Numerical solution of conservation laws applied to the Shallow

Water Wave Equations

Stephen G Roberts

Mathematical Sciences Institute, Australian National University

Updated January 17, 2013

(based on notes by SGR and Chris Zoppou, and report by Angus Griffith)

AMSI Summer School, 2013

1

Contents

1 Shallow Water Wave Equations 5

1.1 Geometry of the problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2 Eulerian Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.3 Lagrangian Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.4 Conservation of mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.5 Other Forcing Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.6 1D Saint Venant equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.7 Conservation and Balance Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.8 Diagonalization and Riemann Invariants . . . . . . . . . . . . . . . . . . . . . . . . . 12

2 First order PDE’s in 2 variables 14

2.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.2 Very Simple Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.3 Constant coefficient Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.4 Variable coefficient Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.5 More complicated Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.6 And an even more complicated Example . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.7 General Semi-linear Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.8 Alternative Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3 Non-Linear Scalar Conservation Laws 24

3.1 Shock-wave Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.2 Advection Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.3 Integral Conservation Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.4 PDE Conservation Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

3.5 Rarefaction Fan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

3.6 Weak Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3.7 Vanishing Viscosity Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.8 Entropy Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3.9 Entropy Inequality and vanishing viscosity . . . . . . . . . . . . . . . . . . . . . . . . 37

3.10 Other Entropy Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.11 Total Variation, Monotonicity, L1 contraction . . . . . . . . . . . . . . . . . . . . . . 39

3.12 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

4 The Riemann Problem 42

4.1 The Riemann Problem for Constant Coefficient System . . . . . . . . . . . . . . . . 42

2

5 Riemann Problem for The Shallow water wave equations 44

5.1 The Dam Break Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

5.2 Stoker’s Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

5.3 Ritter’s Dry Bed Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

6 Conservative Numerical Schemes 64

7 Modified Equivalent Partial Differential Equation 66

7.1 Advective-diffusion equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

7.2 Advective-dispersion equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

7.3 Higher-order equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

7.4 Accuracy in the presence of a discontinuity . . . . . . . . . . . . . . . . . . . . . . . 76

7.5 First-order schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

7.6 Second-order schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

8 Naive Finite Difference Schemes 80

8.1 Lax-Friedrichs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

8.2 Richtmyer Scheme . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

8.3 Lax-Wendroff Scheme . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

8.4 Beam-Warming Scheme . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

8.5 MacCormack Scheme . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

9 Artificial Viscosity and Limited Schemes 101

9.1 Limiters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

9.2 Flux Corrected Transport . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

10 Upwind Schemes 121

10.1 Bermudez and Vazquez Scheme . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

10.2 Steger and Warming Scheme . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

10.3 Yang et al. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

10.4 Harten Scheme . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

11 Godunov-Type Schemes 130

11.1 Averaging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

11.2 Re-averaging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

11.3 Evolution Step . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

11.4 Roe’s Solver . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

11.5 Osher’s scheme . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142

11.6 Harten, Lax and van Leer Scheme . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

11.7 Weighted Average Flux Scheme . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

3

11.8 Piecewise Parabolic Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152

12 Splitting Techniques 157

12.1 Fractional splitting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158

12.2 Strang splitting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158

13 Boundary Conditions 160

13.1 Transmissive Boundary Condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

13.2 Dirichlet Boundary Condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

13.3 Non-Reflective Boundary Condition . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

14 Treatment of the Source Term using Fractional Steps 170

14.1 The Ordinary Differential Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176

15 The Finite Volume Method 178

15.1 Basic Scheme . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178

15.2 Side Channel Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182

15.3 Wetting and Drying of Cells . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183

4

1 Shallow Water Wave Equations

There are several ways to derive the shallow water wave equations (SWWEs). Often they are

derived by assuming incompressible, inviscid, irrotational flow and applying depth averaging to the

Navier-Stokes equations; see for example [55, pp. 12–21].

The methods presented here are more natural in that the SWEE is derived from first principles

using conservation of mass and conservation of momentum.

Unlike other equations describing fluid flow such as Navier-Stokes equation, the shallow water

wave equations are two dimensional equations. The solutions to the shallow water wave equations

give the height of water h(x, y) above the ground level, along with the velocity field (u(x, y), v(x, y)).

1.1 Geometry of the problem

Begin by considering water flowing over some terrain, which has been discretised into two-

dimensional cells. Let h(x, y) be the height of water above some terrain of height b(x, y) Functions,

u(x, y) and v(x, y) describe the rate of flow of water in the horizontal x and y directions respectively.

x

b

h

w

z

u

Figure 1: Nomenclature for shallow water wave equations; depth of water, h; elevation of the bed, b; the

stage, w = h+ b; and the depth averaged horizontal velocity, u.

The total height (stage) w of the water above some reference point is hence given by

w ≡ h+ b

5

1.2 Eulerian Formulation

The Eulerian formulation considers the problem from the perspective of a single cell. The volume

of water in a particular cell ((x0, x1), (y0, y1)) at time t is given by∫ x1x0

∫ y1y0

h(x, y, t) dy dx (1)

As time passes the amount of water changes according to the flows coming in and out of the cell

(see Figure 2). When discretising in time, the volume of water in a cell will evolve according to the

(hu)(x0, y) (hu)(x1, y)

(hv)(x, y0)

(hv)(x, y1)

x0 x1y0

y1

Figure 2: Flow in and out of a cell.

flows u and v, ∫ x1x0

∫ y1y0

h(x, y, t1) dx dy =

∫ x1x0

∫ y1y0

h(x, y, t0) dx dy

+

∫ t1t0

∫ y1y0

h(x0, y, t)u(x0, y, t) dy dt−∫ t1t0

∫ y1y0

h(x1, y, t)u(x1, y, t) dy dt

+

∫ t1t0

∫ x1x0

h(x, y0, t)v(x, y0, t) dx dt−∫ t1t0

∫ x1x0

h(x, y1, t)v(x, y1, t) dx dt

(2)

Applying the fundamental theorem of calculus1 in variables t, x and y,∫ t1t0

∫ x1x0

∫ y1y0

ht dy dx dt+

∫ t1t0

∫ x1x0

∫ y1y0

(uh)x dy dx dt

+

∫ t1t0

∫ x1x0

∫ y1y0

(vh)y dy dx dt = 0.

(3)

Since this is true for any choice of cell in space and time,

ht + (uh)x + (vh)y = 0. (4)

Equation (4) is the mass conservation equation of shallow water wave equations.

1An equivalent derivation can be formulated by applying the divergence theorem to the domain in Figure 2.

6

1.3 Lagrangian Formulation

In contrast to the Eulerian formulation, the Lagrangian formulation takes the perspective of a

domain of water which propagates in time (see Figure 3). We will use a notation which is dimension

independent.

x

y

Ωt0

Ωt1

u

Figure 3: Propagation of a domain Ω of water.

Let x ∈ R2 define an arbitrary position, and let u : R2 × R+ → R2 define the velocity field offluid. That is,

u(x, t) := (u(x, t), v(x, t)) (5)

Now consider the evolution of particles under the influence of the velocity field u. The position

x(s, t) ∈ R2 of a particle flowing in the velocity field satisfies

dx

dt(s, t) = u(x(s, t), t). (6)

Theorem 1.1 (Flowing Integrals). Let f(x, t) be the density of some arbitrary quantity (for exam-

ple, the concentration of a dissolved material). Consider the amount of the quantity in the region

Ωt, moving with the fluid, i.e. consider

F (t) =

∫Ωt

f(x, t) dx. (7)

The rate of change of this quantity is given by,

d

dtF (t) =

∫Ωt

(ft + div(fu)) dx.

7

Proof

dF

dt=

d

dt

∫Ωt

f(x, t), t)dx =

∫Ω0

d

dt(f(x(s, t), t) det xs) ds

=

∫Ω0

d

dt(f(x(s, t), t)) det xsds (8)

+

∫Ω0

f(x(s, t), t)d

dt(det xs) ds (9)

Well we need to differentiate the determinant of a matrix. Jacobi’s formula provides the result

we need.

Theorem 1.2 (Jacobi’s formula: differential of a determinant).

d (detA) = Trace (adj(A) dA)

Proof When considering the determinant as a function of all the matrix elements, it is clear that

the chain rule can be applied to yield,

d(detA) =∑i

∑j

∂ det(A)

∂aijdaij

where aij denotes the ij-th component of the matrix A.

By Laplace’s formula

detA =∑k

(−1)i+kaik det(Aik) =∑k

aik adj(A)ki

where Aij is the block matrix obtained by removing row i and column j. Note that adj(A)ji are

elements in the transpose of the adjoint of A. We have,

∂ det(A)

∂aij=∑k

∂aik∂aij

adj(A)ki = adj(A)ji

since adj(A)ki does not depend on elements of A in the ith row. Therefore,

d (detA) =∑

i

∑j adj(A)ji daij

= Trace (adj(A) dA) .

We can now apply Jacobi’s formula, to equation (9).

d

dt(det xs) = Trace

(adj(xs)

dxsdt

).

8

Nowdx

dt= u and so

dxsdt

= uxxs. So, using Jacobi’s formula yields

d

dt(det xs) = Trace (adj(xs)uxxx)

= Trace (det xsIux)

= det xs Trace(ux)

= det xs div u.

So the rate of change of area form is proportional to the divergence of the velocity field.

This can be used to simplify equation (9), namely,∫Ω0

f(x(s, t), t) div(u) det(xs) ds =

∫Ωt

f(x, t) div(u) dx.

Equation (8) satisfies∫Ω0

d

dt(f(x(s, t), t)) det xs ds =

∫Ω0

(ft(x(s, t), t) +∇xf(x(s, t), t) ·

dx(s, t)

dt

)det xs ds

=

∫Ωt

(ft(x, t) +∇xf(x, t) · u(x, t)) dx.

Taken together we have

d

dt

∫Ωt

f(x, t)dx =

∫Ωt

(ft(x, t) +∇xf(x, t)u(x, t) + f(x, t) div(u)) dx

=

∫Ωt

(ft(x, t) + div(f(x, t)u(x, t))) dx

.

1.4 Conservation of mass

Conservation of mass can be expressed as

d

dt

∫Ωt

h(x, t)dt = 0

By Theroem 1.1

d

dt

∫Ωt

h(x, t) dx =

∫Ωt

(ht +∇ · (uh)) dx.

Conservation of mass hence takes the form∫Ωt

(ht +∇ · (uh)) dx = 0. (10)

This conservation equation holds for all Ωt. In differential form,

ht +∇ · (uh) = 0

ht + (uh)x + (vh)y = 0(11)

This is exactly the conservation of mass equation (4) introduced earlier.

9

Conservation of momentum and the hydrostatic assumption

The Shallow water wave equations are completed by considering so called conservation of mo-

mentum, actually Newton’s second law, the rate of change of momentum with respect to time is

equal to the applied forces.

The applied forces that we need to worry about are the pressure forces. First note that the

pressure force on a unit surface area is normal to that surface.

The shallow water assumption contends that the vertical flows are negligible and that the

magnitude of the pressure depends linearly with water depth (the hydrostatic assumption).

Consider the momentum in a moving region Ωt,∫Ωt

p dx

where p = uh = (uh, vh).

The horizontal pressure term applied to the region Ωt is given by

−∫∂Ωt

1

2gh2n dσ −

∫Ωt

gh∇b dx,

the first term is the horizontal pressure applied around the boundary of the domain Ω and second

term is the horizontal component of the pressure due to the sloping bed.

Newton’s second law then gives

d

dt

∫Ωt

p dx = −∫∂Ωt

1

2gh2n dσ −

∫Ωt

gh∇b dx.

Applying Theorem 1.1 to the first term and Stokes theorem to the first term on the LHS gives∫Ωt

(pt +∇ · (u⊗ p)) dx = −∫

Ωt

∇(12gh2) dx−

∫Ωt

gh∇b dx

and since this integral equation is true for all Ωt), we conclude

pt +∇ · (u⊗ p) +∇(1

2gh2) = −gh∇b.

Specifically, for the x component,

(uh)t +

(uuh+

1

2gh2)x

+ (v uh)y = −ghbx (12)

and for the y component,

(vh)t + (u vh)x +

(v vh+

1

2gh2)y

= −ghby (13)

Equation (11), together with equations (12) and (13) constitute the Shallow Water Wave Equations

(SWWE).

10

1.5 Other Forcing Terms

In practical situations other forcing terms are applied to model particular physical situations.

There will be friction between the fluid and the bed. This is very often modelled by Man-

nings/Chevy friction, where the retarding force term is given by

− gη2

h7/3p|p|.

The Mannings resistance coefficient η is usually determined experimentally and is tabulated for

standard surfaces, but tends to have values between 0.01 (over smooth concrete) to 0.05 (thick

grass), depending on the roughness of the surface.

In storm surge, the atmospheric pressure should be added to the hydrostatic pressure terms to

produce a term1

ρwh∇Pa

where ρw is the density of water (this factor has been divided out in the other terms). This can

be derived by combining the pressure terms applied to a fluid cell Ωt, on the sides and bottom as

before, as well as the influence of the atmospheric pressure on the surface of the cell.

−∫∂Ωt

(Pah+1

2ρwgh

2)n dσ −∫

Ωt

(Pa + ρwgh)∇b dx +∫

Ωt

Pa∇w dx

=

∫Ωt

−∇(Pah+1

2ρwgh

2)− (Pa + ρwgh)∇b+ Pa∇w dx

= ρw

∫Ωt

−∇(12gh2)− gh∇b+ 1

ρwh∇Pa dx

1.6 1D Saint Venant equations

Let’s consider a one-dimensional flow problem. That is, a flow which only has one component.

A rotation of coordinates can always make it so that this component is x, such that the mass

conservation equation reduces to

ht + (uh)x = 0 (14)

and the momentum equations reduce to just one equation,

(uh)t +

(uvh+

1

2gh2)x

= −ghbx (15)

Together these two equations are know as the one dimensional shallow Water Wave Equations, or

as the Saint Venant equations.

Often the momentum equation is written in primitive variables (u instead of uh), as

ut +

(1

2u2 + gh

)x

= −gbx (16)

Often equations (14) and (16) are easier to work with, particularly when finding analytic solutions

to the shallow water wave equation.

11

1.7 Conservation and Balance Laws

The SWW equations in one (spatial) dimension has the form

Ut + F(U)x = S

where

U =

[h

uh

], F(U) =

[uh

uuh+ 12gh2

]and S =

[0

−ghbx

].

U is the conserved quantity vector, F(U) is the associated flux vector, and S is the source term.

The SWW equations in two (spatial) dimensions has the form

Ut + F(U)x + G(U)y = S

where now

U =

huhvh

, F(U) = uhuuh+ 12gh2

u vh

, G(U) = v hv uhv vh+ 12gh

2

, and S = 0−ghbx−ghby

.where F(U) is the flux vector in x direction, G(U) is the flux vector in y direction.

These equations are written in conservation form, and the equations are called conversation laws

(well actually conservation laws if the RHS is zero and called balance laws if the RHS is non-zero).

1.8 Diagonalization and Riemann Invariants

Let’s consider the homogeneous case, i.e. when ∇b = 0. Then, provided the solutions to theseequations are smooth, the equations can be written in the equivalent non-conservation form in

terms of primitive variables [h

u

]t

+

[u h

g u

][h

u

]x

=

[0

0

]This form immediately directs us to try to diagonalize the matrix[

u h

g u

].

The eigenvalues of this matrix are

λ− = u−√gh and λ+ = u+

√gh.

Curves with speeds given by these two eigenvalues will be characteristic curves for this problem.

The quantity√gh is know as the celerity, as well as the sound speed, an is denoted c =

√gh. The

corresponding (left) eigenvectors are

l− =

[h/c

1

]and l+ =

[h/c

−1

].

12

Consequently [c/h 1

c/h −1

][u h

g u

]=

[u+ c 0

0 u− c

][c/h 1

c/h −1

].

Applying the matrix [c/h 1

c/h −1

]to the PDE implies[

c/h 1

c/h −1

][h

u

]t

+

[u+ c 0

0 u− c

][c/h 1

c/h −1

][h

u

]x

=

[0

0

].

It is simple to verify (though more difficult to discover) that[c/h 1

c/h −1

][h

u

]t

=

[u+ 2c

u− 2c

]t

.

Similarly [c/h 1

c/h −1

][h

u

]x

=

[u+ 2c

u− 2c

]x

.

Together we conclude that [u+ 2c

u− 2c

]t

+

[u+ c 0

0 u− c

][u+ 2c

u− 2c

]x

.

Hence we have in some sense decoupled the system into two scalar equations

(u+ 2c)t + (u+ c)(u+ 2c)x = 0,

(u− 2c)t + (u− c)(u− 2c)x = 0.

The quantities u ± 2c are called Riemann Invariants and are constant along the correspondingcharacteristics (curves with slopes u± c).

To continue we must investigate the properties and solution techniques for first order equations

of the form

ut + a(u)ux = 0,

or in conservation form, the scalar conservation law

ut + f(u)x = 0.

We want to understand the properties of the solutions of the SWW equations, but in the next

section we study methods for solving simple non-linear scalar conservation laws.

13

2 First order PDE’s in 2 variables

2.1 Definition

In this section we show how to solve the most general semi-linear 1st order PDE in two variables,

i.e. a PDE of the form

a(x, y)ux + b(x, y)uy + c(x, y, u) = 0 . (17)

[If a = a(x, y, u), and b = b(x, y, u) then the equation is called quasi-linear.] More precisely, we will

show how we can obtain the solution to (17) provided we can solve an associated 1st order ordinary

differential equation (ODE). Let us first discuss some examples to illustrate the method.

2.2 Very Simple Example

Consider the PDE ux = 0. Clearly the most general solution is given by u(x, y) = f(y), where f is

an arbitrary function. Thus solutions to this simple PDE are constant along lines parallel to the

x-axis, i.e. lines of the form y = C, for C ∈ R. The lines are called the characteristic lines of thePDE ux = 0. To find the value of u(x, y) on a characteristic line y = C we only have to know the

value of u(x, y) in one particular point on the line. Thus, a solution to the PDE ux = 0 is completely

determined by giving for example the values of u on the y-axis, i.e. u(0, y). Alternatively, we might

give as initial conditions the value of u on any line y = ax+ b for a 6= 0, or even on monotonicallyincreasing, or decreasing, curves y = g(x). However, for generic curves the problem only admits a

solution if at all intersection points of the curve with all lines y = C the prescribed value of u(x, y)

is the same.

The treatment of the generic 1st order PDE in (17) proceeds along the same lines, in fact can

be reduced to the problem above. Before we get to that let us discuss a few, more complicated,

examples.

2.3 Constant coefficient Example

Consider aux + buy = 0, with a, b ∈ R, and a2 + b2 6= 0. We will discuss two ways of solving thisPDE.

Coordinate Method: Upon defining new coordinates

ξ = ax+ by , η = bx− ay ,

we have

ux = uξξx + uηηx = auξ + buη ,

uy = uξξy + uηηy = buξ − auη .

14

Thus

aux + buy = (a2 + b2)uξ ,

hence the PDE in new coordinates reads uξ = 0. The general solution, as in the example

before, is given by

u(x, y) = f(η(x, y)) = f(bx− ay) .

Geometric Method: We can think of the equation as the statement that the derivative of u in

the direction of the vector

(a

b

)vanishes. Recall that the directional derivative Dnu of a

multivariable function u(x) in the direction of a vector n, is given by

Dnu(x) ≡ limt→0

u(x + tn)− u(x)t

=∑i

ni∇iu(x) = n · ∇u ,

where the gradient ∇u is the vector with components (ux, uy, . . .). Thus we conclude thatsolutions to the PDE are precisely those functions which are constant in the direction of the

vector

(a

b

), i.e. functions which are constant on the lines bx− ay = C, parametrized by C.

The value of u on these so-called characteristic lines is thus determined by the value of C.

As before we can then introduce a characteristic variable η(x, y) = C = bx− ay.

Summarizing, as before, we find the most general solution to be of the form

u(x, y) = f(η(x, y)) = f(bx− ay) .

As a concrete example, suppose we want to solve the PDE and initial condition

4ux − 3uy = 0 , u(0, y) = y3 .

As we have seen the most general solution of the PDE is given by u(x, y) = f(−3x− 4y). Settingx = 0 gives the condition f(−4y) = y3. Hence, by setting w = −4y we find f(w) = −w364 , and thusthe solution is given by u(x, y) = 164(3x+ 4y)

3 .

2.4 Variable coefficient Example

Consider the transport equation

ux + yuy = 0 . (18)

Again, geometrically this equation tells us that the function u(x, y) is constant along curves (x, y(x))

which at every point has a tangent vector in the direction (1, y), i.e. has slope y/1 = y. Such curves

are determined as solutions of the ODE

d

dxy(x) = y(x) , (19)

15

which, by separation of variables, can easily be seen to have as a general solution

y(x) = Cex , (20)

where C is an arbitrary constant. I.e.

ye−x = η(x, y) .

defines the characteristic variable η.

As η varies, these curves sweep out the entire two-dimensional (x, y)-plane. Thus, we conclude

that the most general solution to the PDE (18) is given by

u(x, y) = f(η(x, y)) = f(ye−x) . (21)

A particular solution, e.g. for initial condition u(0, y) = y3 is easily found by putting x = 0 in (21),

i.e. u(0, y) = f(y) = y3, hence u(x, y) = (ye−x)3 = y3e−3x.

2.5 More complicated Example

Consider the PDE

ux + 2xy2uy = 0 .

Supposing that the characteristic curves can be parameterised as (x, y(x)), then y(x) satisfies the

ODEdy

dx(x) = 2xy(x)2 .

Separating the variablesdy

y2= 2xdx ,

leads to the solution

−1y

= x2 − C ,

where C is an arbitrary constant. Hence along a characteristic curve,

x2 +1

y= η ,

(C = η is the characterisitic variable).

Hence the general solution of the PDE is given by

u(x, y) = f(η(x, y)) = f(x2 +1

y) .

It can be easily verified that this is indeed a solution (check!).

16

2.6 And an even more complicated Example

Now consider the non-homogeneous PDE

ux + 2xy2uy = 2xy .

Now u will NOT be constant along a characteristic, but will satisfy an ODE.

The characteristic curves can be parameterised as (x, y(x)), to obtain a characteristic variable

x2 +1

y= η .

Now, from the equation for the characteristic curve, and the fact that u satisfies the PDE, we

have

d

dxu(x, y(x)) = ux(x, y(x)) +

d

dxy(x)uy(x, y(x))

= ux(x, y(x)) + 2xy(x)2uy(x, y(x))

= 2xy(x) .

So on a characteristic curve (η constant), u(x, y(x)) satisfies

d

dxu(x, y(x)) = 2xy(x) .

We have an expression for y(x) (for a given value of η)

y(x) =1

η − x2.

So u(x, y(x) satisfies (for η constant)

d

dxu(x, y(x)) =

2x

η − x2.

Integrating this ODE we obtain (the constant of integration can be written as a function of η)

u(x, y(x)) = − ln |η − x2|+ f(η)

= − ln∣∣∣∣x2 + 1y(x) − x2

∣∣∣∣+ f(x2 + 1y(x))= ln |y(x)|+ f(x2 + 1

y(x)) .

Hence the general solution of the PDE is given by

u(x, y) = ln |y|+ f(x2 + 1y

) .

It can be verified that this is a solution (check!).

Note that it is a particular solution of the non-homogeneous PDE, plus the general solution of

the homogeneous PDE.

17

2.7 General Semi-linear Case

Now let us return to the problem of solving the most general semi-linear first order PDE

a(x, y)ux + b(x, y)uy + c(x, y, u) = 0 . (22)

Motivated by the examples, let us define a characteristic curve as a curve along which the solution

of the PDE is given by an ODE (in fact, the ODE in the most of the examples was just a zeroth

order ODE stating that u was constant along those curves).

First let us parameterise the characteristic curve as a function of x, i.e. (x, y(x)). In this case

it is useful to consider the (essentially) equivalent PDE

ux +b(x, y)

a(x, y)uy +

c(x, y, u)

a(x, y)= 0 . (23)

The equation for the characteristic curves are then given by

dy

dx(x) =

b(x, y)

a(x, y).

which should provide a solution for y(x) which depends on a constant. This is often written as

η(x, y(x)) = C ,

which defines a characteristic variable η.

The function u(x, y(x)) (which is just a function of x) satisfies the ODE

d

dxu(x, y(x)) = −c(x, y(x), u(x, y(x)))

a(x, y(x)).

The solution of this ODE will introduce a constant of integration which will be constant for a

specific value of η. So the constant of integration can be different for each value of η. So it is a

function of η.

Example Let us review this for the equation

xux + 2x2y2uy = 2x

2y u .

The equation is in form (22). Put in form (23).

ux + 2xy2uy = 2xy u .

This is similar to equation we have already investigated. We look for the characteristic curve

of the form (x, y(x)). The equation for the characteristic curve is

dy

dx(x) = 2x y(x)2 .

18

The characteristic curves can be parameterised as (x, y(x)), to obtain a characteristic variable

x2 +1

y= η .

Now, from the equation for the characteristic curve, and the fact that u satisfies the PDE, we

have

d

dxu(x, y(x)) = ux(x, y(x) +

d

dxy(x)uy(x, y(x))

= ux(x, y(x) + 2xy(x)2uy(x, y(x))

= 2xy(x)u(x, y(x)) .

So on a characteristic curve (η constant), u(x, y(x)) satisfies

d

dxu(x, y(x)) = 2xy(x)u(x, y(x)) .

We have an expression for y(x) (for a given value of η)

y(x) =1

η − x2.

So u(x, y(x) satisfies (for η constant)

d

dxu(x, y(x)) =

2x

η − x2u(x, y(x)) .

Integrating this ODE we obtain (the constant of integration can be written as a function of η)

ln |u(x, y(x))| = − ln |η − x2|+ f(η)

= − ln |x2 + 1y− x2|+ f

(x2 +

1

y(x)

)= ln |y(x)|+ f

(x2 +

1

y(x)

).

Hence the general solution of the PDE is given by

ln |u(x, y)| = ln |y|+ f(x2 +

1

y

).

Leads to the solution

u(x, y) = yg

(x2 +

1

y

),

where the minus signs from the absolute values have been absorbed into the function g.

As usual we should check that this function does indeed satisfy the PDE.

ux + 2xy2uy = yg

′(x2 +

1

y

)2x+ 2xy2

(g

(x2 +

1

y

)+ yg′

(x2 +

1

y

)−1y2

)= 2xy2g

(x2 +

1

y

)= 2xyu .

19

2.8 Alternative Formulation

Often it is useful to parametrize a characteristic curve in the two-dimensional plane by the more

general form (x(s), y(s)), and let u(s) = u(x(s), y(s)) be the solution of (22) along the curve. Since,

du

ds= ux

dx

ds+ uy

dy

ds, (24)

it is clear that the characteristic curves, and the solution u(s) along the curve, are determined by

the following set of characteristic equations

dx

ds= a(x, y) ,

dy

ds= b(x, y) ,

du

ds= −c(x, y, u) . (25)

This set of ODE’s has a unique solution, at least for s in a neighborhood of s = 0, for given initial

conditions (x(0), y(0), u(0) = u(x(0), y(0))).

To make contact with the previous examples, note that if we eliminate the parameter s, and

describe the characteristic curve by an relation y(x) we have

dy

ds=dy

dx

dx

ds. (26)

I.e.dy

dx=b(x, y)

a(x, y). (27)

This reduces the problem of solving the PDE (22) to a system of ODE’s.

Sometimes it is useful to adopt a different, but equivalent, point of view. Namely, in case we

are not only given a PDE (22), but also a specific initial condition on a curve γ, parametrized

by γ(r) = (x(r), y(r)), i.e. u(x(r), y(r)) = u(r). In that case it is useful to consider replacing the

coordinates (x, y) by coordinates (s, r), where s is a coordinate along the characteristic curves, and

r a coordinate along the initial curve, normalized such that s = 0 on the curve γ(r). I.e. we get the

following system of equations

d

dsx(s, r) = a(x(s, r), y(s, r)) ,

d

dsy(s, r) = b(x(s, r), y(s, r)) ,

d

dsu(s, r) = −c(x(s, r), y(s, r), u(s, r)) ,

x(0, r) = x(r) ,

y(0, r) = y(r) ,

u(0, r) = u(r) . (28)

20

We find solutions x(s, r), y(s, r) and u(s, r), and provided the transformation x = x(s, r), y = y(s, r)

is invertible, i.e. we can solve s = s(x, y), r = r(x, y) we find the solution to our original problem

as u(x, y) = u(s(x, y), r(x, y)).

2.9 Problems

Problem 2.1 (Systems of ODEs.). Solve the following systems of ODEs

Hint: Try using different methods, e.g, for (a) and (b) one might try turning the systems into

second order equations and solve via the characteristic equation, or by the method of matrix exponen-

tiation. In (c) one might try an integrating factor or perhaps the method of variation of parameters.

(a)

dx

ds= y

dy

ds= x

(b)

dx

ds= y

dy

ds= −x.

(c)

du

dt+Au(t) = f(t)

where:

(i) u and f are real valued functions and A a constant.

(ii) u and f are vector valued functions (in Rn) and A is a constant n× n matrix.

Problem 2.2 (Constant coefficient PDE). Solve the first-order equation 3ux + 4uy = 0 with the

auxiliary condition u = sin y when x = 0.

Problem 2.3 (First order linear PDE with constant coefficients.). Consider the equation aux +

buy + cu = 0, where a, b, c ∈ R.

(a) First, use the method characteristics to find the general solution of this equation in terms of

some given function.

21

(b) Second, use the coordinate transformation

ξ(x, y) = ax+ by , η(x, y) = bx− ay ,

to find the general solution of this equation.

(c) Show that both methods provide the same solution (though on first glance they look quite

different).

Problem 2.4 (Homogeneous variable coefficient PDE). Solve the equation yux + xuy = 0.

Problem 2.5 (Inhomogeneous variable coefficient PDE). Obtain the general solution of the first

order PDE

yux + xuy − yu = xex . (29)

If we prescribe u(x, y) = φ(x, y) on the upper portion of the hyperbola y2 − x2 = 1, y ≥ 1, showthat no solution exists unless φ(x, y) is of a special form. Find this form and show that in such a

case there are infinitely many solutions.

Problem 2.6 (Inhomogeneous variable coefficient PDE). Consider the First Order PDE

yux + xuy − yu = y .

(a) What are the characteristic equations for this equation?

(b) Solve for the characteristic curves (for x and y). Either provide a parameterisation of the

characteristic curves, or as expressions constant on the curve.

(c) Find the general solution of the PDE.

(d) Find the solution when the condition u(x, 1) = ex is specified.

(e) For what region of the x-y plane does the solution in part (d) cover.

Problem 2.7 (First order inhomogeneous linear PDE). Solve ux+uy+u = ex+2y, with u(x, 0) = 0.

Problem 2.8 (First order linear PDE.). (a) Solve the equation yux+xuy = 0 with the condition

u(0, y) = e−y2

(b) In which region of the xy-plane is the solution uniquely determined?

Problem 2.9 (First order inhomogeneous linear PDE.). Solve ux+uy+u = ex+2y, with u(x, 0) = 0.

Problem 2.10 (Heuristic method for first order PDEs). (a) Consider the most general 1st or-

der PDE in two variables

F (x, y, u, ux, uy) = 0 (30)

with initial condition u(x, 0). Assuming we can solve for uy, give a heuristic derivation of the

solution to (30) using Taylor series.

22

(b) Apply your formula to find the solution to the equation aux + buy = 0 with initial condition

u(x, 0) = x3.

(c) Consider the semi-linear 1st order PDE

a(x, y)ux + b(x, y)uy + c(x, y, u) = 0 .

Show that under a coordinate transformation ξ = ξ(x, y), η = η(x, y), the equation is trans-

formed to an equation of a similar form where, in particular, the transformed b is given by

β(x, y) = a(x, y)ηx + b(x, y)ηy, and derive the equations for a characteristic curve in analogy

with the discussion in the lectures on 2nd order PDEs. Show that this notion is equivalent

to the notion of a characteristic curve introduced in the lectures on first order PDEs.

Problem 2.11 (A ‘Counterexample’.). Are all solutions of the PDE

x∂u

∂x= y

∂u

∂y(31)

functions of xy?

23

3 Non-Linear Scalar Conservation Laws

3.1 Shock-wave Example

Burgers’ equation models the time evolution of the concentration u of a quantity.

Consider the shock-wave equation, Burgers equation

ut + uux = 0

u(x, 0) =

1, x < −1,

−x, −1 ≤ x ≤ 1,

−1, x > 1.

Figure 4: First example of initial condition.

General solution by characteristics

Consider initial conditions u(x, 0) = u0(x).

The characteristics equations are

dx

dt= z(t)

dz

dt= 0,

where z(t) = u(x(t), t). The second equation follows from

u(x(t), t)

dt= ux(x(t), t)

dx

dt+ ut(x(t), t) = 0.

as u(x, t) is assumed to satisfy the PDE.

Initial conditions for the ODE x(t), z(t) are

x(0) = x0

z(0) = u0(x0)

24

Along a characteristic z (i.e u) is constant. So u(x(t), t) = u0(x0). Now we can solve for x(t)

dx

dt= u0(x0)

So

x(t) = x(0) + u0(x0)t

Characteristics are straight lines emanating from the x-axis with speed u0(x0). The solution

u(x, t) can now be obtained.

u(x(t), t) = u0(x0).

Hence, eliminating x0, we obtain,

u(x(t), t) = u0(x(t)− u(x(t), t) t).

This is true for each characteristic curve (i.e. starting from a different initial position x0), and so

we conclude that

u(x, t) = u0(x− u(x, t) t).

This is a non-linear equation for u(x, t) given an initial condition u0.

Often the solution is easier to obtain by just drawing the characteristics on the x, t plane, as

presented in Figure 5.

So our method doesn’t seem to work for t > 1, at least around the t-axis.

3.2 Advection Equations

Where do equation of this form come from? Consider the equation

ut + vux = 0 .

The characteristics are lines of the form

x(t) = x0 + v(x, t)t .

Solution of the PDE is given by

u(x, t) = f(x− vt) .

The solution profile is advected with speed v.

So the equation models advection of marker particles (but does not model conservation of

chemical species, pollutant etc.)

25

Figure 5: Solution obtained by characteristics (up to time t = 1).

3.3 Integral Conservation Laws

Consider a pipe or river with velocity of fluid given by v(x, t). Consider some chemical which has

a concentration u(x, t) kg/m at position x and time t. See Figures 6 and 7.

The amount of chemical in the region from x0 to x1 at time t0 is given by∫ x1x0

u(x, t0) dx .

26

Figure 6: Flow in a pipe or river.

Figure 7: Flux diagram.

At a later time t1, the amount of chemical in the section of river x0 to x1 is∫ x1x0

u(x, t1) dx .

The amount of chemical flowing past a point per second will be

{concentration} × {velocity} = uv (kg/s as u kg/m × v m/s) .

So we have (in words) the conservation law

27

{Amount of Chemical in

interval [x0, x1] at time t1

}=

{Amount of Chemical in

interval [x0, x1] at time t0

}

+

Amount of Chemical which

flows across x0 from time t0

to t1

−

Amount of Chemical which

flows across x1 from time t0

to t1

In mathematical notation (which allows us to manipulate using all our skills from Calculus) is∫ x1

x0

u(x, t1) dx =

∫ x1x0

u(x, t0) dx

+

∫ t1t0

v(x0, t)u(x0, t) dt

−∫ t1t0

v(x1, t)u(x1, t) dt .

This expression must hold for all choices of x0, x1, t0 and t1. We call this the integral form of

the conservation law.

3.4 PDE Conservation Laws

Rearranging the conservation law gives∫ x1x0

(u(x, t1)− u(x, t0)) dx+∫ t1t0

(v(x1, t)u(x1, t)− v(x1, t)u(x1, t)) dt = 0 .

If u and v are smooth enough, these differences can be rewritten using the Fundamental theorem

of Calculus to obtain ∫ x1x0

∫ t1t0

ut dt dx+

∫ t1t0

∫ x1x0

(vu)x dx dt = 0 .

Interchanging the order of integration we obtain∫ x1x0

∫ t1t0

ut + (vu)x dt dx = 0 ,

for all choices of x0, x1, t0 and t1. So surely for smooth enough u and v the quantity u must satisfy

the PDE

ut + (vu)x = 0 .

28

For more general problems the flux uv may be more complicated. But if f(u) is a flux function

for a conserved quantity u, then u (if smooth enough) satisfies the PDE

ut + f(u)x = 0

f is the flux function and u is the conserved quantity.

A PDE of this form is called a Conservation Law.

Example: Burgers equation Our previous example

ut + uux = 0

is not a PDE in conservation form. But the equation can be rewritten in the form

ut +

(1

2u2)x

= 0

is a conservation equation (for the quantity u).

As long as the solution is smooth, either form produces the same solution. But the non-

conservative form doesn’t make sense for non-smooth functions, but the conservative form is es-

sentially a short hand for the integral conservation equations which do make sense for non-smooth

functions.

You do have to be careful, the equation(1

2u2)t

+

(1

3u3)x

= 0

is another conservation law, where 12u2 is the conserved quantity. It turns out that solutions to this

conservation equation can be different to the conservation law for the quantity u.

Example: Burgers’ Equation Burgers equation is a model for the non-linear conservation

laws that arise in the equations for fluid flow, in particular the Navier-Stokes equation and the

momentum equation given by Euler equations.

ut +

(1

2u2)x

= 0.

For smooth solutions this is equivalent to the quasi-linear equation

ut + uux = 0

We can solve the conservation equation using characteristics.

29

But what happens when the characteristics intersect. (i.e. when we get discontinuities). Go

back to the derivation of the conservation law using the integral formulation∫ x1x0

u(x, t1) dx =

∫ x1x0

u(x, t0) dx

+

∫ t1t0

f(u(x0, t)) dt−∫ t1t0

f(u(x1, t)) dt

Figure 8: Domain associated with integral formulation of conservation law.

Suppose we start with an initial condition which is a step function, as shown in Figure 9.

u(x, 0) =

uL x ≤ 0uR x > 0Can we find a solution to our integral equations with this initial condition. This type of problem,

with a step function as initial conditions is called a Riemann problem.

Make a guess that the solution will remain a step function, but moves with speed s, as shown

in Figure 10.

Choose a very simple region to apply integral equation. let x0 = 0, x1 = s, t0 = 0 and t1 = 1,

see Figure 11

Then must have ∫ s0u(x, 1) dx =

∫ s0u(x, 0) dx (32)

+

∫ 10f(u(0, t)) dt−

∫ 10f(u(s, t)) dt (33)

Assumptions

30

Figure 9: Step function initial condition as found in Riemann problems.

• u(x, 1) = uL for 0 ≤ x ≤ s,

• u(x, 0) = uR for 0 ≤ x ≤ s,

• u(0, y) = uL for 0 ≤ y ≤ 1,

• u(s, y) = uR for 0 ≤ y ≤ 1.

So equation (33) implies

suL = suR + f(uL)− f(uR)

So

s =f(uL)− f(uR)

uL − uRThese are known as the Rankine Hugoniot conditions.

Example: Burgers Equation

ut +

(1

2u2)

= 0

So

s =

(12u

2L

)−(

12u

2R

)uL − uR

=1

2

(uL + uR)(uL − uR)(uL − uR)

=1

2(uL + uR)

Burgers equation with initial condition

u(x, 0) =

1 x ≤ 00 x > 031

Figure 10: Left of shock u = uL, right of shock u = uR.

s =1

2(uL + uR) =

1

2

What about

u(x, 0) =

0 x ≤ 01 x > 0Again the shock speed is

s =1

2(uL + uR) =

1

2

This shock solution does solve the integral equations, but it is a non physical shock (information

is generated from nowhere and it is unstable to slight perturbation in the initial conditions).

But there is another physically sensible solution, the rarefaction fan.

3.5 Rarefaction Fan

When the characteristics fan out (as opposed to collide as in the shock situation) it is reasonable

to assume that the solution takes the form u(x, t) = g(x/t) (where we have taken the origin to be

a the location of the initial jump in the initial conditions.

Substituting the assumption u(x, t) = g(x/t) into the quasi-linear form of the PDE, i.e.

ut + f′(u)ux = 0

leads to the requirement that g must satisfy

f ′(g(ξ)) = ξ.

32

Figure 11: Simple domain used to derive Rankine-Hugoniot condition.

Example: Burgers equation In the case of Burgers equation, where f ′(u) = u, this implies

that g(ξ) = ξ. Consequently the rarefaction fan solution for Burgers equation is u(x, t) = xt , in the

region joining up the constant states.

u(x, 0) =

0 x ≤ 01 x > 0In particular the solution is

u(x, t) =

0 x/t ≤ 0

x/t 0 < x/t < 1

1 x > 0.

3.6 Weak Solutions

Though the integral equations are the equations which are derived from the physical problem,

another formulation is useful, and in some sense easy to manipulate mathematically. This is the so

called weak formulation. Formally.

Definition 3.1 (Weak Solution). A function u ∈ L∞(R × R+) is a weak solution of the scalarconservation equation with initial condition u0 ∈ L∞ if∫ ∞

0

∫Ruφt + f(u)φx dx dt =

∫Ru0(x)φ(x, 0) dx

for all φ ∈ C10 (R× R+). (R+ = [0,∞))

33

Figure 12: A physically sensible shock is formed when characteristics intersect.

Here C10 (R × R+) is the space of C1 functions with support in the set R × R+. In the sequelwe will from time to time neglect the initial condition by considering test functions whose support

does not intersect t = 0.

3.7 Vanishing Viscosity Solution

We need a way to identify those weak solutions of the conservation law that are somehow physically

sensible. This is provided by the idea of vanishing viscosity solutions and its link to so called entropy

solutions.

Since the solutions of the conservation law are non smooth it is useful to consider a related

parabolic problem,

u�t + f(u�)x = �u

�xx,

the so called vanishing viscosity equation, in each the solutions u� will be smooth. We will be

investigating what happens as �→ 0.First it is clear that if u is the limit as � → 0 of the u�, then u satisfies the conservation law.

I.e. ∫∫u�φt + f(u

�)φt dx dt = �

∫∫u�xxφdx dt

= �

∫∫u�φxx dx dt→ 0,

34

Figure 13: Non physical shock solution of Riemann Problem.

for fixed φ as � → 0. This is possible since the term∫∫

u�φxx dx dt is well behaved for fixed φ,

because we can assume that u� is bounded for all �. We do need to ensure the u� → u but this usesa non trivial result that a set of bounded variation is compact in L1, so we just assume that we

have convergence. By the way, we have dropped the initial conditions by using test functions with

support disjoint from t = 0.

So ∫∫uφt + f(u)φt dx dt = 0.

Since this solution is the limit of unique solutions of the viscosity equations, we conclude that

this type of weak solution is unique.

3.8 Entropy Solutions

Another way to characterize unique weak solutions is via so-called entropy inequalities.

In the scalar case, we can consider any convex function U(u) as an entropy function. In the

smooth case, if u is the solution of the conservation law, then U(u) will also satisfy a conservation

law, in particular

U(u)t + F (u)x = 0

where F ′(u) = U ′(u)f ′(u) (The conservation law for U(u) is obtained by multiplying the non-

conservative form of the PDE by U ′(u). )

35

Figure 14: Rarefaction fan solution of Riemann Problem for Burgers Equation.

Now the equality can not be expected in the case of weak solutions. Amazingly, what we do

get is the following entropy inequality,

U(u)t + F (u)x ≤ 0

for any convex function U(u). So the quantity U(u) decreases (cannot increase) over time.

We are being a little perverse with the nomenclature, as surely if we call this an entropy

inequality, it should be increasing, but in the mathematical literature it is common to work with

convex “entropies” and so get a ≤ in the inequality.It turns out that if u satisfies the weak form of the conservation law and a weak entropy

inequality (for any convex U(u), then the solution is unique.

So what is the relationship between vanishing viscosity solutions and entropy solutions. For

scalar problems they are the same, as will be demonstrated in the next section.

There will be a corresponding Rankine Hugoniot type inequality in the case of piecewise smooth

solutions. Using the same argument as we used to derive the Rankine Hugoniot condition, we have

that ∫ x1x0

U(x, t1) dx ≤∫ x1x0

U(x, t0) dx+

∫ t1t0

F (x0, t) dt−∫ t1t0

F (x1, t) dt.

using the situation presented in Figure 11 we conclude that

sUl ≤ sUr + Fl − Fr,

or

s(Ul − Ur) ≤ Fl − Fr.

36

Here is is important to get the order left to right correct, as other wise the inequality changes.

3.9 Entropy Inequality and vanishing viscosity

Let u be a vanishing viscosity solution, i.e. a function u which is the limit is a suitable sense of

solutions u� of the corresponding parabolic problems

u�t + f(u�)x = �u

�xx.

Now (for a convex U(u)), multiply this equation by U ′(u) to obtain

U(u�)t + F (u�)t = �U

′(u�)u�xx

To derive useful estimates we need to “move” derivatives from u�xx. As in the weak case, we multiply

the equation with a suitable test function and integrate by parts. As we are expecting an inequality,

the test function should be non-negative, i.e. φ ≥ 0 and as usual it should be smooth enough andcompactly supported. Then∫∫

U(u�)tφ+ F (u�)tφdx dt = �

∫∫U ′(u�)u�xxφdx dt.

Integration by parts produces∫∫−U(u�)φt − F (u�)φt dx dt = �

∫∫U ′(u�)u�xxφdx dt

= −�∫∫

U ′′(u�)(u�x)2φdx dt− �

∫∫U ′(u�)(u�x)φx dx dt

≤ −�∫∫

U(u�)xφx dx dt

= �

∫∫U(u�)φxx dx dt

where the inequality is due to the positiveness of the integral∫∫

U ′′(u�)(u�x)2φdx dt. The last term

(for fixed φ) is well behaved, and so we can let �→ 0 to conclude that a vanishing viscosity solutionsatisfies the entropy inequality ∫∫

−U(u)φt − F (u)φt dx dt ≤ 0.

Indeed the converse is also true. By a compactness argument, it can be shown that the entropy

solutions are unique, and so must also be vanishing viscosity solutions.

3.10 Other Entropy Conditions

In this scalar case, if we want to test whether a weak solution is an entropy solution, we can choose

our favorite entropy function U(u) and test the entropy inequality. Let’s see what w can do by

choosing some special entropy functions.

37

In 1970 Kružkov [67] provided the definitive proof of existence of solutions of scalar conservation

laws. As part of that he used extensively an interesting class of entropy functions,

U(u) = |u− k|

for any scalar k, along with the corresponding Entropy fluxes,

F (u) = sign(u− k)(f(u)− f(k)).

With these functions we can derive some simple entropy conditions.

Rankine Hugoniot type entropy condition

Consider the case of a shock with speed s separating two constant states, ul and ur. Will this

constitute an entropy solution? Well let’s test it with these “Kružkov” entropy functions. We must

have

s(|ul − k| − |ur − k|) ≤ sign(ul − k)(f(ul)− f(k))− sign(ul − k)(f(ur)− f(k)). (34)

The regular Rankine Hugoniot condition is obtained by considering in turn, k > max(ul, ur) and

then k < min(ul, ur).

So for something new we need to consider k between ul and ur. Let’s consider the case ul >

k > ur. Then equation (34) reduces to

s(ul + ur − 2k) ≤ f(ul) + f(ur)− 2f(k). (35)

This implies

s(ul − ur + 2ur − 2k) ≤ f(ul)− f(ur) + 2f(ur)− 2f(k),

and using the expression for s given by the Rankine Hugoniot condition we conclude that

s(ur − k) ≤ f(ur)− f(k).

As ur < k we obtain

s ≥ f(ur)− f(k)ur − k

.

Working at the other end we conclude that if ul > ur then

f(ur)− f(k)ur − k

≤ s ≤ f(ul)− f(k)ul − k

(36)

for all k between ur and ul. Indeed since we are working also with convex flux functions, f(u), the

velocity function a(u) = f ′(u) is an increasing function of u and so condition (36) is equivalent to

a(ur) ≤ s ≤ a(ul)

38

which is just that the characteristics converge into the shock. Actually in this case of convex flux

is equivalent to ul > ur!

If we consider the case ur > ul we actually get the same result, condition (36). But for convex

flux f this is not possible if ur > ul.

So we have proved:

Theorem 3.2. A entropy satisfying shock separating state ul on the left, and state ur on the right,

must satisfy

f(ur)− f(k)ur − k

≤ s ≤ f(ul)− f(k)ul − k

for all k between ul and ur.

For convex flux functions this is equivalent to convergence of characteristics, i.e.

a(ur) ≤ s ≤ a(ul)

and indeed this is equivalent to ul > ur.

3.11 Total Variation, Monotonicity, L1 contraction

Based on Kružkov’s formulation of entropy, it is possible to prove the following, [40]

Theorem 3.3. Let u and v be L∞(R× R+) weak solutions of the conservation law, which satisfythe entropy conditions Ut + Fx ≤ 0 for all convex functions of the form U(u) = |u − k|, and fluxfunction F (u) = sign(u− k)(f(u)− f(k)) for all k ∈ R.

Then (in the weak sense)

∂

∂t|u− v|+ ∂

∂xsign(u− v)(f(u)− f(v)) ≤ 0.

This is proved by applying the entropy conditions, replacing k = v in the condition for u and

replacing k = u in the condition for v. And then integrating over a double integral.

This theorem then implies that∫R|u(x, t)− v(x, t)| dx ≤

∫R|u(x, 0)− v(x, 0)| dx

If we think of an evolution operator S(t) : L1(R) → L1(R) which maps u(·, 0) 7→ u(·, t), thenthis can be written as ∫

R|S(t)u0 − S(t)v0| dx ≤

∫R|u0 − v0| dx

So S(t) is a contraction mapping. We also have∫RS(t)u0 dx ≤

∫Ru0 dx

by conservation.

There is an amazing general theorem for L1 conservative mappings, which connects contraction,

and order preservation. Here it is.

39

Theorem 3.4 (Crandall-Tartar). Suppose C is a set of functions in L1 such that if f, g ∈ C thenmax(f, g) ∈ C. Suppose T is a mapping from C into L1 such that∫

RT f dx =

∫Rf dx.

Then the following three properties are equivalent:

1. T is order preserving, i.e. if f ≤ g then Tf ≤ Tg.

2.∫R(T f − T g)+ dx ≤

∫R(f − g)+ dx. (Here f+ = max(f, 0))

3.∫R |T f − T g| dx ≤

∫R |f − g| dx.

Our evolution operator S(t) satisfies the requirements of theorem 3.4 and satisfies property 3,and so must also satisfy properties 1 and 2. In particular the operator S(t) is order preserving.The contraction property also implies that the Total variation of the solution is bounded,

TV (u(·, t)) ≤ TV (u(·, 0))

by substituting v(x, t) = u(x+ ∆x, t) in the contraction property and then letting ∆x→ 0. Here

TV (f) = sup

{∫Rfφx dx, for all φ ∈ C10 (R), ‖φ‖L∞(R) ≤ 1

}but has the simpler form for smoother f

TV (f) =

∫R|fx| dx.

This means that the initial condition u0 is monotone, then the solution at later time must be

monotone. Essentially the solution can not form oscillations as time progresses.

3.12 Problems

Problem 3.1 (Shock waves.). Shock waves are often modeled by a first order (quasi-linear) PDE

of the form

a(u)ux + uy = c(x, y) . (37)

In this problem we will see that the method of characteristics can be applied to such equations as

well.

(a) Consider the equation

a(u)ux + uy = 0 . (38)

Obviously, any constant function u(x, y) = u0 is a solution of (38). Let u(x, y) = u0 +�v(x, y),

where � � 1. Find the most general solution of this form to (38), to first order in �, andargue that these represent waves moving with velocity a(u0) (where the variable y has the

interpretation of time).

40

(b) Use the method of characteristics to show that the most general solution of (38) is implicitly

given by

u = F (x− a(u)y) . (39)

Find the (implicitly defined) solution to (38) for the initial condition u(x, 0) = φ(x).

(c) Solve the initial value problem (for small y ≥ 0)

uux + uy = 0 , u(x, 0) = −x . (40)

Draw the lines in the (x, y)-plane along which the solution is constant. Do you see the

development of a ‘shock’ for y ≥ 0?

(d) Solve

uux + uy = 1 , u(x, x) =x

2. (41)

[Hint: It is convenient to choose the parameter s along the characteristic curves such that for

s = 0 we are at the intersection of the characteristic curve with the curve (x(t), y(t)) = (t, t)

for which u(x(t), y(t)) = t2 .]

Problem 3.2 (Riemann Problem for Fourth Power Flux.). Consider the conservation law

ut +

(1

32u4)x

= 0

with initial condition

u(x, 0) =

uL, x < 0,uR, x ≥ 0.For uL > uR we would expect a solution with a shock and for uL < uR a rarefaction fan.

(a) Find the Rankine-Hugoniot shock speed condition for this problem.

(b) Hence find the solution to the problem with the initial condition uL = 2, uR = −1. Draw apicture of the x-t plane showing representative characteristic curves and the shock curve for

this solution.

(c) Find the expression for the rarefaction fan for this problem. In particular find the function g

such that u(x, t) = g(x/t) is a solution of this PDE.

(d) Hence find the solution to the problem with the initial condition uL = −1, uR = 2. Draw apicture of the x-t plane showing representative characteristic curves and the rarefaction fan

for this solution.

41

4 The Riemann Problem

4.1 The Riemann Problem for Constant Coefficient System

The Riemann problem for the hyperbolic, constant coefficient system is a special initial value

problem∂U

∂t+ A

∂U

∂x= 0 −∞ < x 0, (42)

with

U(x, 0) = U(0)(x) =

{Ul if x < 0

Ur if x > 0.

The system is strictly hyperbolic with ordered real distinct eigenvalues

λ1 < λ2 < . . . < λm.

The structure of the solution of the Riemann problem (42) in the x− t plane is depicted in Figure68. It consists of m waves emanating from the origin, one for each eigenvalue λi. Each wave i

carries a jump discontinuity in U propagating with speed λi. Naturally, the solution to the left

of the λ1 wave is simply the initial data Ul and to the right of the λm wave is Ur. The task

at hand is to find the solution in the wedge between the λ1 and λm waves. As the eigenvectors

M(1),M(2), . . . ,M(m) are linearly independent, we can expand the data Ul, constant left state, and

Ur, constant right state, as linear combinations of the set M(1),M(2), . . . ,M(m), that is

Ul =m∑i=1

αiM(i), Ur =

m∑i=1

βiM(i) (43)

with constant coefficients αi, βi for i = 1, 2, . . . ,m. Formally, the solution of the initial value

problem (42) is given by (153) in terms of the initial data R(0)i (x) for the characteristic variables

and the right eigenvectors M(i). Note that each of the expansions in (43) is a special case of (153).

In terms of the characteristic variables we have m scalar Riemann problems for

∂Ri∂t

+ λi∂Ri∂x

= 0

with the initial data obtained by comparing (43) with (153), that is

R(0)i (x) =

{αi if x < 0

βi if x > 0

for i = 1, 2, . . . ,m. From the previous result, see equation (152), we know that the solution of these

scalar Riemann problems are given by

Ri(x, t) = R(0)i (x− λit) =

{αi if x− λit < 0βi if x− λit > 0

. (44)

42

We can thus write the final solution in terms of the original variables as

U(x, t) =m∑

i=I+1

αiM(i) +

I∑i=I

βiM(i) (45)

where the integer I = I(x, t) is the maximum value of the sub-index i for which x− λit > 0.The solution for a 2 × 2 system. As an example consider the Riemann problem for a general

2 × 2 linear system. From the origin (0, 0) in the (x, t) plane there will be two waves travelingwith speeds that are equal to the characteristic speeds λ1 and λ2 (λ1 < λ2) see Figure 67. The

solution to the left of dx/dt = λ1 is simply the data state Ul = α1M(1) + α2M

(2) and to the right

of dx/dt = λ2 the solution is the constant data state Ur = β1M(1) + β2M

(2). The wedge between

the λ1 and λ2 waves is usually called the star region and the solution there is denoted by U∗ ; its

value is due to the passage of two waves emerging from the origin of the initial discontinuity. From

the point P ∗(x, t) we trace back the characteristics with speed λ1 and λ2. These are parallel to

those passing through the origin. The characteristics through P ∗ pass through the initial points

x(2)0 = x − λ2t and x

(1)0 = x − λ1t. The coefficients in the expansion (45) for U(x, t) are thus

determined. The solution at point P ∗ has the form (45). It is a question of choosing the correct

coefficients αi or βi. Select a time t∗ and a point xl to the left of the slowest wave so U(xl, t

∗) = Ul,

see Figure 69. The solution at the starting point (xl, t∗) is obviously

Ul = Ul =2∑i=1

αiMi = α1M(1) + α2M

(2)

i.e. all coefficients are α’s, that is, the point (xl, t∗) lies to the left of every wave. As we move to the

right of (xl, t∗) on the horizontal line t = t∗ we cross the wave dx/dt = λi, hence x − λ1t changes

from negative to positive, see (44), and therefore the coefficient α1 above changes to β1. Thus the

solution in the entire star region, between the λ1 and λ2 waves, is

U∗(x, t) = β1M(1) + α2M

(2). (46)

As we continue moving right and cross the λ2 wave the value x − λ2t changes from negative topositive and hence the coefficient α2 in (45) and (46) change to β2, i.e. the solution to the right of

the fastest wave of speed λ2 is

U∗(x, t) = β1M(1) + β2M

(2).

From equation (43) it is easy to see that the jump in U across the whole wave structure in the

solution of the Riemann problem is

∆U = Ur −Ul = (β1 − α1)M(1) + · · ·+ (βm − αm)M(m).

It is an eigenvector expansion with coefficients that are strengths of the waves present in the

Riemann problem. The wave strength of wave i is βi−αi and the jump in U across wave i, denoted

43

by

(∆U)i = (βi − αi)M(i).

When solving the Riemann problem, sometimes it is more useful to expand the total jump ∆U =

Ur −Ul in terms of the eigenvectors and unknown wave strengths βi − αi.

Let ∆R = Rr −Rl or ∆U = Ur −Ul. Let ∆Ri = Rri −Rli be the jump in the ith characteristicvariable. The quantity ∆Ri is often called the strength or amplitude of the ith wave. The ith wave

creates a jump ∆Ri in the ith characteristic variable. Note that ∆R1 + ∆R2 + · · ·+ ∆Rn = ∆R =Rr −Rl.

5 Riemann Problem for The Shallow water wave equations

In the exact Riemann solver use is made of the fact that all the possible waves for the shallow water

wave equation that can be connected to state Ul are shown in Figure 15. The region surrounding Ul

is divided into four disjoint open regions; I, II, III and IV. Depending on the status of the adjoining

state, Ur it can be connected to state, Ul by either a 1-shock, S1, a 2-shock, S2, a 1-rarefaction

fan, R1 or a 2-rarefaction fan, R2. Because S1 and R1 represent back waves and S2 and R2 are

u

h

III

I

II

IV

U

lS u u s h: - = ( ;U )2 l 2

lR u u r h: - = ( ;U )1 l 1

lS u u s h: - = ( ;U )1 l 1

lR u u r h: - = ( ;U )2 l 2

l

Figure 15: Waves connecting state Ul.

front waves, the Riemann problem will always involve either an S1 or R1 and either an S2 or R2

wave. Depending in which region Ur lies, a different Riemann problem must be solved. These are

depicted in Figure 18 for zone I, Figure 19 for zone II, Figure 20 for zone III, and in Figure 21 for

zone IV. The characteristics and the depth and velocity profile for each case is also shown in their

respective figure for subcritical flows.

Denoting all the waves associated with the eigenvalue, λ1, S1 and R1 as back waves and those

associated with the other eigenvalue for the shallow water wave equations, λ2, S2 and R2 as front

44

Table 1: Wave structure connecting the left state, Ul with the right state, Ur located in the four zones

representing the possible solution to the Riemann problem.

Zone Left Wave Right Wave

I shock (S1) shock (S2)

II shock (S1) rarefaction (R2)

III rarefaction (R1) shock (S2)

IV rarefaction (R1) rarefaction (R2)

waves. For convenience, the solution to the the Riemann problem involves travelling from the left

state clockwise to the right state. Since λ2 < λ1 then the solution of the Riemann problem involves

travelling along S1 or R1 then along S2 and R2.

For an arbitrary flux function F(U) the following relationship between the shock speed, S and

the states Ul and Ur across the shock, called the Rankine-Hugoniot shock condition condition must

be satisfied[76]

F(Ul)− F(Ur) = S(Ul −Ur). (47)

For (??) with ST = [0, 0] this gives the following shock speeds

S =u2l hl − u2rhr + g/2

(h2l − h2r

)ulhl − urhr

and

S =ulhl − urhrhl − hr

.

Eliminating S between these expression, provides the relationship

ur − ul = ±

√g

2

[h2r − h2l

][hr − hl]

hrhl

across the shock that joins the two states, ul and ur.

Using Ul, Ur and Um, the intercell flux fj+1/2 at x = (j + 1/2)∆x can be calculated.

This is used in (123), which is the re-averaging step, to advance the solution in time.

In defining the Hugonoit locus above, the question of whether a given discontinuity is physically

relevant has not been considered. Lax[72] proposed a simple entropy condition for system of

equations that are genuinely nonlinear. This condition is known as Lax entropy condition, which

45

(a) (b)

(c) (d)

Figure 16: The family of shocks and rarefaction fans that are possible for the shallow water wave equations

(a) backward shock S1, (b) forward shock S2, (c) backward rarefaction fan R1 and (d) forward

rarefaction fan R2.

46

will ensure that non-physical solutions will not appear states that a jump in the p-th field is

admissible only if

λp(Ur) < S1 < λp(Ul)

Therefore, shocks must satisfy the entropy conditions, see Figure 17; for the 1-shock

S1 < λ1(Ul) λ1(Ur) < S1 < λ2(Ur) (49)

while for the 2-shock

λ1(Ur) < S2 λ1(Ul) < S2 < λ2(Ul). (50)

If these conditions are not satisfied then a non-physical solution is obtained. Violation of these

Figure 17: Entropy satisfying shock.

conditions will result in a rarefaction shock or an expansion shock.

Equations (47), (49), (50) and the Riemann invariants

um + 2√ghm = ul + 2

√ghl (51a)

and

um − 2√ghm = ur − 2

√ghr (51b)

provide the necessary conditions to solve all possible Riemann problems that could be encountered

for the shallow water wave equations. There are 16 possible combinations to connect state Ul with

state Ur and therefore to calculate the intercell flux. Most of these will involve the solution of a

non-linear equation.

This set of states is called the Hugoniot locus. There is a Hugoniot locus through any point

Um that gives the set of all points Ul that can be connected to Ur by an elementary p-wave. For

a general Riemann problem with arbitrary Ul and Ur, the solution consists of two discontinuities

travelling with speeds λ1 and λ2, with a new constant state, Um in between. The location of Um in

the pase plane must be where the 1-wave locus through Ul intersects the 2-wave locus through Ur.

In each case we travel from Ul to Ur by first going in the direction r1 and then in the direction r2.

47

u

h

Ul

Ul Ur

Ur

Um

Um

S1 R2

t

x

h

u

hl

ul

hm

um

hr

ur

t+ tD

x

x0

0

0

Figure 18: Waves connecting state Ul when Ur lies in zone I.

u

h

Ul

Ul

Ur

Ur

Um

Um

S1 S2

t

x

x

x

h

u

hl

ul

hm

um

hr

ur

t+ tD

0

0

0

Figure 19: Waves connecting state Ul when Ur lies in zone II.

48

u

h

Ul

Ul Ur

Ur

Um

Um

R1 S2

t

x

x

x

h

u

hl

ul

hm

um

hr

ur

t+ tD

0

0

0

Figure 20: Waves connecting state Ul when Ur lies in zone III.

u

h

Ul

Ul

Ur

Ur

Um

Um

R1

t

x

x

x

h

u

hl

ul

hm

um

hr

ur

t+ tD

0

0

0

R2

Figure 21: Waves connecting state Ul when Ur lies in zone IV.

49

This is required by the fact that λ1 < λ2 since clearly the jump between Ul and Um must travel

slower than the jump between Um and Ur if we are to obtain a unique solution.

The simple wave region, which is characterized by a rarefaction fan consists of straight charac-

teristics and curved characteristics emanating from the constant state region surrounding the simple

wave region. Along the curved characteristics u± 2c = α±. For the R1 wave along the curved C+

characteristic u + 2c = α+, see Figure 22. For the R2 wave along the curved C− characteristic

u− c = constant, see Figure 23.

Figure 22: Characteristics in zone III.

0x

t

R2 simple

wave regionS1

constantstate region

constantstate region

constantstate region

C+

C+

C-

C+

C- C+

C-

Figure 23: Characteristics in zone I.

Two shocks occur in zone II. The characteristics for this case are shown in Figure 24.

In zone IV there are two rarefaction fans, see Figure 25. In the constant state region the

quantities, u+ 2c along C+ and u− 2c along C− are equal.If Ur lies in zone I then from Figure 18 the waves separating the states are

UlS1→ Um

R2→ Ur

The Riemann problem involves solving the 1-shock, S1 wave followed by the R2 rarefaction fan.

Ur lies in zone I if hr ≥ hl and

−2(√

ghr −√ghl

)≤ ul − ur <

√g

2

[h2l − h2r

][hl − hr]

hlhr

50

t

x

S1

S2

C

C

C

C

0

Figure 24: Characteristics in zone II.

Figure 25: Characteristics in zone IV.

In this region

ul − um =

√g

2

[h2m − h2l

][hm − hl]

hmhl

and

ur − um = 2(√

ghr −√ghm

)should be solved.

Ur lies in zone II when

ur − ul ≥

√g

2

[h2m − h2l

][hm − hl]

hmhl

then from Figure 19 the states

UlS1→ Um

S2→ Ur

and the following equations must be solved

ul − um =

√g

2

[h2m − h2l

][hm − hl]

hmhl(52)

with

um − ur =

√g

2

[h2m − h2r ] [hm − hr]hmhr

(53)

51

for hm using some nonlinear solver. Substituting hm into (52) or (53) will give hm and the inter-

mediate state, Um.

If Ur lies in zone III if hr < hl and

−2(√

ghl −√ghr

)≤ ul − ur <

√g

2

[h2l − h2r

][hl − hr]

hlhr.

Then from Figure 20 the states

UlR1→ Um

S2→ Ur

and the following equations must be solved

um − ur =

√g

2

[h2m − h2r ] [hm − hr]hmhr

with

ul − um = −2(√

ghl −√ghm

).

In zone IV, then the states are

UlR1→ Um

R2→ Ur.

from Figure 21. Ur lies in zone IV when hr > hl and ul − ur < −2(√ghl −

√ghr)

or hr ≤ hl andul−ur < −2

(√ghr −

√ghl). If either of these conditions are satisfied, then the following equations

must be solved

ul − um = −2(√

ghl −√ghm

)with

ur − um = 2(√

ghr −√ghm

).

There are four possible outcomes for zone I and III, three for zone II and five for zone IV. For zone

I, the four possibilities are depicted in Figure ??. The cases are;

• fj+1/2 = f(Um),

• S1 ≥ 0→ fj+1/2 = f(Ul),

• ur +√ghr ≤ 0→ fj+1/2 = f(Ur) and

• um +√ghm < 0 < ur +

√ghr → u = (ur − 2cr + 2x/t)/3, c = (−ur + 2cr + x/t)/3 then

fj+1/2 = f(u, h).

For zone II the three possibilities are shown in Figure ??. These result in;

52

(a) (b)

(c) (d)

Figure 26: Estimating the correct flux fj+1/2 in zone I for the Riemann problem.

• fj+1/2 = f(Um),

• S1 ≥ 0→ fj+1/2 = f(Ul) and

• S2 ≤ 0→ fj+1/2 = f(Ur).

For zone III, there are four possible outcomes. From Figure ??;

• fj+1/2 = f(Um),

• ul −√ghl ≥ 0→ fj+1/2 = f(Ul),

• S2 ≤ 0→ fj+1/2 = f(Ur) and

• ul −√ghl < 0 < um −

√ghm → u = (ul + 2cl + 2x/t)/3, c = (ul + 2cl − x/t)/3 then

fj+1/2 = f(u, h).

Finally in zone IV there are five possible outcomes, which are shown in Figure ??. For

53

(a) (b)

(c)

Figure 27: Estimating the correct flux fj+1/2 in zone II for the Riemann problem.

2

1

j+1/2 S

R

x

t

f

l

m

r

0

2

1

j+1/2 S

R

x

t

f

lm

r

0

(a) (b)

2

1

j+1/2

S

R

x

t

f

l

m

r

0

2

1

j+1/2 S

R

x

t

f

l

m

r

0

(c) (d)

Figure 28: Estimating the correct flux fj+1/2 in zone III for the Riemann problem.

54

2

1

j+1/2R

R

x

t

f

l

m

r

0

2

1j+1/2

R

R

x

t

f

l

m

r

0

(a) (b)

2

1

j+1/2 R

R

x

t

f

l

m

r

0

2

1

j+1/2 R

R

x

t

f

l

m

r

0

(c) (d)

2

1 j+1/2

R

R

x

t

f

l

m r

0

(e)

Figure 29: Estimating the correct flux fj+1/2 in zone IV for the Riemann problem.

• fj+1/2 = f(Um),

• ur +√ghr ≤ 0→ fj+1/2 = f(Ur),

• ul −√ghl ≥ 0→ fj+1/2 = f(Ul),

• ul −√ghl < 0 < um −

√ghm → u = (ul + 2cl + 2x/t)/3, c = (ur + 2cr − x/t)/3 then

fj+1/2 = f(u, h).

55

Figure 30: Piecewise linear MUSCL reconstruction using three consecutive cells.

5.1 The Dam Break Problem

The dam-break problem is an idealized instantaneous collapse of a dam in a wide horizontal fric-

tionless rectangular channel. The resulting wave is considered to be one-dimensional.

5.2 Stoker’s Solution

Stoker (1905-1992)[114], an American Engineer, extended Ritter’s dam-break solution, which is

described in this Section, to the propagation of a dam-break wave into a region of still tail-water.

He derived analytical expressions for the free-surface profile and for the wave-front speed and depth

in terms of the initial upstream and downstream water depths. The water on both sides of the dam

is assumed to be at stagnant. At time t = 0s, the dam is suddenly removed.

Figure 31: Stoker’s solution to the behaviour of the dam break problem at time t > 0 with stagnant water

downstream.

56

Figure 32: Behaviour of the dam break problem at time t = ti with stagnant water downstream.

For t > 0s, the resulting water surface profile has four distinct regions, see Figure 31.

Each region is separated either by a (i) shock wave or (ii) rarefaction fan, also known as a

rarefaction fan. The region of undisturbed water in region (1), where u1 = 0m/s is separated from

the simple wave region (3), which has a parabolic free-surface by a characteristic given by

x = −c1t. (54)

The simple wave region is separated from the constant wave region (2), were the water depth

remains constant. The characteristic separating these regions is given by

x = (u2 − c2)t. (55)

Finally a sharp front connects the third region to the downstream undisturbed region. The dis-

continuity between states (0) and (2) is a surge moving at the speed ξ. The moving surge can be

reduced to a stationary surge or hydraulic jump by having the co-ordinate system moving at the

shock speed ξ. In this case, the conservation of mass through a hydraulic jump is given by

h2(ξ − u2) = h0(ξ − u0). (56)

Since u0 = 0m/s and c =√gh then

c20c22

=h0h2

=ξ − u2ξ

. (57)

The conservation of momentum across a hydraulic jump is

1

2gh22 −

1

2gh20 = h2(ξ − u2)(u2 − u0).

57

Figure 33: Behaviour of the dam break problem at time t = ti with stagnant water downstream.

Making use of (56) then

2ξ(ξ − u2) = (c22 + c20). (58)

Eliminating c2 between (57) and (58) and solving the resulting quadratic equation for the velocity

of the fluid in region (2) u2, then

u2 = ξ −c204ξ

(1 +

√1 +

8ξ2

c20

). (59)

Eliminating u2 between (57) and (58) and solving the resulting quadratic equation for c2, then

c22c20

=h2h0

=1

2

√1 +

8ξ2

c20− 1

2. (60)

To solve these equations for c2 (or h2) and u2, ξ or a relationship between c2 (or h2) and u2 must be

known. The additional relationship is obtained by considering the characteristic originating from

the simple wave region (2). Along these characteristics

u2 + 2c2 = 2c1. (61)

Using this relationship with (57) and (58) provides the following non-linear expression for the speed

of the surge, ξ in terms of the known quantities c0 and c1.

c1 =ξ

2− c

20

8ξ

[1 +

√1 +

8ξ2

c20

]+

[c202

√1 +

8ξ2

c20− c

20

2

]1/2. (62)

58

Solving this equation for ξ then (59) and (60) can be used to obtain u2 and c2 (and h2) respectively.

In the centered simple wave region or rarefaction fan, region (3)

dx

dt=x

t= u3 − c3 (63)

combined with

u3 + 2c3 = 2c1

gives

u3 =2

3

(c1 +

x

t

)(64)

and

c3 =2

3c1 −

x

3t. (65)

Given the initial condition h0, h1, u0 and u1 at t = 0s, then the water surface profile and the velocity

at any time t > 0s, can be determined using (54) to (65). When h0 = 0m/s there is no hydraulic

jump. At x = 0m, from (65) c3 = 2c1/3 or h3 = 4h1/9. This can only occur if the rarefaction

fan straddles the t−axis, so that u2 − c2 > 0. At x = 0m, from (64) and (65) u3 = c3 andFr = 1, which corresponds to critical flow conditions. The water depth at the location of the dam

remains constant for all time and the depth only depends on the upstream quiescent flow depth,

h1. Equation (65) indicates that the water surface profile for −c1t < x < (u2 − c2)t is also onlydependent on the upstream water depth. This is the water profile within the simple wave region

or rarefaction fan, see Figure 31.

More recently, Wu et al.[133] derived a single algebraic expression for the depth of the shock,

h2 which depends only on the ratio of the initial upstream and downstream water depths, h0 and

h1.

Recalling that u0 = 0m/s, then from (56)

ξ =h2u2h2 − h0

. (66)

Substituting (61), c1 =√gh1 and c2 =

√gh2 into (66) so that

ξ =2h2

h2 − h0

(√gh1 −

√gh2

)which is used to eliminate ξ in (60) so that

h2h0

=1

2

√1 + 8

(2h2

h2 − h0

√gh1 −

√gh2√

gh0

)2− 1

2

where c0 =√gh0. Rearranging and dividing by h

31 to yield(

h2h1

)3− 9

(h2h1

)2(h0h1

)+ 16

(h2h1

)3/2(h0h1

)−(h2h1

)(h0h1

)(h0h1

+ 8

)+

(h0h1

)3= 0 (67)

59

which shows that the height of the shock, h2 is only dependent on the upstream and downstream

flow depths, h1 and h2 respectively. This equation is much simpler to solve than the iterative

solution of (62) followed by the solution of (60) for h2. The monotone relationship between h2/h1

and h0/h1, given by (67) is illustrated in Figure 34.

Figure 34: Relationship between the ratio of the water depth at the shock and the upstream water depth,

h2/h1 to the ratio of the upstream water depth and the downstream water depth, h1/h0 and

to the shock Froude number, Fr2 for the dam break problem.

Equation (67) can be used to establish the maximum height of the shock wave and the ratio

h0/h1 that this will occur. These quantities have been plotted in Figure 35. The maximum height of

the shock wave h2−h0 ≈ 0.133043h1 and this occurs when the ratio of the upstream to downstreamwater depth h0/h1 ≈ 0.175692. This is a useful results because it indicates the water level in thereservoir upstream of a dam that should be avoided in the event of a breach of a dam.

Figure 35: Relationship between the ratio of the shock height and the upstream water depth, (h2−h0)/h1to the ratio of the upstream water depth and the downstream water depth, h1/h0 for the dam

break problem.

60

Although the dam break will result in a shock traveling downstream when there is a finite water

depth downstream, the shock does not necessarily result in supercritical flows downstream of the

breach. From (61)

Fr2 =u2c2

= 2c1c2− 2 = 2

√h1h2− 2.

The Froude number for a range of water depth ratio, h2/h1 are also plotted in Figure 34. When

Fr2 = 1, h2/h1 = 4/9 or h0/h1 ≈ 0.1379 and critical flow exist below the breach. When h0/h1 <0.1379 the flow downstream of the dam is supercritical and when h0/h1 > 0.1379 the downstream

flow remains subcritical.

Figure 34 can be used to establish the upstream to downstream water depth that is required

to produce downstream flow with the desired Froude number. It is interesting to note that the

maximum loss in energy across the jump, shown in Figure ?? does not coincide with the maximum

shock height shown in Figure 35.

5.3 Ritter’s Dry Bed Solution

Ritter, a German Engineer in 1892[101] considered a channel of infinite length, horizontal bott