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SOLUTIONS MANUAL FOR ADVANCED TOPICS IN APPLIED MATHEMATICS

Sudhakar Nair

Contents

1 GREEN’S FUNCTIONS

2 INTEGRAL EQUATIONS

3 FOURIER TRANSFORMS

4 LAPLACE TRANSFORMS

149

Chapter 1

GREEN’S FUNCTIONS 1.1 The deflection of a beam is governed by the equation d4 v = −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 preceding calculations. EI

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 ′′′ = − [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

v ′′′′ = −

CHAPTER 1. GREEN’S FUNCTIONS Using the boundary conditions v(0) = 0,

v ′ (0) = 0,

C4 = 0,

C3 = 0.

Using v(1) = 0,

1 1 v ′ (1) = 0, C1 + C2 = − , 2 6

1 1 1 C1 + C2 = − . 6 2 24

Solving 1 C1 = − , 2

C2 =

1 . 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 δ(x − ξ), v ′′′ = − [hx − ξi0 + C1 ], EI EI P0 ℓ4 v ′′ = − [hx − ξi1 + C1 x + C2 ], EI 4 P 1 0ℓ 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

v ′′′′ = −

Using v(0) = 0 = v ′ (0),

C4 = C3 = 0.

Using v(1) = 0 = v ′ (1),

1 1 C1 +C2 = − (1−ξ)2 , 2 2

1 1 1 C1 + C2 = − (1−ξ)3, 6 2 6

Then C1 = −(1 + 2ξ)(1 − ξ)2 ,

C2 = ξ(1 − ξ)2.

v(x) = −

P0 ℓ4 hx − ξi3 − (1 + 2ξ)(1 − ξ)2 x3 + 3ξ(1 − ξ)2 x2 . 6EI

(c) M0 ℓ3 M0 ℓ3 ′ δ (x − ξ), v ′′′ = − [δ(x − ξ) + C1 ], EI EI M0 ℓ3 v ′′ = − [hx − ξi0 + C1 x + C2 ], EI 3 M 1 0ℓ [hx − ξi1 + C1 x2 + C2 x + C3 ], v′ = − EI 2 M0 ℓ3 1 1 1 [ hx − ξi2 + C1 x3 + C2 x2 + C3 x + C4 ]. v = − EI 2 6 2

v ′′′′ = −

Using v(0) = 0 = v ′ (0),

C3 = C4 = 0,

Using v(1) = 0 = v ′ (1),

1 C1 + C2 = −(1 − ξ), 2

1 1 1 C1 + C2 = − (1 − ξ)2 . 6 2 2

Then C1 = −6ξ(1 − ξ),

v(x) = −

C2 = (3ξ − 1)(1 − ξ).

M0 ℓ3 hx − ξi2 − 2ξ(1 − ξ)x3 + (3ξ − 1)(1 − ξ)x2 . 2EI

The Green’s function in terms of non-dimensional coordinate x/ℓ corresponds to P0 = −1; ℓ4 hx − ξi3 − (1 − ξ)2x2 (3ξ + x + 2ξx) . 6EI ( (1 − ξ)2 x2 (3ξ − x − 2ξx), x < ξ g(x, ξ) = (1 − x)2 ξ 2(3x − ξ − 2xξ), x > ξ

g(x, ξ) =

CHAPTER 1. GREEN’S FUNCTIONS 1.2 Solve the preceding problem when the beam is simply supported. That is, v(0) = v(ℓ) = 0,

v ′′ (0) = v ′′ (ℓ) = 0.

Solution Let us make the substitution x → x/ℓ. The derivative of the delta function will have a 1/ℓ in front when the non-dimensional x is used. The beam equation becomes (a) p0 ℓ 4 p0 ℓ 4 , v ′′′ = − [x + C1 ], EI EI p0 ℓ4 x2 v ′′ = − [ + C1 x + C2 ], EI 2 p0 ℓ4 x3 x2 [ + C1 + C2 x + C3 ], v′ = − EI 3 2 p0 ℓ4 x4 x3 x2 [ + C1 + C2 + C3 x + C4 ]. v = − EI 24 6 2

v ′′′′ = −

Using the boundary conditions

v(0) = 0,

v ′′ (0) = 0,

C2 = 0,

C4 = 0.

1 C1 = − , 2

C3 =

Using v(1) = 0,

v ′′ (1) = 0,

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

1 . 24

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

v ′′′′ = −

Using v(0) = 0 = v ′′ (0),

C4 = C2 = 0.

Using v(1) = 0 = v ′′ (1), v(x) = − (c)

C1 = ξ − 1,

1 C2 = ξ(1 − ξ)(2 − ξ), 6

P0 ℓ4 hx − ξi3 − (1 − ξ)(1 − ξ)2 x3 + ξ(1 − ξ)(2 − ξ)x . 6EI

M0 ℓ3 ′ M0 ℓ3 ′′′ δ (x − ξ), v = − [δ(x − ξ) + C1 ], v = − EI EI Note: δ ′ is now the derivative with respect to the non-dimensional variable x. ′′′′

M0 ℓ3 v = − [hx − ξi0 + C1 x + C2 ], EI 3 M 1 0ℓ [hx − ξi1 + C1 x2 + C2 x + C3 ], v′ = − EI 2 M0 ℓ3 1 1 1 v = − [ hx − ξi2 + C1 x3 + C2 x2 + C3 x + C4 ]. EI 2 6 2 ′′

Using v(0) = 0 = v ′′ (0),

C2 = C4 = 0,

Using v(1) = 0 = v ′′ (1),

C1 = −1,

1 C3 = [1 − 3(−ξ)2 ]. 6

CHAPTER 1. GREEN’S FUNCTIONS

v(x) = −

M0 ℓ3 [3hx − ξi2 − x3 + [1 − 3(1 − ξ)2 ]x . 6EI

The Green’s function in terms of the non-dimensional coordinate x/ℓ corresponds to P0 = −1; ℓ4 hx − ξi3 − (1 − ξ)2x3 + ξ(1 − ξ)(2 − ξ)x . 6EI ( x(1 − ξ)(2ξ − x2 − ξ 2 ), x < ξ g(x, ξ) = ξ(1 − x)(2x − ξ 2 − x2 ), x > ξ

g(x, ξ) =

11 1.3 Obtain the derivative of the function g(x) = |f (x)|, in a < x < b, assuming f (x) has a simple zero at the point c inside the interval (a, b). Use the Signum function and/or delta function to express the result. Solution Let g(x) = f (x) sgn[f (x)]. (a) Assuming f ′ (c) > 0, sgn[f (x)] = sgn(x − c). (b) Assuming f ′ (c) < 0, sgn[f (x)] = − sgn(x − c). Then g(x) = f (x) sgn[f ′ (c)] sgn(x−c),

g ′(x) = [f ′ (x) sgn(x−c)+2f (x)δ(x−c)] sgn[f ′ (c)].

CHAPTER 1. GREEN’S FUNCTIONS

1.4 Assuming a function f (x) has simple zeros at xi , i = 1, 2, . . . , n, find an expression for δ(f (x)). Solution Near a zero, x = xi , f (x) = f ′ (xi )(x − xi ) + · · · ,

For a simple zero, f ′ (xi ) 6= 0.

Using a test function φ, Z ∞

δ(f (x))φ(x)dx =

n Z xi +ǫ X xi −ǫ

−∞

i=1 n Z ǫ X i=1 n X

φ(xi + ξ)δ[f ′(xi )ξ]dξ,

−ǫ

1 = ′ f (xi ) i=1 = Then δ(f ) =

φ(x)δ[f ′ (xi )(x − xi )]dx

n X

f ′ (xi ) i=1

Z ǫ

−ǫ

φ xi +

φ(xi ).

δ(x − xi ).

η δ(η)dη f ′ (xi )

13 1.5 Convert the equation Lu =

d2 u du + x2 + 2u = f 2 dx dx

into the Sturm-Liouville form. Solution The Sturm-Liouville form is (pu′ )′ + qu = f ∗ . Expanding u′′ +

f∗ p′ ′ q u + u= . p p p

Comparing with the given equation p′ = x2 , p log p = x3 /3,

q = 2, p 3

f∗ = f. p 3

p = ex /3 ,

q = 2ex /3 ,

(ex /3 u′ )′ + 2ex /3 u = ex /3 f.

f ∗ = ex /3 f.

CHAPTER 1. GREEN’S FUNCTIONS

1.6 Find the adjoint system for xu′′ + u′ + u = 0,

u(2) + u′ (2) = 0.

u(1) = 0,

Solution hv, Lui − hu, L∗ vi = 0.

hv, Lui =

Z 2

v[xu′′ + u′ + u]dx

= [xvu

′

2 − (xv) u + vu] 1 + ′

Z 2 1

u[(xv)′′ − v ′ + v]dx,

L∗ v = (xv)′′ − v ′ + v = xv ′′ + 2v ′ − v ′ + 1. d2 d L∗ = x 2 + + 1. dx dx The adjoint boundary conditions are found from the bilinear concomitant, 2 [xvu′ − (xv)′ u + vu] 1 = 0. x = 1, x = 2,

u′ = −u,

xv = 0,

v(1) = 0,

−2v − v − 2v ′ + v = 0,

v ′ (2) + v(2) = 0.

15 1.7 Solve the differential system u′′ + u′ − 2u = x2 ,

u′(1) = 0.

u(0) = 0,

Solution For this constant coefficient equation, we try u = eαx , to get

1 3 α2 + α − 2 = 0, α = − ± . 2 2 α1 = 1, α2 = −2.

For a particular solution, we assume up = Cx2 + Dx + E, which, upon substituting in the differential equation gives 2C + 2Cx + D − 2Cx2 − 2Dx − 2E = x2 . Matching the coefficients of equal powers of x, C = −1/2,

D = −1/2,

E = −3/4.

The general solution is 1 1 3 u = Aex + Be−2x − x2 − x − . 2 2 4 The boundary conditions give u(0) = 0, u′ (1) = 0, Then A=

A+B−

3 = 0, 4

Ae − 2Be−2 −

3 e2 + 1 , 2 e3 + 2

3 = 0. 2

3 e3 − 2e2 . 4 e3 + 2

CHAPTER 1. GREEN’S FUNCTIONS

1.8 Solve the differential system (x2 u′ )′ − n(n + 1)u = 0,

u(0) = 0,

u(1) = 1.

Solution This equation is of variable coefficients of the Euler type. We try u = xα . The indicial equation is α(α + 1) − n(n + 1) = 0, α2 + α − n(n + 1) = 0. r 1 1 α=− ± + n(n + 1), 2 4 α1 = n, α2 = −(n + 1). So, our solution is u = Axn + Bx−(n+1) . (a) n > 0 u(0) = 0,

B = 0,

u(1) = 1,

A = 1.

u = xn . (b) n = 0 u = A + B/x,

u(0) = 0, A = B = 0,

and u(1) = 1 cannot be satisfied. Thus, there is no solution. (c) n = −1

u = A/x + B,

u(0) = 0, A = B = 0,

again, no solution. (d) n < −1 u(0) = 0,

A = 0,

u(1) = 1,

u = x−(n+1) .

B = 1.

17 1.9 Convert the following system to one with homogeneous boundary conditions: (x2 u′ )′ − n(n + 1)u = 0,

u(0) = 0,

u(1) = 1.

Solution Let u = v + Ax + B. u(0) = v(0) + B = 0,

v(0) = 0,

u(1) = 1 = v(1) + A,

A = 0, V (1) = 0.

u = v + x,

B = 0.

[x2 (v ′ + 1)]′ − n(n + 1)[v + x] = 0.

(x2 v ′ )′ − n(n + 1)v = [n(n + 1) − 2]x,

v(0) = v(1) = 0.

CHAPTER 1. GREEN’S FUNCTIONS

1.10 Obtain the Green’s function for the equation 9 (xu′ )′ − u = f (x), x

u(0) = 0,

u(1) = 0.

Using the Green’s function explicitly, find the solution when f (x) = xn . Check if there are any values for the integer n for which your solution does not satisfy the boundary conditions. Solution Let u = xα . α2 − 9 = 0,

u = x3 , x−3 .

To satisfy u(0) = 0, we choose u1 = x3 . To satisfy u(1) = 0, we choose u2 = x3 − x−3 . Then ( x3 (ξ 3 − ξ −3), g=C ξ 3 (x3 − x−3 ), 1 [[g ′ ]]x=ξ = , ξ

x<ξ x>ξ

1 C[ξ 3 (3ξ 2 + 3ξ −4) − 3ξ 2(ξ 3 − ξ −3 )] = . ξ 1 C= . 6

1 g= 6 For f = xn

( x3 (ξ 3 − ξ −3), ξ 3 (x3 − x−3 ),

(1.1) x<ξ x>ξ

1 u = 6

Z x 0

−3

(x − x )ξ

n+3

dξ +

Z 1

x (ξ

n+3

−ξ

n−3

)dξ

" # n+4 n+4 n−2 1 x ξ ξ 1 (x3 − x−3 ) + x3 ( − ) = 6 n+4 n+4 n−2 x 1 6 xn+4 xn−2 1 n+7 n+1 3 (x − x ) + x (− − + ) = 6 n+4 (n + 4)(n − 2) n + 4 n − 2 n+1 1 = x − x3 (n + 4)(n − 2)

We restrict n 6= 2 for this solution to apply. When n = 2

1 1 9 x3 3 6 3 u = (x − x ) + (1 − x ) + x log x 6 6 6 1 3 x log x. = 6 Note that the boundary conditions cannot be satisfied if n ≤ −1.

CHAPTER 1. GREEN’S FUNCTIONS

1.11 Find the Green’s function for u′′ + ω 2 u = f (x), Examine the special cases ω = n,

u(0) = 0,

u(π) = 0.

n = 0, 1, · · · .

Solution The Green’s function satisfies g ′′ + ω 2 g = δ(x − ξ). g1 = sin ωx, g2 = sin ω(π − x). ( g1 (x)g2 (ξ), x < ξ g=C g2 (x)g1 (ξ), x > ξ The jump condition gives C[g2′ (ξ)g1(ξ) − g1′ (ξ)g2(ξ)] = 1. That is Cω[− cos ω(π − ξ) sin ωξ − sin ω(π − ξ) cos ωξ] = 1. 1 C=− ω sin πω ( sin ωx sin ω(ξ − π), 1 g= ω sin πω sin ωξ sin ω(x − π),

x<ξ x>ξ

When ω = n, g = sin nx satisfies both the boundary conditions and a regular Green’s function does not exist. We have to look for a generalized Green’s function.

21 1.12 Express the equation u′′ − 2u′ + u = f (x),

u(0) = 0,

u(1) = 0,

in the self-adjoint form. Obtain the solution using the Green’s function when f (x) = ex . Solution Comparing with the Sturm-Liouville form, Our equation becomes

p′ /p = −2,

p = e−2x .

(e−2x u′)′ + e−2x u = e−2x f. Trying a solution of the form u = eαx , α2 − 2α + 1 = 0,

α = 1, 1.

To find the Green’s function, let g1 = xex , g2 = (x − 1)ex . ( x(ξ − 1)ex+ξ , x < ξ g=C ξ(x − 1)ex+ξ , x > ξ

The jump condition gives

C[ξ + (ξ − 1)ξ − (ξ − 1) − ξ(ξ − 1)]e2ξ = e2ξ .

Then C = 1 and

( x(ξ − 1), g = ex+ξ ξ(x − 1),

When f = ex , u =

Z x 0

x+ξ−ξ

ξ(x − 1)e

= (x − 1)ex

dξ +

Z 1 x 2

x (x − 1) − xex 2 2

x−1 x 2 e [x − x(x − 1)] 2 1 = x(x − 1)ex . 2 =

x<ξ x>ξ

x(ξ − 1)ex+ξ−ξ dξ

CHAPTER 1. GREEN’S FUNCTIONS

1.13 Transform the equation xu′′ + 2u′ = f (x);

u′ (0) = 0,

u(1) = 0,

into the self-adjoint form. Find the Green’s function, and express the solution in terms of f (x). State the restrictions on f (x) for the solution to exist. Solution Comparing with the Sturm-Liouville form, p′ /p = 2/x,

p = x2 .

The self-adjoint form is (x2 u′ )′ = xf. The homogeneous equation is (x2 u′ )′ = 0, which can be integrated to get A A , u = − + B. 2 x x ′ To satisfy g1 (0) = 0, we choose A = 0 and B = 1. To satisfy g2 (1) = 0, we choose A = 1 and B = 1. x2 u′ = A,

u′ =

( 1 − 1ξ , g=C 1 − x1 ,

x<ξ x>ξ

C/ξ 2 = 1/ξ 2 ,

C = 1.

The jump condition gives Then

( 1 − 1ξ , g= 1 − x1 ,

x<ξ x>ξ

For a given function f , Z x Z 1 1 1 u= 1− ξf dξ + 1− ξf dξ ′ x xi 0 x

Solution exists if ξf is integrable in (0, x) and (1 − ξ)f is integrable in (x, 1)

23 1.14 Find the Green’s function for x2 u′′ − xu′ + u = f (x),

u(0) = 0,

u(1) = 0.

Solution With the solution u = xn , n(n − 1) − n + 1 = 0,

n = 1, 1.

For this repeated index, u1 = x,

u2 = x log x.

These satisfy the left and right boundary conditions, respectively. g=C

(

xξ log ξ, ξx log x,

x<ξ x>ξ

The jump condition gives C[ξ(log ξ + 1) − ξ log ξ] = 1/ξ 2, C = 1/ξ 3 . ( xξ −2 log ξ, x < ξ g= . ξ −2x log x, x > ξ

CHAPTER 1. GREEN’S FUNCTIONS

1.15 Using the self-adjoint form of the differential equation x2 u′′ + 3xu′ − 3u = f (x),

u(0) = 0,

u(1) = 0,

find the Green’s function and obtain an explicit solution when f (x) = x. Solution The self-adjoint form for this equation is (x3 u′ )′ − 3xu = xf. Using u = xn , we find n(n − 1) + 3n − 3 = 0, u1 = x,

n = 1, −3.

and u2 = x − x−3

satisfy the required boundary conditions. ( x(ξ − ξ −3 ), g=C ξ(x − x−3 ),

x<ξ x>ξ

The jump condition gives C[ξ(1 + 3ξ −4) − (ξ − ξ −3 )] = ξ −3 ,

1 C= . 4

Then, 1 g= 4

(

x(ξ − ξ −3 ), ξ(x − x−3 ),

x<ξ x>ξ

For f = x, Z x Z 1 1 −3 3 −3 2 u = (x − x )ξ dξ + x(ξ − ξ )ξ dξ 4 0 x 4 1 1 − x4 −3 x = (x − x ) + x + x log x 4 4 4 1 = x log x. 4

25 1.16 Solve the equation x2 u′′ + 3xu′ = x2 ,

u(1) = 1,

u(2) = 2,

using the Green’s function. Solution This equation can be written as (x3 u′ )′ = x3 . To obtain homogeneous boundary conditions, let u = v + x, where the function x satisfies the non-homogeneous boundary conditions. Then, u′ = v ′ + 1 and (x3 v ′ )′ = x3 − 3x2 . The homogeneous equation can be integrated to get (x3 v ′ )′ = 0,

x3 v ′ = A,

v′ =

A , x3

v=−

To satisfy v1 (1) = 0 we choose v1 = 1 −

1 x2

and to satisfy v2 (2) = 0 we choose v2 = 1 −

4 . x2

With these the Green’s function can be written as   1 − 12 1 − 42 , x < ξ x ξ g=C 4  1 − 2 1 − 12 , x > ξ x ξ

A + B. 2x2

CHAPTER 1. GREEN’S FUNCTIONS Using the jump condition 1 8 4 2 1 C 1− 2 − 1− 2 = 3. 3 3 ξ ξ ξ ξ ξ 1 C= . 6 The solution of the non-homogeneous equation is Z x 1 4 1 v = 1− 2 1 − 2 (ξ 3 − 3ξ 2)dξ 6 x ξ 0 Z 1 4 1 3 2 1 − 2 (ξ − 3ξ )dξ + 1− 2 x ξ x x2 5 41 = −x− 2 + . 8 6x 24 and u=

x2 5 41 − 2+ . 8 6x 24

27 1.17 For the problem Lu = u′′ + u′ = 0,

u′ (1) = 0,

u(0) = 0,

obtain the adjoint system. Solve the eigenvalue problems, L∗ v = λv,

Lu = λu,

and show that their eigenfunctions are bi-orthogonal. Solution To obtain the adjoint system, we use integration by parts. hv, Lui = =

Z 1

v[u′′ + u′ ]dx

1 (vu − v u + vu) 0 + ′

′

Z 1 0

u[v ′′ − v ′ ]dx.

Then L∗ v = v ′′ − v ′ ,

v(0) = 0,

v ′ (1) − v(1) = 0.

The two eigenvalue problems are: u′′ + u′ − λu = 0, u = eαx , α2 + α − λq = 0,

α = − 21 ± 14 + λ, p Let iµ = λ + 1/4

v ′′ − v ′ − λv = 0 v = eβx β 2 − β −q λ=0 β = 12 ±

1 +λ 4

u = e−x/2 [A sin µx + B cos µx], v = ex/2 [C sin µx + D cos µx]

Using u(0) = 0, u′ (1) = 0 and v(0) = 0, v ′ (1) − v(1) = 0 we find B = 0, tan µ = 2µ, D = 0, tan µ = 2µ. −x/2 ui = e sin µi x, vj = ex/2 sin µj x, where µi and µj are solutions of tan µ = 2µ.

CHAPTER 1. GREEN’S FUNCTIONS When µi 6= µj , I =

Z 1

sin µi x sin µj xdx

µ2j 1− 2 µi

I = = =

Thus hui , vj i = 0.

1 − cos µi x sin µi x sin µj x + cos µj x µi µ2i 0 Z µ2j 1 + 2 sin µi x sin µj xdx µi 0 cos µi µj − sin µj + 2 sin µi cos µj µi µ i µj tan µj tan µi cos µi cos µj − + µi µj µi 0

29 1.18 By solving the nonhomogeneous problem u′′ = δǫ (x − ξ), where δǫ (x) =

u(0) = 0,

0, 1 , 2ǫ

u(1) = 0,

|x| > ǫ, |x| < ǫ,

in three parts: a) 0 < x < ξ − ǫ, b) ξ − ǫ < x < ξ + ǫ, and c) ξ + ǫ < x < 1, show that, in the limit ǫ → 0, we recover the Green’s function. Solution Let u1 , u2 , and u3 represent the solutions in the three domains. u′′1 = 0,

x < ξ − ǫ.

u1 = A1 x + B1 ,

u1 (0) = 0,

B = 0.

u1 = A1 x u′′3 = 0, u3 = A3 (1 − x) + B3 , u′′2 =

1 , 2ǫ

x > ξ + ǫ. u3 (1) = 0,

B3 = 0.

ξ − ǫ < x < ξ + ǫ.

x2 + A2 x + B2 . 4ǫ Using continuity of the solutions u2 =

u1 (ξ − ǫ) = u2 (ξ − ǫ), ξ−ǫ A1 = A2 + , 2ǫ

(ξ − ǫ)2 A1 (ξ − ǫ) = A2 (ξ − ǫ) + B2 + . 4ǫ

u3 (ξ + ǫ) = u2 (ξ + ǫ), −A3 = A2 +

(ξ + ǫ) , 2ǫ

u′1 (ξ − ǫ) = u′2 (ξ − ǫ),

u′3 (ξ + ǫ) = u′2 (ξ + ǫ),

A3 (1 − ξ − ǫ) = A2 (ξ + ǫ) + B2 +

(ξ + ǫ)2 . 4ǫ

CHAPTER 1. GREEN’S FUNCTIONS

B2 =

(ξ − ǫ)2 , 4ǫ

(ξ − ǫ)2 (ξ + ǫ)2 (ξ + ǫ)2 + = −A2 (1 − ξ − ǫ) − (1 − ξ − ǫ), 4ǫ 4ǫ 2ǫ ξ+ǫ (ξ − ǫ)2 + (ξ + ǫ)2 A2 = − (1 − ξ − ǫ) − , 2ǫ 4ǫ ξ+ǫ ξ − ǫ (ξ − ǫ)2 + (ξ + ǫ)2 A1 = − (1 − ξ − ǫ) + − , 2ǫ 2ǫ 4ǫ ξ+ǫ ξ + ǫ (ξ − ǫ)2 + (ξ + ǫ)2 (1 − ξ − ǫ) − + . A3 = − 2ǫ 2ǫ 4ǫ

A2 (ξ + ǫ) +

As ǫ → 0, u2 (ξ) can be evaluated as 2 x u2 = lim + A2 x + B2 ǫ→0 4ǫ 2 ξ ξ2 2ξ 3 ξ 2 = lim + − (1 − ξ) − + + ξ(ξ − 1) ǫ→0 4ǫ 2ǫ 4ǫ 4ǫ = ξ(ξ − 1), where ǫ2 terms have been neglected from the outset. A1 = (ξ − 1),

A3 = −ξ.

u1 = x(ξ − 1),

u3 = ξ(x − 1).

(

x<ξ x>ξ

Finally, we have g=

x(ξ − 1), ξ(x − 1),

31 1.19 Expanding g(x, ξ) = in terms of un =

√

x(ξ − 1), x < ξ, ξ(x − 1), x > ξ,

2 sin nπx as a Fourier series, show that

g(x, ξ) =

X un (x)un (ξ) n=1

λn

λn = −π 2 n2 .

Solution Let g(x, ξ) =

∞ X

An sin nπx,

n=1

An = =

An = 2

Z 1

g(x, ξ) sin nπxdx.

Z ξ Z 1 2 (ξ − 1) x sin xdx + ξ (x − 1) sin nπxdx 0 ξ ( ξ − cos nπx − sin nπx − 2 (ξ − 1) x nπ (nπ)2 0 1) − cos nπx − sin nπx − + ξ (x − 1) nπ (nπ)2 ξ cos nπξ sin nπξ + 2 (ξ − 1) −ξ nπ (nπ)2 cos nπξ sin nπξ + ξ (ξ − 1) − nπ (nπ)2 sin nπξ −2 2 2 . nπ

Then g(x, ξ) = −

∞ X

∞ X 2 1 sin nπx sin nπξ = un (x)un (ξ). 2π2 n λ n n=1 n=1

CHAPTER 1. GREEN’S FUNCTIONS

1.20 Obtain the Green’s function for κ∇2 u(x1 , x2 ) = v1

∂u ∂u + v2 , ∂x1 ∂x2

where v1 and v2 are constants, by transforming the dependent variable. Solution ∂ 2 u v1 ∂ 2 u v2 ∂u f − + 2− = 2 ∂x1 κ ∂x2 κ ∂x2 κ Let u = eλ1 x1 +λ2 x2 φ(x1 , x2 ). Then

∂u ∂φ = eλ1 x1 +λ2 x2 [λ1 φ + ], ∂x1 ∂x1 ∂2u ∂φ ∂2φ λ1 x1 +λ2 x2 2 =e λ1 φ + 2λ1 + . ∂x21 ∂x1 ∂x21

The given equation becomes ∂2φ ∂2φ λ1 v1 λ2 v2 2 2 + + λ1 + λ2 − − φ ∂x21 ∂x22 κ κ v1 ∂φ v2 ∂φ f e−(λ1 x1 +λ2 x2 ) + 2λ2 − = . + 2λ1 − κ ∂x1 κ ∂x2 κ We choose v1 v2 λ1 = , λ2 = , 2κ 2κ to get ∇2 φ − k 2 φ = F, where

f e−(λ1 x1 +λ2 x2 ) v12 + v22 , F = . 4κ2 κ The solutions of the homogeneous equation are the modified Bessel functions, K0 (kr) and I0 (kr), with q r = x21 + x22 . k2 =

33 For a bounded solution for r → ∞, we choose φ = AK0 (kr). As kr → 0, K0 behaves as − log r and we choose A=−

1 , 2π

by comparison to the two dimensional Laplace operator. For the original equation with f on the right hand side, g=− where ρ =

1 −[v1 (x1 −ξ1 )+v2 (x2 −ξ2 )]/(2κ) e K0 (kρ) 2π

(x1 − ξ1 )2 +)x2 − ξ2 )2 .

CHAPTER 1. GREEN’S FUNCTIONS

1.21 The anisotropic Laplace equation in a two dimensional infinite domain is given by ∂2u ∂2u k12 2 + k22 2 = 0. ∂x1 ∂x2 Find the Green’s function for this equation. SolutionThe Green’s function satisfies k12

2 ∂2u 2∂ u + k = δ(x1 , x2 ), 2 ∂x21 ∂x22

if the source is at the origin. Let y1 = x1 /k1 ,

y2 = x2 /k2 .

Then, u satisfies the Laplace equation ∂2u ∂2u + = 0, ∂y12 ∂y22 away from the origin. 1 g(y1, y2 ) = log 2π

q y12 + y22.

In terms of the original variables, with the source being at (ξ1 , ξ2) 1/2 1 (x1 − ξ1 )2 (x2 − ξ2 )2 log + . g(x1 , x2 , ξ1 ξ2 ) = 2π k12 k12 Note: When g is used in an integral, the original area element dy1 dY2 = dx1 dx2 /(k1 k2 ).

35 1.22 In a semi-infinite medium, −∞ < x < ∞, 0 < y < ∞, the pressure fluctuations satisfy the wave equation

1 ∂2p ∇ p= 2 2, c ∂t 2

where c is the wave speed and t is time. If the boundary, y = 0, is subjected to a pressure p = P0 δ(x)h(t), show that

t y h(t − kr), p(x, y, t) = A 2 √ 2 r t − k2r2 p where r = x2 + y 2 and k = 1/c, is a solution of the wave equation. Evaluate the constant A using equilibrium of the medium in the neighborhood of the applied load. Hint: Use polar coordinates.

Solution In polar coordinates, the wave equation becomes

2 ∂ 2 p 1 ∂p 1 ∂2p 2∂ p + + = k . ∂r 2 r ∂r r 2 ∂θ2 ∂t2

Away from r = 0, the given solution,

p=A

t sin θ , rβ

β=

√

t2 − k 2 r 2 ,

CHAPTER 1. GREEN’S FUNCTIONS has the derivatives ∂p sin θ 1 t2 = A − ∂t r β β3 sin θ k 2 r 2 = −A r β3 3k 4 rt ∂2p k 2 2 = A sin θ 5 ∂t β 2 sin θ t 1∂ p = −A 3 2 2 r ∂θ r β 2 ∂p k t t = A sin θ − 2 ∂r β3 βr t sin θ = A 2 3 k2r2 − β 2 r β t sin θ = A 2 3 2k 2 r 2 − t2 r β t2 t sin θ 2 = A 3 2k − 2 β r 2 ∂ p t2 3k 2 r 2t2 2 = At sin θ 2k − 2 + 3 3 ∂r 2 r β5 r β t sin θ = A 3 5 6k 4 r 4 − 3k 2 t2 r 2 + 2t2 (t2 − k 2 r 2 ) r β t sin θ = A 3 5 6k 4 r 4 − 5k 2 t2 r 2 + 2t4 r β

Now, the left hand side of the wave equation becomes = At sin θ

1 1 1 (6k 4 r 4 − 5k 2 t2 r 2 + 2t4 ) + 3 3 (2k 2 r 2 − t2 ) − 3 3 5 r β r β r β

t sin θ 4 4 6k r − 5k 2 t2 r 2 + 2t4 + 2k 2 r 2 (t2 − k 2 r 2 ) r3β 5 −t2 (t2 − k 2 r 2 ) − (t4 − 2k 2 r 2 t2 + k 4 r 4 ) t sin θ = 3A 5 k 4 r. β = A

This is identical to the right hand side quantity.

37 To find the value of A, we draw a semi-circle under the concentrated load P0 and balance the vertical force. We let r → 0. Z π/2 P0 = 2 p sin θrdθ 0 Z π/2 = 2A sin2 θdθ = Aπ/2. 0

Then A=

2P0 π

CHAPTER 1. GREEN’S FUNCTIONS

1.23 For the two dimensional wave equation ∇2 u =

1 ∂2u , c2 ∂t2

steady state solutions are obtained using u(x, y, t) = v(x, y)e−iΩt where v satisfies the Helmholtz equation, Lv = 0,

L = ∇2 + k 2 ,

k = Ω/c.

(1)

(2)

Show that the Hankel p functions H0 (kr) and H0 (kr) satisfy Lg = δ(x, y) when r = x2 + y 2 6= 0. Examine their asymptotic forms for kr << 1 and for kr >> 1, using the results shown in Abramowitz and Stegun (1965) and select multiplication constants A and B to have (1) (2) g = AH0 or g = BH0 by comparing the asymptotic form with the (1) Green’s function for the Laplace operator (k → 0). Show that H0 (2) corresponds to an outgoing wave and H0 to an incoming wave. Solution The Helmholtz equation, (∇2 + k 2 )v = δ(x − ξ, y − η) with a source at the origin, has the polar form (rv ′ )′ + k 2 rv = δ(r), where we assume axisymmetry. Solutions are the Hankel functions as (1)

(2)

v = H0 (kr), H0 (kr). When kr << 1, (1)

H0 (kr) = J0 + iY0 ∼

2 log r, π

2 (2) H0 (kr) = J0 − iY0 ∼ − log r, π

39 Comparing with the 2D Laplace operator g=

−i (1) H (kr), 4 0

i (2) or g = H0 (kr). 4

When kr >> 1, (1) H0 (kr) ∼

r r

2 i(kr−π/4) e , πkr

2 −i(kr−π/4) e , πkr Combining these with exp(−iΩt), we find terms of the form (r − ct) for (1) (2) H0 (kr) and (r + ct) for H0 (kr), with former being an outgoing wave and the latter an incoming wave. (2) H0 (kr) ∼

CHAPTER 1. GREEN’S FUNCTIONS

1.24 Show that g=−

e±ikr , 4πr

satisfies the Helmholtz equation ∇2 g + k 2 g = δ(x − ξ), in a 3D infinite domain, with r = |x − ξ|. Assuming the Helmholtz equation is obtained from the wave equation by separating the time dependence using a factor e−iΩt , show that (±) signs correspond to outgoing and incoming waves, respectively. Solution In spherical coordinates the spherically symmetric Green’s function satisfies (r 2 g ′)′ + k 2 r 2 g = 0, away from the source point. Substituting the given solution and neglecting the factor −1/(4π), ′ 1 ik ±ikr 2 r − 2± e + k 2 re±ikr r r = {(−1 ± ikr)′ + (−1 ± ikr)(±ik)} + k 2 r = ±ik ∓ ik − k 2 r + k 2 r = 0. Integrating the equation for the Green’s function over a sphere of radius r << 1,

lim 4πr 2

r→0

4πr 2 g ′ |r = 1 1 ik ∓ e±ikr = 1 4πr 2 4πr

For the (+) sign, we find the exponential term (r − ct) representing an outgoing wave and for the (−) sign, (r + ct) representing an incoming wave. Here c = Ω/k.

41 1.25 Consider a volume V enclosed by the surface S in 3D. From ∇2 u + k 2 u = f,

∇2 g + k 2 g = δ(x − ξ),

obtain the solution Z Z ∂u ∂g u(ξ) = g(ξ, x)f (x)dV − g −u dS. ∂n ∂n V S Solution Using the inner products of the first equation with g and the second equation with u and subtracting hg, Lui − hu, Lgi = hg, f i − hu, δi. For a self-adjoint operator, the left hand side can be integrated by parts (Gauss theorem) to get Z Z ∂u ∂g u(ξ) = g(ξ, x)f (x)dV − g −u dS. ∂n ∂n V S

CHAPTER 1. GREEN’S FUNCTIONS

1.26 In the previous problem, assuming V is a sphere of radius R centered at x = 0 and x is a point on its surface, and g is the Green’s function for the 3D infinite space, show that Z ikr 1 e ∂u ∂ eikr u(ξ) = −u dS, 4π S r ∂n ∂n r if f = 0. If the included angle between x and x − ξ is ψ, show that dr/dn = cos ψ. Also show that as R → ∞, r → R and R2 (1 − cos ψ) is finite and for the surface integral to exist ∂u − iku → 0, and u → 0. r ∂r These are known as the Sommerfeld radiation conditions. Solution For an infinite 3D space the Green’s function for the Helmholtz operator is 1 ikr g=− e . 4πr When there is no forcing function Z ikr 1 e ∂u ∂ eikr u(ξ) = −u dS 4π S r ∂n ∂n r From Fig. 1.1, dr = cos ψ dn Then Z 1 1 ∂u ik u u(ξ) = − u cos ψ + 2 cos ψ eikr dS 4π S r ∂n r r Z 1 1 ∂u iku ik u = − + (1 − cos ψ) + 2 cos ψ eikr dS 4π S r ∂n r r r From the triangle in Fig. 1.1, with R = |x|, we find ξ 2 = R2 + r 2 − 2Rr cos ψ.

43 As R → ∞ and r → ∞, dS ∼ 4πR2 and 2R2 (1 − cos ψ) = ξ 2 is finite and the condition for the existence of the integral is ∂u r − iku → 0, and u → 0. ∂n dndr ψ r

Figure 1.1: A source inside a spherical domain