Numerical methods shallow water wave equations · 2010-11-18 · Numerical methods for shallow water wave equations Th. Katsaounis Dept. of Applied Math. Univ. of Crete, IACM, FORTH,

Dept. of Applied Mathematics Univ. of Crete Numerical methods for shallow water waves – 1 / 60

Numerical methodsfor

shallow water wave equations

Th. Katsaounis

Dept. of Applied Math. Univ. of Crete,IACM, FORTH, Crete, GREECE

Supported by ACMAC project EU-FP7

joint work withA. Delis TUC, Greece

Overview


Finite volume schemes for Conservation Laws

Relaxation Model for Conservation Laws

– Scalar and Systems of Cons. Laws

– Cons. Laws with Source Terms

Shallow Water Equations

– Relaxation Model - Schemes

– Numerical Results

Dispersive water wave models

– Finite volume schemes

– Numerical Results

Finite volume schemes for Conservation Laws


Scalar Conservation Laws


We consider the initial value problem for scalar conservation laws

ut + f(u)x = 0, x ∈ R, t > 0

u(x, 0) = u0(x), x ∈ R

Consider a uniform partition of R× R+ in cells [xi− 1

2

, xi+ 1

2

]× [tn, tn+1]

xi− 1

2

xi+ 1

2

tn

tn+1

xi−1 xi xi+1 xi = i∆x, i ∈ Z, xi+ 1

2

=xi+xi+1

2 , tn = n∆t


0 =

∫ tn+1

tn

∫ xi+1

2

xi− 1

2

[ut + f(u)x] dx dt

=

∫ xi+1

2

xi− 1

2

[u(x, tn+1)− u(x, tn)

]dx+

∫ tn+1

tn

[f(u(xi+ 1

2

, t))− f(u(xi− 1

2

, t))]dt

= Change of Mass+ Difference of Fluxes in cell [xi− 1

2

, xi+ 1

2

]× [tn, tn+1]

Uni ∼ 1

∆x

∫ xi+1

2

xi− 1

2

u(x, tn)dx, Fni+ 1

2

:= F (U+,ni , U−,n

i+1 ) ∼1

∆t

∫ tn+1

tnf(u(xi+ 1

2

, t))dt

Uni approximates the average of u in Ci = [xi− 1

2

, xi+ 1

2

] at time tn

Fni+ 1

2

approximates the average in [tn, tn+1] at x = xi+ 1

2

F is a numerical flux functionU+,ni some approximation of u(xi+ 1

2

− 0, tn)

U−,ni+1 some approximation of u(xi+ 1

2

+ 0, tn)

Basic FV scheme


Un+1i = Un

i − ∆t

∆x

(Fni+ 1

2

− Fni− 1

2

)

F numerical flux function:

1. Consistency : F (u, u) = f(u)

2. Monotonicity : ∂uF > 0, ∂vF < 0

3. Upwind, Lax-Friedrichs, Lax-Wendroff, Godunov, Central, Roe’s, ...

CFL condition : supu |f ′(u)|∆t∆x ≤ 1

U+i , U−

i+1 ? Reconstruction process :

1. Piecewise constants : U+i = Ui, U−

i+1 = Ui+1

2. Piecewise linear : U+i = Ui +

∆x2 Si, U−

i+1 = Ui+1 − ∆x2 Si+1 where

Si ∼ ux(xi+ 1

2

), x ∈ Ci Si+1 ∼ ux(xi+ 1

2

), x ∈ Ci+1. Limiters.

3. Higher order polynomials are constructed using the cell averages Ui.

Reconstruction process


Constant xi− 3

2

xi− 1

2

xi+ 1

2

xi+ 3

2

Ui−1Ui U

+i

Ui+1U−i+1

Linear

b

b

b

xi− 3

2

xi− 1

2

xi+ 1

2

xi+ 3

2

Ui

U+i

Ui+1

U−i+1

Relaxation model for Conservation Laws


Relaxation Model for Scalar CL


ut + f(u)x = 0, x ∈ R, t > 0,

u(x, 0) = u0(x), x ∈ R.(1)

Relaxation system proposed by Jin & Xin 1995

ut + vx = 0,

vt + c2ux = −1

ǫ(v − f(u)), ǫ → 0

(2)

This system can be viewed as a regularization of (1) by the wave operator

ut + f(u)x = −ǫ(utt − c2uxx) +O(ǫ2).

Applying the Champan-Enskog expansion we get

ut + f(u)x = ǫ∂x((c2 − f ′(u)2

)∂xu

)+O(ǫ2).

If the subcharacteristic condition : |f ′(u)| < c holds then a rigorousconvergence analysis, for 1D scalar case, can be applied yielding at therelaxation limit ǫ → 0 the conservation law (1). (JX, 1995)

Relaxation Model with Source term


For a conservation law with a source term

ut + f(u)x = q(u), x ∈ R, t > 0,

u(x, 0) = u0(x), x ∈ R,(3)

a relaxation system considered takes the form

ut + vx = q(u),

vt + c2ux = −1

ǫ(v − f(u)),

(4)

yielding the following regularization of (3),

ut + f(u)x = q(u) + ǫq(u)t − ǫ(utt − c2uxx).

Remarks1) Same subcharacteristic condition2) Extra term : ǫq(u)t.3) In general (4) does not preserve the steady states.4) The time discretization of (4)


Un+1 − Un

∆t+ V n

x = q(Un+1),

V n+1 − V n

∆t+ c2Un

x = −1

ǫ(V n+1 − f(Un+1)),

(5)

is fully coupled system, not the case for the corresponding time discetizationof (2).An alternative approach : we consider the following relaxation system

ut + vx = 0,

vt + c2ux = −1

ǫ(v − f(u))− 1

ǫR(u),

(6)

where R(u) is an antiderivative of q(u),

R(u(x)) =

∫ x

q(u(s))ds .


In this case (6) provides exactly a wave-type regularization of (3),

ut + f(u)x = q(u)− ǫ(utt − c2uxx). (7)

Also an implicit-explicit time discretization is now possible when we treat thesource terms implicitly:

Un+1 − Un

∆t+ V n

x = 0,

V n+1 − V n

∆t+ c2Un

x =− 1

ǫ

(V n+1 − f(Un+1)

)− 1

ǫR(Un+1).

(8)

If |f ′(u)| < c from (7) we recover formally (3)

System (6) preserves steady states

Initial, Boundary Cond. : v0 = f(u0), vb = f(ub)

System of CL


∂tu+d∑

j=1

∂xjFj(u) = 0, x ∈ R

d, u = u(x, t) ∈ Rn, t > 0

u(·, 0) = u0(·)

Relaxation model

∂tu+d∑

j=1

∂xjvj = 0,

∂tvi +Ai∂xiu = −1

ǫ(vi − Fi(u)) , i = 1, . . . , d

it’s a regularization by a wave operator of order ǫ, and Ai are symmetricpositive definite matrices with constant coefficients that are selected to satisfythe corresponding sub-characteristic conditions.

Shalllow Water Equations (SWE)


Shallow water eqns (1D)


ht + (hu)x = 0,

(hu)t + (hu2 +g

2h2)x = −ghZ ′,

(9)

General steady states :Q = hu = Cnst u2

2 + g(h+ Z) = CnstSWE is a hyperbolic system with source term

SW Relaxation Models


Relaxation Model A

ht + vx = 0

Qt + wx = −ghZ ′

vt + c21hx = −1

ǫ(v −Q)

wt + c22Qx = −1

ǫ(w − (

Q2

h+

g

2h2))

Relaxation Model B

ht + vx = 0

Qt + wx = 0

vt + c21hx = −1

ǫ(v −Q)

wt + c22Qx = −1

ǫ(w − (

Q2

h+

g

2h2)) +

1

ǫR(Z;h)


R(Z;h)(x) =

∫ x

g(hZ ′)(y)dy

c1, c2 are chosen according to sub-characteristic condition :

|λi(F′)| < ci, i = 1, 2, F ′ = Jacobian of flux vector

For Z ≡ 0, (A) ≡ (B)

For ǫ → 0 we recover the original SW system

Both relaxation systems have linear principal part

Implicit-explicit time discretizations for (B)

System (B) have same steady states as the continuous problem


We consider the Relaxation Model B and let

u =

[hQ

], v =

[vw

],

our system can be rewritten as

ut + vx = 0,

vt +C2ux = −1

ǫ(v − F(u))− 1

ǫS(u),

F(u) = (Q,Q2

h+

g

2h2)T , S(u) = (0,−

∫ x

gh(y)Z ′(y)dy)T

where u,v ∈ R2 and C

2 ∈ R2×2 is a positive matrix.

Relaxation Schemes for SWE (1D)


Upwind Scheme


We assume a uniform spaced grid with ∆x = xi+ 1

2

− xi− 1

2

and a uniform time

step ∆t = tn+1 − tn, n = 0, 1, 2, . . ..

uni ∼ 1

∆x

∫ xi+1

2

xi− 1

2

u(x, tn)dx, uni+ 1

2

∼ u(xi+ 1

2

, tn)

We start by considering the following one-step conservative system for thehomogeneous case (no source term present)

∂

∂tui +

1

∆x(vi+ 1

2

− vi− 1

2

) = 0,

∂

∂tvi +

1

∆xC

2(ui+ 1

2

− ui− 1

2

) = −1

ǫ(vi − F(ui)).

The linear hyperbolic part has two Riemann invariants (characteristic speeds)v ±Cu associated with the characteristic fields ±C respectively. The firstorder upwind approximation of v ±Cu is

(v +Cu)i+ 1

2

= (v +Cu)i, (v −Cu)i+ 1

2

= (v −Cu)i+1.


Hence,

ui+ 1

2

=1

2(ui + ui+1)−

1

2C

−1(vi+1 − vi),

vi+ 1

2

=1

2(vi + vi+1)−

1

2C(ui+1 − ui).

First order upwind semi-discrete approximation of the relaxation scheme :

∂

∂tui +

1

2∆x(vi+1 − vi−1)−

1

2∆xC(ui+1 − 2ui + ui−1) = 0,

∂

∂tvi +

1

2∆xC

2(ui+1 − ui−1)−1

2∆xC(vi+1 − 2vi + vi−1) = −1

ǫ(vi − F(ui))

− 1

ǫS(ui),

where

S(ui) =

[0

−∫ x

i+12 gh(y)Z ′(y)dy

].

MUSCL Scheme


We replace the piecewise constant approximation by a MUSCL piecewiselinear interpolation : for the k−th component of v ±Cu we have:

(v + cku)i+ 1

2

= (v + cku)i +1

2∆xs+i ,

(v − cku)i+ 1

2

= (v − cku)i+1 −1

2∆xs−i+1,

where u, v are the k−th components of v,u and the slopes s± in the i−thcell :

s±i =1

∆x(vi+1 ± ckui+1 − vi ∓ ckui)φ(θ

±i )

θ±i =vi ± ckui − vi−1 ∓ ckui−1

vi+1 ± ckui+1 − vi ∓ ckui,

where φ is a limiter function satisfying 0 ≤ φ(θ) ≤ minmod(2, 2θ).MinMod (MM) : φ(θ) = max(0,min(1, θ)),

VanLeer (VL) : φ(θ) = |θ|+θ1+|θ| ,

Monotonized Central (MC) : φ(θ) = max(0,min((1 + θ)/2, 2, 2θ))


Second order semi-discrete relaxation scheme (componentwise form)

∂

∂tui +

1

2∆x(vi+1 − vi−1)−

ck2∆x

(ui+1 − 2ui + ui−1)

− 1

4(s−i+1 − s−i + s+i−1 − s+i ) = 0,

∂

∂tvi +

c2k2∆x

(ui+1 − ui−1)−ck

2∆x(vi+1 − 2vi + vi−1)

+ck4(s−i+1 − s−i − s+i−1 + s+i ) = −1

ǫ(vi − Fk(ui))−

1

ǫSk(ui),

with Sk, Fk being the k−th components of S,F respectively.

Fully Discrete Schemes


A first order in time RK-type scheme, (Z ≡ 0)

(A) Given un,vn apply a finite volume method to update u,v over time ∆t

by solving the homogeneous linear hyperbolic system

[u

v

]

t

+

[0 I

C2 0

] [u

v

]

x

=

[00

],

and obtain new values u(1),v(1).

(B) Update u(1),v(1) to u

n+1,vn+1 by solving the equations,

ut = 0,

vt = −1

ǫ(v − F(u)),

over time ∆t.


A second order in time RK-type scheme,

un,1 = u

n, vn,1 = v

n +∆t

ǫ(vn,1 − F(un,1)) +

∆t

ǫS(un,1);

u(1) = u

n,1 −∆tD+vn,1, v

(1) = vn,1 −∆tC2D+u

n,1;

un,2 = u

(1),

vn,2 = v

(1) − ∆t

ǫ(vn,2 − F(un,2))− 2∆t

ǫ(vn,1 − F(un,1))

− ∆t

ǫS(un,2)− 2∆t

ǫS(un,1);

u(2) = u

n,2 −∆tD+vn,2, v

(2) = vn,2 −∆tC2D+u

n,2;

un+1 =

1

2(un + u

(2)), vn+1 =

1

2(vn + v

(2)).

where

D+wi =1

∆x(wi+ 1

2

−wi− 1

2

).

Choice of parameters


CFL Condition

1st order scheme : maxc1, c2∆t

∆x≤ 1

2nd order scheme : maxc1, c2∆t

∆x≤ 1

2

Choice of c1, c2 : based on rough estimates of the eigenvalues :u±

√gh and satisfy the subcharacteristic condition

c1 ≥ sup |u+√

gh| and c2 ≥ sup |u−√

gh|

c1 = c2 = maxsup |u+

√gh|, sup |u−

√gh|

Choise of ǫ : ǫ << ∆x, ǫ << ∆t

SWE(1D) : Numerical Results


No source term : Z ≡ 0: Dam Break problem

– Subcritical, supercritical, strongly supercritical

– Dry bed problem

Source term : Z 6= 0 :

– Flow at Rest

– Nontrivial Steady States

– Drain on Non-Flat bottom

Dam Break Flow Z ≡ 0


We consider a channel of length L = 2000m. A dam is located atx0 = 1000m and at time t = 0 the dam collapses. We compute the solutionfor time T = 50s with initial conditions:

u(x, 0) = 0, h(x, 0) =

h1 x ≤ 1000,

h0 x > 1000,

with h1 > h0. This is the Riemann problem for the homogeneous problem.The flow consists of a shock wave (bore) travelling downstream and ararefaction wave (depression wave) travelling upstream. The upstream depthh1 was kept constant at 10m, while the downstream depth h0 was differentfor each problem.

h0/h1 > 0.5 : subcritical flow

h0/h1 < 0.5 : subcritical upstream, supercritical downstream

h0/h1 << 0.5 : strongly supercritical downstream

CFL = 0.5m, ∆x = 20m, c1 = 5, c2 = 12, ǫ = 1.E − 4

Dam Break Flow, Subcritical


0 200 400 600 800 1000 1200 1400 1600 1800 20004

5

6

7

8

9

10h

(m)

x (m)

Dam-break flow, h0/h1=0.5, (x) Upwind and (o)MUSCL with MC limiter.

Dam Break Flow, Supercritical


0 200 400 600 800 1000 1200 1400 1600 1800 20000

1

2

3

4

5

6

7

8

9

10

h (m

)

x (m)

Dam-break flow, h0/h1=0.05, (x)Upwind and (o)MUSCL with MC limiter.

Dam Break Flow, Strongly Supercritical


0 200 400 600 800 1000 1200 1400 1600 1800 20000

1

2

3

4

5

6

7

8

9

10

h (m

)

x (m)

Dam-break flow, h0/h1=0.005, (x)Upwind and (o)MUSCL with MC limiter.

Dry Bed problem, h0 = 0


This a challenging problem as a result of the singularity that occurs at thetransition point of the advancing front. We compute the solution at T = 40s

No modifications to the scheme to incorporate the dry area

Globally accurate results free of oscillations

The water height and discharge remain positive

The transition point between the wet and the dry zone is close to theexact one, but some difficulties appear on the velocity.

Overall the solution is stable, monotone with no special front trackingtechniques

CFL = 0.5 ∆x = 10m, c1 = 18, c2 = 16, ǫ = 1.E − 4

Dry Bed problem, Water Height


0 200 400 600 800 1000 1200 1400 1600 1800 20000

1

2

3

4

5

6

7

8

9

10

x (m)

h (m

)

Dry bed dam-break flow (h), (o)MUSCL with MM limiter.

Dry Bed problem, Discharge


0 200 400 600 800 1000 1200 1400 1600 1800 20000

5

10

15

20

25

30

x (m)

q (m

2 /s)

Dry bed dam-break flow (q), (o)MUSCL with MM limiter.

Dry Bed problem, Velocity


0 200 400 600 800 1000 1200 1400 1600 1800 20000

2

4

6

8

10

12

14

16

18

20

u (m

/s)

x (m)

Dry bed dam-break flow (u), (o)MUSCL with MM limiter.

Flow at Rest, Z 6= 0


We consider at channel of length L = 25m with a non-trivial bathymetry Z,with initial conditions

u(x, 0) = 0, ∀x ∈ R,

h(x, 0) + Z(x) = H, ∀x ∈ R,

Exact solution

u(x, t) = 0, ∀x ∈ R, t ≥ 0,

h(x, t) + Z(x) = H, ∀x ∈ R, t ≥ 0,

Z(x) =

0.2− 0.05(x− 10)2, 8 ≤ x ≤ 12,

0, otherwise,

with H = 2m, ǫ = 1.E − 5, c1 = 4, c2 = 4.5, CFL = 0.5,T = 200s, ∆x = 0.125

Flow at Rest, Water Height


0 5 10 15 20 250

0.5

1

1.5

2

2.5

Hei

gh t(

m)

Distance (m)

Flow at rest (water height): (+) standard source, (o) integral source

Flow at Rest, Discharge


0 5 10 15 20 25−0.1

−0.08

−0.06

−0.04

−0.02

0

0.02

0.04

0.06

0.08

0.1

x (m)

q (m

2 /s)

Flow at rest (discharge): (+) standard source, (o) integral source

Flow at Rest, Discharge (zoom)


5 10 15 20

−0.01

−0.005

0

0.005

0.01

x (m)

q (m

2 /s)

Flow at rest: Magnified view of the discharge.

Non trivial steady states


We consider with the convergence towards steady flow over the parabolichump in a channel of length L = 25m.Depending on the boundary conditions the flow maybe subcritical, transcriticalwith a shock or without a shock. In all cases we use MUSCL scheme with

CFL = 0.5, ∆x = 0.125m, T = 200s, ǫ = 1.E − 5, c1 = 5, c2 = 7

u(x, 0) = 0, ∀x ∈ R,

h(x, 0) + Z(x) = H0, ∀x ∈ R,

where H0 water level downstream.

Subcritical Flow : qup = 4.42m2/s, H0 = 2m

Transcritical Flow without shock : qup = 1.53m2/s, H0 = 0.66m

Transcritical Flow with shock : qup = 0.18m2/s, H0 = 0.33m

Transcritical Flow with shock : Water Level


0 5 10 15 20 250

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

h (m

)

x (m)

Transcritical flow with shock (h)

Transcritical Flow with shock : Discharge


0 5 10 15 20 250

0.05

0.1

0.15

0.2

0.25

0.3

x (m)

q (m

2 /s)

Transcritical flow with shock (q)

Drain on a non-flat bottom.


Difficult problem since it involves the calculation of dry areas.

BC’s : Upstream reflective, Downstream dry bed

IC’s : h+ Z = 0.5m and q = 0m3/s

Solution : a state at rest, on the left part of the hump withh+ Z = 0.2m with q = 0m3/s and a dry state (i.e. h = 0 andq = 0m3/s) on the right of the hump.

MUSCL scheme with∆x = 0.1m, CFL = 0.5, ǫ = 1.E − 6, c1 = c2 = 3.5

No modification of the method to overcome the dry area problem ofzero depth and discharge.

Drain on a non-flat bottom: Water Level


0 5 10 15 20 250

0.1

0.2

0.3

0.4

0.5h

(m)

x (m)

t=0s

t=10s

t=20s

t=100s

t=1000s

Drain on a non-flat bottom (h)

Drain on a non-flat bottom: Discharge


0 5 10 15 20 250

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

x (m)

q (m

2 /s)

t=10s

t=20s

t=100st=1000st=0s

Drain on a non-flat bottom (q)

The 2D Shallow Water Equations


Ut + F(U)x +G(U)y = S(U); (x, y) ∈ Ω, t ≥ 0

U =

hhu1hu2

=

hq1q2

, S(U) =

0

−gh∂Z

∂x(x, y)− ghSx

f

−gh∂Z

∂y(x, y)− ghSy

f

,

F(U) =

q1

q21h

+1

2gh2

q1q2h

, G(U) =

q2

q1q2h

q22h

+1

2gh2

.

Sxf = n2

mu1√

u21 + u22h−4/3 Sy

f = n2mu2

√u21 + u22h

−4/3, where nm is theManning roughness coefficient.

Relaxation System for 2D SWE


u

v

w

t

+

0 I 0C

2 0 00 0 0

u

v

w

x

+

0 0 I

0 0 0D

2 0 0

u

v

w

y

=

0

−1ǫ (v − F(u) + S(u))

−1ǫ (w −G(u) +

˜S(u))

S(u) =

0

−1

2

∫ x

gh(s, y)∂Z

∂x(s, y)ds

−1

2

∫ x

gh(s, y)∂Z

∂y(s, y)ds

˜S(u) =

0

−1

2

∫ y

gh(x, s)∂Z

∂x(x, s)ds

−1

2

∫ y

gh(x, s)∂Z

∂y(x, s)ds

Subcharacteristic condition:

λ2i

c2i+

µ2i

d2i≤ 1, ∀ i = 1, 2, 3,

with λi, µi the eigenvalues of ∂F(u)/∂u and ∂G(u)/∂u respectively.

Fully Discrete Schemes


Upwind : 1st order; MUSCL : 2nd order

CFL condition, guarantees the TVD property of both schemes

CFL = max

((max

ici)

∆t

∆x, (max

idi)

∆t

∆y

)≤ 1

2.

I.B. Cond. : v0 = F(u0), w0 = G(u0), vb = F(ub), wb = G(ub)

Choice of ck, dk, k = 1, 2, 3 :

1. rough estimates of the eigenvalues (u1, u1 ±√gh) and

(u2, u2 ±√gh)

2. calculate c and d locally at every cell asci+ 1

2,j = maxu∈u

i+12,j,u

i− 12,j |∂F(u)/∂uk|

di,j+ 1

2

= maxu∈ui,j+1

2

,ui,j− 1

2

|∂G(u)/∂uk|

3. global choice : ck = dk = maxi,j(ci+ 1

2,j , di,j+ 1

2

)

∆t ≫ ǫ and ∆y,∆x ≫ ǫ

Numerical Results for 2D-SWE


No source Z ≡ 0 : Partial Dam Break, Circular Dam Break

Dam break in a channel with topography

2D Partial Dam-Break


The dam, located in the center of a channel breaks instantaneously.

No friction (nm = 0). hu = 10m and hd = 5, 0.1, 0m

Channel : 200m× 200m, 41× 41 square grid.

The breach is 75m in length, 30m from the left bank, 95m from theright.

BC’s : x = 0 and x = 200m transmissive and all other boundaries arereflective.

2nd order MUSCL scheme

ǫ = 10−6 and c1 = 10, c2 = 6, c3 = 11, d1 = 10, d2 = 5, d3 = 11.

T = 7.2s

2D Partial Dam-Break, hd = 5


0

50

100

150

200

0

50

100

150

200

5

6

7

8

9

10

Y(m)X(m)

Wat

er D

epth

(m)

2D Partial Dam-Break, hd = 0.1m


0

50

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150

200

0

50

100

150

200

0

2

4

6

8

10

Y(m)X(m)

Wat

er D

epth

(m)

2D Partial Dam-Break, hd = 0m


0

50

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150

200

0

50

100

150

200

0

2

4

6

8

10

Y(m)

X(m)

Wat

er D

epth

(m)

Circular Dam Break


A two dimensional Riemann problem for the 2D SWEs

Two regions of still water separated by a cylindrical wall with radius 11mcentered in a channel. The water depth within the cylinder is 10m and1m outside.

The wall is removed instantaneously, the bore waves will spread andpropagate radially and symmetrically

There is a transition from subcritical to supercritical flow.

2nd order MUSCL scheme

Channel : 50× 50m, 51× 51 square grid

ǫ = 10−6 and c1 = c3 = 12, c2 = 7, d1 = d3 = 12, d2 = 7, T = 0.69s

Circular Dam Break


0

10

20

30

40

50

010

2030

4050

0

1

2

3

4

5

6

7

8

9

10

X(m)

Y(m)

Wate

r D

epth

(m)

0 5 10 15 20 25 30 35 40 45 500

5

10

15

20

25

30

35

40

45

50

X(m)

Y(m

)

2

2

2

2

2

2

3.1

3.1

3.1

3.1

3.1

3.8 3.8

3.8

3.8

4.7

4.7

4.7

5.5

5.5

5.5

6.3

6.3

7

7

7.8

7.8

8.7

at t = 0.69s, Water depth, depth contours, velocity field, (MM, VL limiter)

Circular Dam Break


0

10

20

30

40

50

010

2030

4050

0

1

2

3

4

5

6

7

8

9

10

X(m)

Y(m)

Wate

r D

epth

(m)

0 5 10 15 20 25 30 35 40 45 500

5

10

15

20

25

30

35

40

45

50

X(m)

Y(m

)

2

2

2

2

2

3.1

3.1

3.1

3.1

3.1

3.1

3.1

3.1

3.8

3.8

3.8

4.7

4.7

4.7

5.5

5.5

6.3

6.3

7

7

7.8

7.8

8.7

at t = 0.69s, Water depth, depth contours, velocity field, (MC, SB limiters)

Circular Dam Break, Dry Bed


0

10

20

30

40

50

0

10

20

30

40

50

0

2

4

6

8

10

X(m)Y(m)

Wate

r D

epth

(m)

0 5 10 15 20 25 30 35 40 45 500

5

10

15

20

25

30

35

40

45

50

X(m)

Y(m

)

0.1

0.10.1

0.1

0.1

0.1

0.5

0.50.5

0.5

0.5

0.5

1

1

1

1

1

2

2

2

2

2

3.1

3.1

3.1

3.1

3.8

3.8

3.8

4.7

4.7

4.7

5.5

5.5

5.5

6.3

6.3

7

7

7.8

7.8

8.7

8.7

at t = 0.69s, Water depth, depth contours, velocity field, (VL limiter)

Dam Break in channel with topography


Consider a channel 75m long and 30m wide

A dam is situated at x = 16m with initial water depth h+ Z = 1.875mwhile the rest of the channel is considered dry.

The topography consists of three mounds located in the channelbottom.

Manning coefficient nm = 0.018, ci = di = 5, ǫ = 1.E − 8

Conclusions


Relaxation Schemes for SW which combine

– Simplicity

– Robustness

– Efficiency

– Riemann solver free

Novel ways to incorporate source terms

Small errors of order of ǫ while preserving steady states.

The benchmark tests show that the schemes provide accurate solutionsin good agreement with well known analytical solutions.

Comparable solutions with well known solvers

Can be considered for practical applications?

References


Jin S, Xin Z. The relaxing schemes of conservations laws in arbitrary

space dimensions. Comm. Pure Appl. Math, 1995

Delis AI, Katsaounis Th., Relaxation schemes for the shallow water

equations, Int. J. Numer. Meth. Fluids, 2003

Delis AI, Katsaounis Th., Numerical solution of the two-dimensional

shallow water equations by the application of relaxation methods, App.Math. Model., 2005

Ch. Arvanitis, Ch. Makridakis, A. Tzavaras, Stability and convergence

of a class of finite element schemes for hyperbolic systems of

conservation laws, SINUM, 2004

Ch. Arvanitis and Th. Katsaounis, Ch. Makridakis, Adaptive finite

element relaxation schemes for hyperbolic systems of conservation laws,M2AN, 2001

Th. Katsaounis, Ch. Makridakis, Relaxation models and finite element

schemes for the shallow water equations, HYP2002, 2003

Numerical methods shallow water wave equations · 2010-11-18 · Numerical methods for shallow water wave equations Th. Katsaounis Dept. of Applied Math. Univ. of Crete, IACM, FORTH,

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