Fluid Dynamics - Math 6750Waves - Weeks 6 and 7

1 The wave equation

A classical model for the small transversal vibrations h(x, t) of a tightly stretched horizontalstring is the one-dimensional wave equation

htt = c2hxx, where c2 =T

ρ

and T is the tensile force and ρ is the linear string density. D’Alembert solution says that foran infinite string the solution has the form h(x, t) = F (x− ct) +G(x− ct) where the functionsF,G are related to the initial position and speed of the string. Alternatively, using the Fouriertransform defined as f = 1√

2π

∫∞−∞ f(x)e−ikxdx, the solution is

h(x, t) =1√2π

∫ ∞−∞

[A(k)eik(x−ct) +B(k)eik(x+ct)

]dk.

Here, k is the wavenumber and it relates to the wavelength λ = 2π/k. The quantity ω = ckis the temporal frequency and it relates to the period T = 2π/ω. The expression for h(x, t)represents the superposition of two travelling waves moving at constant speed c > 0 in thepositive and negative x− direction respectively and these waves are not dispersive, i.e. alldisturbances travel at a constant speed.

In contrast with these small amplitude waves on a string, water waves are dispersive, i.e.different Fourier components that make up a general disturbance travels at different speeds,depending on their wavelength. Water waves problem involve a free surface that is part of theboundary and that varies in time. This means that we need to solve for both the velocity fieldu and the moving free surface.

2 Surface Waves on Deep Water

Consider an irrotational flow of an incompressible, inviscid fluid occupying the two-dimensionaldomain Dt with infinite depth, defined by

Dt = {(x, y) ∈ R2 : −∞ < x <∞,−∞ < y < η(x, t)}.

We assume that the mean free surface is located at y = 0 and the interface between fluid andair, i.e. the free surface, is the graph of some unknown displacement function y = η(x, t).This is equivalent to assuming that there is no out-plane component of velocity and that thevelocity doesn’t depend on z.

1

y = 0Water

WaterAir

Air

x

y

gravity

y = η(x, t)

Figure 1: Surface waves on deep water.

2.1 Existence of velocity potential

Since the fluid is assumed to be inviscid (ν = 0), it follows from the two-dimensional vorticityequation that

Dω

Dt= ω · ∇u = 0

and the vorticity ω = (0, 0, ω) remains unchanged. In particular, ω ≡ 0 since the flow isassumed to be irrotational. Consequently, there exists a function φ(x, y, t), called the velocitypotential, such that u = ∇φ. We verify that φ satisfies the zero vorticity condition:

ω =∂v

∂x− ∂u

∂y=

∂

∂x

∂φ

∂y− ∂

∂y

∂φ

∂x= 0.

Remark 1. The existence of velocity potential is a result from multivariable calculus: AnyC1 vector field u on a simply-connected domain D ⊂ R3 is conservative if and only if it isirrotational on D.

By virtue of the incompressibility condition the velocity potential φ satisfies Laplace’sequation in Dt:

0 = ∇ · u = ∇ · (∇φ) = ∆φ.

2.2 Kinematic boundary condition

In the water-waves problem, water and air are two immiscible fluids. Define the free surfaceas the zero level set of the implicit function

F (x, y, t) = y − η(x, t).

The no-penetration boundary condition for a moving boundary is that

u · n = umoving boundary · nmoving boundary.

This condition is equivalent to the requirement that fluid particles on the free surface mustremain on the free surface, i.e. the material derivative of F must be zero:

DF

Dt=∂F

∂t+∇F · u = 0. (1)

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This is called the kinematic boundary condition because it only involves the flow field. Rewrit-ing (1) using ∇F = n|∇F |, where n is the unit outward normal (pointing in the y direction),we obtain

0 =∂F

∂t+ |∇F |n · u ⇒ 0 =

1

|∇F |∂F

∂t+ u · n.

In particular, this asserts that the zero level set of F is convected in the normal direction nwith the flow field u. Finally, computing and substituting all the derivatives of F into (1)yields the kinematic equation for the free surface:

ηt + uηx = v on y = η(x, t).

2.3 Dynamic boundary condition

For an irrotational flow of an ideal fluid with gravitational body force F b = g = −gy, thelinear momentum equation (see the derivation of Bernoulli streamline theorem) reduces to

∂u

∂t= −∇

(p

ρ+ gy +

1

2|u|2

)on y ≤ η(x, t) (2)

Using u = ∇φ and integrating (2) with respect to x gives

φt +p

ρ+ gy +

1

2|∇φ|2 = G(t) on y ≤ η(x, t) (3)

where G(t) is an arbitrary function of time t. In particular, G(t) may be chosen arbitrarily byabsorbing a suitable function of t into φ using the transformation

Φ(x, y, t) = φ(x, y, t)−∫ t

t0

G(τ) dτ (4)

in which (3) reduces to

Φt +p

ρ+ gy +

1

2|∇Φ|2 = 0 on y ≤ η(x, t). (5)

This does not affect the relation of φ to the flow velocity since u = ∇φ = ∇Φ. Eq. (3) is theBernoulli equation for unsteady potential flow.

Since the flow is inviscid, there are no tangential stresses and the tangential stress is bal-anced at the free surface, provided we neglect the surface tension. On the other hand, sincethe only normal stress exerted on the free surface is the pressure, the normal stress boundarycondition translates to no pressure jump across the free surface, i.e.

p(x, η(x, t)−, t) = p0 at y = η(x, t)

with p0 the (uniform) atmospheric pressure. By virtue of (4), evaluating (3) at y = η(x, t)and choosing G(t) = p0/ρ to eliminate the constant term we obtain the dynamic boundarycondition:

φt + gη +1

2|∇φ|2 = 0 on y = η(x, t).

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2.4 Linearisation of the surface boundary condition

The two-dimensional water-waves problem, in terms of the velocity potential φ, takes the form

∆φ = 0 for y < η(x, t)

ηt + φxηx = φy on y = η(x, t)

φt + gη +1

2|∇φ|2 = 0 on y = η(x, t).

This problem is difficult to solve in general since the boundary conditions are nonlinear. Weassume that the free surface displacement η(x, t) and the fluid velocities u, v are small, in asense to be made precise later. This allows us to linearise the problem by dropping quadratic(and higher terms) of u, v, η and the two surface conditions on y = η(x, t) reduce to

ηt = φy = v

φt + gη = 0.

Expanding v(x, η, t) and φ(x, η, t) around the mean free surface y = 0, i.e. letting ∂yφ(x, η, t) =∂yφ(x, 0, t)+η∂yy(x, 0, t)+O(η2) and neglecting quadratic (and higher terms) again, we obtain

ηt(x, t) = v(x, 0, t)

φt(x, 0, t) + gη(x, t) = 0.

This allows us to rewrite the surface boundary conditions at y = η(x, t) as surface boundaryconditions on y = 0. Consequently, we obtain the linear water-waves problem:

∆φ = 0 for y < 0 (6a)

ηt = φy on y = 0 (6b)

φt + gη = 0 on y = 0. (6c)

2.5 Travelling wave solution and dispersion relation

We look for sinusoidal travelling wave solution for the free surface displacement

η(x, t) = A cos(kx− ωt),

where A is the amplitude, k is the wavenumber, and ω is the angular frequency. Further, welet λ = 2π

k be the spatial wavelength and T = 2πω be the temporal frequency. The boundary

conditions (6b) and (6c) suggest that the velocity potential should be of the form

φ(x, y, t) = f(y) sin(kx− ωt).

Plugging into Laplace’s equation (6a), the function f(y) must satisfy

f ′′(y)− k2f(y) = 0 for y < η(x, t)

which has general solutionf(y) = Ceky +De−ky.

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We take k > 0. Since the water is of infinite depth, we require the velocity to be bounded asy −→ −∞ and this is not possible unless D = 0. Thus

φ(x, y, t) = Ceky sin(kx− ωt).

Substituting this, together with the ansatz of η, into both (6b) and (6c) and cancelling thecommon sinusoidal term, we obtain{

ωA− kC = 0

−ωC + gA = 0⇐⇒

[ω −kg −ω

] [AC

]=

[00

].

This system has a nontrivial solution if and only if the matrix is singular, that is if

ω2 = gk. (7)

This equation relating the frequency ω and the wavenumber k is called the dispersion relation.It takes the form ω2 = g|k| if no restriction is placed on the sign of k. For any given A > 0 thesolution of the linear water-waves problem (6) is

φ(x, y, t) =Aω

keky sin(kx− ωt) (8a)

η(x, t) = A cos(kx− ωt) (8b)

ω2 = gk. (8c)

The dynamic pressure P for the linear water-waves can be recovered from Bernoulli equationfor unsteady potential flow (5). Neglecting higher-order terms we find that

P = p+ ρgy = −ρφt = ρAω2

keky cos(kx− ωt) = ρgAeky cos(kx− ωt).

For k > 0, the surface waves travel to the right with phase velocity

c =ω

k=

√g

k=

√gλ

2π

and they are dispersive since waves with different wavenumbers move at different speed. More-over, the phase velocity is an increasing function of the wavelength λ so longer waves propagatesfaster.

2.6 Small amplitude assumption

The linear water-wave problem is obtained by assuming η, u, v are small, but small compared towhat? The standard procedure to compare magnitudes of different terms is to nondimension-alise, but here we take advantage of the fact that we have the explicit solution. The expressionfor φ shows that that u and v are of the same order of magnitude. To obtain (6b), we neglectedthe term uηx compared with the term v. This is reasonable provided

|uηx| � |v| ⇐⇒ |ηx| � 1 ⇐⇒ |Ak| � 1 ⇐⇒ |A| � λ.

5

Thus, we require that the free surface displacement is small compared to the wavelength of thewaves and this is sometimes called the small-slope approximation. To obtain (6c), we neglectedthe term |∇φ|2 = u2 + v2 compared with the term gη. This is reasonable provided

|u2 + v2| � g|η| ⇐⇒ A2ω2 � gA ⇐⇒ A2gk � gA ⇐⇒ |Ak| � 1

and we again recover the small-slope approximation.

2.7 Particle paths

As a wave train travels across the surface of an otherwise quiescent liquid, the individualparticles of the liquid undergo small cyclical motions. From (8c) the velocity components are

u(x, y, t) = φx = Aωeky cos(kx− ωt) v(x, y, t) = φy = Aωeky sin(kx− ωt).

At a particular time t > 0, a fluid particle initially at its mean position (x, y) reaches (x, y)and its particle velocity at that point must satisfy

dx

dt= u(x, y, t) and

dy

dt= v(x, y, t)

and in principle we can integrate these to obtain the particle path. In the linear water-wavesproblem, any particle deviates only a small amount from its mean position, i.e.

x(t) = x+ x(t) y(t) = y + y(t).

Since x(t)� x(t) and y(t)� y(t), we can approximate the RHS of the ODEs as

dx

dt= u(x, y, t) = Aωeky cos(kx− ωt) dy

dt= v(x, y, t) = Aωeky sin(kx− ωt)

which can be easily integrated. Hence

x(t) = −Aeky sin(kx− ωt) and y(t) = Aeky cos(kx− ωt)

and the particle paths are circular with radius Aeky since

x2 + y2 =(Aeky

)2.

In particular, the radius and the fluid velocities both decrease exponentially with the depthand so the fluid motion is limited to the free surface of depth y = O(1/k) = O(λ).

2.8 Group Velocity

If we have a dispersion relation, ω = ω(k), then the group velocity is defined as cg = dωdk .

In a dispersive system, cg depends on k. In the free water wave problem, we have cg =√

g4k

(assuming k > 0). This means that the wavetrain will be locally sinusoidal in the neighborhoodof a distance x = cgt from the initial disturbance region. Equivalently, it is the velocity atwhich we must travel to continuously see waves of the same frequency 2π/k. The group velocity

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is also the velocity at which a wave packet travels as a whole of the velocity at which energyis transported by waves of wavelength 2π/k. The group velocity is the velocity with whichthe overall envelope of the waves amplitudes propagate. For example, the example ring ofwaves after a stone is thrown in a pond is the wave group. The phase speed is the speedof propagation of individual wavelets, i.e. the velocity at which the phase of any frequencypropagates.

3 Surface Tension Effects

Since the flow is assumed to be inviscid, there are no tangential stresses and the tangentialstress is balanced on F (t). Because fluid surface tension is present only at the liquid-airinterface, adding the effect of surface tension will only alter the dynamic boundary condition.In this case, the pressure jump across F (t) is balanced by the curvature pressure, i.e.

p(x, η(x, t)−, t)− patm = T (∇ · n) on y = η(x, t), (9)

where T is the fluid surface tension, n is the unit outward normal to the surface and ∇ · nis the curvature of the fluid free surface {y = η(x, t)}.Evaluating the Bernoulli’s equation aty = η(x, t) and choosing G(t) = patm/ρ as before, the dynamic boundary condition now takesthe form

φt +1

2|∇φ|2 + gη = −T

ρ∇ · n on y = η(x, t).

It is easy to check that ∇ · n has the form:

∇ · n =−ηxx

(1 + η2x)3/2

,

so that linearization givesφt + gη = Tηxx on y = η(x, t).

Expanding around y = 0 and neglecting quadractic and higher order terms again yields

φt + gη =T

ρηxx at y = 0. (10)

3.1 Dispersion relation

Since we are looking for solutions of the form η = A cos(kx−ωt), the new term due to surface

tension has the form Tk2

ρ cos(kx− ωt), therefore we can find the dispersion relation simply by

replacing g in the case without surface tension by g + Tk2/ρ. We find

ω2 = gk +Tk3

ρc =

ω

k=

√g

k+Tk

ρ.

The group velocity is then

cg =g + 3Tk2/ρ

2√gk + Tk3/ρ

.

7

Figure 2: Wave patterns produced by a raindrop (capillary waves, left) and a large stone(gravity waves, right) ( from Acheson).

3.2 Capillary waves

The importance of surface tension is measured by the parameter S = Tk2

ρg . If S � 1, then theeffects of surface tension are negligible, so called gravity waves. When S � 1, the waves aredominated by surface tension effects and are called capillary waves. In this case,

ω2 =Tk3

ρc =

√Tk

ρcg =

3

2c.

For capillary waves, short waves propagates faster and the group velocity is larger than thephase velocity. Capillary waves can be observed when raindrops fall on a pound. For surfacewater, S = 1 when the wavelength is about 1.7 cm. Since k = 2π

λ , S large means λ (wavelength)small.

4 Finite depth effects

Suppose that the fluid is bounded below by a plane y = −h (solid, no penetration). Neglectingsurface tension again, we must now impose a boundary condition at the bottom:

∂φ

∂y= 0 at y = −h.

4.1 Dispersion relation

The analysis is exactly the same has in the infinite case, except that the function f(y) mustnow satisfy f ′(−h) = 0, i.e. 0 = Cke−kh − Dkekh or C = De2kh. The velocity potential isthen

φ(x, y, t) = D(e2kheky + e−ky) sin(kx− ωt)

so that plugging in (6b) and (6c) and cancelling the common sinusoidal term give{ωA− kD(e2kh − 1) = 0

−ωD(e2kh + 1) + gA = 0⇐⇒

[ω −k(e2kh − 1)g −ω(e2kh + 1)

] [AD

]=

[00

].

8

This system has a nontrivial solution if and only if the matrix is singular, that is if

ω2 = gke2kh − 1

e2kh + 1= gk tanh(kh) c2 =

g

ktanh(hk). (11)

As a result, the phase speed c = ω/k cannot exceed√gh. If kh is large, i.e. h � λ, then

tanh(hk) ≈ 1 and c2 ≈ gk , while for shallow water, i.e. h � λ

2π , tanh(hk) ≈ hk and c2 ≈ gh,which is independent of k. So gravity waves in shallow water are non-dispersive.

Figure 3: Phase speed as a function of wavelength for capillary-gravity waves (from Acheson).This is the in-plane picture of the full 3D problem.

4.2 Traveling waves solution

The solution of the linearized water wave problems is

φ(x, y, t) =ωA

k

ekh

e2kh − 1(ek(y+h) + e−k(y+h)) sin(kx− ωt) =

ωA

k

cosh(k(y + h))

sinh(kh)sin(kx− ωt).

Since u, v can also be expressed in terms of a stream function ψ, i.e. u = ∂yψ, v = −∂xψ, wealso have

∂ψ

∂y=∂φ

∂x= ωA

cosh(k(y + h))

sinh(kh)cos(kx− ωt),

so that

ψ(x, y, t) =ωA

k

sinh(k(y + h))

sinh(kh)cos(kx− ωt) + F (x).

Plugging into v, we find

−∂ψ∂x

=∂φ

∂y⇒ ωA

sinh(k(y + h))

sinh(kh)sin(kx− ωt)− F ′(x) = ωA

sinh(k(y + h))

sinh(kh)sin(kx− ωt)

9

and F ′(x) = 0 and F = 0. The corresponding complex potential is F = φ+ iψ: let z = x+ iy

k

ωAF (z, t) sinh(kh) = cosh(k(y + h)) sin(kx− ωt) + i sinh(k(y + h)) cos(kx− ωt)

= [cosh(ky) cosh(kh) + sinh(ky) sinh(kh)] sin(kx− ωt)+ i[sinh(ky) cosh(kh) + cosh(ky) sinh(kh)] cos(kx− ωt)

= cosh(kh) [cosh(ky) sin(kx− ωt) + i sinh(ky) cos(kx− ωt)]sinh(kh) [sinh(ky) sin(kx− ωt) + i cosh(ky) cos(kx− ωt)]

= cosh(kh) [cos(iky) sin(kx− ωt) + sin(iky) cos(kx− ωt)]+ sinh(kh) [−i sin(ky) sin(kx− ωt) + i cos(iky) cos(kx− ωt)]

= cosh(kh) sin(kx− ωt+ iky) + sinh(kh)i cos(kx− ωt+ iky)

= cos(ikh) sin(kz − ωt) + sin(ikh) cos(kz − ωt)= sin(kz − ωt+ ikh)

or F (z, t) = Aωk sinh(kh) sin(kz − ωt + ikh). In the above, we have used trigonometric identities

(sum of angles) and the identities sin(iα) = i sinh(α) and cos(iα) = cosh(α).

4.3 Particle paths

The instantaneous position of a particle will be indicated by a fixed-position vector z0 and anadditional vector z1 that varies with time, i.e. ,the length and orientation of z0 remain fixedwhile both the length and inclination of z1 vary with time. Consider the complex conjugatevariable position vector z1 = x1 − iy1. It follows that

dz1

dt= u− iv = W =

dF

dz.

This comes from the definition of the complex velocity W and complex potential F . Thus wehave

dz

dt=

Aω

sinh(kh)cos(kz − ωt+ ikh)

and integrating gives

z1 = − A

sinh(kz)sin(kz − ωt+ ikh) = −F (z, t)k

ω=−kω

(φ+ iψ).

Therefore, we find x1 = −φ(x, y, t)k/ω and x2 = ψ(x, y, t)k/ω. Hence, the coordinates aregiven by

x1 = − A

sinh(kh)cosh(k(y + h)) sin(kx− ωt) x2 =

A

sinh(kh)sinh(k(y + h)) cos(kx− ωt).

We note that (x1 sin(kh)

− cosh(k(y + h))

)2

+

(x2 sinh(kh)

sinh(k(y + h))

)2

= 1

and that the trajectories are ellipses with axis depending on the depth. Each particle expe-riences the same waves passing above it, irrespective of its x coordinate. Thus the motionexperienced by two particles that are separated in the x direction only will be the same, butthe phasing will be different.

10

5 Internal gravity waves

In the absence of a free surface, fluid elements of different densities, e.g. when salt concen-trations differ, can lead to internal gravity waves. Suppose that the stratification in the fluid,ρ0(y), is decreasing with height in a way that the more salty water is further down when thefluid is at rest, i.e. ρ′0(y) < 0. The hydrostatic pressure equation is

0 = −dp0

dy− ρ0g

Since density is not constant, we need to rethink the conservation of mass equation. Recallthat the most general form is

Dρ

Dt+ ρ∇ · u = 0.

If a flow is incompressible, the above reduces to DρDt = 0. Therefore away from the rest state,

the equations of motion are

ρ

[∂u

∂t+ u · ∇u

]= −∇p+ ρg (12)

∂ρ

∂t+ u · ∇ρ = 0 (13)

∇ · u = 0 (14)

Again, assuming for simplicity that the motion is two-dimensional (with y the vertical direction)and further assuming small oscillations, i.e. expanding and neglecting quadratic terms, weobtain from (12)-(14)

ρ0∂u1

∂t= −∂p1

∂xρ0∂v1

∂t= −∂p1

∂y− ρ1g

∂u1

∂x+∂v1

∂y= 0

∂ρ1

∂t+ v1

∂ρ0

ρy.

In the above, we set

u = u1(x, y, t)e1 + u2(x, y, t)e2 p = p0(y) + p1(x, y, t) ρ = ρ0(y) + ρ1(x, y, t).

5.1 Fourier series solution

In order to look for traveling wave solutions, we look at a particular Fourier mode:

v1 = v1(y)ei(kx−ωt) u1 = u1(y)ei(kx−ωt) p1 = p1ei(kx−ωt) ρ1 = ρ1e

i(kx−ωt)

and take the real part. Since ρ0 = ρ0(y), we find after plugging in

ρ0ωu1 = kp1 ρ0iωv1 = p′1 + ρ1g iku1 + v′1 = 0 − iωρ1 + v1dρ0

dy= 0.

Eliminating all variables except v1(y) yields the second order ODE

v′′1(y) +ρ′0(y)

ρ0(y)v′1(y) + k2

(N2

ω2− 1

)v1(y) = 0,

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where N2 = − gρ0(y)ρ

′0(y) is the buoyancy frequency. We can only solve the ODE if we know

the functional form of ρ0(y). Assume that ρ ∝ e−y/H , thenρ′0ρ0

is constant and the solution tothe ODE are wave like

v1(y) ∝ e−y/(2H)eil

where −4l2 = 1H2 − 4k

(N2

ω2 − 1)

or equivalently

ω2 =N2k2

k2 + l2 + 14H2

.

This is an example of an anisotropic dispersion relation. Note that the other solutions iseliminated because of bounded solutions. Plugging back, we have

v1 ∝ ey/(2H)ei(kx+ly−ωt) ρ1 ∝ e−y/(2H)ei(kx+ly−ωt).

In practice, the wavelength λ = 2π√k2+l2

is small compared to the scale height H, so that

the dispersion relation reduces to

ω2 =N2k2

k2 + l2cg =

(∂ω

∂k,∂ω

∂l

)=

ωl

k(k2 + l2)(l,−k).

Suppose that all numbers are positive. The line of constant phase kx+ ly − ωt =constant arecrests and move in the direction (k, l). On the other hand, the group velocity is perpendicularto it and the packet as a whole moves in the direction (l,−k).

6 Nonlinear waves: Finite amplitude waves in shallow water

In the shallow water assumption, we don’t assume that the amplitude of the waves is smallcompared to the depth, rather we assume that h0 � L, where h0 is some typical scale of thefree surface y = η(x, t) and L is some horizontal length scale. In this context, we assume thatthe bottom is at y = 0 and is flat. The Navier-Stokes governing equations are

∂u

∂t+ u

∂u

∂x+ v

∂u

∂y= −1

ρ

∂p

∂x(15)

∂v

∂t+ u

∂v

∂x+ v

∂v

∂y= −1

ρ

∂p

∂y− g (16)

∂u

∂x+∂v

∂y= 0. (17)

Let h0 = εL, where ε is a small parameter. Introducing the following nondimensional quantities

x = Lx y = h0y u = Uu v = V v t = T T p = P p,

plugging in and dropping the tilde, we have

U

T

∂u

∂t+U2

L

∂u

∂x+UV

h0v∂u

∂y= −1

ρ

P

L

∂p

∂x

V

T

∂v

∂t+UV

L

∂v

∂x+V 2

h0v∂v

∂y= −1

ρ

P

h0

∂p

∂y− g

U

L

∂u

∂x+V

h0

∂v

∂y= 0.

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Since there is only one time scale and no other time in the problem, we have U = L/T .Furthermore, rearranging the incompressibility condition gives

∂u

∂x+V

U

1

ε

∂v

∂y= 0,

which only gives sensible and nontrivial solution if V = εU . Plugging in, the second equationbecomes

εU2

L

(∂v

∂t+ u

∂v

∂x+ v

∂v

∂y

)= −1

ρ

P

εL

∂yp

∂y− g.

Therefore, for the equation to make sense and for the terms to balance, we must have that Pscales like ε so that the terms on the right hand side are the dominant terms. This is sayingthat the vertical pressure is basically hydrostatic or going back to dimensional quantities,∂p∂y = −ρg. Integrating and assuming constant pressure at the free surface, we find

p = p0 − ρg(y − η(x, t)).

Substituting the pressure back in yields

∂u

∂t+ u

∂u

∂x+ v

∂u

∂y= −g ∂η

∂x.

Since the left hand side is the Eulerian derivative of u and the right hand side is independentof y, we see immediately that if u is independent of y initially, it will remain so for all time.Assuming thus u = u(x, t) gives

∂u

∂t+ u

∂u

∂x= −g ∂η

∂x. (18)

Integrating the incompressibility condition yields

v = −∂u∂y

+ f(x, t).

Since there is no penetration at y = 0, f(x, t) = 0. Using the kinematic boundary conditionat the free surface, i.e. v = ∂η

∂t + u∂η∂x , we find

∂η

∂t+ u

∂η

∂x+ η

∂u

∂x= 0 (19)

Equations (18)-(19) from the shallow-water equations. To solve these equations, we introducea new variable c(x, t) =

√gη(x, t). Plugging in, (18)-(19) become

∂u

∂t+ u

∂u

∂x+ 2c

∂c

∂x= 0

∂(2c)

∂x+ u

∂(2c)

∂x+ c

∂u

∂x= 0.

Adding and subtracting yield the system

[∂t + (u+ c)∂x] (u+ 2c) = 0 [∂t + (u− c)∂x] (u− 2c) = 0.

These PDEs are solved using the method of characteristics. Let x = x(s) and t = t(s) starting(x0, t0) such that dt

ds = 1 and dxds = u+ c. Then the first equation reduces to(

dt

ds∂t +

dx

ds∂x

)(u+ 2c) = 0 or

d

ds(u+ 2c) = 0.

Thus, u± 2c is constant along positive and negative characteristic curves.

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