2 Linear Differential Equations General Theory Linear Equations with Constant Coefficients Operator Methods...

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1 MA322 Ordinary Differential Equations Wong Yan Loi

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3 Contents First Order Differential Equations 5 Introduction 5 2 Exact Equations, Integrating Factors 8 3 First Order Linear Equations 4 First Order Implicit Equations 4 5 Further Exercises 6 2 Linear Differential Equations 7 2 General Theory 7 22 Linear Equations with Constant Coefficients Operator Methods 3 3 Second Order Linear Differential Equations 33 3 Exact 2nd Order Equations The Adjoint Differential Equation and Integrating Factor Lagrange s Identity and Green s Formula Regular Boundary Value Problem Regular Sturm-Liouville Boundary Value Problem Nonhomogeneous Boundary Value Problem Oscillations and the Sturm Separation Theorem Sturm Comparison Theorem 48 4 Linear Differential Systems 5 4 Linear Systems 5 42 Homogeneous Linear Systems Non-Homogeneous Linear Systems Homogeneous Linear Systems with Constant Coefficients Higher Order Linear Equations Introduction to the Phase Plane 7 47 Linear Autonomous System in the Plane 7 48 Appendix : Proof of Lemma

4 4 CONTENTS 5 Power Series Solutions 83 5 Power Series Series Solutions of First Order Equations Second Order Linear Equations and Ordinary Points Regular Singular Points and the Method of Frobenius Bessel s Equation Appendix 2: Some Properties of the Legendre Polynomials 2 6 Fundamental Theory of ODEs 5 6 Existence-Uniqueness Theorem 5 62 The Method of Successive Approximations 6 63 Convergence of the Successive Approximations 9 64 Non-local Existence of Solutions 2 65 Gronwall s Inequality and Uniqueness of Solution 5 66 Existence and Uniqueness of Solutions to Systems 9

5 Chapter First Order Differential Equations Introduction Ordinary differential equations An ordinary differential equation (ODE for short) is a relation containing one real variable x, the real dependent variable y, and some of its derivatives y, y,, y (n),, with respect to x The order of an ODE is defined to be the order of the highest derivative that occurs in the equation Thus, an n-th order ODE has the general form F (x, y, y,, y (n) ) = () We shall always assume that () can be solved explicitly for y (n) in terms of the remaining n + quantities as where f is a known function of x, y, y,, y (n ) An n-th order ODE is linear if it can be written in the form y (n) = f(x, y, y,, y (n ) ), (2) a (x)y (n) + a (x)y (n ) + + a n (x)y = r(x) (3) The functions a j (x), j n are called coefficients of the equation We shall always assume that a (x) in any interval in which the equation is defined If r(x), (3) is called a homogeneous equation If r(x), (3) is said to be a non-homogeneous equation, and r(x) is called the non-homogeneous term 2 Solutions A functional relation between the dependent variable y and the independent variable x that satisfies the given ODE in some interval J is called a solution of the given ODE on J A general solution of an n-th order ODE depends on n arbitrary constants, ie the solution y depends on x and n real constants c,, c n A first order ODE may be written as F (x, y, y ) = (4) 5

6 6 CHAPTER FIRST ORDER DIFFERENTIAL EQUATIONS In this chapter we consider only first order ODE The function y = φ(x) is called an explicit solution of (4) in the interval J provided F (x, φ(x), φ (x)) = for all x in J (5) A relation of the form ψ(x, y) = is said to be an implicit solution of (4) provided it determines one or more functions y = φ(x) which satisfy (5) The pair of equations is said to be a parametric solution of (4) if ( F x(t), y(t), ẏ(t) ) = ẋ(t) x = x(t), y = y(t) (6) Example Consider the ODE: x + yy = for x (, ) x 2 + y 2 = is an implicit solution while x = cos t, y = sin t, t (, π) is a parametric solution 3 Integral curves The solutions of a first order ODE y = f(x, y) (7) represent a one-parameter family of curves in the xy-plane These are called integral curves In other words, if y = y(x) is a solution to (7), then vector field F(x, y) =, f(x, y) is tangent to the curve r(x) = x, y(x) at every point (x, y) since r (x) = F(x, y) 4 Elimination of constants: formation of ODE Given a family of functions parameterized by some constants, a differential equation can be formed by eliminating the constants of this family and its derivatives Example 2 The family of functions y = Ae x + B sin x satisfies the ODE: y y = when the constants A and B are eliminated using the derivatives Exercise Find the differential equation satisfied by the family of functions y = x c for x >, where c is a parameter Ans: y = y ln y x ln x 5 Separable equations Typical separable equation can be written as y = f(x), or g(y)dy = f(x)dx (8) g(y) The solution is given by g(y)dy = f(x)dx + c

7 INTRODUCTION 7 Exercise 2 Solve y = 2xy, y() = Ans: y = e x2 The equation y = f( y x ) can be reduced to a separable equation by letting u = y x, ie y = xu So f(u) = y = u + xu, du dx f(u) u = x + c Exercise 3 Solve 2xyy + x 2 y 2 = Ans: x 2 + y 2 = cx 6 Homogeneous equations A function is called homogeneous of degree n if f(tx, ty) = t n f(x, y) for all x, y, t For example x 2 + y 2 and x + y are homogeneous of degree, x 2 + y 2 is homogeneous of degree 2 and sin(x/y) is homogeneous of degree The ODE M(x, y) + N(x, y)y = is said to be homogeneous of degree n if both M(x, y) and N(x, y) are homogeneous of degree n If we write the above DE as y = f(x, y), where f(x, y) = M(x, y)/n(x, y) Then f(x, y) is homogeneous of degree To solve the DE y = f(x, y), where f is homogeneous of degree, we use the substitution y = zx Then dy dx = z + x dz dx Thus the DE becomes z + x dz dx = f(x, zx) = x f(, z) = f(, z) Consequently, the variables can be separated to yield and integrating both sides will give the solution dz f(, z) z = dx x, Exercise 4 Solve y = x+y x y Ans: tan (y/x) = ln x 2 + y 2 + c Example 3 An equation in the form y = a x + b y + c a 2 x + b 2 y + c 2 can be reduced to a homogeneous equation by a suitable substitution x = z + h, y = w + k when a b 2 a 2 b, where h and k are solutions of the system of linear equations a h + b k + c =, a 2 h + b 2 k + c 2 =

8 8 CHAPTER FIRST ORDER DIFFERENTIAL EQUATIONS Exercise 5 Solve y = x+y 2 x y ( ) Ans: tan y x = ln (x ) 2 + (y ) 2 + c Exercise 6 Solve (x + y + ) + (2x + 2y + )y = Ans: x + 2y + ln x + y = c, x + y = 2 Exact Equations, Integrating Factors Exact equations We can write a first order ODE in the following form (2) is called exact if there exists a function u(x, y) such that M(x, y)dx + N(x, y)dy = (2) M(x, y)dx + N(x, y)dy = du = u u dx + x y dy Once (2) is exact, the general solution is given by u(x, y) = c, where c is an arbitrary constant Theorem Assume M and N together with their first partial derivatives are continuous in the rectangle S: x x < a, y y < b A necessary and sufficient condition for (2) to be exact is M y = N for all (x, y) in S (22) x When (22) is satisfied, a general solution of (2) is given by u(x, y) = c, where and c is an arbitrary constant u(x, y) = x x M(s, y)ds + y y N(x, t)dt (23) Proof Let u(x, y) = x x M(s, y)ds + y y N(x, t)dt We have to show that u x = M(x, y) and u y = N(x, y) The first equality is immediate by the fundamental theorem of calculus For the second equality, we have u y = x x y M(s, y)ds + N(x, y) = x x s N(s, y)ds + N(x, y) = N(x, y) N(x, y) + N(x, y) = N(x, y) Remark In Theorem, the rectangle S can be replaced by any region which does not include any hole In that case, the proof is by Green s theorem Exercise 7 Solve (x 3 + 3xy 2 )dx + (3x 2 y + y 3 )dy = Ans: x 4 + 6x 2 y 2 + y 4 = c 2 Integrating factors

9 2 EXACT EQUATIONS, INTEGRATING FACTORS 9 A non-zero function µ(x, y) is an integrating factor of (2) if the equivalent differential equation µ(x, y)m(x, y)dx + µ(x, y)n(x, y)dy = (24) is exact If µ is an integrating factor of (2) then (µm) y = (µn) x, ie Nµ x Mµ y = µ(m y N x ) (25) One may look for an integrating factor of the form µ = µ(v), where v is a known function of x and y Plugging into (25) we find dµ µ dv = M y N x (26) Nv x Mv y If M y N x Nv x Mv y is a function of v alone, say, φ(v), then is an integrating factor of (2) µ = e v φ(v)dv Let v = x If My Nx N of (2) is a function of x alone, say, φ (x), then e x φ (x)dx is an integrating factor Let v = y If My Nx M of (2) is a function of y alone, say, φ 2(y), then e y φ 2(y)dy is an integrating factor Let v = xy If My Nx yn xm is a function of v = xy alone, say φ 3(xy), then e xy φ 3(v)dv is an integrating factor of (2) Exercise 8 Solve (x 2 y + y + ) + x( + x 2 )y = Ans: xy + tan x = c Exercise 9 Solve (y y 2 ) + xy = Ans: y = ( cx), y = Exercise Solve (xy 3 + 2x 2 y 2 y 2 ) + (x 2 y 2 + 2x 3 y 2x 2 )y = Ans: e xy (/x + 2/y) = c, y = 3 Find integrating factors by inspection The followings are some differential formulas that are often useful d( x ydx xdy ) = y y 2

10 CHAPTER FIRST ORDER DIFFERENTIAL EQUATIONS d(xy) = xdy + ydx d(x 2 + y 2 ) = 2xdx + 2ydy d(tan x ydx xdy ) = y x 2 + y 2 d(ln x ydx xdy ) = y xy We see that the very simple ODE ydx xdy = has /x 2, /y 2, /(x 2 +y 2 ) and /xy as integrating factors Example 4 Solve xdy + ydx = x cos xdx Solution The differential xdy + ydx can be written as d(xy) Thus the DE can be written as d(xy) = x cos xdx Integrating, we have xy = x sin x + cos x + c Exercise Solve (x 2 y 3 + y)dx + (x x 3 y 2 )dy = Ans: ln x/y = /(2x 2 y 2 ) + c and y = 3 First Order Linear Equations Homogeneous equations A first order homogeneous linear equation is of the form y + p(x)y =, (3) where p(x) is a continuous function on an interval J Let P (x) = x p(s)ds Multiplying (3) a by e P (x), we get d dx [ep (x) y] =, so e P (x) y = c The general solution of (3) is given by y(x) = ce P (x), where P (x) = 2 Non-homogeneous equations Now consider a first order non-homogeneous linear equation x a p(s)ds (32) y + p(x)y = q(x), (33) where p(x) and q(x) are continuous functions on an interval J Let P (x) = x a p(s)ds Multiplying (33) by e P (x) we get d dx [ep (x) y] = e P (x) q(x)

11 3 FIRST ORDER LINEAR EQUATIONS Thus e P (x) y(x) = The general solution is given by y(x) = e P (x) [ P (x) = x Exercise 2 Solve y y = e 2x a x a p(s)ds x a e P (t) q(t)dt + c e P (t) q(t)dt + c], where (34) Ans: y = ce x + e 2x 3 The Bernoulli equation An ODE in the form y + p(x)y = q(x)y n, (35) where n,, is called the Bernoulli equation The functions p(x) and q(x) are continuous functions on an interval J Let u = y n Substituting into (35) we get This is a first order linear ODE u + ( n)p(x)u = ( n)q(x) (36) Exercise 3 Solve xy + y = x 4 y 3 Ans: y 2 = x 4 + cx 2, or y = 4 The Riccati equation An ODE of the form y = P (x) + Q(x)y + R(x)y 2 (37) is called the Riccati equation The functions P (x), Q(x), R(x) are continuous on an interval J In general, the Riccati equation cannot be solved by a sequence of integrations However, if a particular solution is known, then (37) can be reduced to a linear equation, and thus is solvable Theorem 2 Let y = y (x) be a particular solution of the Riccati equation (37) Set x H(x) = [Q(t) + 2R(t)y (t)]dt, x Z(x) = e H(x)[ x ] c e H(t) R(t)dt, x where c is an arbitrary constant Then the general solution is given by (38) y = y (x) + Z(x) (39)

12 2 CHAPTER FIRST ORDER DIFFERENTIAL EQUATIONS Proof In (7) we let y = y (x) + u(x) to get y + u = P + Q(y + u) + R(y + u) 2 Since y satisfies (37), we have y = P + Qy + Ry 2 From these two equalities we get u = (Q + 2Ry )u + Ru 2 (3) This is a Bernoulli equation with n = 2 Set Z = u and reduce (3) to Z + (Q + 2Ry )Z = R (3) (3) is a linear equation and the solution is given by (38) Exercise 4 Solve y = y/x + x 3 y 2 x 5 Note y = x is a solution Ans: ce 2x5 /5 = y x y+x From (38), (39), the general solution y of the Riccati equation (37) can be written as y = cf (x) + G(x) cf(x) + g(x), (32) where f(x) = e H(x), x g(x) = e H(x) e H(t) R(t)dt, x F (x) = y (x)f(x), G(x) = y g(x) + Given four distinct functions p(x), q(x), r(x), s(x), we define the cross ratio by (p q)(r s) (p s)(r q) Property The cross ratio of four distinct particular solutions of a Riccati equation is independent of x Proof From (32), the four solutions can be written as Computations show that y j (x) = c jf (x) + G(x) c j f(x) + g(x) (y y 2 )(y 3 y 4 ) (y y 4 )(y 3 y 2 ) = (c c 2 )(c 3 c 4 ) (c c 4 )(c 3 c 2 )

13 3 FIRST ORDER LINEAR EQUATIONS 3 The right-hand side is independent of x As a consequence we get Property 2 Suppose y, y 2, y 3 are three distinct particular solutions of a Riccati equation (37) Then the general solution is given by where c is an arbitrary constant (y y 2 )(y 3 y) = c, (33) (y y)(y 3 y 2 ) Property 3 Suppose that y and y 2 are two distinct particular solutions of a Riccati equation (37), then its general solution is given by where c is an arbitrary constant Proof y and y j satisfy (37) So ln y y y y 2 = [y (x) y 2 (x)]r(x)dx + c, (34) y y j = (y y j )[Q + R(y + y j )], y y j y y j = Q + R(y + y j ) Thus Integrating yields (34) y y y y y y 2 y y 2 = R(y y 2 ) Exercise 5 Solve y = e x y 2 Note y = e x and y 2 = are 2 solutions Ans: y = ex Ae, y = x Exercise 6 A natural generalization of Riccati s equation is Abel s equation y = P (x) + Q(x)y + R(x)y 2 + S(x)y 3, where P (x), Q(x), R(x) and S(x) are continuous functions of x on an interval J Using the substitution z = y/x, solve the equation y = y x + xy2 3y 3 Ans: x/y + 3 ln x/y 3 = x3 3 + c, y = x/3 and y =

14 4 CHAPTER FIRST ORDER DIFFERENTIAL EQUATIONS 4 First Order Implicit Equations In the above we discussed first order explicit equations, ie equations in the form y = f(x, y) In this section we discuss solution of some first order explicit equations which are not solvable in y F (x, y, y ) = (4) Method of differentiation Consider an equations solvable in y: y = f(x, y ) (42) Let p = y Differentiating y = f(x, p) we get [f x (x, p) p]dx + f p (x, p)dp = (43) This is a first order explicit equation in x and p If p = φ(x) is a solution of (43), then y = f(x, φ(x)) is a solution of (42) Example 5 Clairaut s equation is the equation of the form y = xy + f(y ), (44) where f has continuous first order derivative Let p = y We have y = xp + f(p) Differentiating we get [x + f (p)]p = When p = we have p = c and (44) has a general solution y = cx + f(c) When x + f (p) = we get a solution of (44) given by parameterized equations x = f (p), y = pf (p) + f(p) Exercise 7 Solve Clairaut s equation y = xy y 2 /4 Ans: y = cx c 2 /4, y = x 2 Exercise 8 Let C be the curve with parametric equation x = f (p), y = pf (p) + f(p) Show that the tangent to C at the point p = c has the equation y = cx + f(c)

15 4 FIRST ORDER IMPLICIT EQUATIONS 5 2 Method of parameterization This method can be used to solve equations where either x or y is missing Consider where x is missing Let p = y and write (45) as F (y, y ) =, (45) F (y, p) = It determines a family of curves in yp plane Let y = g(t), p = h(t) be one of the curves, ie F (g(t), h(t)) = Since we have x = t dx = dy y = dy p = g (t)dt h(t), g (t) h(t) dt + c The solutions of (45) are given by x = t g (t) dt + c, h(t) y = g(t) This method can also be applied to the equations F (x, y ) =, where y is missing Exercise 9 Solve y 2 + y 2 = Ans: y = cos(c x) For the equation F (y, y ) = or F (x, y ) =, if y can be solved in terms of y or x, then the equation becomes explicit Exercise 2 Solve y 2 (2x + e x )y + 2xe x = Ans: y = x 2 + c, y = e x + c 2 3 Reduction of order Consider the equation F (x, y, y ) =, (46) where y is missing Let p = y Then y = p Write (46) as F (x, p, p ) = (47) It is a first order equation in x and p If p = φ(x, c ) is a general solution of (47), then the general solution of (46) is y = x x φ(t, c )dt + c 2

16 6 CHAPTER FIRST ORDER DIFFERENTIAL EQUATIONS Exercise 2 Solve xy y = 3x 2 Ans: y = x 3 + c x 2 + c 2 Consider the equation F (y, y, y ) =, (48) where x is missing Let p = y Then y = dp dx = dp dy dy dx = dp dy p Write (48) as F (y, p, p dp ) = (49) dy It is a first order equation in y and p If p = ψ(y, c ) is a general solution of (49), then we solve the equation y = ψ(y, c ) to get a general solution of (48) Exercise 22 Solve y + k 2 y =, where k is a positive constant Ans: y = c sin(kx) + c 2 cos(kx) 5 Further Exercises Exercise 23 Solve the differential equation y 2 dx + (xy x 3 )dy = by using the substitution u = x 2 Ans: 3y = 2x 2 + cx 2 y 3 and y = Exercise 24 Solve the differential equation y 2 dx + (xy + tan(xy))dy = by using (i) the substitution u = xy, (ii) the integrating factor cos(xy) Ans: y sin(xy) = c Exercise 25 Solve the differential equation (x ln y)ydx xdy = (i) by using the substitution y = e z, (ii) by finding an integrating factor using the formula (26) Ans: ln y = x 2 + cx 3 Exercise 26 Solve y + y = Ans: y = { { if x sin(x ) + x if x > if x x if x >, y() =, y () =

17 Chapter 2 Linear Differential Equations 2 General Theory Consider n-th order linear equation y (n) + a (x)y (n ) + + a n (x)y + a n (x)y = f(x), (2) where y (k) = dk y dx k Throughout this section we assume that a j (x) s and f(x) are continuous functions defined on the interval (a, b) When f(x), (2) is called a non-homogeneous equation The associated homogeneous equation is y (n) + a (x)y (n ) + + a n (x)y + a n (x)y = (22) Let us begin with the initial value problem: y (n) + a (x)y (n ) + + a n (x)y = f(x), y(x ) = y, y (x ) = y, y (n ) (x ) = y n (23) Theorem 2 (Existence and Uniqueness Theorem) Assume that a (x),, a n (x) and f(x) are continuous functions defined on the interval (a, b) Then for any x (a, b) and for any numbers y,, y n, the initial value problem (23) has a unique solution defined on (a, b) Especially if a j (x) s and f(x) are continuous on R then for any x and y,, y n, the initial value problem (23) has a unique solution defined on R Proof of this theorem will be given in later chapter Corollary 22 Let y = y(x) be a solution of the homogeneous equation (22) in an interval (a, b) Assume that there exists x (a, b) such that Then y(x) on (a, b) y(x ) =, y (x ) =,, y (n ) (x ) = (24) 7

18 8 CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS Proof y is a solution of the initial value problem (22), (24) From Theorem 2, this problem has a unique solution Since φ(x) is also a solution of the problem, we have y(x) on (a, b) In the following we consider the general solutions of (2) and (22) Given continuous functions a j (x), j =,,, n and f(x), define an operator L by L[y] = a (x)y (n) + a (x)y (n ) + + a n (x)y (25) Property L[cy] = cl[y] for any constant c Property 2 L[u + v] = L[u] + L[v] Proof Let s verify Property 2 L[u + v] = a (x)(u + v) (n) + a (x)(u + v) (n ) + + a n (x)(u + v) = ( a (x)u (n) + a (x)u (n ) + + a n (x)u ) + ( a (x)v (n) + a (x)v (n ) + + a n (x)v ) = L[u] + L[v] Definition An operator satisfying Properties and 2 is called a linear operator The differential operator L defined in (23) is a linear operator Note that (2) and (22) can be written as L[y] = f(x), (2 ) and L[y] = (22 ) From Properties and 2 we get the following conclusion Theorem 23 () If y and y 2 are solutions of the homogeneous equation (22) in an interval (a, b), then for any constants c and c 2, is also a solution of (22) in the interval (a, b) y = c y + c 2 y 2 (2) If y p is a solution of (2) and y h is a solution of (22) on an interval (a, b), then is also a solution of (2) in the interval (a, b) y = y h + y p Proof () As L[c y + c 2 y 2 ] = c L[y ] + c 2 L[y 2 ] =, we see that c y + c 2 y 2 is also a solution of (22) (2) L[y h + y p ] = L[y h ] + L[y p ] = + f(x), we see that y = y h + y p is a solution of (2)

19 2 GENERAL THEORY 9 In order to discuss structures of solutions, we need the following definition Definition Functions φ (x),, φ k (x) are linearly dependent on (a, b) if there exists constants c,, c k, not all zero, such that c φ (x) + + c k φ k (x) = for all x (a, b) A set of functions are linearly independent on (a, b) if they are not linearly dependent on (a, b) Lemma 24 Functions φ (x),, φ k (x) are linearly dependent on (a, b) if and only if the following vector-valued functions φ (x) φ k (x) φ (x),, φ k (x) (26) φ (n ) (x) φ (n ) k (x) are linearly dependent on (a, b) Proof = is obvious To show =, assume that φ,, φ k are linearly dependent on (a, b) There exists constants c,, c k, not all zero, such that, for all x (a, b), c φ (x) + + c k φ k (x) = Differentiating this equality successively we find that Thus c c φ (x) + + c k φ k(x) =, c φ (n ) (x) + + c k φ (n ) k (x) = φ (x) φ k (x) φ (x) + + c φ k k (x) = φ (n ) (x) φ (n ) k (x) for all x (a, b) Hence the k vector-valued functions are linearly dependent on (a, b) Recall that, n vectors in R n are linearly dependent if and only if the determinant of matrix formed by these vectors is zero Definition The Wronskian of n functions φ (x),, φ n (x) is defined by W (φ,, φ n )(x) = φ (x) φ n (x) (27) (x) φ (n ) n (x) φ (n )

20 2 CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS Note that Wronskian of φ,, φ n is the determinant of the matrix formed by the vector-valued functions given in (26) Theorem 25 Let y (x),, y n (x) be n solutions of (22) on (a, b) and let W (x) be their Wronskian () y (x),, y n (x) are linearly dependent on (a, b) if and only if W (x) on (a, b) (2) y (x),, y n (x) are linearly independent on (a, b) if and only if W (x) does not vanish on (a, b) Corollary 26 () The Wronskian of n solutions of (22) is either identically zero, or nowhere zero (2) n solutions y,, y n of (22) are linearly independent on (a, b) if and only if the set of vectors y (x ) y (x ),, y (n ) (x ) are linearly independent for some x (a, b) y n (x ) y n(x ) y (n ) n (x ) Proof of Theorem 25 Let y,, y n be solutions of (22) on (a, b), and let W (x) be their Wronskian Step We first show that, if y,, y n are linearly dependent on (a, b), then W (x) Since these solutions are linearly dependent, from Lemma 24, n vector-valued functions y (x) y n (x) y (x),, y n(x) y (n ) (x) y n (n ) (x) are linearly dependent on (a, b) Thus for all x (a, b), the determinant of the matrix formed by these vectors, namely, the Wronskian of y,, y n, is zero Step 2 Now, assume that the Wronskian W (x) of n solutions y,, y n vanishes at x (a, b) We shall show that y,, y n are linearly dependent on (a, b) Since W (x ) =, the n vectors y (x ) y (x ),, y (n ) (x ) y n (x ) y n(x ) y (n ) n (x ) are linearly dependent Thus there exist n constants c,, c n, not all zero, such that y (x ) y n (x ) y c (x ) + + c y n(x ) n = (28) y (n ) (x ) y n (n ) (x )

21 2 GENERAL THEORY 2 Define y (x) = c y (x) + + c n y n (x) From Theorem 23, y is a solution of (22) From (28), y satisfies the initial conditions y(x ) =, y (x ) =,, y (n ) (x ) = (29) From Corollary 22, we have y, namely, c y (x) + + c n y n (x) = for all x (a, b) Thus y,, y n are linearly dependent on (a, b) Example 2 Consider the differential equation y x y = on the interval (, ) Both φ (x) = and φ 2 (x) = x 2 x 2 are solutions of the differential equation W (φ, φ 2 )(x) = 2x = 2x for x > Thus φ and φ 2 are linearly independent solutions Theorem 27 () Let a (x),, a n (x) and f(x) be continuous on the interval (a, b) The homogeneous equation (22) has n linearly independent solutions on (a, b) (2) Let y,, y n be n linearly independent solutions of (22) defined on (a, b) The general solution of (22) is given by where c,, c n are arbitrary constants y(x) = c y (x) + + c n y n (x), (2) Proof () Fix x (a, b) For k =, 2,, n, let y k be the solution of (22) satisfying the initial conditions y (j) k (x if j k, ) = if j = k The n vectors y (x ) y n (x ) y (x ),, y n(x ) y (n ) (x ) y n (n ) (x ) are lineally independent since they form the identity matrix From Corollary 26, y,, y n are linearly independent on (a, b) From Theorem 23, for any constants c,, c n, y = c y + + c n y n is a solution of (22) (2) Now let y,, y n be n linearly independent solutions of (22) on (a, b) We shall show that the general solution of (22) is given by y = c y + + c n y n (2)

22 22 CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS Given a solution ỹ of (22), and fix x (a, b) Since y,, y n are linearly independent on (a, b), the vectors y (x ) y n (x ) y (x ),, y n(x ) y (n ) (x ) y n (n ) (x ) are linearly independent vectors They form a basis for R n Thus the vector ỹ(x ) ỹ (x ) ỹ (n ) (x ) can be represented as a linear combination of the n vectors, namely, there exist n constants c,, c n such that ỹ(x ) y (x ) y n (x ) ỹ (x ) = c y (x ) + + c y n(x ) n ỹ (n ) (x ) y (n ) (x ) y n (n ) (x ) Let φ(x) = ỹ(x) [ c y (x) + + c n y n (x)] φ(x) is a solution of (22) and satisfies the initial conditions (24) at x = x By Corollary 22, φ(x) on (a, b) Thus ỹ(x) = c y (x) + + c n y n (x) So (2) gives a general solution of (22) Any set of n linearly independent solutions is called a fundamental set of solutions Now we consider the non-homogeneous equation (2) We have Theorem 28 Let y p be a particular solution of (2), and y,, y n be a fundamental set of solutions for the associated homogeneous equation (22) The general solution of (2) is given by y(x) = c y (x) + + c n y n (x) + y p (x) (22) Proof Let y be a solution of the non-homogeneous equation Then y y p is a solution of the homogeneous equation Thus y(x) y p (x) = c y (x) + + c n y n (x) 22 Linear Equations with Constant Coefficients Let us begin with second order linear equation with constant coefficients y + ay + by =, (22)

23 22 LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 23 where a and b are constants We look for a solution of the form y = e λx Plugging into (22) we find that, e λx is a solution of (22) if and only if λ 2 + aλ + b = (222) (222) is called the auxiliary equation or characteristic equation of (22) The roots of (222) are called characteristic values (or eigenvalues): λ = 2 ( a + a 2 4b), λ 2 = 2 ( a a 2 4b) If a 2 4b >, (222) has two distinct real roots λ, λ 2, and the general solutions of (22) is y = c e λx + c 2 e λ2x 2 If a 2 4b =, (222) has one real root λ (we may say that (222) has two equal roots λ = λ 2 ) The general solution of (22) is y = c e λx + c 2 xe λx 3 If a 2 4b <, (222) has a pair of complex conjugate roots λ = α + iβ, λ 2 = α iβ The general solution of (22) is y = c e αx cos(βx) + c 2 e αx sin(βx) Example 22 Solve y + y 2y =, y() = 4, y () = 5 Ans: λ =, λ 2 = 2, y = e x + 3e 2x Example 23 Solve y 4y + 4y =, y() = 3, y () = Ans: λ = λ 2 = 2, y = (3 5x)e 2x Example 24 Solve y 2y + y = Ans: λ = + 3i, λ 2 = 3i, y = e x (c cos 3x + c 2 sin 3x) Now we consider n-th order homogeneous linear equations with constant coefficients where a,, a n are real constants y (n) + a y (n ) + + a n y + a n y =, (223)

24 24 CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS y = e λx is a solution of (223) if and only if λ satisfies λ n + a λ n + + a n λ + a n = (224) The solutions of (224) are called characteristic values or eigenvalues for the equation (223) Let λ,, λ s be the distinct eigenvalues for (223) Then we can write λ n + a λ n + + a n λ + a n = (λ λ ) m (λ λ 2 ) m2 (λ λ s ) ms, (225) where m,, m s are positive integers and m + + m s = n We call them the multiplicity of the eigenvalues λ,, λ s respectively Lemma 29 Assume λ is an eigenvalue of (223) of multiplicity m (i) e λx is a solution of (223) (ii) If m >, then for any positive integer k m, x k e λx is a solution of (223) (iii) If λ = α + iβ, then x k e αx cos(βx), x k e αx sin(βx) are solutions of (223), where k m Theorem 2 Let λ,, λ s be the distinct eigenvalues for (223), with multiplicity m,, m s respectively Then (223) has a fundamental set of solutions e λx, xe λx,, x m e λx ; ; e λsx, xe λsx,, x ms e λsx (226) Proof of Lemma 29 and 2 Consider the n-th order linear equation with constant coefficients y (n) + a y (n ) + + a n y + a n y =, (A) where y (k) = dk y dx k Let L(y) = y (n) + a y (n ) + + a n y + a n y and p(z) = z n + a z n + + a n z + a n Note that p is a polynomial in z of degree n Then we have L(e zx ) = p(z)e zx (A2) Before we begin the proof, let s observe that 2 z x ezx = 2 x z ezx by Clairaut s theorem because e zx is differentiable in (x, z) as a function of two variables and all the higher order partial derivatives exist and continuous That means we can interchange the order of differentiation with respect to x and z as we wish Therefore d dz L(ezx ) = L( d dz ezx ) For instance, one may verify directly that d d k dz dx k (ezx ) = xz k e zx + kz k e zx = dk dx k ( d dz ezx )

25 22 LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 25 Here one may need to use Leibniz s rule of taking the k-th derivative of a product of two functions: (u v) (k) = k i= ( ) k u (i) v (k i) i More generally, d k dz k L(e zx ) = L( dk dz k e zx ) (Strictly speaking, partial derivative notations should be used) Now let s prove our results () If λ is a root of p, then L(e λx ) = by (A2) so that e λx is a solution of (A) (2) If λ is a root of p of multiplicity m, then p(λ) =, p (λ) =, p (λ) =,, p (m ) (λ) = Now for k =,, m, differentiating (A2) k times with respect to z, we have L(x k e zx ) = L( dk dz k ezx ) = dk dz k L(ezx ) = dk dz k (p(z)ezx ) = k i= ( ) k p (i) (z)x k i e zx i Thus L(x k e λx ) = and x k e λx is solution of (A) (3) Let λ,, λ s be the distinct roots of p, with multiplicity m,, m s respectively Then we wish to prove that e λx, xe λx,, x m e λx ; ; e λsx, xe λsx,, x ms e λsx are linearly independent over R To prove this, suppose for all x in R c e λx + c 2 xe λx + + c m x m e λx + + c s e λsx + c s2 xe λsx + + c sms x ms e λsx = (A3) (A4) Let s write this as P (x)e λx + P 2 (x)e λ2x + + P s (x)e λsx =, for all x in R, where P i (x) = c i + c i2 x + + c imi x mi We need to prove P i (x) for all i By discarding those P i s which are identically zero, we may assume all P i s are not identically zero Dividing the above equation by e λx, we have P (x) + P 2 (x)e (λ2 λ)x + + P s (x)e (λs λ)x =, for all x in R Upon differentiating this equation sufficiently many times (at most m times since P (x) is a polynomial of degree m ), we can reduce P (x) to Note that in this process, the degree of the resulting polynomial multiplied by e (λi λ)x remains unchanged Therefore, we get Q 2 (x)e (λ2 λ)x + + Q s (x)e (λs λ)x =, where deg Q i = deg P i Canceling the term e λ x we have Q 2 (x)e λ2x + + Q s (x)e λsx = For instance, if deg P i =, then P i is a nonzero constant The resulting Q i is equal to (λ i λ ) α P i for some positive integer α Since λ i λ, Q i is also a nonzero constant Thus deg Q i =

26 26 CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS Repeating this procedure, we arrive at R s (x)e λsx =, where deg R s = deg P s Hence R s (x) which is a contradiction Thus all the P i (x) s are identically zero That means all c ij s are zero and the functions in (A4) are linearly independent Remark If (223) has a complex eigenvalue λ = α + iβ, then λ = α iβ is also an eigenvalue Thus both x k e (α+iβ)x and x k e (α iβ)x appear in (226), where k m In order to obtain a fundamental set of real solutions, the pair of solutions x k e (α+iβ)x and x k e (α iβ)x in (226) should be replaced by x k e αx cos(βx) and x k e αx sin(βx) In the following we discuss solution of the non-homogeneous equation y + P (x)y + Q(x)y = f(x), (227) The associated homogeneous equation is y + P (x)y + Q(x)y = The method applies to higher order equations Methods of variation of parameters Let y and y 2 be two linearly independent solutions of the associated homogeneous equation y + P (x)y + Q(x)y = and W (x) be their Wronskian We look for a particular solution of (227) in the form y p = u y + u 2 y 2, where u and u 2 are functions to be determined Suppose u y + u 2y 2 = Differentiating this equation once, we get u y + u 2y 2 = u y u 2y 2 Plugging y p into (227) we get u y + u 2y 2 = f Hence u and u 2 satisfy Solving it, we find that Integrating yields { u y + u 2y 2 =, u y + u 2y 2 = f u = y 2 W f, u 2 = y W f x y 2 (t) u (x) = W (t) f(t)dt, u 2 (x) = x x x y (t) W (t) f(t)dt (228) (229)

27 22 LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 27 Example 25 Solve the differential equation y + y = sec x Solution A basis for the solutions of the homogeneous equation consists of y = cos x and y 2 = sin x Now W (y, y 2 ) = cos x cos x ( sin x) sin x = Thus u = sin x sec x dx = ln cos x + c and u 2 = cos x sec x dx = x + c 2 From this, a particular solution is given by y p = cos x ln cos x + x sin x Therefore, the general solution is y = c cos x + c 2 sin x + cos x ln cos x + x sin x The method of variation of parameters can also be used to find another solution of a second order homogeneous linear differential equation when one solution is given Suppose z is a known solution of the equation We assume y = vz is a solution so that That is y + P (x)y + Q(x)y = = (vz) + P (vz) + Q(vz) = (v z + 2v z + vz ) + P (v z + vz ) + Qvz = (v z + 2v z + P v z) + v(z + P z + Qz) = v z + v (2z + P z) v v = 2 z z P An integration gives v = z 2 e P dx and v = z 2 e P dx dx We leave it as an exercise to show that z and vz are linearly independent solutions by computing their Wronskian Example 26 Given y = x is a solution of x 2 y + xy y =, find another solution Solution Let s write the DE in the form y + x y x y = Then P (x) = /x Thus a second 2 linearly independent solution is given y = vx, where v = x 2 e /x dx dx = x 2 x dx = 2x 2 Therefore the second solution is y = 2 x and the general solution is y = c x + c 2 x 2 Method of undetermined coefficients Consider the equation y + ay + by = f(x), where a and b are real constants Case f(x) = P n (x)e αx, where P n (x) is a polynomial of degree n We look for a particular solution in the form y = Q(x)e αx, where Q(x) is a polynomial Plugging it into y + ay + by = f(x) we find Q + (2α + a)q + (α 2 + aα + b)q = P n (x) (22) Subcase If α 2 + aα + b, namely, α is not a root of the characteristic equation, we choose

28 28 CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS Q = R n, a polynomial of degree n, and y = R n (x)e αx The coefficients of R n can be determined by comparing the terms of same power in the two sides of (22) Note that in this case both sides of (22) are polynomials of degree n Subcase 2 If α 2 + aα + b = but 2α + a, namely, α is a simple root of the characteristic equation, then (22) is reduced to Q + (2α + a)q = P n (22) We choose Q to be a polynomial of degree n + Since the constant term of Q does not appear in (22), we may choose Q(x) = xr n (x), where R n (x) is a polynomial of degree n y = xr n (x)e αx Subcase 3 If α 2 + aα + b = and 2α + a =, namely, α is a root of the characteristic equation with multiplicity 2, then (22) is reduced to Q = P n (222) We choose Q(x) = x 2 R n (x), where R n (x) is a polynomial of degree n y = x 2 R n (x)e αx Example 27 Find the general solution of y y 2y = 4x 2 Ans: y = c e 2x + c 2 e x 3 + 2x 2x 2 Example 28 Find a particular solution of y + 2y y = 3x 2 2x + Ans: y = 27x 5x 2 x 3 Example 29 Solve y 2y + y = xe x Ans: y = c e x + c 2 xe x + 6 x3 e x Case 2 f(x) = P n (x)e αx cos(βx) or f(x) = P n (x)e αx sin(βx), where P n (x) is a polynomial of degree n We first look for a solution of Using the method in Case we obtain a complex-valued solution y + ay + by = P n (x)e (α+iβ)x (223) z(x) = u(x) + iv(x),

29 22 LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 29 where u(x) = R(z(x)), v(x) = I(z(x)) Substituting z(x) = u(x) +iv(x) into (223) and taking the real and imaginary parts, we can show that u(x) = R(z(x)) is a solution of and v(x) = I(z(x)) is a solution of y + ay + by = P n (x)e αx cos(βx), (224) y + ay + by = P n (x)e αx sin(βx) (225) Example 2 Solve y 2y + 2y = e x cos x Ans: y = c e x cos x + c 2 e x sin x + 2 xex sin x Alternatively to solve (224) or (225), one can try a solution of the form Q n (x)e αx cos(βx) + R n (x)e αx sin(βx) if α + iβ is not a root of λ 2 + aλ + b =, and xq n (x)e αx cos(βx) + xr n (x)e αx sin(βx) if α + iβ is a root of λ 2 + aλ + b =, where Q n and R n are polynomials of degree n The following conclusions will be useful Theorem 2 Let y and y 2 be particular solutions of the equations y + ay + by = f (x) and y + ay + by = f 2 (x) respectively, then y p = y + y 2 is a particular solution of Proof Exercise y + ay + by = f (x) + f 2 (x) Example 2 Solve y y = e x + sin x Solution A particular solution for y y = e x is given by y = 2 xex Also a particular solution for y y = sin x is given by y 2 = 2 sin x Thus 2 (xex sin x) is a particular solution of the given differential equation The general solution of the corresponding homogeneous differential equation is given by c e x + c 2 e x Hence the general solution of the given differential equation is c e x + c 2 e x + 2 (xex sin x)

30 3 CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 23 Operator Methods Let x denote independent variable, and y dependent variable Introduce Dy = d dx y, Dn y = dn dx n y = y(n) We define D y = y Given a polynomial L(x) = n j= a jx j, where a j s are constants, we define a differential operator L(D) by n L(D)y = a j D j y Then the equation can be written as j= n a j y (j) = f(x) (23) j= Let L(D) f denote any solution of (232) We have However, L(D) f is not unique L(D)y = f(x) (232) D D = DD = D, L(D) L(D) = L(D)L(D) = D To see the above properties, first recall that D f means a solution of y = f Thus D f = f Hence it follows that D D = DD = identity operator D For the second equality, note that a solution of L(D)y = L(D)f is simply f Thus by definition of L(D), we have L(D) (L(D)f) = f This means L(D) L(D) = D Lastly, since L(D) f is a solution of L(D)y = f(x), it is clear that L(D)(L(D) f) = f In other words, L(D)L(D) = D More generally, we have: D f(x) = f(x)dx + C, 2 (D a) f(x) = Ce ax + e ax e ax f(x)dx, 3 L(D)(e ax f(x)) = e ax L(D + a)f(x), 4 L(D) (e ax f(x)) = e ax L(D + a) f(x) (233) Proof Property 2 is just the solution of the first order linear ODE To prove Property 3, first observe that (D r)(e ax f(x)) = e ax D(f(x)) + ae ax f(x) re ax f(x) = e ax (D + a r)(f(x)) Thus (D s)(d r)(e ax f(x)) = (D s)[e ax (D + a r)(f(x))] = e ax (D + a s)(d + a r)(f(x)) Now we may write L(D) = (D r ) (D r n ) Then L(D)(e ax f(x)) = e ax L(D + a)f(x)

31 23 OPERATOR METHODS 3 This says that we can move the factor e ax to the left of the operator L(D) if we replace L(D) by L(D + a) To prove Property 4, apply L(D) to the right hand side We have L(D)[e ax L(D + a) f(x)] = e ax [L(D + a)(l(d + a) f(x))] = e ax f(x) Thus L(D) (e ax f(x)) = e ax L(D + a) f(x) Let L(x) = (x r ) (x r n ) The solution of (232) is given by y = L(D) f(x) = (D r ) (D r n ) f(x) (234) Then we obtain the solution by successive integration Moreover, if r j s are distinct, we can write L(x) = A + + A n, x r x r n where A js can be found by the method of partial fractions Then the solution is given by y = [A (D r ) + + A n (D r n ) ]f(x) (235) Next consider the case of repeated roots Let the multiple root be equal to m and the equation to be solved is (D m) n y = f(x) (236) To solve this equation, let us assume a solution of the form y = e mx v(x), where v(x) is a function of x to be determined One can easily verify that (D m) n e mx v = e mx D n v Thus equation (236) reduces to D n v = e mx f(x) (237) If we integrate (237) n times, we obtain v = e mx f(x) dx dx + c + c x + + c n x n (238) Thus we see that (D m) n f(x) = e mx [ e mx f(x)dx dx + c + c x + + c n x n ] (239) Example 22 Solve (D 2 3D + 2)y = xe x Solution First D 2 3D+2 = D 2 D Therefore y = (D 2 3D + 2) (xe x ) = (D 2) (xe x ) (D ) (xe x ) = e 2x D (e 2x xe x ) e x D (e x xe x ) = e 2x D (e x x) e x D (x) = e 2x ( xe x e x + c ) e x ( 2 x2 + c 2 ) = e x ( 2 x2 + x + ) + c e 2x + c 2 e x

32 32 CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS Example 23 Solve (D 3 3D 2 + 3D )y = e x Solution The DE is equivalent to (D ) 3 y = e x Therefore, y = (D ) 3 e x = e x [ e x e x dx + c + c x + c 2 x 2 ] = e x [ 6 x3 + c + c x + c 2 x 2 ] If f(x) is a polynomial in x, then ( D)( + D + D 2 + D 3 + )f = f Thus ( D) (f) = ( + D + D 2 + D 3 + )f Therefore, if f is a polynomial, we may formally expand (D r) into power series in D and apply it to f If the degree of f is n, then it is only necessary to expand (D r) up to D n Example 24 Solve (D 4 2D 3 + D 2 )y = x 3 Solution We have y = (D 4 2D 3 + D 2 ) f = D 2 ( D) 2 x 3 = D 2 ( + 2D + 3D 2 + 4D 3 + 5D 4 + 6D 5 )x 3 = D 2 (x 3 + 6x 2 + 8x + 24) = D ( x x3 + 9x x) = x5 2 + x x3 + 2x 2 Therefore, the general solution is y = (c + c 2 x)e x + (c 3 + c 4 x) + x5 2 + x x3 + 2x 2 Exercise 2 For all real x, the real-valued function y = f(x) satisfies y 2y + y = 2e x (a) If f(x) > for all real x, must f (x) > for all real x? Explain (b) If f (x) > for all real x, must f(x) > for all real x? Explain Exercise 22 Find the general solution of y 2y 3y = x(e x + e x ) Ans: y = c e x + c 2 e 3x x 4 ex x2 8 e x x 6 e x

33 Chapter 3 Second Order Linear Differential Equations 3 Exact 2nd Order Equations The general 2nd order linear differential equation is of the form The equation can be written as It is said to be exact if p (x)y + p (x)y + p 2 (x)y = f(x) (3) (p y p y) + (p y) + (p p + p 2 )y = f(x) (32) p p + p 2 (33) In the event that the equation is exact, a first integral to (3) is p (x)y p (x)y + p (x)y = f(x) dx + C Example 3 Find the general solution of the DE x y + ( x 2 x 2 )y ( x 2 2 x 3 )y = ex Solution Condition (33) is fulfilled The first integral is That is x y + x 2 y + ( x 2 x 2 )y = ex + C y + ( x )y = xex + C x From the last equation, the general solution is found to be y = 2 xex + C x + C 2 xe x 33

34 34 CHAPTER 3 SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS 32 The Adjoint Differential Equation and Integrating Factor If (3) is multiplied by a function v(x) so that the resulting equation is exact, then v(x) is called an integrating factor of (3) That is This is a differential equation for v, which is, more explicitly, (p v) (p v) + p 2 v = (324) p (x)v + (2p (x) p (x))v + (p (x) p (x) + p 2 (x))v = (325) Equation (325) is called the adjoint of the given differential equation (3) A function v(x) is thus an integrating factor for a given differential equation, if and only if it is a solution of the adjoint equation Note that the adjoint of (325) is in turn found to be the associated homogeneous equation of (3), thus each is the adjoint of the other In this case, a first integral to (3) is v(x)p (x)y (v(x)p (x)) y + v(x)p (x)y = v(x)f(x) dx + C Example 32 Find the general solution of the DE Solution The adjoint of this equation is (x 2 x)y + (2x 2 + 4x 3)y + 8xy = (x 2 x)v (2x 2 )v + (4x 2)v = By the trial of x m, this equation is found to have x 2 as a solution Thus x 2 is an integrating factor of the given differential equation Multiplying the original equation by x 2, we obtain (x 4 x 3 )y + (2x 4 + 4x 3 3x 2 )y + 8x 3 y = x 2 Thus a first integral to it is (x 4 x 3 )y (4x 3 3x 2 )y + (2x 4 + 4x 3 3x 2 )y = x 2 dx + C After simplification, we have y + 2x x y = 3(x ) + C x 3 (x ) An integrating factor for this first order linear equation is e 2x (x ) 2 Thus the above equation becomes e 2x (x ) 2 y = e (x )e 2x 2x (x ) dx + C 3 x 3 dx + C 2 That is e 2x (x ) 2 y = [ x ] e 2x + C e2x 4 2x 2 + C 2 Thus the general solution is ( x y = (x ) C ) x 2 + C 2e 2x

35 33 LAGRANGE S IDENTITY AND GREEN S FORMULA 35 Exercise 3 Solve the following differential equation by finding an integrating factor of it Ans: y = C 2x + C2x 2x e 2x y + 4x 2x y + 8x 8 (2x ) 2 y = Exercise 32 The equation (p(x)y ) + q(x)y + λr(x)y = (326) is called a Sturm-Liouville equation, where λ is a real parameter Show that the adjoint of a Sturm- Liouville equation is itself 33 Lagrange s Identity and Green s Formula Let L be the differential operator given by the left hand side of (3), that is L[y] = p (x)y + p (x)y +p 2 (x)y The formal adjoint of L is the differential operator defined by L + [y] = (p (x)y) (p (x)y) +p 2 (x)y, where p, p and p 2 are continuous on an interval [a, b] Let u and v be functions having continuous second derivatives on the interval [a, b] Direct simplification gives the following identity relating L and L + Theorem 3 (Lagrange s identity) where P (u, v) = up v u(p v) + u p v L[u]v ul + [v] = d [P (u, v)], dx Proof Expanding both sides, we get p u v + p u v p uv 2p uv p uv + p uv + p uv Integrating Lagrange s identity leads to Green s formula Corollary 32 (Green s formula) b ( L[u]v ul + [v] ) dx = P (u, v)(x) a Let s define an inner product for continuous real-valued functions on [a, b] by (f, g) = b a f(x)g(x) dx With this notation, Green s formula becomes (L[u], v) = (u, L + [v]) + P (u, v)(x) b a b a (337) Though the operators L and L + are meaningful when applied to any function y having a continuous second derivative on [a, b], it is usually more advantageous to further restrict the domains they act In general, it is evidently more useful to restrict L and L + so that (L[u], v) = (u, L + [v]) (338)

36 36 CHAPTER 3 SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS holds for all u in the domain of L and all v in the domain of L + This is the case when we try to solve boundary value problems in which L and L + are restricted to those functions that satisfy the given boundary conditions When (338) holds, L + is called the adjoint operator of L (ie without the adjective formal ) For the special case when L + = L and (338) holds, we say that L is self-adjoint As an example, let L be the differential operator given by the first two terms on the left hand side of the Sturm-Liouville equation (326), that is L[y] = (p(x)y ) + q(x)y By the exercise for the equation in (326), it follows that L + = L Then Lagrange s identity and Green s formula reduce to the followings Theorem 33 (Lagrange s identity for L[y] = (p(x)y ) + q(x)y) L[u]v ul[v] = d [pw (u, v)], dx where W (u, v) = uv vu is the Wronskian of u and v Corollary 34 (Green s formula for L[y] = (p(x)y ) + q(x)y) b (L[u], v) (u, L[v]) = (L[u]v ul[v]) dx = pw (u, v)(x) a b a (339) Exercise 33 Let L[y] = p (x)y + p (x)y + p 2 (x)y Prove that L + = L if and only if p (x) = p (x) Exercise 34 Prove theorem 33 Exercise 35 Prove corollary 34 Exercise 36 Prove that if u and v are solutions of (p(x)y ) + q(x)y =, then pw (u, v)(x) is a constant for all x [a, b] 34 Regular Boundary Value Problem The problem of finding a solution to a second order linear differential equation satisfying the boundary conditions y + p(x)y + q(x)y = f(x), a < x < b; (34) a y(a) + a 2 y (a) + b y(b) + b 2 y (b) = d, a 2 y(a) + a 22 y (a) + b 2 y(b) + b 22 y (b) = d 2, (34) is called a two-point boundary value problem When d = d 2 =, the boundary conditions are said to be homogeneous; otherwise we refer them as nonhomogeneous For the homogeneous equation with homogeneous boundary condition one can verify the following properties: y + p(x)y + q(x)y =, a < x < b; a y(a) + a 2 y (a) + b y(b) + b 2 y (b) =, a 2 y(a) + a 22 y (a) + b 2 y(b) + b 22 y (b) =, (342)

37 34 REGULAR BOUNDARY VALUE PROBLEM 37 If φ(x) is a nontrivial solution to (342), then so is cφ(x) for any constant c Thus (342) has a one-parameter family of solutions 2 If (342) has two linearly independent solutions φ and φ 2, then c φ +c 2 φ 2 is also a solution of (342) for any constants c and c 2 Thus (342) has a two-parameter family of solutions 3 The remaining possibility is that φ(x) is the unique solution to (342) For nonhomogeneous equation (34), there is an additional possibility that it has no solution The following examples illustrate some of these cases Example 33 Find all solutions to the boundary value problem y + 2y + 5y = ; y() = 2, y(π/4) = e π/4 Solution The general solution to the equation y +2y +5y = is y = c e x cos 2x+c 2 e x sin 2x Substituting the boundary conditions, we have y() = c = 2 and y(π/4) = c 2 e π/4 = e π/4 Thus c = 2, c 2 =, and so the boundary value problem has the unique solution y = 2e x cos 2x + e x sin 2x Example 34 Find all solutions to the boundary value problem y + y = cos 2x; y () =, y (π) = Solution The general solution to the equation y + y = cos 2x is y = c cos x + c 2 sin x (/3) cos 2x Thus y = c sin x + c 2 cos x + (2/3) sin 2x Substituting the boundary conditions, we have y () = c 2 = and y (π) = c 2 = Thus the boundary value problem has a oneparameter family of solutions y = c cos x (/3) cos 2x, where c is any real number Example 35 Find all solutions to the boundary value problem y + 4y = ; y( π) = y(π), y ( π) = y (π) Solution The general solution to the equation y +4y = is y = c cos 2x+c 2 sin 2x Since cos 2x and sin 2x are periodic functions of period π, y = c cos 2x + c 2 sin 2x and y = 2c sin 2x + 2c 2 cos 2x automatically satisfy the boundary conditions y( π) = y(π), y ( π) = y (π) Hence the boundary value problem has a two-parameter family of solutions y = c cos 2x + c 2 sin 2x, where c and c 2 are any real numbers Example 36 Find all solutions to the boundary value problem y + 4y = 4x; y( π) = y(π), y ( π) = y (π) Solution The general solution to the equation y + 4y = 4x is y = x + c cos 2x + c 2 sin 2x Since y( π) = π + c and y(π) = π + c, there are no solutions satisfying y( π) = y(π) Hence the boundary value problem has no solution Exercise 37 Find all solutions to the boundary value problem Ans y = ( 2 + eπ ) sin x 2 cos x + 2 ex y + y = e x ; y() =, y(π) + y (π) =

38 38 CHAPTER 3 SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS 35 Regular Sturm-Liouville Boundary Value Problem Let L[y] = (p(x)y ) + q(x)y Consider the regular Sturm-Liouville boundary value problem: L[y] + λr(x)y =, a < x < b; a y(a) + a 2 y (a) =, b y(b) + b 2 y (b) =, (353) where p(x), p (x), q(x) and r(x) are continuous functions on [a, b] and p(x) > and r(x) > on [a, b] We only consider separated boundary conditions Also we exclude the cases when a = a 2 = or b = b 2 = Let u and v be functions that have continuous second derivatives on [a, b] and satisfy the boundary conditions in (353) The boundary conditions in (353) imply that W (u, v)(b) = W (u, v)(a) = This is because the system of equations a u(a) + a 2 u (a) =, a v(a) + a 2 v (a) = has nontrivial solutions in a and a 2 since a and a 2 are not both zero Therefore the determinant of the system W (u, v)(a) must be zero Similarly W (u, v)(b) = Thus by Green s formula (339), (L[u], v) = (u, L[v]) Therefore L is a self-adjoint operator with domain equal to the set of functions that have continuous second derivatives on [a, b] and satisfy the boundary conditions in (353) Self-adjoint operators are like symmetric matrices in that their eigenvalues are always real Remark Here we only require u and v satisfy the boundary conditions in (353) but not necessarily the differential equation L[y] + λr(x)y = However, if u and v satisfy the differential equation, then L[u]v = ul[v] = λruv and hence (L[u], v) = (u, L[v]) too The regular Sturm-Liouville boundary value problem in (353) involves a parameter λ The objective is to determine for which values of λ, the equation in (353) has nontrivial solutions satisfying the given boundary conditions Such problems are called eigenvalue problems The nontrivial solutions are called eigenfunctions, and the corresponding number λ an eigenvalue If all the eigenfunctions associated with a particular eigenvalue are just scalar multiple of each other, then the eigenvalue is called simple Theorem 35 All the eigenvalues of the regular Sturm-Liouville boundary value problem (353) are (i) real, (ii) have real-valued eigenfunctions and (iii) simple Proof (i) Let λ be a possibly complex eigenvalue with eigenfunction u for (353) Also u may be complex-valued Thus u, L[u] = λru, and u satisfies the boundary conditions in (353) Then L[u] = L[u] = λru = λru Also u satisfies the boundary conditions in (353) as a, a 2, b, b 2 are real Thus λ is an eigenvalue with eigenfunction u for (353) We have By Green s formula, (L[u], u) = ( λru, u) = λ(ru, u) = λ b (L[u], u) = (u, L[u]) = (u, λru) = λ(u, ru) = λ a r u 2 dx b a r u 2 dx