Week 42018학년도 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 16:30-17:45/ 목 15:00-16:15), 반도체관 400102호 담당교수: 김응기 박사
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주차 |
주교재 |
부교재 |
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4 |
(briefly) 12.3: Solution by Separating Variables. Use of Fourier Series 12.6: Heat Equation: Solution by Fourier Series. Steady Two-Dimensional Heat Problems. Dirichlet Problem 12.7: Heat Equation: Solution by Fourier Integrals and Transforms |
12.3 파동방정식 12.4 확산방정식 |
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web |
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Week 4
12.3. Solution by Separating Variables. Use of Fourier Series
Model of a vibrating elastic string
The model of a vibrating elastic string (a violin string) consists of the One-dimensional wave equation
(1) 
, 
for the known deflection 
of the string, a PDE that we have just obtained, and some additional conditions, which we shall now derive.
|
Two Boundary Conditions |
|
|
(2) (a) (b) |
|
|
Two Initial Conditions |
|
|
(3) |
|
To find a solution of the PDE 
satisfying the condition 
and 
, we shall do three steps, as follows.
Step 1. “Method of separating variables” or product method,
Setting 
,
we obtain from 
ODEs, one for 
and the other one for 
.
Step 2. Determine solutions of these ODEs that satisfy the boundary condition 
.
Step 3. Using Fourier series, for a solution
we compose the solutions gained in Step 2 to obtain a solution of 
satisfying both 
and 
, that is, the solution of our model of the vibrating string.
Step 1. Two ODEs from the Wave Equation (1)
In the method of separating variables or product method,
we determine solutions of the wave equation 
of the form
(4) 
which are a product of two functions, each depending only on one of variables 
and 
.
Differentiating 
(dots denote derivatives with respect to 
)
(primes denote derivatives with respect to 
)
By inserting this into the wave equation
.
Dividing by 
.
Multiplying by the denominators gives immediately two ODEs
The left side depending only on 
.
(5) 
.
The right side depending only on 
.
(6) 
.
Here, the separation constant 
is still arbitrary.
Step 2. Satisfying the Boundary Condition(2)
Determine solutions 
and 
of 
and 
so that 
satisfies the boundary conditions 
,
(7) 
, 
for all 
.
Find solution 
and 
of 
and 
We first solve 
.
If 
, then 
, which is of no interest.
Hence 
and then by 
,
(8) (a) 
, (b) 
.
Case 1 For positive 
.
A general solution of 
is
.
imply 
imply 
imply 
(No interest).
Cases 2 For 
.
A general solution of 
is
.
imply 
(No interest).
Cases 3 For negative 
.
A general solution of 
is
.
(No interest).
since otherwise 
. Hence 
. Thus
(9) 
, so that 
.
Setting 
, we thus obtain infinitely many solutions 
, where
(10) 
.
These solution satisfy 
.
We now solve 
with 
resulting from 
, that is,
(11*) 
where 
.
A general solution is
.
(11) 
where 
.
Function 
are the eigenfunctions or characteristic functions.
Values 
are the eigenvalues, or characteristic values, of the vibrating string.
Set 
is the spectrum.
Discussion of Eigenfunctions.
The motion 
is the 
normal mode of the string.
Each 
represents a harmonic motion having the frequency 
cycles per unit time.
First normal mode is the fundamental mode 
.
Since in 
at 
,
the 
normal mode has 
nodes, that is, points of the string that do not move.
Fig. 287. Normal modes of the vibrating string
Fig, 288. Second normal mode for various values of 
Fig 
shows the second normal mode for variables of 
.
At any instant the string has the form of a sine wave.
When the left part of the string is moving down, the other half is moving up, and conversely.
For the other modes the situation is similar.
Tuning is done by changing the tension 
. Our formula for the frequency 
of 
with 
confirms that effect because it shows that the frequency is proportional to the tension. 
cannot be increased indefinitely, but can you see what to do to get a string with a high fundamental mode? (Think of both 
and 
.)
Why is a violin smaller than a double-bass?
Step 3. Solution of the Entire Problem. Fourier Series
The eigenfunctions 
satisfy the wave equation 
and the boundary conditions 
. A single 
will generally not satisfy the intial condition 
. But since the equation 
is linear and homogeneous, it follows from Fundamental Theorem 
in Sec.
that the sum of finitely many solution 
is a solutions 
.
We consider the infinite series(with 
as before)
(12) 
.
Satisfying Initial Condition 
(Given Initial Displacement).
From 
and 
we obtain
(13) 
(
)
Hence we must choose the 
’s so that 
becomes the Fourier sine series of 
. Thus, by 
in Sec.
,
(14) 
.
.
Satisfying Initial Condition 
(Given Initial Velocity).
Similarly, by differentiating 
with respect to 
and using 
, we obtain
.
Hence we must choose the 
’s so that for 
the derivative 
becomes the Fourier sine series of 
. Thus, again by 
in Sec.
,
.
Since 
, we obtain by division
(15) 
.
Solution
Established
According to our derivation to the solution 
is at first a purely formal expression, but we shall now establish it.
The initial velocity 
.
Then 
, and 
reduces to
(16) 
, 
.
.
, 
(17) 
where 
where 
is the odd periodic extension of 
with the period 
(Fig.
).
Odd periodic extension of 
Since the initial deflection 
is continuous on the interval 
and zero at the endpoints, it follows from 
that 
is a continuous function of both variables 
and 
for all values of the variables. By differentiating 
we see that 
is a solution of 
, provided 
is twice differentiable on the interval 
, and has one-sided second derivatives at 
and 
, which are zero. Under the conditions 
is established as a solution of 
, satisfying 
and 
with 
.
Generalized Solution.
If 
and 
are merely piecewise continuous (Sec. 
), or if those one-sided derivatives are not zero, then for each 
there will be finitely many value of 
at which the second derivatives of 
appearing in 
do not exist. Except at these points the wave equation will still be satisfied.
We may then regard 
as a “Generalized Solution,” as it is called, that is, as a solution in a broader sense. For instance, a triangular initial deflection as in Example 
(below) leads to a generalized solution.
Physical Interpretation of the Solution 
The graph of 
is obtained from the graph of 
by shifting the latter 
units to the right (Fig.
).
This means that 
represents a wave that is traveling to the right as 
increases. Similarly, 
represents a wave that is traveling to the left, and 
is the superposition of these two waves.
Interpretation of (17)
Example 1 Vibrating String if the Initial Deflection Is Triangular
Find the solution of the wave equation 
, 
corresponding to the triangular initial deflection
and initial velocity zero 
.(Fig 291 shows 
at the top.)
Solution
Since 
in 
From Example 
in Sec. 11.3 
.
For graphing the solution we may use 
and the above interpretation of two function in the representation 
.
This leads to graph shown in Fig. 291.
Solution 
in Example 1 for various values of t (right part of the figure) obtained as the superposition of a wave traveling to the right (dashed) and a wave traveling to the left (left part of the figure)
12.6 Heat Equation: Solution by Fourier Series
The heat equation
, 
gives the temperature 
in a body of homogeneous material.
Here 
is the thermal diffusivity,
is the thermal conductivity,
the specific heat,
the density of the material of the body.
is the Laplacian of 
and with respect to Cartesian coordinates 
,
diffusion equation.
Then 
depends only on 
and time 
, and the heat equation becomes the one-dimensional heat equation
(1) 
.
|
Boundary Conditions |
|
|
(2) |
|
|
Initial Conditions |
|
|
(3) |
|
Here we must have 
and 
because of 
.
Step I. Two ODEs from the heat equation 
.
Differentiating 
,
and 
where 
and 
.
.
Dividing by 
gives
(4) 
The variables are now separated.
The right side depending only on 
.
(5) 
.
The left side depending only on 
.
(6) 
.
Here, the separation constant 
is still arbitrary.
Step 2. Satisfying the boundary conditions 
.
We determine solutions 
and 
of 
and 
so that 
satisfies the boundary conditions 
, that is,
(*) 
, 
for all 
.
We first solve 
.
If 
, then 
, which is of no interest.
Hence 
and then by (*),
(**) 
, 
.
Case 1 For positive 
.
A general solution of 
is
.
, 
(No interest).
Cases 2 For 
.
A general solution of 
is
.
(No interest).
Cases 3 For negative 
.
A general solution of 
is
.
(No interest).
since otherwise 
. Hence 
. Thus
, hence 
.
Setting 
, a solution of 
is
.
We now solve 
with 
, that is,
where 
.
A general solution is
, 
.
where 
is a constant. Hence the functions
(8) 
are solutions of the heat equation 
, satisfying 
.
These are the eigenfunctions of the problem, corresponding to the eigenvalues 
.
Step 3. Solution of the entire problem. Fourier series.
A series of these eigenfunctions,
(9) 
.
From this and 
we have
Case 
: initial condition
.
is coefficients of the Fourier series.
(10) 
Example 1 Sinusoidal Initial Temperature
Find the temperature 
in a laterally insulated copper bar 
long if the initial temperature is 
and the ends are kept at 
. How long will it take for the maximum temperature in the bar to drop to 
? First guess, then calculate. Physical data for copper; density 
, specific heat 
, thermal conductivity 
.
Solution
The initial condition is
.
Case 
.
Hence we obtain
The solution of 
is
. 
.
Example 2 Speed of Decay
Find the temperature 
in a laterally insulated copper bar 
long if the initial temperature is 
and the ends are kept at 
. How long will it take for the maximum temperature in the bar to drop to 
? First guess, then calculate. Physical data for copper; density 
, specific heat 
, thermal conductivity 
.
Solution
The initial condition is
.
Case 
.
Hence we obtain
.
The solution of 
is
.
.
Hence the maximum temperature drops to 
in 
, which is much faster.
Had we chosen a bigger 
, the decay would have been still faster,
and in a sum or series of such terms, each term has its own rate of decay,
and terms with large 
are practically 
after a very short time.
The curve corresponding to 
looks almost like a sine curve; that is, it is practically the graph of the first term of the solution.
Decrease of temperature with time 
for 
and 
Example 3 Triangular Initial Temperature in a Bar
Find the temperature in a laterally insulated bar of the length 
whose ends are kept at temperature 
, assuming that and initial temperature is
.
Solution
From 
we get
Integration gives 
if 
is even
and 
The solution is
The temperature decreases with increasing 
, because of the heat loss due to the cooling of the ends.
Example 4 Bar with Insulated Ends. Eigenvalue 
Find a solution formula of 
, 
with 
replaced by the condition that both ends of the bar are insulated.
Solution
The rate of heat flow is proportional to the gradient of the temperature.
Hence if the ends 
and 
of the bar are insulated, so that no heat can flow thought the ends, we have 
and the boundary condition
(2*) 
, 
for all 
.
Since 
The solution is 
Differentiating of 
is
.
Case 
, 
, 
, 
.
From this and 
with 
and 
we get 
, 
.
With 
, 
The eigenfunctions
(11) 
corresponding to the eigenvalues 
.
We now have the additional eigenvalue 
and eigenfunction 
, which is the solution of the problem if the limit temperature 
is constant.
A separation constant can very well be zero, and zero can be an eigenvalue.
Furthermore, whereas 
gave a Fourier sine series,
, 
we now get from 
a Fourier cosine series
(12) 
Its coefficients result follow the initial condition 
.
in the from 
, Sec. 
, that is,
(13) 
Example 5 Triangular Initial Temperature in a Bar which Insulated Ends
Find the temperature in the bar in Example 
, assuming that the ends are insulated.
Solution
For the triangular initial temperature, 
gives 
and
The solution 
is
.
, this is the mean value of the initial temperature.
This is plausible no heat can escape from this totally insulated bar.
In constant, the cooling of the ends in Example 
led to heat and 
, the temperature at which the ends were kept.
Steady Two-Dimensional Heat Problems. Laplace’s Equation
The two-dimensional heat equation
for steady problems.
Then 
and the heat equation reduces to Laplace’s Equation
(14) 
A heat problem then consists of this PDE to be considered in some region 
of the 
plane and a given boundary condition on the boundary curve 
of 
.
This is a boundary value problem (BVP).
First BVP or Dirichlet Problem if 
is prescribed on 
(“Dirichlet boundary condition”)
Second BVP or Neumann Problem if the normal derivative 
is prescribed on 
(“Neumann boundary condition”)
Third BVP, Mixed BVP or Robin Problem if 
is prescribed on a portion of 
(“Robin boundary condition”).
Dirichlet Problem in a Rectangle 
.
We consider a Dirichlet Problem for Laplace’s Equation 
in a rectangle 
, assuming that the temperature 
equals a given function 
on the upper side and 
on the other three sides of the rectangle.
Substituting 
into 
written as
dividing by 
,
and equating both sides to a negative constant, we obtain
.
.
,
and the left and the right boundary condition imply
and 
.
Case 1 For negative 
.
A general solution of the second-order ODE is
.
imply 
imply 
(No interest).
Cases 2 For 
.
A general solution of 
is
.
imply 
(No interest).
Cases 3 For positive 
.
A general solution of the second-order ODE is
.
imply 
(No interest).
since otherwise 
. Hence 
. Thus
, hence 
.
Setting 
, a solution of the second-order ODE is
(15) 
, 
.
,
The ODE for 
with 
then becomes
.
Solution are
.
Now the boundary condition 
on the lower side of 
implies that 
; that is,
.
This gives
.
We obtain as the eigenfunctions of our problem
(16) 
.
These solution satisfy the boundary condition 
on the left, right, and lower sides.
To get a solution also satisfy the boundary condition on the upper side, we consider the infinite series
.
From this and 
with 
we obtain
.
We can write this in the form
.
The Fourier coefficients 
of 
.
From this and 
we see that the solution of our problem is
where
.
The series for 
, 
, and 
have the right sums. 
and 
are continuous and 
is piecewise continuous on the interval 
.
12.7 Heat Equation:
Solution by Fourier Integrals and Transforms
Model of bars of infinite length
Heat equation
(1) 
|
Initial Conditions |
|
|
(2) where |
|
Substituting 
into 
.
The two ODEs
(3) 
and
(4) 
The solutions of 
is
where 
and 
are any constants.
The solutions of 
is
.
Hence a solution of 
is
.
Use of Fourier Integrals
Heat equation is linear and homogeneous
(6) 
is then a solution of 
.
Determination of 
and 
from the Initial Condition
(7) 
where 
and 
.
Using the formula 
We choose 
as a new variable of integration and set 
.
Then 
and 
, so that 
becomes
.
where 
If 
is bounded for all values of 
and integrable in every finite interval.
Example 1 Temperature in an Infinite Bar
Find the temperature in the infinite bar if the initial temperature is
Solution
, 
Example 2 Temperature in the Infinite Bar in Example 1
Solve Example 1 using the Fourier transform.
Solution
: Fourier transform of 
, regarded as a function of 
Heat equation : 
Interchange the order of differentiation and integration
General solution : 
Initial condition 
Example 3 Solution in Example 
by the Method of Convolution
Solve the heat problem in Example 
by the method of convolution.
Solution
(18) 
*
with a suitable 
. With 
or 
, using 
we obtain
Hence 
has the inverse
.
Replacing 
with 
and substituting this into 
we finally have
*
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
, 
String is fastened at the left end is 
.
, 
String is fastened at the right end is 
.
, initial deflection (deflection at time 
) 
initial velocity (velocity at time 
) 
where 
.
, 
for all 
.
[
given]
is the given initial temperature of the bar.