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 16:30-17:45/ 목 15:00-16:15), 반도체관 400102호 담당교수: 김응기 박사

 주차 주교재 부교재 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 web

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) ,        String is fastened at the left end is .     (b) ,        String is fastened at the right end is .

 Two Initial Conditions (3)  , initial deflection (deflection at time )       initial velocity (velocity at time )  where .

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)                      ,        for all .

 Initial Conditions (3)                            [ given]

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

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 is the given initial temperature of the bar.

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

*

[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