Solution of the St Venant
Equations / Shallow-Water
equations of open channel flow
Dr Andrew Sleigh
School of Civil Engineering
University of Leeds, UK
www.efm.leeds.ac.uk/CIVE/UChile
Shock Capturing Methods


Ability to examine extreme flows
Changes between sub / super critical




Other techniques have trouble with trans-critical
Steep wave front
Front speed
Complex Wave interactions


Alternative – shock fitting
Good, but not as flexible
More recent



Developed from work on Euler equations in
the aero-space where shock capturing is very
important (and funding available)
1990s onwards
Euler equations / Numerical schemes:


Roe, Osher, van Leer, LeVeque, Harten, Toro
Shallow water equations

Toro, Garcia-Navarro, Alcrudo, Zhao
Books



E.F. Toro. Riemann Solvers and Numerical Methods
for Fluid Dynamics. Springer Verlag (2nd Ed.) 1999.
E.F. Toro. Shock-Capturing Methods for FreeSurface Flows. Wiley (2001)
E.F. Toro. Riemann Problems and the WAF Method
for Solving Two-Dimensional Shallow Water
Equations. Philosophical Transactions of the Royal
Society of London. A338:43-68 1992.
Dam break problem

The dam break problem can be solved

It is in fact integral to the solution technique
Conservative Equations

As earlier, but use U for vector
U t  F U
 A
U  
Q 
Q



F U    Q 2
 gI 1 

 A

x
 S U

0

S U   
 gI  gA S  S
o
f
 2



I1 and I2

Trapezoidal channel

Base width B, Side slope SL= Y/Z
h dS L 
 1 dB
I2  h 


2
dx
3
dx


S 
B
I1  h   h L 
3 
 2
2
2

Rectangular, SL = 0
2
I1 

h B
2
Source term

A
2
I2 
2B
Sf 
QQn
2
A R
2
(4 / 3)
A
2
2B
dB
2
dx
Rectangular Prismatic

Easily extendible to 2-d
 h 
U  
 hu 

0

S U   
 gh S  S
o
f

hu


F U    2 1
2
hu  gh


2
Integrate over control volume V
 U dx  F U dt    S U dU
V
V



2-dimensions

In 2-d have extra term:
U t  F U
 h 
 
U  hu
 
 hv 



hu
 2 1
2
F U    hu  gh 
2


huv


friction
S fx 
n
h
 x  G U  y



hu



G U   
huv
 2 1
2
hv

gh


2


2
(1 / 3 )
 S U
u u v
2
2

0

S U    gh S o x  S f x

 gh S o  S f
y
y











Normal Form

Consider the control volume V, border Ω
n

V
x

U

t
V
dV 
 H U d    S U dU

V
H U   cos  F U   sin  G U   n1 F U   n 2 G U

Rotation matrix

H(U) can be expressed
H U   n1 F U   n 2 G U   T F T U
1
1

T  0
0

0
cos 
 sin 


sin  
cos  
0
T
1
1

 0
0

0
cos 
sin 


 sin  
cos  
0

 h 
 
T U   h uˆ
 
 h vˆ 
Finite volume formulation

Consider the homogeneous form

i.e. without source terms
 U dx  F U dt   0
V

And the rectangular control volume in x-t space
t
tn + 1
F i+ 1 /2
F i-1 /2
x i-1 /2
i
x i+ 1 /2
x
Finite Volume Formulation

The volume is bounded by


xi+1/2 and xi-1/2
tn+1 and tn
The integral becomes
t n 1

xi 1 / 2 U  x , t n 1 dx  xi 1 / 2 U  x , t n dx   tn F U  x i 1 / 2 , t dt 
xi 1 / 2
xi 1 / 2

t n 1
tn
F U  x i 1 / 2 , t dt 

Finite Volume Formulation

We define the integral averages
U
n
i

F i 1 / 2 

1
x 
x i 1 / 2
1
t 
t n 1
tn
U  x , t n dx
xi 1 / 2
n 1
Ui

Fi  1 / 2 
F U  x i 1 / 2 , t dt
1
x
1
t
U  x , t n 1 dx

xi 1 / 2

t n 1
x i 1 / 2
tn
F U  x i 1 / 2 , t dt
And the finite volume formulation becomes
U
n 1
i
U
n
i

t
x
 F i  1 / 2  F i 1 / 2 
Finite Volume Formulation

Up to now there has been no approximation

The solution now depends on how we
interpret the integral averages

In particular the inter-cell fluxes

Fi+1/2 and F1-1/2
Finite Volume in 2-D

The 2-d integral equation is

U

t
V

dV 
 H U d    S U dU

V
H(u) is a function normal to the volume
Fn s 

A s 1
As
H U d  
 n F U   n G U d 
A s 1
As
1
2
Finite Volume in 2-D

Using the integral average
Ui 


1
V
 U dV
V
Where |V| is the volume (area) of the volume
then
Fn
U
n 1
i
U
n
i

t
V
N
 Fn
s 1
s
Finite Volume in 2-D

If the nodes and sides are labelled as :
A2
L1
L2
A1
V
A3
L4
L3
U

n 1
i
U
n
i

t
V
A4
 Fn s1 L1  Fn s 2 L 2
 Fn s 3 L 3  Fn s 4 L 4 
Where Fns1 is normal flux for side 1 etc.
FV 2-D Rectangular Grid
y

For this grid
G j+ 1 /2
i+ 1 /2 ,j+1 /2
i-1 /2 ,j+ 1/2
F i-1 /2
F i+ 1 /2
i, j
i+ 1 /2 ,j-1 /2
i-1 /2 ,j-1 /2
G j-1 /2
x

Solution reduces to
n 1
Ui
Ui 
n
t
x
F
i 1 / 2 , j

 Fi  1 / 2 , j 
t
y
G
i , j 1 / 2
 G i , j 1 / 2

Flux Calculation

We need now to define the flux
Many flux definitions could be used to that
satisfy the FV formulation

We will use Godunov flux (or Upwind flux)

Uses information from the wave structure of
equations.

Godunov method

Assume piecewise linear data states
n+ 1
F i-1 /2
F i+ 1 /2
C ells
n
i-1
i-1
i
D ata state s
U (0 ) i-1 /2
U (0 ) i+ 1 /2
n+ 1
n

Means that the flux calc is solution of local
Riemann problem
Riemann Problem

The Riemann problem is a initial value
problem defined by
U t  F U
x
0
 U in if x  x i  1 / 2
U x, tn    n
U i  1 if x  x i  1 / 2

Solve this to get the flux (at xi+1/2)
FV solution

We have now defined the integral averages
of the FV formulation
U

n 1
i
U
n
i

t
x
 F i  1 / 2  F i 1 / 2 
The solution is fully defined

First order in space and time
Dam Break Problem

The Riemann problem we have defined is a
generalisation of the Dam Break Problem
D a m w all
D eep w a ter at rest
S hallo w w a ter at rest
Dam Break Solution

Evolution of solution
h
W ater le v els at tim e t= t*
x
v
V elo city at tim e t= t*
x
R are factio n
t
S ho ck
t*
x

Wave structure
Exact Solution

Toro (1992) demonstrated an exact solution

Considering all possible wave structures a
single non-linear algebraic equation gives
solution.
Exact Solution

Consider the local Riemann problem
U t  F U  x  0
 h 
 
U  hu
 
 hv 
U L if x  0
U  x ,0   
U R if x  0



hu
 2 1
2
F U    hu  gh 
2


huv


Wave structure
S tar regio n, h *, u *
L e ft w a ve ( u -c)
S hea r w a v e
t (u)
vL
R ight w a ve, ( u+ c)
vR
h L, u L, v L
hR, uR, vR
0
x
Possible
Wave structures

Across left and
right wave h, u
change v is
constant
S hea r w a v e
L e ft R are fa ctio n
R ight S ho c k
t
x
L e ft S ho c k
S hea r w a v e
R ight R are fa ctio n
t
x
L e ft R are fa ctio n
S hea r w a v e
R ight R are fa catio n
t

Across shear
wave v changes,
h, u constant
x
L e ft S ho c k
R ight S ho c k
S hea r w a v e
t
x
Determine which wave

Which wave is present is determined by the
change in data states thus:




h* > hL
h* ≤ hL
left wave is a shock
left wave is a rarefaction
h* > hR
h* ≤ hR
right wave is a shock
right wave is a rarefaction
Solution Procedure

Construct this equation
f h   f L h , hL   f L h , hL    u

And solve iteratively for h (=h*).

The functions may change each iteration
f(h)

The function f(h) is defined
f h   f L h , hL   f L h , hL    u
u  uR  uL


 2 gh  gh L

fL  
1  h  hL


h

h
g 
L

2  hh L



And u*
u 
*

 if



 2 gh  gh R

fR  
1  h  hR


h

h
g
R

2  hh R

1
2
u L
 uR  
h  h L ( rarefactio n )
if
h  h L ( shock )
h  h R ( rarefactio n )
if

 if


1
2
h  h R ( shock )
 f h
R
*


, hR  f L h , hL
*

Iterative solution

The function is well behaved and solution by
Newton-Raphson is fast




(2 or 3 iterations)
One problem – if negative depth calculated!
This is a dry-bed problem.
Check with depth positivity condition:
 u  u R  u L  2 c L  c R 
Dry–Bed solution



Dry bed on one side
of the Riemann
problem
Dry bed evolves
Wave structure is
different.
t
D ry b ed
W et b ed
x
t
D ry b ed
W et b ed
x
D ry b ed
t
W et b ed
W et b ed
x
Dry-Bed Solution

Solutions are explicit


Need to identify which applies – (simple to do)
Dry bed to right
u* 

3
u L
 2cL 
c* 
1
c* 
1
3
Dry bed to left
u* 

1
1
3
u R
 2cR 
Dry bed evolves

u L
3
 2cL 
 u R
h*  c / g
 2cR 
h* = 0 and u* = 0
Fails depth positivity test
2
*
Shear wave


The solution for the shear wave is straight
forward.

If vL >

Else
0
v* = vL
v* = vR
Can now calculate inter-cell flux from h*, u*
and v*

For any initial conditions
Complete Solution


The h*, u* and v* are sufficient for the Flux
But can use solution further to develop exact
solution at any time.

i.e. Can provide a set of benchmark solution

Useful for testing numerical solutions.

Choose some difficult problems and test your
numerical code again exact solution
Complete Solution
h
W ater le v els at tim e t= t*
x
v
V elo city at tim e t= t*
x
R are factio n
t
S ho ck
t*
x
Difficult Test Problems

Toro suggested 5 tests
Test No.
hL (m)
uL (m/s)
hR (m)
uR (m/s)
1
1.0
2.5
0.1
0.0
2
1.0
-5.0
1.0
5.0
3
1.0
0.0
0.0
0.0
4
0.0
0.0
1.0
0.0
5
0.1
-3.0
0.1
3.0
Test 1: Left critical rarefaction and right shock
Test 2: Two rarefactions and nearly dry bed
Test 3: Right dry bed problem
Test 4: Left dry bed problem
Test 5: Evolution of a dry bed
Exact Solution

Consider the local Riemann problem
U t  F U  x  0
 h 
 
U  hu
 
 hv 
U L if x  0
U  x ,0   
U R if x  0



hu
 2 1
2
F U    hu  gh 
2


huv


Wave structure
S tar regio n, h *, u *
L e ft w a ve ( u -c)
S hea r w a v e
t (u)
vL
R ight w a ve, ( u+ c)
vR
h L, u L, v L
hR, uR, vR
0
x
Returning to the Exact Solution


We will see some other Riemann solvers that
use the wave speeds necessary for the exact
solution.
Return to this to see where there come from
Possible
Wave structures

Across left and
right wave h, u
change v is
constant
S hea r w a v e
L e ft R are fa ctio n
R ight S ho c k
t
x
L e ft S ho c k
S hea r w a v e
R ight R are fa ctio n
t
x
L e ft R are fa ctio n
S hea r w a v e
R ight R are fa catio n
t

Across shear
wave v changes,
h, u constant
x
L e ft S ho c k
R ight S ho c k
S hea r w a v e
t
x
Conditions across each wave
t

Left Rarefaction wave
t
Right bounding characteristic
Left bounding characteristic
dx / dt  u *  c *
dx / dt  u L  c L
hL, uL
h*, u*
x



Smooth change as move in x-direction
Bounded by two (backward) characteristics
Discontinuity at edges
Crossing the rarefaction

We cross on a forward characteristic
u  2 c  constant

States are linked by:
u L  2 c L  u *  2 c*

or
u *  u L  2 c L  c * 
Solution inside the left rarefaction
dx

The backward characteristic equation is dt  u  c
For any line in the direction of the rarefaction

Crossing this the following applies: u L  2 c L  u  2 c

1
dx 
c   u L  2cL 

3
dt 

Solving gives

On the t axis dx/dt = 0
c
1
3
u L
 2cL 
u 
u
1
3
u L
1
dx 
u

2
c

 L

L
3
dt 
 2cL 
Right rarefaction


dx
Bounded by forward characteristics
Cross it on a backward characteristic
u R  2 c R  u *  2 c*
In rarefaction

If Rarefaction crosses axis
c
1
3
c
 u R
 2cR 
dt
u *  u R  2 c *  c R 
1
dx 

u

2
c



R
R
3
dt 

uc
u 
1
3
u 
u R
1
dx 
 u R  2cR  2

3
dt 
 2cR 
Shock waves



Two constant data states are separated by a
discontinuity or jump
Shock moving at speed Si
Using Conservative flux for left shock
UL
 hL 


h
u
 L L
 h* 
U*  

h
u
 * *
Conditions across shock

Rankine-Hugoniot condition
F U *   F U L   S i U *  U L 

Entropy condition
 i U L   S i   i U * 

λ1,2 are equivalent to characteristics.

They tend towards being parallel at shock
Shock analysis


Change frame of reference, add Si
uˆ *  u *  S L
uˆ L  u L  S L
 hL 
Uˆ L  

ˆ
h
u
 L L
 h* 
ˆ
U*  

ˆ
h
u
 * *
Rankine-Hugoniot gives
h* uˆ *  h L uˆ L
h* uˆ 
2
*
1
2
gh
2
*
 h L uˆ 
2
L
1
2
2
gh L
Shock analysis

Mass flux conserved

From eqn 1
M
L
Using
uˆ *  M
L
/ h*
uˆ L  M

L
L
M
L
also
uˆ *  uˆ L  u *  u L
M
 h*2  h L2 

  g 
2  uˆ *  uˆ L 
1
M

 h* uˆ *  h L uˆ L
/ hL

1
2

 h*2  h L2 

  g 
2  u *  u L 
1
L

g h*  h L h* h L
Left Shock Equation

Equating gives
u *  u L  f L  h* , h L 

Also
S L  uL  aLqL
f L  h* , h L    h*  h L 
qL 


1  h*  h L h* 


2


2
hL

 h  hL
g  *
2  h* h L
1




Right Shock Equation

Similar analysis gives
u *  u R  f R  h* , h R 

Also
S R  uR  aRqR
f R  h* , h R    h*  h R 
qR 


1  h*  h R h* 


2

2 
hR

 h  hR
g  *
2  h* h R
1




Complete equation

Equating the left and right equations for u*
u *  u L  f L  h* , h L 
u *  u R  f R  h* , h R 
u R  u L  f L h , hL   f R h , hL   0

Which is the iterative of the function of Toro
f h*   f L h* , h L   f R h* , h L    u  0
Approximate Riemann Solvers


No need to use exact solution
Expensive

Iterative

When other equations, exact may not exist

Many solvers
Some more popular than others

Toro Two Rarefaction Solver


Assume two rarefactions
Take the left and right equations
u *  u L  2 c L  c * 

u *  u R  2 c *  c R 
Solving gives
c* 

uL  uR
4

cL  cR
2
For critical rarefaction use solution earlier
Toro Primitive Variable Solver

Writing the equations in primitive variables

Non conservative
h
W   
u 
W t  A  W W x  0

h

u
Approximate A(W) by a constant matrix
W 
1
2

u
A W   
g
W R
 WL 
Gives solution
h* 
h L
 hR 
2

u R
 u L  h L  h R 
4 c L  c R 
u* 
u L
 uR 
2

h R
 h L c L  c R 
4 h L  h R 
Toro Two Shock Solver

Assuming the two waves are shocks
h* 
u* 
qL 

1
2
q L hL  q R hR  u L  u R
qL  qR
1
 u L  u R    h*  h R q R   h*  h L q L 
2
g  ho  h L 
2 ho h L
qR 
g  ho  h R 
2 ho h R
Use two rarefaction solver to give h0
Roe’s Solver


Originally developed for Euler equations
Approximate governing equations with:
U t  F U

Where
uL
u~ 
~
A
hL  u R
hL 
hL
x
 U t  AU
x
~
 U t  AU x
is obtained by Roe averaging
hR
~
h 
hL hR
c~ 
1
2
c
2
L
 cR
2

Roe’s Solver

Properties of matrix

Eigen values

Right eigen vectors


Wave strengths
Flux given by
~
1  u~  c~
~
1  u~  c~
~ (2)  1 
R
 ~ ~
u  c 
~ (1 )  1 
R  ~ ~
u  c 
~


1
h
~1    h  ~  u 

2 
c

Fi  1 / 2 
1
2
F
n
i
F
~


1
h
~ 2    h  ~  u 

2 
c

n
i 1
2
   ~
2
1
j 1
~ ~  j
j
j R
 h  hR  hL
HLL Solver




Harten, Lax, van Leer
Assume wave speed
Construct volume
Integrate round
SL 
t
xL
SR 
t
xR
t
U*
FL
FR
F*
t
UL
UR
x
xL
 xR
 x LU L   x RU R    x L   x R U *   tU R   tU L  0
U *   S R U R  S LU L  F L  F R  /  S R  S L 

Alternative gives flux
F*   S R F L  S L F R  S L S R U R  U L  /  S R  S L 
HLL Solver

What wave speeds to use?


Free to choose
One option:
S L  min u L  c L , u TR  c TR

For dry bed (right)

Simple, but robust

S R  min u R  c R , u TR  c TR
S L  u L  cL
S R  u R  2cR

Higher Order in Space



Construct Riemann problem using cells
further away
Danger of oscillations
Piecewise
reconstruction
L
R
i-1

i
Limiters
i+ 1
i+ 2
i+ 1
i+ 2
L
R
i-1
i
Limiters

Obtain a gradient for variable in cell i, Δi
UL  Ui 


2
 x i
U R  U i 1 
1
2
 x  i 1
Gradient obtained from Limiter functions
Provide gradients at cell face
 i 1 / 2  a 

1
u i 1  u i
xi 1 / 2  xi
Limiter Δi =G(a,b)
 i 1 / 2  b 
u i  u i 1
xi  xi 1
Limiters

A general limiter
 max 0 , min   a , b , min  a ,  b 
G a , b   
 min 0 , max   a , b , max  a ,  b 




for a  0
for a  0
β=1 give MINMOD
β=2 give SUPERBEE
van Leer
G a , b  
ab  ab
a  b
Other SUPERBEE expression
G  a , b   s max 0 , min 2 b , s a , min  b , 2 s a 
 s = sign
Higher order in time


Needs to advance half time step
MUSCL-Hancock



Primitive variable
W  A  W W
Limit variable
Evolve the cell face values 1/2t:
t
Wi

L,R
 Wi
L,R
x

0
1 t
2 x

A Wi
n
W
Update as normal solving the Riemann
problem using evolved WL, WR
n 1
Ui
Ui 
n
t
x
 F i  1 / 2  F i 1 / 2 
L
i
 Wi
R

Alternative time advance

Alternatively, evolve the cell values using
n 1 / 2
Ui

 Ui 
1 t
2 x
F W   F W 
R
i
L
i
Solve as normal
n 1
Ui

n
Ui 
n
t
x
 F i  1 / 2  F i 1 / 2 
Procedure is 2nd order in space and time
Boundary Conditions

Set flux on boundary





Directly
Ghost cell
Wall u, v = 0.
Transmissive
Ghost cell un+1=-un
Ghost cell hn+1 = hn
un+1 = un
Wet / Dry Fonts

Wet / Dry fronts are difficult



Source of error
Source of instability
Common


near tidal boundaries
Flooding - inundation
Dry front speed

We examined earlier dry bed problem
S *L  u  c
SL
t
u L  2cL  u  2c
W et b ed
D ry b ed
x



c0
Front is fast
S *L  u L
Faster than characteristic.
Can cause problem with time-step / Courant
 2cL
Solutions

The most popular way is to artificially wet bed


Give a small depth, zero velocity
Loose a bit of mass and/or momentum

Can drastically affect front speed


E.g. a 1.0 dambeak with 1cm gives 38% error
1mm gives 25% error – try it!
Conservation Errors

The conserved variable are h and hu

Often require u

Need to be very careful about divide by zero

Artificial dry-bed depths could cause this
Source Terms



“Lumped” in to one term and integrated
Attempts at “upwinding source”
Current time-step


Could use the half step value
E.g.
0

S     z u u
 2
 gh 
   x K

  2 B 

 2 B x 


K  CR
K 
R
1/ 2
2/3
n
Main Problem is Slope Term


Flat still water over uneven bed starts to
move.
z
Problem with discretisation of
gh
i-1
i
x
i+ 1
w ater surfa ce
hi 1
hl
hi
hr
hi  1
bed level
gh i
z, h
zi 1
zl
zi
zr
z i 1
x
datum le vel
zr
 zl 
x
Discretisation

Discretised momentum eqn
 hu i
n 1

  hu i
2
2
t 
gh l
gh r
2
2
 hu l  hu r 



x 
2
2
For flat, still water
 hu i
n 1

 t

  x gh i  z r  z l 

Require
2
 t   hl

 g
 hi z l


x   2
2
hl
2
  h r2

  2  hi z r
 
2
 hi z l 
hr
2
 hi z r

  0


A solution

Assume a “datum” depth, measure down
2
g hi
z
x
  g hi
2
 hi
x
1  hi
2
 g
2 x
and
g
1 h i  z i  z r
 h i  z i  z l
2
x
2

2
Momentum eqn:

hl  hi  z i  z l
Flat surface
hu i
'

g t
2 x
h
2
l
hr  hi  z i  z r
 h r   hi  z i  z r    hi  z i  z l 
2
2
2

Some example solutions
Weighted mesh gives more detail for same number of cells
Dam Break - CADAM
Channel with 90° bend
Secondary Shocks