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July 2013

Chapter 27

فيرجى العام للنفع مجانية عن المساهمةالنوتات خطأ باإلبالغ ضرورية مالحظات أوأي نصية تراها برسالةاإللكتروني أو 9 4444 260 بالبريد

Physics I/II, English 123, Statics, Dynamics, Strength, Structure I/II, C++, Java, Data, Algorithms, Numerical, Economyشعبان. حمادة محلولة info@ eng-hs.com 260 4444 9م ومسائل , eng-hs. com بالموقعين مجاناشرح

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Boundary-Value and Eigenvalue Problems

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Ordinary differential equation is accompanied by auxiliary conditions.

These conditions are used to evaluate the constants of integration that

result during the solution of the equation. For an nth order equation,

n conditions are required. If all the conditions are specified at the same

value of the independent variable, then we are dealing with an

initial-value problem (next figure a).

In contrast, there is another application for which the conditions are not

known at a single point, but rather, are known at different values of the

independent variable. Because these values are often specified at the

extreme points or boundaries of a system, they are customarily referred to

as boundary-value problems (previous figure b).



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بال صديق يبحثعن منبال يبقى عيب،

تركي) ( مثل صديق


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We will discuss two general approaches for obtaining a solution:

the shooting method and the finite-difference approach.

Additionally, we present techniques to approach a special type of

boundary-value problem: the determination of eigenvalues.

27.1 General Methods for Boundary-Value Problems

The conservation of heat can be used to develop a heat balance for a

long, thin rod. If the rod is not insulated along its length and the system is

at a steady state, the equation that result is

d2Td x2 +h' (T a−T )=0 (27.1)

where h’ is a heat transfer coefficient (m-2) that parameterizes the rate of

heat dissipation to the surrounding air and Ta is the temperature of the

surrounding air (⁰C).

To obtain a solution for the previous equation, there must be

appropriate boundary conditions. A simple case is where the temperatures

at the ends of the bar are held at fixed values. These can be expressed

mathematically as

T(0) = T1

T(L) = T2



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كل من أفضل تكون أن يمكنك: طريقين بأحد أصدقائك

الحمقى تصادق أن أو االجتهاد


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With these conditions, the previous equation can be solved analytically

using calculus. For a 10-m rod with Ta = 20, T1 = 40, T2 = 200 and h’ = 0.01,

the solution is

T = 73.4523 e0.1x – 53.4523 e-0.1x + 20

27.1.1 The Shooting Method

The shooting method is based on converting the boundary-value

problem into an equivalent initial-value problem. A trial-and-error

approach is then implemented to solve the initial-value version.

The approach can be illustrated by an example.

EXAMPLE 27.1 The Shooting Method

Problem Statement:

Use the shooting method to solve Eq.(27.1) for a 10-m rod with h’=0.01

m-2, Ta = 20, and the boundary conditions

T(0) = 40 T(10) = 200

Solution:

The second-order equation can be expressed as two first-order ODEs:dTdx

=z

dzdx

=h' (T−T a)



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دمـي مـن تقطـر الهنـد


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Continue:

To solve these equations, we require an initial value for z.

For the shooting method, we guess a value- say, z(0) = 10 and we can then

obtain the solution by integration. For example, using a fourth-order RK

method with a step size of 2, we obtain a value at the end of the interval

of T(10) = 168.3797 (next figure a), which differs from the boundary

condition of T(10) = 200. Therefore, we make another guess, z(0) = 20,

and perform the computation again. This time, the result of

T(10) = 285.8980 is obtained (next figure b).

Now, because the original ODE is linear, the values

z(0) = 10 T(10) = 168.3797

and

z(0) = 20 T(10) = 285.8980

are linearly related. As such, they can be used to compute the value of z(0)

that yields T(10) = 200. A linear interpolation formula can be employed for

this purpose:

z (0 )=10+ 20−10285.8980−168.3797 (200 – 168.3797) = 12.6907

This value can then be used to determine the correct solution, as shown in

the next figure (c).



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: عجبت إبليس كالم منالله يحب آدم، البن

ويبغضني ويعصاه،


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Nonlinear Two-Point Problems

For nonlinear boundary-value problems, linear interpolation or

extrapolation through two solution points will not necessarily result in an

accurate estimate of the required boundary condition to attain an exact

solution. An alternative is to perform three applications of the shooting

method and use a quadratic interpolating polynomial to estimate the

proper boundary condition.

However, it is unlikely that such approach would yield the exact

answer, and additional iterations would be necessary to obtain the

solution.



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الذي الناسالجرح يرى قدال لكنهم رأسك في

الذي باأللم يشعرون


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Another approach for a nonlinear problem involves recasting it as a

roots problem. Recall that the general form of a root problem is to find the

value of x that makes the function f(x) = 0. Now, let us use Example 27.1

to understand how the shooting method can be recast in this form.

First, recognize that the solution of the pair of differential equations is

also a “function” in the sense that we guess a condition at the left-hand

end of the bar, z0, and the integration yields a prediction of the

temperature at the right-hand end, T10.

Thus, we can think of the integration as

T10 = f(z0)

That is, it represents a process whereby a guess of z0 yields a

prediction of T10. Viewed in this way, we can see that what we desire is the

value of z0 that yields a specific value of T10. If, as in the example,

we desire T10 = 200, the problem can be posed as

200 = f(z0)

By bringing the goal of 200 over to the right-hand side of the

equation, we generate a new function, g(z0), that represents the difference

between what we have, f(z0), and what we want, 200.

g(z0) = f(z0) – 200



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if we drive this new function to zero, we will obtain the solution.

The next example illustrates the approach.

EXAMPLE 27.2 The Shooting Method for Nonlinear Problems

Problem Statement:

Suppose that the following nonlinear ODE is used to simulate the

temperature of the heated bar:

d2Td x2 +h' ' (T a−T )4=0

where h’’ = 5 × 10-8. The reaming problem conditions are as specified in

Example 27.1

Solution:

The second-order equation can be expressed as two first-order ODEs:dTdx

=z

dzdx

=h' ' (T−T a )4

Now, these equations can be integrated using any of the methods

described in the previous chapter. We used the constant step-size version

of the fourth-order RK and we implemented this approach as an Excel

macro function written in Visual BASIC. The function integrated the

equations based on an initial guess for z(0) and returned the temperature

at x = 10. The difference between this value and the goal of 200 was then

placed in a spreadsheet cell. The Excel Solver was then invoked to adjust

the value of z(0) until the difference was driven to zero.



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: نوع أنواع الناسثالثةونوع دائما، إليه تحتاج كالغذاء

أحيانا، إليه تحتاج كالدواءيضركفقط كالداء والثالث

أكثر الناسدائما امنحيتوقعون مما

. عليه الحصول


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The result is shown in the next figure along with the original linear case.

As might be expected, the nonlinear case is curved more than the linear

model. This is due to the power of four term in heat transfer relationship.

27.1.2 Finite-Difference Methods

The most common alternatives to the shooting method are

finite-difference approaches. In these techniques, finite divided

differences are substituted for the derivatives in the original equation.

Thus, a linear differential equation is transformed into a set of

simultaneous algebraic equations that can be solved using other methods.

The finite divided-difference approximation for the second derivative is

d2Td x2 =

T i+1−2T i+T i−1

∆ x2

This approximation can be substituted into Eq.(27.1) to giveT i+1−2T i+T i−1

∆ x2 −h' (T i−T a )=0



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Collecting terms gives

−T i−1+( 2+h'∆x2 )T i−T i+1=h'∆ x2T a

This equation applies for each of the interior nodes of the rod.

The first and last interior nodes, T i−1 and T i+1, respectively, are specified by

the boundary conditions. Therefore, the resulting set of linear algebraic

equations will be tridiagonal.

EXAMPLE 27.3 Finite Difference Approximation of Boundary-Value

Problems

Problem Statement:

Use the finite-difference approach to solve the same problem as in

Example 27.1

Solution:

Employing the parameters in Example 27.1, we can write the above

equation for the rod.

Using four interior nodes with a segment length of ∆ x=2m results in the

following equations:

[2.04−100

−12.04−10

0−12.04−1

00

−12.04 ]{T1

T2

T3

T 4}={ 40.8

0.80.8

200.8}



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مالبس أعرفأنوموبايلك وسيارتك

لكن عالية، قيمتهاأنت تكون أن حاول


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which can be solved for

{T }T=⌊65.9698 93.7785 124.5382159.4795 ⌋

The next table provides a comparison between the analytical solution and

the numerical solutions obtained in Examples 27.1 and 27.3

For both numerical methods, these errors can be omitted by decreasing

their respective step sizes. Although both techniques perform well for the

present case, the finite-difference approach is preferred because of the

ease with which it can be extended to more complex cases.

27.2 Eigenvalue Problems

Eigenvalue, or characteristic-value, problems are a special class of

boundary-value problems that are common in engineering problem

contexts involving vibrations, elasticity, and other oscillating systems.

27.2.1 Mathematical Background



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هو ما تصنع أن يمكنك التجلس أن من حماقة أكثر

حتى الطريق جانب علىويحاول أحدهم يأتيك


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We dealt with methods for solving sets of linear algebraic equations

of the general form

[A ] {X }={B }

If the equations comprising such a system are linearly independent

(that is, have a nonzero determinant), they will have a unique solution.

In contrast, a homogenous linear algebraic system has the general

form

[A]{X} = 0

Although nontrivial solutions (that is, solutions other than all x’s = 0)

of such systems are possible, they are generally not unique.

Rather, the simultaneous equations establish relationships among the x’s

that can be satisfied by various combinations of values.

Eigenvalue problems associated with engineering are typically of the

general form

(a11− λ ) x1+a12 x2+…+a1n xn=0

a21 x1+(a22−λ )x2+…+a2n xn=0

. .

. .

. .

an1 x1+an2 x2+…+(ann−λ )xn=0



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تـنـتـهـي ال صــداقــاتبــمـرآتـهـا)) .. الــمـرأة

بـكـتـابـة .. الـقـارئمـن بـحـذاء الـمـنـافـق


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where λ is an unknown parameter called the eigenvalue, or characteristic

value. A solution {X} for such a system is referred to as an eigenvector.

The above set of equations may also be expressed concisely as

[ [A ]−λ [1 ] ] {X }=0

The solution of the previous equation depends on determining λ.

One way to accomplish this is based on the fact that the determinant of

the matrix [ [A ]−λ [1 ] ] must be equal to zero for non-trivial solutions to be

possible. Expanding the determinant yields a polynomial in λ.

The roots of this polynomial are the solutions for the eigenvalues.

27.2.2 Physical Background

The mass-spring system in the next figure (a) is a simple context to

illustrate how eigenvalues occur in physical problem settings.



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حلتفي واحدة روحالصداقة هي تلك جسدين،

...... أرسطو عنها سألتني إن


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Assume that each mass has no external or damping forces acting on

it. In addition, assume that each spring has the same natural length l and

the same spring constant k.

Finally, assume that the displacement of each spring is measured

relative to its own local coordinate system (previous figure a).

Under these assumptions, Newton’s second law can be employed to

develop a force balance for each mass

m1d2 x1

d t2=−k x1+k (x2−x1)

and

m2d2 x2

d t2=−k (x¿¿2−x1)+k x2¿

where x i is the displacement of mass i away from its equilibrium position

(previous figure b).

These equations can be expressed as

m1d2 x1

d t2−k (x2−2 x1)=0



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غيرك عند يغتابك أن عندك


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m2d2 x2

d t2−k (x1−2 x2 )=0

From vibration theory, it is known that solutions to the previous equations

can take the form

x i=Ai sin (ωt)

where Ai = the amplitude of the vibration of mass i and ω = the frequency

of the vibration, which is equal to

ω=2πT p

where T p is the period. It follows that

x i' '=−A iω

2 sin (ωt )

The previous two equations can be transformed to

( 2km1

−ω2)A1−km1A2=0

−km2A1+( 2k

m2−ω2)A2=0

at this point, the solution has been reduced to an eigenvalue problem.



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المسائل مناقشة تستغرقألن طويال وقتا التافهةأكثر يعرفعنها بعضنا

المسائل يعرفعن مما


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EXAMPLE 27.4 Eigenvalues and Eigenvectors for a Mass-Spring System

Problem Statement:

Evaluate the eigenvalues and the eigenvectors of the previous equations

for the case where m1=m2=40 kg and k=200 N/m.

Solution:

Substituting the parameter values into the equations yields

(10−ω2 ) A1−5 A2=0

−5 A1+(10−ω2 ) A2=0

The determinant of this system is

(ω¿¿2)2−20ω2+75=0¿

which can be solved by the quadratic formula for ω2=15 and 5 s-2.

Therefore, the frequencies for the vibrations of the masses are

ω=3.873 s−1 and 2.236 s−1, respectively. These values can be used to determine

the periods for the vibrations. For the first mode, Tp = 1.62 s, and for the

second, Tp = 2.81 s.



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تعتذر أن الحكمة من....... كنتمخطئا إذا لرجل

حتى المرأة تعتذر وأنكنتعلىصواب ولو


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As stated before, a unique set of value cannot be obtained for the

unknowns. However, their ratios can be specified by substituting the

eigenvalues back into the equations. For example, for the first mode

(ω¿¿2=15 s−2) , A1=−A2 ¿. For the second mode (ω¿¿2=5 s−2) , A1=A2¿.

Continue:

This example provides valuable information regarding the behavior of

the system in the previous figure. Aside from its period, we know that if

the system is vibrating in the first mode, the amplitude of the second mass

will be equal but of opposite sign to the amplitude of the first.

As in the next figure (a), the masses vibrate apart and then together

indefinitely.

In the second mode, the two masses have equal amplitudes at all

times. Thus, as in the figure (b), they vibrate back and forth in unison.

It should be noted that the configuration of the amplitudes provides

guidance on how to set their initial values to attain pure motion in either

of the two modes. Any other configurations will lead to superposition of

the modes.



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Problem 27.4

Use the shooting method to solve

7 d2 yd x2 −2 dy

dx− y+ x=0

with the boundary conditions y(0) = 5 and y(20) = 8

Solution

The second-order ODE can be expressed as the following pair of first-order

ODEs,dydx

=z

dzdx

=2 z+ y−x7



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These can be solved for two guesses for the initial condition of z.

For our cases we used –1 and 0.5. We solved the ODEs with the Heun

method without iteration using a step size of 0.125. The results are

z(0) 1 0.5

y(20) 11,837.64486 22,712.34615

Clearly, the solution is quite sensitive to the initial conditions.

These values can then be used to derive the correct initial condition,

z (0 )=−1+ −0.5+122712.34615−(−11837.64486 )

(8−(−11837.64486 ) )

¿−0.82857239

Continue:

The resulting fit is displayed below:

x y0 5

2

4.151

601

4

4.461

229

6

5.456

0478 6.852



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العالم طبيبفي أفضلال فهو البيطري هو

مريضه يسأل أن يستطيعيكتشف بل عنشكواه


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2431

0

8.471

4741

2

10.17

8131

4

11.80

2771

6

12.97

9421

8

12.69

8962

0 8



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.. بعد نرها لم األنهار أجملنقرأها لم الكتب أجمل

لم.. حياتنا أيام أجمل بعد ........ حكمت ناظم بعد تأت


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Problem 27.3

Use the finite-difference approach with ∆ x=1 to solve

d2Td x2 −0.15T=0

Solution

A centered finite difference can be substituted for the second derivative to

give,T i+1−2T +T i−1

h2 −0.15T i=0

or for h = 1,−T i−1+2.15T i– T i+1=0

The first node would be2.15T1 –T 2=240

and the last node would be−T 9+2.15T 10=150

The tridiagonal system can be solved with the Thomas algorithm or

Gauss-Seidel for (the analytical solution is also included)



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الذي المشط هي الخبرةلإلنسان الطبيعة تقدمه

رأسه يتساقطشعر عندما


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Continue:

x T Analytical0 240 2401 165.7573 165.32902 116.3782 115.76893 84.4558 83.79244 65.2018 64.54255 55.7281 55.09576 54.6136 54.01717 61.6911 61.14288 78.0223 77.55529 106.0569 105.746910 150 150



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لنفسكمن انتقمت إذاإليكساويت المسيء

.. صفحت وإذا به نفسك( . فرنسيس استعبدته عنه


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