2019학년도 1학기

반도체 공학과 공학수학1

주교재: Erwin Kreyszig, Engineering Mathematics, 10th Edition

부교재: 이상구 외 4인, 최신공학수학 I, 1st Edition

실습실: http://www.hanbit.co.kr/EM/sage/ http://matrix.skku.ac.kr/LA/

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

담당교수: 김응기 박사

Week 5

5주차

2.10: Solution by Variation of Parameters

4.5 매개변수변화법(2계)

http://www.hanbit.co.kr/EM/sage/1_chap4.html

*4.1: Systems of ODEs as Models in Engineering Applications

6.1 연립미분방정식이란

6.2 동차 선형연립미분방정식

6.4 기본행렬과 지수행렬을 이용한 연립미분방정식의 풀이

6.6 연립미분방정식의 분리

http://www.hanbit.co.kr/EM/sage/1_chap6.html

http://matrix.skku.ac.kr/2018-EM/EM-2-W7-lab.html

5.1. Power Series Method

7.2 급수해법

http://www.hanbit.co.kr/EM/sage/1_chap7.html

2.10 Solution by Variation of Parameters

Nonhomogeneous linear ODEs

(1) .

A general solution of is the sum of a general solution of the corresponding homogeneous ODE and any particular of . It is called the method of variation of parameters and is credited to Lagrange. Here , , in may be variable(given function ), but we assume that they are continuous on some open interval . Lagrange’s method gives a particular solution of on in the form

(2)

where , form a basis of solution of the corresponding homogeneous ODE

(3)

on , and is the Wronskian of , .

(4) .

The general solution of is

.

Proof

A general solution

of the homogeneous ODE on open interval and to replace the constants and by functions and . We shall determine and so that the resulting function

(5)

is a particular solution of the homogeneous ODE .

Differentiation , we obtain

We can determine and such that satisfies and and satisfy as a second condition the equation

(6)

Hence, by differentiation

(7)

Differentiation , we obtain

(8)

We now substitute and its derivatives according to , , into .

Collecting terms in and term in , we obtain

Since are solution of the homogeneous ODE , this reduces to

(9a)

Equation is

(9b)

This is a linear system of two algebraic equations for the unknown functionand .

By Cramer’s rule

Since and from a basis, we have

Hence, by integration

A particular solution is

The general solution of is

Example 1 Method of Variation of Parameters

Solve the nohomogeneous ODE

.

Solution

The ODE has the general solution .

The characteristic equation is .

The characteristic roots are and .

The general solution is .

Wronskian

.

From (2), choosing zero constants of integration, The particular solution of given ODE

.

From and the general solution of the homogeneous ODE,

we obtain the general solution is

Fig. 70. Particular solution and its first term in Example 1

Sage Coding

http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/

var('x') y = function('y')(x) de = desolve(diff(y, x, 2) + y == sec(x), [y, x]) print de |

Evaluate

k1*sin(x) + k2*cos(x) + x*sin(x) + 1/2*log(1/2*cos(2*x) + 1/2)*cos(x)

*Chapter 4. Systems of ODEs

*4.1 Systems of ODEs as Models in Engineering Applications

Example 1 Mixing Problem Involving Two Tanks

Tank and in contain initially of water each. In the water is pure, whereas of fertilizer are dissolved in . By circulating liquid at a rate of and stirring (to keep the mixture uniform) the amounts of fertilizer in and in change with time . How long should we let the liquid circulate so that will contain at least half as much fertilizer as there will be left in ?

Solution

Step 1. Setting up the model.

For tank , the time rate of change of equals inflow minus outflow.

For tank , the time rate of change of equals inflow minus outflow.

The mathematical model of our mixture problem is the system of first-order ODEs

(Tank )

(Tank )

A vector equation with column vector .

Matrix this becomes

Step 2. General solution.

We try an exponential function of ,

. Then

Dividing the equation by , we obtain

.

Find the eigenvalues and eigenvectors of .

The eigenvalues are the solutions of the characteristic equation

The eigenvalues of is , .

Find the eigenvector of corresponding .

Hence , we can take .

Eigenvector of corresponding to is

Find the eigenvector of corresponding .

Hence , we can take .

Eigenvector of corresponding to is

From and the superposition principle (which continues to hold for systems of homogeneous linear ODEs) we thus obtain a solution

(3)

where and are arbitrary constants.

Step 3. Use of initial conditions.

The initial conditions are (no fertilizer in tank ).

The initial conditions are

From this and (3) with we obtain

.

,

(Tank , lower curve)

(Tank , upper curve)

The exponential increase of and the exponential decrease of to the common limit .

Did you expect this for physical reasons? Can you physically explain why the curves look “symmetric”? Would the limit change if initially contained of fertilizer and contained ?

Step 4. Answer.

contains half the fertilizer amount of if it contains of the total amount, that is,

. Thus

Hence the fluid should circulate for at least about half an hour.

Sage Coding

http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/

t = var('t') x = function('x')(t);y = function('y')(t) de1 = diff(x, t) == -0.02*x + 0.02*y de2 = diff(y, t) == 0.02*x - 0.02*y desolve_system([de1, de2], [x, y], ics=[0, 0, 150]) |

Evaluate

[x(t) == -75*e^(-1/25*t) + 75, y(t) == 75*e^(-1/25*t) + 75]

Example 2 Electrical Network

Find the currents and in the network. Assume all currents and charges to be zero at , the instant when the switch is closed.

Solution

Step 1. Setting up the mathematical model.

The model of this network is obtained from Kirchhoff’'s Voltage Law, (where we considered single circuits).

Let and be the currents in the left and right loops, respectively.

In the left loop, the voltage drops are over the inductor and over the resistor, the difference because and flow through the resistor in opposite directions. By Kirchhoff’'s Voltage Law the sum of these drops equals the voltage of the battery; that is, , hence

(4a) .

In the right loop, the voltage drops are and over the resistors and over the capacitor, and their sum is zero,

or .

Division by and differentiation gives .

To simplify the solution process, we first get rid of , which by (4a) equals

.

Substitution into the present ODE gives

and by simplification

(4b) .

In matrix form, (4) is (we write since is the unit matrix)

(5) where , , .

Step 2. Solving (5).

Because of the vector this is a nonhomogeneous system, and we try to proceed as for a

single ODE, solving first the homogeneous system (thus ) by substituting . This gives

, hence .

Hence, to obtain a nontrivial solution, we again need the eigenvalues and eigenvectors. For the present matrix they are derived in Example 1 in Sec. 4.0:

, : , .

Hence a “general solution” of the homogeneous system is

.

For a particular solution of the nonhomogeneous system (5), since is constant, we try a constant column vector with components , . Then , and substitution into (5) gives ; in components,

.

The solution is , ; thus . Hence

(6) ;

in components,

.

The initial conditions give

.

Hence and . As the solution of our problem we thus obtain

(7) .

In components (Fig. 80b),

.

Now comes an important idea, on which we shall elaborate further, beginning in Sec. 4.3. Figure 80a shows and as two separate curves. Figure 80b shows these two currents as a single curve in the -plane. This is a parametric representation with time as the parameter. It is often important to know in which sense such a curve is traced. This can be indicated by an arrow in the sense of increasing , as is shown. The -plane is called the phase plane of our system (5), and the curve in Fig. 80b is called a trajectory. We shall see that such “phase plane representations” are far more important than graphs as in Fig. 80a because they will give a much better qualitative overall impression of the general behavior of whole families of solutions, not merely of one solution as in the present case.

Sage Coding

http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/

t = var('t') x = function('x')(t);y = function('y')(t) de1 = diff(x, t) == -4.0*x + 4.0*y + 12.0 de2 = diff(y, t) == -1.6*x + 1.2*y + 4.8 desolve_system([de1, de2], [x, y], ics=[0, 0, 0]) |

Evaluate

[x(t) == -8*e^(-2*t) + 5*e^(-4/5*t) + 3, y(t) == -4*e^(-2*t) + 4*e^(-4/5*t)]

Theorem 1 Conversion of an ODE

An th-order ODE

can be converted to a system of n first-order ODEs by setting

This system is of the form

Example 3 Mass on a Spring

To gain confidence in the conversion method, let us apply it to an old friend of ours, modeling the free motions of a mass on a spring (see Sec. 2.4)

or

For this ODE (8) the system (10) is linear and homogeneous,

Setting , we get in matrix form

The characteristic equation is

.

For an illustrative computation, let , and . Then

.

This gives the eigenvalues and .

Eigenvectors follow from the first equation in which is .

For this gives , say, , .

For it gives , say, , .

These eigenvectors

, give

.

This vector solution has the first component

which is the expected solution.

The second component is its derivative

.

Sage Coding

http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/

c = 2;m = 1;k = 0.75 t = var('t') x = function('x')(t);y = function('y')(t) de1 = diff(x, t) == 1.0*y de2 = diff(y, t) == -k/m*x -c/m*y desolve_system([de1, de2], [x, y]) |

Evaluate

[x(t) == -1/2*(x(0) + 2*y(0))*e^(-3/2*t) + 1/2*(3*x(0) + 2*y(0))*e^(-1/2*t),

y(t) == 3/4*(x(0) + 2*y(0))*e^(-3/2*t) - 1/4*(3*x(0) + 2*y(0))*e^(-1/2*t)]

Chapter 5. Series Solution of ODEs. Special Functions

5.1 Power Series Method

Power Series (in powers of ) is

is a variable. are constants, called the coefficient of the series.

is a constant. is the center of the series.

If , we obtains a power series in powers of

Maclaurin series is

.

Example 1 Familiar Power Series are the Maclaurin series

(, geometric series)

Idea of the Power series Method

For a given ODE

we first represent and by power series in powers of . Often and are polynomials.

(3)

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

Example 2 Power Series Solution.

Solve .

Solution

The recursion formula is

.

We obtain successively

Sage Coding

http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/

y = function('y')(x) de = desolve(diff(y, x) - y == 0, y) print "y(x) =", de |

Evaluate

y(x) = c*e^x

Example 3 A Special Legendre Equation

The ODE

occurs in models exhibiting spherical symmetry. Solve it.

Solution

The recursion formula is

.

We obtain successively

and so on and remain arbitrary. With these coefficients the series (3) becomes

Sage Coding

http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/

y = function('y')(x) de = (1- x^2)*diff(y, x, 2) - 2*x*diff(y, x) + 2*y == 0 desolve(de, y, contrib_ode = True) |

Evaluate

[y(x) == k1*gauss_a(-1, 2, 1, -1/2*x + 1/2) + k2*gauss_b(-1, 2, 1, -1/2*x + 1/2)]

Theory of the Power Series Method

The partial sum of is

(6)

where .

If we omit the terms of from , the remaining expression

.

This expression is called the remainder of after the term .

We have now associated with the sequence of the partial sums

.

If for some this sequence converges,

then the series is called convergent at , the number is the value or sum of at , and we write

.

Then we have for every ,

(4) .

If that sequence diverges at , the series is divergent at .

Convergence Interval. Radius of Convergence

There are three cases with respect to the convergence of the power series

For a given power series there are only three possibilities

(i) The series converges only when .

(ii) The series converges for all .

(iii) There is a positive number such that the series converges if and diverges if .

is called the radius of convergence of the power series

is interval of convergence.

, is an endpoint of the interval.

The interval of convergence of a power series is the interval that consists of all values of for which the series converges

is converge at both endpoints

is converge at once endpoints

,

is diverge at both endpoints

The Absolute Ratio Test

as ,

(1) is convergent.

(2) is divergent

(3) by the Ratio is inconclusive.

Example 4 Convergence Radius

In the case of the series .

Solution

By Ratio Test.

is converges for all .

Radius of convergence is .

Sage Coding

http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/

var('m') a(m) = 1/factorial(m) limit(a(m + 1)/a(m), m = +oo) |

Evaluate

0

In the case of the series

.

Solution

By Ratio Test.

Geometric series converges when .

Radius of convergence is .

Sage Coding

http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/

var('m') a(m) = 1 rho = limit(a(m + 1)/a(m), m = +oo) R = 1/ rho R |

Evaluate

1

In the case of series .

Solution

is convergent when

By the Ratio Test

The series diverges when .

Thus, the series converges only when .

The radius of convergence is .

Sage Coding

http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/

var('m') a(m) = factorial(m) rho = limit(a(m + 1)/a(m), m = +oo) R = 1/ rho R |

Evaluate

0

Definition Real Analytic Function

A real function is called analytic at a point if it can be represented by a power series in powers of with radius of convergence .

Theorem 1 Existence of Power Series Solutions

If , and in are analytic at , then every solution of is analytic at and can thus be represented by a power series in powers of with radius of convergence .

Operations on Power Series

1. Termwise Differentiation

A power series may be differentiated term by term. If

converges for , where .

2. Termwise Addition

Two power series may be added term by term. If

and

have positive radii of convergence and their sums and , then the series

converges and represents for each that lies in the interior of the convergence interval of each of the two given series.

3. Termwise Multiplication

Two power series may be multiplied term by term. If

and

have positive radii of convergence and their sums and , then the series

converges and represents for each that lies in the interior of the convergence interval of each of the two given series.

4. Vanishing of All Coefficients

If a power series has a positive radius of convergence and a sum that is identically zero throughout its interval of convergence, then each coefficient of the series must be zero.

http://matrix.skku.ac.kr/sglee/

[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