Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 1 pps

We apply Fourier transforms in x and y to the Green function problem... We substitute the solution u and the adjoint Green function G∗ into the generalized Green’s Theorem.... We take th

The jump condition is

ˆG(ω, 0; ξ, τ+) = F [δ(x − ξ)]

We write the solution for ˆG and invert using the convolution theorem

ˆ

G = F [δ(x − ξ)] e−κω2(t−τ )H(t − τ )ˆ

G = 12π

Solution 45.2

1 We apply Fourier transforms in x and y to the Green function problem

Gtt− c2(Gxx+ Gyy) = δ(t − τ )δ(x − ξ)δ(y − η)ˆ

Next we introduce polar coordinates for x and y.

G(x, t|ξ, τ ) = H(c(t − τ ) − |x − ξ|)

2πcp(c(t − τ ))2− |x − ξ|2

2 To find the 1D Green function, we consider a line source, δ(x)δ(t) Without loss of generality, we have taken the

source to be at x = 0, t = 0 We use the 2D Green function and integrate over space and time.

2cH (ct − |x|)g(x, t|ξ, τ ) = 1

2cH (c(t − τ ) − |x − ξ|)Solution 45.3

1

Gtt = c2Gxx, G(x, 0) = 0, Gt(x, 0) = δ(x − ξ)ˆ

Gtt = −c2ω2G,ˆ G(ω, 0) = 0,ˆ Gˆt(ω, 0) = F [δ(x − ξ)]

ˆ

G = F [δ(x − ξ)] 1

cω sin(cωt)ˆ

Z ∞

−∞

δ(x − ξ − η)H(ct − |η|) dηG(x, t) = 1

2cH(ct − |x − ξ|)

2 We can write the solution of

Rx+t

−1 (1 − |ξ|) dξ, x − t < −1 < x + t

1 2

Rx+t x−t(1 − |ξ|) dξ, −1 < x − t, x + t < 1

1 2

R1 x−t(1 − |ξ|) dξ, x − t < 1 < x + t

1

2(2t − t2− x2) x − t < 0 < x + t(1 − x)t 0 < x − t, x + t < 1

1

4(1 + t − x)2 x − t < 1 < x + t

Next we consider the case 1/2 < t < 1.

Rx+t

−1 (1 − |ξ|) dξ, x − t < −1 < x + t

1 2

Rx+t x−t(1 − |ξ|) dξ, −1 < x − t, x + t < 1

1 2

R1 x−t(1 − |ξ|) dξ, x − t < 1 < x + t

Finally we consider the case 1 < t.

Rx+t

−1 (1 − |ξ|) dξ, −1 < x + t < 1

1 2

R1

−1(1 − |ξ|) dξ, x − t < −1, 1 < x + t

1 2

R1 x−t(1 − |ξ|) dξ, −1 < x − t < 1

0.1 0.2 0.3 0.4 0.5

Figure 45.1: The solution at t = 1/2 and t = 2

Figure 45.2 shows the behavior of the solution in the phase plane There are lines emanating form x = −1, 0, 1showing the range of influence of these points

u=0 u=0

u=1

Figure 45.2: The behavior of the solution in the phase plane

Solution 45.4

We define

L[u] ≡ ∇2u + a(x) · ∇u + h(x)u

We use the Divergence Theorem to derive a generalized Green’s Theorem

(u∇v − v∇u + uva) · n dA

We define the adjoint operator L∗

L∗[u] = ∇2u − ∇ · (au) + hu

We substitute the solution u and the adjoint Green function G∗ into the generalized Green’s Theorem.

2 We take c = D/ and consider the limit  → 0.

u = lim

4π

1p(x − /2)2+ y2+ z2 − 1

|x − ξ|



We write this in polar coordinates Denote x = r eıθ and ξ = ρ eıϑ Let φ = θ − ϑ be the difference in angle between

x and ξ

G(x|ξ) = 1

2πln

pr2+ ρ2 − 2rρ cos φρpr2+ 1/ρ2− 2(r/ρ) cos φ

Z 2π

0

f (ϑ)Gρ(r, θ|1, ϑ) dϑ

Gρ= 12π

ρ − r4ρ + r(r2− 1)(ρ2+ 1) cos φ(r2+ ρ2 − 2rρ cos φ)(r2ρ2+ 1 − 2rρ cos φ)

Gρ(r, θ|1, ϑ) = 1

2π

1 − r2

1 + r2− 2r cos φu(r, θ) = 1 − r

1

∆G = δ(x − ξ)δ(y − η)δ(z − ζ)1

We find the homogeneous solutions.

G = − 1

4πr

3 We write the Laplacian in circular coordinates.

∆G = δ(x − ξ)δ(y − η)1

Thus we see that G = c ln r We determine the constant by integrating ∆G over a ball about the origin, R.

Z Z

R

∆G dx = 1Z

∂R

∇G · n ds = 1Z

G = 12π ln r

u = − 14π

Z ∞

−∞

r(r2+ ζ2)3/2 dζ

ur = 14π

2r

u = 12πln rSolution 45.8

1 We take the Laplace transform of the differential equation and the boundary conditions in x

Gt− νGxx = δ(x − ξ)δ(t − τ )

s ˆG − ν ˆGxx = δ(x − ξ)ˆ

Now we have an ordinary differential equation Green function problem We find homogeneous solutions whichrespectively satisfy the left and right boundary conditions and compute their Wronskian

y1 = sinhr s

νx

, y2 = sinhr s

ν(L − x)



W =

= −2r s

ν

sinhr s

νx

coshr s

ν(L − x)

+ coshr s

νx

sinhr s

We use the expansion of the hyperbolic cosecant in our expression for the Green function.

3 We take the inverse Laplace transform to find the Green function for the diffusion equation.

2√πνt

in the sine series decay exponentially Again, a small number of terms could be used for an accurate approximation.Solution 45.9

1 We take the Fourier cosine transform of the differential equation

Gt− νGxx = δ(x − ξ)δ(t − τ )ˆ

Gt+ νω2G = Fˆ c[δ(x − ξ)]δ(t − τ )ˆ

G = Fc[δ(x − ξ)] e−νω2(t−τ )H(t − τ )ˆ

We do the inversion with the convolution theorem.

G = 12π

2 The fundamental solution on the infinite domain is

Now we have an initial value problem with no forcing.

ut= νuxx, for t > 0, u(x, 0) = δ(x − ξ)

2 We take the Laplace transform of the initial value problem

sˆu − u(x, 0) = ν ˆuxxˆ

uxx− s

νu = −ˆ

1

νδ(x − ξ), u(±∞, s) = 0ˆThe solutions that satisfy the left and right boundary conditions are, respectively,

= −2r s

νˆ

u = −1νe

Solution 45.11

Gtt− c2Gxx = δ(x − ξ)δ(t − τ ),G(x, t; ξ, τ ) = 0 for t < τ

We take the Fourier transform in x

ˆ

Gtt+ c2ω2G = F [δ(x − ξ)]δ(t − τ ), G(ω, 0; ξ, τˆ −) = ˆGt(ω, 0; ξ, τ−) = 0Now we have an ordinary differential equation Green function problem for ˆG We have written the causality condition,the Green function is zero for t < τ , in terms of initial conditions The homogeneous solutions of the ordinary differentialequation are

{cos(cωt), sin(cωt)}

It will be handy to use the fundamental set of solutions at t = τ :

cos(cω(t − τ )), 1

cω sin(cω(t − τ ))

.The continuity and jump conditions are

ˆG(ω, 0; ξ, τ+) = 0, Gˆt(ω, 0; ξ, τ+) = F [δ(x − ξ)]

We write the solution for ˆG and invert using the convolution theorem

ˆ

G = F [δ(x − ξ)]H(t − τ ) 1

cω sin(cω(t − τ ))ˆ

Z ∞

−∞

δ(y − ξ)H(c(t − τ ) − |x − y|) dy

G = 12cH(t − τ )H(c(t − τ ) − |x − ξ|)

G = 12cH(c(t − τ ) − |x − ξ|)

The Green function for ξ = τ = 0 and c = 1 is plotted in Figure 45.3 on the domain x ∈ (−1 1), t ∈ (0 1) TheGreen function is a displacement of height 2c1 that propagates out from the point x = ξ in both directions with speed c.The Green function shows the range of influence of a disturbance at the point x = ξ and time t = τ The disturbanceinfluences the solution for all ξ − ct < x < ξ + ct and t > τ

-1 -0.5

0 0.5

1 x

0 0.2 0.4 0.6 0.8 1

t

0 0.2 0.4

-1 -0.5

0 0.5

1 x

0 0.2 0.4 0.6 0.8 1

t

Figure 45.3: Green function for the wave equation

Now we solve the wave equation with a source

u = 12c

 (2n − 1)πx2L

Lsin

 (2n − 1)πx2L

Lsin

 (2n − 1)πξ2L

 r 2

Lsin

 (2n − 1)πx2L

Lsin

 (2n − 1)πξ2L

δ(y − ψ)From the boundary conditions at y = 0 and y = H, we obtain boundary conditions for the an(y)

2L

.The Wronskian of these solutions is

W = −(2n − 1)π

 (2n − 1)π2



Thus the solution for an(y) is

an(y) =

r2

Lsin

 (2n − 1)πξ2L

 (2n − 1)π2

cosh (2n − 1)πy<

2L

.This determines the Green function

G(x; xi) = −2

√2Lπ

∞

X

n=1

12n − 1csch

 (2n − 1)π2

cosh (2n − 1)πy<

2L

sin (2n − 1)πx

amn(z)√2

LH sin

mπxL

sinnπyH



We substitute this into the differential equation

∇2

∞

X

m=1 n=1

amn(z)√2

LH sin

mπxL

sinnπyH



= δ(x − ξ)δ(y − ψ)δ(z − ζ)

+nπH

2

amn(z)

2

√

LH sin

mπxL

sinnπyH

2

√

LH sin

 mπξL

sin nπψH

2

√

LH sin

mπxL

sinnπyH



a00mn(z) − π



mL

2

+nH

2

amn(z) = √2

LH sin

 mπξL

sin nπψH

δ(z − ζ)From the boundary conditions on G, we obtain boundary conditions for the amn

amn(0) = amn(W ) = 0The solutions that satisfy the left and right boundary conditions are

amn1 = sinh

r

mL

2

+nH

2

πz

!, amn2 = sinh

r

mL

2

+nH

2

π(W − z)

!.The Wronskian of these solutions is

W = −

r

mL

2

+nH

2

π sinh

r

mL

2

+nH

2

πW

!.Thus the solution for amn(z) is

amn(z) = √2

LH sin

 mπξL

sin nπψH



sinh

q

m L

q

m L

2

+ Hn
2π sinh

q

m L

2

+ Hn
2πW



sinh (λmnπz<) sinh (λmnπ(W − z>)) ,where

λmn =

r

mL

2

+nH

2

.This determines the Green function

1

λmncsch (λmnπW ) sin

 mπξL

sinmπxL



sin nπψH

sinnπyH

sinh (λmnπz<) sinh (λmnπ(W − z>))

3 First we write the problem in circular coordinates

From the boundary conditions on G, we obtain boundary conditions for the gn.

gn(0) = gn(a) = 0The solutions that satisfy the left and right boundary conditions are

gn1 = rn, gn2 =r

a

n

−ar

n

.The Wronskian of these solutions is

W = 2na

n

r .Thus the solution for gn(r) is

4 First we write the problem in circular coordinates

We substitute the series into the differential equation.

gn(0) = g0n(a) = 0The solutions that satisfy the left and right boundary conditions are

gn1 = r2n, gn2 =r

a

2n

+ar

2n

.The Wronskian of these solutions is

W = −4na

2n

r .Thus the solution for gn(r) is

Solution 45.13

1 The set

{Xn} =

sin (2m − 1)πx

are eigenfunctions of ∇2 and satisfy the boundary conditions Yn0(0) = Yn0(H) = 0 The set

sin (2m − 1)πx

2L

cosnπyH

∞ m=1,n=0

are eigenfunctions of ∇2 and satisfy the boundary conditions of this problem We expand the Green function in

a series of these eigenfunctions

LH sin

 (2m − 1)πx2L

+

∞

X

m=1 n=1

gmn√2

LH sin

 (2m − 1)πx2L

cosnπyH



We substitute the series into the Green function differential equation

∆G = δ(x − ξ)δ(y − ψ)

gmn  (2m − 1)π

2L

2

+nπyH

2! 2

√

LH sin

 (2m − 1)πx2L

cosnπyH

LH sin

 (2m − 1)πξ2L

 r2

LH sin

 (2m − 1)πx2L

2

√

LH sin

 (2m − 1)πξ2L

cos nπψH

2

√

LH sin

 (2m − 1)πx2L

cosnπyH



We equate terms and solve for the coefficients gmn

gm0 = −

r2LH

2L(2m − 1)π

2

+ Hn
2 sin

 (2m − 1)πξ2L

cos nπψH

, sinmπy

H

, sinnπzW

: k, m, n ∈ Z+

 kπxL

sinmπyH

sinnπzW



We substitute the series into the Green function differential equation.

2

+

mπH

2

+

nπW

2! r 8

 kπxL

sin

mπyH

sin

nπzW

 kπξL

sin mπψ

H

sin nπζW



r8

 kπxL

sinmπyH

sinnπzW

H
sin nπζ

W

π2 k L

2

+ mH
2+ Wn
2This determines the Green function

3 The Green function problem is

Now we look for eigenfunctions of the laplacian.

, m, n ∈ Z+here jn,m is the mth root of Jn

Note that

sin(nθ)Jn jn,mr

a

: m, n ∈ Z+

: m, n ∈ Z+

sin(nθ)

We substitute the series into the Green function differential equation.

2

π a|Jn+1(jn,m)|Jn

 jn,mra

sin(nθ)



π a|Jn+1(jn,m)|Jn

 jn,mra

sin(nθ)

We equate terms and solve for the coefficients gmn

gnm = −

a

sin(nϑ)This determines the green function

4 The Green function problem is

Now we look for eigenfunctions of the laplacian.

a , Rnm = J2n

 j02n,mra

, m, n ∈ Z+here j0n,m is the mth root of J0n

Note that

sin(2nθ)J02n j

0

a

: m, n ∈ Z+







We expand the Green function in a series of these eigenfunctions.

sin(2nθ)

We substitute the series into the Green function differential equation

2

0 2n,m

We equate terms and solve for the coefficients gmn

gnm = −

a

j0 2n,m

This determines the green function

Solution 45.14

We start with the equation

∇2G = δ(x − ξ)δ(y − ψ)

We do an odd reflection across the y axis so that G(0, y; ξ, ψ) = 0.

∇2G = δ(x − ξ)δ(y − ψ) − δ(x + ξ)δ(y − ψ)Then we do an even reflection across the x axis so that Gy(x, 0; ξ, ψ) = 0

∇2G = δ(x − ξ)δ(y − ψ) − δ(x + ξ)δ(y − ψ) + δ(x − ξ)δ(y + ψ) − δ(x + ξ)δ(y + ψ)

We solve this problem using the infinite space Green function

2+ (y + ψ)2
− 1

4πln (x + ξ)

2 + (y + ψ)2

G = 14πln

 ((x − ξ)2+ (y − ψ)2) ((x − ξ)2+ (y + ψ)2)((x + ξ)2+ (y − ψ)2) ((x + ξ)2+ (y + ψ)2)

∂n − G∂u(x, y)

∂n



dS +Z

V

G∆u dVu(ξ, ψ) =

G = F [δ(x − ξ)]H(t − τ ) 1

cω sin(cω(t − τ ))ˆ

Z ∞

−∞

δ(y − ξ)H(c(t − τ ) − |x − y|) dy

G = 12cH(t − τ )H(c(t − τ ) − |x − ξ|)

G = 12cH(c(t − τ ) − |x − ξ|)

1 So that the Green function vanishes at x = 0 we do an odd reflection about that point

Gtt− c2Gxx = δ(x − ξ)δ(t − τ ) − δ(x + ξ)δ(t − τ )

G = 12cH(c(t − τ ) − |x − ξ|) −

12cH(c(t − τ ) − |x + ξ|)

2 Note that the Green function satisfies the symmetry relation

G(x, t; ξ, τ ) = G(ξ, −τ ; x, −t)

This implies that

Gxx = Gξξ, Gtt = Gτ τ

We write the Green function problem and the inhomogeneous differential equation for u in terms of ξ and τ

Gξ = 1

2c (δ(c(t − τ ) − |x − ξ|)(−1) sign(x − ξ)(−1) − δ(c(t − τ ) − |x + ξ|)(−1) sign(x + ξ))

Gξ(x, t; 0, ψ) = 1

cδ(c(t − τ ) − |x|) sign(x)

We are interested in x > 0.

Gξ(x, t; 0, ψ) = 1

cδ(c(t − τ ) − x)Now we can calculate the solution u

Z t +

0

h(τ )δ

(t − τ ) − x

c

dτ

gtt+ c2ω2gˆxx = 0, ˆg(x, 0; ξ, τ ) = F [δ(x − ξ)], ˆgt(x, 0; ξ, τ ) = 0

ˆ

g = F [δ(x − ξ)] cos(cωt)ˆ

g = F [δ(x − ξ)]F [π(δ(x + ct) + δ(x − ct))]

g = 12π

Z ∞

−∞

δ(ψ − ξ)π(δ(x − ψ + ct) + δ(x − ψ − ct)) dψg(x, t; ξ) = 1

2(δ(x − ξ + ct) + δ(x − ξ − ct))

Gρ(r, θ |1, ϑ) = 1< /sup>

2π

1 − r2

1 + r2− 2r cos φu(r, θ) = 1 − r

1

∆G = δ(x − ξ)δ(y − η)δ(z − ζ )1