Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 2 potx

We write the solution as the difference of infinite space Green functions.. We write the solution as the sum of infinite space Green functions... The series expansion of the Green functi

This solution demonstrates the domain of dependence of the solution The first term is an integral over the triangledomain {(ξ, τ ) : 0 < τ < t, x − cτ < ξ < x + cτ } The second term involves only the points (x ± ct, 0) The thirdterm is an integral on the line segment {(ξ, 0) : x − ct < ξ < x + ct} In totallity, this is just the triangle domain This

is shown graphically in Figure 45.4

x-ct x+ct

Domain of Dependence x,t

Figure 45.4: Domain of dependence for the wave equation

which has the solutions,

λn = nπ

a , Xn= sin

nπxa

, n ∈ N

We expand the solution u in a series of these eigenfunctions

2

cn(y) + c00n(y) − k2cn(y)

sinnπxa

δ(x − ξ)δ(y − ψ) sinnπx

a

dx

dn(y) = 2

asin

 nπξa

δ(y − ψ)

The the equations for the cn(y) are

δ(y − ψ), cn(0) = cn(b) = 0

The homogeneous solutions are {cosh(σny), sinh(σny)}, where σn = pk2(nπ/a)2 The solutions that satisfy theboundary conditions at y = 0 and y = b are, sinh(σny) and sinh(σn(y − b)), respectively The Wronskian of these

solutions is,

W (y) =

sinh(σny) sinh(σn(y − b))

σncosh(σny) σncosh(σn(y − b))

= σn(sinh(σny) cosh(σn(y − b)) − sinh(σn(y − b)) cosh(σny))

 sinh(σny<) sinh(σn(y>− b))

σnsinh(σnb) .The Green function for the partial differential equation is

sin nπξa



Gyy(α, y) − (k2+ α2) ˆG(α, y) == 1

π cos(αξ)δ(y − ψ),

ˆG(α, 0) = 0

Then we take the Fourier sine transform in y

−β2ˆG(α, β) + β

π

ˆG(α, 0) − (k2+ α2)G(α, β) =ˆ 1

π2cos(αξ) sin(βψ)ˆ

G = − cos(αξ) sin(βψ)

π2(k2+ α2+ β2)

We take two inverse transforms to find the solution For one integral representation of the Green function we take theinverse sine transform followed by the inverse cosine transform.

ˆ

G = − cos(αξ)sin(βψ)

π

1π(k2+ α2+ β2)ˆ

G = − cos(αξ)Fs[δ(y − ψ)]Fc

1

δ(z − ψ)√ 1

k2+ α2

exp−√k2+ α2|y − z|− exp−√k2+ α2(y + z) dzˆ

G(α, y) = − cos(αξ)

2π√

k2+ α2

exp−√k2+ α2|y − ψ|− exp−√k2+ α2(y + ψ)

G(x, y; ξ, ψ) = −1

π

Z ∞ 0

cos(αξ)

√

k2+ α2

exp−√k2+ α2|y − ψ|− exp−√k2+ α2(y + ψ) dα

For another integral representation of the Green function, we take the inverse cosine transform followed by the inversesine transform

ˆG(α, β) = − sin(βψ)cos(αξ)

π

1π(k2+ α2+ β2)ˆ

2π

Z ∞ 0

G(0, θ, ρ, ϑ) = 0G(r, θ, ρ, ϑ) → 0 as r → ∞.

Let w = r eiθ and z = x + iy We use the conformal mapping, z = wπ/α to map the sector to the upper half z plane.The problem is (x, y) space is

Gxx+ Gyy = δ(x − ξ)δ(y − ψ), −∞ < x < ∞, 0 < y < ∞,G(x, 0, ξ, ψ) = 0,

G(x, y, ξ, ψ) → 0 as x, y → ∞

We will solve this problem with the method of images Note that the solution of,

Gxx+ Gyy = δ(x − ξ)δ(y − ψ) − δ(x − ξ)δ(y + ψ), −∞ < x < ∞, −∞ < y < ∞,

G(x, y, ξ, ψ) → 0 as x, y → ∞,satisfies the condition, G(x, 0, ξ, ψ) = 0 Since the infinite space Green function for the Laplacian in two dimensions is

14π ln (x − ξ)

2+ (y − ψ)2
,the solution of this problem is,

 (x − ξ)2+ (y − ψ)2

(x − ξ)2 + (y + ψ)2



Now we solve for x and y in the conformal mapping.

 r2π/α+ ρ2π/α− 2rπ/αρπ/αcos(π(θ − ϑ)/α)

r2π/α+ ρ2π/α− 2rπ/αρπ/αcos(π(θ + ϑ)/α)



= 14πln

 (r/ρ)π/α/2 + (ρ/r)π/α/2 − cos(π(θ − ϑ)/α)(r/ρ)π/α/2 + (ρ/r)π/α/2 − cos(π(θ + ϑ)/α)



= 14πln





Now recall that the solution of

∆u = f (x),subject to the boundary condition,

u(x) = g(x),is

u(x) =

Z Z

f (xi)G(x; xi) dAξ+

Ig(xi)∇ξG(x; xi) · ˆn dsξ

The normal directions along the lower and upper edges of the sector are −ˆθ and ˆθ, respectively The gradient in polarcoordinates is

∇ξ = ˆρ ∂

∂ρ +

ˆϑρ

sin(π(θ − ϑ)/α)4αρcoshπαlnrρ− cos(π(θ + ϑ)/α)

Along ϑ = 0, this is

1

ρGϑ(r, θ, ρ, 0) =

sin(πθ/α)2αρcoshπαlnrρ− cos(πθ/α)

Along ϑ = α, this is

1

ρGϑ(r, θ, ρ, α) = −

sin(πθ/α)2αρcoshπαlnrρ+ cos(πθ/α)

The solution of our problem is

2αρcoshαπlnrρ− cos(πθ/α) dρu(r, θ) =

Z ∞ c

− sin(πθ/α)2αρcoshπαlnρr− cos(πθ/α) +

sin(πθ/α)2αρcoshπαlnrρ+ cos(πθ/α) dρu(r, θ) = −1

αsin

 πθα

cos πθα

 Z ∞ c

cos πθα

 Z ∞ ln(c/r)

1cosh2 πxα
− cos2 πθ

cos πθα

 Z ∞ ln(c/r)

1cosh 2πxα
− cos 2πθ

The solution for u(x, t) is the linear combination of the Green functions that satisfies the initial condition u(x, 0) =

f (x) This linear combination is

0 1 2

0.1

0.2

Figure 45.5: G(x, t; 1) and G(x, t; −1)

Now we consider the problem

ut= κuxx, u(x, 0) = f (x) for x > 0, u(0, t) = 0

Note that the solution of

Gt= κGxx, x > 0, t > 0,G(x, 0; ξ) = δ(x − ξ) − δ(x + ξ),satisfies the boundary condition G(0, t; ξ) = 0 We write the solution as the difference of infinite space Green functions

Next we consider the problem

ut= κuxx, u(x, 0) = f (x) for x > 0, ux(0, t) = 0

Note that the solution of

Gt= κGxx, x > 0, t > 0,G(x, 0; ξ) = δ(x − ξ) + δ(x + ξ),satisfies the boundary condition Gx(0, t; ξ) = 0 We write the solution as the sum of infinite space Green functions

0.1

0.15

0.2 0.250

0.2 0.4 0.6

0.05

0.1 0.15

0.2 0.250

0.2 0.4 0.6

0 1 2

0.05

0.1 0.15

0.2 0.25

Figure 45.6: Green functions for the boundary conditions u(0, t) = 0 and ux(0, t) = 0

Solution 45.22

a) The Green function problem is

Gtt− c2Gxx = δ(t − τ )δ(x − ξ), 0 < x < L, t > 0,

G(0, t; ξ, τ ) = Gx(L, t; ξ, τ ) = 0,G(x, t; ξ, τ ) = 0 for t < τ

The condition that G is zero for t < τ makes this a causal Green function We solve this problem by expanding G in

a series of eigenfunctions of the x variable The coefficients in the expansion will be functions of t First we find theeigenfunctions of x in the homogeneous problem We substitute the separation of variables u = X(x)T (t) into the

homogeneous partial differential equation.

λn = (2n − 1)π

2L , Xn = sin

 (2n − 1)πx2L

, n ∈ N

The series expansion of the Green function has the form,

We determine the coefficients by substituting the expansion into the Green function differential equation

δ(x − ξ)δ(t − τ ) sin (2n − 1)πx

2L

dx

dn(t) = 2

Lsin

 (2n − 1)πξ2L

δ(t − τ )

By equating coefficients in the sine series, we obtain ordinary differential equation Green function problems for the gn’s.

δ(t − τ )From the causality condition for G, we have the causality conditions for the gn’s,



A set of homogeneous solutions of the ordinary differential equation are

cos (2n − 1)πct

2L

, sin (2n − 1)πct

2L



Since the continuity and jump conditions are given at the point t = τ , a handy set of solutions to use for this problem

is the fundamental set of solutions at that point:

cos (2n − 1)πc(t − τ )

sin (2n − 1)πc(t − τ )

2L

H(t − τ )

Substituting this into the sum yields,

 (2n − 1)πξ2L

sin (2n − 1)πc(t − τ )

2L

sin (2n − 1)πx

2L



We use trigonometric identities to write this in terms of traveling waves

λ0 = 0, X0 = 1,

λn = nπ

L , Xn= cos

nπxL

, n = 1, 2, The series expansion of the Green function for t > τ has the form,

(Note the factor of 1/2 in front of g0(t) With this, the integral formulas for all the coefficients are the same.) We

determine the coefficients by substituting the expansion into the partial differential equation.

Gtt− c2Gxx = δ(x − ξ)δ(t − τ )1

δ(x − ξ)δ(t − τ ) cosnπx

L

dx

dn(t) = 2

Lcos

 nπξL

δ(t − τ )

By equating coefficients in the cosine series, we obtain ordinary differential equations for the gn

δ(t − τ ), n = 0, 1, 2, From the causality condition for G, we have the causality condiions for the gn,

.The homogeneous solutions of the ordinary differential equation for n = 0 and n > 0 are respectively,

{1, t},

cos nπct

L

, sin nπct

L



Since the continuity and jump conditions are given at the point t = τ , a handy set of solutions to use for this problem

is the fundamental set of solutions at that point:

{1, t − τ },

cos nπc(t − τ )

L

, Lnπcsin

 nπc(t − τ )L



.The solutions that satisfy the causality condition and the continuity and jump conditions are,

sin nπc(t − τ )

L

H(t − τ )

Substituting this into the sum yields,

G(x, t; ξ, τ ) = H(t − τ ) t − τ

2πc

sin nπc(t − τ )

L

cosnπxL



!

We can write this as the sum of traveling waves

G(x, t; ξ, τ ) = t − τ

L H(t − τ ) +

12πcH(t − τ )

Solution 45.23

First we derive Green’s identity for this problem We consider the integral of uL[v] − L[u]v on the domain 0 < x < 1,

0 < t < T

Z T 0

Z 1 0

(uL[v] − L[u]v) dx dt

Z T 0

Z 1 0

u(vtt− c2vxx− (utt− c2uxx)v
dx dt

Z T 0

Z 1 0

I

∂Ω

−c2(uvx− uxv), uvt− utv
· n dsThe domain and the outward normal vectors are shown in Figure 45.7

Writing out the boundary integrals, Green’s identity for this problem is,

(uvt− utv)t=0dx+

Z 0 1

(uvt− utv)t=Tdx − c2

Z T 0

(uvx− uxv)x=1dt + c2

Z 1 T

(uvx− uxv)x=0dtThe Green function problem is

x=0 x=1

t=0 t=T

n=(0,-1)

n=(1,0)

n=(0,1)

n=(-1,0)

Figure 45.7: Outward normal vectors of the domain

Now we apply Green’s identity for u = u(ξ, τ ), (the solution of the wave equation), and v = G(x, t; ξ, τ ), (the Greenfunction), and integrate in the (ξ, τ ) variables The left side of Green’s identity becomes:

Z T 0

Z 1 0

u(Gτ τ − c2Gξξ) − (uτ τ − c2uξξ)G
dξ dτ

Z T 0

Z 1 0

(u(δ(x − ξ)δ(t − τ )) − (0)G) dξ dτ

u(x, t)

Since the normal derivative of u and G vanish on the sides of the domain, the integrals along ξ = 0 and ξ = 1 inGreen’s identity vanish If we take T > t, then G is zero for τ = T and the integral along τ = T vanishes The oneremaining integral is

−

Z 1 0

(u(ξ, 0)Gτ(x, t; ξ, 0) − uτ(ξ, 0)G(x, t; ξ, 0) dξ

Thus Green’s identity allows us to write the solution of the inhomogeneous problem.

u(x, t) =

Z 1 0

(uτ(ξ, 0)G(x, t; ξ, 0) − u(ξ, 0)Gτ(x, t; ξ, 0)) dξ

With the specified initial conditions this becomes

u(x, t) =

Z 1 0

Gτ(x, t; ξ, 0) = −1 − 2

∞

X

n=1

cos(nπξ) cos(nπct) cos(nπx)

The integral of the first term is,

Z 1 0

For c = 1, the solution at x = 3/4, t = 7/2 is,

u(3/4, 7/2) = 53

15− 316π4

−7π4

720u(3/4, 7/2) = 12727

3840

Chapter 46

Conformal Mapping

on the quarter plane x, y > 0 with u(x, 0) = u(0, y) = 0 (and ξ, ψ > 0).

1 Use image sources to find u(x, y; ξ, ψ)

2 Compare this to the solution which would be obtained using conformal maps and the Green function for the upperhalf plane

3 Finally use this idea and conformal mapping to discover how image sources are arrayed when the domain is nowthe wedge bounded by the x-axis and the line y = x (with u = 0 on both sides)

Exercise 46.3

ζ = ξ + ıη is an analytic function of z, ζ = ζ(z) We assume that ζ0(z) is nonzero on the domain of interest u(x, y)

is an arbitrary smooth function of x and y When expressed in terms of ξ and η, u(x, y) = υ(ξ, η) In Exercise 8.15

dζdz

2

φ(ξ, η),where φ(ξ, η) = f (x, y)

4 Show that if in the z-plane, u satisfies the Green function problem,

uxx+ uyy = δ(x − x0)δ(y − y0),then in the ζ-plane, υ satisfies the Green function problem,

υξξ+ υηη = δ(ξ − ξ0)δ(η − η0)

46.3 Solutions

Solution 46.1

We map the wedge to the upper half plane with the conformal transformation ζ = z4

1 We map the wedge to the upper half plane with the conformal transformation ζ = z4 The new problem is

uξξ+ uηη = 0, u(ξ, 0) = 0

This has the solution u = η We transform this problem back to the wedge

u(x, y) = = z4
u(x, y) = = x4+ ı4x3y − 6x2y2 − ı4xy3+ y4

u(x, y) = 4x3y − 4xy3u(x, y) = 4xy x2− y2

2 We don’t need to use conformal mapping to solve the problem with Neumman boundary conditions u = c is asolution to

uxx+ uyy = 0, du

dn = 0

on any domain

Solution 46.2

1 We add image sources to satisfy the boundary conditions

uxx+ uyy = δ(x − ξ)δ(y − η) − δ(x + ξ)δ(y − η) − δ(x − ξ)δ(y + η) + δ(x + ξ)δ(y + η)

u = 14πln

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

2 The Green function for the upper half plane is

G = 14πln

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

ax ay

bx by

=

2x −2y2y 2x

...

720 u(3/4, 7 /2) = 127 27

3840

Chapter 46< /h2>

Conformal... ξ2< /small>+ η2< /small>)2< /small>+ (2xy − 2? ?η)2< /small>)((x2< /small>− y2< /small> − ξ2< /small>+ η2< /small>)2< /small>+ (2xy + 2? ?η)2< /small>)...

0 .2 0 .25 0

0 .2 0.4 0 .6< /h3>

0 2< /h3>

0.05

0.1 0.15

0 .2 0 .25

Figure 45 .6: Green functions for the boundary