Shallow water

Finite Volume Method for Shallow WaterEquations

Oto Havle, Jir Felcman

Workshop Dresden-Prague on Numerical Analysis 2008

Contents

Shallow water equations

SWE without source term

SWE with topographic source term

Conclusion

Shallow water equationsA system of PDEs

ht

+ div(hv) = 0,

t(hvs) + div(hvsv) +

12

g

xsh2 = gh z

xs, s = 1,2.

z ... given function, g > 0 ... given constant

z

h v

Shallow water equations - a hyperbolic system

wt

+2

i=1

xif i(w) = r(x ,w)

where

w (

hq

)(

hhv

)

hhv1hv2

f 1(w) =

q1h1q21 + 12gh2h1q1q2

= hv1hv21 + 12gh2

hv1v2

f 2(w) =

q2h1q1q2h1q22 +

12gh

2

= hv2hv1v2

hv22 +12gh

2

r(x ,w) =

(0

ghz(x)

)


Difficulties with Shallow Water EquationsI existence / uniqueness theorems not availableI dry areas - SWEs do not have sense for h = 0I source term


Flat bottom case z = const .I system of conservation laws

I standard FVM discretizationI we need a numerical flux

wt

+2

s=1

xsf s(w) = 0


Flat bottom case z = const .I system of conservation lawsI standard FVM discretization

I we need a numerical flux

Di

w(x , tk+1) dx

Diw(x , tk ) dx

+

tk+1tk

Di

2i=1

ni f i(w) H(w |Di ,w |Dj ,nij )

dS = 0


Flat bottom case z = const .I system of conservation lawsI standard FVM discretizationI we need a numerical flux

wki w on the cell Diand time interval (tk , tk+1)

wk+1i = wki

k

|Di |

jS(i)

|ij |H(wki ,wkj ,nij)

Numerical flux

How to derive a numerical flux H(wL,wR,n)I Choose a new coordinate system such that n = (1,0).I Linearize and solve a linear 1D Riemann problem.

wt

+ Awx1

= 0

w(x1, t = 0) =

{wL, x1 < 0wR, x1 > 0

I Take the linear flux Aw at x1 = 0 and rewrite in the originalcoordinate system.

How to linearize the Riemann problem

We adapt the Vijayasundaram flux,

gV (wL,wR) = A+(w?)wL + A(w?)wR (?)

where w? =wL + wR

2A1(w) = D f 1(w)

As A1(w)w 6= f 1(w), the 1D flux (?) is not consistent with theShallow Water Equations. We fix it with additional term

gV1(wL,wR) = A+(w?)wL + A(w?)wR

12

gh2??

010

where h?? =

hL + hR2

SWE with no source term - Numerical resultsElementary Riemann problems in 1D

I continuous solution (rarefaction wave) - EOC 0.75I discontinuous solution (contact discontinuity) - EOC 0.5

2D test problem - circular dam break

-200 -150 -100 -50 0 50 100 150 200-200-150

-100-50

0 50

100 150

200

0 1 2 3 4 5 6 7 8 9

10

-200 -150 -100 -50 0 50 100 150 200-200-150

-100-50

0 50

100 150

200

0

1

2

3

4

5

6

Source term - stationary solution

Shallow water equations

ht

+ div(hv) = 0,

t(hvs) + div(hvsv) +

12

g

xsh2 = gh z

xs, s = 1,2.

Stationary solution (lake at rest)

h(x , t) = H0 z(x), v(x , t) = 0.

Discrete stationary solution

wki =

H0 zi00

, zi zDi

Discretization of source terms

Rewrite to a system of conservation laws with a nonzero RHS

wt

+2

s=1

xsf s(w) = r(x ,w)

Operator splitting

wt

+2

s=1

xsf s(w) = 0

wt

+ 0 = r(x ,w)

would not work. Stationary solutions are different.

Discretization of source terms

Difficulty with the source termI We use a piecewise-constant approximation of the

topography function z.

hi = H0 zi

The source term is now a distribution.

r(x ,w) =(

0ghz(x)

)I Source term must be taken into account when

approximating convective term.

Linearised Riemann problem with a source termWe construct a linear Riemann problem with nonzero RHS

wt

+ Awx

= (x)r , t > 0

w(x ,0) =

{wL, x < 0wR, x > 0

The matrix A is the same as before

A = A1(w?) = Df 1(w?) w? =12(wL + wR)

The right hand side has special form

r =

0gh?(zL zR)0

Linear Riemann problem with a source term

To solve the Riemann problemI Use diagonal form A = UU1 to decouple equations.

I Use characteristics to solve scalar problems.I The solution is a function, unless v1 = c.I For the purposes of deriving a numerical flux, ignore the

distributional case.

ut

+ ux

= (x), t > 0

u(x ,0) = u0(x)


To solve the Riemann problemI Use diagonal form A = UU1 to decouple equations.I Use characteristics to solve scalar problems.

I The solution is a function, unless v1 = c.I For the purposes of deriving a numerical flux, ignore the


= 0 = u(x , t) = u0(x) + t(x),

6= 0 = u(x , t) = u0(x t) +

||K(x , t),

where K(x , t) =

{1, x(x t) < 0,0, x(x t) 0.


To solve the Riemann problemI Use diagonal form A = UU1 to decouple equations.I Use characteristics to solve scalar problems.I The solution is a function, unless v1 = c.

I For the purposes of deriving a numerical flux, ignore thedistributional case.

|v | < c ... subcritical|v | = c ... critical|v | > c ... supercritical

c =

gh

w =

hhv1hv2


To solve the Riemann problemI Use diagonal form A = UU1 to decouple equations.I Use characteristics to solve scalar problems.I The solution is a function, unless v1 = c.I For the purposes of deriving a numerical flux, ignore the


The linear flux Aw at x = 0 is

A(12

limx0+

w(x , t) +12

limx0

w(x , t))

= A+wL + AwR +12(sgnA)r?.

New numerical fluxThe numerical flux (discretization of convective term)

gconv (wL,wR, zL, zR)

= A+1 (w?)wL + A1 (w?)wR

12

gh2??e2

12

gh?(zR zL)(sgnA1)e2

In fact, the numerical flux does not depend on sgn2.

1 = v1 c, 2 = v1, 3 = v1 + c, c =

gh

Therefore, it is continuous in the subcritical case |v | < c,

sgnA1(w?)e2 =

1/cv1/cv2/c

.

Discretizing the source term

We discretize the source term proper

Di

tk+1tk

r(x ,w(x , t)) dx dt ghiDi

tk+1tk

(0

z(x)

)dx dt

= kghiDi

z(x)(0nij

)dS

kghi

jS(i)

|ij |z?ij nij

where z?ij =zi + zj

2is an approximation to z at the interface ij .

The numerical scheme

Both convective and source have a quasi-flux form

k

jS(i)

|ij |(. . . )

We can write

wk+1i = wki

k

|Di |

jS(i)

|ij |H total(

wki ,wkj , zi , zj ,nij

)Theoretical properties

I The scheme is conservative with respect to h.I The discrete "lake at rest" is a stationary solution.

Numerical resultsThe scheme seems to work in the subcritical case |v | < c, butnot in the supercritical case. It gets better if we replace

h?? =hL + hR

2

with

h?? =

hR, < 1,

1+2 h

2L +

12 h

2R, 1 < < 1,

hL, > 1,

=v1,?c?

gconv (wL,wR, zL, zR)

= A+1 (w?)wL + A1 (w?)wR

12

gh2??e2

12

gh?(zR zL)(sgnA1)e2

Summary

Numerical scheme for Shallow Water Equations.I Simple first order Finite Volume scheme.I Handles topographic source term.

Future workI Enhance with mesh adaptivity.I Investigate the case h = 0.

Thank you for your attention.

Summary

Numerical scheme for Shallow Water Equations.I Simple first order Finite Volume scheme.I Handles topographic source term.

Future workI Enhance with mesh adaptivity.I Investigate the case h = 0.

Thank you for your attention.

Shallow water equationsSWE without source termSWE with topographic source termConclusion

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Shallow water