MMMS

Mathematical Methods for Mechanical Sciences

M. S. Howe iv


PREFACE

A mathematical (as opposed to computational) model of a physical system can provide theengineer with the necessary insight and intuitive understanding generally required to makeefficient system design changes or other system modifications. A simple formula will often revealconnections between different control parameters that might otherwise take hours or weeks todeduce from a full scale computer simulation. This book is intended to supply the undergraduateengineer with the basic mathematical tools for developing and understanding such models. A firmgrasp of the topics covered will also enable the working engineer (educated to Bachelor degreelevel) to understand, write and otherwise make sensible use of industrial technical reports.

The book is written for students taking the Boston University senior level course in EngineeringMathematics for Mechanical and Aerospace Engineers. This course represents the final exposure ofthese students to formal mathematical training prior to graduation. The intention is to consolidateearlier courses in ordinary differential equations, vector calculus, Fourier series and transforms andlinear algebra, and to introduce more advanced topics, including applications of complex variabletheory, partial differential equations and elementary generalised functions leading to Green’sfunctions. It is not possible to cover in class all topics with which an ‘educated’ engineer mightreasonably be expected to be familiar; some additional material is included in the text, mainly forreference, on conformal transformations, special functions and variational methods. However, anoverriding objective has been compactness of presentation, and to avoid the currently fashionabletrend of attempting to achieve encyclopedic coverage with a text that typically runs to a thousandor more pages.

M. S. Howe

M. S. Howe v Preface


GREEK ALPHABET

alpha α, A nu ν, N

beta β, B xi ξ, Ξ

gamma γ, Γ omicron o, O

delta δ, ∆ pi π, Π

epsilon ε, E rho ρ, P

zeta ζ, Z sigma σ, Σ

eta η, H tau τ, T

theta θ, Θ upsilon υ, Υ

iota ι, I phi φ, Φ

kappa κ, K chi χ, X

lambda λ, Λ psi ψ, Ψ

mu µ, M omega ω, Ω

MATHEMATICAL CONSTANTS

Euler’s γ = 0.5772 15665

exponential e = 2.7182 81828

π = 3.1415 92654

M. S. Howe vi


1. LINEAR ORDINARY DIFFERENTIAL EQUATIONS

1.1 First order equations

General form:dy

dx+ p(x)y = r(x), or y′ + p(x)y = r(x), where y′ =

dy

dx.

Homogeneous form: y′ + p(x)y = 0.

Solve by separating the variables:

∫dy

y= −

∫p(x)dx + C1, C1 = constant

. ˙ . ln y = −∫

p(x)dx + C1

. ˙ . The general solution is y = Ce−∫

p(x)dx, C = eC1 = arbitrary constant

Example: Find the general solution of y′ + x2y = 0.

∫dy

y= −

∫x2dx + C1,

. ˙ . ln y = −1

3x3 + C1

. ˙ . y = Ce−x3/3.

If y = 2 when x = 0, then C = 2 and y = 2e−x3/3.

Inhomogeneous form: y′ + p(x)y = r(x).

This is solved by multiplying by the integrating factor f(x) ≡ e∫

p(x)dx:

fy′ + fpy ≡ d

dx

(y(x)e

∫p(x)dx

)= r(x)e

∫p(x)dx

. ˙ . y(x)e∫

p(x)dx =∫

r(x)e∫

p(x)dx dx + C

. ˙ . y = e−∫

p(x)dx∫

r(x)e∫

p(x)dx dx + Ce−∫

p(x)dx

= particular integral + solution of the homogeneous equation

M. S. Howe 1 §1.1 First order equations


Example: Find the general solution of y′ + x2y = x2.

Integrating factor = e

∫x2dx

= ex3/3

. ˙ .d

dx

(y(x)ex3/3

)= x2ex3/3

. ˙ . y(x)ex3/3 =

∫x2ex3/3 dx + C

. ˙ . y = 1 + Ce−x3/3

If y = 2 when x = 0, then C = 1 and y = 1 + e−x3/3.

Problems 1A

Find the general solution of:

1. y′ − 4y = 2x − 4x2 [y = x2 + Ce4x]

2. xy′ + 2y = 4ex2[y = (C + 2ex2

)/x2]

3. y′ + 2y tan x = sin2 x [y = sin x cos x + (C − x) cos2 x]

4. y′ + y cot x = sin 2x [y = 23

sin2 x + Ccosecx]

5. sin xy′ − y cos x = sin 2x [y = 2 sin x ln(sinx) + C sin x]

6. x lnxy′ + y = 2 ln x [y = lnx + C/ ln x]

7. y′ + 2yx

= ex [y = C/x2 + (1 − 2/x + 2/x2)ex]

8. (x − 1)y′ + 3y = x2 [(x − 1)3y = C + x5/5 − x4/2 + x3/3]

9. (x + 1)y′ + (2x − 1)y = e−2x [e2xy = C(x + 1)3 − 13]

10. y′ + yx

= 12

sin(

x2

)[y = − cos x

2+ 2

xsin x

2+ C

x]

11. (1 − x2)y′ + x(y − a) = 0 [y = a + C(1 − x2)12 ]

12. y′ − (1 + cot x)y = 0 [y = Cex sin x]

13. (1 + x2)y′ + xy = 3x + 3x3 [y = 1 + x2 + C(1 + x2)−12 ]

14. sinx cos xy′ + y = cotx [y = (C + ln tan x)/ tan x]

Solve:

15. y′ + 2xy = 4x, y(0) = 3 [y = 2 + e−x2]

16. y′ coth 2x = 2y − 2, y(0) = 0 [y = 1 − cosh 2x]

17. y′ + ky = e−kx, y(0) = 1 [y = (1 + x)e−kx]

18. y′ = a(y − g), y(0) = b [y = g + (b − g)eax]

19. yy′ = 2a, y(0) = 0 [y2 = 4ax]



20. yy′ + x = 0, y(0) = a [x2 + y2 = a2]

21. yy′ + b2xa2 = 0, y(0) = b [x2

a2 + y2

b2= 1]

22. (x + 1)y′ = y − 3, y(0) = 8 [y = 5x + 8]

23. 2xy′ + y = 0, y(1) = 1 [xy2 = 1]

24. (1 + x2)y′ =√

y, y(0) = 0 [y = 14 (tan−1 x)2]

25. didt

+ 3i = sin 2t, i = 0 when t = 0 [i = sin(2t − α) + e−3t sinα/√

13, where tan α = 23]

26. Water runs out through a hole in the base of a circular cylindrical tank at speed√

2gh ft/sec, where g = 32

ft/sec2 and h is the water depth. If the tank is 2 ft in height, 1 ft in diameter and is full at time t = 0, calculate

the time at which half the water has run out when the effective area of the hole is 0.25 in2. [47 secs]

27. The current i in a circuit satisfies Ldi/dt + Ri = E, where L, R, E are constants. Show that when t is large

the current is approximately equal to E/R.

If, instead, E = Eo cos ωt, where Eo, ω are constants, show that when t is large

i ≈ Eo cos(ωt − ε)√R2 + ω2L2

, where tan ε =ωL

R.



1.2 Second order equations with constant coefficients

Homogeneous form: y′′ + ay′ + by = 0, a, b = real constants.

Inhomogeneous form: y′′ + ay′ + by = r(x).

General solution:

y = Ay1(x) + By2(x) + yp(x), A, B = constant

where y1, y2 are any two linearly independent solutions of the homogeneous equation, called

basis functions or complementary functions, and yp is a particular integral that yields r(x) when

substituted into the equation.

Solution of the homogeneous equation

Because d(eλx)/dx = λeλx, y = eλx will be a solution of the homogeneous equation if λ is a

solution of the characteristic equation

λ2 + aλ + b = 0, i.e. for λ =−a ±

√a2 − 4b

2= λ1, λ2. (1.2.1)

Case 1: λ1 6= λ2

y1 = eλ1x and y2 = eλ2x are linearly independent and the general solution is therefore

y = Aeλ1x + Beλ2x. (1.2.2)

The values of the constants A, B are fixed by the boundary conditions.

Example: Solve y′′ + 2y′ − 8y = 0, y(0) = 1, y′(0) = 0.

Characteristic equation : λ2 + 2λ − 8 = 0

. ˙ . λ = −4, 2

. ˙ . y(x) = Ae−4x + Be2x.

At x = 0 : y = 1, and y′ = 0

. ˙ . A + B = 1,

and − 4A + 2B = 0.

. ˙ . y =e−4x + 2e2x

3.

M. S. Howe 4 §1.2 Second order equations


Case 2: λ1 = λ2 ≡ λ. The two solutions in (1.2.2) are not independent. The differential equation

can now be written in the factored form

y′′ + ay′ + b ≡(

d

dx− λ

) (d

dx− λ

)y = 0.

If z =dy

dx− λy, then z′ − λz = 0, i.e. z = Beλx, B = constant,

. ˙ . y′ − λy = Beλx.

An integrating factor is : e−λx

. ˙ .d

dx

(y(x)e−λx

)= B,

i.e. the general solution is

y = (A + Bx)eλx, A, B = arbitrary constants. (1.2.3)

Case 3: Complex roots λ of the characteristic equation.

When a2 − 4b < 0 the roots (1.2.1) of the characteristic equation are complex conjugates

λ = −a/2 ± i√

4b − a2/2 ≡ −a/2 ± iΩ, say, where i =√−1, and the general solution assumes

either of the forms

y = e−ax/2(A′eiΩx + B′e−iΩx

), Ω =

√4b− a2

2, A′, B′ = constants,

= e−ax/2 (A cos(Ωx) + B sin(Ωx)) , A, B = constants, (1.2.4)

Example The two forms of the solution (1.2.4) are related by Euler’s formula

eix = 1 + ix +(ix)2

2!+

(ix)3

3!+

(ix)4

4!+

(ix)5

5!+

(ix)6

6!+

(ix)7

7!+ . . .

=

(1 − x2

2!+

x4

4!− x6

6!+ . . .

)+ i

(x − x3

3!+

x5

5!− x7

7!+ . . .

)

≡ cos x + i sinx. (1.2.5)

Example Simple harmonic motion is described by the equation

d2y

dt2+ ω2y = 0, where t denotes time.

The roots of the characteristic equation are λ = ±iω, with the general solution

y = A cos(ωt) + B sin(ωt) ≡ A′eiωt + B′e−iωt.



Problems 1B


1. y′′ + 10y′ + 25y = 0 [y = (A + Bx)e−5x]

2. y′′ + 4y′ + 9y = 0 [y = (A cos√

5x + B sin√

5x)e−2x]

3. y′′ − 6y′ + 8y = 0 [y = Ae4x + Be2x]

4. y′′ − 6y′ + 25y = 0 [y = (A cos 4x + B sin 4x)e3x]

5. y′′ − 4y = 0 [y = Ae2x + Be−2x]

6. y′′ + 4y = 0 [y = A cos 2x + B sin 2x]

7. y′′ − y′ + y = 0 [y = (A cos√

32

x + B sin√

32

x)e12 x]

8. y′′ + 3y′ = 0 [y = A + Be−3x]

9. Transform the equation y′′ + x2 + y + 2 = 0 by making the substitution y(x) = z(x) − x2, and hence find the

general solution. [y = −x2 + A cos x + B sinx]

Solve:

10. 4(y′′ − y′) + y = 0, y(0) = 0, y(2) = 2 [y = xe12 x−1]

11. y′′ − 16y = 0, y(0) = 1, y′(0) = 20 [y = 3e4x − 2e−4x]

12. y′′ + 6y′ + 9y = 0, y(0) = −4, y′(0) = 14 [y = (2x − 4)e−3x]

13. y′′ − 16y = 0, y(0) = 5, y( 14 ) = 5e [y = 5e4x].

14. y′′ + 9y = 0, y(π) = −2, y′(π) = 3, [y = 2cos 3x − sin 3x].

15. y′′ − 2y′ + 2y = 0, y(0) = −3, y(π/2) = 0, [y = −3ex cos x].



1.3 Euler’s homogeneous equation

The equation

x2y′′ + axy′ + by = r(x), a, b = constant, (1.3.1)

is equivalent to

xd

dx

(x

d

dx

)y + (a − 1)x

dy

dx+ by = r(x),

which is reduced to a constant coefficient equation by the substitution

x = ez , which implies that xd

dx=

d

dz.

Thus,d2y

dz2+ (a − 1)

dy

dz+ by = r(ez).

The homogeneous form of this equation is solved by the method of §1.2 using the characteristic

equation

λ2 + (a − 1)λ + b = 0.

Example Find the general solution of x2y′′ + 9xy′ + 16y = 0.

The substitution x = ez reduces the equation to

d2y

dz2+ 8

dy

dz+ 16y = 0, with characteristic equation λ2 + 8λ + 16 = 0 → λ = −4, −4.

. ˙ . y = (A + Bz)e−4z =(A + B lnx)

x4.

Problems 1C


1. x2y′′ + 6.2xy′ + 6.76y = 0 [y = (A + B ln x)/x2.6]

2. x2y′′ + xy′ + y = 0 [y = A cos(lnx) + B sin(lnx)]

3. x2y′′ + xy′ − 9y = 0 [y = Ax3 + B/x3]

4. x2y′′ − 2xy′ + 2y = 0 [y = Ax + Bx2]

5. (x + 1)2y′′ − 2(x + 1)y′ − 10y = 0 [y = A(x + 1)5 + B(x + 1)−2]

6. x2y′′ − 3xy′ + 4y = 0 [y = x2(A + B lnx)]

7. x2y′′ + xy′ − 4y = 0 [y = Ax2 + B/x2]

8. x2y′′ − 2xy′ − 4y = 0 [y = Ax4 + B/x]

M. S. Howe 7 §1.3 Euler’s homogeneous equation


9. x2y′′ − 20y = 0 [y = Ax5 + B/x4]

10. x2y′′ − xy′ + 2y = 0 [y = xA cos(lnx) + B sin(ln x)]

11. y′′ + 2xy′ = 0 [y = A + B/x]

12. x2y′′′ + 3xy′′ + y′ = 0 [y = A(lnx)2 + B ln x + C]

13. x2y′′ + 9xy′ + 25y = 0 [y = A cos(3 ln x) + B sin(3 ln x)/x4]

14. (1 + 2x)2y′′ − 6(1 + 2x)y′ + 16y = 0 [y = (1 + 2x)2A ln(1 + 2x) + B]

15. (1 + x)2y′′ + (1 + x)y′ + y = 0 [y = A cosln(1 + x) + α]

Solve:

16. 4x2y′′ + 4xy′ − y = 0, y(4) = 2, y′(4) = − 14

[y = 4/√

x].

17. x2y′′ − xy′ + 2y = 0, y(1) = −1, y′(1) = −1 [y = −x cos(lnx)].

M. S. Howe 8 §1.3 Euler’s homogeneous equation


Problems 2A

1. Find the value of x given that a = 3i − 2j and b = 4i + xj are perpendicular. [6].

2. Find the lengths of a = 2i − j + 2k, b = 5i + 3j − k. What is the angle between the directions of a and b?

[3,√

35, cos−1(5/3√

35].

3. Solve for x the vector equation x + a(b · x) = c, where a, b, c are constant vectors. What happens when

a · b = −1? [x = c − a(b · c)/(1 + a · b)].

4. Solve Ax + a × x = b, where A 6= 0 is a constant. [x = A2b + a(a · b) + Ab× a/A(A2 + a2)].

5. Solve the simultaneous equations x + y × p = a, y + x × p = b.

[x = (p · a)p + a − b× p/(1 + p2), y = (p · b)p + b− a × p/(1 + p2) ].

6. Calculate r and r, where the dot denotes differentiation with respect to the time t, when

r = (t+ sin t)i+ (t− sin t)j+√

2(1− cos t)k. Show that r and r are perpendicular and have constant magnitudes.

[r = (1 + cos t)i + (1 − cos t)j +√

2 sin tk; r = − sin ti + sin tj +√

2 cos tk; r·r = 0; |r| = 2; |r| =√

2].

7. Show that (a × b) · (a × c) × d = (a · d)(a · b× c).

8. If r = (x,y, z), show that the equation of the straight line through the point ro in the direction of the unit vector

t is r = ro + λt, −∞ < λ < ∞.

9. If r = (x, y, z) lies on the straight line through ro in the direction of the unit vector t, show that (r− ro)× t = 0.

10. Show that the equation of the plane whose unit normal is n (so that |n| = 1) and which passes through

the point ro is (r − ro) · n = 0. Show that the perpendicular from the point r1 intersects the plane at

r = r1 − n[(r1 − ro) · n].

11. Show that the equation to the perpendicular line from the point b to the straight line r = a + λt is

r = b + µt× (a − b) × t, −∞ < µ < ∞.

12. Show that the straight lines r = a + λu and r = b + µv will intersect if v · b×u = v · a×u, and that the point

of intersection can be written in either of the forms

a +a · b× v

v · a × uu = b +

a · b × u

v · b× uv.

13. If the straight lines r = a + λu and r = b + µv do not intersect show that the length of the common

perpendicular joining them is |(b− a)·n|, where n = u× v/|u× v|.

14. Show that the equation of the plane through the points r1, r2, r3 can be written

r · r2 × r3 + r · r3 × r1 + r · r1 × r2 = r1 · r2 × r3.

15. Show that the equation of the sphere of radius a and center ro is (r − ro) · (r − ro) = a2.

16. The foci of an ellipse of major axis 2a are at the points ±b. Show that the point r lies on the ellipse if

a4 − a2(r2 + b2) + (b · r)2 = 0.

M. S. Howe 51 §2.1 Vectors


Example A vector field whose divergence vanishes is called a solenoidal vector. The velocity v(x, t) of an

incompressible fluid is solenoidal, since the volume occupied by a moving fluid element is invariant, i.e., its rate of

volumetric expansion is zero, although the mean fluid density may vary with position in the fluid.

In an ideal fluid in irrotational motion the velocity v = ∇ϕ, where ϕ(x, t) is a velocity potential. When the motion

is incompressible div(∇ϕ) = 0. Now

div(∇) = div grad = ∇ · ∇ ≡ ∇2 =∂2

∂x2+

∂2

∂y2+

∂2

∂z2. (2.3.4)

Hence, for incompressible flow ϕ satisfies Laplace’s equation

∇2ϕ ≡ ∂2ϕ

∂x2+

∂2ϕ

∂y2+

∂2ϕ

∂z2= 0. (2.3.5)

Problems 2C

Find the divergence of

1. F = yi + zj + xk [0].

2. F = x [3].

3. F = (x, y2, z3) [1 + 2y + 3z2].

4. F = 4x2i + 4y2j− z2k [8x + 8y − 2z].

5. F = xyz(i + j + k) [yz + xz + xy].

6. F = r/r3 , r = xi + yj + zk [0].

7. F = r(r · a), r = xi + yj + zk, a = constant. [4r · a].

Prove that

8. ∇2(1/r) = 0, r > 0, r = xi + yj + zk.

9. ∇2(rn) = n(n + 1)rn−2, r = xi + yj + zk.

10. For scalar and vector fields ϕ(x) and a(x), div(ϕa) = ϕdiv a + ∇ϕ · a.

11. div(f∇g) − div(g∇f) = f∇2g − g∇2f .

12.∮S

ϕn · ∇ϕdS =∫V(∇ϕ)2 dV, provided ∇2ϕ = 0.

13. V = 16

∮Sn·∇(r2)dS, where r = xi + yj + zk and V is the volume enclosed by S.

14.∮S(x3i + y3j + z3k)·dS = 12

5πR5, where S is a sphere of radius R with center at the origin.

Evaluate∮Sn · FdS when

15. F = (x,x2y,−x2z) and S is the surface of the tetrahedron with vertices (0, 0, 0), (1, 0, 0), (0, 1, 0) (0, 0, 1) [ 16 ].

16. F = 13(x3, y3, z3) and S is the surface of the sphere |x| = 2 [128π/5].

17. F = axi + byj + czk where a, b, c are constants, and S is the unit sphere |x| = 1 [ 43π(a + b + c)].

M. S. Howe 58 §2.3 The divergence


Problems 2G

Establish the following formulae:

1. ∇2ϕ = ∂2ϕ∂xi∂xi

= ∂2ϕ

∂x2i

.

2. (curl F)i = (∇×F)i ≡ eijk∂Fk∂xj

.

3. a · b× c = eijkaibjck =

∣∣∣∣∣∣∣∣

a1 a2 a3

b1 b2 b3

c1 c2 c3

∣∣∣∣∣∣∣∣.

4. δii = 3.

5. δijeijk = 0.

6. eijkeipq = δjpδkq − δjqδkp

7. eijkeljk = 2δil

8. eijkeijk = 6

9.(a × (b× c)

)i≡ eijkajekpqbpcq = biajcj − ciajbj ≡ bia · c− cia · b.

10. If n is the unit outward normal to the surface S of a sphere of unit radius, show that

∮

S

ninjdS =4π

3δij ,

∮

S

ninjnknldS =4π

15

(δijδkl + δikδjl + δilδjk

).

11. Let e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1). Show that ei · ej × ek = eijk.

Deduce that if a = (a1, a2, a3) ≡ aj ej and b = (b1, b2, b3) ≡ bk ek , then

(a × b)i = ajbk ei · ej × ek = eijkajbk

M. S. Howe 77 §2.8 Suffix notation

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