Fowler A. Mathematical Geoscience

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790 11 Jökulhlaups

and, assuming that s|

=s|

−

and h|

=h|

−

γβ



−

ξN dξ =



−



2T +ν

U −

β(s −h)



dξ,

and deduce that if ν is sufﬁciently small, then



−

ξN dξ ≈0.

Interpret these results in terms of overall force and torque on the ice beam.

11.10 The subsidence rate w(X,t) of an ice cauldron on an ice sheet is governed

by the beam equation

XXXX

=−N,

where the effective load N(X,t) satisﬁes

+w,

and N

(t) is a function of t (

being its derivative). The model is to be

solved on 0 <X<∞, and we prescribe

w =w

=0,N=N

, on X =0,

w →−

as X →∞.

(i) By writing w =−

+W and integrating repeatedly, show that

W =



∞

(ξ −X)

N(ξ,t)dξ,

and deduce that



∞

N(ξ,t)dξ.

Hence show that N satisﬁes the integro-differential equation



∞

G(X, ξ )N(ξ, t) dξ,

and give the deﬁnition of G.

Show that



∞

ξN(ξ,t)dξ = 0, and deduce that G can be written in the

symmetric form

G(X, ξ ) =



−

X −

,ξ>X,

−

ξ −

,ξ<X.

(ii) Show directly from the governing equations that if

−t)

there is a similarity solution of the form

N =N

ψ(η), W =

φ(η),

11.9 Exercises 791

where

η =mX(t

−t)

and β should be determined. Show that, by choosing the value of m suitably,

the equation for φ can be written in the form (after eliminating ψ)

εηφ

−φ

−4φ =0,

where the Roman numeral superscripts indicate the number of derivatives.

Give the value of ε, and write down suitable boundary conditions for φ.

Appendix A

The Schwarzschild–Milne Integral Equation

The exact solution of (2.15)–(2.17) is obtained as follows. We deﬁne, as before, the

local average intensity

J(τ)=



−1

I(τ,μ)dμ, (A.1)

and the formal solution of (2.15)is

I =





∞

−(t−τ)/μ

J(t)

,μ>0,



−(t−τ)/μ

J(t)

(−μ)

,μ<0,

(A.2)

providing J does not grow exponentially as τ →∞(speciﬁcally, J = o(e

)). Sub-

stituting this expression back into (A.1), we ﬁnd, after some algebra, that J satisﬁes

the Schwarzschild–Milne integral equation

J(τ)=



∞



|t −τ|



J(t)dt, (A.3)

and the ﬂux conservation law (2.17) can be written in the form

Φ =2π





∞

J(t)E

(t −τ)dt −



J(t)E

(τ −t)dt



. (A.4)

The exponential integrals E

and E

are deﬁned by

(y) =y



∞

−s

ds, E

(y) =



∞

−s

ds; (A.5)

(A.4) acts as a normaliser for the linear equation (A.3).

Equation (A.3) is amenable to treatment by the Wiener–Hopf technique. It de-

ﬁnes J for τ>0, and we extend the deﬁnition of J so that

J =0,τ<0, (A.6)

A. Fowler, Mathematical Geoscience, Interdisciplinary Applied Mathematics 36,

DOI 10.1007/978-0-85729-721-1, © Springer-Verlag London Limited 2011

793

794 A The Schwarzschild–Milne Integral Equation

and we deﬁne a function h(τ ), h =0forτ>0, so that

J(τ)=



∞

−∞



|t −τ |



J(t)dt +h(τ), (A.7)

for all values of τ . Write K(t) =

(|t|), so that, if we take Fourier transforms of

(A.7), we get

−

, (A.8)

where

(z) is the transform of J and the + indicates that

(z) is analytic in an

upper half plane (since J =0forτ<0). Since J =o(e

) as τ →∞, this is at least

Im z>1. Similarly

−

is analytic in a lower half-plane.

The solution of (A.8) is now effected through the splitting of (1−

K) into factors

analytic in upper and lower half planes, and this can be done by solution of an

appropriate Hilbert problem. The transform

K is deﬁned as

K(z) =



∞

−∞

K(s)e

isz

ds, (A.9)

and we ﬁnd that

K =

2iz



1 +iz

1 −iz



tan

−1

z. (A.10)

We will now strengthen our assumption on J so that J does not grow exponentially

as τ →∞, i.e., J =o(e

ατ

) for any α>0; then

is analytic in Im z>0. Our aim

now is to ﬁnd a function G analytic in Im z

0 such that G

−

= 1 −

K on R,

and this is done by solving the Hilbert problem ln G

−ln G

−

=ln(1 −

K).Todo

this we wish to have 1 −

K =0, in order that ln(1 −

K) be Hölder continuous. On

the other hand we want ln{1−

K(t)}→0ast ∈R →±∞. These concerns motivate

the modiﬁcation of 1 −

K(t) by a factor (t

+1)/t

, since 1 −

K =O(t

) as t →0

(and is non-zero for t =0), so that we seek a function G such that

(t)

−

(t)





1 −

2it



1 +it

1 −it



, (A.11)

for t ∈ R. Clearly G is only determined up to a multiplicative analytic function,

and to be speciﬁc we will suppose G

→ 1asz →∞. We take the branches of

ln(1 ±it) to be such that ln 1 =0. The solution of (A.11)is

G(z) =exp



2πi



∞

−∞





1 −

tan

−1



t −z



, (A.12)

and with this deﬁnition of G(z) (and thus G

(t) and G

−

(t)), Eq. (A.8)for

can

be written in the form, for t ∈R,

z +i

=(z −i)

−

. (A.13)

A The Schwarzschild–Milne Integral Equation 795

Fig. A.1 Inversion contour

for (A.16)

Clearly the left hand side deﬁnes the limit on Imz = 0+ of a function analytic

in the upper half plane Im z>0, while the right hand side is the limit on Im z =0−

of a function analytic in Imz<0 (since (A.7) implies that h grows no faster than

J(−τ)). We infer that each function can be analytically continued into its opposite

half plane, thus deﬁning an entire function E(z), so that

(z) =

(z +i)E(z)

(z)

. (A.14)

The deﬁnition of

as a Fourier transform requires

→ 0asz →∞, while

also G

→ 1asz →∞. It follows that

∼ E/z, which requires that E = ic is

constant, i.e.,

ic(z +i)

(z)

, (A.15)

and the constant c is determined by the normalising condition (A.4). (The factor i

is inserted for later convenience.)

Some information on the structure of

can be gleaned from (A.11). Evidently

can be extended to Imz<0, and G

−

to Im z>0 by the reciprocal relationship

(z)

−

(z)





1 −

2iz



1 +iz

1 −iz



. (A.16)

Care needs to be used in interpreting (A.16). If Imz<0, then (A.16) provides an

analytic continuation for G

there, which shows that the continuation of G

Im z<0 (very deﬁnitely not equal to G

−

) has a logarithmic branch point at z =−i.

Similarly G

−

, extended to Im z>0, has a logarithmic branch point at z =+i.

Therefore

, extended via (A.15)toImz<0, has a double pole at z = 0(as

(0) =

√

=0) and a branch cut which we may take from −i to −i∞.

The inverse transform of (A.15)is

J(τ)=

2π



∞

−∞

(z)e

−izτ

dz, (A.17)

796 A The Schwarzschild–Milne Integral Equation

where the contour is indented above the origin. If τ<0, we complete the contour in

the upper half plane, whence we have J =0 (as we assumed). If τ>0, we complete

the contour as shown in Fig. A.1. The result of this is that

J(τ)=−i



Res



−izτ





z=0

2π



∞

−τ(1+x)



−





, (A.18)

where

(x) =

[−i +xe

−iπ/2

−

(x) =

[−i +xe

3iπ/2

]. Calculation of the

residue yields the result

Res|

z=0

=ic

√

3(1 +τ −j), (A.19)

where

j =



∞



(1 −t

−1

tan

−1

−1 −



1 +t

. (A.20)

We use (A.16) to substitute for G

in (A.15), and then we ﬁnd

(x) =

−c

(2 +x)G

−

[−i(1 +x)]l

(x)

, (A.21)

where

(x) =1 −

2(2 +x)





2 +x



±iπ



. (A.22)

It follows that

−

iπc

−

(x)



2 +x −



2+x





, (A.23)

where g

−

(x) =G

−

(−i −ix), and from (A.12), we ﬁnd

−

(x) =exp



−

(1 +x)

2π



∞

−∞





1 −

tan

−1



+(1 +x)

}



(A.24)

Finally, therefore, J =cJ

(τ ), where

(τ ) =

√

3(1+τ −j)+

−τ



∞

−xτ

−

(x)



2 +x −



2+x





. (A.25)

Evidently J ≈c

√

3(1−j +τ)+o(e

−τ

) as τ →∞, which conﬁrms the assumption

of non-exponential growth.

It only remains to compute c (which is evidently real, hence the choice of con-

stant ic in (A.15)), and there seems no obvious short cut other than laborious sub-

stitution of the expression (A.25)forJ into (A.4), which can be written in the form

c =

2π



∞

(t)H (τ −t)dt

, (A.26)

A.1 Exercises 797

where

H(θ)=



(−θ), θ <0,

−E

(θ), θ > 0.

(A.27)

A.1 Exercises

A.1 What is wrong with the following argument? To determine c in (A.26), write

(A.4) in the form (since J = 0forτ<0)

Φ =2π



∞

−∞

J(t)H(τ −t)dt,

where

H(θ)=



(−θ), θ <0,

−E

(θ), θ > 0.

A Fourier transform yields, via the convolution theorem,

2πiz

(z)

H(z),

where

H(z)=−2i



∞

(θ) sin zθ dθ.

Show that

−



∞

(θ)e

izθ

dθ =

ln(1 −iz) +iz

so that

2πiz

=2i



2iz −ln



1+iz

1−iz



Since also

ic(z +i)

(z)

this implies

(z) =

A(z +i)



1 −

2iz



1+iz

1−iz



where A =

8πc

; but this is not analytic in Im z>0.

Appendix B

Turbulent Flow

Shear ﬂows become turbulent if the Reynolds number Re is sufﬁciently large. Usu-

ally, this means Re ∼ 10

. For ﬂow in a cylindrical pipe, the Reynolds number is

conventionally chosen to be

Re =

, (B.1)

where U is the mean velocity, d is the pipe diameter, and ν is the kinematic viscosity.

With this deﬁnition, the onset of turbulence occurs at Re = 2,300, although the

details of the transition process are complicated (Fowler and Howell 2003), and

occur over a range of Reynolds number.

Most obviously, one might suppose that turbulence arises because of an insta-

bility of the uniform (laminar) ﬂow, and for half a century this motivated the study

of the famous Orr–Sommerfeld equation (one version of which is studied in Ap-

pendix C), which describes normal modes of the linearised Navier–Stokes equations

describing perturbations about a steady uniform ﬂow. Commonly such studies are

done in two dimensions, for example for plane Poiseuille ﬂow, when the Reynolds

number is deﬁned in terms of the maximum (centre-line) speed of the laminar ﬂow

and the half-width. This leads to a deﬁnition which is

of that which would arise

using the mean velocity and width. For plane Poiseuille ﬂow, it is found that the

steady ﬂow is linearly unstable if Re > 5,772; on the other hand, turbulence sets in

at Re ≈1,000 (Orszag and Patera 1983). For pipe ﬂow, the ﬂow is linearly stable

at all Reynolds numbers, although the decay rate of disturbances tends to zero as

Re →∞.

It appears that the transition to turbulence is only vaguely related to the stabil-

ity of the uniform state. The story is most simply told in the plane Poiseuille case.

The instability at Re =Re

=5,772 is subcritical, and an (unstable) branch of ﬁnite

amplitude stationary solutions bifurcates for Re < Re

, and exists down to about

Re =2,900 before bending back on to a higher amplitude stable branch. Crucially,

the (two-dimensional) stability or instability occurs on a long viscous time scale.

However, these stationary solutions are subject to a three-dimensional instability

which occurs on the fast convective time scale, and it is this which appears to cause

the transition. Its occurrence at Re ≈1,000 is associated with the fact that while the

A. Fowler, Mathematical Geoscience, Interdisciplinary Applied Mathematics 36,

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