﻿ 2019학년도 1학기

2019학년도 1학기

반도체 공학과 공학수학1

주교재: Erwin Kreyszig, Engineering Mathematics, 10th Edition

부교재: 이상구 외 4인, 최신공학수학 I,  1st Edition

강의시간:  공학수학1, (화09:00-10:15)   (목10:30-11:45)

담당교수:  김응기 박사

Week 6

6주차

5.2: Legendre's Equation. Legendre Polynomials

7.4 르장드르 방정식

5.3: Extended Power Series Method: Frobenius Method

7.3 프로베니우스 해법

5.2 Legendre’s Equation. Legendre Polynomial

Legendre’s Equation

where is a given constant. The parameter in (1) is a given real number. Any solution of (1) is called a Legendre function.

Diving (1) by the coefficient of ,

or

We see that the coefficient and of the new equation are analytic at .

Legendre’s Equation has power series solution of the form

.

and insert this series and the series obtained by termwise differentiation.

(2)        ,

Substituting into , we obtain

We obtain the general formula

(3)

or

(4)

This is called a recurrence relation or recursion formula.

It gives each coefficient in terms of the second one preceding it, except for and , which are left as arbitrary constants. We find successively

is even function is odd function

and are linearly independent solutions.

is a general solution of the Legendre’s equation on the interval .

Legendre Polynomials

For Legendre’s equation this happens when the parameter is a nonnegative integer because then the right side of is .

(i) is even, reduces to polynomial of degree .

(ii) is odd reduces to polynomial of degree .

is a Legendre polynomial of degree .

for every .

Thus .

is a Legendre polynomial of degree .

is a particular solution of the Legendre’s differential equation.

Sage Coding

 P. = QQ[] print "P_1(x) =", legendre_P(1,x) print "P_2(x) =", legendre_P(2,x) print "P_3(x) =", legendre_P(3,x) print "P_4(x) =", legendre_P(4,x) print "P_5(x) =", legendre_P(5,x) p1 = plot(legendre_P(1, x)) p2 = plot(legendre_P(2, x), color = 'red') p3 = plot(legendre_P(3, x), color = 'orange') p4 = plot(legendre_P(4, x), color = 'green') p5 = plot(legendre_P(5, x), color = 'black') p1 + p2 + p3 + p4 + p5

Evaluate

P_1(x) = x

P_2(x) = 3/2*x^2 - 1/2

P_3(x) = 5/2*x^3 - 3/2*x

P_4(x) = 35/8*x^4 - 15/4*x^2 + 3/8

P_5(x) = 63/8*x^5 - 35/4*x^3 + 15/8*x

Rodrigue’s Formula

Legendre polynomials is .

Then is

(O. Rodrigue’s formula)

Proof

Let then   or  .

Differential

When

This equation is Legendre differential equation of degree .

satisfies Legendre differential equation.

is a Legendre polynomial of degree ,

is the constant times of .

hence

.

For instance

Sage Coding

 var('n') n = 3      # n을 바꿔가며 르장드르 다항식을 구할 수 있다. f(x) = (x^2 - 1)^n P(x) = (1/(factorial(n)*2^n)*diff(f(x), x, n)).expand() print "P(x) =", P(x)

Evaluate

P(x) = 5/2*x^3 - 3/2*x

Legendre Polynomials

(I)

Proof

Using Rodrigues’ formula

By integration by parts

.

Hence

(i) when and

By ① when .

.

By ① when

Hence

(ii)

By ①

Thus

or

Let , then .

Hence

.

(II) If , , show that

.

Proof

hence

5.3 Extended Power Series Method: Frobenius Method

Theorem 1  Frobenius Method

Let and be any functions that are analytic at . Then the ODE

has at least one solution that can be represented in the form

where the exponent may be any (real or complex) number.

Regular and Singular Points

A regular point of

is a point at which the coefficients and are analytic. If is not regular, is called singular point.

A regular point of the ODE

is a point at which are analytic and . If is not regular, is called singular point.

Indicial Equation, Indicating the Form of Solutions

Multiplication of (1) by gives the more convenient form

.

We first expand and in power series,

,

or we do nothing if and are polynomials. Then we differentiate (2) term by term, finding

By inserting all these series into () we readily obtain

.

We now equate the sum of the coefficients of each power to .

This yields a system of equation involving the unknown coefficients . The equation corresponding to the power is

The quadratic equation is called the indicial equation of the ODE (1).

Theorem 2  Frobenius Method. Basis of Solution. Three Cases

Suppose that the ODE (1) satisfies the assumptions in Theorem 1. Let and be the roots of the indicial equation (4). Then we have the following three cases.

Cases 1. Distinct Roots Not Differing by an Integer.

A basis is

and

with coefficients obtained successively from (3) with and , respectively.

Cases 2. Double Root

A basis is

and

Cases 3. Roots Differing by an Integer.

A basis is

and

where the roots are so dented that and may turn out to .

Example 1

For Euler-Cauchy equation

( constant)

substitution of gives the auxiliary equation

.

which is the indicial equation.

For different roots and we get a bassis , .

For a double root we get a basis . . Accordingly, for this simple ODE.

Example 2

Solve the ODE .

Solution

Given ODE .

is a point at which the coefficients are analytic.

is called singular point.

The solution of ODE is

Substituting this formula into given ODE, we obtain

.

This gives the recursion formula

or .

Hence , we obtain the solution

By choosing , we obtain the solution

.

We get a second independent solution by the method of reduction of order.

Let .

Let , then .

From this, using partial fractions and integrating, we get

,

imply

so , so

,

The general solution is

.

and are linearly independent and thus form a basis on interval (as well as on ).

Sage Coding

 var('x') y = function('y')(x) de = x*(x - 1)*diff(y, x, 2) + (3*x - 1)*diff(y,x) + y == 0 desolve(de, [y, x])

Evaluate

k1*log(x)/(x - 1) + k2/(x - 1)

Example 3

Solve the ODE .

Solution

Given ODE is

is a point at which the coefficients are analytic.

is called singular point.

The solution of ODE is

Substituting this formular into given ODE, we obtain

.

This gives the recursion formula

.

The roots are and . They differ by an integer.

From we have

Hence , we obtain the solution

By choosing , we obtain the solution

.

We get a second independent solution by the method of reduction of order.

Let .

Let , then .

From this, using partial fractions and integrating, we get

so

so

so , so

so

The general solution is

and are linearly independent, and has a logarithmic term.

Hence and constitute a basis of solutions for positive .

Sage Coding

 var('x') y = function('y')(x) de = x*(x - 1)*diff(y, x, 2) - x*diff(y,x) + y == 0 desolve(de, [y, x], contrib_ode = True)

Evaluate

[y(x) == (x*log(x) + 1)*k2 + k1*x]

[한빛 아카데미] Engineering Mathematics with Sage:

[저자] 이상 구, 김영 록, 박준 현, 김응 기, 이재 화

Contents

A. 공학수학 1 – 선형대수, 상미분방정식+ Lab

Chapter 01 벡터와 선형대수 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-1.html

Chapter 02 미분방정식의 이해 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-2.html

Chapter 03 1계 상미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-3.html

Chapter 04 2계 상미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-4.html

Chapter 05 고계 상미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-5.html

Chapter 06 연립미분방정식, 비선형미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-6.html

Chapter 07 상미분방정식의 급수해법, 특수함수 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-7.html

Chapter 08 라플라스 변환 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-8.html

B. 공학수학 2 - 벡터미적분, 복소해석 + Lab

Chapter 09 벡터미분, 기울기, 발산, 회전 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-9.html

Chapter 10 벡터적분, 적분정리 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-10.html

Chapter 11 푸리에 급수, 적분 및 변환 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-11.html

Chapter 12 편미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-12.html

Chapter 13 복소수와 복소함수, 복소미분 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-13.html

Chapter 14 복소적분 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-14.html

Chapter 15 급수, 유수 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-15.html

Chapter 16 등각사상 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-16.html

Made by Prof. Sang-Gu LEE  sglee at skku.edu

http://matrix.skku.ac.kr/sglee/   with Dr. Jae Hwa LEE