fluid - terrapub · fluid 9.2 stream function for two-dimensional problems, the scalar function,...

Chapter 9

Fluid

Rocks behave like fluids at depths in geologic timescales. In this chapter, basic equa-tions for a simple fluid are introduced. Then equations are used to consider verticalmovements. Linear stability analysis of the motions of the fluid is applied to periodicgeological structures.

9.1 Fluids

9.1.1 Definition

Fluids are distinguished from solids by their unlimited deformations [276]. Solids ‘remember’ theirinitial shape—they tend to return to their original shapes once they are relieved of the force todeform. By contrast, fluids change their shapes anyway. Water conforms to the shape of its container.Accordingly, the constitutive equation of fluids does not contain a strain tensor such as V. The fasterwe swim, the greater the resistance we feel from the surrounding water. The equation should havea relationship with velocity instead of strain. However, there is no resistance if we move at thesame velocity as the water velocity. Difference in velocity causes resistance, i.e., velocity gradientis important.

We will consider only isotropic fluid for simplicity. This allows us to use the representationtheorem (Eq. (5.5)), but we should replace V by the velocity gradient tensor D in the constitutiveequation

S = φ0I + φ1D + φ2D2,

where φ0, φ1, and φ2 are the scalar functions of the basic invariants of D. As S is negative forcompression, the hydrostatic pressure is −pI. Hydrostatic pressure is often separated from the firstterm of the above equation,

S = (−p + φ0)I + φ1D + φ2D2 (9.1)

Materials that have this constitutive equation are called Reiner-Rivlin fluids. In this case, φ0 vanishesfor stationary fluids.

205

206 CHAPTER 9. FLUID

If the constitutive equation of a material is written as

S = −pI + f(D), (9.2)

then the material is called a Stokesian fluid. The function f has a value of a second-order tensor,possibly anisotropic. If f is linear, the fluid is called Newtonian fluid. The fluid should be relaxed ifit is at rest. Therefore, f(O) = O. If it is at rest, stress in the fluid is just the pressure. However, dueto the term f(D), the fluid parcels feel different forces in motion, even if p is constant. Accordingly,f is called dynamic pressure. Due to this pressure, flow in the mantle can push up or pull down thesurface (§9.5).

9.1.2 Newtonian fluid

Newtonian fluids have a linear and homogeneous function f in Eq. (9.2). For simplicity, we assumeincompressibility. The constitutive equation of incompressive Newtonian fluid is

S = −pI + 2ηD, (9.3)

where η is a material constant representing resistance against movement and is called viscosity.Viscosity is half the constant of proportion between the velocity gradient and stress1. Table 9.1shows the viscosity of familiar and geologic fluids.

Let us derive the divergence of stress, ∇ · S, from Eq. (9.3). The ith component is

∑j

∂

∂xj

[−pδij + η

(∂vi∂xj

+∂vj

∂xi

)]= − ∂p

∂xi+ η

∑j

∂2vi

∂x2j

+ η∑j

∂2vj

∂xi∂xj. (9.4)

We have assumed incompressibility so that Eq. (2.29) holds and density is constant everywhere.Differentiating the equation by xi, we have

∑j ∂

2vj/∂xj∂xi = 0. The last term in Eq. (9.4) vanishesaccordingly. Substituting this ∇ · S into the equation of motion (3.23), we obtain the Navier-Stokesequation

ρa = −∇p + η∇2v + ρX. (9.5)

Long-term tectonic movements are very slow and the inertia term ρa is neglected. So we have theStokes equation

0 = −∇p + η∇2v + ρX. (9.6)

The Navier-Stokes equation is rewritten with vorticity instead of velocity. Vorticity is defined asω = ∇ × v so that we apply ∇× to both sides of Eq. (9.5), and we have

∇ × ρa = −∇ × ∇p + ∇ × (η∇2v)+ ∇ × (ρX) . (9.7)

1Viscosity is measured with the unit Pa s = Nm−2s. In the CGS system, the viscosity unit is called Poise (P), and 1 P =10 Pa s.

9.1. FLUIDS 207

Table 9.1: Viscosity of several fluids. As viscosity depends on temperature, rough values are shown.Glacier, crust, and mantle are of the Earth.

Viscosity (Pa s) Reference

Air 10−5

Water 10−3

Olive oil 10−1

Alcohol 100

Asphalt (15◦ C) 107

Columbia River Basalt (1250◦ C) 101 [210]Lunar mare basalt (1200◦ C) 100 [159]Glacier 1013 − 1015 [123]Crust 1018 − 1030 [231]Upper mantle 1020 [112, 113, 114]Lower mantle 1021 [112, 113, 114]

This differential equation holds in every small parcel of fluid within which the continuum hypothesisverifies that viscosity can be regarded as being constant. Density is also constant because we haveassumed incompressibility. The left-hand side of Eq. (9.7) becomes

ρa = ∇ ×(ρ

Dv

Dt

)= ρ

D∇ × v

Dt= ρ

Dω

Dt.

The term ∇×∇p vanishes as Eq. (C.57), and second term of the right-hand side of Eq. (9.7) becomes∇ × (η∇2v

)= η∇2ω. Accordingly, Eq. (9.7) is simplified to the equation

Dω

Dt=η

ρ∇2ω + ∇ ×X.

This is equivalent to the Navier-Stokes equation. In the case of ∇ ×X = 0, we have

Dω

Dt= ν∇2ω. (9.8)

This is the diffusion equation of vorticity and the diffusion coefficient ν ≡ η/ρ is called kinematicviscosity. To emphasize the difference from this, η is called dynamic viscosity. Heavy liquids tendto keep their velocity due to inertia, so that vorticity is difficult to change, resulting in a low diffusioncoefficient. In the case that inertia is negligible, Eq. (9.8) become

0 = η∇2ω + ρ(∇ ×X). (9.9)


9.2 Stream function

For two-dimensional problems, the scalar function, ψ (x1, x2), that is related to velocity componentsby the equations

v1 = − ∂ψ∂x2

, v2 =∂ψ

∂x1. (9.10)

is useful, and is called the stream function. The velocity field that is described by the stream functionis always incompressible, because

∇ · v = − ∂

∂x1

∂ψ

∂x2+

∂

∂x2

∂ψ

∂x1= 0.

Two-dimensional, incompressible, Newtonian fluids have a stream function.The function is useful because the contour lines of ψ drawn on the O-12 plane show the stream

lines. In addition, flux is indicated by the difference in the value of ψ . To show these facts, supposea smooth curve C between points P and R on the plane (Fig. 9.1). A point Q is on the curve. Let(x1, x2) be the coordinates of the point and s be its distance from P along the curve. The unit vectortangent to the curve at Q is s = (dx1/ds, dx2/ds)T. The unit vector perpendicular to s and points tothe left of s is n = R · s, where

R =

(cos 90◦ − sin 90◦

sin 90◦ cos 90◦

)=

(0 −11 0

).

This is the orthogonal tensor which rotates vectors counterclockwise by 90◦ on a plane (Eq. (C.9)).The components of n are (

n1

n2

)=

(0 −11 0

)(dx1/dsdx2/ds

)=

(−dx2/dsdx1/ds

).

Given the flow velocity v at the point Q, the flux crossing the curve C at the point is v · n, thereforethe quantity of fluid that crosses the entire length of the curve in a unit of time is

F =∫

OPv · n ds =

∫OP

(v1n1 + v2n2) ds

=∫

OP

(∂ψ

∂x2

dx2

ds+∂ψ

∂x1

dx1

ds

)ds =

∫OP

∂ψ

∂sds = ψ (P) − ψ (O).

Accordingly F = 0 if the points P and R are on the same contour line of ψ , indicating that thecontour line lies along a stream line. Note that the flow direction is to the left2 of the vector ∇ψ . Asthe above equation suggests, only the relative ψ is meaningful. Therefore, we can assume ψ = 0 atconvenient points, such as at infinitely far points or at the origin of the coordinates.

2There are textbooks in which the stream function is defined with opposite signs to Eq. (9.10). In that case, the flowdirection is to the right of ∇ψ .

9.2. STREAM FUNCTION 209

Figure 9.1: Schematic picture showing the contour lines of stream function ψ .

The stream function has an important relationship with vorticity. Substituting Eq. (9.10) intothe equation that defines vorticity in two-dimensional flow, we have

ω =∂

∂x2

(− ∂ψ∂x2

)− ∂

∂x1

(∂ψ

∂x1

)= −

(∂2ψ

∂x21

+∂2ψ

∂x22

)= −∇2ψ.

Hence, if ω = 0, ψ is a harmonic equation satisfying ∇2ψ = 0.

Biharmonic equation

The stream function is a harmonic equation if vorticity vanishes. However, the function is always abiharmonic equation that satisfies the equation

∂4ψ

∂x4+ 2

∂4ψ

∂x2∂z2+∂4ψ

∂z4= 0 or ∇4ψ = 0. (9.11)

Let us consider that gravity is the only body force and let us derive this equation. We use the Carte-sian coordinates O-xz with the z-axis taken vertically downward. Gravitational force is representedby X = (0, 0, g)T, where g stands for the acceleration due to gravity that does not change horizon-tally.

First, we will write D using the stream function:

Dxx =∂vx∂x

= − ∂2ψ

∂x∂z, Dyy =

∂vy

∂y= 0, Dzz =

∂vz∂z

=∂2ψ

∂z∂z.

Dxy =12

(∂vx∂y

+∂vy

∂x

)= 0, Dyz =

12

(∂vy

∂z+∂vz∂y

)= 0,

Dxz =12

(− ∂

∂z

∂ψ

∂z+∂

∂x

∂ψ

∂x

)=

12

(∂2ψ

∂x2− ∂2ψ

∂z2

),

Dyx = Dxy, Dzy = Dyz, Dzx = Dxz.


In deriving these equations, we used the fact that vx and vz do not depend on y and that vy is zero orconstant anywhere. Substituting these equation into Eq. (9.3), we obtain

Sxx = −p − 2η∂2ψ

∂x∂z, Sxz = η

(∂2ψ

∂x2− ∂2ψ

∂z2

),

Szx = η

(∂2ψ

∂x2− ∂2ψ

∂z2

), Szz = −p + 2η

∂2ψ

∂x∂z.

(9.12)

Other stress components are zero: Syy = Sxy = Szy = 0. Here, we neglect the inertia term. Hence,combining X = (0, 0, g)T and the force balance equation (Eq. (3.24)), we have

0 =∂Sxx∂x

+∂Sxz∂z

, 0 =∂Szx∂x

+∂Szz∂z

+ ρg.

Substituting Eq. (9.12), we obtain

0 = − ∂p∂x

− 2η∂3ψ

∂x2∂z+ η

(∂3ψ

∂x2∂z− ∂3ψ

∂z3

), (9.13)

0 = −∂p∂z

+ 2η∂3ψ

∂x∂z2+ η

(∂3ψ

∂x3− ∂3ψ

∂x∂z2

)+ ρg. (9.14)

Gravity is supposed not to depend on x, so that ∂(ρg)/∂x = 0. Differentiating Eqs. (9.13) and (9.14)by z and x, respectively, and subtracting both sides, we obtain the biharmonic equation (Eq. (9.11))that describes very slow flows of Newtonian fluids. In the above derivation, we take gravity intoaccount, but the gravity term has been dropped.

We will often use the Fourier transform of the biharmonic equation. The nth-order differentia-tion ∂nf (x)/∂xn is transformed into (ik)nf (k), so that the Fourier transform can simplify differentialequations—ordinary and partial differential equations are replaced by algebraic and ordinary differ-ential equations, respectively.

We use the Cartesian coordinates O-xz, of which the z-axis is taken as being downward. Letψ (k, z) be the Fourier transform of a stream function φ(x, z). They are transformed from eachother by the equations

ψ (k) =1√2π

∫∞−∞

ψ (x)e−ikxdx, (9.15)

ψ (x) =1√2π

∫∞−∞

ψ (k)eikxdk, (9.16)

where i and k stand for the imaginary unit and horizontal wave number. The biharmonic equation istransformed to the equation

k4ψ − 2k2 ∂2ψ

∂z2+∂4ψ

∂z= 0. (9.17)

9.2. STREAM FUNCTION 211

Figure 9.2: Velocity profile in Newtonian fluid flowing between parallel walls.

The general solution of this equation has several forms such as

ψ = Aekz + Bzekz + Ce−kz +Dze−kz, (9.18)

ψ = A sinh kz + Bx sinh kz + C cosh kz +Dx cosh kz, (9.19)

ψ = (A + Bkz) cosh kz + (C +Dkz) sinh kz, (9.20)

where the coefficients A, B, C and D are the functions of k and z (and sometimes t), and aredetermined under given boundary conditions. The forms (Eqs. (9.19) and (9.20)) are convenientwhen boundary conditions are given at z = 0, because A is determined by the condition as ψ

∣∣z=0 =

A.

Simple velocity field

With a simple example, let us see how Newtonian fluids behave. Suppose that a Newtonian fluid isflowing between parallel walls that are separated by a distance 2h (Fig. 9.2). Due to the viscosity,the fluid clings to the walls so that we have the boundary condition for the velocity v = 0 at z = ±h.We assume that the walls are very long in the x direction so that v does not change in direction.Namely, ∂v/∂x = 0 and vz = 0. The question is how vx acts. In this case, the biharmonic equation(Eq. (9.11)) becomes the ordinary differential equation d4ψ/dz4 = 0. Its general solution is

ψ = Az3 + Bz2 + Cz +D.

Hence, we have

vx = −∂ψ∂z

= −3Az2 − 2Bz − C.

The system is symmetric with respective to z = 0, therefore vx(z) should be an even function. Thisis satisfied when B = 0. Let us replace −3A by A′, then the solution becomes vx = A′z2 −C. Using


Figure 9.3: Post-glacial uplift of Fennoscandia illustrated by the altitudes of the ancient strandlinebetween central Norway and Estonia [156].

the boundary condition, we obtain the velocity profile

vx = A′(h2 − z2),

a parabola with the maximum velocity at the center. The integral∫ h−h vx dz is the volume of fluid that

passes across a plane perpendicular to the x-axis in a unit time. Given the volume, A′ is determined.

9.3 Viscous relaxation

Continental ice sheets depress the continent by their sheer weight. Once an ice sheet disappearedafter the last glacial period, the underlying continental block rose rapidly to recover isostatic com-pensation without the sheet. This is the phenomenon known as post-glacial rebound. The Holoceneuplift of Fennoscandia is a well-known example (Fig. 9.3).

Isostatic compensation is an important process for the deformation of large impact craters.Shortly after large impact cratering, a crater tends to have a smooth, bowl-shaped interior with araised rim. Figure 1.14 shows fresh craters with such rims in the Mare Selenitatis on the Moon.Fresh craters show a diameter-depth ratio of about 2:5. If the ambient rocks are hard, they can sup-port a bowl-shaped slope. However, rocks have finite strengths. A large impact crater is isostaticallyunstable so that the depression is gravitationally relaxed. The relaxation induces flows in a deep-seated ductile layer if the impact basin is large compared to the surface brittle layer3. Ice is softerthan rocks, so that impact craters on icy satellites are more easily relaxed than those on rocky planetsand satellites. Figure 9.4 shows craters on the icy satellite Ganymede. Among them, large cratershave flat or even upbowed floors (Fig. 9.5).

3Large impact cratering fluidizes the shallow part of the target body, and gravitational collapse of the impact basin beginswhile the target is excavated by the impact [146]. The above explanation gives a simplified picture of impact cratering. See[145] for further reading.

9.3. VISCOUS RELAXATION 213

Figure 9.4: Large impact craters on the icy satellite Ganymede. Voyager Image 0550J2-001. Thetopographic profile along the white line is shown in Fig. 9.5.

The speed of isostatic rebound depends on the horizontal dimension (wavelength) of a depres-sion. Let us see this by the order of magnitude estimation by assuming the Newtonian viscosity forthe resistance for the gravitational collapse. Consider a bowl-shaped basin with a diameter D andcentral depth H (Fig. 9.6). The stress under the basin may be in the order of

τ ≈ ρgH. (9.21)

If the center subsides by ΔH , the corresponding engineering shear strain for the subsidence may berepresented by ΔH/(D/2), where D/2 is the radius of the basin. Infinitesimal shear strain is half


Figure 9.5: Topographic profile of a crater along the line shown in Fig. 9.4 showing its upbowedfloor (dotted line). After [225].

Figure 9.6: Diagram showing the depth and diameter of a basin to explain its viscous relaxation.

the engineering shear strain, so that we have ε ≈ ΔH/D. Therefore, the representative strain ratemay be in the order of

ε ≈ −H/D, (9.22)

where −H is the uplifting rate of the basin center. Substituting Eqs. (9.21) and (9.22) into theconstitutive equation of Newtonian fluid τ = 2ηε, we have the differential equation on H , H/H =−2η/ρgD. Consequently, we obtain the solution H = H0 exp(−t/τ), where H0 is the initial depthof the basin and τ = 2η/ρgD is the time constant for the relaxation. Therefore, τ is inverselyproportional to D, indicating that the amplitude of a horizontally large depression is attenuatedmore rapidly than that of a small one. This is contrary to diffusion phenomena including thermalconduction where short-wavelength temperature anomalies disappear faster than those with longwavelengths.

Large impact craters on Ganymede are examples (Fig. 9.4). A raised crater rim accompanied bya fresh crater is a short-wavelength topography so that the circular mountain range holds out againstgravitational spreading for a much longer time than the bowl-shaped depression of the crater itself.Morphology of still larger craters have been subdued on Ganymede; only circular bright low-reliefcircular patches called palimpsests are left (Fig. 3.6).

A degree of viscous relaxation was used to estimate the chemical composition of the icy crust ofthe Neptunian satellite Triton, where impact craters smaller than 1.5 km in diameter have not beenrelaxed. The reflectance spectrum indicates abundant CH4 and N2 ice on the surface of the satellite,but they have too low viscosity to maintain the crater morphology. Therefore, the icy crust probablyconsists of H2O ice [206].

On the other hand, there are impact craters on Mars that show different modes of relaxation. Thesubdued impact craters have a raised rim, whereas some of the Martian impact craters have pro-

9.4. SELF-ORGANIZING GEOLOGICAL STRUCTURES 215

Figure 9.7: Gravitational relaxation of surface undulations of a thin layer of viscous fluid on a solidbasement. (a) Model configuration of Squyres’ finite element simulation for the relaxation of acrater with a diameter of 20 km. (b) Results of the simulation for different thicknesses H of thelayer [226].

nounced rounding of sharp topographic features. Short wavelength topographic features have beenalready erased around those craters. Original crater morphology seems to have creeped, the phe-nomenon called terrain softening [226]. Mars has a near-surface permafrost layer with a thicknessof 102–103 m. The layer is underlain by a less viscous regolith layer or solid basement. Namely, ter-rain softening is thought to be a near-surface phenomenon. A large depression relaxes by gatheringfluid from distant places around it. If a flat basement is sheeted by a viscous fluid, the gravitationalrelaxation of a large depression must carry the fluid through a thin conduit (Fig. 9.7(a)). There-fore, it takes longer to erase long wavelength topographic features than those with short wavelengthfeatures4. Figure 9.7(b) shows the results of finite element simulation by Squyres for the softeningof a crater with a different thickness of its viscous layer. The simulation demonstrates that terrainsoftening occurs for a thin viscous layer lying on a solid basement. In addition, the layer is thick andthe relaxation is shown to lead to subdued craters with an upbowed floor like those on Ganymede.

9.4 Self-organizing geological structures

Stability analysis

Tectonic deformations include mechanisms of self-organization. There are multiple geological struc-tures with spatial periodicity. The joint sets in Fig. 6.15, load cast in Fig. 8.16, and boudins5 in Fig.9.8 are examples of mesoscale periodic structures. It is often observed that macroscale faults with

4See Exercise 9.4 and its answer.5Boudins are the linear segments of a layer that has been pulled apart along periodically spaced lines of separation called

boudin lines [248].


Figure 9.8: Periodically changing thickness like sausages, which are known as boudins, of a sand-stone bed. Quartz veins fill fractures at boudin lines (arrows). Melange zone in the Shimanto Belt,Southwest Japan.

the same trend have constant intervals. The grooves on the icy crust of Ganymede are macroscalesamples (Fig. 9.4). Let us consider such a pattern formation associated with tectonic deformation.The stability of fluid motions is a clue to the pattern formation by ductile deformations.

The theory of stability separates a fluid motion into a mean flow and perturbation around it. Theformer is also called a basic flow. For example, flow velocity is expressed as v = v + v, where v

is the velocity of the basic flow and v is the velocity of the perturbation. Likewise, other variablesare separated into two parts associated with the basic one and perturbation. For example, we dealwith pressure p = p + p, stream function ψ = ψ + ψ , stress S = S + S, and so on. If the basic flowis unstable, one or more perturbations grow to overwhelm the basic flow. Therefore, we investigatethe rate of growth or extinction of each perturbation using basic equations such as the Navier-Stokesequation and given boundary conditions. If the fluid is incompressible, each mean and perturbationsatisfies the incompressibility condition as follows. The fluid is incompressible, so we have

∇ · v = ∇ · (v + v)= 0.


Figure 9.9: Deformation of a high viscosity layer between low viscosity fluids. The system isunstable—even if pure-shear deformation is exerted, the high viscosity layer is folded by layer-parallel compression (a) or is broken into boudins (b). For both cases, deformation proceeds in thenumbered order.

The mean flow must rigorously satisfy the condition: ∇ · v = 0. Therefore, we obtain

∇ · v = 0, (9.23)

which indicates that the incompressibility condition also holds for the perturbation. Likewise,

∇4ψ = 0 (9.24)

is obtained from Eq. (9.11). As for the Navier-Stokes equation (Eq. (9.5)), we have

ρa = −∇ (p + p) + η∇2 (v + v)+ ρX

andρa = −∇p + η∇2v + ρX.

Therefore, the pressure and velocity perturbations must satisfy

0 = −∇p + η∇2v.

Formation of boudins and folds

The ductility of rocks depends on the lithology. When a sedimentary pile of alternating consolidatedsand- and mudstones is subject to ductile deformation, their ductility contrast leads to instabilityto form folds when they are compressed and boudins when they are extended in the orientationparallel to the bedding (Fig. 9.9). Let us investigate the development of a spacing of bodins and foldwavelength assuming incompressible Newtonian fluids [215].

Consider a thin layer of fluid with a viscosity of η(2) and a thickness of H is embedded in verythick fluid layers with a viscosity η(1) (Fig. 9.10(a)). The layer is initially flat with infinitesimalundulations. In what follows, we discriminate the plate and ambient fluids by the superscripts (1)


and (2). The origin of the rectangular Cartesian coordinates O-xz is placed on the middle surfaceof the thin layer, and the x- and z-axes are defined as being parallel and perpendicular to the layer.In order to develop a unified theory for the formation of boudins and folds, we assume pure shear inthe range −L ≤ z ≤ L as the basic flow, where L � H . The perturbation in vz leads to folds orboudins.

The existence of the layer causes disturbance in the pure shear. Hence, it is appropriate thatthe velocity field far from the layer equals that of the basic flow. For this reason, we assume theboundary condition for the velocity to be

vz∣∣z=±L = ±V, (9.25)

where V is a constant. Correspondingly, D = V/L is the representative velocity gradient in thissystem. Pure shear is the basic flow so that we have the condition for the shear stress

Szx∣∣z=±L = 0. (9.26)

We neglect both inertia force and surface tension at the interfaces, so that stress is continuous acrossthem:

S(1)zz

∣∣z=H/2 = S

(2)zz

∣∣z=H/2, (9.27)

S(1)zx

∣∣z=H/2 = S

(2)zx

∣∣z=H/2. (9.28)

Velocity is also assumed to be continuous there. Namely, we have

v(1)∣∣z=H/2 = v(2)

∣∣z=H/2. (9.29)

The other side of the layer at z = −H/2 has the same conditions. Equation (9.29) represents thelack of gaps or overlaps of the two fluids. Force balance between them results in Eqs. (9.27) and(9.28). It should be noted that Sxx can be discontinuous across the interface6. It is shown later thatthe difference in Sxx raises the undulation of the interface that eventually forms boudins or folds.

Let hT(x) = H/2 + h(x) be the top surface of the thin layer. It is difficult to deal with the layerif the interfaces have undulations with a large amplitude. Hence, we consider this system only whilethe condition |h(x)| � 1 holds. The interface moves at the velocity

∂hT

∂t≈ DH

2. (9.30)

The right-hand side of this equation designates the mean flow of the interface. Under the condition|h(x)| � 1, perturbations are small. Namely, the velocity field is approximated by that of pure shear

vx ≈ −Dx, vz ≈ Dz. (9.31)

6Imagine that water and barley sugar consist of the fluids (1) and (2). The latter is much more viscous than water, so thata layer of barley sugar carries much larger tensile stress or compression than water.


Figure 9.10: (a) Initial configuration of the model. L is much larger than H , the thickness of a thinlayer. The interfaces at z = ±H/2 have infinitesimal undulations. (b) Close-up of the top interface.Forces acting at the top interface and force balance for small rectangle.

The basic flow has the constant rate of strain

E = D =

(−D 00 D

). (9.32)

Substituting Eq. (9.32) into the constitutive equation of Newtonian fluid, we have

Sxx = −p + 2ηDxx = −p − 2ηD, (9.33)

Szz = −p + 2ηDzz = −p + 2ηD, (9.34)

Szx = 2ηDxz = 0. (9.35)

Equation (9.35) indicates that this system always satisfies the boundary condition in Eq. (9.28).Combining Eqs. (9.27) and (9.34), we obtain

−p(1) + 2η(1)D = −p(2) + 2η(2)D.

Therefore, there is a pressure difference between the fluids on the other side of the interface:

Δp ≡ p(2) − p(1) = 2[η(2) − η(1)]D. (9.36)

The fluids have different Sxx. Equation (9.33) holds of each of the fluids:

S(1)xx = −p(1) − 2η(1)D, S

(2)xx = −p(2) − 2η(2)D. (9.37)

Combining Eqs. (9.36) and (9.37), we have the difference in the normal stress in the fluids:

ΔSxx ≡ S(2)xx − S (1)

xx = −Δp − 2[η(2) − η(1)]D

= −4[η(2) − η(1)]D. (9.38)


In this section, we use overlines and tildes to distinguish the basic flow and perturbation aroundit. Namely, velocity is divided as

v = v + v. (9.39)

The basic flow of this system has the velocity

v =(vx, vz

) T = (−Dx, Dz) T. (9.40)

The continuity of the velocity field across the interface leads to the boundary condition for thevelocity perturbation

v(1)∣∣z=hT

= v(2)∣∣z=hT

. (9.41)

The interface is not only driven by vz but also by the lateral flow vx. If the interface with aninclination to the left (∂h/∂x > 0) moves to the right, the interface goes down (Fig. 9.10). Therefore,we have

∂h

∂t= −∂h

∂x· vx∣∣∣z=hT

+ vz∣∣∣z=hT

. (9.42)

Hence, substituting Eqs. (9.39) and (9.40), we obtain

∂hT

∂t= − (vx + vx) ∂h

∂x+(vz + vz

)= − (−Dx + vx) ∂h

∂x+(Dz + vz

)= D

(x∂h

∂x+ hT

)︸︷︷︸

basic flow

+

[−∂h∂x

· vx∣∣∣z=H/2+h

+ vz∣∣∣z=H/2+h

]︸︷︷︸

perturbation

. (9.43)

The initial undulations of the interface is extended vertically and constricted laterally by the firstterm in the last line. Therefore, the first term is not responsible for pattern formation. The secondterm gives rise to instability.

Let us consider the force balance in the x-direction of the small rectangle shown in Fig. 9.10(b),using the stress tensor

S = S + S. (9.44)

The fluid (1) includes the left side of the rectangle so that the force −S (1)xx dz = −S (1)

xx (∂h/∂x)dxacts there. Since we are assuming that the amplitude |h(x)| is very small, the stress perturbation is

also very small∣∣∣Sx∣∣∣ �

∣∣∣Sxx∣∣∣. Hence, the force is approximated by −S (1)xxdz = −S (1)

xx (∂h/∂x)dx.

The right side of the rectangle is in the fluid (2), so that the normal force at the side is S(2)xxdz =

S(1)xx (∂h/∂x)dx+ΔSxx(∂h/∂x)dx, where ΔSxx is given by Eq. (9.38). Because of (9.32) and (9.35),

we have Szx = Szx = 0. However, the perturbation in those shear stresses does not necessarilyvanish. The top and base of the rectangle feel the shear forces S (1)

zx dx and −S (2)zx dx, respectively.

Therefore, the total force balances if

left︷︸︸︷−S (1)

xx

∂h

∂x+

right︷︸︸︷S

(1)xx

∂h

∂x+ ΔSxx

∂h

∂x+

top︷︸︸︷S

(1)zx −

base︷︸︸︷S

(2)zx = 0.


We have a similar equation for the z-direction. Using the difference in the stress components betweenthe fluids

ΔSzx ≡ S(2)zx − S (1)

zx , ΔSzz ≡ S(2)zz − S (1)

zz ,

we have the force balance in the x- and z-directions:

ΔSzx = ΔSxx∂h

∂x, (9.45)

ΔSzz = ΔSzx∂h

∂x. (9.46)

The governing equations for the development of undulations have higher-order terms such as(∂h/∂x) · v

∣∣z=H/2+h in Eq. (9.43) and the right-hand sides of Eqs. (9.45) and (9.46). We employ

linear stability analysis to neglect non-linear terms with the assumption that the amplitude is verysmall |h| � 1. Under this condition, we have |∂h/∂x| � 1, and all perturbation terms designatedby a tilde are infinitesimal. Therefore, the second-order term vx(∂h/∂x) is neglected in Eq. (9.43)to give

∂h

∂t= vz

∣∣∣z=H/2+h

. (9.47)

This boundary condition is defined at the level H/2 + h, which includes the unknown variable hitself. Therefore, we have to linearize this condition also. The right-hand side of Eq. (9.47) isexpanded as

vz

∣∣∣z=H/2+h

≈ vz

∣∣∣z=H/2

+ h · ∂∂zvz

∣∣∣z=H/2

,

whereas the last term is a second-order term and is therefore negligible. Consequently, we have

∂h

∂t= vz

∣∣∣z=H/2

. (9.48)

Likewise, neglecting non-linear terms, Eq. (9.41) is rewritten as a condition at z = H/2:

v(1)∣∣∣z=H/2

= v(2)∣∣∣z=H/2

. (9.49)

The discontinuity of the normal stress is also divided into mean and the infinitesimal fluctuation,ΔSxx = ΔSxx + ΔSxx. The jump conditions for stress components (Eqs. (9.45) and (9.46)) arelinearized as

S(2)zx

∣∣∣z=H/2

− S (1)zx

∣∣∣z=H/2

= ΔSxx∣∣∣z=H/2

· ∂h∂x, (9.50)

S(2)zz

∣∣∣z=H/2

− S (1)zz

∣∣∣z=H/2

= 0. (9.51)


We investigate the wavenumber-dependence of the growth rate of fluctuations. To this end, weassume solutions of the form

h(x, t) = a(t) sin kx, (9.52)

ψ (x, z, t) = ϕ(z, t) sin kx (9.53)

for the linearized governing equations (Eqs. (9.24) and (9.48)–(9.51)). If an even function ϕ(2) (z) =ϕ(2) (−z) is the result, v(2)

z = ∂ϕ(2)/∂x is also an even function of z. In this case, the top and basalinterfaces of the thin layer approach or depart to each other, leading to the formation of boudins. Onthe other hand, if an odd function ϕ(2) (z) = −ϕ(2) (−z) is the result, the upwarping portions of thetop interface are associated with the upwarping portions of the basal interface. Therefore, foldingoccurs in this case. Substituting Eq. (9.53) into the biharmonic equation (Eq. (9.24)), we have(

∂4

∂z4+ 2

∂2

∂x2∂z2+∂4

∂x4

)ϕ(z, t) sin kx

= ϕzzzz sin kx − 2k2ϕzz sin kx + k4ϕ sin kx = 0,

where the subscripts indicate partial derivatives such as ϕzz = ∂2ϕ/∂z2. Each of the fluids satisfythis equation so that we have

ϕ(i)zzzz − 2k2ϕ

(i)zz + k4ϕ(i) = 0 (9.54)

for the ith layer. This equation has the same form as Eq. (9.17). Hence, the general solution of Eq.(9.54) is

ϕ(i) = a(i)e−kz + b(i)ze−kz + c(i)ekz + d(i)zekz. (9.55)

The condition for the formation of boudins and folds is indicated by the conditions among the coef-ficients of this equation:

Boudins a(2) = c(2) and b(2) = d(2), (9.56)

Folds a(2) = c(2) and b(2) = d(2). (9.57)

From the stream function in Eq. (9.53), we have the velocity components

v(i)x = −∂ψ

(i)

∂z= −ϕ(i)

z sin kx, v(i)z =

∂ψ (i)

∂x= kϕ(i) cos kx. (9.58)

Substituting Eq. (9.58) into (9.49), we have

ϕ(1) = ϕ(2), (9.59)

ϕ(1)z = ϕ

(2)z . (9.60)

Equation (9.58) allows us to write the stretching tensor as

D =

(∂vx/∂x ∂vx/∂z∂vz/∂x ∂vz/∂z

)=

(−kϕ(i)

z cos kx −ϕ(i)zz sin kx

−ϕ(i)zz sin kx kϕ

(i)z cos kx

).


Hence, the constitutive equation S(i)

= −p(i)I + 2η(i)D(i)

becomes

S = −(p 00 p

)+ 2η

(−kϕ(i)

z cos kx −ϕ(i)zz sin kx

−ϕ(i)zz sin kx kϕ

(i)z cos kx

).

Equations (9.50) and (9.51) are, therefore, rewritten as

−2[η(2)ϕ

(2)zz − η(1)ϕ

(1)zz

]sin kx = ΔSxxka cos kx, (9.61)[

η(2)ϕ(2)z − η(1)ϕ

(1)z

]2k cos kx = Δp. (9.62)

Combining Eqs. (9.48) and (9.58), we obtain

dadt

= kϕ(i) tan kx. (9.63)

Now we introduce the following parameters:

E ≡ η(2)

η(1), G ≡ 1

a

dadt, R ≡ 1

2Sxx

2η(1)G. (9.64)

E stands for the viscosity ratio, G is the growth rate, and 2η(1)G represents the viscous resistance todrive the fluid (1) for the growth. Using these parameters, Eqs. (9.59)–(9.62) are rewritten as

ϕ(1) = ϕ(2), (9.65)

ϕ(1)z = ϕ

(2)z , (9.66)

E(ϕ

(2)zz + k2ϕ(2)

)−(ϕ

(1)zz + k2ϕ(1)

)= 4kRϕ(1), (9.67)

E(ϕ

(2)zzz + 3k2ϕ

(2)z

)=(ϕ

(1)zzz + 3k2ϕ

(1)z

). (9.68)

Assuming the boundary condition that perturbations vanish at infinity (z → ±∞), we readily obtain

c(1)∣∣∣z→±∞

= d(1)∣∣∣z→±∞

= 0 (9.69)

from the general solution (Eq. (9.55)).In the case of the formation of boudins (Eq. (9.56)), we have

ϕ(1) = a(1)e−kz + b(1)ze−kz, ϕ(2) = 2b(2)ze−kz.

Substituting these equations into Eqs. (9.65)–(9.68), we obtain the simultaneous equations⎛⎜⎝k � −fk −g�−k 1 − � gk −g + f�

−k − Rk 1 − � − R� Efk −Ef + Eg�k � −Egk −Ef�

⎞⎟⎠⎛⎜⎝a(1)

b(1)

a(2)

b(2)

⎞⎟⎠ =

⎛⎜⎝0000

⎞⎟⎠ . (9.70)


Let A be the square matrix in this left-hand side. The newly introduced parameters �, f and gdepend on k and H as

� =Hk

2, f =

e−� − e�

e−�, g =

e−� + e�

e−�. (9.71)

The simultaneous equations (Eq. (9.70)) have non-trivial solutions if

det(A) = 0.

Solving this equation, we obtain the growth rate G for the formation of boudins as a function of �and E,

Rb =1

2E�

[sinh2 �(E + � − E2�) + sinh � cosh �(1 + E2) + cosh2 �(E − � + E2�)

]. (9.72)

On the other hand, in the case of folding (Eq. (9.57)), we obtain simultaneous equations with thematrix

A =

⎛⎜⎝k � −gk −f�−k 1 − � fk −f + g�

−k − Rk 1 − � − R� Egk −Eg + Ef�k � −Efk −Eg�

⎞⎟⎠ . (9.73)

Consequently, we have the rate of folding

Rf =1

2E�

[cosh2 �(E + � − E2�) + cosh � sinh �(1 + E2) + sinh2 �(E − � + E2�)

]. (9.74)

Combining Eqs. (9.38), (9.38), (9.72) and (9.74), the growth rate G is obtained as

G = [1 − E] [1/R] [D],

where Smith’s notation is used [215], i.e., the brackets indicate the different signs for the followingconditions:

[1 − E] ={

+(1 − E) (boudins),−(1 − E) (folds) ,[

1/R]=

{+1/R

(η(2) > η(1)

),

−1/R(η(2) < η(1)

),

[D] ={

+D (Basic flow is layer-parallel shortening,−D (Basic flow is layer-parallel extension).

The growth rate G is a function of k and H through the parameter � (Eq. (9.71)). Boudins orfolds with a wavenumber that maximizes the growth rate are expected to emerge. The dimensionlesswavelength λ/H = 1/kH chosen by this theory is plotted in Fig. 9.11(a), indicating that thewavelength gradually increases with the viscosity contrast E = η(2)/η(1). This model predicts thatthere is no single layer that folds with a λ/H ratio smaller than 6. However, field observationsindicate the dominance of the single-layer folding with a ratio of around 6 (Fig. 9.11(b)). Smithexplains the high frequency folding with non-Newtonian viscosity (Fig. 10.19).

9.5. DYNAMIC TOPOGRAPHY 225

Figure 9.11: (a) Dominant non-dimensional wavelength versus viscosity ratio for single layer fold-ing predicted from Smith’s model [216]. (b) Histogram of observed wavelengths [211].

9.5 Dynamic topography

Mantle convection has surface expressions such as subduction zones, oceanic ridges, and hotspots.They are the loci of large vertical movements that reflect sublithospheric flow. The Hawaiian Swellis an example where the oceanic lithosphere is uplifted by ∼1 km by the mantle upwelling under thehotspot (Fig. 8.8(a)). A vertical velocity gradient in viscous fluid leads to a dynamic pressure, whichraises or pulls down the lid of the fluid. The depression or uplift supported by the dynamic pressurein the sublithospheric mantle is known as dynamic topography. When the flow dies away, dynamictopography vanishes to leave isostatic compensation. If the vertical movement can be grasped fromstratigraphic records, we have a clue to the patterns of mantle convection over billions of years.

In the vicinity of mantle up or downwelling, various parameters, including velocity and tem-perature, tend to have large variations. Therefore, elaborated numerical calculations are needed tounderstand actual dynamic topography. However, the purpose of the following discussion is to inves-tigate whether dynamic topography has enough amplitudes to infer the convection from stratigraphicdata so that we assume a very simple thermal convection in the mantle to estimate the amplitude.

9.5.1 Linearized boundary condition

First of all, it is demonstrated that the overburden stress can be derived from the Navier-Stokesequation if the inertia term is omitted. To this end, we take the rectangular Cartesian coordinatesO-xz with the z-axis pointing downward. The body force significant for this problem is the gravity,i.e., X = (0, 0, g)T. Consider that all forces are balanced and the fluid is at rest, v = 0. Then, Eq.(9.6) becomes 0 = −∇p + (0, 0, ρg)T. Therefore, we obtain

dpdz

= ρg.


Figure 9.12: Linearization of boundary condition.

Indeed, integrating both sides of this equation results in the equation of the overburden stress (Eq.(3.29)), i.e., we have the pressure

p =∫ z

surfaceρ(ζ)g dζ.

Next, we consider dynamic pressure. From the equation for incompressible Newtonian fluid (Eq.(9.3)), we have the vertical stress

Szz = −p + 2η∂vz

∂z. (9.75)

When the local isostasy was introduced, overburden was assumed to be supported solely by p. Thesecond term in the right-hand side of Eq. (9.75) represents the dynamic pressure, through whichthe vertical velocity gradient has an additional effect for topography. The term has a positive signbecause the z-axis is defined downward (Fig. 9.12). Equation (9.75) designates that the amplitudeof dynamic topography has a positive correlation with the viscosity η. For simplicity we assume aconstant viscosity.

Consider the origin of the z-axis is placed at the surface without dynamic topography (Fig.9.12), and let z = h(x) be the surface topography. If dynamic topography leads to a depressionand a bulge, we have h > 0 and < 0, respectively. In addition, we assume that the surface is a freeboundary. Namely, we have the boundary condition S

∣∣surface = O. However, this equation is not

convenient for further calculation, because this condition is defined at the surface of which level his unknown. In addition, the surface generally has an inclination dh/dx, which also includes theunknown. Therefore, we have to simplify the boundary condition. Fortunately, dynamic topographyproduces long wavelength undulations, so that the inclination is neglected for brevity. Then, thehorizontal gradient of S becomes negligible, and we have the approximate boundary condition forthe remaining stress component

Szz

∣∣∣z=h

= 0. (9.76)

The left-hand side of this equation is expanded as

Szz

∣∣∣∣z=h

= Szz

∣∣∣∣z=0

+ h∂Szz∂z

∣∣∣∣z=0

+ · · · ,

where the gradient ∂Szz/∂z can be approximated by Δρg. Δρ represents the density differencebetween the mantle and overlying substance, seawater or sediments. Therefore, from Eq. (9.76) we


obtain the linearized boundary condition

Szz

∣∣∣z=0

= −Δρgh. (9.77)

The right-hand side represents topographic load (Fig. 9.12). Once the vertical stress at z = 0 isobtained, dynamic topography h is calculated with this equation.

9.5.2 Simple convection model

We assume that a Newtonian fluid with a constant viscosity η and thickness d is convected by basalheating. Parsons and Daly [173] presented the dynamic topography for this simple thermal convec-tion.

Let ρm be the mantle density at the equilibrium state T = 0. Then, the density at the temperatureanomaly T is given by ρm(1 − αT ). Therefore, omitting the inertia term, we have the equation ofmotion

−∇p + η∇2v + ρm (1 − αT ) g = 0. (9.78)

The acceleration due to gravity is assumed to be constant throughout the convecting cell, g =(0, 0, g)T. Temperature perturbation in the mantle may be of the order of 100 K. Density varia-tion in the cell is on the order of αΔT , i.e., ∼10−5 K−1 ×102 K = 10−3 � 1. For this reason,we employ the Boussinesq approximation to simplify governing equation; variations in density isneglected except insofar as they are coupled to the gravitational acceleration and buoyancy forces[65, §16.1]. We assume incompressibility of the fluid which allows us to use the stream function ψ .When the function was introduced in Section 9.2, the buoyancy term ∂ρg/∂x vanished. However,this gradient drives the convection. Therefore, the buoyancy term is indispensable here. Instead ofEq. (9.11), the stream function satisfies(

∂4

∂x4+ 2

∂4

∂x2∂z2+∂4

∂z4

)ψ =

ρmgα

η

∂T

∂x. (9.79)

Taking the Fourier transform of both sides of Eq. (9.79) we obtain(d2

dz2− k2

)2ψ = ik

(ρmgα

η

)T , (9.80)

where i =√−1 is the imaginary unit and k is horizontal wavenumber.

Dynamic topography is calculated from the vertical stress at z = 0 (Eq. (9.77)). According tothe constitutive equation of Newtonian fluid, this stress component satisfies

Szz = −p0 − p1 + 2η∂vz

∂z, (9.81)

where hydrostatic pressure is divided into p0 = ρgz that is hydrostatic pressure when the fluid at restand p1 that is the correction term for the hydrostatic pressure due to convection. The former has no


horizontal variation so that Eq. (9.78) becomes

−dp1

dx+ η

(∂2vx

∂x2+∂2vx

∂z2

)= 0.

Taking the Fourier transform of both sides of this equation, we have

p1 =iη

k∇2 dψ

dz. (9.82)

If the lithosphere is water-loaded, we have Δρ = ρm−ρw. Then, the boundary condition (Eq. (9.77))becomes Szz

∣∣z=0 = (ρw − ρm)gh. Substituting Eq. (9.81) into this boundary condition, we have

ρwgh = ρmgh − p1

∣∣∣∣z=0

− 2η∂vz∂z

∣∣∣∣z=0. (9.83)

Combining Eqs. (9.82) and (9.83), we obtain

h =1

(ρm − ρw)giη

k

(d2

dz2− 3k2

)dψdz

∣∣∣∣z=0. (9.84)

Let T1 be a representative temperature fluctuation. Using the dimensionless variables (Fig. 9.13),

x′ =x

d, z′ =

z

d, k′ = dk, T ′ =

T

T1, ψ ′ =

η

ρmgαT1d3ψ,

The equation of motion (Eq. (9.80)) is rewritten as(d2

dz′2− k′2

)2ψ ′ = ik′T ′. (9.85)

T ′ is a function of z. The solution of Eq. (9.85) includes the integral∫T ′dz, i.e., the flow ψ ′ is af-

fected by the temperature distribution over the entire convecting cell. Thus, the dynamic topographydetermined through (Eq. (9.84)) is affected by the entire temperature distribution, unlike the isostaticcompensation which bears little relation to variations under the depth of compensation. However,the effect of temperature depends on depth z; shallow temperature anomalies are more importantthan deep ones. Accordingly, let us use Green’s function7

ψ ′(k′, z′) = ik′∫ 1

0Ψ(k′, z′, ζ)T ′(k′, ζ) dζ. (9.86)

Ψ(k′, z′, ζ) describes how the temperature distribution at a depth of ζ affects the flow ψ at the depthz′, and is the solution of the equation(

d2

dz′2− k′2

)2ψ ′ = δ(z′ − ζ), (9.87)

7See [173] for detail.


Figure 9.13: Diagrams for the dynamic topography induced by two-dimensional thermal convection[173].

which describes the flow by a point source at z′ = ζ. In this equation, δ( ) is the delta function. Thedimensionless dynamic topography h′ is also related to the entire temperature distribution throughGreen’s function H (k′, z′) as

h′(k′) =ρmαT1

ρm − ρw

∫ 1

0H (k′, ζ)T ′(k′, ζ) dζ. (9.88)

H is obtained from the equation

H (k′, ζ) = −(

d2

dz′2− 3k′2

)dΨdz′

∣∣∣∣z′=0

, (9.89)

and depends on the top and basal boundary conditions. If the convecting cell has a fluid substratumwith a density of ρb and the interface is deformed by the convection, the interface has dynamictopography hb and also by its Green’s function Hb. They are related to the temperature distributionas

h′b(k′) =ρmgαT1

ρb − ρm

∫ 1

0Hb(k′, ζ)T (k′, ζ) dζ, (9.90)

where

Hb(k′, ζ) =(

d2

dz′2− 3k′2

)dΨdz′

∣∣∣∣z′=1

. (9.91)

The dynamic topography of the surface h′(k′) is obtained from Eq. (9.88), which needs theGreen’s function given in Eq. (9.89). To this end, Eq. (9.87) has to be solved with tectonically


appropriate boundary conditions. If we think of dynamic topography around an oceanic ridge, thetop of the convecting cell may be a free boundary because the oceanic plate is a part of the cell. Incontrast, the base of the lithosphere may be a rigid boundary for a mantle plume under a hotspot. Ifthe lid of the cell is a stable continent, a rigid boundary may be appropriate, also. The base of theconvecting cell can be either a free or rigid boundary.

In case of the free boundary, we have the velocity and shear stress the conditions vz = 0 andSzx = 0 at z = 0. The latter condition indicates that the flow is parallel to the x-axis at z = 0, i.e.,the top surface coincides with the contour line of ψ so that ψ |z=0 is constant there. The absolutevalue of ψ is meaningless, so that we define ψ to have this constant at zero. The other condition, theshear stress Szx satisfies

Szx = 2η∂vz∂x

= 2η∂2ψ

∂x2.

Consequently, the free boundary has the conditions

ψ∣∣∣z=0

= 0,∂2ψ

∂x2

∣∣∣∣z=0

= 0 (9.92)

for the stream function at z = 0.In the case of the rigid boundary, the stream function has the boundary condition ψ = 0 at the

base of the lithosphere z = 0, also. In addition, if the lithosphere is stationary, the horizontal velocitycomponent of the asthenospheric flow vanishes at z = 0. Consequently, the rigid boundary conditionis described by the equations

ψ∣∣∣z=0

= 0,∂ψ

∂z

∣∣∣∣z=0

= 0. (9.93)

Equation (9.87) has the solution

ψ ′ = A1 sinh k′z′ + B1 cosh k′z′ + C1z′ sinh k′z′ +D1z

′ cosh k′z′ (z′ < ζ), (9.94)

ψ ′ = A2 sinh k′(1 − z′) + B2 cosh k′(1 − z′)+ C2(1 − z′) sinh k′(1 − z′) +D2(1 − z′) cosh k′(1 − z′) (z′ > ζ). (9.95)

The coefficients A1, B1, etc., are the functions of k′ and z′, and are determined from the boundaryconditions. We readily obtain B1 = B2 = 0 from the boundary condition ψ ′ = 0 at the base and topof the cell.

The pressure perturbation p1 has discontinuity (Fig. 9.14) at z′ = ζ by the point source that isrepresented by the delta function in Eq. (9.87). The discontinuity corresponds to the discontinuousd3ψ ′/dz′3 in Eq. (9.14) at z′ = ζ. The first- and second-order derivatives are still continuous thereso that we integrate both sides of Eq. (9.87) over the infinitesimal interval

[ζ − ε, ζ + ε

], where

0 < ε � 1. The result isd3ψ ′

dz′3

∣∣∣∣z′=ζ+ε

− d3ψ ′

dz′3

∣∣∣∣z′=ζ−ε

= 1. (9.96)


Figure 9.14: Schematic picture showing pressure perturbation induced by a sinking sphere. Theperturbation has a discontinuity at the sphere.

The other boundary conditions constrain the coefficients of Eqs. (9.94) and (9.95). For example, ifthe convecting cell has free boundaries at its top and base (Eq. (9.92)), we obtain

A1 =1

2k′3 sinh2 k′

[k′z′ sinh k′ cosh k′(z′ − 1) − k′ sinh k′z′ − sinh k′ sinh k′(z′ − 1)

], (9.97)

A2 =1

2k′3 sinh2 k′

[k′(z′ − 1) sinh k′ cosh k′z′ − k′ sinh k′(z′ − 1) − sinh k′ sinh k′z′

], (9.98)

C1 = C2 = 0, (9.99)

D1 =sinh k′(z′ − 1)

2k′2 sinh k′, D2 =

sinh k′z′

2k′2 sinh k′. (9.100)

Figure 9.13(c) shows H for this case, where H is plotted for three different k′. The dynamic topog-raphy for other boundary conditions has similar amplitudes so that we omit the other cases8.

Dynamic topography is controlled by the temperature distribution in the convecting mantle,which is affected by the distribution of heat sources. Neglecting radioactive heating in the uppermantle and assuming the basal heating only, the dimensionless Rayleigh number

Ra =ρmgαΔTd3

ηκ

represents the relative magnitude of advective and conductive heat, where ΔT is the temperaturedifference of the top and base of the convecting layer. The basal heating of constant viscosity fluidinduces thermal convection whose temperature is proportional to sin πz′, and Eq. (9.85) becomes(

d2

dz′2− k′2

)2

ψ ′ = ik′ sinπz′.

8See [173] for further reading.


Figure 9.15: Dynamic topography resulting from the thermal convection in the upper mantle (d =600 km) with a constant viscosity corresponding to an assumed Rayleigh number of Ra = 1.4× 106

[173].

This equation has the general solution

ψ ′ =ik′

(π2 + k′2)2

[sinπz′ + A(z′ − 1) sinh k′z′ + Bz′ sinh k′(z′ − 1)

].

Substituting this into Eq. (9.84), we obtain

h′ =ρmαT1

ρm − ρw

π3 + 3πk′2 − 2k′3A(π2 + k′2)2

, h′b =ρmαT1

ρb − ρm

π3 + 3πk′2 − 2k′3B(π2 + k′2)2

. (9.101)

Figure 9.15 shows the dynamic topography with an amplitude of the order of 100 m generated bythe upper mantle convection with an appropriate Rayleigh number9. This amplitude is large enoughto leave stratigraphic evidence such as regression or transgression if a shallow marine basin is abovethe mantle convection.

9.5.3 Epeirogeny

Subduction zones are the surface expressions of the downgoing mantle flow, so that island arcs havenegative dynamic topography. The amplitude of the topography depends on the density anomalyand downgoing velocity of the slab and the viscosity of the asthenosphere under the arc, The densityanomaly depends on the oceanic age of the slab and the velocity. A subduction zone is so com-plex that numerical simulations are needed to estimate the dynamic topography. Among those, thesimulation by Zhong et al. [280] demonstrate that the dynamic topography can be more than 1 kmover an area of 1000 km from the trench axis. The amplitude has a positive correlation with slabage. Dynamic topography is an important factor for the secular vertical movement of island arcs. Itaffects vertical movements in vast backarc regions if an old slab subducts with a low angle.

9Mantle convection induces vertical stresses at the base of the lithosphere, resulting in the dynamic topography. Thestresses act as vertical loads and bend the lithosphere. Therefore, the lithosphere masks the dynamic topography with awavelength much shorter than the flexural parameter of the lithosphere. In this respect, the lithosphere acts as a low-passfilter for the dynamic topography.


Figure 9.16: Cretaceous Western Interior Basin in North America. (a) Isopach map of theCampanian–Maastrichitian (65–83 Ma) deposits in the basin [46]. The contour interval is 1000feet. Dot-bar line indicates the western border of the present distribution of the deposits. Distri-bution of igneous rocks produced by island-arc magmatism is shown by hatching. (b) Slab modelfor the calculation of dynamic topography at 65 Ma. Stable continent is exhibited by hatching. (c)Dynamic topography at 65 Ma. Panels (b) and (c) are after [26].

The Cretaceous Western Interior Basin was produced not only by the topographic load of theCordilleran orogen (Fig. 8.3) but also by the negative dynamic topography by a shallow dipping slabsubducting from the Pacific [46, 152]. Figure 9.16(a) is the isopach map of Cretaceous strata. Thewestern part of the sedimentary basin has been eroded away by the upheaval of the Rockies. Thebasin subsided with little tectonic deformation except for its western margin which was involvedthrough Cordilleran orogeny. The sedimentary basin was wider than 1000 km, too extensive forthe flexural subsidence of the lithosphere, even if the effective elastic thickness was as great as100 km. Island-arc magmatism occurred at that time far away from the western coast of NorthAmerica, suggesting that the dip of the slab was shallow. Therefore, the extensive negative dynamictopography left the stratigraphic record [46, 152]. Figures 9.16(c) and (d) show the modelled slabsand the calculated dynamic topography from [26]. The result shows that the shallow subduction ofthe Farallon Plate produced a depression deeper than 1 km in the vast backarc region.

Horizontally extensive vertical movements occur with little tectonic deformation. Those phe-nomena are known as epeirogenic movements. The Western Interior Basin is an example. TheRussian platform is another one, where the Precambrian basement continued from the Baltic Shieldis blanketed by the latest Precambrian to the Mesozoic sedimentary rocks more than 3 km thick. Asingle mechanism cannot explain the lengthy subsidence [88, 124]. However, the Devonian–Permiansubsidence was rapid for this region but accompanied by little magmatism and deformation, so thatit is attributed to the negative dynamic topography induced by the the subduction at the eastern andsouthern margins of the Russian platform [153]. It is suggested that Devonian subsidence in theRussian platform and in Canada was attributed to the avalanche of cold subducted materials piled upon the 660 km boundary into the lower mantle [49, 187].


Figure 9.17: Hypothetical tectonic evolution around Japan from 20 to 15 Ma [272]. Note that slabage abruptly changed from to Early Miocene for the Southwest Japan Arc. The Pacific Plate hasCretaceous lithosphere in this region. The gross form of the paleotopography in these periods isillustrated in Fig. 3.14. IBA, Izu-Bonin Arc; KP, Kyushu-Palau Ridge.

The sudden uplift of the Southwest Japan at 15 Ma (Fig. 3.14) was presumably due to a jump inthe dynamic topography brough about by the abrupt change in the slab age from Cretaceous to EarlyMiocene (Fig. 9.17). The young Philippine Sea Plate began subduction at that time. The arc wasraised above sea-level simultaneously with the onset of widespread magmatism and the beginningof the arc-perpendicular compressional stress field [272].


Figure 9.18: Surface topography of a viscous fluid layer with an average thickness of H . The layeris underlain by a rigid basement with a horizontal interface.

Exercises

9.1 Show that gravity force X satisfies ∇ ×X = 0. Due to this identity, the diffusion equation ofvorticity (Eq. (9.8)) holds if gravity is the only body force for Newtonian fluid.

9.2 Estimate the viscosity of the mantle under Fennoscandia from the ancient strandlines shown inFig. 9.3 with the assumption that the uplift began at 13 kyr ago. [156].

9.3 Assuming the crust as a viscous fluid, consider the relationship between the Argand number(p. 86) and the viscosity.

9.4 Consider the relaxation of surface depression of a layer of viscous fluid covering a rigid base-ment with a flat interface. Let H be the average thickness of the layer, and z = h0(x) be the initialsurface topography, where z-axis is defined downward with the level z = 0 at the average altitude(Fig. 9.18). We assume

∣∣h0(x)∣∣ � H , indicating gentle undulations, and

∣∣h0(±∞)∣∣ = 0. Then,

the Fourier transform

h0(k) =1

2π

∫∞∞h0(x)eikxdx (9.102)

exists, where k is an horizontal wavenumber. The dimensionless number kH is regarded as thedimensionless thickness of the layer. Namely, the two cases kH � 1 and kH � 1 correspondto very thick and very thin layers relative to a wavenumber k, respectively. The time constant forviscous relaxation τ can be nondimensionalized as τ = (ρgH/2η) τ. Show τ ∝ kH and τ ∝ (kH)−2

for very thick (kH � 1) and very thin (kH � 1) viscous layers, respectively [123].

9.5 Show that Argand number for the situation in Fig. 3.19 is given by Ar = ρgΔh/ηV , where Δhis the difference in surface heights and V is the velocity of the plate under the fluid [29].

fluid - terrapub · fluid 9.2 stream function for two-dimensional problems, the scalar function,...

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