odd function is symmetric with respect to origin. Half Range Extensions: Let f be a function deﬁned on [0, L]. If we want to expand it in terms of sine and cosine functions, we can think of it as periodic with period 2L. Now we need to deﬁne f on the interval [−L, 0]. There are inﬁnitely many possibilities, but for simplicity, we are interested in making f an even or an odd function. If we deﬁne f for negative x values as f (x) = f (−x), we obtain the even periodic extension of f , which is represented by a Fourier cosine series. If we deﬁne f for negative x values as f (x) = −f (−x), we obtain the odd periodic extension of f , which is represented by a Fourier sine series. Half-Range Cosine Expansion: (or Fourier cosine series) ∞ f (x) = a0 + 11.1 n=1 Fourier Cosine and Sine Series If f (−x) = f (x), f is an even function. If f (−x) = −f (x), f is an odd function. We can easily see that, for functions: even × even = even, odd × odd = even, even × odd = odd For example |x|, x2 , x4 , cos x, cos nx, cosh x are even functions. x, x3 , sin x, sin nx, sinh x are odd functions. ex is neither even nor odd. L L If f is even: f (x) dx f (x) dx = 2 −L L If f is odd: (11.1) 0 f (x) dx = 0 (11.2) −L Using the above equations, we can see that in the Fourier expansion of an even function, bn = 0, and in the expansion of an odd function, an = 0. This will cut our work in half if we can recognize the given function as odd or even. 87 an cos where a0 = 1 L nπx , (0 < x < L) L L f (x) dx, 0 an = 2 L L f (x) cos 0 nπx dx L (11.3) (11.4)
50. 50. 11.1. FOURIER COSINE AND SINE SERIES 89 ∞ bn sin n=1 nπx , (0 < x < L) L CHAPTER 11. FOURIER ANALYSIS II 11.2 Half-Range Sine Expansion: (or Fourier sine series) f (x) = 90 (11.5) Complex Fourier Series Consider the Fourier series of f (x): ∞ ∞ f (x) = a0 + where L 2 L bn = f (x) sin 0 nπx dx L 0 f (x) = bn sin nx (11.7) n=1 Using Euler’s formula eix = cos x + i sin x we can express the sine and cosine functions as: 0<x< π 2 π <x<π 2 if if π 2 n=1 (11.6) Example 11.1 Find the half-range cosine and sine expansions of an cos nx + cos nx = einx + e−inx , 2 sin nx = einx − e−inx 2i (11.8) an + ibn 2 (11.9) Therefore Here, L = π, therefore 1 a0 = π 2 π an = π π 2 π π 2 sin nx = n π 2 einx + e−inx If we deﬁne c0 = a0 and π cos nx dx 2 π an − ibn 2 an cos nx + bn sin nx = π π dx = 2 4 cn = nπ 2 sin =− n an + ibn an − ibn , c−n = , n = 1, 2, 3, . . . 2 2 (11.10) We will obtain ∞ cn einx (11.11) f (x)e−inx dx n = 0, ±1, ±2, . . . (11.12) f (x) = Therefore half-range cosine series of f is n=−∞ ∞ sin nπ π π 1 1 2 f (x) = − cos nx = − cos x − cos 3x + cos 5x − · · · 4 n=1 n 4 3 5 where cn = On the other hand, 2 bn = π π π 2 π sin nx dx 2 − cos nx = n π π 2 cos nπ − cos nπ 2 = n Therefore half-range sine series of f is ∞ f (x) = n=1 cos nπ − cos nπ 1 1 2 sin nx = sin x − sin 2x + sin 3x + sin 5x + · · · n 3 5 1 2π π −π For a function of period 2L we have ∞ cn einπx/L , f (x) = cn = n=−∞ 1 2L L f (x)e−inπx/L dx −L Example 11.2 Find the complex Fourier series of f (x) = x if −π < x < π and f (x + 2π) = f (x). We have to evaluate the integral cn = 1 2π π xe−inx dx −π (11.13)
51. 51. 11.2. COMPLEX FOURIER SERIES 91 For n = 0 this integral is zero, so we have c0 = 0. For n = 0 cn = −inx π xe −in 1 2π π − −π inπ −π 1 2π Therefore CHAPTER 11. FOURIER ANALYSIS II 11.3 Fourier Integral Representation −inx e dx −in πe−inπ + πe −0 −in 1 einπ + e−inπ cos nπ =− =− in 2 in i n = (−1) n = 92 In this section, we will apply the basic idea of the Fourier series to nonperiodic functions. Consider a periodic function with period= 2L and its Fourier series. In the limit L → ∞, the summation will be an integral, and f will be a nonperiodic function. Then we will obtain the Fourier integral representation: ∞ i (−1)n einx , n n=−∞ n=0 where A(u) = Note that we can obtain the real Fourier series from the complex one. If we add nth and −nth terms we get cos nx + i sin nx cos(−nx) + i sin(−nx) sin nx i(−1) + i(−1)−n = (−1)n+1 n −n n B(u) = n ∞ n+1 sin nx (−1) x= n n=1 1 π 1 π ∞ f (x) cos ux dx (11.15) f (x) sin ux dx (11.16) −∞ ∞ −∞ Like the Fourier series, we have A(u) = 0 for odd functions and B(u) = 0 for even functions. Theorem 11.1: If f and f are piecewise continuous in every ﬁnite interval ∞ This is the real Fourier series. |f | dx is convergent, then the Fourier integral of f converges to: and if −∞ Example 11.3 Find the complex Fourier series of f (x) = k • f (x) if f is continuous at x. π 1 ke−inx dx 2π −π π k e−inx = (n = 0) 2π −in −π cn = inπ • f (x+) + f (x−) if f is discontinuous at x. 2 Example 11.4 Find the Fourier integral representation of −inπ k e −e nπ 2i k = sin nπ nπ =0 = If n = 0 we have (11.14) 0 ∞ x= [A(u) cos ux + B(u) sin ux] du f (x) = 1 c0 = 2π =k f (x) = π/2 0 if if |x| < 1 1 < |x| Note that f is even therefore B(u) = 0 π k dx A(u) = −π ∞ 1 π f (x) cos ux dx = −∞ 1 = cos ux dx = 0 sin ux u 1 π 1 = 0 1 −1 π cos ux dx 2 sin u u
52. 52. 11.3. FOURIER INTEGRAL REPRESENTATION 93 Therefore, Fourier integral representation of f is ∞ f (x) = 0 sin u cos ux du u 94 CHAPTER 11. FOURIER ANALYSIS II Exercises For the following functions deﬁned on 0 < x < L, ﬁnd the half-range cosine and half-range sine expansions: Example 11.5 Prove the following formulas using two diﬀerent methods: eax e cos bx dx = 2 (a cos bx + b sin bx) a + b2 eax eax sin bx dx = 2 (a sin bx − b cos bx) a + b2 We can obtain the formulas using integration by parts, but this is the long way. A better method is to express the integrals as a single complex integral using eibx = cos bx + i sin bx, then evaluate it at one step, and then separate the real and imaginary parts. ax Example 11.6 Find the Fourier integral representation of f (x) = −ex cos x e−x cos x if if x<0 0<x This function is odd therefore A(u) = 0. 2 ∞ −x 1 ∞ f (x) sin ux dx = e cos x sin ux dx π −∞ π 0 2 ∞ −x sin(ux + x) + sin(ux − x) e dx = π 0 2 ∞ 1 e−x = [− sin(u + 1)x − (u + 1) cos(u + 1)x] π 1 + (u + 1)2 0 ∞ e−x 1 [− sin(u − 1)x − (u − 1) cos(u − 1)x] + π 1 + (u − 1)2 0 1 u−1 u+1 = + π 1 + (u + 1)2 1 + (u − 1)2 2 u3 = π u4 + 4 B(u) = So f (x) = 2 π ∞ 0 u3 sin ux du +4 u4 1) f (x) = 2kx/L 2k(L − x)/L if if 0 < x < L/2 2) f (x) = ex L/2 < x < L 3) f (x) = k 4) f (x) = x4 5) f (x) = cos 2x 0 < x < π 6) f (x) = 0 k if if 0 < x < L/2 L/2 < x < L Find the complex Fourier series of the following functions: 7) f (x) = 0 1 if if −π < x < 0 0<x<π 9) f (x) = sin x 8) f (x) = x2 , −L < x < L 10) f (x) = cos 2x Find the Fourier integral representations of the following functions:  π π  cos x, |x| <  π − x, 0 < x < π 2 2 11) f (x) = (f odd) 12) f (x) = π  0, π<x  0, |x| > 2 13) f (x) = e−x , 0 < x ex , x < 0 π 0 14) f (x) = Prove the following formulas. (Hint: Deﬁne a suitable then ﬁnd its Fourier integral representation.)   πx2 /2,    ∞ 2 cos ux 2 15) 1 − 2 sin u + cos u du = π/4,  u u u 0    0,   0, x<0  ∞ cos ux + u sin ux 16) du = π/2, x = 0  −x 1 + u2 0  πe , x > 0 if if 0<x<1 Otherwise function f and 0 x<1 x=1 1<x