Chapter 6: Series Solutions of Linear Equations Department of - PowerPoint PPT Presentation

Review of Power Series Solutions about Ordinary Points Solutions about Singular Points Summary Chapter 6: Series Solutions of Linear Equations Department of Electrical Engineering National Taiwan University ihwang@ntu.edu.tw November 13, 2013 1 / 43 DE Lecture 9 王奕翔 王奕翔

Review of Power Series Solutions about Ordinary Points Solutions about Singular Points Summary Solving Higher-Order Linear Equations In Chapter 4, we learns how to analytically solve two special kinds of higher-order linear differential equations: 1 Linear Differential Equation with Constant Coefficients 2 Cauchy-Euler Equations Essentially only one kind – linear DE with constant coefficients! Question : Is it possible to solve other kinds, like the following? 2 / 43 DE Lecture 9 Because to solve Cauchy-Euler DE, we substitute x = e t ! ( x 2 + 2 x − 3) y ′′ − 2( x + 1) y ′ + 2 y = 0 王奕翔

Review of Power Series Solutions about Ordinary Points Solutions about Singular Points Summary Idea : Express the solution function as a power series! c n x n 3 / 43 DE Lecture 9 ∞ ∑ y ( x ) = n =0 王奕翔

Review of Power Series Solutions about Ordinary Points Solutions about Singular Points Summary Focus on: Linear Second-Order Differential Equations Throughout this lecture, we shall focus on solving homogeneous linear second order differential equations using the method of power series. Standard Form : Frequently throughout the discussions in this lecture: 4 / 43 DE Lecture 9 a 2 ( x ) y ′′ + a 1 ( x ) y + a 0 ( x ) = 0 , y ′′ + P ( x ) y ′ + Q ( x ) y = 0 . 王奕翔

Review of Power Series Solutions about Ordinary Points Solutions about Singular Points Summary 1 Review of Power Series 2 Solutions about Ordinary Points 3 Solutions about Singular Points 4 Summary 5 / 43 DE Lecture 9 王奕翔

Review of Power Series series of the following form: 6 / 43 Some Examples : Solutions about Ordinary Points DE Lecture 9 Definition Power Series Summary Solutions about Singular Points A power series in ( x − a ) (or a power series centered at a ) is an infinite ∞ ∑ c n ( x − a ) n , n =0 where { c n } ∞ 0 is a sequence of real numbers. ∞ ∞ x n = 1 + x + x 2 + · · · , 2 n x n = 1 + 2 x + 4 x 2 + · · · . ∑ ∑ n =0 n =0 王奕翔

Review of Power Series lim 7 / 43 not sure divergence absolute convergence c n Solutions about Ordinary Points lim convergence of the series: exists. N DE Lecture 9 exists. N Convergence, Divergence, Absolute Convergence Summary lim Solutions about Singular Points Convergence : A power series ∑ ∞ n =0 c n ( x − a ) n converges at x = x 0 if ∑ c n ( x 0 − a ) n N →∞ n =0 Otherwise, the power series diverges at x = x 0 . Absolute Convergence : A power series ∑ ∞ n =0 c n ( x − a ) n converges absolutely at x = x 0 if ∑ | c n ( x 0 − a ) n | N →∞ n =0 Ratio Test : Suppose c n ̸ = 0 for all n , then the following test tells us about the < 1 � � � � c n +1 ( x 0 − a ) n +1 � � � � c n +1 � = | x 0 − a | lim > 1 � � � � � c n ( x 0 − a ) n � � n →∞ n →∞ = 1 王奕翔

Review of Power Series Interval of Convergence : Every power series has an interval of 8 / 43 Solutions about Ordinary Points DE Lecture 9 Interval of Convergence Summary Solutions about Singular Points convergence I = ( a − R , a + R ) , in which he power series ∑ ∞ n =0 c n ( x − a ) n converges absolutely. R > 0 is called the radius of convergence . absolute divergence divergence convergence x a − R a a + R series may converge or diverge at endpoints ∞ ∑ A power series defines a function of x , f ( x ) := c n ( x − a ) n for x ∈ I . n =0 王奕翔

Review of Power Series Define the function ( I : interval of convergence) 9 / 43 Solutions about Ordinary Points Differentiation DE Lecture 9 Summary Solutions about Singular Points Function Defined by a Power Series ∑ ∞ n =0 c n ( x − a ) n ∞ ∑ y ( x ) := c n ( x − a ) n , x ∈ I . n =0 ∞ y ′ ( x ) = c 1 + 2 c 2 x + 3 c 3 x 2 + · · · = ∑ nc n ( x − a ) n − 1 , x ∈ I n =1 ∞ y ′′ ( x ) = 2 c 2 + 6 c 3 x + 12 c 4 x 2 + · · · = ∑ n ( n − 1) c n ( x − a ) n − 2 , x ∈ I n =2 王奕翔

Review of Power Series Examples 10 / 43 x n Solutions about Ordinary Points DE Lecture 9 Taylor’s Series represented by Taylor’s Series as follows, with a radius of convergence Solutions about Singular Points Summary If a function f ( x ) is infinitely differentiable at a point a , then it can be R > 0 . ∞ f ( n ) ( a ) ∑ f ( x ) = ( x − a ) n . n ! n =0 ∞ 1! + x 2 e x = 1 + x ∑ 2! + · · · = n ! , x ∈ R n =0 ∞ 1 1 − x = 1 + x + x 2 + · · · = ∑ x n , x ∈ ( − 1 , 1) n =0 王奕翔

Review of Power Series Solutions about Ordinary Points 11 / 43 DE Lecture 9 Summary Solutions about Singular Points Interval Maclaurin Series of Convergence 1! � x 2 2! � x 3 � 1 e x � 1 � x 3! � . . . � � n ! x n ( �� , � ) n � 0 cos x � 1 � x 2 2! � x 4 4! � x 6 � ( � 1) n 6! � . . . � � (2 n )! x 2 n ( �� , � ) n � 0 sin x � x � x 3 3! � x 5 5! � x 7 � ( � 1) n 7! � . . . � � (2 n � 1)! x 2 n � 1 ( �� , � ) n � 0 tan � 1 x � x � x 3 3 � x 5 5 � x 7 � ( � 1) n 7 � . . . � � 2 n � 1 x 2 n � 1 [ � 1, 1] (2) n � 0 cosh x � 1 � x 2 2! � x 4 4! � x 6 � 1 6! � . . . � � (2 n )! x 2 n ( �� , � ) n � 0 sinh x � x � x 3 3! � x 5 5! � x 7 � 1 7! � . . . � � (2 n � 1)! x 2 n � 1 ( �� , � ) n � 0 ln(1 � x ) � x � x 2 2 � x 3 3 � x 4 � ( � 1) n � 1 4 � . . . � � x n ( � 1, 1] n n � 1 � 1 1 � x � 1 � x � x 2 � x 3 � . . . � � x n ( � 1, 1) n � 0 王奕翔

Review of Power Series Solutions about Ordinary Points Solutions about Singular Points Summary 1 Review of Power Series 2 Solutions about Ordinary Points 3 Solutions about Singular Points 4 Summary 12 / 43 DE Lecture 9 王奕翔

Review of Power Series Definition (Ordinary and Singular Points) 13 / 43 positive radius of convergence. Solutions about Ordinary Points DE Lecture 9 Focus on homogeneous linear 2nd order DE Ordinary and Singular Points Summary Solutions about Singular Points a 2 ( x ) y ′′ + a 1 ( x ) y ′ + a 0 ( x ) y = 0 Rewrite it into its standard form y ′′ + P ( x ) y ′ + Q ( x ) y = 0 . x = x 0 is an ordinary point of the above DE if both P ( x ) and Q ( x ) are analytic at x 0 . Otherwise, x = x 0 is a singular point. Analytic at a Point : a function f ( x ) is analytic at a point x = x 0 if and only if f ( x ) can be represented as a power series ∑ ∞ n =0 c n ( x − x 0 ) n with a In our lecture analytic ≡ infinitely differentiable. 王奕翔

Review of Power Series Solutions about Ordinary Points 14 / 43 DE Lecture 9 Examples: Ordinary and Singular Points Summary Solutions about Singular Points 1 Constant coefficients: a 2 y ′′ + a 1 y ′ + a 0 y = 0 . Every x ∈ R is ordinary. 2 Cauchy-Euler DE: x 2 y ′′ + xy ′ + y = 0 . P ( x ) = 1 x is analytic at x ∈ R \ { 0 } . 1 Q ( x ) = x 2 is analytic at x ∈ R \ { 0 } . Hence, x = 0 is the only singular point. 3 Polynomial Coefficients: a 2 ( x ) y ′′ + a 1 ( x ) y ′ + a 0 ( x ) y = 0 , where a 2 ( x ) ̸ = 0 , a 1 ( x ) , a 0 ( x ) are all polynomials of x . P ( x ) = a 1 ( x ) a 2 ( x ) is analytic at x ∈ R \ { r ∈ R : a 2 ( r ) = 0 } . Q ( x ) = a 0 ( x ) a 2 ( x ) is analytic at x ∈ R \ { r ∈ R : a 2 ( r ) = 0 } . Hence, { r ∈ R : a 2 ( r ) = 0 } are singular points. 4 y ′′ + xy ′ + ( ln x ) y = 0 . P ( x ) = x is analytic at x ∈ R . Q ( x ) = ln x is analytic at x ∈ (0 , ∞ ) . Hence, every x ≤ 0 is singular. 王奕翔

Review of Power Series Theorem 15 / 43 Solutions about Ordinary Points Then, we can find two linearly independent solutions in the form of power DE Lecture 9 The following theorem lays the theoretical foundations of the method. Existence of Power Series Solutions about Ordinary Points Summary Solutions about Singular Points Let x = x 0 be an ordinary point of a homogenous linear 2nd order DE. series centered at x 0 , that is, ∞ ∑ y = c n ( x − x 0 ) n . n =0 Moreover, the radius of convergence ≥ the distance from x 0 to the closest singular point in C . 王奕翔

Review of Power Series Find the minimum radius of convergence of a power series solution about 16 / 43 the DE exists (and converges absolutely). Based on the previous theorem, we obtain the minimum radius of Solutions about Ordinary Points DE Lecture 9 Example: Minimum Radius of Convergence Solutions about Singular Points Summary Example Consider a linear second order DE ( x 2 + 1) y ′′ + xy ′ − y = 0 . the ordinary points x = − 1 and x = 0 . A: The singular points in the complex domain C is ± i . √ √ 1 2 + 1 2 = The distance between − 1 and ± i is 2 . The distance between 0 and ± i is 1 . √ convergence R = 2 and R = 1 respectively. √ In other words, for | x + 1 | < 2 and | x | < 1 , the power series solution of 王奕翔