Advanced Topics in Applied Mathematics: For Engineering and the Physical Sciences (Instructor Solution Manual, Solutions)

189 Pages·2011·0.546 MB·English

by  Sudhakar Nair

SOLUTIONS MANUAL FOR
ADVANCED TOPICS IN
APPLIED MATHEMATICS
Sudhakar Nair

c
(cid:13)
S. Nair
ii
Contents
1 GREEN’S FUNCTIONS 5
2 INTEGRAL EQUATIONS 59
3 FOURIER TRANSFORMS 93
4 LAPLACE TRANSFORMS 149
1
2 CONTENTS
Preface
This is the solutions manual to accompany the text, “Advanced Topics in
Applied Mathematics.” I have attempted to show all the intermediate steps
in the solutions. For the convenience of posting solutions, a new solution
appearsonanewpage. Pleaseletmeknowifanyonenoticesanyerrors. Some
figures were created using the MathematicaTM package and I am indebted
to Wolfram Research Incorporated.
S. Nair ( [email protected])
Illinois Institute of Technology,
Chicago, Illinois
3
4 CONTENTS
Chapter 1
GREEN’S FUNCTIONS
1.1 The deflection of a beam is governed by the equation
d4v
EI = p(x)
dx4 −
where EI is the bending stiffness and p(x) is the distributed loading
on the beam. If the beam has a length ℓ, and at both the ends the
deflection and slope are zero, obtain expressions for the deflection by
direct integration, using the Macaulay brackets when necessary, if
a) p(x) = p ,
0
b) p(x) = P δ(x ξ),
0
−
c) p(x) = M δ (x ξ).
0 ′
−
Obtain the Green’s function for the deflection equation from the pre-
ceding calculations.
Solution
Let us make the substitution x x/ℓ. The derivative of the delta
→
function in Part (c) will bring 1/ℓ when the non-dimensional x is used.
The beam equation becomes
(a)
p ℓ4 p ℓ4
0 0
v = , v = [x+C ],
′′′′ ′′′ 1
− EI − EI
p ℓ4 x2
0
v = [ +C x+C ],
′′ 1 2
− EI 2
p ℓ4 x3 x2
0
v = [ +C +C x+C ],
′ 1 2 3
− EI 3 2
p ℓ4 x4 x3 x2
0
v = [ +C +C +C x+C ].
1 2 3 4
− EI 24 6 2
5
6 CHAPTER 1. GREEN’S FUNCTIONS
Using the boundary conditions
v(0) = 0, v (0) = 0, C = 0, C = 0.
′ 4 3
Using
1 1 1 1 1
v(1) = 0, v (1) = 0, C +C = , C + C = .
′ 1 2 1 2
2 −6 6 2 −24
Solving
1 1
C = , C = .
1 2
−2 12
p ℓ4 x4 x3 x2
0
v(x) = + ,
− EI 24 − 12 24
(cid:20) (cid:21)
p ℓ4
= 0 (x 1)2x2.
−24EI −
(b)
P ℓ4 P ℓ4
v = 0 δ(x ξ), v = 0 [ x ξ 0 +C ],
′′′′ ′′′ 1
− EI − − EI h − i
P ℓ4
v = 0 [ x ξ 1+C x+C ],
′′ 1 2
− EI h − i
P ℓ4 1 1
v = 0 [ x ξ 2 + C x2 +C x+C ],
′ 1 2 3
− EI 2h − i 2
P ℓ4 1 1 1
v = 0 [ x ξ 3 + C x3 + C x2 +C x+C ].
1 2 3 4
− EI 6h − i 6 2
Using
v(0) = 0 = v (0), C = C = 0.
′ 4 3
Using
1 1 1 1 1
v(1) = 0 = v (1), C +C = (1 ξ)2, C + C = (1 ξ)3,
′ 1 2 1 2
2 −2 − 6 2 −6 −
Then
C = (1+2ξ)(1 ξ)2, C = ξ(1 ξ)2.
1 2
− − −