2018학년도 2학기
반도체 공학과 공학수학2 (GEDB005) 강의교안 (2018학년도용)
주교재: Erwin Kreyszig, Engineering Mathematics, 10th Edition
부교재: 최신공학수학 I 과 II, 한빛출판사 및 (영문) 다변량 Calculus (이상구, 김응기 et al) (http://www.hanbit.co.kr/EM/sage/)
강의시간: BD1615(화 16:3017:45/ 목 15:0016:15), 반도체관 400102호 담당교수: 김응기 박사
주차 
주교재 
부교재 
3 
11.9: Fourier Transform. Discrete and Fast Fourier Transforms 12.1: Basic Concepts of PDEs 12.2: Modeling: Vibrating String, Wave Equation 
11.6 푸리에 변환 12.1 편미분방정식이란 
web 
Week 3
11.9. Fourier Transform. Discrete and Fast Fourier Transforms:
Complex Form of the Fourier Integral
The (real) Fourier integral is
where
, .
Substitution and into the integral for , we have
(1)
is an even function of
.
.
is an odd function of
Euler formula
complex Fourier integral
is complex Fourier integral.
Fourier Transform and Its Inverse
(5) .
Writing , we have
is the Fourier transform of
is the inverse Fourier transform of .
Remark 

Other notations for the Fourier transform is , so that . 
The process of obtaining the Fourier transform from a given is also called the Fourier transform or the Fourier transform method.
Theorem 1 
Existence of the Fourier Transform 
If is absolutely integrable on the axis and piecewise continuous on every finite interval, then the Fourier transformation of given by exists. 
Example 1 
Fourier Transform 
Find the Fourier transform of if and otherwise. 
Solution
Using Fourier transform and integrating,
.
Example 2 
Fourier Transform 
Find the Fourier transform of if and if ; here . 
Solution
From the definition Fourier transform we obtain by integration
.
Linearity. Fourier Transform of Derivatives
Theorem 2 
Linearity of the Fourier Transform 
The Fourier transform is a linear operation ; that is, for any functions and whose Fourier transforms exist and any constant and , the Fourier transform of exists, and (8) . 
Proof
This is true because integration is a linear operation, so that gives
.
Theorem 3 
Fourier Transform of the Derivative of 
Let be continuous on the axis and as . Furthermore, let be absolutely integrable on the axis. Then (9) . 
Proof
From the definition of the Fourier transform we have
.
Two successive application of give
.
We have for the transform of the second derivative of
(10) .
Similarly for higher derivatives.
Example 3 
Application of the Operational Formula (9) 
Find the Fourier transform of from Table III, Sec. 11. 10. 
Solution
We use , By formula in Table III.
.
Convolution
The convolution * of function and is defined by
(11) *.
Theorem 4 
Convolution Theorem 
Suppose that and are piecewise continuous, bounded, and absolutely integrable on the axis. Then (12) *. 
Proof
By the definition,
*
.
Taking
.
Hence *
This double integral can be written as a product of two integrals and gives the desired result
*
.
By taking the inverse Fourier transform on both sides of , writing and as before, and nothing that in and cancel each other, we obtain
(13) *,
a formula that will help us in solving partial differential equation.
Chapter 12. Partial Differential Equations (PDEs)
12.1 Basic Concepts of PDEs
Partial Differential Equation(PDE)
An equation involving one or more partial derivatives of an (unknown) function that depends on two or more variables.
Order of the PDE is the order of the highest derivative.
PDEs
Linear : The first degree in the unknown function and its partial derivatives
Homogeneous : each of its terms contains either or one of its partial derivatives
Nonhomogeneous
Nonlinear
Example 1 Important SecondOrder PDEs
(1) Onedimension wave equation
(2) Onedimension heat equation
(3) Twodimension Laplace equation
(4) Twodimension Poisson equation
(5) Twodimension Wave equation
(6) Threedimension Laplace equation
Here is positive constant, is time, , , are Cartesian coordinates, and dimension is the number of these coordinates in the equation.
A solution of a PDE in some region of the space of the independent variables is a function that has all the partial derivatives appearing in the PDE in some domain containing , and satisfies the PDE in some everywhere in .
In general, the totality of solution of a PDE is very large.
The unique solution of a PDE corresponding to a given physical problem will be obtained by the use of additional conditions arising from the problem.
Addition Conditions
Boundary Conditions
Initial Conditions
Theorem 1 
Fundamental Theorem on Superposition 
If and are solutions of a homogeneous linear PDE in some region , then
with any constants and is also a solution of that PDE in the region . 
Example 2 Solving like an ODE
Find solution of the PDE depending on and .
Solution
Since no derivatives.
The PDE
, where and are constant.
Here and are function of .
Answer is
with arbitrary function and .
Solve .
Solution
Characteristic equation is
Characteristic roots are and
General solution is .
Example 3 Solving like an ODE
Find solution of this PDE.
Solution
Setting .
separable equation
integration with respect to
integration with respect to .
where ,
here, and are arbitrary.
12.2 Modeling: Vibrating String, Wave Equation
Derive the equation modeling small transverse vibrations of an elastic string.
Place the string along the axis, stretch it to length , and fasten it at the ends and .
Distort the string, and at some instant, call it , we release it and allow it to vibrate.
Problem :
To determine the vibrations of the string, that is, to find its deflection at any point and at any time .
is the solution of a PDE.
is the model of our physical system to be derived.
Physical Assumptions
1. The mass of the string per unit length is constant. The string is perfectly elastic and
does not offer any resistance to bending.
2. The tension caused by stretching the string before fastening it at the ends is so large that the action of the gravitational force on the string can be neglected.
3. The string perform small transverse motions in a vertical plane : that is, every particle of the string moves strictly vertically and so that the deflection and the slope at every point of the string always remain small in absolute value.
Derivation of the PDE of the Model (“Wave Equation”) from Forces
Deflected string at fixed time .
The forces acting on a small portion of the string.
Since the string offers no resistance to bending, the tension is tangential to the curve
of the string at each point.
is the tension at the endpoints of that portion.
is the tension at the endpoints of that portion.
Since the points of the string move vertically, there is no motion in the horizontal direction.
The horizontal components of the tension
Horizontal components of the tension must be constant.
Horizontal components of
Horizontal components of
(1) .
The vertical components of the tension
Vertical components of
Vertical components of
The minus sign appears because the component at is directed downward.
By Newton’s second law
two forces = (the mass of the portion) (the acceleration ),
evaluated at some point between and
here, is the mass of the undeflected string per unit length
is the length of the portion of the undeflected string
(Newton’s second law).
Since the points of the string move vertically
In the vertical direction
Newton’s second law : .
Using (1), we can divide this by , obtaining
(2)
is the slopes of the string at
is the slopes of the string at
Here we have to write partial derivatives because depends also on time .
Dividing (2) by , we thus have
.
(3)
is the onedimension wave equation.
is homogeneous and of the second order.
The physical constant is denoted by (instead of ) to indicate that this constant is positive, a fact that will be essential to the form of the solutions.
“Onedimensional” means that the equation involves only one space variable, .
In the vertical direction
Newton’s second law :
Onedimensional wave equation :
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/EMSage/EMathChapter1.html
Chapter 02 미분방정식의 이해 http://matrix.skku.ac.kr/EMSage/EMathChapter2.html
Chapter 03 1계 상미분방정식 http://matrix.skku.ac.kr/EMSage/EMathChapter3.html
Chapter 04 2계 상미분방정식 http://matrix.skku.ac.kr/EMSage/EMathChapter4.html
Chapter 05 고계 상미분방정식 http://matrix.skku.ac.kr/EMSage/EMathChapter5.html
Chapter 06 연립미분방정식, 비선형미분방정식 http://matrix.skku.ac.kr/EMSage/EMathChapter6.html
Chapter 07 상미분방정식의 급수해법, 특수함수 http://matrix.skku.ac.kr/EMSage/EMathChapter7.html
Chapter 08 라플라스 변환 http://matrix.skku.ac.kr/EMSage/EMathChapter8.html
B. 공학수학 2  벡터미적분, 복소해석 + Lab
Chapter 09 벡터미분, 기울기, 발산, 회전 http://matrix.skku.ac.kr/EMSage/EMathChapter9.html
Chapter 10 벡터적분, 적분정리 http://matrix.skku.ac.kr/EMSage/EMathChapter10.html
Chapter 11 푸리에 급수, 적분 및 변환 http://matrix.skku.ac.kr/EMSage/EMathChapter11.html
Chapter 12 편미분방정식 http://matrix.skku.ac.kr/EMSage/EMathChapter12.html
Chapter 13 복소수와 복소함수, 복소미분 http://matrix.skku.ac.kr/EMSage/EMathChapter13.html
Chapter 14 복소적분 http://matrix.skku.ac.kr/EMSage/EMathChapter14.html
Chapter 15 급수, 유수 http://matrix.skku.ac.kr/EMSage/EMathChapter15.html
Chapter 16 등각사상 http://matrix.skku.ac.kr/EMSage/EMathChapter16.html