SOLUTIONS MANUAL for Advanced Engineering Mathematics with Mathematica 1st Edition by Edward Magrab

SOLUTION MANUAL

SOLUTION MANUAL

Chapter 1 Matrices, Determinants, and Systems of Equations Chapter 2 Introduction to Complex Variables Chapter 3 Fourier Series and Fourier Transforms Chapter 4 Ordinary Differential Equations Part I: Review of First- and Second-Order Equations Chapter 5 Ordinary Differential Equations Part II: Power Series Solutions Chapter 6 Ordinary Differential Equations Part III: Sturm–Liouville Equation Chapter 7 Partial Differential Equations Chapter 8 Laplace Transforms Chapter 9 Putting It All Together

Solutions to Exercises in Chapter 1 Section 1.2 1.1 A matrix is an orthogonal matrix if

XTX = I Is the following matrix an orthogonal matrix?

⎛ −1 −1 ⎞ 1 ⎜ 1 −1 ⎟ ⎟ X= ⎜ 2 ⎜ −1 1 ⎟ ⎜⎝ 1 1 ⎟⎠ Solution: x={{-1.,-1},{1,-1},{-1,1},{1,1}}/2; Transpose[x].x//MatrixForm yields

⎛ 1 0 ⎞ ⎜ ⎟ ⎝ 0 1 ⎠ Therefore, X is an orthogonal matrix.

1.2 If

⎛ 1 −1 ⎞ A=⎜ ⎟ ⎝ 2 −1 ⎠

⎛ 1 1 ⎞ B=⎜ ⎟ ⎝ 4 −1 ⎠

does (A + B)2 = A 2 + B 2?

Solution: a={{1,-1},{2,-1}}; b={{1,1},{4,-1}}; ((a+b).(a+b)-a.a-b.b)//MatrixForm yields

⎛ 0 0 ⎞ ⎜ ⎟ ⎝ 0 0 ⎠ Therefore, the expressions are equal.

2

1.3 Given the two matrices ⎛ 1 4 −3 ⎞ A=⎜ ⎝ 2 5 4 ⎟⎠

and

⎛ 4 1 ⎞ B=⎜ 2 6 ⎟ ⎜ ⎟ ⎜⎝ 0 3 ⎟⎠

Find the matrix products AB and BA. Solution:

⎛ 4 1 ⎞ ⎛ 1 4 −3 ⎞ ⎜ ⎛ 12 16 ⎞ AB = ⎜ 2 6 ⎟ =⎜ ⎟ ⎝ 2 5 4 ⎠ ⎜⎜ 0 3 ⎟⎟ ⎝ 18 44 ⎟⎠ ⎝ ⎠ ⎛ 4 1 ⎞ ⎛ 6 21 −8 ⎞ ⎛ 1 4 −3 ⎞ ⎜ ⎜ ⎟ BA = 2 6 ⎜ = 14 38 18 ⎟ ⎜ ⎟ ⎝ 2 5 4 ⎟⎠ ⎜ ⎟ ⎜⎝ 0 3 ⎟⎠ ⎜⎝ 6 15 12 ⎟⎠ Aa={{1,4,-3},{2,5,4}}; Bb={{4,1},{2,6},{0,3}}; Aa.Bb//MatrixForm Bb.Aa//MatrixForm 1.4 Given the following matrices and their respective orders: A (n´m), B (p´m), and C (n´s).

Show one way in which these three matrices can be multiplied. What is the order of the resulting matrix?

Solution:

C T ABT → (n × s)T (n × m)( p × m)T → (s × n)(n × m)(m × p) → (s × p) 1.5 Given

⎛ ab A=⎜ ⎝ −a 2 Determine A2.

Solution: From Eq. (1.13)

3

b2 ⎞ ⎟ −ab ⎠

⎛ a11 a12 ⎞ ⎛ a11 a12 ⎞ ⎛ a112 + a12 a21 a12 ( a11 + a22 ) ⎞ ⎟ AA = ⎜ ⎟⎜ ⎟ =⎜ 2 a a a a ⎜ a a + a a a + a ⎟⎠ ( ) ⎝ 21 22 ⎠ ⎝ 21 22 ⎠ ⎝ 21 11 22 21 12 22 ⎛ a 2b 2 − a 2b 2 b 2 ( ab − ab ) ⎞ ⎟ =0 =⎜ ⎜⎝ −a 2 ( ab − ab ) −a 2b 2 + a 2b 2 ⎟⎠ Aa={{a b, b^2},{-a^2,-a b}}; Aa.Aa//MatrixForm 1.6 Given the matrix ⎛ −4 −3 −1 ⎞ A=⎜ 2 1 1 ⎟ ⎜ ⎟ ⎝ 4 −2 4 ⎠

Determine the value of 4I - 4A - A2 + A3.

Solution:

⎛ 6 11 −3 ⎞ ⎜ A = −2 −7 3 ⎟ ⎜ ⎟ ⎜⎝ −4 −22 10 ⎟⎠ 2

⎛ −14 −1 −7 ⎞ A3 = ⎜ 6 −7 7 ⎟ ⎜ ⎟ ⎜⎝ 12 −30 22 ⎟⎠ Then,

⎛ 1 0 0 ⎞ ⎛ −4 −3 −1 ⎞ ⎜ ⎟ 4I − 4 A − A + A = 4 0 1 0 − 4 ⎜ 2 1 1 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎝ 0 0 1 ⎟⎠ ⎝ 4 −2 4 ⎠ 2

3

⎛ 6 11 −3 ⎞ ⎛ −14 −1 −7 ⎞ − ⎜ −2 −7 3 ⎟ + ⎜ 6 −7 7 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎝ −4 −22 10 ⎟⎠ ⎜⎝ 12 −30 22 ⎟⎠ ⎛ 0 0 0 ⎞ =⎜ 0 0 0 ⎟ ⎜ ⎟ ⎜⎝ 0 0 0 ⎟⎠ Mathematica verification Aa={{-4,-3,-1},{2,1,1},{4,-2,4}};

4

A2=Aa.Aa; MatrixForm[A2] A3=A2.Aa; MatrixForm[A3] (4 IdentityMatrix[3]-4 Aa-A2+A3)//MatrixForm Section 1.3 1.7 Given the following matrices:

⎧ 1 ⎫ x=⎨ ⎬, 2 ⎩ ⎭

⎛ 2 a ⎞ A=⎜ , ⎝ 3 4 ⎟⎠

⎛ 6 4 ⎞ B=⎜ ⎝ 7 5 ⎟⎠

What is the value of a that satisfies the following equation?

x T Ax = det B Solution:

x T Ax =

{ 1 2 }⎛⎜⎝ 23 a4 ⎞⎟⎠ ⎧⎨⎩ 12 ⎫⎬⎭ = { 1 2 } ⎧⎨⎩ 2 +112a ⎫⎬⎭

= 24 + 2a ⎛ 6 4 ⎞ detB = det ⎜ = 6×5−4×7= 2 ⎝ 7 5 ⎟⎠ Therefore,

24 + 2a = 2 a = −11 Mathematica verification Solve[{1,2}.{{2,a},{3,4}}.{{1},{2}}==Det[{{6,4},{7,5}}],a]

1.8 Show that ⎛ a b+c 1 ⎞ det ⎜ b a + c 1 ⎟ = 0 ⎜ ⎟ ⎝ c a+b 1 ⎠

Solution:

⎛ a b+c 1 ⎞ a+c 1 b 1 b a+c det ⎜ b a + c 1 ⎟ = a − (b + c) + ⎜ ⎟ a+b 1 c 1 c a+b ⎝ c a+b 1 ⎠ = a(c − b) − (b + c)(b − c) + b(a + b) − c(a + c) =0 5

Mathematica verification Det[{{a,b+c,1},{b,a+c,1},{c,a+b,1}}] 1.9 Expand the following determinants and reduce them to their simplest terms. a) ⎛ 1+ a a a ⎞ ⎜ det b 1+ b b ⎟ ⎜ ⎟ b 1+ b ⎠ ⎝ b

Solution:

⎛ 1+ a a a ⎞ 2 2 2 2 det ⎜ b 1+ b b ⎟ = (1+ a ) ⎡⎣(1+ b) − b ⎤⎦ − a ⎡⎣b(1+ b) − b ⎤⎦ + a ⎡⎣b − b(1+ b) ⎤⎦ ⎜ ⎟ b 1+ b ⎠ ⎝ b = (1+ a )[1+ 2b ] − ab − ab = 1+ a + 2b + 2ab − 2ab = 1+ a + 2b Mathematica verification Det[{{1+a,a,a},{b,1+b,b},{b,b,1+b}}]

b) ⎛ x3 + 1 1 1 ⎞ ⎜ ⎟ det ⎜ 1 x3 + 1 1 ⎟ ⎜⎝ 1 1 x 3 + 1 ⎟⎠

Solution: ⎛ x3 + 1 1 1 ⎞ 2 ⎜ ⎟ 3 3 3 3 det ⎜ 1 x3 + 1 1 ⎟ = ( x + 1) ⎡⎣( x + 1) − 1⎤⎦ − ⎡⎣ x + 1− 1⎤⎦ + ⎡⎣1− x − 1⎤⎦ ⎜⎝ 1 1 x 3 + 1 ⎟⎠ 2 = ( x 3 + 1) ⎡( x 3 + 1) − 1⎤ − 2x 3 ⎣ ⎦

= ( x 3 + 1) ⎡⎣ x 6 + 2x 3 ⎤⎦ − 2x 3

= x 3 ( x 6 + 2x 3 ) + x 6 + 2x 3 − 2x 3 = x 9 + 2x 6 + x 6 = x 6 ( x 3 + 3)

Mathematica verification

6

Det[{{x^3+1,1,1},{1,1+x^3,1},{1,1,1+x^3}}]

1.10 Determine if the following determinant a function of a ⎛ ex det ⎜ e x ⎜ ⎜⎝ 1

sin x cos x ⎞ ⎟ cos x sin x ⎟ 1− a a ⎟⎠

a function of a?

Solution: ⎛ ex det ⎜ e x ⎜ ⎜⎝ 1

sin x cos x ⎞ ⎛ x ⎛ cos x sin x ⎞ x ⎟ e − det ⎜ e cos x sin x ⎟ = det ⎜ ⎟ a ⎠ ⎝ 1− a ⎝ 1 1− a a ⎟⎠ ⎛ x + det ⎜ e ⎝ 1

sin x ⎞ sin x ⎟ a ⎠

cos x ⎞ cos x ⎟ 1− a ⎠

= e x ( a cos x − (1− a)sin x ) − sin x ( ae x − sin x ) + cos x ( e x (1− a) − cos x )

= e x ( a cos x − sin x + asin x − asin x + cos x − a cos x ) + sin 2 x − cos 2 x = e x ( cos x − sin x ) + sin 2 x − cos 2 x

which is not a function of a.

Mathematica verification Det[{{Exp[x],Sin[x],Cos[x]},{Exp[x],Cos[x],Sin[x]},{1,1-a,a}}] 1.11 Show that

⎛ 2 ⎜⎜ x1 ⎜ det ⎜⎜⎜ x22 ⎜⎜ ⎜⎜ x32 ⎝

⎞ x1 1 ⎟⎟ ⎟⎟ x2 1 ⎟⎟⎟ = ( x1 − x2 )( x1 − x3 )( x2 − x3 ) ⎟⎟ x3 1 ⎟⎟⎟⎠

Solution:

7

⎛ 2 ⎜⎜ x1 ⎜ det ⎜⎜⎜ x22 ⎜⎜ ⎜⎜ x32 ⎝

⎞ x1 1 ⎟⎟ ⎟⎟ 2 2 2 2 2 x2 1 ⎟⎟⎟ = x1 ( x2 − x3 ) − x1 ( x2 − x3 ) + ( x2 x3 − x3 x2 ) ⎟⎟ x3 1 ⎟⎟⎟⎠ = x12 ( x2 − x3 ) − x1 ( x2 − x3 )( x2 + x3 ) + x2 x3 ( x2 − x3 ) = ( x2 − x3 ) ⎡⎢ x12 − x1 ( x2 + x3 ) + x2 x3 ⎤⎥ ⎣ ⎦ = ( x2 − x3 ) ⎡⎢ x1 ( x1 − x2 ) − x3 ( x1 − x2 )⎤⎥ ⎣ ⎦ = ( x1 − x2 )( x1 − x3 )( x2 − x3 )

Mathematica verification Simplify[Det[{{x1^2,x1,1},{x2^2,x2,1},{x3^2,x3,1}}]]

Section 1.4 1.12 Given

⎛ 17 7 ⎞ A=⎜ ⎟ ⎝ 19 9 ⎠ Determine A-1 and verify your result.

Solution: From Eq. (1.27)

⎛ a 11 ⎜ ⎜⎝ a21

−1

⎛ a a12 ⎞ −a12 ⎞ 1 22 ⎟ = ⎜ ⎟ a11a22 − a12 a21 ⎜⎝ −a21 a11 ⎟⎠ a22 ⎟⎠ −1

⎛ 17 7 ⎞ ⎛ 9 −7 ⎞ 1 ⎜ 19 9 ⎟ = 17 × 9 − 19 × 7 ⎜ −19 17 ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ 9 / 20 −7 / 20 ⎞ =⎜ ⎟ ⎝ −19 / 20 17 / 20 ⎠ Since, from Eq. (1.13)

⎛ a11 a12 ⎞ ⎛ b11 b12 ⎞ ⎛ a11b11 + a12b21 a11b12 + a12b22 ⎞ ⎜ ⎟⎜ ⎟ =⎜ ⎟ ⎝ a21 a22 ⎠ ⎝ b21 b22 ⎠ ⎝ a21b11 + a22b21 a21b12 + a22b22 ⎠ then

8

⎛ 17 7 ⎞ ⎛ 9 / 20 −7 / 20 ⎞ ⎛ 17 × 9 / 20 − 7 × 19 / 20 −17 × 7 / 20 + 7 × 17 / 20 ⎞ ⎜ 19 9 ⎟ ⎜ −19 / 20 17 / 20 ⎟ = ⎜ 19 × 9 / 20 − 9 × 19 / 20 −19 × 7 / 20 + 9 × 17 / 20 ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎛ 1 0 ⎞ =⎜ ⎟ ⎝ 0 1 ⎠ Mathematica verification Aa={{17,7},{19,9}}; Inverse[Aa]//MatrixForm Inverse[Aa].Aa//MatrixForm

Section 1.5 1.13 Given the two matrices ⎛ 3 1 4 ⎞ A=⎜ 2 1 2 ⎟ ⎜ ⎟ ⎜⎝ 4 2 3 ⎟⎠

⎛ 2 1 3 ⎞ B=⎜ 1 2 5 ⎟ ⎜ ⎟ ⎜⎝ 0 2 1 ⎟⎠

and

Show that (AB)T = BTAT. Solution: T

⎛⎛ 3 1 4 ⎞ ⎛ 2 1 3 ⎞⎞ ⎛ 7 13 18 ⎞ ⎜ ⎟ T ⎜ ⎟ ⎜ ⎟ ( AB) = ⎜ 2 1 2 1 2 5 ⎟ = ⎜ 5 8 13 ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜⎝ 10 14 25 ⎟⎠ ⎜⎝ ⎜⎝ 4 2 3 ⎟⎠ ⎜⎝ 0 2 1 ⎟⎠ ⎟⎠

T

⎛ 7 5 10 ⎞ = ⎜ 13 8 14 ⎟ ⎜ ⎟ ⎜⎝ 18 13 25 ⎟⎠ and T

T

⎛ 2 1 3 ⎞ ⎛ 3 1 4 ⎞ ⎛ 2 1 0 ⎞⎛ 3 2 4 ⎞ T T ⎜ ⎟ ⎜ ⎟ B A = 1 2 5 2 1 2 =⎜ 1 2 2 ⎟⎜ 1 1 2 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜⎝ 0 2 1 ⎟⎠ ⎜⎝ 4 2 3 ⎟⎠ ⎜⎝ 3 5 1 ⎟⎠ ⎜⎝ 4 2 3 ⎟⎠ ⎛ 7 5 10 ⎞ = ⎜ 13 8 14 ⎟ ⎜ ⎟ ⎜⎝ 18 13 25 ⎟⎠ Mathematica verification

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Aa={{3,1,4},{2,1,2},{4,2,3}}; Bb={{2,1,3},{1,2,5},{0,2,1}}; Aa.Bb//MatrixForm Transpose[Aa.Bb]//MatrixForm Transpose[Bb]//MatrixForm Transpose[Aa]//MatrixForm Transpose[Bb].Transpose[Aa]//MatrixForm Section 1.7 1.14 Does the following system of equations have a solution?

100 y1 + 420 y2 + 486 y3 = 17 400 y1 +1050 y2 + 972 y3 = 18 700 y1 +1680 y2 +1458y3 = −3

Solution: It is seen from Eq. (1.30) that a = 300, b = 630, c = 486, and k = 2. Therefore, the determinant equals zero.

Mathematica verification Det[{{100,420,486},{400,1050,972},{700,1680,1458}}] 1.15 Without solving, determine whether the following system of equations has a solution. a1 + 2a2 + 3a3 = 1 4a1 + 5a2 + 6a3 = 0 7a1 + 8a2 + 9a3 = −7

Solution: It is seen from Eq. (1.28) that c = 3 and d = 6. Therefore, the determinant equals zero.

Mathematica verification Det[{{1,2,3},{4,5,6},{7,8,9}}] 1.16 Given the following system of equations

é7 2ù ì x1 ü é1 0ù ì x1 ü ì0ü l í ý ê5 1 ú x ê0 1 ú í x ý = í0ý ë ûî 2þ ë ûî 2þ î þ When l = 4, what are the values of x1 and x2? Justify your answer.

Solution: When

10

2 ù ì x1 ü ì0ü é3 2 ù ì x1 ü ì0ü é7 - l é3 2 ù í ý=í ý®ê í ý = í ý ® det ê ê 5 ú ú ú = -19 1 - l û î x2 þ î0þ ë5 -3û î x2 þ î0þ ë ë5 -3û Since determinant is not equal to zero, x1 and x2 must equal zero.

1.17 Under what conditions does the following system of equations have a solution?

(1− λ ) y1 + 2 y2 + 3y3 = 0 4 y1 + (6− λ ) y2 + 5y3 = 0 7 y1 + 8y2 + (9− λ ) y3 = 0

Solution:

⎛ 1− λ 2 3 ⎞⎟⎟ ⎜⎜ ⎛ 4 5 ⎞⎟⎟ 5 ⎞⎟⎟ ⎟⎟ = (1− λ )det ⎛⎜⎜ 6− λ ⎜⎜ det ⎜⎜ 4 − 2det 6− λ 5 ⎟ ⎟ ⎜⎜ 8 ⎜⎜ 7 9− λ ⎟⎟ ⎟⎠ ⎜⎜ ⎟⎟ 9− λ ⎝ ⎠ ⎝ ⎜⎝ 7 8 9− λ ⎟⎠ ⎛ ⎞ + 3det ⎜⎜⎜ 4 6− λ ⎟⎟⎟ ⎜⎝ 7 8 ⎟⎠ = (1− λ )((6− λ )(9− λ )− 40) − 2(4(9− λ )−35) + 3(32−7(6− λ )) = −λ 3 +16λ 2 − 29λ +14 + 8λ − 2 + 21λ −30 = −λ 3 +16λ 2 −18 Thus, this system of equation has a solution for those values of l that satisfy

−λ 3 +16λ 2 −18 = 0 That is, when l = -1.02814, l = 1.09908, and l = 15.9291.

Mathematica verification N[Eigenvalues[{{1,2,3},{4,6,5},{7,8,9}}]] 1.18 Given the following system of equations

(5 − λ )x1 + x2 = 2 −2x1 + (2 − λ )x2 = 1 Under what specific conditions can one solve for x1 and x2.

Solution The determinant of the coefficients must not equal zero; that is, we can solve for x1 and x2 provided that

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⎛ 5− λ det ⎜ ⎝ −2

1 ⎞ ⎟ ≠0 2−λ ⎠

the values of x1 and x2 are zero. This will be true for all values of l except when l has the following values

⎛ 5− λ det ⎜ ⎝ −2

1 ⎞ ⎟ =0 2−λ ⎠

(5 − λ )(2 − λ ) + 2 = 0

λ 2 − 7 λ + 12 = 0 λ1,2 = 3,4 Therefore, when l = 4, values of x1 and x2 are unknown.

Mathematica verification Eigenvalues[{{5,1},{-2,2}}]

1.19 Determine the eigenvalues and eigenvectors for the following matrix and verify that the eigenvectors are not orthogonal. ⎛ 1 0 −1 ⎞ A = ⎜ −2 −1 2 ⎟ ⎜ ⎟ ⎝ −1 2 1 ⎠

Solution:

⎛ 1− λ det ⎜ −2 ⎜ ⎝ −1

0 −1 ⎞ ⎛ −1− λ ⎛ −2 −1− λ ⎞ 2 ⎞ − det −1− λ 2 ⎟ = (1− λ ) det ⎜ ⎜⎝ −1 ⎟⎠ ⎟ 2 1− λ ⎟⎠ 2 ⎝ 2 1− λ ⎠ = − (1− λ ) ⎡⎣(1+ λ ) (1− λ ) + 4 ⎤⎦ − ( −4 − 1− λ ) = −4 (1− λ ) − (1− λ ) (1− λ 2 ) + 5 + λ = −4 + 4 λ − 1+ λ + λ 2 − λ 3 + 5 + λ = 6λ + λ 2 − λ 3

Therefore, the eigenvalues are: l1 = -2, l2 = 0, l3 = 3. The eigenvector corresponding to l1 = -2 is obtained as

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⎛ 1− (−2) 0 −1 ⎞ ⎧ c1 ⎫ ⎪ ⎜ ⎟⎪ −2 −1− (−2) 2 ⎜ ⎟ ⎨ c2 ⎬ = 0 ⎜⎝ −1 2 1− (−2) ⎟⎠ ⎪ c3 ⎪ ⎩ ⎭ ⎛ 3 0 −1 ⎞ ⎧ c1 ⎫ ⎧ c ⎫ ⎪ ⎪ ⎜ −2 1 2 ⎟ ⎨ c ⎬ → ⎛ 1 2 ⎞ ⎪⎨ 2 ⎪⎬ = ⎧⎨ 2 ⎫⎬ ⎜ ⎟ ⎪ 2 ⎪ ⎜⎝ 2 3 ⎟⎠ ⎪ c3 ⎪ ⎩ 1 ⎭ ⎩ ⎭ −1 2 3 ⎝ ⎠ c3 ⎩ ⎭

Therefore,

⎛ 2 det ⎜ ⎝ 1 c2 = ⎛ 1 det ⎜ ⎝ 2

2 ⎞ 3 ⎟⎠ 4 = = −4 2 ⎞ −1 3 ⎟⎠

⎛ 1 det ⎜ ⎝ 2 c3 = ⎛ 1 det ⎜ ⎝ 2

2 ⎞ 1 ⎟⎠ −3 = =3 2 ⎞ −1 3 ⎟⎠

and x(1) = {1, -4,3}T. For l2 = 0, ⎛ 1 0 −1 ⎞ ⎧ c1 ⎫ ⎪ ⎪ ⎜ −2 −1 2 ⎟ ⎨ c ⎬ = 0 2 ⎜ ⎟ ⎝ −1 2 1 ⎠ ⎪ c3 ⎪ ⎩ ⎭ ⎛ 1 0 −1 ⎞ ⎧ c1 ⎫ ⎧ c ⎫ ⎪ ⎪ ⎜ −2 −1 2 ⎟ ⎨ c ⎬ → ⎛ −1 2 ⎞ ⎪⎨ 2 ⎪⎬ = ⎧⎨ 2 ⎫⎬ 2 ⎜⎝ 2 1 ⎟⎠ c ⎜ ⎟ ⎪⎩ 3 ⎪⎭ ⎩ 1 ⎭ ⎝ −1 2 1 ⎠ ⎪ c3 ⎪ ⎩ ⎭

Therefore,

⎛ 2 2 ⎞ det ⎜ ⎝ 1 1 ⎟⎠ 0 c2 = = =0 ⎛ −1 2 ⎞ −5 det ⎜ ⎝ 2 1 ⎟⎠

⎛ −1 det ⎜ ⎝ 2 c3 = ⎛ −1 det ⎜ ⎝ 2

2 ⎞ 1 ⎟⎠ −5 = =1 2 ⎞ −5 1 ⎟⎠

and x(2) = {1, 0,1}T. For l3 = 3, ⎛ 1− (3) 0 −1 ⎞ ⎧ c1 ⎫ ⎪ ⎜ ⎟⎪ c −2 −1− (3) 2 ⎨ ⎬=0 2 ⎜ ⎟ ⎜⎝ −1 2 1− (3) ⎟⎠ ⎪ c3 ⎪ ⎩ ⎭ ⎛ −2 0 −1 ⎞ ⎧ c1 ⎫ ⎧ c ⎫ ⎪ ⎪ ⎜ −2 −4 2 ⎟ ⎨ c ⎬ → ⎛ −4 2 ⎞ ⎪⎨ 2 ⎪⎬ = ⎧⎨ 2 ⎫⎬ ⎜ ⎟ ⎪ 2 ⎪ ⎜⎝ 2 −2 ⎟⎠ ⎪ c3 ⎪ ⎩ 1 ⎭ ⎩ ⎭ ⎝ −1 2 −2 ⎠ c3 ⎩ ⎭

13

Therefore,

⎛ 2 2 ⎞ det ⎜ ⎝ 1 −2 ⎟⎠ −6 −3 c2 = = = ⎛ −4 2 ⎞ 4 2 det ⎜ ⎟ ⎝ 2 −2 ⎠

⎛ −4 det ⎜ ⎝ 2 c3 = ⎛ −4 det ⎜ ⎝ 2

2 ⎞ 1 ⎟⎠ −8 = = −2 4 2 ⎞ −2 ⎟⎠

and x(3) = {1, -3/2, -2}T. Thus, ⎛ 1 1 1 ⎞ ⎜ Φ = −4 0 −3 / 2 ⎟ ⎜ ⎟ −2 ⎠ ⎝ 3 1

( ) x = 0, i ≠ j; therefore,

For the eigenvectors are linearly independent x (i)

(x ) x (1) T

(2)

=

T

{ 1 −4 3 }

( j)

⎧ 1 ⎫ ⎪ ⎪ ⎨ 0 ⎬ = 1+ 3 = 4 ≠ 0 ⎪ 1 ⎪ ⎩ ⎭

has to be zero, which is not the case. Therefore, the eigenvectors are not orthogonal. Without actually solving, we also know that because A is not a symmetric matrix the eigenvectors are not orthogonal.

Mathematica verification {lam,xn}=Eigensystem[{{1,0,-1},{-2,-1,2},{-1,2,1}}]; (Transpose[Transpose[{xn[[All,3]]}]].xn[[All,1]])[[1]]

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Solutions to Exercises in Chapter 2 Section 2.1 2.1 Show that

a + jb =1 b + ja Solution: a + jb a + jb b − ja 2ab + j ( b − a ) = × = b + ja b + ja b − ja a2 + b2 2

2

Therefore,

a + jb 1 1 = 2 4a 2b 2 + (b 2 − a 2 )2 = 2 4a 2b 2 + a 4 + b 4 − 2a 2b 2 2 2 b + ja a + b a +b =

1 1 a 4 + b 4 + 2a 2b 2 = 2 2 a +b a + b2 2

(a + b ) = 1 2 2

2

Mathematica verification z=(a+I b)/(b+I a); Simplify[ComplexExpand[Re[z]^2+Im[z]^2]] 2.2 If z = x + jy, then verify the following relations

(a) z + 5 j = z − 5 j (b) jz = − jz (c) (1+ j)4 = −4 (d)

1+ j = j 1− j

Solution:

(a) z + 5 j = x + j( y + 5) = x − jy − j5 = z − 5 j (b) jz = j(x + jy) = − y + jx = − y − jx = − j(x + y / j) = − jz (c) (1+ j)4 = (d)

( 2e ) = ( 2e ) = 4e = −4 j tan −1 1

4

jπ /4

4

1+ j (1+ j) (1+ j) 1+ 2 j − 1 = = = j 1− j (1− j) (1+ j) 2

Mathematica verification z=x+I y; Conjugate[z+5 I] 15

jπ

Conjugate[I z] ComplexExpand[(1+I)^4] ComplexExpand[(1+I)/(1-I)] Section 2.2 2.3 Show that

(a)

j=±

1+ j 2

⎛ −1+ j 3 ⎞ (b) ⎜ ⎟ 2 ⎝ ⎠

3/2

= ±1

Solution: Since

z n =r n exp ⎡⎣ jn (θ + 2mπ ) ⎤⎦

m = 0,±1,±2,...

then

j = e j( π /2+2mπ )/2 = e jπ /4 e jmπ = ± ( cos π / 4 + j sin π / 4 ) =± ⎛ −1+ j 3 ⎞ ⎜ ⎟ 2 ⎝ ⎠

3/2

1+ j 2

⎛ 1 3 j tan−1 3/(−1) ⎞ =⎜ + e ⎟ ⎝ 4 4 ⎠

3/2

(

= e j(2π /3+2mπ )

) =e 3/2

Mathematica verification ComplexExpand[Sqrt[I]] ((-1+I Sqrt[3])/2)^(3/2) 2.4 Show that

(−1+ j)10 = −32 j Solution: We note that

−1+ j = 2e jθ where θ = tan −1 Then

16

1 3π = −1 4

jπ +3mπ

= ±1

(−1+ j)10 = 25 e j(15π /2) = 25 e j(3π /2+6π ) = 25 e j3π /2 = 25 ⎡⎣cos(3π / 2) + j sin(3π / 2) ⎤⎦ = − j25 Mathematica verification ComplexExpand[(-1+I)^10] 2.5 Use Euler’s formula to obtain expressions for cos(a + b) and sin(a + b) in terms of sin(a), sin(b), cos(a), and cos(b). Hint: note that ej(a + b) = eja ej b.

Solution:

e j(α +β ) = cos(α + β ) + j sin(α + β )

= e jα e jβ = ( cos α + j sin α ) ( cos β + j sin β )

= cos α cos β − sin α sin β + j ( sin α cos β + cos α sin β ) Therefore, upon equating the real and imaginary parts, we obtain

cos(α + β ) = cos α cos β − sin α sin β sin(α + β ) = sin α cos β + cos α sin β Mathematica verification ComplexExpand[Exp[I (a+b)]] ComplexExpand[ComplexExpand[Exp[I a]] ComplexExpand[ Exp[I b]]] 2.6 Show that the four roots of

z 4 − 2cos(2θ )z 2 + 1 = 0 are ±ejq and ±e-jq.

Solution:

(

)

1 2cos(2θ ) ± 4cos 2 (2θ ) − 4 = cos(2θ ) ± cos 2 (2θ ) − 1 2 = cos(2θ ) ± j sin(2θ ) = e ±2 jθ

z2 =

Therefore,

z = ±e± jθ Mathematica verification PowerExpand[FullSimplify[Solve[ z^4-2 Cos[2 th] z^2+1==0,z],Assumptions->th Î Reals]]

17

2.7 If ∞

1

∑ q = 1− q n

q <1

n=0

then show that

1− ε cost

∞

∑ ε cos(nt) = 1+ ε − 2ε cost n

2

ε <1

n=0

Hint: Let q = eejt.

Solution: We note that

q = ε e jt → q n = ε n e jnt Then ∞

∞

∞

n=0

n=0

n=0

∑ q n = ∑ ε ne jnt = ∑ ε n ( cos nt + j sin nt ) =

1 1− ε e jt

Therefore, ∞

1

1

1− ε cost + jε sint

∑ ε ( cos nt + j sin nt ) = 1− ε e = 1− ε cost − jε sint × 1− ε cost + jε sint n

jt

n=0

=

1− ε cost + jε sint 1− ε cost + jε sint = 2 2 2 (1− ε cost ) + ε sin t 1− 2ε cost + ε 2 cost + ε 2 sin 2 t

=

1− ε cost + jε sint 1+ ε 2 − 2ε cost

Equating the real and imaginary parts gives

1− ε cost

∞

∑ ε cos nt = 1+ ε − 2ε cost n

2

n=0

ε sint

∞

∑ ε sin nt = 1+ ε − 2ε cost n

2

n=0

Mathematica verification q=e Exp[I t]; cq=ComplexExpand[1/(1-q)]; Simplify[Re[cq],Assumptions->{e,t} Î Reals] (*Imag part = 0*) Simplify[Im[cq],Assumptions->{e,t} Î Reals] (*Imag part = 0*) 2.8 Using the appropriate results of Exercise 2.7, show that ∞

2∑ e−α n sin(nt) = n=0

sint cosh α − cost

18

α >0

Solution:

ε sint

∞

∑ ε sin nt = 1+ ε − 2ε cost n

2

n=0

We let e = e-a and obtain ∞

∑ e−α n sin nt = n=0

e−α sint sint sint = α −α = −2 α −α 1+ e − 2e cost e + e − 2 cost 2 ( cosh α − cost )

2.9 If z = x + jy, then show that

sinh x + j sin y cosh x + cos y sinh x − j sin y (b) coth(z / 2) = cosh x − cos y (a) tanh(z / 2) =

Solutions: (a)

tanh(z / 2) =

e z/2 − e− z/2 e z − 1 e x+ jy − 1 e x (cos y + j sin y) − 1 = = = e z/2 + e− z/2 e z + 1 e x+ jy + 1 e x (cos y + j sin y) + 1

e x cos y − 1+ je x sin y e x cos y + 1− je x sin y = x × e cos y + 1+ je x sin y e x cos y + 1− je x sin y

( e cos y ) − 1+ ( e sin y ) + je sin y ( e cos y + 1− e cos y + 1) = ( e cos y + 1) + ( e sin y ) e − 1+ j2e sin y ( e − e ) / 2 + j sin y = = e + 2e cos y + 1 ( e + e ) / 2 + cos y 2

x

2

x

x

2

x

2x

2x

=

x

x

x

−x

x

−x

sinh x + j sin y cosh x + cos y

(b)

19

x

x

x

2

coth(z / 2) = =

e z/2 + e− z/2 e z + 1 e x+ jy + 1 e x (cos y + j sin y) + 1 = = = e z/2 − e− z/2 e z − 1 e x+ jy − 1 e x (cos y + j sin y) − 1 e x cos y + 1+ je x sin y e x cos y − 1− je x sin y × e x cos y − 1+ je x sin y e x cos y − 1− je x sin y

( e cos y ) − 1+ ( e sin y ) + je sin y ( e cos y − 1− e cos y − 1) = ( e cos y − 1) + ( e sin y ) e − 1− j2e sin y ( e − e ) / 2 − j sin y = = e − 2e cos y + 1 ( e + e ) / 2 − cos y 2

x

2

x

x

2

x

2x

2x

=

x

x

x

−x

x

−x

x

x

x

2

sinh x − j sin y cosh x − cos y

Mathematica verification z=x+I y; ComplexExpand[Tanh[z/2]] ComplexExpand[Coth[z/2]] Section 2.3 2.10 If z = x + jy, which of the following functions are analytic and for those that are analytic, what is their region of analyticity? (a) f (z) = e2 x cos y + je2 x sin y (b) f (z) = sinh z (c) f (z) =| z | sin z (d) f (z) = z (e) f (z) = jz + z (f) f (z) =

z− j z+ j

Solutions: (a) Not analytic

f ( z ) = e 2 x cos y + je 2 x sin y ¶u ¶v = 2e 2 x cos y ¹ = e 2 x cos y ¶x ¶y ¶v ¶u = 2e 2 x sin y ¹ = -e 2 x sin y ¶x ¶y (b) Analytic everywhere

20

f ( z ) = sinh z ® cos y sinh x + j sin y cosh x u ( x, y ) = cos y sinh x v( x, y ) = sin y cosh x ¶u ¶v = cos y cosh x = = cos y cosh x ¶x ¶y ¶v ¶u = sin y sinh x = = - sin y sinh x ¶x ¶y

(c) Not analytic

f (z) =| z | sin z → x 2 + y 2 ( sin x cosh y + j cos x sinh y ) → u(x, y) = x 2 + y 2 sin x cosh y → v(x, y) = x 2 + y 2 cos x sinh y ∂u x sin x cosh y] = x 2 + y 2 cos x cosh y + ∂x x2 + y2 ∂v y cos x sinh y] = x 2 + y 2 cos x cosh y + ∂y x2 + y2 ∂v x cos x sinh y = − x 2 + y 2 sin x sinh y 2 2 ∂x x +y −

∂u y sin x cosh y =− − x 2 + y 2 sin x sinh y 2 2 ∂y x +y

(d) Not analytic

u(x, y) = x, v(x, y) = −y ∂u ∂v =1≠ = −1 ∂x ∂y ∂v ∂u =0=− ∂x ∂x (e) Not analytic

f (z) = jz + z = − y + jx + x 2 + y 2 u = − y + x2 + y2 , ∂u = ∂x

x x +y 2

2

,

(f) Analytic

21

v=x ∂v =0 ∂y

f (z) =

z − j x + j( y − 1) x + j( y − 1) x − j( y + 1) = = z + j x + j( y + 1) x + j( y + 1) x − j( y + 1)

x2 + y2 − 1 2x = 2 −j 2 2 x + ( y + 1) x + ( y + 1)2 u=

x2 + y2 − 1 , x 2 + ( y + 1)2

v=−

∂u 4x( y + 1) = , 2 ∂x x 2 + ( y + 1)2

(

(

)

∂v 4x( y + 1) = 2 ∂y x 2 + ( y + 1)2

(

)

∂v 2 x − ( y + 1) = , 2 2 2 ∂x x + ( y + 1)

(

2

2

)

2x x + ( y + 1)2 2

−

(

)

)

∂u 2 x − ( y + 1) = 2 ∂y x 2 + ( y + 1)2

(

2

2

)

Mathematica verification (* u and v must be determined first *) cr[u_,v_]:=Simplify[{D[u,x]==D[v,y],D[v,x]==-D[u,y]}] cr[Exp[2 x] Cos[y],Exp[2 x] Sin[y]] cr[Sinh[x] Cos[y],Cosh[x] Sin[y]] cr[Sqrt[x^2+y^2] Sin[x] Cosh[y],Sqrt[x^2+y^2] Cos[x] Sinh[y]] cr[x,-y] cr[-y+Sqrt[x^2+y^2],x] cr[(x^2+y^2-1)/(x^2+(y+1)^2),-2 x/(x^2+(y+1)^2)] 2.11 Express the following expressions in terms of trigonometric and hyperbolic functions and them determine the value(s) of the real quantities a and b that make them analytic: (a) cos(ax + jby) (b) sin(ax - jby) Solutions: (a)

1 j ( ax + jby ) - j ( ax + jby ) +e (e ) 2 1 = ( e - by ( cos ax + j sin ax ) + eby ( cos ax - j sin ax ) ) 2 1 = ( eby + e - by ) cos ax - j ( eby - e -by ) sin ax 2 = cosh by cos ax - j sinh by sin ax

cos(ax + jby ) =

(

)

Therefore,

u ( x, y ) = cosh by cos ax v( x, y ) = - sinh by sin ax

22

Consequently

¶u ¶v = - a cosh by sin ax = -b cosh by sin ax ¶x ¶y ¶v ¶u = - a sinh by cos ax = -b sinh by cos ax ¶x ¶y Hence, to be analytic, we see that the C-R conditions are satisfied if a = b. (b)

(

)

1 j( ax− jby ) − j( ax− jby ) e −e 2j 1 by = e ( cos ax + j sin ax ) − e− by ( cos ax − j sin ax ) 2j 1 = eby − e− by cos ax + j eby + e− by sin ax 2j = cosh by sin ax − j sinh by cos ax

sin(ax − jby) =

(

((

)

(

)

Therefore,

u(x, y) = cosh by sin ax v(x, y) = − sinh by cos ax Consequently

∂u = acosh by cos ax ∂x ∂v = asinh by sin ax ∂x

∂v = −bcosh by cos ax ∂y ∂u = bsinh by sin ax ∂y

Thus, the Cauchy-Riemann conditions are satisfied when a = -b.

Mathematica verification (* u and v must be determined first *) crr[u_,v_]:={D[u,x]==D[v,y],D[v,x]==-D[u,y]} crr[Cosh[b y] Cos[a x],-Sinh[b y] Sin[a x]] crr[Cosh[b y] Sin[a x],-Sinh[b y] Cos[a x]] 2.12 Given the analytic function

f (z) = u(x, y) + jv(x, y) Determine f(z) when: (a) u(x,y) = x3 - 3xy2 (b) u(x,y) = excosy Solutions:

23

)

)

(a)

u(x, y) = x 3 − 3xy 2 →

∂u = 3x 2 − 3y 2 ∂x

Since the function is analytic,

∂u ∂v = = 3x 2 − 3y 2 → v = 3x 2 y − y 3 + c ∂x ∂ y As a check, we note that

∂v ∂u = 6xy = − = 6xy ∂x ∂y (b)

u(x, y) = e x cos y →

∂u = e x cos y ∂x

Since the function is analytic,

∂u ∂v = = e x cos y → v = e x sin y + c ∂x ∂ y 2.13 Determine the value of the following integrals. Assume that the path C includes all singularities. (a) (b)

e−3π z ∫ 2z + j dz C

∫ z − π / 2 dz

C

(c)

e z cos z

(

zn

∫ z−a

C

(

)

n+1

)

2

dz

Solutions:

e-3p z 1 -3p z -3 jp /2 òC 2 z + j dz = 2p j 2 e z =- j /2 = p je = p j (- j ) = p e z cos z

d

ò ( z - p / 2 ) dz = 2p j dz ( e cos z ) p = 2p j ( e {cos z - sin z}) p = -2p je 2

ò ( z - a)

C

dz = n +1

p /2

z

z = /2

z = /2

C

zn

z

2p j d n n ( z )z =a = 2p j n ! dz n

Mathematica verification

24

2 Pi I Residue[Exp[-3 Pi z]/(2 z+I),{z,-I/2}] 2 Pi I Residue[Exp[z] Cos[z]/(z-Pi/2)^2,{z,Pi/2}] 2 Pi I Residue[z^3/(z-a)^4,{z,a}] 2.14 Evaluate the following integral

1

∫ z (z + 4) dz 3

C

for (a) inside the circle |z| = 2, and (b) inside the circle |z + 2| = 3.

Solutions: (a) Third order pole at z = 0. Therefore,

⎡ d2 ⎛ ⎡ d2 −1 ⎤ 1 1 z3 ⎞ ⎤ dz = 2π j lim ⎢ 2 ⎜ 3 ⎥ = π j lim ⎢ 2 ( z + 4) ⎥ ∫ 3 ⎟ z→0 z→0 2! ⎢⎣ dz ⎝ z (z + 4) ⎠ ⎥⎦ ⎣ dz ⎦ |z|=2 z (z + 4) −3 2π j π j = π j lim ⎡⎢ 2 ( z + 4 ) ⎤⎥ = = z→0 ⎣ ⎦ 64 32

(b) The center of the circle is at z = -2 and the radius is 3; therefore, both z = 0 and z = -4 are inside this contour and the residue is

⎡ d2 ⎛ ⎡ (z + 4) ⎤ 1 1 z3 ⎞ ⎤ dz = 2 π j lim + 2π j lim ⎢ 3 ⎢ ⎥ ⎥ ∫ 3 2 3 ⎜ ⎟ z→−4 z (z + 4) 2! z→0 ⎢⎣ dz ⎝ z (z + 4) ⎠ ⎥⎦ ⎣ ⎦ |z+2|=3 z (z + 4) =

π j 2π j π j π j + = − =0 32 (−4)3 32 32

Mathematica verification 2 Pi I Residue[1/(z+4)/z^3,{z,0}] 2 Pi I Residue[1/(z+4)/z^3,{z,0}]+ 2 Pi I Residue[1/(z+4)/z^3,{z,-4}] 2.15 Determine the value of the following integral when the contour includes all singularities.

etz dz 2 2 C z ( z + 2z + 2 )

I=∫ Solution: The poles are at z = 0 and at

1 z1,2 = ⎡ −2 ± 4 − 8 ⎤ = −1± j ⎦ 2⎣ Then

25

⎧⎪ d ⎡ ⎤ ⎡ ( z − z1 ) etz ⎤ etz dz z 2 etz = 2 π j + ⎢ ⎥ ⎨ ⎢ 2 ⎥ ∫C z 2 ( z 2 + 2z + 2 ) dz z 2 z 2 + 2z + 2 ) ⎥⎦ ⎣ z ( z − z1 ) ( z − z2 ) ⎦ z=z1 z=0 ⎩⎪ ⎢⎣ ( ⎡ ( z − z2 ) etz ⎤ ⎫⎪ +⎢ 2 ⎥ ⎬ ⎣ z ( z − z1 ) ( z − z2 ) ⎦ z=z2 ⎭⎪ ⎧⎡ ⎡ ⎤ 2z + 2 ) etz ⎤ tetz etz ( ⎪⎢ ⎥ = 2π j ⎨ 2 − +⎢ 2 ⎥ 2 2 ⎪⎩ ⎢⎣ ( z + 2z + 2 ) ( z + 2z + 2 ) ⎥⎦ z=0 ⎣ z ( z − z2 ) ⎦ z=z1 ⎡ ⎤ ⎫⎪ etz +⎢ 2 ⎥ ⎬ ⎣ z ( z − z1 ) ⎦ z=z2 ⎪⎭ ⎧t − 1 ⎫ etz1 etz2 = 2π j ⎨ + 2 − 2 ⎬ z1 ( z1 − z2 ) z2 ( z1 − z2 ) ⎭⎪ ⎩⎪ 2 But

z1 − z2 = −1+ j − (−1− j) = 2 j

( ) = 2e = 2(cos(3π / 2) + j sin(3π / 2)) = −2 j z = ( −1− j ) = ( 2e ) = 2e = 2(cos(3π / 2) − j sin(3π / 2)) = 2 j z12 = ( −1+ j ) = 2 2

2

2e3π j/4

2

−3π j/4

2

3π j/2

2

−3π j/2

Therefore,

⎧t − 1 etz dz etz1 etz2 ⎫ 1 tz ⎧ tz ⎫ = 2 π j + − ⎨ ⎬ = π j ⎨t − 1+ ( e 1 + e 2 ) ⎬ ∫C z 2 ( z 2 + 2z + 2 ) 2 (−2 j)2 j (2 j)2 j 2 ⎩ ⎭ ⎩ ⎭

(

)

1 1 ⎧ ⎫ ⎧ ⎫ = π j ⎨t − 1+ et( −1+ j ) + et( −1− j ) ⎬ = π j ⎨t − 1+ e−t ( e jt + e− jt ) ⎬ 2 2 ⎩ ⎭ ⎩ ⎭ = π j {t − 1+ e−t cost } Mathematica verification Simplify[ComplexExpand[ 2 Pi I Residue[Exp[t z]/(z^2+2 z +2)/z^2,{z,0}]+ 2 Pi I Residue[Exp[t z]/(z^2+2 z +2)/z^2,{z,-1+I}]+ 2 Pi I Residue[Exp[t z]/(z^2+2 z +2)/z^2,{z,-1-I}]]] 2.16 Evaluate the following integral around the unit circle.

∫ z e dz 2 1/z

|z|<1

Hint: Express the exponential function as a series using Eq. (A.3) of Appendix A.

26

Solution: In this case, we expand the argument in series form to get

1 1 1 ⎛ 1 ⎞ z 2 e1/z = z 2 ⎜ 1+ + + + + ...⎟ 2 3 4 ⎝ ⎠ z 2!z 3!z 4!z = z2 + z +

1 1 1 + + + ...

2! 3!z 4!z 2

Upon integrating this expansion term by term, the only nonzero residue is that associated with 1/z.

Therefore, from Eq. (2.50), we see that

⎛ 1⎞

πj

∫ z e dz = 2π j ⎜⎝ 3!⎟⎠ = 3 2 1/z

|z|<1

Mathematica verification 2 Pi I Residue[Expand[x^2 Normal[Series[Exp[z],{z,0,4}]] /.z->1/x],{x,0}] 2.17 Determine the order of the pole for the following expressions: (a)

etz z sinh(az) (b) 2

ez z4 (c)

sin 2 (z) z3 Solution: (a) Order equals 2. etz 1+ tz + t 2 z 2 / 2 +!

1+ tz + t 2 z 2 / 2 +! = = z sinh(az) z az + a 3 z 3 / 3!+! az 2 1+ a 2 z 2 / 3!+!

(

)

(

)

(b) Order equals 4.

2

(

)

ez 1 = 4 1+ z 2 + z 4 / 2 +!

4 z z (c) Simple pole.

27

⎞ 1 z 2z 3 sin 2 (z) 1 ⎛ 2 z 4 2z 6 = z − + +! ⎟⎠ = z − 3 + 45 +!

3 45 z3 z 3 ⎜⎝

Mathematica verification Series[Exp[t z]/z/Sinh[a z],{z,0,1}] Series[Exp[z^2]/z^4,{z,0,1}] Series[Sin[z]^2/z^3,{z,0,1}]

28

Solutions to Exercises in Chapter 3 Section 3.1 3.1 Referring to Figure E3.1, consider a sine wave of magnitude Fo and frequency wb that has a duration tb = Ntp, where tp = 2p/wb and N is an integer. This sine wave is repeated every period T, where T = Mtp = 2p/wo, M ³ N, and M is an integer. Determine the Fourier coefficients ai and bi and express them in terms of i, N, and M. To simplify subsequent calculations, it is suggested that the following notation be introduced prior to performing any integrations:

τ = ω ot,

ω b = Mω o ,

ω otb =

2π N ω o 2π N = , ωb M

and

T ω o = 2π

Solution: For ai ω t

t

2F b 2Fo o b ai = o ∫ sin(ω bt)cos(iω ot) dt = sin(τω b / ω o )cos(iτ ) d τ T 0 ω oT ∫0

where we have set wot = t. To simplify the subsequent integrations, we introduce the following notation:

w b = M wo ,

wotb =

2p Nwo

wb

2p N , M

=

and

T wo = 2p

Then, for ai and bi, we have

Fo π

2π N M

F bi = o π

2π N M

ai =

∫ sin( Mτ )cos(iτ ) dτ

i≠ M

0

∫ sin( Mτ )sin(iτ ) dτ

i≠ M

0

From a table of integrals, we find that for i ¹ M, 2 π N/ M

F ⎡ cos[( M + i)τ ] cos[( M − i)τ ] ⎤ ai = o ⎢ − − π ⎣ 2( M + i) 2( M − i) ⎥⎦0 =

Fo ⎡ 1 1 cos[( M + i)(2π N/ M )] cos[( M − i)(2π N/ M )] ⎤ + − − ⎢ ⎥ 2π ⎣ M + i M − i M +i M −i ⎦

=

Fo ⎡ 2 M 2 M cos(2π iN/ M ) ⎤ − ⎢ ⎥ 2 2 2π ⎣ M − i M 2 − i2 ⎦

=

(

Fo M

π M −i 2

2

)

⎡⎣1− cos(2π iN/ M ) ⎤⎦ =

2Fo M

(

π M −i

and for i = M, 29

2

2

)

sin 2 (π iN/ M )

i≠ M

aM = =

Fo π

2π N M

Fo 1

∫ sin( Mτ )cos( Mτ ) dτ = π 2 M sin ( Mτ ) 2

0

Fo sin 2 (2π N ) = 0 2π M

2π N M 0

i= M

For i = 0, Fo π

a0 =

2π N M

2Fo

∫ sin( Mτ ) dτ == n sin (nπ ) = 0 2

0

Similarly, for i ¹ M, 2 π N/ M

F ⎡ sin[( M − i)τ ] sin[( M + i)τ ] ⎤ bi = o ⎢ − π ⎣ 2( M − i) 2( M + i) ⎥⎦0 =

Fo ⎡ sin[( M − i)(2π N/ M )] sin[( M + i)(2π N/ M )] ⎤ − ⎥ 2π ⎢⎣ M −i M +i ⎦

=

Fo ⎡ sin(2π iN/ M ) sin(2π iN/ M ) ⎤ − − ⎥ 2π ⎢⎣ M −i M +i ⎦

=

− Fo sin(2π iN/ M ) ⎡ 1 1 ⎤ + ⎢ ⎥ 2π ⎣ M −i M +i⎦

=

− Fo M sin(2π iN/ M )

(

π M 2 − i2

)

i≠ M

and for i = M,

F bi = o π =

2π N M

∫ 0

2π N M

Ï ⎡ τ sin( M τ )cos( M τ ) ⎤ sin ( M τ ) d τ = o ⎢ − ⎥ π ⎣2 2M ⎦0 2

Fo ⎛ 2π N ⎞ Fo N = π ⎜⎝ 2 M ⎟⎠ M

i= M

Mathematica verification ai=Fo/Pi Simplify[Integrate[Sin[M t] Cos[i t], {t,0,2 Pi Nn/M}],Assumptions->{M,Nn} Î Integers] aM=Fo/Pi Simplify[Integrate[Sin[M t] Cos[M t], {t,0,2 Pi Nn/M}],Assumptions->{M,Nn}Î Integers] a0=Fo/Pi Simplify[Integrate[Sin[M t], {t,0,2 Pi Nn/M}],Assumptions->{M,Nn}Î Integers] bi=Fo/Pi Simplify[Integrate[Sin[M t] Sin[i t], {t,0,2 Pi Nn/M}],Assumptions->{M,Nn}Î Integers] bM=Fo/Pi Simplify[Integrate[Sin[M t] Sin[M t], {t,0,2 Pi Nn/M}],Assumptions->{M,Nn}Î Integers] (*Visual verification: a plot*)

30

Ai[Nn_,Mm_,n_]:=If[n==Mm,0, 2 Mm Sin[Pi n Nn/Mm]^2/(Mm^2-n^2)/Pi] Bi[Nn_,Mm_,n_]:=If[n==Mm,Nn/Mm, Mm Sin[2 Pi n Nn/Mm]/(Mm^2-n^2)/Pi] ser[t_,Nn_,Mm_]:=Total[Table[Ai[Nn,Mm,n] Cos[n t]+ Bi[Nn,Mm,n] Sin[n t],{n,1,31}]] Plot[ser[t,4,8],{t,0,2 Pi}] 3.2 Expand the following periodic function in complex notation and use the results to express the series as a real quantity.

−π < x <π

f (t) = x Solution:

It is seen that the period T = 2p. Then the Fourier series coefficients are obtained from Eq. (3.2) as

cn =

−2 n 2π =0 =

π

1 2 ⎛ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎞ ⎛ nπ ⎞ x e− jnx dx = 2 ⎜ nπ cos ⎜ − sin ⎜ sin ⎟ ∫ ⎝ 2 ⎠ ⎝ 2 ⎟⎠ ⎟⎠ ⎜⎝ 2 ⎟⎠ 2π − π nπ⎝

n>0

n = odd n = even

Thus, cn is a real quantity. The same results are obtained for n < 0. When n = 0, π

1 π c0 = x dx = ∫ 2π − π 2

Then the final result in complex notation is f (x) =

π 2 −∞ e jnx 2 ∞ e jnx − − ∑ ∑ 2 π n=−1,−3,−5, n2 π n=1,3,5, n2

To express this result as a real quantity, since f(x) is real, we use Eqs. (3.14) and arrive at ∞

∞

n=1

n=1,3,5...

f (x) = c0 + 2∑ cn cos[2nπ x / T − ϕ n ] → c0 + 2 ∑ cn cos[nx − ϕ n ] where

π 2 2 cn = 2 nπ c0 =

ϕ n = tan −1

n = odd 0 =π −2 / n 2π

(

Therefore, 31

)

f (x) =

π 4 ∞ 1 π 4 ∞ 1 + cos(nx − π ) = − ∑ ∑ cos nx 2 π n=1,3,5... n2 2 π n=1,3,5... n2

Mathematica verification cn=Simplify[ComplexExpand[Integrate[Abs[x] Exp[-I n x], {x,-Pi,Pi},Assumptions->n Î Integers&&n>0]/(2 Pi)]] cn=Simplify[ComplexExpand[Integrate[Abs[x] Exp[-I n x], {x,-Pi,Pi},Assumptions->n Î Integers&&n<0]/(2 Pi)]] c0=Integrate[Abs[x],{x,-Pi,Pi}]/(2 Pi) 3.3 (a) Obtain the Fourier series expansion of the following periodic function

f (τ ) = t 2

−π < t <π

(b) Using the results in (a), chose appropriate values for t and show that

(−1)n+1 π 2 ∑ n 2 = 12 n=1 ∞

1 π2 ∑ n2 = 6 n=1 ∞

(c) Add the results in part (b) and show that (see also Exercise 3.2)

1 π2 ∑ 2= 8 n=1,3,5 n ∞

Solutions: (a) We note that T = 2p and, therefore, wo = 2p/T = 1. Then, the Fourier coefficients are π

π

1 1 ⎡t3 ⎤ 2π 3 2 2 a0 = ∫ t 2 dt = ⎢ ⎥ = = π π −π π ⎣ 3 ⎦ − π 3π 3 π

1 1 4nπ 4 an = ∫ t 2 cos(nt)dt = (−1)n = 2 (−1)n 3 π −π π n n bn =

π

1 2 t sin(nt)dt = 0 π −∫π

Therefore,

f (t) = t 2 =

∞ π2 (−1)n + 4 ∑ 2 cos(nt) 3 n=1 n

(b) If t = 0, then we obtain ∞ π2 (−1)n π 2 ∞ (−1)n+1 = −4 ∑ 2 → =∑ 3 12 n=1 n 2 n=1 n

32

If t = p, then we obtain

π2 =

∞ π2 (−1)n + 4 ∑ 2 cos(nπ ) 3 n=1 n ∞

(−1)2n n2 n=1

2π 2 = 12∑

π2 ∞ 1 =∑ 6 n=1 n 2 (c) Expanding each series, we find

π2 1 1 1 = 1− + − + ...

12 4 9 16 2 π 1 1 1 = 1+ + + + ...

6 4 9 16 Therefore, adding these two expressions, we obtain ∞ π2 π2 1 + =2 ∑ 2 12 6 n=1,3,5 n ∞ π2 1 = ∑ 2 8 n=1,3,5 n

Mathematica verification FourierCoefficient[t^2,t,n] (* Gives coefficients cn only *) 3.4 Obtain the Fourier series expansion of the following periodic function

f (τ ) = t 4

−π < t <π

and choose appropriate values for t and the results of Exercise 3.3 to show that

(−1)n+1 7π 4 ∑ n 4 = 720 n=1 ∞

1 π4 ∑ n 4 = 90 n=1 ∞

Solutions: (a) We note that T = 2p and, therefore, wo = 2p/T = 1. Then, the Fourier coefficients are

33

π

π

1 4 1 ⎡t5 ⎤ 2π 5 2 4 a0 = ∫ t dt = ⎢ ⎥ = = π π −π π ⎣ 5 ⎦ − π 5π 5 an =

π

1 4 1 8π 8 t cos(nt)dt = (−1)n n 2π 2 − 6 = 4 (−1)n n 2π 2 − 6 4 ∫ π −π π n n

(

)

(

)

π

1 bn = ∫ t 4 sin(nt)dt = 0 π −π

Therefore, ∞ 1 4 (−1)n 2 2 f (t) = t = π + 8∑ 4 n π − 6 cos(nt) 5 n=1 n

(

4

)

If t = 0, then we obtain ∞ ∞ ∞ 1 4 (−1)n+1 2 2 (−1)n+1 (−1)n+1 2 π = 8∑ n π − 6 = 8π ∑ − 48∑ 5 n4 n2 n4 n=1 n=1 n=1

(

)

∞ ⎛ π2 ⎞ (−1)n+1 = 8π 2 ⎜ ⎟ − 48∑ n4 ⎝ 12 ⎠ n=1

(−1)n+1 1 ⎛ 2π 4 π 4 ⎞ 7π 4 ∑ n 4 = 48 ⎜⎝ 3 − 5 ⎟⎠ = 720 n=1 ∞

If t = p, then we obtain ∞ ∞ ∞ 1 (−1)2n 2 1 1 π 4 = π 4 + 8∑ 4 ( n 2π 2 − 6 ) = π 4 + 8π 2 ∑ 2 − 48∑ 4 5 n 5 n=1 n=1 n n=1 n 2 ⎞ π4 ⎛ 1 4 ⎞ 1 1 ⎛1 4 π4 2⎛π ⎞ 4 = π + 8 π − π = + − 1 = ∑ n 4 48 ⎜⎝ 5 ⎜⎝ 6 ⎟⎠ ⎟⎠ 48 ⎜⎝ 5 3 ⎟⎠ 48 × 15 ( 3 + 20 − 15 ) n=1 ∞

8π 4 π 4 = = 720 90

Mathematica verification FourierCoefficient[t^4, t, n]

(* Gives coefficients cn only *)

3.5 Determine the Fourier series expansion for the periodic function f(t) = 2Aot/T, -T/2 £ t £ T/2 using Eqs. (3.2) and (3.14) and Eqs. (3.18) and (3.19).

Solutions: From Eq. (3.2)

34

cn =

T /2 2 Aot − j 2nπ t/T Ao nπ − jy jAo (−1) n 1 e dt = ye dy = T −T∫/2 T nπ 2n2π 2 −∫nπ

T /2 2 Aot 2 Ao T /2 2 Ao 2 T /2 1 c0 = dt = t dt = t =0 T −T∫/2 T T 2 −T∫/2 T 2 −T /2

and since the real part of cn is zero

−(−1) n π ϕ n = tan = (−1) n+1 0 2 −1

Then, from Eq. (3.14) ∞

2 Ao ∞ 1 f (t) = c0 + 2∑ cn cos[2nπ t / T − ϕ n ] = cos 2nπ t / T − (−1) n+1π / 2 ∑ π n=1 n n=1

(

)

2 Ao ∞ (−1) n+1 = sin ( 2nπ t / T ) ∑ π n=1 n From Eq. (3.19), an =

T /2 2 Aot Ao nπ 2 cos(2n π t / T ) dt = y cos y dy = 0 T −T∫/2 T n2π 2 −∫nπ

4 Ao T /2 2 A T /2 a0 = 2 ∫ t dt = 2o t 2 =0 −T /2 T −T /2 T bn =

T Ao nπ 2 Ao (−1) n+1 2 2 Aot sin(2n π t / T ) dt = y sin y dy = T ∫0 T nπ n2π 2 −∫nπ

and from Eq. (3.18) ∞

f (t) = ∑ bn sin(nω ot) = n=1

2Ao ∞ (−1)n+1 sin(2nπ t / T ) ∑ π n=1 n

Mathematica verification FourierCoefficient[2 t/T,t,n,FourierParameters->{1,2 Pi/T}, Assumptions->T>0] (*Verification for cn only*) 3.6 Determine the Fourier series expansion for f(t) = Aoe-at/T, a > 0, over the interval -T/2 £ t £ T/2.

Solutions: From Eq. (3.2)

35

c0 =

T /2 Ao T /2 −α t/T A 2 e dt = − o e−α t/T = sinh(α / 2) ∫ −T /2 T −T /2 α α

2sinh ( (α + j2nπ ) / 2 ) Ao T /2 −α t/T − j 2nπ t/T Ao T /2 −(α + j 2nπ )t/T cn = e e dt = e dt = T −T∫/2 T −T∫/2 α + j2nπ =

(−1) n sinh (α / 2 )

=

α / 2 + jnπ

(−1) n sinh (α / 2 ) (α / 2)2 + (nπ )2

(α / 2 − jnπ )

Then,

cn =

sinh (α / 2 )

n = 0,1,2,3,...

(α / 2)2 + (nπ )2

ϕ n = tan −1

nπ α /2

n = 1,2,3,...

Therefore, ∞

f (t) = c0 + 2∑ cn cos[nω ot − ϕ n ] n=1

=

sinh (α / 2 )

α /2

(

∞

cos 2nπ t − tan −1 ( nπ / (α / 2 )

n=1

(α / 2) + (nπ )

+ 2sinh (α / 2 ) ∑

2

)

2

3.7 For the wave forms shown in Figure E3.2, use Eq. (3.2) to determine the magnitude and phase of the Fourier coefficients cn.

Solutions: (a) Case 2: From Eq. (3.2) t=0

t /2

0

t=td /2

1 1 d − j 2 pπ t/T je− j 2 pπ t/T je− j 2 pπ t/T − j 2 pπ t/T cp = e dt − e dt = − T −t∫/2 T ∫0 2 pπ t=−t /2 2 pπ t=0 d

d

(

)

(

ϕ n = tan

−1

)

j j 1− e jpπ td /T + 1− e− jpπ td /T 2 pπ 2 pπ j 2j = 2 − 2cos( pπ td / T ) ) = sin 2 ( pπ td / (2T ) ) ( 2 pπ pπ c0 = 0 =

and from Eq. (3.11)

sin 2 ( pπ td / (2T ) ) 0

Mathematica verification A=Sequence[FourierParameters->{1,wo},

36

=

π 2

Assumptions->wo>0&&td>0&&td<Pi/wo]; FourierCoefficient[(UnitStep[t+td/2]-UnitStep[t](UnitStep[t]-UnitStep[t-td/2]))/(td/T),t,n,A]/.wo->2 Pi/T (b) Case 3: From Eq. (3.2) and Mathematica t /2 2t cos ( pπ td / T ) 1 d cp = cos(π t / td )e− j 2 pπ t/T dt = d ∫ T −t /2 π T 1− (2 ptd / T )2 d

c0 =

2td πT

⎡ 2t cos ( pπ td / T ) ⎤ td c p = lim ⎢ d ⎥= p→T /(2td ) π T 1− (2 pt / T ) 2 ⎢⎣ ⎥⎦ 2T d

and from Eq. (3.11) 0 ϕ n = tan −1 , g

g=

cos ( pπ td / T ) 1− (2 ptd / T )2

Mathematica verification A=Sequence[FourierParameters->{1,wo}, Assumptions->wo>0&&td>0&&td<Pi/wo]; FourierCoefficient[Sin[Pi t/td+Pi/2] (UnitStep[t+td/2]-UnitStep[t-td/2])/(td/T),t,n,A] /.wo->2 Pi/T (c) Case 4: From Eq. (3.2) and Mathematica t /2 − jtd sin ( pπ td / T ) 1 d − j 2 pπ t/T cp = sin(2 π t / t )e dt = d T −t∫/2 π T 1− ( ptd / T )2 d

c0 = 0

⎡ − jt sin ( pπ td / T ) ⎤ − jtd c p = lim ⎢ d ⎥= p→T /(td ) π T 1− ( pt / T ) 2 ⎢⎣ ⎥⎦ 2T d and from Eq. (3.11)

ϕ n = tan −1

−g , 0

g=

sin ( pπ td / T ) 1− ( ptd / T )2

Mathematica verification A=Sequence[FourierParameters->{1,wo}, Assumptions->wo>0&&td>0&&td<Pi/wo]; FourierCoefficient[Sin[2 Pi t/td] (UnitStep[t+td/2]-UnitStep[t-td/2])/(td/T),t,n,A] /.wo->2 Pi/T 37

(d) Case 5: From Eq. (3.2) and Mathematica t /2

0

1 1 d − j 2 pπ t/T cp = 2t / t + 1 e dt + ( ) (1− 2t / td ) e− j 2 pπ t/T dt d T −t∫/2 T ∫0 d

t sin ( pπ td / (2T ) ) T = 2 2 1− cos ( pπ td / T ) = d 2 p π td 2T ( pπ t / (2T ) )

(

2

)

d

c0 =

td 2T

and from Eq. (3.11)

ϕ n = tan −1

0 = 0, g

g >0

Mathematica verification A=Sequence[FourierParameters->{1,wo}, Assumptions->wo>0&&td>0&&td<Pi/wo]; Simplify[ComplexExpand[FourierCoefficient[ Piecewise[{{2 (t/td+1/2),-td/2<=t<0}, {2(1/2-t/td),0<=t<td/2}}]/(td/T),t,n,A]]]/.wo->2 Pi/T Section 3.3 3.8 Find the Fourier transform of the following waveform f (t) = h(1+ at) −1/ a ≤ t ≤ 0 = h(1− at) 0 ≤ t ≤1/ a =0 | t | >1/ a

What happens when h = a and a ® ¥? What does this result signify?

Solution: The Fourier transform is obtained from Mathematica as 0

1/a

−1/a

0

F(ω ) = h ∫ (1+ at)e− jω t dt + h ∫ (1− at)e− jω t dt 2ah = 2 (1− cos(ω / a)) ω When h = a, we note that

38

2 4 ⎞⎞ 2a 2 ⎛ ⎛ 1 ⎛ ω ⎞ 1 ⎛ω ⎞ F(ω ) ≈ 2 ⎜ 1− ⎜ 1− ⎜ ⎟ + ⎜ ⎟ −!⎟ ⎟ ω ⎝ ⎝ 2 ⎝ a ⎠ 24 ⎝ a ⎠ ⎠⎠

1 ⎛ω ⎞ ≈ 1− ⎜ ⎟ +!

24 ⎝ a ⎠ 2

Then,

lim F(ω ) → 1 a→∞

This is the Fourier transform of a delta function.

Mathematica verification sol=Simplify[ComplexExpand[FourierTransform[ h (1+a t) (UnitStep[t+1/a]-UnitStep[t]),t,w, FourierParameters->{1,-1}]+ FourierTransform[h (1-a t) (UnitStep[t]-UnitStep[t-1/a]), t,w,FourierParameters->{1,-1}]]] Limit[sol/.h->a,a->Infinity] 3.9 Determine the Fourier transform of the following expression

f (τ ) =

ω n2 e−ζω nτ sin (ω dτ ) u(τ ) ωd

where u(t) is the unit step function and

ω d = ω n 1− ζ 2 Present you results in terms of the non-dimensional quantity W = w/wn.

Solution: The Fourier transform is obtained as ∞

ω n2 −ζω nτ F(ω ) = e sin (ω dτ ) e− jωτ dτ ∫ ωd 0 =

ω n2 ω n2 = ω d2 + (ζω n + jω )2 ω d2 − ω 2 + ω n2ζ 2 + 2 jζωω n

ω n2 ω n2 − ω n2ζ 2 − ω 2 + ω n2ζ 2 + 2 jζωω n 1 F(Ω) = 2 1− Ω + 2 jζΩ =

In polar form, we have that

1− Ω2 + 2 jζΩ = r(Ω)e jϕ (Ω)

39

where

r(Ω) =

(1− Ω ) + ( 2ζΩ) 2 2

ϕ (Ω) = tan −1

2

2ζΩ 1− Ω2

Therefore,

F(Ω) =

1 − jϕ (Ω) e r(Ω)

Mathematica verification sol=FourierTransform[ wn^2/wd Exp[-z wn t] Sin[wd t]* UnitStep[t],t,w,FourierParameters->{1,-1}]; den=ComplexExpand[Denominator[sol]]; den=Simplify[den/.wd->wn Sqrt[1-z^2]]; fw=Simplify[(Numerator[sol]/den)/.w->Om wn] 3.10 Find the Fourier transform of the following waveform

f (τ ) = e−α |t|

α >0

Solution: The Fourier transform is ∞

F(ω ) = ∫ e

− α |t| − jω t

e

−∞

0

(α − jω )t

dt = ∫ e −∞

∞

dt + ∫ e−(α + jω )t dt 0

=

∞ 1 1 ⎡ e(α − jω )t ⎤ − ⎡ e−(α + jω )t ⎤ ⎦ −∞ (α + jω ) ⎣ ⎦0 (α − jω ) ⎣

=

1 1 α + jω + α − jω + = α2 +ω2 (α − jω ) (α + jω )

=

2α α +ω2

0

2

Mathematica verification FourierTransform[ Exp[-a Abs[t]],t,w, FourierParameters->{1,-1},Assumptions->a>0] 3.11 Find the Fourier transform of the following waveform

f (τ ) = cos α t −1 < t <1 f (τ ) = 0 otherwise Solution: The Fourier transform is

40

1

F(ω ) = ∫ cos(α t)e− jω t dt = −1

1

1

(

)

1 jα t − jα t − jω t ∫ e + e e dt 2 −1

1

1 1 = ∫ e j(α −ω )t dt + ∫ e− j(α +ω )t dt 2 −1 2 −1 +1

1 ⎡ e j(α −ω )t e− j(α +ω )t ⎤ = ⎢ − ⎥ 2 ⎣ j (α − ω ) j (α + ω ) ⎦ −1 = =

1 ⎡ e j(α −ω ) − e− j(α −ω ) e− j(α +ω ) − e j(α +ω ) ⎤ − ⎢ ⎥ 2j ⎣ α −ω α +ω ⎦

sin (α − ω ) sin (α + ω ) + α −ω α +ω

Mathematica verification sol=FourierTransform[Cos[a t] (UnitStep[t+1]-UnitStep[t-1]), t,w,FourierParameters->{1,-1}] 3.12 For the wave forms shown in Figure E3.3, use Eq. (3.34) to determine the magnitude of their Fourier transforms. Compare these results with the corresponding results obtained in Exercise 3.7 Solutions: (a) Case 2: From Eq. (3.34) 0

F(ω ) =

∫e

− jω t

−td /2

td /2

dt − ∫ e

t=0

− jω t

0

t=td /2

e− jω t e− jω t dt = − − jω t=−t /2 − jω t=0 d

(

)

(

j 2j 1− e jω td /2 + 1− e− jω td /2 = 1− cos (ω td / 2 ) ω ω 4j = sin 2 (ω td / 4 ) ω F(0) = 0 =

)

Comparison with Exercise 3.7:

F(ω ) sin 2 (ω t d / 4) = td ω td / 4

ω ≠0

F(ω ) =0 td

c0 =0 td / T

ω =0

cp td / T

=

sin 2 ( pπ t d / (2T )) p≠0 pπ t d / (2T )

Mathematica verification B=Sequence[FourierParameters->{1,-1}]; FourierTransform[(UnitStep[t+td/2]-UnitStep[t](UnitStep[t]-UnitStep[t-td/2]))/td,t,w,B]

41

(b) Case 3: From Eq. (3.34) and Mathematica

F(ω ) =

td /2

∫ cos (π t / t ) e

− jω t

d

dt = 2π td

−td /2

cos ( tdω / 2 )

π 2 − ω 2td2

2 π

F(0) = Comparison with Exercise 3.7:

cos ( t dω / 2 ) F(ω ) = 2π td π 2 − ω 2t d2 F(ω ) 2 = td π

ω =0

F(ω ) 1 = td 2

ω = π / td

cp

ω ≠ π / td

td / T

=

2 cos ( pπ t d / T ) π 1− ( 2 pt d / T )2

t d / T ≠ 1 / (2 p), p ≠ 0

c0 2 = td / T π cp td / T

=

1 2

t d / T = 1 / (2 p),

p≠0

Mathematica verification B=Sequence[FourierParameters->{1,-1}]; Simplify[FourierTransform[(Sin[Pi t/td+Pi/2]* (UnitStep[t+td/2]-UnitStep[t-td/2]))/td,t,w,B]] (c) Case 4: From Eq. (3.34) and Mathematica td /2

F(ω ) =

− jω t ∫ sin ( 2π t / td ) e dt = −4 jπ td

−td /2

sin ( tdω / 2 ) 4π 2 − ω 2td2

F(0) = 0 Comparison with Exercise 3.7:

sin ( t ω / 2 ) F(ω ) =π 2 d 2 2 td π − ω td / 4 F(ω ) =0 td

ω =0

F(ω ) 1 = td 2

ω = 2π / t d

ω ≠ 2π / t d

c0 =0 td / T cp td / T

cp td / T

=

1 sin ( pπ t d / T ) π 1− ( pt d / T )2

t d / T ≠ 1 / p,

p=0 =

1 2

t d / T = 1 / p,

p≠0

Mathematica verification B=Sequence[FourierParameters->{1,-1}]; Simplify[FourierTransform[Sin[2 Pi t/td]* (UnitStep[t+td/2]-UnitStep[t-td/2])/td,t,w,B]]

42

p≠0

(d) Case 5: From Eq. (3.34) and Mathematica 0

td /2

−td /2

0

F(ω ) = =

− jω t − jω t ∫ ( 2t / td + 1) e dt + ∫ (1− 2t / td ) e dt

(

)

4 8 1− cos (ω td / 2 ) = sin 2 (ω td / 4 ) 2 tdω tdω 2

t ⎛ sin (ω td / 4 ) ⎞ = d⎜ ⎟ 2 ⎝ ω td / 4 ⎠

2

t F(0) = d 2 Comparison with Exercise 3.7:

F(ω ) 1 ⎛ sin ( t dω / 4 ) ⎞ = ⎜ td 2 ⎝ t dω / 4 ⎟⎠ F(ω ) 1 = td 2

ω =0

2

cp

ω ≠0

td / T

=

1 sin ( pπ t d / (2T )) p≠0 2 ( pπ t d / (2T ))2 2

c0 1 = td / T 2

Mathematica verification B=Sequence[FourierParameters->{1,-1}]; Simplify[ComplexExpand[FourierTransform[ Piecewise[{{2 (t/td+1/2),-td/2<=t<0}, {2(1/2-t/td),0<=t<td/2}}]/td,t,w,B,Assumptions->td>0]]] Section 3.4.6 3.13 Consider the pulse

{

f (t) = Ap ⎡⎣u(t) − u(t − t p ) ⎤⎦ − ⎡⎣u(t − t p ) − u(t − 2t p ) ⎤⎦

}

where tp is the duration of the positive and the negative portions of the pulse and Ap is the magnitude of each portion. (a) Determine an expression for the magnitude of the Fourier transform of this pulse when a rectangular window function is used; that is, one with no weighting function. The results can be normalized to 2Aptp. (b) Determine an expression for the magnitude of the Fourier transform of this pulse when a Hamming window function is used.

Solutions: (a) The Fourier transform of the signal with the rectangular window is obtained from

43

tp

FR (ω ) = Ap ∫ e 0

2 ⎛1 ⎞ dt − Ap ∫ e− jω t dt = Apt p ⎜ ∫ e− jΩτ dτ − ∫ e− jΩτ dτ ⎟ ⎝0 ⎠ tp 1 2t p

− jω t

4 jApt p e− jΩ 2 = sin (Ω / 2) Ω where W = wtp. Therefore, FR (Ω) sin 2 (Ω / 2) sin 2 (Ω / 2) FR (Ω) = 2Apt p → = Ω/2 2Apt p Ω/2

Mathematica verification sol=FourierTransform[(UnitStep[t]-UnitStep[t-tp])(UnitStep[t-tp]-UnitStep[t-2 tp]),t,w, FourierParameters->{1,-1}] csol=ComplexExpand[w sol] PowerExpand[Sqrt[Simplify[Re[csol]^2+Im[csol]^2, Assumptions->{tp,w} Î Reals]]/w] (b) The Fourier transform of the signal with the Hamming window is obtained from tp

{

}

FH (ω ) = Ap ∫ a − b cos[2π t / (2t p )] e 0

2t p

− jω t

{

}

dt − Ap ∫ a − b cos[2π t / (2t p )] e− jω t dt tp

2 ⎛1 ⎞ − jΩτ = Apt p ⎜ ∫ {a − b cos[πτ ]} e dτ − ∫ {a − b cos[πτ ]} e− jΩτ dτ ⎟ ⎝0 ⎠ 1

Upon carrying out the integrations, we obtain

44

2 2 FH (ω ) aj − jΩ bjΩ =− e −1 + 2 e− jΩ + 1 2 Apt p Ω Ω −π

(

)

(

)

(−A (e − 1) + B (e + 1) ) Ω (Ω − π ) j = (−A e (e − e ) + B e (e Ω (Ω − π ) j

=

2

− jΩ

2

2

o

2

2

o

2

− jΩ

o

− jΩ

jΩ/2 2

− jΩ/2

− jΩ

o

)(

− jΩ/2

+ e jΩ/2

=

je− jΩ 2 2 −Ao ( −2 j sin(Ω / 2)) + Bo ( 2 cos(Ω / 2)) 2 2 Ω Ω −π

=

4 je− jΩ Ω Ω2 − π 2

=

2 je− jΩ ( Ao (1− cosΩ) + Bo (1+ cosΩ)) Ω Ω2 − π 2

=

( ( (

)) 2

)

A sin (Ω / 2) + B cos (Ω / 2)) )( 2

2

o

o

)

− jΩ

2 je ( Ao + Bo + ( Bo − Ao ) cosΩ) Ω Ω2 − π 2

(

)

where

(

Ao = a Ω2 − π 2 Bo = bΩ

)

2

Therefore, FH (ω ) ( Ao + Bo + ( Bo − Ao ) cosΩ ) = 2Apt p Ω Ω2 − π 2

(

)

Mathematica verification (*Integrations only*) a Simplify[Together[(Integrate[Exp[-I Om t],{t,0,1}]Integrate[Exp[-I Om t],{t,1,2}])]] -b (Integrate[Exp[-I Om t] (Exp[I Pi t]+Exp[-I Pi t]), {t,0,1}]-Integrate[Exp[-I Om t] (Exp[I Pi t]+ Exp[-I Pi t]),{t,1,2}])

45

Solutions to Exercises in Chapter 4 Section 4.1.1 4.1 Obtain the solution to the following first-order differential equations: (a)

2 x2

dy = x2 + y 2 dx

(b)

+ ye - cos y = 0 ( e + x sin y ) dy dx x

x

(c) 1/2

⎡ y⎤ dy y = − k ⎢1− ⎥ dx x ⎣ x⎦ (d)

(3y + 6xe + sin(2x)) dy + ( 4x + 6e + 2 y cos(2x)) dx = 0 2

y

3

y

(a)

2 x2

dy = x2 + y 2 dx

Solution Rearranging terms, we find dy 1 1 æ y ö = + ç ÷ dx 2 2 è x ø

2

If we let v = y/x,

dy dv =v+x dx dx then

v+x

dv 1 1 2 dv 1 1 1 = + v ® x = + v 2 - v = (v - 1)2 dx 2 2 dx 2 2 2

or

2dv dx 2dv dx -2 = ®ò =ò ® = ln x + C 2 2 (v - 1) x (v - 1) x v -1

46

Solving for v gives

v = 1-

2 ln x + C

y = x-

2x ln x + C

and, therefore,

Mathematica verification DSolveValue[2 x^2 y'[x]==x^2+y[x]^2,y[x],x] (b)

+ ye - cos y = 0 ( e + x sin y ) dy dx x

x

Solution Rearranging terms, we find

(e + x sin y ) dy + ( ye - cos y ) dx = 0 x

x

Therefore,

M ( x, y ) = ye x - cos y N ( x, y ) = e x + x sin y To determine if this equation is exact, we note that

¶M = e x + sin y ¶y ¶N = e x + sin y ¶x Therefore, the equation is exact and from Eq. (4.12),

u ( x, y ) = ò M ( x, y )dx + f ( y )

= ò ( ye x - cos y ) dx + f ( y ) = ye x - x cos y + f ( y ) = co

To determine f(y), we note that

47

¶u ( x, y ) = N ( x, y ) ¶y df ( y ) e x + x sin y + = e x + x sin y dy df ( y ) = 0 ® f ( y ) = c1 dy Then,

ye x - x cos y + c1 = co ye x - x cos y = C which is an implicit form.

Mathematica verification DSolve[(Exp[x]+x Sin[y[x]]) y'[x]+(y[x] Exp[x]-Cos[y[x]])==0 ,y,x]

(c) 1/2

⎡ ⎛ y⎞2⎤ dy y = − k ⎢1+ ⎜ ⎟ ⎥ dx x ⎢⎣ ⎝ x ⎠ ⎥⎦ Solution Setting v = y/x, we find that

dy dv =v+x dx dx Then, the differential equation becomes

dv = v − k 1+ v 2 dx 1 dv k =− x 1+ v 2 dx v+x

dv

k = − dx x 1+ v 2 Therefore,

48

dv

dx

∫ 1+ v = −k ∫ x 2

sinh −1 v = −k ln x + lnC

( ) x v + 1+ v ) = 1 C(

ln v + 1+ v 2 = −k ln x + lnC k

2

and

1⎛ xk ⎞ v + 1+ v 2 = Cx − k → v = ⎜ Cx − k − ⎟ 2⎝ C⎠ Then,

x⎛ xk ⎞ 1 ⎛ x1+k ⎞ y = ⎜ Cx − k − ⎟ = ⎜ Cx1−k − 2⎝ C ⎠ 2⎝ C ⎟⎠ Mathematica verification TrigToExp[DSolveValue[y'[x]==y[x]/x-k (1+(y[x]/x)^2)^(1/2), y[x],x]/.C[1]->Log[c]]

(d)

(3y + 6xe + sin(2x)) dy + ( 4x + 6e + 2 y cos(2x)) dx = 0 2

y

3

y

Solution We see that N (x, y) = 3y 2 + 6xe y + sin(2x) M (x, y) = 4x 3 + 6e y + 2 y cos(2x)

and, therefore,

∂N ∂M = 6e y + 2cos(2x) = ∂x ∂y Thus, we have an exact equation. Then, from Eq. (4.12), we have that

(

)

u(x, y) = ∫ 4x 3 + 6e y + 2 y cos(2x) dx + f ( y) = x 4 + 6xe y + y sin(2x) + f ( y) From Eq. (4.13),

49

6xe y + sin(2x) +

df ( y) = 3y 2 + 6xe y + sin(2x) dy df ( y) = 3y 2 dy

Upon integration, we obtain

f ( y) = y 3 + C and the solution is

x 4 + 6xe y + y sin(2x) + y 3 + C = 0 As a check, we differentiate this equation with respect to x and arrive at

dy dy dy + sin(2x) + 2 y cos(2x) + 3y 2 =0 dx dx dx 4x 3 + 6e y + 2 y cos(2x) dx + 6xe y + sin(2x) + 3y 2 dy = 0

4x 3 + 6e y + 6xe y

(

)

(

)

Mathematica verification DSolve[(2 y[x]^2+6 x Exp[y[x]]+Sin[2 x])y'[x]+ 4 x^3+6 Exp[y[x]]+2 y[x] Cos[2 x]==0,y[x],x] 4.2 For e ¹ 0, find the solution to

dθ + θ 1+ε = 0 dτ when the initial condition is q(0) = 1.

Solution: We rewrite the equation as −1/ ε dθ θ −ε = −d τ → − = −τ + C → θ = ⎡⎣ε (τ − C ) ⎤⎦ 1+ε ε θ

Using the initial conditions gives

θ (0) = 1 = ⎡⎣ −ε C ⎤⎦

−1/ ε

→ 1 = −ε C → C = −

1 ε

Thus,

θ = ⎡⎣ε (τ − C ) ⎤⎦

−1/ ε

→ ( ετ + 1)

−1/ ε

ε ≠0

Mathematica verification Simplify[DSolveValue[{y'[t]+y[t]^(1+e)==0, y[0]==1},y[t],t]]//Quiet

50

4.3 Given

⎛ dw 2 ⎞ − w cot ϕ = k ⎜ cosϕ − ⎝ dϕ 1+ cosϕ ⎟⎠ with the boundary condition w(jo) = 0. The parameter k is a constant. Show that

⎛ 1+ cosϕ ⎞ 1 1 w(ϕ ) = k ⎜ ln − + ⎟ sin ϕ ⎝ 1+ cosϕ o 1+ cosϕ 1+ cosϕ o ⎠ Solution: The solution is given by

w(ϕ ) =

1 ⎡ ρ (ϕ )q(ϕ ) dϕ + C ⎤⎦ ρ (ϕ ) ⎣ ∫

ρ (ϕ ) = exp ⎡⎣ ∫ p(ϕ ) dϕ ⎤⎦ where

p(ϕ ) = − cot ϕ ⎛ 2 ⎞ q(ϕ ) = k ⎜ cosϕ − ⎝ 1+ cosϕ ⎟⎠ Then

1 ρ (ϕ ) = exp ⎡⎣ ∫ p(ϕ ) dϕ ⎤⎦ = exp ⎡⎣ − ∫ cot ϕ dϕ ⎤⎦ = exp ⎡⎣ − lnsin ϕ ⎤⎦ = sin ϕ and

⎡ ⎤ 1 ⎛ 2 ⎞ w(ϕ ) = sin ϕ ⎢ k ∫ cos ϕ − d ϕ + C ⎥ ⎜ 1+ cosϕ ⎟⎠ ⎣ sin ϕ ⎝ ⎦ Performing the integration yields ⎛ sin ϕ ⎞ 1 ⎛ 2 ⎞ 1 cos ϕ − d ϕ = ln − ∫ sinϕ ⎜⎝ ⎟ ⎜ ⎟ 1+ cosϕ ⎠ ⎝ tan(ϕ / 2) ⎠ 1+ cosϕ

However,

tan(ϕ / 2) =

sin ϕ 1+ cosϕ

Therefore,

⎛ 1 ⎞ w(ϕ ) = k ⎜ ln (1+ cosϕ ) − sin ϕ + C sin ϕ 1+ cosϕ ⎟⎠ ⎝

51

Using the boundary condition, it is found that

⎛ ⎞ 1 w(ϕ o ) = k ⎜ ln (1+ cosϕ o ) − sin ϕ o + C sin ϕ o = 0 1+ cosϕ o ⎟⎠ ⎝ ⎛ ⎞ 1 C = −k ⎜ ln (1+ cosϕ o ) − 1+ cosϕ o ⎟⎠ ⎝ Then, the final result is

⎛ 1+ cosϕ ⎞ 1 1 w(ϕ ) = k ⎜ ln − + ⎟ sin ϕ ⎝ 1+ cosϕ o 1+ cosϕ 1+ cosϕ o ⎠ Mathematica verification Simplify[DSolveValue[{y'[x]-y[x] Cot[x]== k (Cos[x]-2/(1+Cos[x])),y[xo]==0},y[x],x]]

Section 4.1.2 4.4 Obtain the solution to the following equations: (a)

y2

dy = ex − y3 dx

(b)

dy − 2xy = 2x 3 y 2 dx (a)

y2

dy = ex - y3 dx

Solution When the terms are rearranged, this is a Bernoulli equation of the form

dy + y = e x y -2 dx Therefore, n = -2 and

p ( x) = 1

and

Consequently,

52

q ( x) = e x

P( x) = (1 - n) p( x) = 3 Q( x) = (1 - n)q( x) = 3e x and the equation to be solved is

du + 3u = 3e x dx where u = y1-n = y3.

The solution to this equation is obtained with the use of an integrating factor as determined by Eq. (4.22). Thus,

(

)

(

)

ρ (x) = exp ∫ P(x) dx = exp ∫ 3dx = exp(3x) Then from Eq. (4.29),

u = y3 =

1 é 1 r ( x)Q( x)dx + C ùû = 3 x éë ò e3 x 3e x dx + C ùû ò ë r ( x) e

é3 ù 3 = e -3 x ê e 4 x + C ú = e x + Ce -3 x ë4 û 4 1/3

æ3 ö y = ç e x + Ce-3 x ÷ è4 ø Mathematica verification

DSolveValue[y[x]^2 y'[x]==Exp[x]-y[x]^3,y[x],x]//Quiet

(b)

dy − 2xy = 2x 3 y 2 dx for the boundary condition y(0) = 1.

Solution This is a Bernoulli equation with n = 2 and with

p(x) = −2x q(x) = 2x 3 Then, from Eq. (4.29),

ρ (x) = exp ⎡⎣(1− n) ∫ p(x) dx ⎤⎦ = exp ⎡⎣ 2 ∫ x dx ⎤⎦ = e x and

53

2

u=

1 ⎡ (1− n) ∫ ρ (x)q(x) dx + C ⎤⎦ ρ (x) ⎣

(

)

2 2 2 2 = e− x ⎡ −2 ∫ e x x 3 dx + C ⎤ = e− x ⎡ e x 1− x 2 + C ⎤ ⎣ ⎦ ⎣ ⎦

= 1− x 2 + Ce− x

2

Then, since y = u1/(1-n), y = 1/u. Therefore,

y=

1 1− x + Ce− x 2

2

When x = 0, then y = 1 so that

y=

1 1− x + Ce− x 2

2

→ 1+ C = 1 ∴ C = 0

Therefore,

y=

1 1− x 2

x <1

Mathematica verification Simplify[DSolveValue[{y'[x]-2 x y[x]==2 x^3 y[x]^2,y[0]==1}, y[x],x]]

Section 4.2.2 4.5 Find the general solution to

d 2 ⎛ − βη d 2W ⎞ e − Ω4 e− βηW = 0 2 ⎜ 2 ⎟ dη ⎝ dη ⎠ in terms of trigonometric, hyperbolic, and exponential functions when W2 > b2/4.

Solution: Expand the equation and cancel e-bh to obtain 2 d 4W d 3W 2 d W − 2 β + β − Ω4W = 0 dη 4 dη 3 dη 2

Assume a solution of the form W = elh. Then

54

λ 4 − 2βλ 3 + β 2 λ 2 − Ω4 = 0

(

)

λ 2 λ 2 − 2βλ + β 2 − Ω4 = 0 λ 2 ( λ − β ) − Ω4 = 0 → λ ( λ − β ) = Ω2 2

and, therefore, 2

⎛β⎞ β λ − βλ − Ω = 0 → λ = ± ⎜ ⎟  Ω2 2 ⎝ 2⎠ 2

2

With the following definitions

δ=

β 2

ε p = Ω2 + δ 2 ε m = Ω2 − δ 2

Ω2 > δ 2

the four solutions are

λ1,2 = δ ± ε p λ3,4 = δ ± jε m Therefore,

(δ +ε )η + A e(δ −ε )η + A e(δ + jε )η + A e(δ − jε )η

W (η ) = A1e

(

p

p

2

ε η

= eδη A1e p + A2 e

(

p

3

− ε pη

p

4

jε η

+ A3e p + A4 e

− jε pη

)

= eδη C1 cosh ε pη + C2 sinh ε pη + C3 cos ε mη + C4 sin ε mη

)

Mathematica verification DSolveValue[D[Exp[-b x] D[w[x],x,x],x,x]-Om^4 Exp[-b x]* w[x]==0,w[x],x]/.{Sqrt[b^2-4 Om^2]->2 I em, Sqrt[b^2+4 Om^2]->2 I ep} Section 4.2.3 4.6 Find the solution to

d2y dy ( x - 1) + 3( x - 1) + y = 0 2 dx dx 2

Hint: Assume a solution of the form y = (x - 1)r to find the first solution and then use the reduction of order to find the complete solution.

Solution: 55

Assume a solution of the form y = (x - 1)r. Then

dy = r ( x - 1) r -1 dx d2y = r (r - 1)( x - 1) r - 2 2 dx Using these relations in the original equation gives

( x - 1)2 r (r - 1)( x - 1) r -2 + 3( x - 1)r ( x - 1) r -1 + ( x - 1) r = 0 r (r - 1) + 3r + 1 = 0 r 2 - r + 3r + 1 = 0 (r + 1)2 = 0 Thus, we have two equal roots: r = -1. Thus, the first solution is

y1 = C1 (x − 1)−1 To find the second solution, we use reduction of order given by Eq. (5.66). For our case,

y1 = ( x - 1) -1 a1 = 3( x - 1) a2 = ( x - 1) 2 Noting that

é a ( x) ù é ( x - 1) ù 1 exp ê -ò 1 dx ú = exp ê-3ò dx ú = exp [ -3ln( x - 1)] = 2 ( x - 1) ( x - 1)3 ë û ë a2 ( x) û Eq. (5.66) becomes

é a ( x) ù y = C1 y1 ( x) ò y1-2 ( x) exp ê - ò 1 dx ú dx + C2 y1 ( x) ë a2 ( x) û ( x - 1)2 = C1 ( x - 1) -1 ò dx + C2 ( x - 1) -1 3 ( x - 1) = C1 ( x - 1) -1 ln( x - 1) + C2 ( x - 1) -1 Mathematica verification DSolveValue[(x-1)^2 y''[x]+3 (x-1) y'[x]+y[x]==0,y[x],x] 4.7 Find the solution to

4x 2

(

)

d2y dy − 4x + 3− 16x 2 y = 0 2 dx dx

56

when given that y1 (x) = x1/2 e2 x .

Solution: Since a1(x) = -4x and a2(x) = 4x2, from Eq. (4.51) we find that the second independent solution is

y2 (x) = C2 x1/2 e2 x ∫

1

(x e ) 1/2 2 x

2

⎛ −4x ⎞ exp ⎜ − ∫ 2 dx ⎟ dx ⎝ ⎠ 4x

⎛ e−4 x ⎞ 1 ln x 1/2 2 x −4 x 1/2 2 x = C2 x e ∫ 4 x e dx = C2 x e ∫ e dx = C2 x e ⎜ − xe ⎝ 4 ⎟⎠ 1/2 2 x

= C3 x1/2 e−2 x Mathematica verification DSolveValue[4 x^2 y''[x]-4 x y'[x]+(-16 x^2+3) y[x]==0,y[x],x] 4.8 Find the solution to

(2x + 1)x

d2y dy − 2(2x 2 − 1) − 4(x + 1) y = 0 2 dx dx

when given that y1 (x) = x −1.

Solution: Since a1(x) = -2(2x2 - 1) and a2(x) = (2x+1)x, from Eq. (4.51) we find that the second independent solution is

⎡ (2x 2 − 1) ⎤ y2 (x) = C2 x −1 ∫ x 2 exp ⎢ 2 ∫ dx ⎥ dx x(2x + 1) ⎣ ⎦ ⎡ 2x + 1 ⎤ = C2 x −1 ∫ x 2 exp ⎢ 2x + ln 2 ⎥ dx x ⎦ ⎣ ⎛ 2x + 1⎞ = C2 x −1 ∫ x 2 e2 x ⎜ 2 ⎟ dx = C2 x −1 ∫ (2x + 1)e2 x dx ⎝ x ⎠

( )

= C2 x −1 xe2 x = C2 e2 x Mathematica verification DSolveValue[(2 x+1) x y''[x]-2 (2x^2-1) y'[x]-4 (x+1) y[x]==0, y[x],x]

Section 4.2.4

57

4.9 Find the solution to

x2

d2y − 2 y = −2 dx 2

with the boundary conditions y(e) = 0 and y(1) = 1 when e ¹ 0.

Solution: The equation is a Cauchy-Euler equation; therefore, we assume a solution of the form

yh = x r Substituting this expression into the homogeneous differential equation, we obtain

x 2 r(r − 1)x r−2 − 2x r = 0 → r 2 − r − 2 = 0 → r = −1,2 The particular solution can be determined by inspection to give yp = 1. Then, the total solution is

y = C1x −1 + C2 x 2 + 1 To determine the constants, we substitute this expression into the boundary conditions to arrive at

y(1) = 1 = C1 + C2 + 1→ C1 = −C2 y(ε ) = 0 = C1ε −1 + C2ε 2 + 1→ C1 =

−1 −ε = 2 ε −ε 1− ε 3 −1

Then

y = C1x −1 + C2 x 2 + 1

(

)

= C1 x −1 − x 2 + 1 =

ε x 2 − x −1 + 1 1− ε 3

(

)

Mathematica verification DSolveValue[{x^2 y''[x]-2 y[x]==-2,y[e]==0,y[1]==1},y[x],x] 4.10 Find the solution to

x2

d 2u du − 3x + 13u = 0 2 dx dx

when u(1) = -1 and du(1)/dx = 1.

Solution: We see that this is a Cauchy-Euler equation with a2 = 1, a1 = -3, and a0 = 13. Then, from Eq. (4.80)

58

r1,2 =

1⎡ 4 ± 16 − 4 × 13 ⎤ = 2 ± −9 = 2 ± 3 j ⎦ 2⎣

Using Eq. (4.60), the solution is

u = x 2 ⎡⎣ D1 cos(3ln x) + D2 sin(3ln x) ⎤⎦ Substituting this result into the boundary conditions

u(1) = −1 = D1 and

du = 2x ⎡⎣ D1 cos(3ln x) + D2 sin(3ln x) ⎤⎦ + 3x ⎡⎣ D2 cos(3ln x) − D1 sin(3ln x) ⎤⎦ dx = x ( 2D1 + 3D2 ) cos(3ln x) + ( 2D2 − 3D1 ) sin(3ln x)

{

}

Therefore,

du(1) = 2D1 + 3D2 = 1 → D2 = 1 dx The final result is

u = x 2 ⎡⎣sin(ln x 3 ) − cos(ln x 3 ) ⎤⎦ Mathematica verification DSolveValue[{x^2 u''[x]-3 x u'[x]+13 u[x]==0,u[1]==-1, u'[1]==1},u[x],x] 4.11 Find the solution to

⎤ d2y dy ⎡⎢ 2 ⎥y=0 + 2 + 1+ dx ⎢ (1+ 3x )2 ⎥ dx 2 ⎣ ⎦ [Hint: Assume a solution of the form y = ue-x and then use the change of variables 1 + 3x = et.]

Solution: We note that when y = ue-x.

⎛ du ⎞ dy du − x = e − ue− x = e− x ⎜ − u⎟ dx dx ⎝ dx ⎠ d 2 y d ⎛ du − x d 2u du − x du − x −x ⎞ = e − ue = − e − e + ue− x 2 2 ⎜ ⎟ dx ⎝ dx dx dx ⎠ dx dx ⎛ d 2u ⎞ du = e− x ⎜ 2 − 2 + u ⎟ dx ⎝ dx ⎠

59

Then

⎤ ⎛ d 2u ⎞ −x ⎛ du ⎞ − x ⎡⎢ du 2 ⎥ ue− x = 0 − 2 + u e + 2 − u e + 1+ 2 ⎜ ⎟ ⎜⎝ dx 2 ⎟ dx ⎝ dx ⎠ ⎢ (1+ 3x ) ⎥ ⎠ ⎣ ⎦ 2 d u 2 + u=0 2 dx (1+ 3x )2 Using the change of variable, we find that

dt 3 = dx 1+ 3x du du dt 3 du = = dx dt dx 1+ 3x dt d 2u d ⎛ du ⎞ dt du d 2t ⎛ d 2u dt ⎞ dt du d 2t = + = + dx 2 dx ⎜⎝ dt ⎟⎠ dx dt dx 2 ⎜⎝ dt 2 dx ⎟⎠ dx dt dx 2 t = ln (1+ 3x ) →

2

2

2

d 2u ⎛ dt ⎞ du d 2t ⎛ 3 ⎞ d 2u ⎛ 3 ⎞ du = 2⎜ ⎟ + = − dt dx 2 ⎜⎝ 1+ 3x ⎟⎠ dt 2 ⎜⎝ 1+ 3x ⎟⎠ dt dt ⎝ dx ⎠ Then,

d 2u 2 + u=0 2 dx (1+ 3x )2 2

2

⎛ 3 ⎞ d 2u ⎛ 3 ⎞ du 2 u=0 2 ⎜⎝ 1+ 3x ⎟⎠ dt 2 − ⎜⎝ 1+ 3x ⎟⎠ dt + 1+ 3x ( ) d 2u du 9 2 − 9 + 2u = 0 dt dt To solve this equation, we assume a solution of the form ert, and obtain

9r 2 − 9r + 2 = 0 The roots of this equation are

1⎡ 1 9 ± 81− 4 × 9 × 2 ⎤ = ( 9 ± 3) ⎣ ⎦ 18 18 1 2 r1 = , r2 = 3 3

r1,2 =

Therefore,

u = C1er1t + C2 er2t = C1et/3 + C2 e2t/3 Converting back to our original variables, we arrive at

60

(

)

y = ue− x = C1et/3 + C2 e2t/3 e− x

(

= C1e (

ln 1+3x )/3

= ⎛C1e ( ⎝

ln 1+3x )

(

1/3

+ C2 e

2/3ln(1+3x )

+ C2 e (

ln 1+3x )

2/3

)e

−x

⎞e− x ⎠

= C1 (1+ 3x ) + C2 (1+ 3x ) 1/3

2/3

)e

−x

Mathematica verification DSolveValue[y''[x]+2 y'[x]+(1 + 2/(1+3 x)^2) y[x]==0,y[x],x]

Section 4.2.5 4.12 Obtain a solution to

d 2w + w = −k1 − k2 cosϕ dϕ 2 and the boundary conditions w¢(0) = w¢(p/2) = 0.

Solution: The homogeneous solution is

wh = C1 cosϕ + C2 sin ϕ For the particular solution, we use Eq. (4.65) and assume a solution of the form

w p = A + Bϕ sin ϕ where we have chosen s = 1 since the homogeneous trigonometric function is also a solution to the homogeneous equation. Substituting this solution into the differential equation gives

B ( 2 cosϕ − ϕ sin ϕ ) + A + Bϕ sin ϕ = −k1 − k2 cosϕ 2B cosϕ + A = −k1 − k2 cosϕ

A + k1 + ( 2B + k2 ) cosϕ = 0 Therefore, A = -k1 and B = -k2/2 and the solution is

w = C1 cosϕ + C2 sin ϕ − k1 − (k2 / 2)ϕ sin ϕ Substituting this result into the boundary conditions gives

61

dw = −C1 sin ϕ + C2 cosϕ − (k2 / 2) ( sin ϕ + ϕ cosϕ ) dϕ dw(0) = C2 = 0 dϕ dw(π / 2) k k = −C1 − 2 = 0 → C1 = − 2 dϕ 2 2 Then, the final result is

w(ϕ ) = −k1 −

k2 ( cosϕ + ϕ sinϕ ) 2

Mathematica verification Simplify[DSolveValue[{y''[x]+y[x]==-k1-k2 Cos[x],y'[0]==0, y'[Pi/2]==0},y[x],x]] 4.13 Find the solution to the following boundary value problem.

d2y dy +2 + y= x 2 dx dx subject to y(0) = 0 and y(2) = 3.

Solution: We assume that the solution to the homogeneous equation is elx. Thus,

λ 2 + 2λ + 1 = 0 → λ1,2 =

(

)

1 −2 ± 4 − 4 = −1 2

Therefore, since we have repeating roots,

yh = C1e− x + C2 xe− x The particular solution is determined from inspection as yp = x - 2. Then

y = yh + y p = x − 2 + C1e− x + C2 xe− x From the boundary conditions at x = 0, we obtain

y(0) = −2 + C1 = 0 → C1 = 2 and from the boundary conditions at x = 2, we obtain

3 y(2) = C1e−2 + 2C2 e−2 = 3 → 2e−2 (1+ C2 ) = 3 → C2 = e2 − 1 2 Then the solution is

62

⎛3 ⎞ y = x − 2 + 2e− x + ⎜ e2 − 1⎟ xe− x ⎝2 ⎠ ⎛ ⎛3 ⎞ ⎞ = x − 2 + e− x ⎜ 2 + ⎜ e2 − 1⎟ x ⎟ ⎝2 ⎠ ⎠ ⎝ Mathematica verification Collect[Expand[DSolveValue[{y''[x]+2 y'[x]+y[x]==x, y[0]==0,y[2]==3},y[x],x]],{Exp[-x],x}] 4.14 Find the solution to

(

d 2w + β 2 w = −α η − η 2 2 dη

)

subject to the boundary conditions w(0) = w(1) = 0.

Solution: The solution to the homogeneous equation is

wh = Acos ( βη ) + Bsin ( βη ) To determine the particular solution, we use Eq. (4.64) and assume that

wp = c0 + c1η + c2η 2 Upon substitution into the differential equation, we arrive at

( ) 2c + β c + ( β c + α )η + ( β c − α )η = 0 2c2 + β 2 c0 + c1η + c2η 2 + αη − αη 2 = 0 2

2

2

0

2

1

2

2

Equating the coefficients of the powers of h to zero, we obtain

α β2 α β 2 c1 + α = 0 → c1 = − 2 β 2c 2α 2c2 + β 2 c0 = 0 → c0 = − 22 = − 4 β β β 2 c2 − α = 0 → c2 =

Then

wp = −

2α α α − 2 η + 2 η2 4 β β β

63

The solution is

w = wh + wp = Acos ( βη ) + Bsin ( βη ) −

2α α α − 2 η + 2 η2 4 β β β

Substituting the solution in to the boundary condition at h = 0 gives

w(0) = A −

2α 2α =0→ A= 4 4 β β

and that at h = 1 gives

2α α α − + =0 β4 β2 β2 2α 2α 2α Bsin ( β ) = − 4 cos ( β ) + 4 = 4 1− cos ( β ) β β β w(1) = Acos ( β ) + Bsin ( β ) −

(

B=

)

2α 1− cos ( β ) β 4 sin ( β )

The final result is

w=

⎞ 2α α 1− cos ( β ) 2α ⎛ α 2 cos βη + sin βη ( ) ( ) ⎜ ⎟ − 4 − 2η+ 2η 4 β ⎝ β β sin ( β ) ⎠ β

Mathematica verification Simplify[DSolveValue[{y''[x]+b^2 y[x]==-a (x-x^2),y[0]==0, y[1]==0},y[x],x]] 4.15 Determine the solution to

d2y dy + 6 + 9 y = e3 x 2 dx dx when the initial conditions are y(0) = 0 and y¢(0) = 6.

Solution: Assume a solution to the homogeneous equation as elx. Then,

1 λ 2 + 6λ + 9 = 0 → λ1,2 = ⎡ −6 ± 36 − 4 × 9 ⎤ = −3 ⎦ 2⎣ Since we have repeated roots, the solution to the homogeneous equation is

yh = C1e -3 x + C2 xe -3 x For the inhomogeneous solution, we assume a solution of the form

64

y p = Ae3x Thus, from Eq. (a)

A ( 9 + 3 ´ 6 + 9 ) e3 x = e3 x ® A =

1 36

Then the solution is

y = yh + y p = C1e

-3 x

+ C2 xe

-3 x

e3 x + 36

Substituting this solution into the boundary conditions, we obtain

1 1 = 0 ® C1 = 36 36 1 1 1 35 y¢(0) = -3C1 + C2 + = 6 ® C2 = 6 - - = 12 12 12 6 y (0) = C1 +

Therefore, the final solution is

y=-

e-3 x 35 xe-3 x e3 x + 36 6 36

Mathematica verification DSolveValue[{y''[x]+6 y'[x]+9 y[x]==Exp[3 x],y[0]==0, y'[0]==6},y[x],x] 4.16 Find the solution to 1 d ⎛ du ⎞ r ⎟ = −r k ⎜ r dr ⎝ dr ⎠

0<r

Solution: The homogeneous solution is

d ⎛ duh ⎞ r =0 dr ⎜⎝ dr ⎟⎠ duh c1 = dr r uh = c1 ln r + c2 The particular solution is obtained by assuming a solution of the form up = Arg. Then substitution yields

65

d ⎛ du ⎞ r = −r k+1 dr ⎜⎝ dr ⎟⎠

(

)

d rγ Ar γ −1 = −r k+1 dr d γ Ar γ = −r k+1 dr Aγ 2 r γ −1 = −r k+1

( )

Thus, we that for these quantities to be equal

Aγ 2 = −1 and γ − 1 = k + Therefore,

γ = k + 2 and A =

−1 −1 = 2 γ (k + 2)2

The complete solution is u = uh + u p = c2 + c1 ln r +

−r k+2 (k + 2)2

Mathematica verification DSolveValue[D[r u'[r],r]==-r^(k+1),u[r],r] Section 4.2.6 4.17 Use the variation of parameters to find the particular solution to

d2y dy 2 + 2 + y = 2 e− x 2 dx dx x Solution: We shall use the variation of parameters method. The homogeneous solution is obtained by assuming a solution of the form yh = elx. Then substitution yields

l 2 + 2l + 1 = 0 ® (l + 1)2 = 0 Thus, we have repeating roots so that

y1 = e- x y2 = xe- x yh ( x) = C1e- x + C2 xe- x and

66

y1

y2

e- x dy2 = - x -e dx

W [ y1 , y2 ] = dy1 dx

xe - x = e -2 x e - x (1 - x)

Then from Eq. (68) with a2 = 1

y ( x) = C1 y1 ( x) + C2 y2 ( x) - y1 ( x) ò

y2 ( x)Q( x) y1 ( x)Q( x) dx + y2 ( x) ò dx a2 ( x)W [ y1 , y2 ] a2 ( x)W [ y1 , y2 ]

-x -x xe - x e - x -x e e dx + 2 xe ò x2e-2 x dx x 2 e -2 x 1 1 = C1e - x + C2 xe - x - 2e - x ò dx + 2 xe - x ò 2 dx x x æ -1 ö = C1e - x + C2 xe - x - 2e - x ln x + 2 xe - x ç ÷ è x ø -x -x -x = C1e + C2 xe - 2e ( ln x + 1)

= C1e - x + C2 xe - x - 2e - x ò

Mathematica verification DSolveValue[y''[x]+2 y'[x]+y[x]==2 Exp[-x]/x^2,y[x],x] 4.18 Use the variation of parameters to find the particular solution to

d2y dy − 2 + y = 14x 3/2 e x 2 dx dx Solution: Assuming a solution of the y = elx, we find that l is determined from

1 λ1,2 = ⎡ 2 ± 4 − 4 ⎤ = 1 ⎦ 2⎣ Since we have repeating roots, the homogeneous solution is

yh = C1e x + C2 xe x and therefore,

y1 (x) = e x ,

y2 (x) = xe x

We shall use Eq (4.79) to find the particular solution. From the differential equation, we see that a2(x) = a0(x) = 1 and a1(x) = -2. First, we compute W using Eq. (4.77). Then,

W[ y1 , y2 ] =

ex ex

xe x = e2 x (1+ x) − xe2 x = e2 x x e (1+ x)

From Eq. (4.79), we have

67

x 3/2 x xe x x 3/2 e x x e x e dx + 14xe ∫ e2 x dx e2 x = −14e x ∫ x 5/2 dx + 14xe x ∫ x 3/2 d ξ

y p (x) = −14e x ∫

= −4x 7/2 e x +

28 7/2 x 8 7/2 x x e = x e 5 5

Mathematica verification DSolveValue[y''[x]-2 y'[x]+y[x]==14 x^(3/2) Exp[x],y[x],x] 4.19 Use the variation of parameters to find the particular solution to

x

d2y dy + 2(1− x) + (x − 2) y = e2 x 2 dx dx

where the independent solutions to the homogeneous equation are

y1 (x) = x −1e x

y1 (x) = e x , Solution:

We shall use Eq (4.79) to find the particular solution. From the differential equation, we see that a2(x) = x, a1(x) = 2(1-x), and a0(x) = x-2. First, we compute W using Eq. (4.77). Then,

ex W[ y1 , y2 ] = x e

x −1e x e x −x −2 + x −1

(

)

(

)

= e2 x −x −2 + x −1 − x −1e2 x = −x −2 e2 x

From Eq. (4.79), we have

x −1e x e2 x e x e2 x −1 x dx − x e ∫ xx −2e2 x dx xx −2 e2 x = e x ∫ e x dx − x −1e x ∫ xe x dx = e2 x − x −1e2 x (x − 1)

y p (x) = e x ∫

= x −1e2 x Mathematica verification DSolveValue[x y''[x]+2(1-x)y'[x]+(x-2) y[x]==Exp[2 x],y[x],x] 4.20 Use the variation of parameters to find the particular solution to

4x 2

(

)

d2y dy − 4x + 4x 2 + 3 y = x 7/2 2 dx dx

where the independent solutions to the homogeneous equation are

y1 (x) = x sin x,

y2 (x) = x cos x 68

Solution: We shall use Eq (4.79) to find the particular solution. From the differential equation, we see that a2(x) = 4x2, a1(x) = -4x, and a0(x) = 4x2+3. First, we compute W using Eq. (4.77). Then,

x sin x sin x x cos x + 2 x

W[ y1 , y2 ] =

x cos x cos x − x sin x + 2 x

⎛ ⎛ cos x ⎞ sin x ⎞ = x sin x ⎜ − x sin x + − x cos x x cos x + ⎟ ⎜⎝ ⎟ ⎝ 2 x⎠ 2 x⎠ = −x From Eq. (4.79), we have

y p (x) = − x sin x ∫ =

xx 7/2 cos x xx 7/2 sin x dx + x cos x ∫ 4x 2 (−x) dx 4x 2 (−x)

x x sin x ∫ x cos x dx − cos x ∫ x sin x dx 4 4

x x sin x ( cos x + x sin x ) − cos x ( sin x − x cos x ) 4 4 x 3/2 = 4 =

Mathematica verification DSolveValue[4 x^2 y''[x]-4 x y'[x]+(4 x^2+3) y[x]== x^(7/2),y[x],x]

Section 4.2.7 4.21 Convert the following system of nonlinear equations and boundary conditions to a system of first-order differential equations.

2

⎛ df ⎞ d3 f d2 f + 3 f − 2⎜ ⎟ + T * = 0 3 2 ⎝ dη ⎠ dη dη d 2*T * dT * + 3Pr f =0 dη dη 2 The initial conditions at h = 0 are

69

d2 f =a dη 2

f =0

df =0 dη

T* =1

dT * = −b dη

where a and b are constants.

Solution: We introduce the following set of dependent variables

y1 = f

y4 = T *

df y2 = dη

dT * y5 = dη

d2 f y3 = 2 dη Then, the system of first-order equations for these new variables is

dy1 dy4 = y2 = y5 dη dη dy2 dy5 = y3 = −3Pr y1 y5 dη dη dy3 = 2y22 − 3y1 y3 − y4 dη and the corresponding initial conditions become

y1 (0) = 0 y2 (0) = 0 y3 (0) = a

y4 (0) = 1 y5 (0) = −b

Section 4.2.10 4.22 Determine a solution to the nonlinear equation 2

2 ( x − 1) x ⎛⎜⎝ ddxy2 ⎞⎟⎠ − 1 = 0 2

2

Solution: We use Eq. (4.115) to obtain 2

⎞ ( x − 1) x ⎛⎜⎝ dp ⎟ −1 = 0 dx ⎠ 2

2

70

or

p=∫

dx x x2 − 1

= − tan −1

1 x2 − 1

+ c1

From Eq. (4.111), we find that

y = − ∫ tan −1

1 x2 − 1

dx + c1 x

Upon integrating, we obtain ⎛ 1 ⎞ 1 ⎛ y = c2 + c1 x − x tan −1 ⎜ 2 + ln 1− ⎝ x − 1 ⎟⎠ 2 ⎜⎝

x x2 − 1

⎞ 1 ⎛ ⎟⎠ − 2 ln ⎜⎝ 1+

x x2 − 1

⎞ ⎟⎠

Mathematica verification Simplify[PowerExpand[DSolveValue[(x^2-1) x^2 y''[x]^2-1==0, y[x],x]//Quiet]]

71

Solutions to Exercises in Chapter 5 Section 5.1.2 5.1 Use the power series method to obtain a solution to

(

1− x 2

)

d2y dy − 5x − 4 y = 0 2 dx dx

Solution: We see that this differential equation is of the form given by Eq. (5.22) with x0 = 0, a = -1, b = -5, and g = -4. Then, from Eq. (5.26)

p(n) = −n(n − 1) − 5n − 4 = −(n + 2)2 Then, Eq. (5.31) yields

c2n+2 =

2n + 2 c 2n + 1 2n

n = 0,1,2,... )

c2n+3 =

2n + 3 c 2n + 2 2n+1

n = 0,1,2,...

and Eq. (5.32) yields

Expanding these quantities for a few of the even coefficients, we get

c2 = 2c0 4 8 c4 = c2 = c0 3 3 6 6×8 16 c6 = c4 = c0 = c0 5 3× 5 5 8 16 × 8 128 c8 = c6 = c0 = c 7 5× 7 35 0 and expanding those for the odd coefficients, we get

3 c3 = c1 2 5 3× 5 15 c5 = c3 = c1 = c1 4 2×4 8 7 7 × 15 35 c7 = c5 = c1 = c1 6 6×8 16 9 9 × 35 315 c9 = c7 = c1 = c 8 8 × 16 128 1

72

Using Eq. (5.24), the solution is ⎛ ⎞ ⎛ ⎞ 8 16 128 8 3 15 35 y(x) = c0 ⎜ 1+ 2x 2 + x 4 + x 6 + x +!⎟ + c1 ⎜ x + x 3 + x 5 + x 7 +!⎟ 3 5 35 2 8 16 ⎝ ⎠ ⎝ ⎠

Mathematica verification Use procedure M5.1 5.2 Use the power series method to obtain a solution to

(

)

d2y dy 2 + 4x − 2x − 12(x − 1) − 12 y = 0 2 dx dx 2

Show that the solution can be written as ∞

∞ n n+1 2n + 1 n +1 x − 1 + c x − 1) ( ) 1∑ n n ( 2 n=0 n=0 2

y(x) = c0 ∑ Note that

2 + 4x − 2x 2 = 4 − 2(x − 1)2 Solution: We see that this differential equation is of the form given by Eq. (5.22) with x0 = 1, a = -1/2, b = -3, and g = -3. Then, from Eq. (5.26)

p(n) = −n(n − 1) / 2 − 3n − 3 = −

(

)

1 2 1 n + 5n + 6 = − (n + 3)(n + 2) 2 2

Then, Eq. (5.31) yields

c2n+2 = =

(2n + 3)(2n + 2) c 2(2n + 2)(2n + 1) 2n 2n + 3 c 2(2n + 1) 2n

n = 0,1,2,...

and Eq. (5.32) yields

c2n+3 = =

(2n + 4)(2n + 3) c 2(2n + 3)(2n + 2) 2n+1 n+2 c 2(n + 1) 2n+1

n = 0,1,2,...

n = 0,1,2,...

Expanding these quantities for a few of the even coefficients, we get

73

3 c2 = c0 2 5 5× 3 5 c4 = c2 = c0 = c0 6 6×2 4 7 7×5 7 c6 = c2 = c0 = c0 10 10 × 4 8 !

2n + 1 c2n = c n = 0,1,2,...

2n 0 and expanding those for the odd coefficients, we get

c3 = c1 3 3 c5 = c3 = c1 4 4 4 4×3 4 c7 = c5 = c1 = c1 6 6×4 8 5 5× 4 5 c9 = c7 = c1 = c1 8 8×8 16 ! n +1 c2n+1 = n c1 n = 0,1,2,...

2 Using Eq. (5.24), the solution can be written as ∞

∞ n n+1 2n + 1 n +1 x − 1 + c x − 1) ( ) 1∑ n n ( 2 n=0 n=0 2

y(x) = c0 ∑ or in expanded form

2 3 ⎛ 3 ⎞ 5 7 y(x) = c0 ⎜ 1+ ( x − 1) + ( x − 1) + ( x − 1) +!⎟ 4 8 ⎝ 2 ⎠ 2 3 3 ⎛ ⎞ 3 4 + c1 ⎜ ( x − 1) + ( x − 1) + ( x − 1) + ( x − 1) +!⎟ 4 8 ⎝ ⎠

Mathematica verification coe=ConstantArray[0,12]; coe[[1]]=1; coe[[2]]=3/2; coe[[3]]=1; Do[coe[[2 n+2]]=(2 n+3)/(2 (2 n+1)) coe[[2 n]],{n,1,4}]; Do[coe[[2 n+3]]=( n+2)/(2 (n+1)) coe[[2 n+1]],{n,1,4}]; y=c0 (1+Total[Table[coe[[2 n]] (x-1)^(2 n),{n,1,4}]]) +c1 Total[Table[coe[[2n -1]] (x-1)^(2 n-1),{n,1,4}]]

74

Section 5.1.3 5.3 Use the method of Frobenius to show that the solution to

4x

d2y dy +2 + y =0 2 dx dx

is

y = C1 cos

( x ) + C sin ( x ) 2

Solution: We assume a solution of the form ¥

y ( x) = å cn x n + r n =0

Substitution leads to ¥

¥

¥

n =0

n =0

4 xå cn (n + r )(n + r - 1) x n + r - 2 + 2å cn (n + r ) x n + r -1 + å cn x n + r = 0 n =0

¥

å c [ 4(n + r )(n + r - 1) + 2(n + r )] x n =0

n + r -1

n

¥

å c éë2(n + r ) {2(n + r ) - 1}ùû x n =0

¥

+ å cn x n + r = 0 n =0

n + r -1

n

¥

+ å cn x n + r = 0 n =0

¥

¥

n =1

n =0

2r (2r - 1)c0 x r -1 + å cn éë 2(n + r ) {2(n + r ) - 1}ùû x n + r -1 + å cn x n + r = 0 ¥

2r (2r - 1)c0 x r -1 + å cn +1 éë 2(n + 1 + r ) {2(n + 1 + r ) - 1}ùû + cn x n + r = 0 n=0

Hence, the indicial equation is

2r (2r - 1) = 0 ® r = 0 and r = 1/ 2 The roots are unequal and their difference in not equal to an integer.

The solution for r = 0 is

75

cn+1 = −

cn 2(n + 1) {2(n + 1) − 1}

c1 = −

c0 2(1)

c2 = −

c1 c0 = (4)(3) (4)(3)(2)(1)

c3 = −

c2 c0 =− (6)(5) (6)(5)(4)(3)(2)(1)

n = 0,1,2...

 (−1) n c0 cn = (2n)!

The solution for r = 1/2 is

d n+1 = −

dn dn =− (2n + 3)(2n + 2) 2(n + 3 / 2) {2(n + 3 / 2) − 1}

d1 = −

d0 (3)(2)

d2 = −

d1 d0 = (5)(4) (5)(4)(3)(2)

d3 = −

d2 d0 =− (7)(6) (7)(6)(5)(4)(3)(2)

! dn =

(−1) n d0 (2n + 1)!

Therefore, the complete solution is ¥

¥

y ( x) = å cn x n + å d n x n +1/2 n =0

n =0

¥ (-1) x n (-1) n x n + d 0 x1/2 å n = 0 (2n)! n = 0 (2 n + 1)! ¥

= c0 å

n

But, ∞

(−1) n x 2n+1 sin x = ∑ n=0 (2n + 1)! ∞

(−1) n x 2n cos x = ∑ (2n)! n=0 and, hence,

76

∞

(−1) n x n n=0 (2n + 1)!

sin x = x1/2 ∑ ∞

(−1) n x n n=0 (2n)!

cos x = ∑ and the solution can be written as

y(x) = C1 sin x + C2 cos x Mathematica verification DSolveValue[4 x y''[x]+2 y'[x]+y[x]==0,y[x],x] 5.4 Use the method of Frobenius to obtain the solution to

d2y dy x − 3x + ( 4 − x ) y = 0 2 dx dx 2

To place your solution a compact form, note the following:

3 ⎛ 1⎞ = 1+ , 2 ⎜⎝ 2 ⎟⎠

11 ⎛ 1 1 ⎞ = 1+ + , 6 ⎜⎝ 2 3 ⎟⎠

⎛ 1 1 1⎞ g(n) = ⎜ 1+ + +!+ ⎟ n⎠ ⎝ 2 3

25 ⎛ 1 1 1 ⎞ = 1+ + + !

12 ⎜⎝ 2 3 4 ⎟⎠

(define this function as stated)

Solution: Comparing this equation with Eq. (5.78), it is seen that

α 0 = 1,

α 1 = 0,

α2 = 0

β0 = −3,

β1 = 0,

β2 = 0

γ 0 = 4,

γ 1 = −1,

γ2 =0

Then, from Eq. (5.89)

p0 (r) = r(r − 1) − 3r + 4 = (r − 2)(r − 2) = 0 p1 (r) = −1 p2 (r) = 0 From the indicial equation, p0(r) = 0, we find that r0 = r1 = r2 = 2 and, therefore, we have two equal roots. Hence, the solutions are given by Eqs. (5.101) and (5.104); that is, ∞

y1 (x) = ∑ an (r0 )x n+r0 n=0

∞

y2 (x) = y1 (x)ln x + ∑ an′ (r0 )x n+r0 n=1

77

where, from Eq. (5.97)

a0 (r) = 1 a1 (r) = −

p1 (r) 1 = p0 (1+ r) (r − 1)2

an (r) = −

p1 (n + r − 1)an−1 (r) + p2 (n + r − 2)an−2 (r) p0 (n + r)

=

an−1 (r) an−1 (r) = p0 (n + r) (n + r − 2)2

n≥2

Then

a0 (2) = 1 a1 (2) = 1 a1 (2) 1 = 2 22 2 a (2) 1 a3 (2) = 2 2 = 2 2 3 23 a (2) 1 a4 (2) = 3 2 = 2 2 2 4 234 !

1 an (2) = (n!)2 a2 (2) =

and y1(x) is ∞

y1 (x) = ∑ an (2)x n=0

n+2

∞

x n+2 = a0 ∑ = a0 x 2 I 0 (2 x ) 2 n=0 (n!)

and we have used Eq. (5.152).

The derivatives are

a0′ (r) = 0 a1′(r) =

−2 (r − 1)3

an′ (r) = − =

2an−1 (r) an−1 ′ (r) + 3 (n + r − 2) (n + r − 2)2

1 ′ (r) ) ( −2an−1(r) + (n + r − 2)an−1 (n + r − 2)3

Then

78

n≥2

a0′ (2) = 0 a1′(2) = −2 1 ( −2a1(2) + 2 × a1′(2)) 23 1 3 1 ⎛ 1⎞ = 3 ( −2 × 1+ 2 × (−2) ) = − 2 = − ⎜ 1+ ⎟ 2 ⎝ 2⎠ 2 2

a2′ (2) =

1 ( −2a2 (2) + 3× a2′ (2)) 33 ⎛ 3 ⎞⎞ 1⎛ 1 11 1 ⎛ 1 1⎞ = 3 ⎜ −2 × 2 + 3× ⎜ − 2 ⎟ ⎟ = − 2 3 = − 1+ + ⎝ 2 ⎠⎠ 3 ⎝ 2 23 2 × 32 ⎜⎝ 2 3 ⎟⎠

a3′ (2) =

1 ( −2a3 (2) + 4 × a3′(2)) 43 ⎛ 11 ⎞ ⎞ 1⎛ 1 50 1 ⎛ 1 1 1⎞ = 3 ⎜ −2 × 2 2 + 4 × ⎜ − 2 3 ⎟ ⎟ = − 2 3 3 = − 1+ + + ⎝ 2 3 ⎠⎠ 4 ⎝ 23 2 34 2 × 32 42 ⎜⎝ 2 3 4 ⎟⎠

a4′ (2) =

! ⎛ 1 1 2 1⎞ an′ (2) = − 2 2 2 1+ + +"+ ⎟ 2 ⎜ n⎠ 1 × 2 × 3 "n ⎝ 2 3 =−

2 g(n) (n!)2

where n

1 k=1 k

g(n) = ∑ Then, from Eq. (5.104)

∞

g(n) n+2 x 2 n=1 (n!)

y2 (x) = y1 (x)ln x − 2∑ Mathematica verification

DSolveValue[x^2 y''[x]-3 x y'[x]+(4-x) y[x]==0,y[x],x] 5.5 Use the method of Frobenius to obtain the solution to

d2y dy + ( x + 36x 2 ) y = 0 ( 5 + x + 10x ) x dx 2 + ( 4 + 3x + 48x 2 ) x dx 2

2

Solution: Comparing this equation with Eq. (5.78), it is seen that

79

α 0 = 5,

α 1 = 1,

α 2 = 10

β0 = 4,

β1 = 3,

β 2 = 48

γ 0 = 0,

γ 1 = 1,

γ 2 = 36

Then, from Eq. (5.81)

p0 (r) = 5r(r − 1) + 4r = r 2 − r = r(5r − 1) p1 (r) = r(r − 1) + 3r + 1 = r 2 + 2r + 1 = (r + 1)2 p2 (r) = 10r(r − 1) + 48r + 36 = 10r 2 + 38r + 36 = 10(r + 2)(r + 9 / 5) From the indicial equation, p0(r) = 0, we find that r1 =1/5 and r2 = 0 and since r1 - r2 = 1/5 is not a positive integer the roots will be distinct. Hence, the solutions are given by Eqs. (5.92) and (5.93); that is, ∞

y1 (x) = ∑ an (1/ 5)x n+1/5 n=0 ∞

y2 (x) = ∑ an (0)x n n=0

Using Eq. (5.88), we find that

a0 (r) = 1 a1 (r) = −

p1 (r) (r + 1)2 (r + 1) =− =− p0 (1+ r) (r + 1)(5r + 4) (5r + 4)

an (r) = −

p1 (n + r − 1)an−1 (r) + p2 (n + r − 2)an−2 (r) p0 (n + r)

=−

(n + r)2 an−1 (r) + 10(n + r)(n + r − 1/ 5)an−2 (r) (n + r)(5n + 5r − 1)

=−

(n + r)an−1 (r) + 10(n + r − 1/ 5)an−2 (r) (5n + 5r − 1)

Then, for r = 1/5

80

n≥2

a0 (1/ 5) = 1 a1 (1/ 5) = −

(r + 1) 6/5 6 =− =− (5r + 4) 5 25

a2 (1/ 5) = −

(2 + 1/ 5)(−6 / 25) + 10(2 + 1/ 5 − 1/ 5) (5 × 2 + 5 / 5 − 1)

(11/ 5)(−6 / 25) + 20 1217 =− 10 625 (3+ 1/ 5)(−1217 / 625) + 10(3)(−6 / 25) a3 (1/ 5) = − 15 (16 / 5)(−1217 / 625) − 36 / 5 41972 =− = 15 46875 =−

and for r = 0

a0 (0) = 1 a1 (0) = −

(r + 1) 1 =− (5r + 4) 4

2(−1/ 4) + 10(9 / 5) 35 =− 9 18 3(−35 / 18) + 10(14 / 5)(−1/ 4) −35 / 6 − 7 11 a3 (0) = − =− = 14 14 12

a2 (0) = −

Using Eq. (c), we find that

⎛ 6 1217 2 41972 3 ⎞ y1 (x) = x1/5 ⎜ 1− x − x + x ...⎟ 625 46875 ⎝ 25 ⎠ 1 35 11 y2 (x) = 1− x − x 2 + x 3 ...

4 18 12 Mathematica verification p[j_,r_]:=r (r-1) alp[[j+1]]+r bet[[j+1]]+gam[[j+1]] c[k_,r_]:=(a=Array[0,k]; a[[1]]=1; a[[2]]=-p[1,r]/p[0,r+1]; Do[a[[n+1]]=-(p[1,n+r-1] a[[n]]+p[2,n+r-2] a[[n-1]])/p[0,n+r], {n,2,k-1}];a) alp={5,1,10};

bet={4,3,48};

gam={0,1,36};

rj=Sort[r/.Solve[p[0,r]==0,r],Greater]; k=5;

81

xa=Table[x^n,{n,0,k-1}]; x^rj[[1]] Total[c[k,rj[[1]]] xa] x^rj[[2]] Total[c[k,rj[[2]]] xa] 5.6 Use the method of Frobenius to obtain the solution to

d2y dy + ( −x + 2x 2 ) y = 0 (1+ 2x + x ) x dx 2 + (1+ 3x + 4x 2 ) x dx 2

2

Solution: Comparing this equation with Eq. (5.78), it is seen that

α 0 = 1,

α 1 = 2,

α2 = 1

β0 = 1,

β1 = 3,

β2 = 4

γ 0 = 0,

γ 1 = −1,

γ2 =2

(a)

Then, from Eq. (5.81)

p0 (r) = r(r − 1) + r = r 2 p1 (r) = 2r(r − 1) + 3r − 1 = 2r 2 + r − 1 = (2r − 1)(r + 1)

(b)

p2 (r) = r(r − 1) + 4r + 2 = r + 3r + 2 = (r + 1)(r + 2) 2

From the indicial equation, p0(r) = 0, we find that r1 = r2 = 0 and we have the case of equal roots. Hence, the solutions are given by Eqs. (5.92) and (5.95); that is, ∞

y1 (x) = ∑ an (0)x n n=0

∞

y2 (x) = y1 (x)ln x + ∑ an′ (0)x

(c) n

n=1

where the prime denotes the derivative with respect to r. Using Eq. (5.88), we find that

a0 (r) = 1 a1 (r) = −

p1 (r) (2r − 1)(r + 1) (2r − 1) =− =− 2 p0 (1+ r) (1+ r) (1+ r)

an (r) = −

(2n + 2r − 3)(n + r)an−1 (r) + (n + r − 1)(n + r)an−2 (r) (n + r)2

=−

(2n + 2r − 3)an−1 (r) + (n + r − 1)an−2 (r) (n + r)

Then, for r = 0, Eq. (d) gives

82

n≥2

(d)

a0 (0) = 1 a1 (0) = 1 (4 − 3) + 1 = −1 2 (6 − 3)(−1) + (3− 1) −3+ 2 1 a3 (0) = − =− = 3 3 3

a2 (0) = −

(e)

To determine a¢n, we note that

a0′ (r) = 0 a1′(r) = −

d ⎛ (2r − 1) ⎞ 3 =− ⎜ ⎟ dr ⎝ (1+ r) ⎠ (1+ r)2

an′ (r) = −

d ⎛ (2n + 2r − 3)an−1 (r) + (n + r − 1)an−2 (r) ⎞ ⎟⎠ dr ⎜⎝ (n + r)

=−

(f)

1 ( a (r) + 3an−1(r)) (n + r)2 n−2 −

1 ′ (r) + (n + r − 1) an−2 ′ (r) ) ((2n + 2r − 3)an−1 n+r

Then, from Eq. (f)

a0′ (0) = 0 a1′(0) = −3 an′ (r) = −

1 1 a (0) + 3an−1 (0) ) − ( (2n − 3) an−1 ′ (0) + (n − 1) an−2 ′ (0) ) 2 ( n−2 n n

Therefore,

a0′ (0) = 0 a1′(0) = −3 1 1 a (0) + 3a1 (0) ) − ( (4 − 3) a1′(0) + (2 − 1) a0′ (0) ) 2( 0 2 2 1+ 3 1 3 1 =− − (−3) = −1+ = 4 2 2 2 1 1 a3′ (0) = − 2 ( a1 (0) + 3a2 (0) ) − ( (6 − 3) a2′ (0) + (3− 1) a1′(0) ) 3 3 ⎞ 2 3 31 1 1⎛ ⎛ 1⎞ = − (1+ 3(−1) ) − ⎜ 3⎜ ⎟ + 2(−3)⎟ = + = 9 3⎝ ⎝ 2⎠ ⎠ 9 2 18

a2′ (0) = −

Using Eqs. (e) and (g) in Eq. (c), we obtain

83

(g)

x3 ...

3 ⎛ ⎞ x 31 y2 (x) = y1 (x)ln x + x ⎜ −3+ + x 2 ...⎟ 2 18 ⎝ ⎠ y1 (x) = 1+ x − x 2 +

Mathematica verification p[j_,r_]:=r (r-1) alp[[j+1]]+r bet[[j+1]]+gam[[j+1]] c[k_,r_]:=(a=Array[0,k]; a[[1]]=1; a[[2]]=-p[1,r]/p[0,r+1]; Do[a[[n+1]]=-(p[1,n+r-1] a[[n]]+p[2,n+r-2]* a[[n-1]])/p[0,n+r],{n,2,k-1}]; da=Array[0,k]; da[[1]]=0; da[[2]]=D[-p[1,v]/p[0,v+1],v]/.v->r; Do[da[[n+1]]=-(p[1,n+r-1] da[[n]]+p[2,n+r-2] da[[n-1]])/ p[0,n+r]+D[-(p[1,n+v-1] a[[n]]+p[2,n+v-2] a[[n-1]])/ p[0,n+v],v]/.v->r,{n,2,k-1}]; {a,da}) alp={1,2,1};bet={1,3,4};gam={0,-1,2}; rj=Sort[r/.Solve[p[0,r]==0,r],Greater]; k=5; (*k term solution*) xa=Table[x^n,{n,0,k-1}]; {a,da}=c[k,rj[[1]]]; x^rj[[1]] Total[a xa] x^rj[[1]] Total[da xa] 5.7 Use the method of Frobenius to obtain the solution to

d2y dy (1+ x ) x 2 − ( 3 + 10x ) x + 30xy = 0 dx dx 2

Solution: Comparing this equation with Eq. (5.78), it is seen that

α 0 = 1,

α 1 = 1,

α2 = 0

β0 = −3,

β1 = −10,

β2 = 0

γ 0 = 0,

γ 1 = 30,

γ2 =0

(b)

Then, from Eq. (5.81)

p0 (r) = r(r − 1) − 3r = r 2 − 4r = r(r − 4) p1 (r) = r(r − 1) − 10r + 30 = r 2 − 11r + 30 = (r − 6)(r − 5) p2 (r) = 0

84

(c)

From the indicial equation, p0(r) = 0, we find that r1 = 4 and r2 = 0 and r1 - r2 = 4 is a positive integer. In addition, a2 = b2 = g2 = 0; hence, this is Case A of Case 3. The solutions are given by Eqs. (5.92) and (5.101); that is, ∞

y1 (x) = ∑ an (r1 )x n+r1 n=0

k−1

y2 (x) = ∑ an (r2 )x n=0

n+r2

(e)

∞ ⎛ ⎞ + C ⎜ y1 (x)ln x + ∑ an′ (r1 )x n+r1 ⎟ ⎝ ⎠ n=1

where the prime denotes the derivative with respect to r and from Eqs. (5.100) and (5.102) we have that

a0 (r) = 1 an (r) = −

p1 (n + r − 1)an−1 (r) (n + r − 7)(n + r − 6)an−1 (r) =− p0 (n + r) (n + r)(n + r − 4)

n ≥1

(f)

and

C=−

p1 (r1 − 1) (r − 7)(r1 − 6) (4 − 7)(4 − 6) ak−1 (r2 ) = − 1 a3 (r2 ) = − a3 (0) kα 0 4 4

3 = − a3 (0) 2

(g)

Then, for r = r1 = 4,

a0 (4) = 1 a1 (4) = −

(1+ 4 − 7)(1+ 4 − 6)a0 (4) (−2)(−1)(1) 2 =− =− (1+ 4)(1+ 4 − 4) 5 5

(2 + 4 − 7)(2 + 4 − 6)a2 (r) a2 (4) = − =0 (2 + 4)(2) an (r) = 0

n≥2

and for r = r2 = 0,

a0 (0) = 1 an (0) = −

(n − 7)(n − 6)an−1 (0) n(n − 4)

Therefore,

85

1≤ n ≤ k −1= 3

(h)

a0 (0) = 1 a1 (0) = −

(1− 7)(1− 6)a0 (0) (−6)(−5)(1) =− = 10 (1− 4) −3

a2 (0) = −

(2 − 7)(2 − 6)a1 (0) (−5)(−4)(10) =− = 50 2(2 − 4) −4

a3 (0) = −

(3− 7)(3− 6)a2 (0) (−4)(−3)(50) =− = 200 3(3− 4) 3(−1)

(i)

Thus, from Eq. (g),

3 3 C = − a3 (0) = − (200) = −300 2 2

(j)

We now use eq. (f) to determine a¢n(4) as follows

a0′ (4) = 0 an′ (4) = −

d ⎛ (n + r − 7)(n + r − 6)an−1 (r) ⎞ ⎟⎠ dr ⎜⎝ (n + r)(n + r − 4) r=4

(k)

( n − n − 14n + 24n) a′ (4) + 3a (4) (3n − 4n − 8) =− 4

3

2

2

n−1

n−1

n2 (n + 4)2

n ≥1

Then, using Eqs. (k) and (h)

a0′ (4) = 0 a1′(4) = −

(1− 1− 14 + 24) (0) + 3(1) (3− 4 − 8) = 27 (1+ 4)2

25

( 2 − 2 − (14)2 + 24(2)) (27 / 25) + 3(−2 / 5) (32 − 4(2) − 8) a′ (4) = − 4

3

2

2

22 (2 + 4)2

2

(16 − 8 − 56 + 48) (27 / 25) − (6 / 5) (1024 − 16) = − 1 =− 4(36)

a3′ (4) = − an′ (4) = 0

30

(3 − 3 − (14)3 + 24(3)) (−1/ 30) + 3(0) ((3)3 − 4(3) − 8) = 0 4

3

2

2

32 (3+ 4)2

n≥3

Using Eqs. (g), (i), (j), and (k), Eq. (e) becomes

86

(k)

∞ 2 y1 (x) = ∑ an (4)x n+4 = x 4 − x 5 5 n=0 3 ∞ ⎛ ⎞ y2 (x) = ∑ an (0)x n + C ⎜ y1 (x)ln x + ∑ an′ (4)x n+4 ⎟ ⎝ ⎠ n=0 n=1

⎛ 27 5 1 6 ⎞ = 1+ 10x + 50x 2 + 200x 3 − 300 ⎜ y1 (x)ln x + x − x ⎟ 25 30 ⎠ ⎝ Mathematica verification tes[arg_]:=Expand[(1+x) x^2 D[arg,x,x]-(3+10 x) x D[arg,x] +30 x arg] arg=x^4-2/5 x^5; tes[arg] arg2=1+10 x+50 x^2+200x^3-300(arg Log[x]+27/25 x^5-1/30 x^6); tes[arg2] Section 5.1.8 5.8 A fourth-order differential equation can be factored into the following two second-order differential equations d ⎛ n+1 dW ⎞ ϕ ± λ 2ϕ nW = 0 ⎜ ⎟ dϕ ⎝ dϕ ⎠

Determine the complete solution; that is, one that contains four independent solution functions.

Solution: Expanding the equation gives

ϕ n+1 ϕ2

d 2W dW + (n + 1)ϕ n ± λ 2ϕ nW = 0 2 dϕ dϕ

d 2W dW + (n + 1)ϕ ± λ 2ϕW = 0 2 dϕ dϕ

Comparing this equation with case 2 of Table 5.2, we see that

b 2 k 2 x 2b = ± λ 2ϕ → b = 1 / 2 and k = 2 λ or k = 2 j λ n + 1 = 1− 2ab → a = −n r = 0 → p2 = a2 → p = n and, therefore, from this table

87

W+ λ (ϕ ) = ϕ − n/2 ⎡C1 J n (2λ ϕ ) + C2Yn (2λ ϕ ) ⎤ ⎣ ⎦ W− λ (ϕ ) = ϕ − n/2 ⎡C3 J n (2 jλ ϕ ) + C4Yn (2 jλ ϕ ) ⎤ ⎣ ⎦ = ϕ − n/2 ⎡C3 I n (2λ ϕ ) + C4 K n (2λ ϕ ) ⎤ ⎣ ⎦ Thus,

W (ϕ ) = ϕ − n/2 ⎡C1 J n (2λ ϕ ) + C2Yn (2λ ϕ ) + C3 I n (2λ ϕ ) + C4 K n (2λ ϕ ) ⎤ ⎣ ⎦ Mathematica verification DSolveValue[x^2 w''[x]+(n+1) x w'[x]+lam^2 x w[x]==0,w[x],x] DSolveValue[x^2 w''[x]+(n+1) x w'[x]-lam^2 x w[x]==0,w[x],x] 5.9 Obtain the solution to the following equation that remains finite at x = 0.

x2

d2y dy + 2x + h 2 xy = 0 2 dx dx

Solution: Comparing this equation with case 2 of Table 5.2, we see that

b2 k 2 x 2b = h2 x → b = 1/ 2 and k = 2h 1− 2ab = 2 → 1− a = 2 → a = −1 In addition, since r = 0, p = 1. Hence, from the table, the solution that remains finite at x = 0, is

y(x) = C1 (2h)−1 x −1/2 J1 (2hx1/2 ) To show that this solution is finite at x = 0, we use Eq. (5.148). Thus 1

−1

lim y(x) = C1 (2h) x

−1/2

x→0

C 1 ⎛ 2hx1/2 ⎞ = 1 ⎜ ⎟ 1! ⎝ 2 ⎠ 2

Mathematica verification DSolveValue[x^2 w''[x]+2 x w'[x]+h^2 x w[x]==0,w[x],x] 5.10 Given the equation

x2

d2y dy + x - ( jx 2 + n 2 ) y = 0 2 dx dx

where n is an integer. Show that one solution to this equation is

y = Bern x + jBei n x where

88

(-1) m+ n ( x / 2) 2 m + n cos (p (2m + n) / 4 ) m !(m + n)! m=0 ¥

Bern x = å

(-1) m+ n +1 ( x / 2) 2 m+ n sin (p (2m + n) / 4 ) m !(m + n)! m=0 ¥

Bei n x = å

The quantities Ber and Bei, respectively, are known as the real and imaginary parts of the Kelvin function of the first kind. [Hint: The exponential form of -j will prove beneficial.]

Solution: From Case 2 of TA ble 5.2, we see that k2 = -j, a = 0, b = 1, and p = n. Then,

k 2 = - j = e jp e jp /2 ® k = e jp /2e jp /4 = e j 3p /4 and one solution is

y( x) = xab éëC1 J p (kxb ) + C2Yp (kxb ) ùû ® J n ( xe j 3p /4 ) From Eq. (5.86),

(−1) m (x / 2)2m+n (e j3π /4 )2m+n m!(m + n)! m=0 ∞

J n (xe j3π /4 ) = ∑ However,

e j 3p /4 = e jp e j (3p /4-p ) = e jp e- jp /4

(e

)

j 3p /4 2 m + n

= e jp (2 m+ n ) e- jp (2 m+ n )/4 = e j 2 mp e jnp e- jp (2 m+ n )/4 = (-1)n e- jp (2 m+ n )/4

Therefore,

(-1) m ( x / 2) 2 m + n (-1) n e- jp (2 m + n )/4 m !(m + n)! m=0 ¥

J n ( xe j 3p /4 ) = å

(-1) m+ n ( x / 2) 2 m+ n écos (p (2m + n) / 4 ) - j sin (p (2m + n) / 4 ) ùû m !(m + n)! ë m=0 ¥

=å Hence,

(-1)m+n ( x / 2)2m+n écos (p (2m + n) / 4 ) - j sin (p (2m + n) / 4 )ùû m!(m + n)! ë m=0 ¥

y = Bern x + jBei n x = å or

(-1) m+ n ( x / 2) 2 m + n Bern x = å cos (p (2m + n) / 4 ) m !(m + n)! m=0 ¥

(-1) m+ n +1 ( x / 2) 2 m+ n Bei n x = å sin (p (2m + n) / 4 ) m !(m + n)! m=0 ¥

89

5.11 Consider the equation

x2

(

)

(

)

d2y dy + a + 2bxα x + c + γ 2 x 2δ + b(a + α − 1)xα + b2 x 2α y = 0 2 dx dx

where a, b, c, a, b, and g are real constants. Make the two transformations sequentially α

y = x (1−a)/2 e− bx /α u(x) and z =

γ δ x δ

and show that the resulting equation is a Bessel equation. Use Mathematica to obtain dy/dx and d2y/dx2. It will be found that the order p of the Bessel function is 1/2

2 ⎫⎪ 1 ⎧⎪⎛ a − 1⎞ p = ⎨⎜ − c ⎬ δ ⎪⎝ 2 ⎟⎠ ⎩ ⎭⎪

Solution: From Mathematica,

dy 1 ( -1- a )/2 -bxa /a é du ù = x e 2x + (1 - 2bxa - a ) u ú ê dx 2 ë dx û d 2 y 1 ( -3- a )/2 - bxa /a é 2 d 2u du = x e 4x + (1 - 2bxa - a ) 4 x ê 2 2 dx 4 dx dx ë

+ ( -1 + 4abxa + 4b 2 x 2a + a 2 - 4a bxa ) u ùû

Using this result in the original equation, we obtain

⎡ α 1 d 2u du x 2 x (−3−a)/2 e− bx /α ⎢ 4x 2 2 + 1− 2bxα − a 4x 4 dx dx ⎣

(

)

(

)

+ −1+ 4abxα + 4b2 x 2α + a 2 − 4α bxα u ⎤⎦

(

+ x a + 2bxα

(

) 12 x

(−1−a)/2 − bxα /α

e

⎡ du ⎤ α ⎢ 2x dx + 1− 2bx − a u ⎥ ⎣ ⎦

(

)

)

α

+ c + γ 2 x 2δ + b(a + α − 1)xα + b2 x 2α x (1−a)/2 e− bx /α u = 0

x2

d 2u du ⎡ 1 + ⎡⎣ a + 2bxα + 1− 2bxα − a ⎤⎦ x + ⎢ −1+ 4abxα + 4b2 x 2α + a 2 − 4α bxα 2 dx ⎣ 4 dx

(

+

⎤ 1 a + 2bxα 1− 2bxα − a + c + γ 2 x 2δ + b(a + α − 1)xα + b2 x 2α ⎥ u = 0 2 ⎦

(

)(

) (

)

Performing further simplification, we arrive at

90

)

2 ⎛ a − 1⎞ ⎤ d 2u du ⎡ 2 2δ ⎥u = 0 x + x + ⎢c + γ x − ⎜ dx ⎢ ⎝ 2 ⎟⎠ ⎥ dx 2 ⎣ ⎦ 2

However,

du du dz = dx dz dx 2

d 2u d 2u ⎛ dz ⎞ du d 2 z = + dz dx 2 dx 2 dz 2 ⎜⎝ dx ⎟⎠ Therefore, since

dz = γ xδ −1 dx d 2z = γ (δ − 1)xδ −2 2 dx we find that

du du dz du = = γ xδ −1 dx dz dx dz 2

2 2 d u d 2u d 2u ⎛ dz ⎞ du d 2 z du δ −1 = + = γ x + γ (δ − 1)xδ −2 2 2 ⎜ 2 2 ⎟ dz dx dz dx dz ⎝ dx ⎠ dz

(

)

Hence, 1−2/δ ⎡ ⎛ δ z ⎞ 2−2/δ d 2u ⎛δz⎞ du ⎤ 2 ⎢γ ⎜ ⎟ ⎥ + γ ( δ − 1) ⎜⎝ γ ⎟⎠ dz ⎥ dz 2 ⎢⎣ ⎝ γ ⎠ ⎦ 1/δ 1−1/δ 2 2 ⎛δz⎞ ⎡ ⎛δz⎞ ⎛ a − 1⎞ ⎤ du ⎤ ⎡ 2 ⎛ δ z ⎞ ⎥ + ⎢γ ⎥u = 0 + ⎜ ⎟ ⎢γ ⎜ ⎟ +c−⎜ dz ⎥ ⎢ ⎜⎝ γ ⎟⎠ ⎝ 2 ⎟⎠ ⎥ ⎝γ ⎠ ⎢ ⎝γ ⎠ ⎣ ⎦ ⎣ ⎦

⎛δz⎞ ⎜⎝ γ ⎟⎠

2/δ

Upon simplification, we find that 2 ⎛ a − 1⎞ ⎤ d 2u du ⎡ 2 2 ⎥u = 0 δ z + ⎡δ − 1+ 1⎤⎦ δ z + ⎢δ z + c − ⎜ dz ⎢ ⎝ 2 ⎟⎠ ⎥ dz 2 ⎣ ⎣ ⎦ 2 2 ⎫⎪ ⎤ du ⎡ 2 1 ⎧⎪⎛ a − 1⎞ 2 d u ⎢ z + z + z − − c ⎨ ⎬⎥ u = 0 dz ⎢ dz 2 δ 2 ⎪⎜⎝ 2 ⎟⎠ ⎪⎭ ⎥⎦ ⎩ ⎣ d 2u du z 2 2 + z + z 2 − p2 u = 0 dz dz 2 2

(

where

91

)

2 ⎫⎪ 1 ⎧⎪⎛ a − 1⎞ p = 2 ⎨⎜ − c ⎬ δ ⎪⎝ 2 ⎟⎠ ⎪⎭ ⎩ 2

Then the solution is α

y(x) = x (1−a)/2 e− bx /α ⎡⎣ AJ p (γ xδ / δ ) + AYp (γ xδ / δ ) ⎤⎦ Mathematica verification FullSimplify[PowerExpand[DSolveValue[ x^2 y''[x]+(a+2 b x^alp) x y'[x]+(c+gam^2 x^(2 d)+ b (a+alp-1) x^alp+b^2 x^(2 alp)) y[x]==0, y[x],x] ]]/.{Sqrt[d^2 ((-1+a)^2-4 c)]->2 p d^2} 5.12 Using the appropriate relationship in Section 5.3.3, determine the solution to

d 2 y β o dy α −β + + k 2 x( o o ) y = 0 2 x dx dx Solution: We use Case 6 of Table 5.2; that is,

x2

(

)

(

)

d2y dy + a + 2bxα x + c + γ 2 x 2δ + b(a + α − 1)xα + b2 x 2α y = 0 2 dx dx

Multiplying the original equation by x2, we obtain

x2

d2y dy α − β +2 + βo x + k 2 x( o o ) y = 0 2 dx dx

Upon comparing these two equations, we find that

a = βo ,

b = 0,

c = 0,

γ = k,

2δ = α o − β o + 2

Therefore,

δ=

α o − βo + 2 2

Then, from case 6 of Table 5.2, we obtain

z = kxδ / δ , β −1 p= o 2δ and, therefore,

y(x) = x (1−βo )/2 ⎡⎢ AJ ( β −1)/(2δ ) (kxδ / δ ) + AY( β −1)/(2δ ) (kxδ / δ ) ⎤⎥ o o ⎣ ⎦ Mathematica verification

92

del=(ao-bo+2)/2; w=x^((1-bo)/2) BesselJ[(bo-1)/(2 del),k x^del/del]; FullSimplify[x^2 D[w,x,x]+bo x D[w,x]+k^2 x^(ao-bo+2) w]

93

Solutions to Exercises in Chapter 6 Section 6.1.3 6.1 Place the following equations into Sturm-Liouville form and state whether or not the solutions will be orthogonal functions, subject, of course, to the appropriate boundary conditions. (a) Chebyshev’s equation

+α y = 0 (1− x ) ddxy − x dy dx 2

2

2

2

(b) Euler’s equation

x2

d2y dy + bx + cy = 0 2 dx dx

(c) Hermite’s equation

d2y dy − 2x + 2α y = 0 2 dx dx (d) Laguerre’s equation

x

d2y dy + (1− x) + α y = 0 2 dx dx

Solution (a) Chebyshev’s equation

(

)

d2y dy 1− x − x +α2y = 0 2 dx dx 2

Upon comparing this equation with Eq. (6.33), we see that a0(x) = 1 - x2, a1(x) = -x, a2(x) = 0, a3(x) = 1, and l = a2. Then using Eqs. (6.9) and (6.15), we find that

(

)

⎡ ⎤ x p(x) = exp ⎢ ∫ 2 dx ⎥ = exp ln x 2 − 1 = x 2 − 1 ⎣ x −1 ⎦ q(x) = 0 r(x) =

x2 − 1 1 =− 2 1− x x2 − 1

Then, the Sturm-Liouville form is d ⎛ 2 dy ⎞ x −1 ⎟ − ⎜ dx ⎝ dx ⎠

94

λ x2 − 1

y=0

Since r(x) is less than zero, the functions will not be orthogonal. (b) Euler’s equation

x2

d2y dy + bx + cy = 0 2 dx dx

Upon comparing this equation with Eq. (6.33), we see that a0(x) = x2, a1(x) = bx, a2(x) = 0, a3(x) = 1, and l = c. Then using Eqs. (6.9) and (6.15), we find that

⎡ x ⎤ p(x) = exp ⎢ b ∫ 2 dx ⎥ = exp ln x b = x b ⎣ x ⎦ q(x) = 0

(

r(x) =

)

xb = x b−2 2 x

Then, the Sturm-Liouville form is d ⎛ b dy ⎞ x + x b−2 cy = 0 dx ⎜⎝ dx ⎟⎠

Since r(x) is greater than zero, the functions will be orthogonal with respect to this weight function. (c) Hermite’s equation

d2y dy − 2x + 2α y = 0 2 dx dx Upon comparing this equation with Eq. (6.33), we see that a0(x) = 1, a1(x) = -2x, a2(x) = 0, a3(x) = 1, and l = 2a. Then using Eqs. (6.9) and (6.15), we find that

p(x) = exp ⎡⎣ −2 ∫ x dx ⎤⎦ = e− x

2

q(x) = 0 r(x) = e− x

2

Then, the Sturm-Liouville form is d ⎛ − x 2 dy ⎞ − x2 e + 2e αy=0 dx ⎜⎝ dx ⎟⎠

Since r(x) is greater than zero, the functions will be orthogonal with respect to this weight function. (d) Laguerre’s equation

d2y dy x 2 + (1− x) + α y = 0 dx dx 95

Upon comparing this equation with Eq. (6.33), we see that a0(x) = x, a1(x) = 1 - x, a2(x) = 0, a3(x) = 1, and l = a. Then using Eqs. (6.9) and (6.15), we find that

⎡ 1− x ⎤ p(x) = exp ⎢ ∫ dx ⎥ = exp[ln x − x] = xe− x x ⎣ ⎦ q(x) = 0 xe− x r(x) = = e− x x Then, the Sturm-Liouville form is d ⎛ − x dy ⎞ xe + α e− x y = 0 ⎜ ⎟ dx ⎝ dx ⎠

Since r(x) is greater than zero, the functions will be orthogonal with respect to this weight function.

Section 6.1.2 6.3 Consider the Cauchy-Euler equation

x2

d2y dy + x + ( λ + 2) y = 0 2 dx dx

where it is assumed that l + 2 > 0. The boundary conditions are dy(1)/dx = 0 and dy(2)/dx = 0.

Determine the eigenvalues and eigenfunctions.

Solution We start with Eq. (4.54) with a0 = l + 2, a1 = 1, and a2 = 1. Then, from Eq. (4.57)

r1,2 = ± j λ + 2 This is case three, whose solution is given by Eq. (4.60) with b = 0, and d = λ + 2. Then,

y(x) = D1 cos

( λ + 2 ln x ) + D sin ( λ + 2 ln x ) 2

To satisfy the boundary conditions, we note that

dy(x) λ+2 = − D1 sin dx x

( λ + 2 ln x ) + D λx+ 2 cos( λ + 2 ln x ) 2

Then, from the boundary conditions dy(1)/dx = 0, we find

D2 λ + 2 cos

( λ + 2) = 0 → D = 0 2

and from the boundary condition dy(2)/dx = 0, we find

96

− D1

λ+2 sin 2

( λ + 2 ln 2) = 0 → sin ( λ + 2 ln 2) = 0

Therefore, 2

⎛ nπ ⎞ λ + 2 ln 2 = nπ → λ = λ n = ⎜ −2 ⎝ ln 2 ⎟⎠ where ln is the eigenvalue and the corresponding eigenfunction, is ⎛ nπ ⎞ yn (x) = cos ⎜ ln x ⎟ ⎝ ln 2 ⎠

97

Solutions to Exercises in Chapter 7 Sections 7.2.2 7.1 Solve the Laplace equation

∂ 2 Ψ ∂Ψ ∂ 2 Ψ + + =0 ∂r 2 r ∂r r 2 ∂θ 2 with the boundary conditions on the semicircular annulus that are given by

Ψ(r,0) = Ψ(r, π ) = 0 Ψ(a,θ ) = sin θ Ψ(b,θ ) = 0

b>a

Solution: From Eq. (T7.3c) of Table 7.3, the solution is given by

(

)

Ψ(r,θ ) = Cr m + Dr − m ( Acos mθ + Bsin mθ ) Using the first set of boundary conditions,

(

)

Ψ(r,0) = A Cr m + Dr − m = 0

(

)

Ψ(r, π ) = Cr m + Dr − m ( Acos mπ ) = 0 Therefore, A = 0 and the solution reduces to

(

)

Ψ(r,θ ) = Cr m + Dr − m sin mθ

m≠0

where the constant B has been subsumed in the other constants.

The boundary condition at r = a and r = b give

( ) Ψ(b,θ ) = ( Cb + Db ) sin mθ = 0

Ψ(a,θ ) = Ca m + Da − m sin mθ = sin θ −m

m

Therefore, to satisfy the second equation,

D = −Cb2m To satisfy the first equation, Ca + Da −1 = 1 Ca + Da m

−m

=0

Hence, a solution only exists for m = 1. Then

98

m=1 m = 2,3,...

a a − b2 −ab2 D= 2 2 a −b

C=

2

and the solution is

Ψ(r,θ ) =

ar 2 − ab2

(

r a 2 − b2

)

sin θ

It is seen that the boundary conditions are satisfied.

7.2 Consider the rectangular region 0 £ x £ a and 0 £ y £ b that is governed by the Laplace equation

∂2 u ∂2 u + =0 ∂x 2 ∂ y 2 with the following boundary conditions

∂u(0, y) = 0, ∂x ∂u(x,0) = 0, ∂y

∂u(a, y) =0 ∂x u(x,b) = α + β x

Determine an expression for u.

Solution: From Eq. (T7.3c) of Table 7.3, the solution is given by

u(x, y) = ( a0 + b0 x ) ( c0 + d0 y ) + ( Acos( λ x) + Bsin( λ x) ) ( C cosh( λ y) + D sinh( λ y) ) We note that

∂u = b0 ( c0 + d0 y ) + λ ( − Asin( λ x) + Bcos( λ x) ) ( C cosh( λ y) + D sinh( λ y) ) ∂x ∂u = d0 ( a0 + b0 x ) + λ ( Acos( λ x) + Bsin( λ x) ) ( C sinh( λ y) + D cosh( λ y) ) ∂y Then the boundary conditions yield

∂u(0, y) = b0 ( c0 + d0 y ) + λ B ( C cosh( λ y) + D sinh( λ y) ) = 0 ∂x ∂u(a, y) = b0 ( c0 + d0 y ) + λ ( − Asin( λ a) + Bcos( λ a) ) ( C cosh( λ y) + D sinh( λ y) ) = 0 ∂x ∂u(x,0) = d0 ( a0 + b0 x ) + λ D ( Acos( λ x) + Bsin( λ x) ) = 0 ∂y

99

(a)

Thus, to satisfy these for arbitrary y and x, respectively, we see that b0 = B = d0 = D = 0 and

sin( λ a) = 0 → λ → λ n =

nπ a

n = 1,2,...

Then Eq. (a) reduces to

un (x, y) = a0 + Acos( λ n x)cosh( λ n y) Therefore, the total solution is ∞

u(x, y) = a0 + ∑ An cos( λ n x)cosh( λ n y)

(b)

n=1

To determine An, we use the last boundary condition to obtain ∞

u(x,b) = α + β x = a0 + ∑ An cos( λ n x)cosh( λ n b)

(c)

n=1

We first integrate Eq. (c) with respect to x over the interval {0,a} to obtain a

a

0

0

a

∞

∫ (α + β x ) dx = ∫ a dx + ∑ A cosh(λ b)∫ cos(λ x) dx 0

n

n

n=1

n

0

1 α a + β a 2 = aa0 2 1 a0 = α + β a 2 Next, we multiply Eq. (c) by cos(lmx) and integrate over the region to obtain a

a

∞

a

0

n=1

0

∫ (α + β x ) cos(λm x) dx = a0 ∫ cos(λm x) dx + ∑ An cosh(λnb)∫ cos(λn x)cos(λm x) dx 0

aβ a −1+ (−1) n = ∑ δ mn An cosh( λ n b) 2 2 2 n=1 nπ −4aβ An = 2 2 n = 1,3,5,... n π cosh( λ n b) 2

(

)

∞

The complete solution is 1 4aβ ∞ cos( λ n x)cosh( λ n y) u(x, y) = α + β a − 2 ∑ 2 π n=1,3,5 n2 cosh( λ n b)

7.3 Referring to Figure 7.5, obtain a solution to

∂ 2 u ∂u ∂2 u + + =0 ∂r 2 r ∂r r 2 ∂θ 2

0<a<r <b

100

0 < θ < α < 2π

(a)

over the annular sector indicated and shown in Figure 7.5 with the following boundary conditions

u(r,0) = 0 u(r,α ) = uo

(b)

u(a,θ ) = 0 u(b,θ ) = 0 Solution From Eq. (T7.3c) of Table 7.3, we assume a solution of the form

(

)

u(r,θ ) = a0 + b0θ + Cr m + Dr − m ( Acos mθ + Bsin mθ )

(c)

Upon substituting Eq. (c) into the first boundary conditions given in Eq. (c), we arrive at

(

)

a0 + A Cr m + Dr − m = 0

Therefore, for arbitrary r, ao = A = 0. Then Eq. (c) reduces to

(

)

u(r,θ ) = b0θ + Cr m + Dr − m sin mθ

(d)

Upon substituting Eq. (d) into the second boundary conditions given in Eq. (c), we arrive at

(

)

b0α + Cr m + Dr − m sin mα = uo

Therefore, for arbitrary r, bo = uo/a and

sin mα = 0 → m = nπ / α

n = 1,2,3,...

and the corresponding orthogonal eigenfunction is sin(npq/a). Then, Eq. (d) becomes ∞

(

)

u(r,θ ) = uoθ / α + ∑ Cn r nπ /α + Dn r − nπ /α sin ( nπθ / α ) n=1

(e)

We now substitute Eq. (e) into the last two boundary conditions given by Eq. (b) and obtain ∞

(

)

(

)

uoθ / α + ∑ Cn a nπ /α + Dn a − nπ /α sin ( nπθ / α ) = 0 n=1 ∞

uoθ / α + ∑ Cn bnπ /α + Dn b− nπ /α sin ( nπθ / α ) = 0 n=1

(f)

To determine Cn and Dn, we multiply each equation in Eq. (f) by sin(kpq/a) and integrate from 0 to a and obtain

4uo θ nπ α 4u θ Cn bnπ /α + Dn b− nπ /α = − o nπ α

Cn a nπ /α + Dn a − nπ /α = −

where we have used the facts that

101

(f)

α

2α

∫ sin ( nπθ / α ) dθ = nπ

n = 1,3,5,...

0

α

α

∫ sin ( nπθ / α ) dθ = 2 2

0

Solving for Cn and Dn, we obtain

4uo θ 1 nπ /α nπ α a + bnπ /α 4u θ (ab) nπ /α Dn = − o nπ α a nπ /α + bnπ /α Cn = −

(g)

Substituting Eq. (g) into Eq. (e) gives

u(r,θ ) =

⎫⎪ θ uo ⎧⎪ 4 ∞ (r / a) nπ /α + (r / b)− nπ /α 1− sin n πθ / α ( ) ⎨ ⎬ ∑ α ⎪ π n=1,3,5 n 1+ (b / a) nπ /α ⎪⎭ ⎩

(

)

(e)

7.4 Obtain a solution to the following system 1 ∂ ⎛ ∂u ⎞ ∂ 2 u r + =0 r ∂r ⎜⎝ ∂r ⎟⎠ ∂z 2

0<r <a

0<z<b

(a)

with the boundary conditions

du(a, z) =0 dr u(r,b) = g(r)

u(0, z) = finite, u(r,0) = f (r),

(b)

Solution From Eq. (T7.3d) of Table 7.3, the solution to Eq. (a) is

u(r, z) = ( a0 + b0 z ) + ( C0 J 0 (κ r) + D0Y0 (κ r) ) ( E cosh(κ z) + F sinh(κ z) )

(c)

To satisfy the first boundary condition of Eq. (b), we must set D0 = 0. Then, Eq. (c) simplifies to

u(r, z) = a0 + b0 z + ( E cosh(κ z) + F sinh(κ z) ) J 0 (κ r)

(d)

Upon substituting Eq. (d) into the second boundary condition of Eq. (b), we arrive at

− ( E cosh(κ z) + F sinh(κ z) )κ J1 (κ a) = 0 For Eq. (e) to be satisfied for arbitrary z, ln = ka, n = 1, 2, … must be solutions to

J1 ( λ n ) = 0 Thus, k = ln/a is the eigenvalue and the corresponding orthogonal eigenvector is

J1 ( λ n r / a) 102

(e)

which has a weighting factor equal to r.

Then, Eq. (d) becomes ∞

u(r, z) = a0 + b0 z + ∑ ( En cosh( λ n z / a) + Fn sinh( λ n z / a) ) J 0 ( λ n r / a)

(f)

n=1

Upon substituting Eq. (f) in to the second set of boundary conditions given in Eq. (b), we obtain ∞

∑ E J (λ r / a) = f (r) − a n

0

n

0

n=1

(g)

∞

∑ ( E cosh(λ b / a) + F sinh(λ b / a)) J (λ r / a) = g(r) − a − b b n

n

n

n

0

n

0

0

n=1

To obtain the constants En and Fn, we multiply each equation in Eq. (g) by J1(lmr/a)rdr and integrate from 0 to a to obtain a

∞

∑ E ∫ J (λ r / a)J (λ r / a)r dr = p n

n=1

0

n

0

m

n

0

En =

pn Nn

a ⎡ E cosh( λ b / a) + F sinh( λ b / a) ) ∫ J 0 (λnr / a)J 0 (λmr / a)r dr = sn ∑ ⎢( n n n n n=1 ⎢ 0 ⎣ s En cosh( λ n b / a) + Fn sinh( λ n b / a) = n Nn ∞

(h)

where a

pn = ∫ ( f (r) − a0 ) J 0 ( λ mr / a)r dr 0

a

sn = ∫ ( g(r) − a0 − b0 b) J 0 ( λ mr / a)r dr 0

a

N n = ∫ J 02 ( λ n r / a)r dr 0

Therefore,

Fn =

1 ( s − pn cosh(λnb / a)) N n sinh( λ n b / a) n

and Eq. (f) becomes

103

(i)

u(r, z) = a0 + b0 z + ∞

⎛ pn

n=1

n

∑ ⎜⎝ N cosh(λ z / a) + n

⎞ 1 sn − pn cosh( λ n b / a) ) sinh( λ n z / a)⎟ J 0 ( λ n r / a) ( N n sinh( λ n b / a) ⎠ ∞

= a0 + b0 z + ∑

pn sinh ( λ n (b − z) / a ) + sn sinh( λ n z / a) N n sinh( λ n b / a)

n=1

(j)

J 0 ( λ n r / a)

It is seen from Eq. (i) that a0 and b0 are determined from the form of f(r) and g(r). If neither of these functions has a constant term, then a0 = b0 = 0; otherwise, a0 and b0 are determined from these constants as indicated in the arguments of the integrals in Eq. (i). For example, let f(r) = h(r) + co, where co is known. Then in the argument of the integral for pn becomes h(r) + co - a0.

Therefore, we set a0 = co.

Section 7.2.3 7.5 Consider the equation

∂2 u ∂2 u + 2 + Ω2u = f o sin(απ x / a)sin( βπ x / b) 2 ∂x ∂ y

0<x<a

0< y<b

subject to the boundary conditions

u(0, y) = u(a, y) = 0 u(x,0) = u(x,b) = 0 Using the results of Example 7.8, determine the solution when: (a) a and b are not integers; (b) a = k, an integer and b = l, an integer.

Solution: (a) From Example 7.7, the solution to the homogeneous equation and these boundary conditions is ∞

∞

u(x, y) = ∑ ∑ Anm sin(nπ x / a)sin(mπ y / b) m n=1

Substituting this result into the differential equation, we arrive at

⎛ 2 ⎛ 2 m2 ⎞ ⎞ 2 n A Ω − π ∑ ∑ nm ⎜⎝ ⎜⎝ a 2 + b2 ⎟⎠ ⎟ sin(nπ x / a)sin(mπ y / b) = f o sin(απ x / a)sin( βπ y / b) ⎠ m n=1 ∞

∞

We multiply this equation by sin(kpx/a)sin(lpy/b)dxdy and integrate over the region to obtain

104

⎛ ⎛ n 2 m2 ⎞ ⎞ f ab Anm = o g n hm ⎜ Ω2 − π 2 ⎜ 2 + 2 ⎟ ⎟ 4 b ⎠⎠ ⎝a ⎝

−1

where

gn = hm =

(−1) n+1α nsin(απ )

(

π 2 n2 − α 2

)

β msin( βπ ) π m2 − β 2

(−1)

m+1 2

(

)

(b) When a = k, an integer and b = l, an integer, we find that

a 2 b hm = δ ml 2 g n = δ nk

that is, we have a one term solution.

Section 7.2.4 7.6 Consider the transient symmetrical distribution of non-dimensional temperature Q in a solid sphere of non-dimensional radius 1 that is subject to a constant non-dimensional heat rate g. The governing equation is ∂Θ 1 ∂ ⎛ 2 ∂Θ ⎞ = η +g ∂τ η 2 ∂η ⎜⎝ ∂η ⎟⎠

(a)

The boundary conditions are Q(1,t) = 1 and Q(0,t) is finite and the initial condition is Q(h,0) = 0.

Determine the temperature distribution in the sphere.

Solution: From Eq. (T7.7h) of Table 7.7, the solution to Eq. (a) is

Θ(η ,τ ) = a0 + b0 / η + ( Aj0 ( µη ) + By0 ( µη ) ) T (τ )

(b)

For Eq. (a) to remain finite at h = 0, we set b0 = 0 and B = 0. Then, Eq. (b) reduces to

Θ(η ,τ ) = a0 + j0 ( µη )T (τ ) where the constant A is subsumed in T(t). However, from Eq. (5.113)

j0 ( µη ) =

sin( µη ) µη

Then, Eq. (c) becomes

105

(c)

Θ(η ,τ ) = a0 +

sin( µη ) T (τ ) µη

(d)

Upon substituting Eq. (d) into the boundary condition at h = 1, we get

a0 +

sin( µ ) T (τ ) = 1 µ

(e)

For Eq. (e) to be satisfied for arbitrary t, a0 = 1 and

sin( µ ) = 0 → µ = µ n = nπ

(f)

where µn are the eigenvalues and sin(nph)/(nph) is the eigenfunction. Then, using superposition, the solution so far is

sin(nπη ) Tn (τ ) nπη n=1 ∞

Θ(η ,τ ) = 1+ ∑

(g)

Upon substituting Eq. (g) into Eq. (a) and using the fact that 1 ∂ ⎛ 2 ∂ ⎞ sin(nπη ) sin(nπη ) η = −n2π 2 2 ⎜ ⎟ nπη η ∂η ⎝ ∂η ⎠ nπη

we obtain ⎞ sin(nπη ) ⎛ dTn (τ ) 2 2 + n π T ( τ ) n ⎟⎠ = g nπη ⎜⎝ dτ n=1 ∞

∑

(h)

Noting from Eq. (6.99A) that the weight factor is h2, we multiply Eq. (h) by sin(mph)/(mph)h2dh to arrive at 1 ⎛ dTn (τ ) ⎞ 1 1 g 2 2 ∑ ⎜⎝ dτ + n π Tn (τ )⎟⎠ nmπ 2 ∫ sin(nπη )sin(mπη ) dη = mπ ∫ η sin(mπη ) dη ∞

n=1

0

0

⎛ dTn (τ ) ⎞ 1 g(−1) n+1 2 2 + n π T ( τ ) = n ⎜⎝ dτ ⎟⎠ 2n2π 2 n2π 2 or

dTn (τ ) 2 2 + n π Tn (τ ) = 2(−1) n+1 g dτ

(i)

Using Eq. (4.25), the solution to Eq. (i) is

Tn (τ ) =

2 2 2(−1) n+1 g + Cn e − n π τ 2 2 nπ

Then, Eq. (g) yields ∞ ⎛ 2(−1) n+1 g 2 2 ⎞ sin(nπη ) Θ(η ,τ ) = 1+ ∑ ⎜ + Cn e − n π τ ⎟ 2 2 nπ ⎠ nπη n=1 ⎝

106

(j)

Using Eq. (j) in the initial condition, we find that

⎛ 2(−1) n+1 g ⎞ sin(nπη ) ∑ ⎜⎝ n2π 2 + Cn ⎟⎠ nπη = −1 ∞

(k)

n=1

We multiply Eq. (k) by sin(mph)/(mph)h2dh to arrive at 1 ⎛ 2(−1) n+1 g ⎞ 1 1 1 ∑ ⎜⎝ n2π 2 + Cn ⎟⎠ nmπ 2 ∫ sin(nπη )sin(mπη ) dη = − mπ ∫ η sin(mπη ) dη ∞

n=1

0

0

⎛ 2(−1) g ⎞ 1 (−1) n+1 + C = − n⎟ 2 2 ⎜⎝ n2π 2 n2π 2 ⎠ 2n π n+1

or ⎛ g ⎞ Cn = −2(−1) n+1 ⎜ 1+ 2 2 ⎟ ⎝ nπ ⎠

(l)

Placing the results of Eq.(l) into Eq. (j), we obtain ∞ ⎛ g ⎛ g ⎞ 2 2 ⎞ sin(nπη ) Θ(η ,τ ) = 1+ 2∑ (−1) n+1 ⎜ 2 2 − ⎜ 1+ 2 2 ⎟ e− n π τ ⎟ ⎝ nπ ⎠ ⎝nπ ⎠ nπη n=1

7.7 Consider the transient symmetrical distribution of non-dimensional temperature Q in a solid infinitely long cylindrical rod of non-dimensional radius 1 that is subject to a constant nondimensional heat rate H. The governing equation is ∂Θ α ∂ ⎛ ∂Θ ⎞ = η +H ∂τ η ∂η ⎜⎝ ∂η ⎟⎠

0 ≤η <1 τ > 0

(a)

where a is a known constant. The steady-state temperature at the center of the rod is qc and is an unknown quantity at this point. The boundary condition at h = 1 is

Θ ′(1,τ ) = −hΘ(1,τ ) where the prime indicates the derivative with respect to h and h is a constant. The boundary condition at h = 0 is that Q(0,t) is finite. The initial condition is Q(h,0) = 0. Determine the temperature distribution in the rod. Hint: Assume that Q(h,t) = Qs(h) + Qd(h,t), where Qs(h) is the solution to the nonhomogeneous equation containing H. Then, Qs(0) = qc and qc is subsequently determined from

Θ ′s (1,τ ) = −hΘ s (1,τ ) Solution: To solve this equation with the boundary conditions and initial conditions as stated, we assume that

Θ(η ,τ ) = Θ s (η ) + Θ d (η ,τ )

107

(a)

where the differential equations and boundary and initial conditions that Qs and Qd are specified subsequently. Then the governing equation becomes ∂Θ d α ∂ ⎛ ∂Θ d ⎞ α d ⎛ dΘ s ⎞ = η + η +H ∂τ η ∂η ⎜⎝ ∂η ⎟⎠ η dη ⎜⎝ dη ⎟⎠

and the boundary condition at h = 1 is

Θ ′s (1,τ ) + Θ ′d (1,τ ) = −hΘ s (1,τ ) − hΘ d (1,τ ) The initial condition is

Θ s (η ) + Θ d (η ,0) = 0 We take Qs(h) to be the solution to 1 d ⎛ dΘ s ⎞ H η =− ⎜ ⎟ η dη ⎝ dη ⎠ α

(a)

and the boundary conditions Qs(0) = qc, a finite value, and

Θ ′s (1) = −hΘ s (1) The solution to Eq. (a) is obtained by integration to give

Θs = −

Hη 2 + c1 ln η + c2 4α

Since the result must remain finite at h = 0, we set c1 = 0. Then

Θ s = c2 −

Hη 2 4α

However, at the center of the rod,

Θ s (0) = θ c ∴ c2 = θ c and we obtain

Θs = θ c −

Hη 2 4α

Upon substituting this result into Eq. (b), we obtain

−

⎛ H H⎞ = −h ⎜ θ c − 2α 4α ⎟⎠ ⎝

θc =

H ⎛ 1 1⎞ + 2α ⎜⎝ h 2 ⎟⎠

Therefore,

108

(b)

⎞ H ⎛1 1 + 1− η 2 ⎟ ⎜ 2α ⎝ h 2 ⎠

)

(c)

∂Θ d α ∂ ⎛ ∂Θ d ⎞ = η ∂τ η ∂η ⎜⎝ ∂η ⎟⎠

(d)

Θs =

(

We take Qd(h,t) to be the solution to

with the boundary condition

Θ d (0,τ ) = finite

(e)

Θ′d (1,τ ) = −hΘ d (1,τ ) The initial condition is

Θ d (η ,0) = −Θ s (η ) =

Hη 2 − θc 4α

(f)

From Eq. (T7.7f) of Table 7.7, the solution to Eq. (d) is

Θ d (η ,τ ) = a0 + b0 ln η + ( AJ 0 ( λη ) + BY0 ( λη ) ) e− λ ατ 2

(g)

For Eq. (g) to remain finite at h = 0, we set b0 = 0 and B = 0. Then, Eq. (g) reduces to 2

Θ d (η ,τ ) = a0 + AJ 0 ( λη )e− λ ατ

(h)

Upon substituting Eq. (h) into the second boundary condition given by Eq. (e), we obtain

λ J 0′ ( λ ) + hJ 0 ( λ ) = 0 − λ J1 ( λ ) + hJ 0 ( λ ) = 0 This equation is satisfied when l = ln, where ln are the solutions to

λ n J1 ( λ n ) − hJ 0 ( λ n ) = 0

n = 1,2,3,...

Then, the solution becomes 2

Θ d (η ,τ ) = An J 0 ( λ nη )e− λnατ where, because of the boundary conditions, J0(lnh) is an orthogonal function with a weighting factor h. Using superposition, this portion of the solution is ∞

Θ d (η ,τ ) = ∑ An J 0 ( λ nη )e− λnατ 2

(i)

n=1

Upon placing Eqs. (c) and (i) in Eq. (a), we obtain

Hη 2 ∞ − λ 2ατ Θ(η ,τ ) = θ c − + ∑ An J 0 ( λ nη )e n 4α n=1 To determine An, we use the initial condition given by Eq. (f) in Eq. (j) and obtain

109

(j)

θc −

Hη 2 ∞ + ∑ An J 0 ( λ nη ) = 0 4α n=1

or

Hη 2

∞

∑ A J (λ η ) = 4α − θ n 0

n

(l)

c

n=1

We multiply Eq. (l) by J0(lmh)hdh and integrate over the region to obtain 1

∞

1

⎛ Hη 2

⎞

∑ A ∫ J (λ η )J (λ η )η dη = ∫ ⎜⎝ 4α − θ ⎟⎠ J (λ η )η dη n

n=1

0

n

0

m

c

0

0

m

0

hn Nn

An = where

1 ⎛ Hη 2 ⎞ θ H hn = ∫ ⎜ − θ c ⎟ J 0 ( λ nη )η d η = 2 ⎡⎣ 2J 2 ( λ n ) − λ n J 3 ( λ n ) ⎤⎦ − c J1 ( λ n ) 4α λn 4λ n α ⎠ 0⎝ 1

N n = ∫ J 02 ( λ nη )η d η = 0

1 2 ⎡ J ( λ ) + J12 ( λ n ) ⎤⎦ 2⎣ 0 n

The solution, therefore, is Θ(η ,τ ) = θ c −

2 Hη 2 ∞ hn +∑ J 0 ( λ nη )e− λnατ 4α n=1 N n

(j)

7.8 Obtain a solution to the following non-dimensional system ∂Θ 1 ∂ ⎛ ∂Θ ⎞ = η ∂τ η ∂η ⎜⎝ ∂η ⎟⎠

0 <α ≤η <1 τ > 0

(a)

with the boundary conditions

Θ′(α ,τ ) = −1

(b)

Θ′(1,τ ) + kΘ(1,τ ) = 0 and the initial condition

Θ(η ,0) = γ

(c)

The quantities k and g are constants and the prime (¢) denotes the derivative with respect to h.

Solution From Eq. (T7.7f) of Table 7.7, we assume a solution to Eq. (a) to be

Θ(η ,τ ) = F0 + G0 ln η + ( FJ 0 ( βη ) + GY0 ( βη ) ) e− β τ 2

110

(d)

Upon substituting Eq. (d) into Eq. (b), we obtain 2 G0 − ( FJ1 ( βα ) + GY1 ( βα ) ) e− β τ = −1 α 2 G0 + kF0 + ⎡⎣ F ( kJ 0 ( β ) − J1 ( β ) ) + G ( kY0 ( β ) − Y1 ( β ) ) ⎤⎦ e− β τ = 0

(e)

To satisfy Eq. (e) for arbitrary t,

G0 = −1→ G0 = −α α α G0 + kF0 = 0 → F0 = k

(f)

and

FJ1 ( βα ) + GY1 ( βα ) = 0

F ( kJ 0 ( β ) − J1 ( β ) ) + G ( kY0 ( β ) − Y1 ( β ) ) = 0

(g)

To obtain a non-trivial solution, we require that the determinant of the coefficients of Eq. (g) equal zero. Thus, if we denote those values of b as bn, then bn are the eigenvalues that satisfy

Y1 ( β nα ) ( kJ 0 ( β n ) − J1 ( β n ) ) − J1 ( β nα ) ( kY0 ( β n ) − Y1 ( β n ) ) = 0

(h)

From the first equation of Eq. (g) and the appropriate part of Eq. (d), the corresponding eigenfunction is

ϕ n (η ) = Cn J1 ( β nη ) + Y1 ( β nη )

(i)

where

Y (β α ) Cn = − 1 n J1 ( β nα ) The function jn(h) is an orthogonal eigenfunction whose weight factor is h.

Then the solution that satisfies Eqs. (a) and (b) is

(

)

∞

Θ(η ,τ ) = α k −1 − ln η + ∑ Anϕ n (η )e− β nτ 2

(j)

n=1

To satisfy the initial condition, Eq. (j) in substituted into Eq. (c)

(

)

∞

α k −1 − ln η + ∑ Anϕ n (η ) = γ

(k)

n=1

Since jm(h) is an orthogonal function, we multiply Eq. (k) by jm(h)hdh and integrate from a to 1 to arrive at

111

∞

1

n=1

α

∑ An ∫ ϕ n (η )ϕ m (η )η dη = gn

(l)

g An = n Nn where 1

(

))

(

g n = ∫ γ − α k −1 − ln η ηϕ m (η ) d η α

and we have used the fact that 1

N n = ∫ ϕ n2 (η )η d η α

The complete solution is

(

)

∞

2 gn Cn J1 ( β nη ) + Y1 ( β nη ) ) e− β nτ ( n=1 N n

Θ(η ,τ ) = α k −1 − ln η + ∑

(m)

7.9 Obtain a solution to the following non-dimensional system ∂Θ 1 ∂ ⎛ ∂Θ ⎞ ∂ 2 Θ = η + ∂τ η ∂η ⎜⎝ ∂η ⎟⎠ ∂z 2

0 ≤η ≤α

0 ≤ z <1 τ > 0

(a)

with the boundary conditions

Θ(0, z,τ ) = finite, Θ(η ,0,τ ) = 1,

dΘ(α , z,τ ) + kΘ(α , z,τ ) = 0 dη dΘ(η ,1,τ ) =0 dz

(b)

and the initial condition

Θ(η , z,0) = f (η , z)

(c)

Solution From Eq. (T7.7e) of Table 7.7, we assume a solution to Eq. (a) to be

Θ(η , z,τ ) = ( F0 + G0 ln η ) ( D0 + E0 z ) + ( FJ 0 ( βη ) + GY0 ( βη ) ) ( D cos(γ z) + E sin(γ z) ) e− µ τ 2

(d)

where

β 2 = µ2 − γ 2

(e)

To satisfy the first boundary condition of Eq. (b), we set G = G0 = 0 and Eq. (d) simplifies to

Θ(η , z,τ ) = D0 + E0 z + ( D cos(γ z) + E sin(γ z) ) J 0 ( βη )e− µ τ 2

112

(f)

where F0 and F have been incorporated into the remaining constants. Upon substituting Eq. (f) into the remaining boundary conditions given by Eq. (b), we arrive at

( kJ (βα ) − J (βα ))( D cos(γ z) + E sin(γ z)) e 0

− µ 2τ

1

=0

2

D0 + DJ 0 ( βη )e− µ τ = 1

(g)

E0 + ( − Dγ sin(γ ) + Eγ cos(γ ) ) J 0 ( βη )e− µ τ = 0 2

Thus, for arbitrary h, t, and z, we find that

kJ 0 ( βα ) − J1 ( βα ) = 0 D0 = 1,

D=0

(h)

E0 = 0 cos(γ ) = 0 From the first equation of Eq. (h), b = bn, where bn are the eigenvalues that satisfy

kJ 0 ( λ n ) − J1 ( λ n ) = 0 → β n = λ n / α

(i)

and the corresponding eigenfunction is

J 0 ( λ nη / α )

(j)

which is an orthogonal eigenfunction with a weight factor h. From the last equation of Eq. (h), g = gm, where gm are the eigenvalues that satisfy

cos(γ m ) = 0 → γ m =

2m − 1 π 2

m = 1,2,3,...

(k)

and the corresponding eigenfunction is

sin(γ m z)

(l)

which is an orthogonal eigenfunction.

Then, using Eqs. (h) through (l) in Eq. (f), the solution that satisfies Eqs. (a) and (b) becomes ∞

∞

Θ(η , z,τ ) = 1+ ∑ ∑ Anmθ nm (η , z)e

(

)

− β n2 +γ m2 τ

(m)

m=1 n=1

where

θ nm (η , z) = J 0 ( λ nη / α )sin(γ m z)

(n)

and we have used Eq. (e).

We now determine An from the initial condition given by Eq. (c). Thus, upon substituting Eq. (m) into Eq. (c), we arrive at ∞

∞

∑ ∑ A J (β η / α )sin(γ z) = f (η, z) − 1 nm

0

n

m

m=1 n=1

113

(o)

Since Jn(bnh/a) and sin(gnz) are orthogonal functions, we multiply Eq. (o) by Jp(bph/a)hdh and integrate from 0 to a and additionally multiply Eq. (o) by sin(gsz) integrate from 0 to 1 to arrive at ∞

α

1

m=1 n=1

0

0

∞

∑ ∑ Anm ∫ J 0 (β nη / α )J 0 (β pη / α )η dη ∫ sin(γ m z)sin(γ s z) dz = f ps f nm An = Nn Mm

(p)

where α 1

f ps = ∫ ∫ ( f (η , z) − 1) J 0 ( β pη / α )sin(γ s z)η d η dz 0 0

and α

N n = ∫ J 02 ( β nη / α )η d η = 0

1

M m = ∫ sin 2 (γ m z) dz = 0

α2 2 J ( β )z + J12 ( β n ) 2 0 n

(

)

1 2

The complete solution is ∞ ∞ f θ (η , z) −( β n2 +γ m2 )τ Θ(η , z,τ ) = 1+ 2∑ ∑ nm nm e Nn m=1 n=1

7.10 Obtain a solution to the following non-dimensional system ∂u 1 ∂ ⎛ ∂u ⎞ = η + qo ∂τ η ∂η ⎜⎝ ∂η ⎟⎠

0 ≤η ≤1 τ > 0

(a)

with the boundary conditions

u(0,τ ) = finite u(1,τ ) = 0

(b)

and the initial condition

u(η ,0) = 0

(c)

Solution From Eq. (T7.3f) of Table 7.3 and because of the nature of the differential equation, we assume a solution of the form

u(r,τ ) = F0 + G0 ln η + ( FJ 0 ( βη ) + GY0 ( βη ) ) T (τ )

(d)

It is seen from the first boundary condition of Eq. (b) that we must set G0 = G = 0. Then Eq. (d) reduces to

114

u(η ,τ ) = F0 + FJ 0 ( βη )T (τ )

(e)

Upon substituting Eq. (e) into the second boundary condition of Eq. (b), we find that

F0 + FJ 0 ( β )T (τ ) = 0 Therefore, for arbitrary t, F0 = 0 and b = bn, where bn, n = 1, 2, … are the solutions to

J 0 (β n ) = 0

(f)

Thus, bn is the eigenvalue and the corresponding eigenfunction is

J 0 ( β nη ) This eigenfunction is orthogonal with respect to the weight function h. Equation (d) can now be written as ∞

u(η ,τ ) = ∑ Tn (τ )J 0 ( β nη )

(g)

n=1

We now substitute Eq. (g) into Eq. (a) and obtain ∞

⎛ dTn (τ )

∑ ⎜⎝ dτ

J 0 ( β nη ) −

n=1

⎞ 1 d ⎛ dJ 0 ( β nη ) ⎞ η T ( τ ) ⎟ = qo n η dη ⎜⎝ dη ⎟⎠ ⎠

(h)

However, from Table 7.7, 1 ∂ ⎛ ∂J 0 ( β nη ) ⎞ η = − β n2 J 0 ( β nη ) ⎜ ⎟ η ∂η ⎝ ∂η ⎠

Then, Eq. (h) becomes ∞

⎛ dTn (τ )

∑ ⎜⎝ dτ n=1

⎞ + β n2Tn (τ )⎟ J 0 ( β nη ) = qo ⎠

(i)

We multiply Eq. (i) by J0(bmh)hdh and integrate over the region to obtain 1 ⎛ dTn (τ ) ⎞1 2 ∑ ⎜⎝ dτ + β n Tn (τ )⎟⎠ ∫ J 0 (β nη )J 0 (β mη )η dη = qo ∫ J 0 (β mη )η dη ∞

n=1

0

0

dTn (τ ) 2qo + β n2Tn (τ ) = dτ β n J1 ( β n ) where we have used the relations 1

N n = ∫ J n2 ( β nη )η d η = 0

1

g n = ∫ J 0 ( β nη )η d η = 0

(

)

1 2 1 J 0 ( β n ) + J12 ( β n ) = J12 ( β n ) 2 2 1 J (β ) βn 1 n

and Eq. (f).

115

(i)

The solution to Eq. (i) is 2

Tn (τ ) = An e− β nτ +

2qo β J (β n )

(j)

3 n 1

The initial condition for Eq. (j) is obtained from Eq. (c) as

Tn (0) = 0

(j)

Upon substituting Eq. (i) into Eq. (j), we find that An = −

2qo β J (β n ) 3 n 1

Therefore, Eq. (i) can be written as Tn (τ ) =

(

2 2qo 1− e− β nτ β J (β n )

3 n 1

)

(k)

The complete solution is obtained by substituting Eq. (k) into Eq. (g), which gives

(

J 0 ( β nη ) − β 2τ 1− e n 3 n=1 β n J1 ( β n ) ∞

u(η ,τ ) = 2qo ∑

)

Section 7.2.5 7.11 Given the following partial differential equation

∂2 u ∂u ∂ 2 u + 2 ζ = ∂t ∂x 2 ∂t 2

0 ≤ x ≤1

0 ≤ζ <1

with the boundary conditions

u(0,1) = u(1,t) = 0 and the initial conditions

u(x,0) = 0  u(x,0) = x sin 2π x Obtain the solution for u.

Solution: From Eq. (T7.9b) of Table 7.9 and because of the nature of the differential equation, we assume a solution of the form

u(x,t) = ( Acos λ x + Bsin λ x ) T (t) where T(t) is to be determined Using the boundary conditions, we find that

116

u(0,t) = AT (t) = 0 → A = 0

u(1,t) = ( Acos λ + Bsin λ ) T (t) = Bsin( λ )T (t) = 0 → λ = λ n = nπ Then

un (x,t) = sin( λ n x)Tn (t) and using superposition, the solution becomes ∞

u(x,t) = ∑ Tn (t)sin( λ n x) n=1

Substituting this expression into the differential equation yields

⎛ ∂ 2 Tn ⎞ ∂T ∑ sin λn x ⎜⎝ ∂t 2 + 2ζ ∂tn + λn2Tn ⎟⎠ = 0 n=1 ∞

Thus, this equation is satisfied when

∂ 2 Tn ∂T + 2ζ n + λ n2Tn = 0 2 ∂t ∂t The solution is this equation is given by Eq. (d) of Example 4.10 as

(

)

(

Tn (t) = Cn e−ζ t cos t λ n2 − ζ 2 + Dn e−ζ t sin t λ n2 − ζ 2

)

Thus,

(

∞

)

(

)

u(x,t) = ∑ sin( λ n x)e−ζ t ⎡Cn cos t λ n2 − ζ 2 + Dn sin t λ n2 − ζ 2 ⎤ ⎢⎣ ⎥⎦ n=1 Substituting this result into the first initial conditions gives ∞

u(x,0) = ∑ Cn sin( λ n x) = 0 → Cn = 0 n=1

Hence,

(

∞

u(x,t) = e−ζ t ∑ sin( λ n x)Dn sin t λ n2 − ζ 2 n=1

)

Before using the second initial condition, we note that

(

∞

)

(

)

 u(x,t) = ∑ sin( λ n x)Dn e−ζ t ⎡ −ζ sin t λ n2 − ζ 2 + λ n2 − ζ 2 cos t λ n2 − ζ 2 ⎤ ⎣⎢ ⎦⎥ n=1 Therefore, ∞

! u(x,0) = ∑ λ n2 − ζ 2 sin(nπ x)Dn = x sin 2π x n=1

117

From this relation, we determine Dn by multiplying both sides by sin(mpx) and integrating from 0 to 1. Thus, ∞

1

∑ λ − ζ D ∫ sin(nπ x)sin(mπ x) dx = f 2 n

2

n

n=1

m

0

Dn =

2 fn

λ n2 − ζ 2

where 1

f n = ∫ x sin 2π x sin(nπ x) dx 0

1 n=2 4 = 0 n = 4,6,... −8n = n = 1,3,5,...

2 π 2 n2 − 4 =

(

)

Then, the solution is ∞

u(x,t) = 2e−ζ t ∑ n=1

f n sin(nπ x) (nπ ) − ζ 2

2

(

sin t (nπ )2 − ζ 2

)

7.12 Find the solution to the non-dimensional equation

∂2 u ∂2 u = − hou ∂τ 2 ∂η 2

0 <η <1 τ > 0

for the boundary conditions u(0,t) = u(1,t) = 0 and for the initial conditions

0 ≤ η ≤1/ 2 ⎧2uoη u(η,0) = ⎨ ⎩2uo (−η + 1) 1 / 2 ≤ η ≤ 1  η,0) = 0 u( Solution: Assume the separable solution u(h,t) = X(h)T(t). Then the partial differential equation becomes

∂2 T ∂2 X X = T − ho XT ∂τ 2 ∂η 2 ∂2 T ∂2 X + h = = −k 2 o 2 2 T ∂τ X ∂η Therefore,

118

∂2 T + ( ho + k 2 ) T = 0 → T (τ ) = C sin ατ + D cos ατ 2 ∂τ ∂2 X + k 2 X = 0 → X(η ) = Asin kη + B cos kη ∂η 2

α 2 = ho + k 2

Substituting these results into the boundary conditions yields

u(0, τ ) = T (τ ) ( Asin k0 + B cos k0 ) = 0 → B = 0 u(1, τ ) = T (τ )sin k = 0 → sin k = 0 → k = nπ Thus, the solution that satisfies the boundary conditions is ∞

u(η, τ ) = ∑ sin nπη ( Cn sin α nτ + Dn cos α nτ ) n=1

where

α n = ho + (nπ )2 Using this result in the initial conditions, we obtain ∞

u(η,0) = ∑ Dn sin nπη n=1 ∞

! η,0) = ∑ α n sin nπη ( Cn cos(α n 0) − Dn sin(α n 0)) = 0 u( n=1 ∞

= ∑ α nCn sin nπη = 0 → Cn = 0 n=1

Therefore, ∞

∑ D sin nπη = 2u ⎡⎣η (u(η ) − u(η − 1 / 2)) + (1− η )(u(η − 1 / 2) − u(η − 1))⎤⎦ n

o

n=1

Since sinnph is an orthogonal function over the interval 0 £ h £ 1, ∞

1

1/2

1

∑ D ∫ sin nπη sin mπη dη = 2u ∫ η sin mπη dη + 2u ∫ (1− η )sin mπη dη n

n=1

o

0

o

0

1/2

⎡ 8 cos(nπ / 4)sin (nπ / 4) ⎤ Dm = 2uo ⎢ ⎥ 2 n 2π 2 ⎣ ⎦ 3

Dm =

32uo cos(nπ / 4)sin 3 (nπ / 4) n 2π 2

Then the solution is

32uo ∞ cos(nπ / 4)sin 3 (nπ / 4) u(η, τ ) = 2 ∑ sin nπη cos π n=1 n2

119

( h + (nπ ) τ ) 2

o

7.13 Consider the following circular region that is governed by the wave equation and subjected to an external excitation f(h,t). The governing equation is

∂ 2 u ∂u ∂ 2 u + − = f (η ,τ ) ∂η 2 η ∂η ∂τ 2

0 ≤η <1 0 <τ

subject to the initial conditions

u(r,0) = 0

∂u(r,0) =0 ∂τ

and

and the boundary conditions

u(0,τ ) = finite u(1,τ ) = 0 Obtain an expression for u that satisfies these equations.

Solution: From Eq. (T7.9d) of Table 7.9 and because of the nature of the differential equation, we assume a solution of the form

u(η ,τ ) = ( CJ 0 (kη ) + DY0 (kη ) ) T (τ ) where T(t) is to be determined. For the solution to be finite at h = 0, we must set D = 0. Then,

u(η ,τ ) = J 0 (kη )T (τ ) Substituting this solution in to the remaining boundary condition, we find

u(1,τ ) = J 0 (k)T (τ ) = 0 → J 0 (k) = 0 Therefore, k ® kn, where kn are those values that satisfy

J 0 (kn ) = 0

n = 1,2,3,...

Then, using superposition, ∞

u(η ,τ ) = ∑ J 0 (knη )Tn (τ ) n=1

where J0(knh) is an orthogonal function with weighting factor h.

Substituting this equation in the governing equation, we obtain ∞ ⎛ ∂2 ∂ 2 Tn (τ ) ∂ ⎞ ∑ ⎜⎝ ∂η 2 + η ∂η ⎟⎠ J 0 (knη )Tn (τ ) − ∑ J 0 (knη ) ∂τ 2 = f (η,τ ) ∞

n=1

n=1

However,

⎛ ∂2 ∂ ⎞ 2 + 2 ⎜⎝ ∂η η ∂η ⎟⎠ J 0 (knη ) = −kn J 0 (knη )

120

(a)

Therefore, Eq. (a) can be rewritten as

⎛ ∂ 2 Tn (τ ) ⎞ ∑ ⎜⎝ ∂τ 2 + kn2Tn (τ )⎟⎠ J 0 (knη ) = − f (η,τ ) n=1 ∞

We now multiply this equation by J0(kmh)hdh and integrate over the region to obtain 1 ⎛ ∂ 2 Tn (τ ) 2 ⎞1 + k T ( τ ) J (k η )J (k η ) η d η = − ∑ ⎜⎝ ∂τ 2 n n ∫0 f (η,τ )J 0 (kmη )η dη ⎟∫ 0 n 0 m ⎠0 n=1 ∞

⎛ ∂ 2 Tn (τ ) 2 ⎞ ∑ δ nm N n ⎜⎝ ∂τ 2 + kn Tn (τ )⎟⎠ = −gn (τ ) n=1 ∞

∂ 2 Tn (τ ) 2 g (τ ) + kn Tn (τ ) = − n 2 Nn ∂τ where 1

1 1 N n = ∫ J 02 (knη )η d η = ⎡⎣ J 02 (kn ) + J12 (kn ) ⎤⎦ = J12 (kn ) 2 2 0 1

g n (τ ) = ∫ f (η ,τ )J 0 (kmη )η d η 0

We use Table 4.3 and when the initial conditions are zero and z = 0 and find that

Tn (τ ) = −

τ

1 sin ( kn (τ − ξ ) ) g n (ξ ) d ξ kn N n ∫0

The complete solution is

⎛1 ⎞ J 0 (knη ) τ u(η ,τ ) = − ∑ sin k ( τ − ξ ) f ( η , ξ )J (k η ) η d η ( ) ⎜ ⎟ dξ n 0 m ∫0 ∫ ⎝0 ⎠ n=1 k n N n ∞

Section 7.2.7 7.14 (a) Find the general solution to ⎛ ∂ψ ⎞ ∇ 2 ⎜ ∇ 2ψ − α =0 ∂t ⎟⎠ ⎝

where

∂2 1 ∂ 1 ∂2 ∇ = 2+ + r ∂r r 2 ∂θ 2 ∂r 2

when

121

ψ (r,θ ,t) = Ψ(r)sin θ e jω t Hint: Let Y = Y1 + Y2 and choose the appropriate equations that each will satisfy. (b) Select that portion of the solution in (a) that remains finite as r ® ¥.

Solution (a) Substitution of the assumed solution into the original differential equation gives

(

) (

)

∇ 2r ∇ 2r Ψ − jαωΨ = ∇ 2r − jαω ∇ 2r Ψ = 0 where

∇ 2r =

d2 1 d 1 + − 2 2 r dr r dr

Therefore, we seek solution to the following equations

∇ 2r Ψ1 = 0 →

d 2 Ψ1 1 dΨ1 Ψ1 + − 2 =0 dr 2 r dr r

(a)

and

∇ 2r Ψ 2 − jαωΨ 2 = 0 → r 2

d 2Ψ 2 dΨ 2 +r − β 2r 2 + 1 Ψ 2 = 0 2 dr dr

(

)

(b)

where

β=

jαω

Equation (b) is the modified Bessel equation of order 1. Its solution is

Ψ 2 = C3 I1 ( β r) + C4 K1 ( β r) The solution to Eq. (a) is an Euler equation [Eq. (4.67)] with a2 = 1, a1 = 1, and a1 = -1. The solution is given by Case 1 [i.e., Eq. (4.73)]; thus, [r1 = 1 and r2 = -1]

Ψ1 = C1r + C2 / r Therefore,

Ψ = Ψ1 + Ψ 2 = C1r + C2 / r + C3 I1 ( β r) + C4 K1 ( β r) (b) For the solution to remain finite as r ® ¥, we set C1 = C3 = 0. [Recall Eq. (5.131)].

7.15 Consider the following equation describing a square plate subject to a non-dimensional uniform load P

∇ 4u = P

0 <η <1 0 < ξ <1

where

122

∇2 =

∂2 ∂2 + ∂η 2 ∂ξ 2

On the edges h = 0 and h = 1, the boundary conditions are

u(0, ξ ) = u(1, ξ ) = 0 ∂2 ∂2 u(0, ξ ) = u(1, ξ ) = 0 ∂η 2 ∂η 2 and those on the edges x = 0 and x = 1 are

u(η ,0) = u(η ,1) = 0 ∂ ∂ u(η ,0) = u(η ,1) = 0 ∂ξ ∂ξ Determine the solution to this system.

Solution We start with Eqs. (7.88) and (7.92); that is,

w(η, ξ ) = ( A cos αη + Bsin αη )Y (ξ ) Upon substituting this equation into the boundary conditions at h = 0 and h = 1, we find that

w(0, ξ ) = AY (ξ ) = 0 w(1, ξ ) = ( A cos α + Bsin α )Y (ξ ) = 0 ∂2 w(0, ξ ) = −Aα 2Y (ξ ) = 0 ∂η 2 ∂2 w(1, ξ ) = −α 2 ( A cos α + Bsin α )Y (ξ ) = 0 ∂η 2 Therefore, A = 0 and a = np and the solution becomes

wn (η, ξ ) = sin nπηYn (ξ ) Then, using superposition, ∞

w(η, ξ ) = ∑ Yn (ξ )sin nπη n=1

Substituting this result into the governing differential equation. we arrive at 2 ⎛ d 4Yn ⎞ 2 2 d Yn ∑ ⎜⎝ dy 4 − 2n π dy2 + n 4π 4Yn ⎟⎠ sin nπη = P n=1 ∞

Multiplying this equation by sin(mph)dh and integrating over the interval, we obtain 2 d 4Yn P 2 2 d Yn − 2n π + n 4π 4Yn = 4 2 dy dy 2

123

By inspection, we see that the particular solution is

Y p,n =

P 2n 4π 4

Then, from Eq. (7.98), which is the homogeneous solution,

Yn (ξ ) = C1 cosh(nπξ ) + C2 sinh(nπξ ) + ξ ( C3 cosh(nπξ ) + C4 sinh(nπξ )) +

P 2n 4π 4

Before proceeding, we note that

∂ Yn (ξ ) = C1nπ sinh(nπξ ) + C2 nπ cosh(nπξ ) + C3 ( cosh(nπξ ) + nπξ sinh(nπξ )) ∂ξ + C4 ( sinh(nπξ ) + nπξ cosh(nπξ ))

Substituting this result into the boundary conditions at x = 0 and x = 1, we obtain

⎛ ⎞⎧ 1 0 0 0 C1 ⎫ ⎜ cosh(nπ ) ⎪ sinh(nπ ) cosh(nπ ) sinh(nπ ) ⎟ ⎪ ⎜ ⎟ ⎪ C2 ⎪ 0 nπ 1 0 ⎜ ⎟⎨ ⎬= C ⎜ ⎟ 3 ⎪ ⎪ cosh(nπ ) + sinh(nπ ) + ⎜ nπ sinh(nπ ) nπ cosh(nπ ) ⎟⎪ C ⎪ ⎜⎝ nπ sinh(nπ ) nπ cosh(nπ ) ⎟⎠ ⎩ 4 ⎭ ⎧ −P / (2n 4π 4 ) ⎫ ⎪ ⎪ ⎪ −P / (2n 4π 4 ) ⎪ ⎨ ⎬ 0 ⎪ ⎪ ⎪ ⎪ 0 ⎩ ⎭ Using Mathematica, the solution is

P 2π 4 n 4 P sinh 2 (π n / 2) C2 = 4 4 π n (sinh(π n) + π n) C1 = −

P sinh 2 (π n / 2) π 3n 3 (sinh(π n) + π n) P sinh(π n) C4 = 3 3 2π n (sinh(π n) + π n) C3 = −

Then, doing the appropriate substitutions, we get for the final result

w(η, ξ ) =

P ∞ 1 ⎡1 1 sinh 2 (π n / 2)sinh(nπξ ) − cosh(n πξ ) + ∑ π 4 n=1 n 4 ⎢⎣ 2 2 (sinh(π n) + π n) −

nπ sinh 2 (π n / 2)ξ cosh(nπξ ) nπ sinh(π n)ξ sinh(nπξ ) ⎤ + sin nπη (sinh(π n) + π n) 2(sinh(π n) + π n) ⎥⎦ 124

Solutions to Exercises in Chapter 8 Section 8.1.2 8.1 Obtain the Laplace transform of

cosht sint using Eq. (8.16).

Solution: We express this product as

g(t) =

(

)

1 t −t 1 1 e + e sint = et sint + e−t sint 2 2 2

Then, using Eq. (8.16)

1 1 L ⎡⎣ et sint ⎤⎦ + L ⎡⎣ e−t sint ⎤⎦ 2 2 1 1 1 1 = + 2 2 (s − 1) + 1 2 (s + 1)2 + 1

L[g(t)] = G(s) =

=

s2 + 2

((s + 1) + 1)((s − 1) + 1) 2

2

Mathematica verification LaplaceTransform[Cosh[t] Sin[t],t,s] 8.2 Obtain the Laplace transform of

e−t t cost Solution: From Eq. (8.16),

f (t)ebt ⇔ F(s − b) and from Eq. (8.7)

L ⎡⎣t n f (t) ⎤⎦ ⇔ (−1) n Then, from Eq. (8.13)

125

d n F(s) ds n

⎡ d ⎛ s ⎞⎤ ⎡ d ⎤ L ⎡⎣ e−t t cost ⎤⎦ = ⎢ − L[cost]⎥ = ⎢− ⎜ 2 ⎟ ⎥ ⎣ ds ⎦ s→s+1 ⎣ ds ⎝ s + 1⎠ ⎦ s→s+1 ⎡ ⎤ ⎡ 2 ⎤ 1 2s 2 ⎥ s + 1− 2s 2 ⎥ ⎢ ⎢ =− 2 − =− ⎢ s + 1 s2 + 1 2 ⎥ ⎢ s2 + 1 2 ⎥ ⎣ ⎦ s→s+1 ⎣ ⎦ s→s+1 ⎡ 2 ⎤ s −1 ⎥ (s + 1)2 − 1 ⎢ = = 2 2 ⎢ s2 + 1 2 ⎥ (s + 1) + 1 ⎣ ⎦

(

=

)

(

)

s→s+1

(

(

)

)

s(s + 2)

( s + 2s + 2) 2

2

Mathematica verification LaplaceTransform[Exp[-t] t Cos[t],t,s] 8.3 Use Eqs. (8.6) and (8.7) to obtain the solution to the following equation

d2y dy x 2 − 2(x + 1) + (x + 2) y = 0 dx dx Solution: From Eq. (8.6), we note that

L[ y] = Y (s) ⎡ dy ⎤ L ⎢ ⎥ = sY (s) − y(0) ⎣ dx ⎦ ⎡ d2y ⎤ L ⎢ 2 ⎥ = s 2Y (s) − sy(0) − y′(0) ⎣ dx ⎦

(a)

and from Eq. (8.7) that

d (Y (s)) ds ⎡ dy ⎤ d L ⎢ x ⎥ = − ( sY (s) − y(0) ) ds ⎣ dx ⎦ L[xy] = −

⎡ d2y ⎤ d 2 L ⎢x 2 ⎥ = − s Y (s) − sy(0) − y′(0) ds ⎣ dx ⎦

(

(b)

)

Then the Laplace transform of the differential equation is

⎡ d2y ⎤ ⎡ dy ⎤ ⎡ dy ⎤ L ⎢ x 2 ⎥ − 2L ⎢ x ⎥ − 2L ⎢ ⎥ + L[xy] + 2L[ y] = 0 ⎣ dx ⎦ ⎣ dx ⎦ ⎣ dx ⎦

126

which upon using Eqs. (a) and (b) gives

−

(

)

d 2 d s Y (s) − sy(0) − y′(0) + 2 ( sY (s) − y(0) ) ds ds d − 2 ( sY (s) − y(0) ) − (Y (s) ) + 2Y (s) = 0 ds

Performing the indicated operations and collecting terms, we arrive at

⎛ ⎞ ⎛ dY (s) dY (s) ⎞ − ⎜ 2sY (s) + s 2 − y(0)⎟ + 2 ⎜ Y (s) + s ds ds ⎟⎠ ⎝ ⎠ ⎝ − 2 ( sY (s) − y(0) ) −

dY (s) + 2Y (s) = 0 ds

which reduces to

( s − 2s + 1) dYds(s) + 4 ( s − 1)Y (s) = 3y(0) 2

dY (s) 4 3y(0) + Y (s) = 2 ds s −1 ( s − 1)

(c)

To obtain a solution to Eq. (c), we use Eq. (4.24) with

p(s) = q(s) =

4 s −1 3y(0)

( s − 1)

2

Then, from Eq. (4.22) the integrating factor is ⎡ ds ⎤ ρ (s) = exp ⎢ 4 ∫ = exp ⎡⎣ 4ln(s − 1) ⎤⎦ = (s − 1)4 ⎥ ⎣ s − 1⎦

From Eq. (4.24), we obtain

⎤ 3y(0) 1 ⎡ (s − 1)4 C Y (s) = 3y(0) ds + C (s − 1)2 ds + ⎢ ⎥= 4 ∫ 2 4 ∫ (s − 1) ⎣ (s − 1) (s − 1)4 ⎦ (s − 1) 3y(0) ⎛ s3 ⎞ C 2 = s − s + + 4 ⎜ ⎟ 3 ⎠ (s − 1)4 (s − 1) ⎝

(d)

The inverse of the first term of Eq. (d) requires an extensive table of Laplace transform pairs [see Sanders and Kaufman in the Bibliography] or the use of Mathematica. The second term is obtained from pair 20 of Table 8.2. Then, using Mathematica, the inverse Laplace transform of Eq. (d) is

y(x) =

(

)

y(0) x C e 6 + x 3 + x 3e x 6 6

127

Mathematica verification InverseLaplaceTransform[3 y0/(s-1)^4 (s-s^2+s^3/3)+ c/(s-1)^4,s,x] Section 8.1.3 8.4 Use partial fractions to determine the inverse Laplace transform of

é a3 ù L ê 3ú ë s( s + a) û -1

Solution: We would like to expand this into the form B B2 B3 a3 A = + 1 + + 3 2 s s + a (s + a) (s + a)3 s(s + a)

From Eq. (8.20), we find that

⎡ sa 3 ⎤ A = lim ⎢ ⎥ =1 s→0 s(s + a)3 ⎣ ⎦ From Eq. (8.22), we find that 3 1 d2 ⎛ a3 ⎞ a3 d 2 ⎛ 1⎞ a3 ⎛ 2 ⎞ B1 = ( s + a ) s(s + a)3 ⎟⎠ = 2! ds2 ⎜⎝ s ⎟⎠ = 2 ⎜⎝ s3 ⎟⎠ = −1 2! ds 2 ⎜⎝ s=− a s=− a

B2 = a 3

⎛ −1⎞ d ⎛ 1⎞ = a3 ⎜ 2 ⎟ = −a ⎜ ⎟ ds ⎝ s ⎠ s=− a ⎝ s ⎠ s=− a

⎛ 1⎞ B3 = a 3 ⎜ ⎟ = −a 2 ⎝ s ⎠ s=− a Then,

⎡ a3 ⎤ a ⎤ −1 ⎡ a 2 ⎤ −1 ⎡ 1 ⎤ −1 ⎡ 1 ⎤ −1 ⎡ L−1 ⎢ = L − L − L −L ⎢ ⎢ ⎢s⎥ ⎢s+ a⎥ 3⎥ 2 ⎥ 3⎥ s(s + a) (s + a) ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ (s + a) ⎦

(

)

= 1− 1+ at + a 2t 2 e− at where we have used Eqs. (8.11) and (8.15).

Mathematica verification ap=Apart[a^3/(s (s+a)^3)] InverseLaplaceTransform[ap,s,t]

128

8.5 Use partial fractions to determine the inverse Laplace transform of

6 + (s + 1)(s 2 − 5s + 11) F(s) = s(s − 1)(s − 2)(s + 1) Solution: Since there are no repeating roots, we would like to expand this into the form

F(s) =

A1 A2 A3 A4 + + + s (s − 1) (s − 2) (s + 1)

From Eq. (8.20), we find that

⎡ 6 + (s + 1)(s 2 − 5s + 11) ⎤ 6 + 11 17 A1 = lim ⎢ = ⎥= s→0 2 2 ⎣ (s − 1)(s − 2)(s + 1) ⎦ ⎡ 6 + (s + 1)(s 2 − 5s + 11) ⎤ 6 + 14 A2 = lim ⎢ = −10 ⎥= s→1 s(s − 2)(s + 1) −2 ⎣ ⎦ ⎡ 6 + (s + 1)(s 2 − 5s + 11) ⎤ 6 + 15 7 A3 = lim ⎢ = ⎥= s→2 s(s − 1)(s + 1) 6 2 ⎣ ⎦ ⎡ 6 + (s + 1)(s 2 − 5s + 11) ⎤ 6 A4 = lim ⎢ = −1 ⎥= s→−1 s(s − 1)(s − 2) ⎣ ⎦ −6 Then,

⎡ 6 + (s + 1)(s 2 − 5s + 11) ⎤ 17 −1 ⎡ 1 ⎤ ⎡ 1 ⎤ 7 −1 ⎡ 1 ⎤ −1 ⎡ 1 ⎤ L−1 ⎢ L ⎢ ⎥ − 10L−1 ⎢ ⎥= ⎥+ L ⎢ ⎥ − L ⎢ s + 1⎥ ⎣s⎦ ⎣ s − 1⎦ 2 ⎣ s − 2 ⎦ ⎣ ⎦ ⎣ s(s − 1)(s − 2)(s + 1) ⎦ 2 17 7 = − 10et + e2t − e−t 2 2 Mathematica verification ap=Apart[(6+(s+1) (s^2-5 s+11))/(s (s-1) (s-2) (s+1))] InverseLaplaceTransform[ap,s,t] 8.6 Determine the inverse Laplace transform of

é e -p s ù L-1 ê 2ú ë1 + s û Solution: Using pair 6 of Table 8.2,

129

é e -p s ù L-1 ê = L-1 éëG(s)e-p s ùû Û g (t - p )u (t - p ) = sin(t - p )u (t - p ) 2ú 1 + s ë û since

é 1 ù L-1 ê = sin t 2 ë1 + s úû Mathematica verification InverseLaplaceTransform[Exp[-Pi s]/(1+s^2),s,t] 8.7 Determine the Laplace transform of t

e−t ∫ e−τ τ cos[ω (t − τ )]d τ 0

Solution: We write this equation as

e(t) = e−t f (t) where t

f (t) = ∫ e−τ τ cos[ω (t − τ )]d τ

(a)

0

If we use the convolution relation given by Eq. (8.26), we see that t

L−1[G(s)H (s)] = ∫ g(ξ )h(t − ξ ) d ξ

(b)

⎡t ⎤ G(s)H (s) = L ⎢ ∫ g(ξ )h(t − ξ ) d ξ ⎥ ⎢⎣ 0 ⎥⎦

(c)

0

or

Upon comparing Eq. (a) with Eq. (b), we identify

L−1 ⎡⎣ H (s) ⎤⎦ = e−τ τ L−1 ⎡⎣G(s) ⎤⎦ = cos(ωτ ) Therefore,

H (s) = L ⎡⎣ e−τ τ ⎤⎦ =

1 (s + 1)2

G(s) = L ⎡⎣cos(ωτ ) ⎤⎦ =

130

s s +ω2 2

where we have used Eqs. (8.18) and (8.13). Hence,

⎡t ⎤ F(s) = L[ f (t)] = L ⎢ ∫ e−τ τ cos[ω (t − τ )]d τ ⎥ = H (s)G(s) ⎢⎣ 0 ⎥⎦ s = 2 2 s + ω (s + 1)2

(

)

and from Eq. (8.16),

L[e(t)] = E(s) = L ⎡⎣ e−t f (t) ⎤⎦ = F(s + 1) =

s +1

((s + 1) + ω ) (s + 2) 2

2

2

Mathematica verification Simplify[LaplaceTransform[Exp[-t]Integrate[Exp[-x] x * Cos[w (t-x)],{x,0,t}],t,s]] 8.8 Determine the inverse Laplace transform of

s

(s + k ) 2

2

2

Solution: We write this equation as

s

(s + k ) 2

2

2

=

s 1 2 2 s + k s + k2 2

Therefore, we let

s s + k2 1 G(s) = 2 s + k2 F(s) =

2

and from Eqs. (8.12) and (8.13), we have

L−1 ⎡⎣ F(s) ⎤⎦ = cos kt L−1 ⎡⎣G(s) ⎤⎦ = Using Eq. (8.26),

131

1 sin kt k

t

L−1[F(s)H (s)] = ∫ h(ξ ) f (t − ξ ) d ξ 0

t

⎡ s 1 ⎤ 1 L ⎢ 2 = ∫ cos(kξ )sin ( k(t − ξ ) ) d ξ 2 2 2⎥ ⎣s + k s + k ⎦ k 0 t = sin(kt) 2k −1

Mathematica verification InverseLaplaceTransform[s/(s^2+k^2)^2,s,t] Section 8.1.6 8.9 Obtain the Laplace transform of the following periodic waveform (sawtooth wave) of period T

f (t) = t / T

0≤t ≤T

Solution: From Eq. (8.37)

)

−1

(

Ts 1− e

− sT

F(s) = 1− e = =

2

(

− sT

T

1 1 ⎡1− e− sT (1+ sT ) ⎤⎦ τ e− sτ d τ = 2 ∫ − sT ⎣ T 0 Ts 1− e

1

(

)

)

⎡⎣1− e− sT − sTe− sT ) ⎤⎦

1 e− sT 1 1 − = 2− 2 − sT sT Ts s 1− e Ts s e − 1

(

)

(

)

Mathematica verification Simplify[1/T Integrate[x Exp[-s x],{x,0,T}]/(1-Exp[-s T])] 8.10 Obtain the Laplace transform of the following periodic waveform (triangular wave) of period T

f (t) = 2t / T

0≤t ≤T /2

= 2(1− t / T ) Solution: From Eq. (8.37)

132

T /2≤t ≤T

(

F(s) = 1− e =

− sT

)

−1

T ⎧⎪ 2 T /2 − sτ ⎫⎪ − sτ ⎨ ∫ τ e d τ + 2 ∫ (1− τ / T )e d τ ⎬ ⎪⎩ T 0 ⎪⎭ T /2

{(2 − e s T (1− e ) 1

− sT /2

− sT

2

)

(

(2 + sT ) + e− sT 2 + e− sT /2 (−2 + sT )

)}

( ) = 2 (1− e ) = s T (1− e ) s T (1− e )(1+ e ) 2 (1− e ) ⎛ sT ⎞ 2 = = tanh ⎜ ⎟ s T (1+ e ) sT ⎝ 4⎠ 2e− sT esT /2 − 1

2

− sT /2

− sT

2

− sT /2

2

2

− sT /2

sT /2

− sT /2

2

2

where we have used the results of Example 8.7.

Mathematica verification FullSimplify[(2/T Integrate[x Exp[-s x],{x,0,T/2}]+ 2 Integrate[(1-x/T) Exp[-s x],{x,T/2,T}])/(1-Exp[-s T])] Section 8.1.7 8.11 Use the Bromwich inversion integral to determine the inverse Laplace transform of

F(s) =

1 (s + b)cosh(as)

Solution: The poles of this expression are at s = -b and

cosh(as) = cosh( jβ ) = cos β = 0 → β = β n = ±(2n − 1)π / 2 → s = ± j(2n − 1)π / (2a) Therefore,

f (t) =

∞ ⎡ ⎤ (z + b)e zt e zt + ∑ lim ⎢ ⎥ (z + b)cosh(az) z=− b n=1 z→zn ⎣ (z + b)d cosh(az) / dz ⎦

=

∞ ⎡ ⎤ e− bt e zt + ∑ lim ⎢ ⎥ cosh(ab) n=1 z→± jγ n /(2a) ⎣ a(z + b)sinh(az) ⎦

=

∞ e− bt e jγ nt/(2a) 1 + 2∑ cosh(ab) n=1 ( jγ n + 2ab) ( j sin(γ n / 2))

e− jγ nt/(2a) 1 + 2∑ n=1 (− jγ n + 2ab) (− j sin(γ n / 2)) ∞

=

jγ t/(2a)

− jγ t/(2a)

∞ ∞ e− bt (−1) n+1 e n (−1) n e n + 2∑ + 2∑ cosh(ab) n=1 j(− jγ n + 2ab) k=1 j( jγ n + 2ab)

133

where gn = (2n - 1)p. Upon further manipulation, we find that

f (t) =

∞ ∞ e− bt (−1) n e jγ nt/(2a) (−1) n e− jγ nt/(2a) + 2 j∑ − 2 j∑ cosh(ab) n=1 (− jγ n + 2ab) n=1 ( jγ n + 2ab)

∞ ∞ (−1) n ( jγ n + 2ab)e jγ nt/(2a) (−1) n (− jγ n + 2ab)e− jγ nt/(2a) e− bt = + 2 j∑ − 2 j∑ cosh(ab) γ n2 + 4a 2 b2 γ n2 + 4a 2 b2 n=1 n=1 jγ t/(2a)

= = =

∞ ( jγ n + 2ab)e n − (− jγ n + 2ab)e e− bt + 2 j ∑ (−1) n cosh(ab) γ n2 + 4a 2 b2 n=1

− jγ nt/(2a)

∞ 2 jγ n cos (γ nt / (2a) ) + 4 jabsin (γ nt / (2a) ) e− bt + 2 j ∑ (−1) n cosh(ab) γ n2 + 4a 2 b2 n=1 ∞ 4γ cos (γ nt / (2a) ) + 8absin (γ nt / (2a) ) e− bt − ∑ (−1) n n cosh(ab) n=1 γ n2 + 4a 2 b2

Mathematica verification of one term in series r1=Residue[Exp[z t]/((z+b) Cosh[a z]),{z,-b}]; q=I (2 n-1) Pi/(2 a); r2=FullSimplify[Exp[z t]/(z+b)/D[Cosh[a z],z]/.z->q, Assumptions->nÎIntegers]; r3=FullSimplify[Exp[z t]/(z+b)/D[Cosh[a z],z]/.z->-q, Assumptions->nÎIntegers]; r1+Simplify[ComplexExpand[r2+r3],Assumptions->nÎIntegers] 8.12 Use the Bromwich inversion integral to determine the inverse Laplace transform of

F(s) =

cosh(x s / a ) scosh(k s / a )

Solution: The poles of this expression are at s = 0 and

(

)

cosh k s / a = cosh( jβ ) = cos β = 0 Therefore,

β = β n = ±(2n − 1)π / 2 and

s = −aβ n2 / k 2 Then

134

n = 1,2,3,...

(

)

(

)

⎡ ⎤ zt ⎢ cosh x z / a e ⎥ f (t) = + ∑ lim ⎢ ⎥ z→zn n=1 cosh k z / a ⎢ z d cosh k z / a ⎥ z=0 ⎢⎣ dz ⎥⎦ ⎡ 2a z / a cosh x z / a e zt ⎤ ∞ ⎥ = 1+ ∑ lim ⎢ z→zn ⎢ ⎥ n=1 kz sinh k z / a ⎢⎣ ⎥⎦ z=z n cosh x z / a e zt

(

∞

)

(

(

)

2ajβ n cosh ( jβ n x / k ) e− aβ n t/k 2

∞

= 1+ ∑ n=1

(

)

k 2 −aβ n2 / k 2 sinh ( jβ n )

)

(

2

∞

= 1− 2∑

)

cos ( β n x / k ) e− aβ n t/k

n=1

2

2

β n sin ( β n )

∞

2 2 (−1) n = 1+ 2∑ cos ( β n x / k ) e− aβ n t/k n=1 β n

Mathematica verification of one term in series q=-a b^2/k^2 ; r2=FullSimplify[(Exp[z t]Cosh[x Sqrt[z/a]]/ D[Cosh[k Sqrt[z/a]],z])/z/.z->q,Assumptions->nÎIntegers] 8.13 Use the Bromwich inversion integral to determine the inverse Laplace transform of

F(s) =

sinh(a s ) ssinh( s )

Solution: The poles of this expression are at s = 0 and

sinh

( s ) = sinh( jβ ) = j sin β = 0

Therefore,

β = β n = nπ

n = 1,2,3,...

and

s = sn = −n2π 2 Then

135

f (t) =

=

sinh(a z )e zt sinh( z )

(

⎡ ⎤ ⎢ sinh(a z )e zt ⎥ + ∑ lim2 2 ⎢ ⎥ z=− n π n=1 ⎢ z d sinh( z ) ⎥ z=0 ⎢⎣ dz ⎥⎦ ∞

)

a z 1+ a 2 z / 6 + ... e zt z (1+ z / 6 + ...)

∞ ⎡ 2 z sinh(a z )e zt ⎤ + ∑ lim2 2 ⎢ ⎥ z=− n π z cosh( z ) ⎥⎦ n=1 ⎢⎣ z=0 2 2

2 2

2 jnπ sinh( janπ )e− n π t 2 ∞ sin(anπ )e− n π t = a+∑ = a + ∑ π n=1 ncos(nπ ) −n2π 2 cosh( jnπ ) n=1 ∞

== a +

2 2 2 ∞ (−1) n sin(anπ )e− n π t ∑ π n=1 n

Mathematica verification of one term in series q=-n^2Pi^2 ; a+FullSimplify[(Exp[z t]Sinh[a Sqrt[z]]/D[Sinh[Sqrt[z]],z])/z /.z->q,Assumptions->nÎIntegers] 8.14 Use the Bromwich inversion integral to determine the inverse Laplace transform of

F(s) =

( ) s ( m s cosh ( a s ) + sinh ( a s )) sinh a s

(a)

o

Solution: Using Mathematica, a series expansion of F(s) about s = 0 is

(

s a mo + 6a mo a 3mo a F(s) ≈ − + s(a + mo ) 3(a + mo )2 45(a + mo )3 6

5

2

)

(b)

Thus, we have a simple pole at s = 0 whose residue is a/(a + mo). The other poles are at the roots of

( )

tanh a s = −a sγ o

(c)

where go = a/mo. If we let

jβ n = a s in Eq. (c), Eq. (c) becomes

tan ( β n ) = −γ o β n

(d)

Therefore, for a given go, bn, n = 1, 2, …, are those value that satisfy Eq. (d). The corresponding values of s → sn = − β n2 / a 2 .

136

Then

( ) ( ) ( )) ( ( ) ( ) ( ))

⎡ ⎤ sinh a z e zt ⎢ ⎥ ⎡ zae zt ⎤ ∞ f (t) = lim ⎢ + ∑ lim2 2 ⎢ ⎥ ⎥ z=0 ⎣ z(a + mo ) ⎦ n=1 z=− β n /a ⎢ z d m z cosh a z + sinh a z ⎥ ⎢⎣ dz o ⎥⎦ ⎡ ⎤ ∞ 2 z sinh a z e zt a ⎢ ⎥ = + ∑ lim a + mo n=1 z=− β n2 /a2 ⎢ z ( a + m ) cosh a z + am z sinh a z ⎥ ⎢⎣ ⎥⎦ o o

(

∞ 2 jβ n / a ) sinh ( jβ n ) e− β n t/a ( a = +∑ a + mo n=1 − β n2 / a 2 ( a + mo ) cosh ( jβ n ) + jβ n mo sinh ( jβ n ) 2

2

(

=

∞ sin ( β n ) e a + 2∑ a + mo n=1 ( β n / a ) ( a + mo ) cos ( β n ) − β n mo sin ( β n )

)

− β n2t/a 2

(

)

Equation (e) can be further simplified by using Eq. (d).

Mathematica verification of one term in series q=-b^2/a^2 ; a/(a+mo)+FullSimplify[(Exp[z t]Sinh[a Sqrt[z]]/ D[(mo Sqrt[z]Cosh[a Sqrt[z]]+Sinh[a Sqrt[z]]),z]/z) /.z->q,Assumptions->nÎIntegers] Section 8.2 8.15 Use Laplace transforms to determine the general solution to

d2y dy - 2k + k 2 y = f (t ) 2 dt dt Solution: Taking the Laplace transform of each term, we find that

s 2Y ( s ) - y¢(0) - sy (0) - 2k ( sY ( s ) - y (0) ) + k 2Y ( s ) = F ( s )

( s - 2ks + k ) Y (s) = y¢(0) + ( s + 2k ) y(0) + F (s) 2

2

( s - k ) 2 Y ( s ) = C1 + sC2 + F ( s ) Y ( s) =

C1 sC2 F ( s) + + 2 2 (s - k ) (s - k ) (s - k )2

Taking the inverse Laplace transform using Eqs. (14) and (24) and noting that

137

(e)

s A B A(s - k ) + B = + = 2 2 (s - k ) s - k (s - k ) (s - k )2 and, therefore,

As = s ® A = 1 - kA + B = 0 ® B = k we obtain

é 1 ù é s ù -1 é F ( s ) ù y (t ) = C1 L-1 ê + C2 L-1 ê +L ê 2ú 2ú 2ú ë (s - k ) û ë (s - k ) û ë (s - k ) û t k ù é 1 ù -1 é = C1te + C2 L ê + C2 L ê + ò (t - t )e k (t -t ) f (t )dt 2ú ú ës -k û ë (s - k ) û 0 kt

-1

t

= C1te kt + C2 ( e kt + kte kt ) + ò (t - t )e k (t -t ) f (t )dt 0

t

= C2 e kt + C3te kt + ò (t - t )e k ( t -t ) f (t )dt 0

Mathematica verification with f(t) = 0 lp=LaplaceTransform[y''[t]-2 k y'[t]+k^2 y[t]==0,t,s]/.LaplaceTransform[y[t],t,s]->Ys; yss=Ys/.Solve[lp,Ys][[1]] InverseLaplaceTransform[yss,s,t] InverseLaplaceTransform[1/Denominator[yss],s,t] 8.16 The non-dimensional form of the one-dimensional heat flow equation in a solid is given by

dθ ∂ 2θ = dτ ∂η 2

(a)

It is assumed that the boundary conditions are

∂θ (0,τ ) = 0, ∂η

θ (1,τ ) = θ1

(b)

and the initial condition are

θ (η ,0) = 1

(c)

Use the Laplace transform with respect to t to obtain a solution to Eq. (a) subject to the conditions given by Eqs. (b) and (c).

Solution

138

To obtain a solution, we take the Laplace transform with respect to time of these equations. Then Eq. (a) becomes

¶ 2q = sq - q (h ,0) = sq - 1 ¶h 2

(d)

where we have used Eq. (c). The boundary conditions given by Eq. (b) become

¶q (0, s) = 0, ¶h

q q (1, s) = 1

(e)

s

The solution to the homogeneous portion of Eq. (d); that is,

¶ 2q h - sq h = 0 ¶h 2 is

qh (h , s) = A coshh s + B sinhh s and the solution to inhomogeneous portion of Eq. (d); that is, is

¶ 2q p ¶h 2

- sq p = -1

is, by inspection,

qp =

1 s

Therefore,

q (h , s) = q h (h , s) + q p = A cosh h s + B sinh h s +

1 s

(f)

Since

¶q (h , s) = A s sinhh s + B s coshh s ¶h we find upon substituting Eq. (f) into Eq. (e), the boundary conditions, that

¶q (0, s ) =0®B=0 ¶h

q 1 q q (1, s) = 1 ® A cosh s + - 1 ® A = s

s

s

q1 - 1 s cosh s

The Eq. (f) becomes

1 cosh h s q (h , s) = + (q1 - 1) s s cosh s

139

(g)

To determine the inverse transform, we use Eq. (66) and Eq. (g) to arrive at

q (h ,t ) = u (t ) +

(q1 - 1) g + j¥ e zt cosh h z dz 2p j

ò

g - j¥

z cosh z

We have a simple pole at z = 0 and at the poles 2

æ 2n - 1 ö æ 2n - 1 ö 2 cosh z = 0 ® zn = j ç ÷ p ® zn = - ç ÷ p è 2 ø è 2 ø

(h)

Then

⎧ ⎫ ⎪ ⎪ ⎡ ze zt cosh η z ⎤ ∞ e zk t cosh η z k ⎪ ⎪ θ (η ,τ ) = u(τ ) + (θ1 − 1) ⎨ Lim ⎢ ⎥+∑ ⎬ z→0 z cosh z ⎢⎣ ⎥⎦ k=1 z d cosh z ⎪ ⎪ k ⎪ ⎪ dz z=zk ⎭ ⎩ ⎧ ⎫ ⎪ ⎪ z t ⎧ ∞ ∞ ⎡ z nt e k cosh η z k ⎪ e cosh η z n ⎤ ⎫⎪ ⎪ ⎪ ⎥⎬ = u(τ ) + (θ1 − 1) ⎨1+ ∑ ⎬ = u(τ ) + (θ1 − 1) ⎨1+ 2∑ ⎢ ⎥ z sinh z n=1 ⎢ ⎪ k=1 z 1 sinh z ⎪ ⎪⎩ n ⎦⎪ ⎣ n ⎭ k k ⎪ ⎪ 2 z k ⎩ ⎭ z t ∞ ⎡ n e cosh η z n ⎤ ⎥ = θ1 + 2 (θ1 − 1) ∑ ⎢ ⎥ z sinh z n=1 ⎢ n ⎦ ⎣ n However, from Eq. (h),

é æ 2n - 1 ö ù æ 2n - 1 ö zn sinh zn = j ç ÷ p sinh ê j ç ÷p ú è 2 ø ë è 2 ø û éæ 2n - 1 ö ù æ 2n - 1 ö æ 2n - 1 ö n -1 = -ç ÷ p sin êç ÷p ú = - ç ÷ p (-1) è 2 ø è 2 ø ëè 2 ø û æ 2n - 1 ö = (-1) n ç ÷p è 2 ø Hence,

(

)

⎡ (−1) n e−( 2n−1)2 π 2t/4 cos η 2n − 1 π / 2 ⎤ ( ) ⎥ θ (η ,τ ) = θ1 + 2 (θ1 − 1) ∑ ⎢ ⎢ ⎥ ( 2n − 1)π / 2 n=1 ⎣ ⎦ ∞

Mathematica verification of one term in series zn=-((2 n-1) Pi/2)^2; r2=FullSimplify[Exp[z t]Cosh[x Sqrt[z]]/z/D[Cosh[Sqrt[z]],z] /.z->zn,Assumptions->nÎIntegers]

140

8.17 Consider a hanging chain of length L that is free at x = 0 and fixed at x = L. The governing equation of motion for this system is

¶ æ ¶u ö 1 ¶ 2u çx ÷= ¶x è ¶x ø g ¶t 2

(a)

where g is the gravity constant. The boundary conditions are

u ( L, t ) = A sin wt u (0, t ) = finite

(b)

that is, one end is oscillating harmonically at a frequency w. The initial conditions are assumed to be zero. Use the Laplace transform with respect to t to obtain a solution to Eq. (a) subject to the conditions given by Eqs. (b) and (c).

Solution: To solve this problem, we take the Laplace transform of Eqs. (a) and (b) to arrive at

¶ æ ¶u ( x, s) ö s 2 çx ÷ - u ( x, s ) = 0 ¶x è ¶x ø g

(c)

¶ 2u ( x, s) ¶u ( x, s) s 2 x + - u ( x, s ) = 0 ¶x 2 ¶x g and

Aw s + w2 u (0, s ) = finite

u ( L, s ) =

(d)

2

We have shown in Case 2 of Table 5.2 that

x2

d2y dy + x(1 - 2ab) + ( b2 k 2 x 2b - b2 p 2 ) y = 0 2 dx dx

(e)

y( x) = xab éëC1 J p (kxb ) + C2Yp (kxb )ùû

(f)

has as a solution of the form

Comparing Eq. (c) with Eq. (e) we see that a = 0, p = 0, b = 1/2, and b2k2 = k2/4 = -s2/g. In addition, we note that Y does not remain finite at x = 0 and, therefore, we must set C2 = 0. Then Eq. (f) is used to obtain a solution to Eq. (c) when its form is

(

)

(

u ( x, s) = C1 J 0 2 js x / g = C1 I 0 2s x / g where we have used the relation

I 0 ( x) = J 0 ( jx) To determine C1, we substitute Eq. (g) into Eq. (d) and find that

141

)

(g)

(

) s A+ww

u ( L, s ) = C1 I 0 2s L / g = C1 =

Aw

2

2

( s + w ) I ( 2s L / g ) 2

2

0

Therefore, the solution in the Laplace transform domain is

u ( x, s) =

(

Aw I 0 2s x / g

)

=

(

Aw I 0 sg x / L

)

(h)

( s + w ) I ( 2s L / g ) ( s + w ) I ( sg ) 2

2

2

2

0

0

where g 2 = 4 L / g. Equation (h) has simple poles at s = ±jw and at the simple poles when

I 0 ( sg ) = 0 ® I 0 ( j b n ) = J 0 (b n ) = 0 ® sn = ± j

bn g

To find the inverse transform, we substitute Eq. (h) into Eq. (66) and use residue theory to obtain g + j¥

u ( x, t ) = Aw ò

g - j¥

(

e zt I 0 zg x / L

) dz

( z + w ) I ( zg ) 2

2

0

(

)

(

)

ì é e zt I 0 zg x / L ù é e zt I 0 zg x / L ù ï ê ú ê ú = Aw í lim + lim z ® jw ê z ®- jw ê ú z + jw ) I 0 2 z L / g z - jw ) I 0 2 z L / g ú ïî ë( û ë( û

(

(

)

(

)

é e zt I 0 zg x / L ù é ù 1 ú lim ê lim ê ú z ® sn ê z 2 + w 2 ) ú z ® sn ë dI 0 ( zg ) dz û ( ë û

¥

+å n =1

(

)

é e zt I 0 zg x / L ù é ù 1 ú lim ê lim ê ú 2 2 z ®- sn ê z + w ) ú z ®- sn ë dI 0 ( zg ) dz û ( ë û

¥

+å n =1

ü ï ý ï þ

We note that

d I 0 ( zg ) = g I1 ( zg ) dz and

(

I1 ( sng ) = I1 ( ± j b n ) = ± jJ1 ( b n )

)

(

)

(

I 0 sng x / L = I 0 ± j b n x / L = J 0 b n x / L Then Eq. (i) becomes

142

)

)

(i)

(

)

(

⎧ e jω t I jωγ x / L e− jω t I − jωγ x / L 0 0 ⎪ u(x,t) = Aω ⎨ − 2 jω I 0 ( − jωγ ) ⎪ 2 jω I 0 ( jωγ ) ⎩

(

e jβ nt/γ I 0 jβ n x / L

∞

+∑

(

)

) +e

− jβ nt/γ

(

)

(

I 0 − jβ n x / L

)

) ⎫⎪⎬

γ ω 2 − β n2 / γ 2 I1 ( jβ n ) γ ω 2 − β n2 / γ 2 I1 ( − jβ n ) ⎪ ⎭

n=1

(

)

(

⎧ e jω t J ωγ x / L e− jω t J ωγ x / L 0 0 ⎪ = Aω ⎨ − 2 jω J 0 (ωγ ) ⎪ 2 jω J 0 (ωγ ) ⎩

(

e jβ nt/γ J 0 β n x / L

∞

+∑

(

)

) − e

(

jγ ω 2 − β n2 / γ 2 J1 ( β n )

n=1

(

) (

− jβ nt/γ

)

(

J0 βn x / L

)

)

jγ ω 2 − β n2 / γ 2 J1 ( β n )

(

)

⎫ ⎪ ⎬ ⎪ ⎭

⎡ J ωγ x / L ⎤ ∞ J0 βn x / L 0 1 jω t − jω t 1 jβ nt/γ − jβ nt/γ ⎥ ⎢ = Aω e −e +∑ e −e 2 2 2 ⎢ ω J (ωγ ) 2 j ⎥ J1 ( β n ) j n=1 γ ω − β n / γ 0 ⎢⎣ ⎥⎦ =A

(

J 0 ωγ x / L J 0 (ωγ )

)

) sinω t + 2 Aω ∑

(

J0 βn x / L

∞

n=1

(

(

γ ω −β /γ 2

2 n

2

)

)

) J (β ) 1

(

)

sin β nt / γ

n

Mathematica verification of one term in series arg[z_,s_]:=Exp[z t] BesselI[0,z g Sqrt[x/L]]/(z+I w)/(z-I w)/ If[s==0,BesselI[0,z g],D[BesselI[0,z g],z]] r1=Expand[(z-I w) arg[z,0]]/.z->I w; r2=Expand[(z+I w) arg[z,0]]/.z->-I w; r3=arg[z,1]/.z->I b/g; r4=arg[z,1]/.z->-I b/g; FullSimplify[w(r1+r2)] FullSimplify[w (r3+r4)] 8.18 Consider the non-dimensional equation governing the static displacement of beam having an in-span elastic support at hs, 0 < hs < 1, and subjected to an externally applied load q(h) = qod(h - ho), 0 < ho < 1

d 4w + K sδ (η − ηs ) y = qoδ (η − ηo ) dη 4 The boundary condition at h = 0 are

143

0 ≤η ≤1

(d)

w¢¢¢(0) = - K L w(0),

w¢¢(0) = KtL w¢(0)

(e)

w¢¢¢(1) = - K R w(1),

w¢¢(1) = KtR w¢(1)

(f)

and those at h = 1 are where the prime indicates the derivative with respect to h and the K’s are constants that can vary between 0 and ¥.

Using the Laplace transform method, obtain the general solution for w(h) in terms of w(hs), Ka, and qo. Judiciously define groups of terms at appropriate solution stages in the solution process; this will greatly simplify the algebra.

Solution: We shall solve Eqs. (d) to (f) by using the Laplace transform with respect to h. Then the Laplace Transform of Eq. (d) is

s 4W (s) = qo e−ηos − K s e−ηss w(ηs ) + s3 w(0) + s 2 w′(0) + sw′′(0) + w′′′(0)

(g)

Using Eq. (e) in Eq. (g), we obtain

1 ⎡ − ηo s q e − K s e−ηss w(ηs ) + s3 w(0) + s 2 w′(0) + sK tL w′(0) − K L w(0) ⎤⎦ 4 ⎣ o s 1 = 4 ⎡⎣ qo e−ηos − K s e−ηss w(ηs ) + s3 − K L w(0) + s 2 + sK tL w′(0) ⎤⎦ s e − ηo s e − ηs s 1 1 = qo 4 − K s w(ηs ) 4 + 4 s3 − K L w(0) + 4 s 2 + sK tL w′(0) s s s s

W (s) =

(

)

(

)

(

)

(

(h)

)

Using Eqs. (8.9) and (8.32), the inverse Laplace transform of Eq. (h) is

w(η ) =

qo K (η − ηo )3 u(η − ηo ) − s w(ηs )(η − ηs )3 u(η − ηs ) 6 6 ⎛ ⎞ ⎛ ⎞ K K + ⎜ 1− L η 3 ⎟ w(0) + ⎜ η + tL η 2 ⎟ w′(0) 6 2 ⎝ ⎠ ⎝ ⎠

(i)

where u(h) is the unit step function.

To determine w(0) and w¢(0), we must substitute Eq. (i) into Eq. (f). Before doing this, we note that

K K 1 w′(η ) = qo (η − ηo )2 u(η − ηo ) − s w(ηs )(η − ηs )2 u(η − ηs ) − L η 2 w(0) + (1+ K tLη ) w′(0) 2 2 2 w′′(η ) = qo (η − ηo )u(η − ηo ) − K s w(ηs )(η − ηs )u(η − ηs ) − K Lη w(0) + K tL w′(0) w′′′(η ) = qou(η − ηo ) − K s w(ηs )u(η − ηs ) − K L w(0) where we have used the fact that du(h - ha)/dh = d(h - ha) = 0 at h = 1. Then

144

⎛ ⎛ K K ⎞ K ⎞ 1 w(1) = (1− ηo )3 − s w(ηs )(1− ηs )3 + ⎜ 1− L ⎟ w(0) + ⎜ 1+ tL ⎟ w′(0) 6 6 6 ⎠ 2 ⎠ ⎝ ⎝ K K 1 w′(1) = qo (1− ηo )2 − s w(ηs )(1− ηs )2 − L w(0) + (1+ K tL ) w′(0) 2 2 2 w′′(1) = qo (1− ηo ) − K s w(ηs )(1− ηs ) − K L w(0) + K tL w′(0)

(j)

w′′′(1) = qo − K s w(ηs ) − K L w(0) Then, substituting Eq. (j) into Eq. (f), we find that

qo − K s w(ηs ) − K L w(0) ⎡q ⎤ ⎛ ⎛ K K ⎞ K ⎞ + K R ⎢ o (1− ηo )3 − s w(ηs )(1− ηs )3 + ⎜ 1− L ⎟ w(0) + ⎜ 1+ tL ⎟ w′(0) ⎥ = 0 6 6 ⎠ 2 ⎠ ⎝ ⎝ ⎣6 ⎦ ⎡q ⎤ K qo − K s w(ηs ) + K R ⎢ o (1− ηo )3 − s w(ηs )(1− ηs )3 ⎥ 6 ⎣6 ⎦

(k)

⎡ ⎛ ⎤ ⎛ K ⎞ K ⎞ + ⎢ K R ⎜ 1− L ⎟ − K L ⎥ w(0) + K R ⎜ 1+ tL ⎟ w′(0) = 0 6 ⎠ 2 ⎠ ⎝ ⎣ ⎝ ⎦ and

qo (1− ηo ) − K s w(ηs )(1− ηs ) − K L w(0) + K tL w′(0) ⎡q ⎤ K K − K tR ⎢ o (1− ηo )2 − s w(ηs )(1− ηs )2 − L w(0) + (1+ K tL ) w′(0) ⎥ = 0 2 2 ⎣2 ⎦ ⎡q ⎤ K qo (1− ηo ) − K s w(ηs )(1− ηs ) − K tR ⎢ o (1− ηo )2 − s w(ηs )(1− ηs )2 ⎥ 2 ⎣2 ⎦

(l)

⎡K K ⎤ + ⎢ tR L − K L ⎥ w(0) + ⎡⎣ K tL − K tR (1+ K tL ) ⎤⎦ w′(0) = 0 ⎣ 2 ⎦ We choose to write Eq. (l) in matrix form as

⎡ a ⎢ 11 ⎢ a21 ⎣

a12 ⎤ ⎧⎪ w(0) ⎫⎪ ⎧⎪ − A1 + K s w(ηs ) A2 ⎫⎪ ⎥⎨ ⎬= ⎨ ⎬ a22 ⎥ ⎪ w′(0) ⎪ ⎪ − B1 + K s w(ηs )B2 ⎪ ⎩ ⎭ ⎦ ⎩ ⎭

where

æ K ö a12 = K R ç1 + tL ÷ 2 ø è

æ K ö a11 = K R ç1 - L ÷ - K L , 6 ø è K K a21 = tR L - K L , 2

a22 = KtL - KtR (1 + KtL )

and

145

(m)

A1 = qo + K R

qo (1− ηo )3 6

KR (1− ηs )3 6 K q B1 = qo (1− ηo ) − tR o (1− ηo )2 2 K tR B2 = (1− ηs ) − (1− ηs )2 2 A2 = 1+

The solution to Eq. (m) is

1 [C11 + K s w(hs )C12 ] D 1 w¢(0) = [C21 + K s w(h s )C22 ] D w(0) =

(n)

where

D = a11a22 - a12 a21 C11 = a12 B1 - a22 A1 C12 = a22 A2 - a12 B2 C21 = a21 A1 - a11 B1 C22 = a11 B2 - a21 A2 Substituting Eq. (n) into Eq. (i), we arrive at

qo K (η − ηo )3 u(η − ηo ) − s w(ηs )(η − ηs )3 u(η − ηs ) 6 6 ⎛ ⎞ 1 ⎛ ⎞ K K 1 + ⎡⎣C11 + K s w(ηs )C12 ⎤⎦ ⎜ 1− L η 3 ⎟ + ⎡⎣C21 + K s w(ηs )C22 ⎤⎦ ⎜ η + tL η 2 ⎟ D 6 2 ⎝ ⎠ D ⎝ ⎠

w(η ) =

=

qo K (η − ηo )3 u(η − ηo ) − s w(ηs )(η − ηs )3 u(η − ηs ) 6 6 ⎞ ⎛ ⎞⎤ K K 1⎡ ⎛ + ⎢C11 ⎜ 1− L η 3 ⎟ + C21 ⎜ η + tL η 2 ⎟ ⎥ D⎣ ⎝ 6 2 ⎠ ⎝ ⎠⎦ +

w(η ) =

⎛ ⎞ ⎫⎪ K s w(ηs ) ⎧⎪ ⎛ KL 3⎞ K η ⎟ + C22 ⎜ η + tL η 2 ⎟ ⎬ ⎨C12 ⎜ 1− D ⎪⎩ ⎝ 6 2 ⎠ ⎝ ⎠ ⎪⎭

qo K (η − ηo )3 u(η − ηo ) − s w(ηs )(η − ηs )3 u(η − ηs ) + K s w(ηs )E1 (η ) + E2 (η ) 6 6

where

146

(p)

E1 (h ) =

K 1 ì æ KL 3 ö æ öü h ÷ + C22 çh + tL h 2 ÷ ý íC12 ç1 Dî è 6 2 ø è øþ

E2 (h ) =

K 1 ì æ KL 3 ö æ öü h ÷ + C21 çh + tL h 2 ÷ ý íC11 ç1 Dî è 6 2 ø è øþ

8.19 Take the Laplace transform with respect to t to obtain a solution to the following nondimensional equation

∂u ∂ 2 u = ∂τ ∂η 2

0 ≤η ≤1 τ > 0

with zero initial conditions and with the boundary conditions

∂u(0, τ ) = −1 ∂η

∂u(1, τ ) =0 ∂η

Solution: Taking the Laplace transform of the governing equation and the boundary conditions and using the initial condition gives

∂ 2 U(η, s) − sU(η, s) = 0 ∂η 2 and

∂U(0, s) 1 =− ∂η s

∂U(1, s) =0 ∂η

The solution to the transformed equation is

U(η, s) = A cosh

( sη ) + Bsinh ( sη )

Substituting this result into the boundary conditions yields

(

)

(

∂U(0, s) 1 = − = A s sinh s 0 + B s cosh s 0 ∂η s 1 →B=− s s ∂U(1, s) = A s sinh s + B s cosh s = 0 ∂η

( ) cosh ( s ) →A= s s sinh ( s )

and, therefore,

147

( )

)

( s ) cosh ( sη ) − sinh ( s ) sinh ( sη ) sinh ( s ) cosh ( s[η − 1]) = s s sinh ( s )

U(η, s) =

1 cosh s s

It is seen that we have an second order pole at s = 0 and an infinite set of simple poles at

sinh

( s ) = 0 → − j sin j s = 0 → − j s = nπ → s = −n π 2

2

Using Bromwich’s integral,

(

)

2 zτ 2π j d ⎛ z cosh z[η − 1] e ⎞ u(η, τ ) = ⎜ ⎟ 2π j dz ⎜⎝ ⎟⎠ z z sinh z

(

( ) )

zτ d ⎛ z cosh z[η − 1] e ⎞ = ⎜ ⎟ dz ⎜⎝ ⎟⎠ sinh z

( )

(

)

zτ 2π j ∞ cosh z[η − 1] e + ∑ 2π j n=1 z z d sinh z z=0 dz z=zk =− n 2π 2

( )

cosh ( jnπ [η − 1]) e− n π τ +∑ −n 2π 2 n=1 cosh ( jnπ ) z=0 2 2 2

∞

η2 1 2 ∞ cosh ( jnπ [η − 1]) e− n π τ = −η + +τ − 2 ∑ 2 3 π n=1 n 2 cosh ( jnπ ) 2 2

η2 1 2 ∞ ⎡ cosh ( jnπ ) cosh ( jnπη ) − sinh ( jnπ ) sinh ( jnπη ) ⎤⎦ e = −η + +τ − 2 ∑ ⎣ 2 3 π n=1 n 2 cosh ( jnπ )

− n 2π 2τ

η2 1 2 ∞ cos ( nπη ) e− n π τ = −η + +τ − 2 ∑ 2 3 π n=1 n2 2 2

Mathematica verification of one term in series Limit[D[Sqrt[z] Cosh[(x-1) Sqrt[z]] Exp[z t]/Sinh[Sqrt[z]],z], z->0] FullSimplify[ Cosh[(x-1) Sqrt[z]] Exp[z t]/z^(3/2)/ D[Sinh[Sqrt[z]],z]/.z->-n^2 Pi^2,Assumptions->nÎIntegers] 8.20 Take the Laplace transform with respect to t to obtain a solution to the following partial differential equation

∂u ∂ 2 u ∂u = + ∂τ ∂η 2 η ∂η

0 ≤η ≤1 τ > 0

with the boundary conditions

148

lim | u(η, τ ) |< ∞ η→0

u(1, τ ) = sinΩτ and with the initial condition u(h,t) = 0.

Solution: Taking the Laplace transform of the partial differential equation and the boundary conditions, respectively, we obtain

η2

∂ 2 U(η, s) ∂U(η, s) +η − sη 2U(η, s) = 0 2 ∂η ∂η

and

lim |U(η, s) |< ∞ η→0

U(1, s) =

Ω s + Ω2 2

The solution to the transformed equation is

U(η, s) = AI 0

( sη ) + BK ( sη ) 0

For this solution to remain finite at h ® 0, B = 0. Then,

U(η, s) = AI 0

( sη )

Substituting this into the second boundary condition yields U(1, s) =

Ω = AI 0 s + Ω2 2

( s ) → A = ( s + ΩΩ) I ( s ) 2

2

0

Thus,

U(η, s) =

( sη ) (s + Ω ) I ( s ) ΩI 0

2

2

0

There are simple poles at s = ±jW and at the roots of

I0

( s) = 0

Using Eq. (5.126),

I0

( s ) = J ( j s ) → I ( − jy) = J ( y) = 0 → J (α ) = 0 0

0

0

0

n

where y = jx. Therefore, since y = an,

x = s → x = − jy → s = − jα n → sn = −α n2 The inverse is obtained by using the Bromwich integral, which gives

149

( z + jΩ ) ΩI 0 ( zη ) ezτ ( z − jΩ ) ΩI 0 ( zη ) ezτ u(η, τ ) = + ( z + jΩ )( z − jΩ ) I 0 ( z ) z=− jΩ ( z + jΩ )( z − jΩ ) I 0 ( z ) z= jΩ

( zη ) e +∑ ( z + Ω ) dzd I ( z ) zτ

ΩI 0

∞

2

n=1

2

0

z=− α n2

ΩI ( −α η ) e ( − jΩη ) e + ΩI ( jΩη ) e + = ∑ − j2ΩI ( − jΩ ) j2ΩI ( jΩ ) (α + Ω ) I ( −α ) / ( 2 −α ) jI ( − jΩη ) e jI ( jΩη ) e jα ΩI ( jα η ) e = − − 2∑ 2I ( − jΩ ) 2I ( jΩ ) (α + Ω ) I ( jα ) − jΩτ

ΩI 0

jΩτ

0

0

4 n

− jΩτ

jΩτ

0

2 n

1

2 n

− α n2τ

4 n

n=1

0

2

∞

n

0

− α n2τ

2 n

0

n=1

0

0

∞

0

n

2

1

n

Again using Eq. (5.126),

I m (x) = j − m J m ( jx) → I m ( jx) = j − m J m (−x) = (− j)− m J m (x) Therefore,

(

)

(

)

− jΩτ ∞ I 0 jΩη e jΩτ ⎞ jα nΩJ 0 (α nη ) e−α nτ j ⎛ I 0 − jΩη e u(η, τ ) = ⎜ − ⎟ − 2∑ 4 2 2 ⎜⎝ ⎟⎠ I 0 − jΩ I 0 jΩ n=1 (α n + Ω ) J1 (α n ) / (− j)

(

(

)

)

(

(

)

2

)

− jΩτ ∞ I 0 jΩη e jΩτ ⎞ α ΩJ (α η ) e−α nτ j ⎛ I 0 − jΩη e = ⎜ − ⎟ − 2∑ n 4 0 2 n 2 ⎜⎝ ⎟⎠ I 0 − jΩ I 0 jΩ n=1 (α n + Ω ) J1 (α n )

(

)

(

2

)

Mathematica verification of one term in series arg[z_,s_]:=Exp[z t] Om BesselI[0,x Sqrt[z]]/(z+I Om)/(z-I Om)/ If[s==0,BesselI[0,Sqrt[z]],D[BesselI[0,Sqrt[z]],z]] r1=Expand[(z-I Om) arg[z,0]]/.z->I Om; r2=Expand[(z+I Om) arg[z,0]]/.z->-I Om; Simplify[(r1+r2)] FullSimplify[PowerExpand[arg[z,1]/.z->-an^2]] 8.21 Take the Laplace transform with respect to t to obtain a solution to the following partial differential equation

∂u ∂2 u = a2 2 ∂t ∂x

0≤x≤L

with the boundary conditions

150

t>0

∂u(0,t) =0 ∂x ∂u(L,t) κ = α (1− β t) ∂x and with zero initial condition.

Solution: Taking the Laplace transform of the governing equation, we obtain

∂2U s − U=0 ∂x 2 a 2

(a)

∂U(0, s) =0 ∂x ∂U(L, s) α ⎛ 1 1⎞ = ⎜ −β 2⎟ ∂x κ ⎝s s ⎠

(b)

and the boundary conditions become

where U = U(x,s) is the Laplace transform of u(x,t).

The solution to Eq. (a) is

U(x, s) = A cosh(qx) + Bsinh(qx)

(c)

where q = s a. Substituting Eq. (b) into the first boundary condition of Eq. (c) gives

dU(x, s) dU(0, s) = Aqsinh(qx) + Bq cosh(qx) → =B=0 dx dx Then Eq. (c) becomes

U(x, s) = A cosh(qx)

(d)

Substituting Eq. (d) into the second boundary condition of Eq. (b) yields

∂U(L, s) α ⎛ 1 1⎞ = ⎜ −β 2⎟ ⎝ ∂x κ s s ⎠

α ⎛1 1⎞ ⎜⎝ − β 2 ⎟⎠ κ s s α 1⎞ ⎛1 A= ⎜⎝ − β 2 ⎟⎠ κ qsinh(qL) s s

Aqsinh(qL) =

(e)

Then, Eq. (d) becomes

U(x, s) =

α ⎛1 1 ⎞ cosh(qx) ⎜⎝ − β 2 ⎟⎠ κ s s qsinh(qL)

(f)

To determine the inverse Laplace transform of Eq. (f), we use the Bromwich inversion integral, which for this case becomes

151

γ +∞j

1 α 1 ⎞ cosh(qx) zt ⎛1 U(x,t) = e dz ⎜⎝ − β 2 ⎟⎠ ∫ 2π j κ γ −∞j z z qsinh(qL)

(g)

where, now, q = z a. It is straightforward to show that zqsinh(qL) has a second order pole at z = 0 and that z2qsinh(qL) has a third order pole at z = 0. The other simple poles are those where sinh(qL) = 0. Thus, if qL = jl, then sinh(jl) = sin(l) = 0 and l = np. Therefore,

qL = jnπ zL = jnπ a z → zn = −

n 2π 2 a 2 L2

Then, using residue theorem,

⎡ ⎤ ⎢ cosh(qx)ezt ⎥ α d ⎡ z 2 cosh(qx)ezt ⎤ α ∞ 1 U(x,t) = lim ⎢ + ∑ lim ⎢ ⎥ ⎥ d κ z→0 dz ⎣ zqsinh(qL) ⎦ κ k=1 z→zn ⎢ zq sinh(qL) ⎥ dz ⎣ ⎦ ⎡ ⎤ ⎢ cosh(qx)ezt ⎥ αβ d 2 ⎡ z 3 cosh(qx)ezt ⎤ αβ ∞ 1 − lim − lim ⎢ ∑ ⎥ ⎢ ⎥ 2 2 2 d κ z→0 2!dz ⎣ z qsinh(qL) ⎦ κ k=1 z→zn ⎢ zq sinh(qL) ⎥ dz ⎣ ⎦

(h)

We use the following Mathematica program to determine the limits of the first and third terms. q=Sqrt[z]/a; der1=Limit[D[z Cosh[q x]Exp[t z]/(q Sinh[q l]),z],z->0] der2=Limit[D[z Cosh[q x]Exp[t z]/(q Sinh[q l]),z,z],z->0] The results are

d ⎡ z 2 cosh(qx)ezt ⎤ a 2 ⎛ x2 L2 ⎞ lim ⎢ ⎥ = ⎜t + 2 − 2 ⎟ z→0 dz ⎣ zqsinh(qL) ⎦ L ⎝ 2a 6a ⎠ d 2 ⎡ z 3 cosh(qx)ezt ⎤ a 2 ⎛ t 2 tx 2 tL2 x4 L2 x 2 7L4 ⎞ lim = + − + − + 2 2 4 4 ⎜ z→0 2!dz 2 ⎢ z 2 qsinh(qL) ⎥ 360a 4 ⎟⎠ ⎣ ⎦ L ⎝ 2 2a 6a 24a 12a The evaluation of the second and fourth terms yields

152

(i)

⎡ ⎤ 2 2 2 2 ⎢ cosh(qx)ezt ⎥ ⎡ ⎤ 1 cosh(qx)ezt 2L cosh( jnπ x / L)e− n π a t /L lim ⎢ =− ⎥ = lim z→zn z→zn ⎢ zL cosh(qL) / (2a 2 ) ⎥ d zq n 2π 2 cosh( jnπ ) ⎣ ⎦ ⎢ sinh(qL) ⎥ dz ⎣ ⎦ ⎡ ⎤ 2 2 2 2 ⎢ cosh(qx)ezt ⎥ ⎡ ⎤ 2L3 cosh( jnπ x / L)e− n π a t /L 1 cosh(qx)ezt lim ⎢ = ⎥ = lim z→zn z→zn ⎢ z 2 L cosh(qL) / (2a 2 ) ⎥ d z2q a 2 n 4π 4 cosh( jnπ ) ⎣ ⎦ ⎢ ⎥ sinh(qL) dz ⎣ ⎦ where we have used the fact that q = jnp/L. Noting that cosh(jnp) = cos(np) = (-1)n and using Eq. (i) , Eq. (h) becomes

U(x,t) =

2 2 2 2 α a2 ⎛ x2 L2 ⎞ 2α L ∞ (−1)n t + − − cos(nπ x / L)e− n π a t /L 2 2⎟ 2 ∑ 2 ⎜ κ L ⎝ 2a 6a ⎠ κπ k=1 n

−

αβ a 2 ⎛ t 2 tx 2 tL2 x4 L2 x 2 7L4 ⎞ + − + − + κ L ⎜⎝ 2 2a 2 6a 2 24a 4 12a 4 360a 4 ⎟⎠

−

2 2 2 2 2αβ L3 ∞ (−1)n cos(nπ x / L)e− n π a t /L 4 ∑ 4 κπ k=1 n

8.22 Take the Laplace transform with respect to t to obtain a solution to the following partial differential equation

∂u ∂2 u = a2 2 ∂t ∂x

0≤x<∞

t>0

with the boundary conditions that u is finite as x ® ¥ and u(0,t) = f(t) and that the initial condition is zero.

Solution: Taking the Laplace transform of the governing equation with respect to t and using the initial condition, we obtain

d 2u (x, s) s − 2 u (x, s) = 0 dx 2 a The Laplace transform of the boundary condition at x = 0 is

u (0, s) = F(s) where F(s) is the Laplace transform of f(t). The solution is

u (x, s) = C1e− qx + C2 eqx where q = s / a. Since the solution must remain finite as x ® ¥, we set C2 = 0. Then

u (x, s) = C1e− qx

153

Substituting this result into the remaining boundary condition, we find that

u (0, s) = F(s) = C1 Then the solution is

u (x, s) = F(s)e− qx Taking the inverse Laplace transform with the aid of convolution, we have first that

L−1 [ F(s)] = f (t) − x /(4ta ) ⎤ = x e 3/2 ⎦ 2a π t 2

L ⎡⎣ e −1

− sx/a

2

Then, from Eq. (8.25) 2

t

2

e− x /(4ηa ) u(x,t) = f (t − η ) dη ∫ η 3/2 2a π 0 x

Mathematica verification InverseLaplaceTransform[Exp[-p Sqrt[s]],s,t]

8.23 Take the Laplace transform with respect to t to obtain a solution to the following partial differential equation

∂u ∂ 2 u = 2 − a 2u ∂t ∂x

−1 ≤ x <1 t > 0

with the boundary conditions

u(−1,t) = 1 u(1,t) = 0 and zero initial condition; that is, u(x,0) = 0.

Solution: Taking the Laplace transform of the governing equation with respect to t and using the initial condition, we obtain

∂ 2 u (x, s) − (s + a 2 )u (x, s) = 0 2 ∂x and the Laplace transform of the boundary conditions are

1 s u (1, s) = 0

u (−1, s) =

The solution to the transformed differential equation is

154

u (x, s) = C1 cosh

( s + a x ) + C sinh ( s + a x ) 2

2

2

Substituting this result into the boundary conditions gives

(

)

(

) 1s C cosh ( s + a ) + C sinh ( s + a ) = 0

C1 cosh − s + a 2 + C2 sinh − s + a 2 = 2

2

1

2

Placing in matrix form gives

⎛ cosh ⎜ ⎜ ⎜ cosh ⎝

( s + a ) − sinh ( s + a ) ⎞⎟ ⎧⎪ C ⎫⎪ = ⎧⎪ 1 ⎫⎪ ⎟⎨ C ⎬ ⎨ s ⎬ s + a sinh s + a ( ) ( ) ⎟⎠ ⎩⎪ ⎭⎪ ⎪⎩ 0 ⎪⎭ 2

2

1

2

2

2

Solving for C1 and C2 yields

( s + a ) = 1 sinh s + a ( ) sD 0 sinh ( s + a ) 1 − sinh s

1 C1 = D

2

2

2

1 s + a ( ) 1 s = − 1 cosh s + a C = ( ) D sD cosh ( s + a ) 0 2

cosh

2

2

2

where

D = 2 cosh

( s + a )sinh ( s + a ) = sinh (2 s + a ) 2

2

2

Therefore,

u (x, s) =

=

{sinh ( s + a ) cosh ( s + a x ) − cosh ( s + a )sinh ( s + a x )} s sinh ( 2 s + a ) 1

2

2

2

2

2

(

sinh (1− x) s + a 2

(

s sinh 2 s + a 2

)

)

To obtain the inverse transform, we note that s is a simple pole and we have an infinite number of simple poles that are obtained from

(

)

sinh 2 s + a 2 = 0 → jnπ = 2 s + a 2 → s = sn = − Then, the inverse transform is given by

155

n 2π 2 − a 2 = − λn 4

(

sinh (1− x) z + a 1 u(x,t) = ∫ z sinh 2 z + a 2 2π j γ −∞j γ +∞j

(

2

)

) e dz zt

Then, the residue at z = 0 is

(

)

⎡ z sinh (1− x) z + a 2 ⎤ sinh ( a(1− x)) zt ⎥ ⎢ Res[z = 0] = lim e = 2 z→0 ⎢ ⎥ sinh ( 2a ) ⎢⎣ z sinh 2 z + a ⎥⎦

(

)

The residues at z = -ln are

(

)

⎡ ⎤ 2 zn t ⎢ sinh (1− x) zn + a e ⎥ 1 Res[z = − λn ] = ∑ ⎢ ⎥ d zn n=1 ⎢ ⎥ sinh 2 z + a 2 z=zn ⎦ dz ⎣ ∞

(

)

(

)

(

)

⎡ ⎤ 2 − λnt ⎢ sinh (1− x) − λn + a e ⎥ 1 = ∑⎢ ⎥ d − λn n=1 ⎢ ⎥ sinh 2 z + a 2 z=zn ⎦ dz ⎣ ∞

(

)

⎡ sinh (1− x) − λ + a 2 e− λnt ⎤ n − λn + a 2 ⎢ ⎥ =∑ 2 ⎥ ⎢ − λ n=1 n cosh 2 − λn + a ⎥ ⎢⎣ ⎦ ∞

(

)

But,

− λn + a 2 = −a 2 − n 2π 2 / 4 + a 2 = jnπ / 2 Therefore,

⎡ j sin ( nπ (1− x) / 2 ) e−(a2 +n2π 2 /4 )t jnπ / 2 ⎤ Res[z = − λn ] = ∑ ⎢ ⎥ 2 2 2 cos n π ( ) − a + n π / 4 ( ) n=1 ⎢ ⎥⎦ ⎣ ∞

π ∞ (−1)n nsin ( nπ (1− x) / 2 ) e−(a +n π /4 )t = ∑ 2 n=1 ( a 2 + n 2π 2 / 4 ) 2

2 2

and the solution is

sinh ( a(1− x)) π ∞ (−1)n nsin ( nπ (1− x) / 2 ) e−(a +n π /4 )t u(x,t) = + ∑ sinh ( 2a ) 2 n=1 ( a 2 + n 2π 2 / 4 ) 2

2 2

Mathematica verification of one term in series arg[z_,s_]:=Exp[z t]Sinh[(1-x) Sqrt[z+a^2]]/z/ If[s==0,Sinh[2 Sqrt[z+a^2]],D[Sinh[2 Sqrt[z+a^2]],z]] PowerExpand[Expand[z arg[z,0]]/.z->0]

156

Simplify[PowerExpand[arg[z,1]/.z->(-n^2 Pi^2/4-a^2)], Assumptions->nÎIntegers]

157