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Instructor Solution Manual for Advanced Topics in Applied Mathematics: For Engineering and the Physical Sciences. 1st Edition. 2011, Sudhakar Nair

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Instructor Solution Manual for Advanced Topics in Applied Mathematics: For Engineering and the Physical Sciences. 1st Edition. 2011, Sudhakar Nair

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SOLUTIONS MANUAL FOR
ADVANCED TOPICS IN
APPLIED MATHEMATICS




Sudhakar Nair

,Contents


1 GREEN’S FUNCTIONS 5

2 INTEGRAL EQUATIONS 59

3 FOURIER TRANSFORMS 93

4 LAPLACE TRANSFORMS 149




1

,Chapter 1

GREEN’S FUNCTIONS


1.1 The deflection of a beam is governed by the equation
d4 v
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) = p0 ,
b) p(x) = P0 δ(x − ξ),
c) p(x) = M0 δ ′ (x − ξ).
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)
p0 ℓ 4 p0 ℓ 4
v ′′′′ = − , v ′′′ = − [x + C1 ],
EI EI
p0 ℓ4 x2
v ′′ = − [ + C1 x + C2 ],
EI 2
p0 ℓ4 x3 x2
v′ = − [ + C1 + C2 x + C3 ],
EI 3 2
p0 ℓ4 x4 x3 x2
v = − [ + C1 + C2 + C3 x + C4 ].
EI 24 6 2
5

, 6 CHAPTER 1. GREEN’S FUNCTIONS

Using the boundary conditions

v(0) = 0, v ′ (0) = 0, C4 = 0, C3 = 0.
Using
1 1 1 1 1
v(1) = 0, v ′ (1) = 0, C1 + C2 = − , C1 + C2 = − .
2 6 6 2 24

Solving

1 1
C1 = − , C2 = .
2 12


p0 ℓ4 x4 x3 x2
 
v(x) = − − + ,
EI 24 12 24
p0 ℓ 4
= − (x − 1)2 x2 .
24EI

(b)
P0 ℓ4 P0 ℓ4
v ′′′′ = − δ(x − ξ), v ′′′ = − [hx − ξi0 + C1 ],
EI EI
P0 ℓ4
v ′′ = − [hx − ξi1 + C1 x + C2 ],
EI
4
P 0ℓ 1 1
v′ = − [ hx − ξi2 + C1 x2 + C2 x + C3 ],
EI 2 2
P0 ℓ4 1 1 1
v = − [ hx − ξi3 + C1 x3 + C2 x2 + C3 x + C4 ].
EI 6 6 2

Using
v(0) = 0 = v ′ (0), C4 = C3 = 0.
Using
1 1 1 1 1
v(1) = 0 = v ′ (1), C1 +C2 = − (1−ξ)2 , C1 + C2 = − (1−ξ)3,
2 2 6 2 6

Then
C1 = −(1 + 2ξ)(1 − ξ)2 , C2 = ξ(1 − ξ)2.

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