# Advanced Engineering Mathematics with Mathematica (Solution manual) 9781000034523, 1000034526

##### Advanced Engineering Mathematics with Mathematica® presents advanced analytical solution methods that are used to solve

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Solution Manual

1

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⎜ X= ⎜ 2⎜ ⎜⎝

−1 −1 ⎞ ⎟ 1 −1 ⎟ −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 ⎝ 21 22 ⎠ ⎝ 21 22 ⎠ ⎜⎝ a21 ( a11 + a22 ) a21a12 + a22 ⎛ a 2b 2 − a 2b 2 b 2 ( ab − ab ) =⎜ ⎜⎝ −a 2 ( ab − ab ) −a 2b 2 + a 2b 2

⎞ ⎟ =0 ⎟⎠

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 }⎛⎜⎝

2 a ⎞⎧ 1 ⎫ ⎨ ⎬= 3 4 ⎟⎠ ⎩ 2 ⎭

{ 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 det ⎜ b 1+ b b ⎜ b 1+ b ⎝ b

⎞ ⎟ = (1+ a ) ⎡(1+ b)2 − b 2 ⎤ − a ⎡b(1+ b) − b 2 ⎤ + a ⎡b 2 − b(1+ 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 ⎜ 3 det ⎜ 1 x +1 1 3 ⎜⎝ 1 1 x +1

⎞ ⎟ ⎟ ⎟⎠

Solution: ⎛ x3 + 1 1 1 ⎜ 3 det ⎜ 1 x +1 1 3 ⎜⎝ 1 1 x +1

⎞ ⎟ = x 3 + 1 ⎡ x 3 + 1 2 − 1⎤ − ⎡ x 3 + 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:

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⎛ 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

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⎛ 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

⎛ 2 1 3 ⎞ T T B A =⎜ 1 2 5 ⎟ ⎜ ⎟ ⎜⎝ 0 2 1 ⎟⎠

T

T

⎛ 3 1 4 ⎞ ⎛ 2 1 0 ⎞⎛ 3 2 4 ⎞ ⎜ 2 1 2 ⎟ =⎜ 1 2 2 ⎟⎜ 1 1 2 ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜⎝ 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 ⎜⎜ ⎜ det ⎜ 4 6− λ 5 ⎜⎜ ⎜⎝ 7 8 9− λ

⎞⎟ ⎛ ⎟⎟ 5 ⎟⎟ = (1− λ )det ⎜⎜ 6− λ ⎜⎜ 8 ⎟⎟ 9− λ ⎝ ⎟⎠ ⎛ ⎞ + 3det ⎜⎜⎜ 4 6− λ ⎟⎟⎟ ⎜⎝ 7 8 ⎟⎠

⎞⎟ ⎛ 5 ⎟⎟ − 2det ⎜⎜ 4 ⎜⎜ 7 9− λ ⎟⎠ ⎝

⎞⎟ ⎟⎟ ⎟⎠

= (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

11

⎛ 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 2 1− λ

⎞ ⎛ −2 −1− λ ⎞ 2 ⎞ ⎟ = (1− λ ) det ⎛ −1− λ − det ⎜ ⎜⎝ −1 ⎟⎠ ⎟ 2 1− λ ⎟⎠ 2 ⎝ ⎠ = − (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

12

⎞⎧ ⎟⎪ ⎟⎨ ⎟⎠ ⎪ ⎩ ⎛ 3 0 −1 ⎞ ⎧ ⎪ ⎜ −2 1 2 ⎟ ⎨ ⎜ ⎟ ⎝ −1 2 3 ⎠ ⎪ ⎩

⎛ 1− (−2) 0 −1 ⎜ −2 −1− (−2) 2 ⎜ ⎜⎝ −1 2 1− (−2)

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

Therefore,

⎛ det ⎜ ⎝ c2 = ⎛ det ⎜ ⎝

2 2 ⎞ 1 3 ⎟⎠ 4 = = −4 1 2 ⎞ −1 2 3 ⎟⎠

⎛ det ⎜ ⎝ c3 = ⎛ det ⎜ ⎝

1 2 ⎞ 2 1 ⎟⎠ −3 = =3 1 2 ⎞ −1 2 3 ⎟⎠

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

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

Therefore,

⎛ det ⎜ ⎝ c2 = ⎛ det ⎜ ⎝

2 2 ⎞ 1 1 ⎟⎠ 0 = =0 −1 2 ⎞ −5 2 1 ⎟⎠

⎛ det ⎜ ⎝ c3 = ⎛ det ⎜ ⎝

−1 2 ⎞ 2 1 ⎟⎠ −5 = =1 −1 2 ⎞ −5 2 1 ⎟⎠

and x(2) = {1, 0,1}T. For l3 = 3, ⎞ ⎧ c1 ⎫ ⎪ ⎟⎪ c ⎨ ⎬=0 2 ⎟ ⎟⎠ ⎪ c3 ⎪ ⎩ ⎭ ⎧ ⎫ −2 0 −1 ⎞ ⎪ c1 ⎪ ⎛ −4 2 ⎞ ⎧⎪ c2 ⎫⎪ ⎧ 2 ⎫ ⎟ c → −2 −4 2 ⎨ 2 ⎬ ⎜ ⎟⎠ ⎨ c ⎬ = ⎨ 1 ⎬ ⎟⎪ 2 −2 ⎝ ⎭ ⎩⎪ 3 ⎪⎭ ⎩ −1 2 −2 ⎠ c3 ⎪ ⎩ ⎭

⎛ 1− (3) 0 −1 ⎜ −1− (3) 2 ⎜ −2 ⎜⎝ −1 2 1− (3) ⎛ ⎜ ⎜ ⎝

13

Therefore,

⎛ det ⎜ ⎝ c2 = ⎛ det ⎜ ⎝

2 2 ⎞ 1 −2 ⎟⎠ −6 −3 = = 4 2 −4 2 ⎞ ⎟ 2 −2 ⎠

⎛ det ⎜ ⎝ c3 = ⎛ det ⎜ ⎝

−4 2 ⎞ 2 1 ⎟⎠ −8 = = −2 4 −4 2 ⎞ 2 −2 ⎟⎠

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

( )

For the eigenvectors are linearly independent x (i)

(x )

(1) T

x

(2)

=

{1

−4

3

T

}

⎞ ⎟ ⎟ ⎠

x ( j ) = 0, i ≠ j; therefore,

⎧ 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]]

14

Solutions to Exercises in Chapter 2 Section 2.1 2.1 Show that

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

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

2

+ b2 ) = 1 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 j tan

−1

1

) =( 4

2e jπ /4

) = 4e 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

= −4

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 =⎜ + e ⎝ 4 4

3/(−1)

⎞ ⎟ ⎠

3/2

(

= e j(2π /3+2mπ )

)

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

= e 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 ∞

∑q

=

n

n=0

1 1− q

q {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

n

sin nt =

n=0

ε sint 1+ ε − 2ε cost 2

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 ) = x

e2 x = 2x e =

2

(

)

(

2

)

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

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

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 ) = x

e2 x = 2x e =

2

(

)

(

2

)

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

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

x

x

x

x

x

x

2

−x

−x

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

∂u 4x( y + 1) = ∂x x 2 + ( y + 1)2

(

(

v=−

)

2

∂v 4x( y + 1) = ∂y x 2 + ( y + 1)2

,

(

)

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

(

)

2x x + ( y + 1)2 2

(

)

2

)

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

C

(c)

e z cos z

( z − π / 2) zn

C ( z − a)

n+1

2

dz

dz

Solutions:

e-3p z 1 -3p z òC 2 z + j dz = 2p j 2 e e z cos z

ò ( z - p / 2)

2

dz = 2p j

C

zn

ò ( z - a)

C

dz = n +1

z =- j /2

= p je -3 jp /2 = p j (- j ) = p

d z e cos z ) = 2p j ( e z {cos z - sin z} ) = -2p jep /2 ( z = p /2 dz z =p /2

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

( = ( −1− j ) = (

z12 = ( −1+ j ) = z22

) = 2e ) = 2e

2

2e3π j/4

2

2e−3π j/4

2

3π j/2

2

= 2 ( cos(3π / 2) + j sin(3π / 2) ) = −2 j

−3π j/2

= 2 ( cos(3π / 2) − j sin(3π / 2) ) = 2 j

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

2 1/z

dz

|z|{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 0

n = odd n = even

Thus, cn is a real quantity. The same results are obtained for n < 0. When n = 0, 1 c0 = 2π

π

π 2

x dx =

−π

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 ∞

f (x) = c0 + 2∑ cn cos[2nπ x / T − ϕ n ] → c0 + 2 n=1

where

π 2 2 cn = 2 nπ c0 =

ϕ n = tan −1

n = odd 0 =π −2 / n 2π

(

Therefore, 31

)

n=1,3,5...

cn cos[nx − ϕ n ]

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&&n0] (*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 =

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

T /2 −T /2

2 sinh(α / 2) 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

+ 2sinh (α / 2 ) ∑

(

cos 2nπ t − tan −1 ( nπ / (α / 2 ) (α / 2) + (nπ ) 2

n=1

)

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) 0

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

(

td /2

∫e 0

)

− j 2 pπ t/T

je− j 2 pπ t/T dt = 2 pπ

(

t=0

t=−td

)

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)

ϕ n = tan

−1

sin 2 ( pπ td / (2T ) ) 0

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

36

=

π 2

je− j 2 pπ t/T − 2 pπ /2

t=td /2

t=0

Assumptions->wo>0&&td>0&&td2 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&&td2 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&&td2 Pi/T 37

(d) Case 5: From Eq. (3.2) and Mathematica 0

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

td /2

∫ (1− 2t / t ) e

− j 2 pπ t/T

d

dt

0

d

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

(

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&&tda,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

dt =

(α − jω )t

∫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,-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

dt −

td /2

∫e

− jω t

0

e− jω t dt = − jω

(

)

t=0

t=−td

e− jω t − − jω /2

t=td /2

t=0

(

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 F(ω ) =0 td

ω =0

ω ≠0

cp td / T

=

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

c0 =0 td / T

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

−td /2

cos (π t / td ) e− jω t dt = 2π td

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 ) =

−td /2

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

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

F(ω ) =

−td /2

=

( 2t / td + 1) e− jωt dt +

(

∫ (1− 2t / t ) e d

− jω t

dt

0

)

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 ⎠ F(0) =

td /2

2

td 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

=

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

c0 1 = td / T 2

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

2t p − jω t

dt − Ap

0

∫ {a − b cos[2π t / (2t )]} e p

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

− jω t

dt

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

(

)

(

(−A (e Ω (Ω − π ) j = (−A e Ω (Ω − π ) j

=

2

2

o

2

2

o

− jΩ

− jΩ

)

)

2

))

(

2

− 1 + Bo e− jΩ + 1

(e

− jΩ/2

− e jΩ/2

)

2

2

(

+ Bo e− jΩ e− 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

A sin (Ω / 2) + B cos (Ω / 2)) )(

=

2 je− jΩ Ω Ω2 − π 2

)

( ( (

2 je− jΩ = Ω Ω2 − π 2

(

2

)) 2

)

2

o

o

( Ao (1− cosΩ) + Bo (1+ cosΩ))

)(

Ao + Bo + ( Bo − Ao ) cosΩ )

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)

(e

x

+ x sin y )

dy + ye x - cos y = 0 dx

(c)

⎡ dy y = − k ⎢1− dx x ⎣

1/2

y⎤ x ⎥⎦

(d)

(3y

2

)

(

)

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

(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)

(e

x

+ x sin y )

dy + ye x - cos y = 0 dx

Solution Rearranging terms, we find

(e

x

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

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 dv k = − dx x 1+ v 2 v+x

Therefore,

48

( x v+ C(

dv 2

= −k ∫

dx x

1+ v sinh −1 v = −k ln x + lnC

) 1+ v ) = 1

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

2

)

(

)

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

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 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 (1, s) =

q1

(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, s) =

q1

q1 - 1 1 q ® A cosh s + - 1 ® A = s s s 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 2p j

ò g

- j¥

z dz 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)

has as a solution of the form

y( x) = xab éëC1 J p (kxb ) + C2Yp (kxb )ùû

(f)

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)

(

)

u ( L, s ) = C1 I 0 2s L / g = C1 =

(s

Aw

+w

2

2

) I ( 2s

Aw s + w2 2

L/g

0

)

Therefore, the solution in the Laplace transform domain is

u ( x, s) =

(

)

Aw I 0 2s x / g

(s

2

(

+ w ) I 0 2s L / g 2

=

(

Aw I 0 sg x / L

) (s

2

)

(h)

+ w ) I 0 ( sg ) 2

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

u ( x, t ) = Aw

g + j¥

ò (z

g - j¥

(

e zt I 0 zg x / L 2

) dz

+ w 2 ) I 0 ( zg )

(

)

ì é e zt I 0 zg x / L ï = Aw í lim ê z ® jw ê z + jw ) I 0 2 z L / g ïî ë(

(

(

é e zt I 0 zg x / L lim ê z ® sn ê ( z2 + w2 ) ë

¥

+å n =1

(

) ùú lim é

é e zt I 0 zg x / L lim ê z ®- sn ê z2 + w2 ) ( ë

¥

+å n =1

)

(

ú û

z ® sn

1 ê ë dI 0 ( zg )

(

ù ú dz û

) ùú lim é

ü ù ï 1 ê ú ý ú z ®- sn ë dI 0 ( zg ) dz û ï û 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

)

ù é e zt I 0 zg x / L ú + lim ê ú z ®- jw ê ( z - jw ) I 0 2 z L / g û ë

)

)

ù ú ú û

(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

+∑

(

)

)

jγ ω 2 − β n2 / γ 2 J1 ( β n )

n=1

(

) (

)

) (

e− jβ nt/γ J 0 β n 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 t/γ = Aω ⎢ e −e +∑ e −e n 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 ⎠ ⎝ ⎠⎦

⎛ ⎞ ⎫⎪ K s w(ηs ) ⎧⎪ ⎛ KL 3⎞ K η ⎟ + C22 ⎜ η + tL η 2 ⎟ ⎬ ⎨C12 ⎜ 1− D ⎪⎩ ⎝ 6 2 ⎠ ⎝ ⎠ ⎪⎭ q K w(η ) = o (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 2π 2

Using Bromwich’s integral,

(

)

(

)

2 zτ zτ 2π j d ⎛ z cosh z[η − 1] e ⎞ 2π j ∞ cosh z[η − 1] e u(η, τ ) = ⎜ ⎟ + ∑ 2π j dz ⎜⎝ 2π j n=1 z z d sinh z ⎟⎠ z z sinh z z=0 dz

(

( ) )

( )

z=zk =− n 2π 2

2 2 zτ ∞ cosh ( jnπ [η − 1]) e− n π τ d ⎛ z cosh z[η − 1] e ⎞ = ⎜ ⎟ +∑ −n 2π 2 dz ⎜⎝ ⎟⎠ sinh z n=1 cosh ( jnπ ) z=0 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

Ω

2

+ Ω2 ) I 0

( s)

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 ) 2

n=1

=

=

ΩI 0

(

(

2

0

)

− jΩη e− jΩτ

− j2ΩI 0 jI 0

ΩI 0

(

− jΩ

)

)

− jΩη e− jΩτ

2I 0

(

− jΩ

)

+

ΩI 0

z=− α n2

(

j2ΩI 0 jI 0

(

)

jΩη e jΩτ

(

)

+ ) ∑ (α

jΩη e jΩτ

2I 0

(

)

n=1 ∞

− 2∑ n=1

) + Ω ) I ( −α ) / ( 2 ΩI 0

4 n

(

−α n2 η e−α nτ 2

2

2 n

1

−α n2

)

jα nΩI 0 ( jα nη ) e−α nτ 2

4 n

+ Ω2 ) I1 ( jα 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≤x0

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) u(x,t) =

t

x 2a π

∫ 0

2

2

e− x /(4ηa ) f (t − η ) dη η 3/2

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 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 2 x + C2 sinh

(

s + a2 x

)

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 ) s + a2

− sinh

2

sinh

(

(

s + a2 s + a2

)

) ⎞⎟ ⎧⎪ C

⎫⎪ ⎧⎪ 1 ⎫⎪ ⎟⎨ C ⎬= ⎨ s ⎬ ⎟ ⎪⎩ 2 ⎭⎪ ⎪⎩ 0 ⎪⎭ ⎠ 1

Solving for C1 and C2 yields

1 − sinh s

1 C1 = D

C2 =

0

sinh

( cosh (

s + a2 s+a

) s+a )

2

)

)

=

1 sinh sD

(

1 s = − 1 cosh sD 0

s + a2

cosh

1 D

(

(

2

s + a2

(

)

s + a2

)

s + a 2 = sinh 2 s + a 2

)

where

D = 2 cosh

(

) (

s + a 2 sinh

)

(

Therefore,

u (x, s) =

=

{sinh ( s sinh ( 2 s + a ) 1

2

(

sinh (1− x) s + a 2

(

s sinh 2 s + a 2

)

) (

s + a 2 cosh

)

s + a 2 x − cosh

(

) (

s + a 2 sinh

s + a2 x

)}

)

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

(

)

γ +∞j sinh (1− x) z + a 2 1 u(x,t) = ezt dz ∫ 2 2π j γ −∞j z sinh 2 z + a

(

)

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 dz ⎣ ∞

(

)

(

)

(

⎡ 2 − λnt ⎢ sinh (1− x) − λn + a e 1 = ∑⎢ d − λn n=1 ⎢ sinh 2 z + a 2 dz ⎣ ∞

(

)

⎡ sinh (1− x) − λ + a 2 e− λnt n − λn + a 2 ⎢ =∑ ⎢ − λn n=1 cosh 2 − λn + a 2 ⎢⎣ ∞

(

)

⎤ ⎥ ⎥ ⎥ z=zn ⎦

⎤ ⎥ ⎥ ⎥ z=zn ⎦

)

⎤ ⎥ ⎥ ⎥⎦

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 = ∑ 2 n=1 ( a 2 + n 2π 2 / 4 ) 2

2 2

/4 )t

and the solution is

sinh ( a(1− x)) π ∞ (−1)n nsin ( nπ (1− x) / 2 ) e−(a u(x,t) = + ∑ sinh ( 2a ) 2 n=1 ( a 2 + n 2π 2 / 4 )

2

+n 2π 2 /4 )t

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