Nonlinear Hyperbolic Systems of Conservation
Laws and Related Applications II
Mapundi K. Banda
Applied Mathematics Division - Mathematical Sciences, Stellenbosch University
[email protected]
Jul 30, 2013
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Section 1: Analysis of Systems
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Entropy/Viscous Solutions
Entropy Condition I (Lax)
A discontinuity with shock velocity s which is given by the
Rankine-Hugoniot condition fulfills the entropy condition if
f 0 (ul ) > s > f 0 (ur )
For convex f this yields ul > ur , i.e. a stable shock.
For ul < ur the shock solution is not possible - the rarefaction wave is
the correct physical solution.
For general cases one considers the viscous equation
ut + f (u)x = εuxx
A viscous solution of the conservation law is given by the weak
solution, which can be recovered in the limit ε → 0
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Unique Solutions - Entropy
Let η and ψ be two convex functions, for which we have for all
smooth solutions u of the conservation law:
η(u)t + ψ(u)x = 0,
i.e. η 0 (u)ut + ψ 0 (u)ux = 0.
For smooth solutions
ut + f 0 (u)ux = 0
or
η 0 (u)ut + η 0 (u)f 0 (u)ux = 0.
Giving
η 0 (u)f 0 (u) = ψ 0 (u)
and such functions (η, ψ) are called entropy–entropy flux pairs
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Unique Solutions - Entropy
Sufficient for existence and uniqueness of weak solutions.
The function u(x, t) is the entropy solution if, for all convex entropy
functions, η(u), and corresponding entropy fluxes, ψ(u), the inequality
η(u)t + ψ(u)x ≤ 0,
is satisfied in the weak sense: i.e.
Z
Z Z∞
(φt η(u) + φx ψ(u)) dt dx + φ(x, 0)η(u)(x, 0) dx ≤ 0
R 0
(1)
R
∀φ ∈ C01 (R × R),
φ ≥ 0.
Definition
Let u be a weak solution of the conservation law. Moreover,suppose u
fulfills (1) for any entropy–entropy flux pair (η, ψ), then the function u is
called an entropy solution.
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Unique Solutions - Entropy
Appropriate function spaces.
Definition
Let u ∈ L∞ (Ω), Ω ⊂ Rn be open. Then the total variation of u is
defined by
Z
1
TV (u) = lim sup
|u(x + ε) − u(x)| dx.
ε→0 ε
Ω
The space of bounded variation is
BV (Ω) := {u ∈ L∞ (Ω) : TV (u) < ∞}.
If u 0 ∈ L1 (Ω) holds, then
Z
TV (u) =
|u 0 | dx.
Ω
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Unique Solutions - Entropy
Theorem (Kruskov)
The scalar Cauchy-Problem
ut + (f (u))x = 0,
f ∈ C 1 (R)
u(x, 0) = u0 (x), u0 ∈ L∞ (R)
has a unique entropy solution u ∈ L∞ (R × R+ ), having the following
properties:
(i) ku(·, t)kL∞ ≤ ku0 (·)kL∞ ,
t ∈ R+
(ii) u0 ≥ v0 ⇒ u(·, t) ≥ v (·, t),
t ∈ R+
(iii) u0 ∈ BV (R) ⇒ u(·, t) ∈ BV (R) and TV (u(·, t)) ≤ TV (u0 )
Z
Z
(iv) u0 ∈ L1 (R) ⇒
u(x, t) dx = u0 (x) dx, t ∈ R+
(i)-(iv) are called
R
∞
L -stability,
Mapundi K. Banda (Maties)
R
monotonicity, TV-stability, conservativity.
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Take Note!
The theorem can be extended to several dimensions x ∈ Rd , d > 1.
The theorem cannot be extended to the case of systems (n > 1).
Till now there is no general proposition proved for the system case.
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System of Conservation Laws
Consider systems of conservation laws in one-space-dimension:
 (1) 
F
 .. 
n
ut + (F (u))x = 0, x ∈ R, u, F ∈ R , F =  . 
F (n)
Also
ut + A(u)ux = 0,
A(u) = F 0 (u) =
∂F (i) (u)
,
∂uj
1 ≤ i, j ≤ n
First consider linear systems:
ut + Aux = 0,
A ∈ Rn×n ,
u(x, 0) = u0 (x)
For a hyperbolic equation A is diagonalizable with e-values λ1 , . . . , λn
and e-vectors r1 , . . . , rn .
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System of Conservation Laws
Let R = (r1 | . . . |rn ), AR = RD, A = RDR −1 diagonalize the
system: Let v = R −1 u (characteristic variables)
Rvt + RDR −1 Rvx = 0
or vt + Dvx = 0,
since R is constant.
Obtain n scalar problems for (v1 , . . . , vp , . . . , vn ) with solution
vp (x, t) = vp (x − λp t, 0)
Given v (x, 0) = R −1 u0 (x) obtain
u(x, t) = Rv (x, t) =
n
X
vp (x, t)rp =
p=1
n
X
vp (x − λp t, 0)rp
p=1
The curves x = x0 + λp t are called characterstics of the p-th family
(x 0 (t) = λp )
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General Nonlinear Systems
Consider A = A(u) with (λ1 (u), . . . , λn (u)) - the e-values.
The characteristics of the p-th family are given by xp (t) and take the
form
xp0 (t) = λp (u(xp (t), t)),
xp (0) = x0 ,
p = 1, . . . , n
Note: Problem depends on u and is strongly coupled due to
R = R(u) - not easy to solve!
Consider the Riemann Problem of the linear system of equations
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Revisit Riemann Problem for Linear Systems
Consider systems of conservation laws in one-space-dimension:
ut + Aux = 0,
with
(
ul , x < 0
u(x, 0) =
ur , x > 0
Solution is given by:
ul =
n
X
αp rp ,
p=1
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ur =
n
X
βp rp ,
p=1
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(
αp , x < 0
vp (x, 0) =
βp , x > 0
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Revisit Riemann Problem for Linear Systems
Now
(
αp , x − λ p t < 0
vp (x, t) =
βp , x − λp t
Hence
u(x, t) =
n0
X
βp rp +
p=1
n
X
αp rp
p=n0 +1
where n0 is the minimal value of p with x − λp t > 0, (λ1 ≤ · · · ≤ λn )
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Revisit Riemann Problem for Linear Systems
State u in terms of jump discontinuities:
u(x, t) =
n
X
αp rp +
p=1
= ul +
= ur −
n0
X
βp rp −
p=1
n0
X
n0
X
αp rp
p=1
(βp − αp )rp
p=1
n
X
(βp − αp )rp
p=n0 +1
and
ur − ul =
X
(βp − αp )rp
p
Hence solution of RP can be considered as a splitting of difference
ur − ul in a sum of jumps, which move with velocity λp in direction rp .
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Example
For n = 2 a phase plot can be used to determine the intermediate
states between ul and ur .
If um is the intermediate state:
ur − ul = (β1 − α1 )r1 + (β2 − α2 )r2 = um − ul + ur − um
The jump um − ul moves with velocity λ1 in direction r1 and the
jump ur − um with λ2 in direction r2
Now λ1 ≤ λ2 , which implies velocity of um − ul is smaller - this must
be the first jump!
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Riemann Problem for Nonlinear System
In the general nonlinear case:
ut + A(u)ux = 0
with e-values λ1 (u), . . . , λn (u) and e-vectors r1 (u), . . . , rn (u).
Definition
The p-th characteristic field is genuinely nonlinear, if
∇λp (u) · rp (u) 6= 0,
or it is linear degenerate, if
∇λp (u) · rp (u) = 0
Note: In the linear case all fields are linear degenerate.
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Riemann Problem for Nonlinear System
Construct solution of RP from solutions in the corresponding
characteristic directions.
Not only jump discontinuities moving with velocity λp are possible
but also rarefaction waves and shocks.
In the genuinely nonlinear case one has rarefaction waves and shocks.
Remark
In the genuinely nonlinear case multiple solutions are possible, uniqueness
can be derived from the entropy condition
λp (ul ) > s > λp (ur )
or some generalisations of entropy conditions.
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Section 2: Numerical Approximations
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Numerical Scheme for a Conservation Law
Consider a scalar equation:
ut + f (u)x = 0,
with
f (u) = au - a linear flux function;
u = u(x, t) - a conserved variable;
and a = const - a wave propagation speed.
Take a uniform grid with mesh-size ∆x, ∆t denote mesh size in x
and t, respectively.
Denote uin as an approximation of u(xi , t n ) at the point
(xi = i∆x, t n = n∆t).
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Numerical Scheme for a Conservation Law
ut + f (u)x = 0
Need explicit conservation schemes to the equation above in the form:
uin+1 = uin +
∆t
[f
− fi+1/2 ],
∆x i−1/2
i ∈ Z,
n≥0
where fi+1/2 is the intercell numerical flux and ui0 is given.
Using upwind schemes:
Example 1: a > 0:
uin+1 = uin +
∆t
[aui−1 − aui ]
∆x
uin+1 = uin +
∆t
[aui+1 − aui ]
∆x
Example 2: a < 0:
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Numerical Scheme for a Conservation Law
For upwind schemes above
1
1
fi+1/2 = (f (ui+1 ) + f (ui )) − |a|(ui+1 − ui );
2
2
In general,
n
n
fi+1/2 = F (ui−k+1
, . . . , ui+k
)
where F is a continuous numerical flux.
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Numerical Scheme for a Conservation Law
A naive method:
uin+1 − uin
a
n
=−
(u n − ui−1
)
∆t
2∆x i+1
Rewrite:
a∆t n
n
(u
− ui−1
)
2∆x i+1
This method is not applicable, it is unstable.
Lax-Friedrichs Method:
uin+1 = uin −
a∆t n
1 n
n
n
+ ui−1
)−
− ui−1
)
(u
uin+1 = (ui+1
2
2∆x i+1
The Lax-Friedrichs Method is stable for
a∆t ≤ 1
∆x
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Linear Schemes and Stability
Discretize space leaving time continuous: semi-discretization or
method of lines, to obtain a system of ordinary differential equations.
Carry out stability analysis on the system of ordinary differential
equations.
For Cauchy-Problems consider a subinterval e.g. [0, 1] and prescribe
one boundary condition.
Prescription of boundary condition depends on the direction of
transport e.g. for a > 0, we need a condition at x = 0.
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Linear Schemes and Stability
Example of boundary condition: periodical boundary condition
u(0, t) = u(1, t),
∀t ≥ 0.
Models e.g. Cauchy–Problem with periodical initial conditions of
periodicity 1.
Thus u0 (t) = um+1 (t) is another unknown.
Define the grid vector


u1 (t)


..
U(t) = 
.
.
um+1 (t)
For 2 ≤ i ≤ m the ODE derived using the Naive method
a
ui0 (t) = −
(ui+1 (t) − ui−1 (t)) ,
2∆x
and only the first and the last equations must be modified on the
basis of periodical boundary condition
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Linear Schemes and Stability
Introducing the boundary conditions gives
a
(u2 (t) − um+1 (t)) ,
2∆x
a
0
um+1
(t) = −
(u1 (t) − um (t)) .
2∆x
u10 (t) = −
We get a system of ODEs of the form U 0 (t) = A U(t), with


0
1
−1
−1 0

1



−1 0
1
a 


A=−

 ∈ R(m+1)×(m+1) .
.
.
.
.
.
.

2∆x 
.
.
.



−1 0
1
1
−1 0
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Linear Schemes and Stability
Matrix A is skew-symmetric (AT = −A) , i.e. it’s eigenvalues are
imaginary:
λp =
ia
sin(2πp∆x),
h
p = 1, 2, . . . , m + 1
and the associated eigenvectors have the following components
ujp = exp(2πipj∆x),
j = 1, 2, . . . , m + 1.
The Eigenvalues lie on the imaginary axes between −ia/h and ia/h.
For absolute stability of time discretization, the stability region S
must contain above interval.
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The explicit Euler–Scheme + Naive Method
For explicit Euler method for ordinary differential equations, the
stability region S is the unit disc at −1.
i.e. the method is stable for |1 + ∆tλp | ≤ 1.
eps = 0
1.5
1
0.5
0
−0.5
−1
−1.5
−2.5
−2
−1.5
−1
−0.5
0
0.5
Figure: Eigenvalue distribution for the naive scheme
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The Lax-Wendroff Theorem
Recall:
ut + f (u)x = 0
can be discretized in the form:
uin+1 = uin +
∆t
[f
− fi+1/2 ],
∆x i−1/2
i ∈ Z,
n≥0
where fi+1/2 is the intercell numerical flux and ui0 is given.
Scheme said to be consistent if
F (u, u, . . . , u) = f (u),
∀u
this is a (2k + 1)-point scheme, k = 1 is a three-point scheme.
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The Lax-Wendroff Theorem
Consider
uin+1 = uin +
∆t
[f
− fi+1/2 ],
∆x i−1/2
i ∈ Z,
n≥0
(2)
where fi+1/2 is the intercell numerical flux and ui0 is given
The scheme (2) is also termed conservative and it is in conservative
form.
Theorem
When a conservative consistent scheme ”converges” to a function u (in
some sensible way), the limit u is a weak solution of
ut + f (u)x = 0
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Some remarks about stability
To study stability of the scheme, use the (discrete) Lp -norms for the
sequences u n = (ujn )
For scalar case monotonicity/TVD:
Scheme monotone if given two sequences v 0 = (vj0 ) and w 0 = (wj0 ),
v 0 ≥ w 0 then v 1 ≥ w 1 , where v ≥ w means for all j, vj ≥ wj and
vj1 = (vj0 )
TVD if ∀v 0 = (vj0 ), TV(v 1 ) ≤ TV(v 0 ), where
TV(v ) =
X
(vj+1 − vj )
j∈Z
In general, monotonicity and thus the TVD property is ensured if
min{ukn } ≤ ujn+1 ≤ max{ukn }
k
k
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Why TVD?
Transforms a monotone sequence, say non-decreasing one, into a
monotone (non-decreasing) sequence - oscillations can not occur.
While monotone schemes are at most first-order accurate, it is
possible to design/derive ”high-order” TVD schemes!
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Example: Lax-Friedrichs Scheme
Takes the form:
λ
1 n
n
n
n
vin+1 = (vi+1
+ vi−1
) − (f (vi+1
− f (vi−1
)
2
2
associated with flux:
1
1
g LxF (u, v ) = (f (u) + f (v )) −
(v − u)
2
2λ
∆t
for λ = ∆x
.
The scheme can be re-written in conservative form with the above
flux.
The scheme is consistent.
The scheme is first-order accurate and in the scalar case, under the so
called CFL stability conditions
λ max |f 0 (u)| ≤ 1,
u
it is monotone
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MUSCL - type Scheme
To obtain fi+1/2 , need an extrapolation of the solution in the cell to the
boundary of the cell i.e. need ui+1/2
Godunov first-order upwind method uses piecewise constant data to
extrapolation solution to cell edges.
MUSCL or variable extrapolation approach modifies piecewise
constant data - replace Godunov approach by some monotone
first-order centred scheme, eliminate the Riemann Problem altogether.
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MUSCL - type Scheme
Data reconstruction:
Consider piecewise constant data {uin }, replace constant states uin ,
understood as integral averages in cells Ii = [xi−1/2 , xi+1/2 ], by
piecewise linear functions ui (x):
ui (x) = uin +
(x − xi )
∆i ,
∆x
x ∈ [0, ∆x]
where ∆i is suitably chosen slope of ui (x) in cell Ii .
The centre of xi in local coordinates is x = 12 ∆x and ui (xi ) = uin .
Boundary extrapolated values are:
1
uiL = ui (0) = uin − ∆i ;
2
1
R
n
ui = ui (∆x) = ui + ∆i ;
2
Choose ∆i .
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Slope Limiter Method
Schemes as discussed above will produce spurious oscillations in the
neighborhood of steep gradients, generally.
Construct nonlinear versions by choosing ∆i in the data
reconstruction step using some TVD constraints.
For example, impose restrictions on the evolved boundary
extrapolated values uiR , uiL .
¯ i = φ∆i where φ is a slope limiter function e.g.
Thus choose ∆
minmod, van Leer, Superbee!
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Extension to Nonlinear Systems
Assume 1d time-dependent nonlinear system.
Consider
Ut + F (U)x = 0
with the explicit conservation formula:
Uin+1 = Uin +
∆t
[F
− Fi+1/2 ]
∆x i−1/2
Example L × F method:
1
1 ∆t L
LF
LF
Fi+1/2
= Fi+1/2
(U R , U L ) = [F (U R ) + F (U L )] +
[U − U R ]
2
2 ∆x
Example Rusanov Scheme:
1
1
Rus
Rus
Fi+1/2
= Fi+1/2
(U R , U L ) = [F (U R ) + F (U L )] + |λ|[U L − U R ]
2
2
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Note on designing Numerical Schemes
In Rusanov Scheme λ can be approximated using %(f 0 (u)), the
spectral radius of the Jacobian of f e.g. λ = max{λn (U L ), λn (U R )}
where λn is the largest eigenvalue.
The conservation form is valid whether the flow is smooth or
discontinuous.
To ensure correct shock speed, solve the equation in conservation
form. Thus the problem reduces to a flux estimate at each interface.
The characteristic variables play an essential role: upwind schemes are
generally derived for a scalar convection equation hence for systems
the quantities that are being convected are the characteristic variables.
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Example: Relaxation Systems
Consider:
ut + vx
vt + a2 ux
= 0;
1
= − (v − f (u));
ε
Rewrite in the form
u
0 1
u
+ 2
v t
a 0
v x
=
1
g (u);
ε
Hence the eigenvalues of this Jacobian matrix are ±a. These are also
referred to as characteristic speeds.
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Example: Relaxation Systems
The corresponding eigenvectors:
1 1/a
r1 =
1
2
1 −1/a
r2 =
1
2
1/2a −1/2a
a 1
Hence R =
;
L=
;
1/2
1/2
−a 1
Hence the characteristic
variablestake the form:
a 1
u
v + au
=
=W
LV =
−a 1
v
v − au
The decoupled system takes the form:
∂W
∂W
1
+Λ
= G
∂t
∂x
ε
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Example: Relaxation Schemes
n
Let ∆x = xi+1/2 − xi−1/2 , ∆t = tn+1 − tn , ωi+1/2
:= ω(xi+1/2 , tn ).
ωin
1
=
∆x
Z
x+1/2
ω(x, tn ) dx
x−1/2
Semi-discrete relaxation System takes the form:
dui
+ Dx vi
dt
dvi
+ A2 Dx ui
dt
= 0;
1
= − (vi − f (ui ));
ε
Now approximate the flux at cell boundaries.
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Example: Relaxation Schemes
Consider interval Ii = [xi−1/2 , xi+1/2 ], denote an approximating
polynomial on cell Ii by pi (x, t) then
X
ũ(x, t) =
pi (x, t : u)Xi (x)
i
where X is a characeteristic function defined on cell Ii .
Denote the values of u at cell boundary point between cell Ii and Ii+1 ,
xi+1/2 , as:
u R (xi+1/2 ; u) = pi+1 (xi+1/2 ; u);
u L (xi+1/2 ; u) = pi (xi+1/2 ; u);
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Example: Relaxation Schemes
To apply or MUSCL approach, characteristic variables are used for
reconstruction:
(v − au)i+1/2 = (v − au)R
i+1/2 = pi+1 (xi+1/2 ; v − au);
(v + au)i+1/2 = (v + au)Li+1/2 = pi (xi+1/2 ; v + au);
Giving:
1
pi (xi+1/2 ; v + au) − pi+1 (xi+1/2 ; v − au) ;
2a
1
=
pi (xi+1/2 ; v + au) + pi+1 (xi+1/2 ; v − au) ;
2
ui+1/2 =
vi+1/2
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Example: Relaxation First-Order Scheme
Polynomial: pi (x, u) = ui ;
Hence (v + au)i+1/2 = (v + au)i and (v − au)i+1/2 = (v − au)i+1
Giving:
ui + ui+1 vi+1 − vi
−
;
2
2a
vi + vi+1
ui+1 − ui
=
−a
;
2
2
ui+1/2 =
vi+1/2
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
43 / 95
Example: Relaxation First-Order Scheme - Time
Integration
Given {uin , vin }, compute {uin+1 , vin+1 } by:
ui∗ = uin ;
vi∗ = vin −
∆t ∗
(v − f (ui∗ ));
ε i
(1)
= ui∗ − ∆tDx vi∗ ;
(1)
= vi∗ − ∆ta2 Dx ui∗ ;
ui
vi
(1)
uin+1 = ui ;
(1)
vin+1 = vi
Note: when ε → 0, the equations reduce to the relaxed scheme
(original conservation law) and the time integration is the explicit
Euler scheme with vi+1/2 projected into the local equilibrium
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
44 / 95
Example: Relaxation Second-Order Scheme
Use a second-order polynomial with slope limiters.
Giving:
−
σ + + σi+1
ui + ui+1 vi+1 − vi
−
+ i
;
2
2a
4a
−
σ + − σi+1
vi + vi+1
ui+1 − ui
=
−a
+ i
;
2
2
4
ui+1/2 =
vi+1/2
Using Sweby’s notation: slopes of v ± au can be defined as:
σ ± = (vi+1 ± aui+1 − vi ∓ aui )φ(θi± );
vi ± aui − vi−1 ∓ aui−1
θi± =
;
vi+1 ± aui+1 − vi ∓ aui
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
45 / 95
Example: Relaxation Second-Order Scheme
Some popular slope limiters:
Minmod Slope Limiter:
φ(θ) = max(0, min(1, θ))
van Leer:
φ(θ) =
|θ| + θ
1 + |θ|
Note: if σi± = 0 or φ = 0 we obtain the first-order discretisation!
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
46 / 95
Relaxation Systems [Chen, Levermore, Liu, Yong]
Relaxing System:
Relaxed Equation:
1
∂t u + ∂x f (u) = 0; u ∈ R .
Initial conditions: u(x, 0) = u0 (x),
v ∈ R1 ,
1
∂t v + λ2 ∂x u = − v − f (u) .
ε
∂t u + ∂x v = 0;
v (x, 0) = v0 (x) = f (u0 (x)).
As ε → 0, using Chapman-Enskog Asymptotic Analysis:
v = f (u) + εv1 + ε2 v2 + ε3 v3 + · · · .
i
h
v = f (u); ∂t u + ∂x f (u) = 0; ∂t u + ∂x f (u) = ε (λ2 − f 0 (u)2 )ux ;
x
The sub-characteristic condition is: −λ ≤ f 0 (u) ≤ λ;
∀u.
A linear hyperbolic system with a stiff source term.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
47 / 95
Relaxation Systems for Balance Laws [Katsaonis,
etc.]
Relaxed Equation: ∂t u + ∂x f (u) = g (u, x); u ∈ R1 .
Relaxing Systems:
1
System 1:
∂t u + ∂x v = g (u, x); v ∈ R1 ,
1
∂t v + a2 ∂x u = − v − f (u) .
τ
2
System 2:
v ∈ R1 ,
1Z
1
∂t v + a2 ∂x u = − v − f (u) +
τ
τ
∂t u + ∂x v = 0;
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
x
g (u, s)ds.
Jul 22 - Aug 3
48 / 95
Semi-Discrete Approach
Consider
dui
+ Dx vi
dt
dvi
+ A2 Dx ui
dt
= 0,
1
= − vi − f(ui ) .
ε
Strategy:
Treat space discretizations separately using a MUSCL-type
formulation.
Treat time by an ODE solver e.g. Implicit-Explicit (IMEX) schemes.
Choice of A2 is wide: ai ’s can be chosen as global characteristic
speeds, local speeds, etc.
Source terms are incorporated accordingly.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
49 / 95
General ODE
A semi-discrete formulation as a system of ODEs
1
dY
= F(Y) − G(Y),
dt
ε
Time Integration - Implicit-Explicit (IMEX) Runge-Kutta:
1
2
3
4
Treat non-stiff stage, F, with an explicit Runge-Kutta scheme.
Treat the stiff stage, G, with a diagonally implicit Runge-Kutta (DIRK)
scheme.
Scheme must be asymptotic-preserving.
The limiting Scheme i.e. as ε → 0 must be SSP.
kY n+1 k ≤ kY n k
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
50 / 95
IMEX for Time Integration [Pareschi, Russo]
∆t is time step, Y n the approx. solution at t = n∆t.
IMEX method is implemented as:
For l = 1, . . . , s,
1
Evaluate K∗l as:
K∗l = Y n + ∆t
l−1
X
ãlm F(Km ) −
m=1
2
Solve for Kl :
Kl = K∗l −
l−1
∆t X
alm G(Km ).
ε m=1
∆t
all G(Kl ).
ε
Update Y n+1 as:
Y
n+1
n
= Y + ∆t
s
X
l=1
Mapundi K. Banda (Maties)
s
∆t X
b̃l F(Kl ) −
bl G(Kl ).
ε
CIMPA/MPE 2013
l=1
Jul 22 - Aug 3
51 / 95
Examples of IMEX Schemes - Butcher Tableau
Second-Order Scheme:
0
0
0
−1
−1
0
1
1
0
2
1
1
1
2
1
2
1
2
1
2
Third-Order Scheme
0
0
0
0
0
0
0
0
γ
γ
0
0
γ
0
γ
0
1−γ
γ−1
2 − 2γ
0
1−γ
0
1 − 2γ
γ
0
1
2
1
2
0
1
2
1
2
CIMPA/MPE 2013
Jul 22 - Aug 3
where γ =
√
3+ 3
6 .
Mapundi K. Banda (Maties)
52 / 95
Remarks on IMEX
Neither linear algebraic nor nonlinear source terms can arise.
As ε −→ 0 the time integration procedure tends to an SSP time
integration scheme of the limit equations.
The only restriction is the usual CFL condition
∆t ∆t
CFL = max
,λ
≤ 1.
∆x ∆x
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
53 / 95
Example: Euler Equations of Gas Dynamics.
∂t ρ + ∂x m = 0;
∂t m + ∂x (ρu 2 + p) = 0;
∂t E + ∂x (u(E + p)) = 0;
ρ
p = (γ − 1)(E − u 2 ).
2
Riemann problems.
uL , x < 0;
u(x, 0) =
uR , x > 0.
Mapundi K. Banda (Maties)
+ transparent boundary conditions
CIMPA/MPE 2013
Jul 22 - Aug 3
54 / 95
Example: Sod’s Riemann initial data.
N = 400,T = 0.1644
N = 400,T = 0.1644
1
1
Exact
Upw
JX
RCWENO
0.9
0.8
0.8
0.7
Pressure
Density
0.7
0.6
0.5
0.4
0.6
0.5
0.4
0.3
0.3
0.2
0.2
0.1
0.1
0
Exact
Upw
JX
RCWENO
0.9
0.2
0.4
0.6
0.8
1
0
0
x
Mapundi K. Banda (Maties)
0.2
0.4
0.6
0.8
1
x
CIMPA/MPE 2013
Jul 22 - Aug 3
55 / 95
Example: Euler Equations of Gas Dynamics in Two
dimensions.
∂t ρ + ∂x m + ∂y n = 0;
∂t m + ∂x (ρu 2 + p) + ∂y (ρuv ) = 0;
∂t n + ∂x (ρuv ) + ∂y (ρv 2 + p) = 0;
∂t E + ∂x (u(E + p)) + ∂y (v (E + p)) = 0;
ρ
p = (γ − 1)(E − (u 2 + v 2 )).
2
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
56 / 95
Double Mach Reflection
Domain Ω = [0, 4] × [0, 1].
Reflecting wall in bottom interval: [1/6, 4].
Mach 10 shock at x = 1/6, y = 0, 600 angle with x-axis.
Impose exact post-shock condition at bottom boundary [0, 1/6].
Reflective boundary for the rest of the boundary.
Top boundary exact motion of a Mach 10 shock is used.
Results at t = 0.2.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
57 / 95
Double Mach Reflection
∆x = ∆y = 1/120
1
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
2.5
3
3.5
4
3
3.5
4
∆x = ∆y = 1/240
1
0.8
0.6
0.4
0.2
0
0
0.5
Mapundi K. Banda (Maties)
1
1.5
2
CIMPA/MPE 2013
2.5
Jul 22 - Aug 3
58 / 95
Example: Forward Facing Step
Right-going Mach 3 uniform flow enters a wide tunnel 1 unit by 3
units long.
Step is 0.2 units high located 0.6 units from left hand end of tunnel.
Initialize by uniform right-going Mach 3 flow.
Reflecting boundary conditions along walls, inflow and outflow
boundary conditions are applied at entrance and exit of tunnel.
Results at t = 4.0.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
59 / 95
Example: Forward Facing Step
∆x = ∆y = 1/80
1
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
2.5
3
2
2.5
3
∆x = ∆y = 1/160
1
0.8
0.6
0.4
0.2
0
0
Mapundi K. Banda (Maties)
0.5
1
1.5
CIMPA/MPE 2013
Jul 22 - Aug 3
60 / 95
Section 3: Kinetic Formulation
Mapundi K. Banda (Maties)
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Jul 22 - Aug 3
61 / 95
The Boltzmann Equation
Consider a rarefied monatomic perfect gas and assume space
dimension, d = 3
The Boltzmann Equation takes the form:
∂f
+ v · ∇f = Q(f , f )
∂t
defines time evolution of one-particle distribution f (x, v, t)
(x, v) ∈ R2d is the phase space
x is position vector, v is the molecular velocity
f considered as expected mass density in phase space hence
Z
ρ = ρ(x, t) =
f (x, v, t) dv
Rd
is the mass per unit volume, i.e. density in physical space
Mapundi K. Banda (Maties)
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Jul 22 - Aug 3
62 / 95
Collision Operator
Q(f , f ) is a quadratic integral operator, acting on the velocity
dependence of f
Q(f , g ) = Q(g , f ) - symmetry
Main Property:
Theorem
The states of thermodynamic equilibrium characterized by Q(f , f ) = 0 are
obtained for the Maxwellian distributions
f (v) = A exp(−β|v − u|2 ),
where A ∈ R+ , u ∈ Rd , β ∈ R+ are arbitrary parameters.
Mapundi K. Banda (Maties)
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Jul 22 - Aug 3
63 / 95
Collision Operator
As a consequence:
h(r ) = r log r
is the microscopic or kinetic entropy, and is strictly convex
It can be found
Z
Z
∂
h(f ) dv +
v · ∇h(f ) dv ≤ 0
∂t Rd
Rd
equality holds iff f is a Maxwellian
R
Let H(f ) = Rd h(f ) dv and Ψ(f ) = (Ψi (f )) where
Z
Z
Ψi (f ) =
vi h(f ) dv =
vi f log f dv
Rd
then
Mapundi K. Banda (Maties)
Rd
∂
H(f ) + ∇ · Ψ(f ) ≤ 0
∂t
CIMPA/MPE 2013
Jul 22 - Aug 3
64 / 95
Entropy and the Maxwellian
It can be proven that
Z
H=
Z
H(f ) dx =
R
h(f ) dvdx
Rd ×R
decreases
dH
≤0
dt
H is constant iff f is a Maxwellian
The Boltzmann Equation describes evolution (”relaxation”) towards a
state of minimum H
Thus the final state is a steady state and thus a Maxwellian i.e. the
distribution function in an equilibrium state is a Maxwellian
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
65 / 95
The Hydrodynamic Limit
Proposition
Assume that f satisfies the Boltzmann Equation. Then the vector U
defined as
 
 
1
Z
ρ
U(x, t) = ρu (x, t) =
f (x, v, t)  v  dv
|v|2
Rd
ρe
2
satisfies the system
d
∂ρ X ∂
+
(ρuj ) = 0;
∂t
∂xj
j=1
∂ρe
+
∂t
d
X
j=1
∂
(ρuj e +
∂xj
Mapundi K. Banda (Maties)
d
X
d
∂ρui X ∂
+
(ρui uj + πij ) = 0;
∂t
∂xj
j=1
πjk uk + Qj ) = 0;
k=1
CIMPA/MPE 2013
Jul 22 - Aug 3
66 / 95
The Hydrodynamic Limit
Specific total energy e satisfies:
Z
Z
|v|2
|u|2
|v − u|2
f (x, v, t)
dv =
dv + ρ
f (x, v, t)
2
2
2
Rd
Rd
Thus the internal energy (per unit volume)
Z
|v − u|2
dv
ρε =
f (x, v, t)
2
Rd
The peculiar velocity C = v − u
The stress tensor π = (πij )1≤i,j≤3 ,
Z
3
2
1X
f (x, v, t)Ci Cj dv; p = ρε =
πij =
πii
3
3
Rd
i=1
The heat flow vector Q = (Qi ):
Z
1
Qi =
f (x, v, t)Ci |C |2 dv,
2 Rd
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
1≤i ≤3
Jul 22 - Aug 3
67 / 95
The Hydrodynamic Limit
Corollary
|v − u|2 3
Assume that f is a Maxwellian f (v) = (2πRT )− 2 ρ exp −
. Then
2RT
the vector U satisfies the Euler equations
d
d
∂ρ X ∂
+
(ρuj ) = 0;
∂t
∂xj
∂ρui X ∂
∂p
+
(ρui uj ) +
= 0;
∂t
∂xj
∂xi
j=1
∂ρe
+
∂t
d
X
j=1
j=1
∂
((ρe + p)uj ) = 0;
∂xj
where, for a monatomic perfect gas in dimension d = 3,
p = ρRT =
Mapundi K. Banda (Maties)
2ρε
3
Boyle’s Law
CIMPA/MPE 2013
Jul 22 - Aug 3
68 / 95
The BGK Model
∂f
M(v) − f
+ v · ∇f =
∂t
τ
where M(v; ρ, u, T ) = M(v) is the (local) Maxwellian given by
|v − u|2
d
M(v) = (2πRT (x, t))− 2 ρ(x, t) exp −
(x, t)
2RT
1
J(f ) = (M(v) − f ) is constructed in order to satisfy the following
ν
properties
Z
J(f )K (v) dv = 0,
∀f ≥ 0
Rd
where K (v) is the vector of collision invariants.
Mapundi K. Banda (Maties)
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Jul 22 - Aug 3
69 / 95
Numerical Methods
0
Ma → 0
- IC. NS
C. NS
∼
Newton
N→∞Boltzmann
ν→0
ν→0
@
→
0
@
R
@
?
C. E
?
- IC. E
Ma → 0
Figure: N: Number of particles, : mean free path, ν: viscosity, Ma: Mach
number
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
70 / 95
Finite Discrete-Velocity Method - FDVM
Consider the continuous velocity Boltzmann-type Equation:
∂t f + v.∇x f = J(f );
x ∈ Ω ⊂ R2 ;
v ∈ R2 .
Discretise the velocity (e.g. D2Q9 Model):
Discrete-Velocity Eqn
Scaling
∂t fi + ci .∇x fi = J(f );
∂t fi +
v ∈ {c0 , . . . , c8 },
1
1
ci .∇x fi = 1+α J(f );
α
ε
ε
ci ∈ R2 .
The simplified BGK Model (isothermal):
1
J(f ) = − (f − f eq (ρ, εα u))
τ
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
71 / 95
Finite Discrete-Velocity Method - FDVM
Consider the continuous velocity Boltzmann-type Equation:
∂t f + v.∇x f = J(f );
x ∈ Ω ⊂ R2 ;
v ∈ R2 .
Discretise the velocity (e.g. D2Q9 Model):
c2
c6
c5
c0
c1
c3
c7
c4
c8
Figure: Links in the D2Q9 lattice Boltzmann method.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
72 / 95
Kinetic Relaxation System ⇒ Fluid Dynamic
Relaxation System
Take velocity moments, obtain macroscopic variables:
Z
Z
X
X
ρ = f dv; ρu = f v dv; ρ =
fi ; ρu =
fi ci . . .
i
i
Conservation of mass and momentum demand:
X
X
0=
J(fi ); 0 =
J(fi )ci . . .
i
i
Transform into an equivalent set of moment equations: a
relaxation-type system:
∂t ρ + div ρu = 0,
- 0th
1
∂t (ρu) + div Θ + 2α ∇ρ = 0,
-1st
3
2
1
1
∂t Θ + 2α S[ρu] + Q[q] = − 1+α Θ − ρu ⊗ u .
3
3
τ
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
73 / 95
Kinetic Relaxation System ⇒ Fluid Dynamic
Relaxation System
1
2∂x u1
∂y u1 + ∂x u2
S[u] =
2∂y u2
2 ∂y u1 + ∂x u2
∂y q2
∂y q1 + ∂x q2
Q[q] =
.
∂y q1 + ∂x q2
∂ x q1
The second-order moment forms a symmetric tensor Θ.
Note ∇ρ → 0 as → 0, hence ρ = ρ(1 + 32α p), set ρ = 1.
The system is closed by third and Fourth order moment equations.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
74 / 95
Relaxed Limit Equations ε → 0
For α = 0, we obtain, in lowest order the isothermal Euler equations
∂t ρ + div (ρu) = 0,
∂t (ρu) + div (ρu ⊗ u) + ∇p = 0.
For α = 1, we obtain the incompressible Navier-Stokes equations
div u = 0,
∂t u + div (u ⊗ u) + ∇p =
τ
∆u;
3
Re =
3
.
τ
For 0 < α < 1, we obtain the incompressible Euler equations
div u = 0,
∂t u + div (u ⊗ u) + ∇p = 0.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
75 / 95
Simplified Relaxing System
Rewrite the relaxation-type system:
∂t U + ∂x V + ∂y W = −(1 − 32α )S(U),
1
2
∂t V + a ∂x U = − 2 V − F(U) ,
τ
1
2
∂t W + b ∂y U = − 2 W − G(U) ;
τ
where U = (p, ρu)T , V = (w1 , Θ1 )T , W = (w2 , Θ2 )T ,
S(U) = ( 312α div ρu, ∇p)T , F(U) = ρu and
G(U) = u ⊗ u − 1−α τ S[ρu] + 32α pI.
Note:
1
2
A linear hyperbolic system with a stiff source term.
Linear characteristic variables given by
V ± aU,
Mapundi K. Banda (Maties)
and
CIMPA/MPE 2013
W ± bU.
Jul 22 - Aug 3
76 / 95
Example of Relaxation Numerical Schemes
A semi-discrete formulation as a system of ODEs
T
dY
1
= F(Y) − 2 G(Y),
Y := Ui,j , Vi,j , Wi,j ;
dt
τ


−(1 − 32α )S(U)i,j − Dx Vi,j − Dy Wi,j
;
F(Y) := 
−a2 Dx Ui,j
−b2 Dy Ui,j
T
G(Y) := 0, Vi,j − F(U)i,j , Wi,j − G(U)i,j .
Dx and Dy are discretised differential operators.
Time Integration - Implicit-Explicit (IMEX) Runge-Kutta:
1
2
Treat non-stiff stage, F, with an explicit Runge-Kutta scheme.
Treat the stiff stage, G, with a diagonally implicit Runge-Kutta (DIRK)
scheme.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
77 / 95
Numerical Results - FDVM
Example 1: Thick Shear Layer
Let (x, y ) ∈ [0, 2π]2 . The initial conditions are
u(x, y , 0) =
(
tanh( ρ1 (y − π/2))
tanh( ρ1 (3π/2
y ≤π
− y )) y > π,
v (x, y , 0) = δ sin(x)
with δ = 0.05 and ρ = π/15.
Transition between incompressible Euler and Navier-Stokes equations.
Choose α = 12 (incompressible Euler limit) and α = 1 (incompressible
Navier-Stokes limit). Below are the vortices at time t = 8 with
= 10−6 .
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
78 / 95
Example 1: Thick Shear Layer
α = 1 and t = 8
α = 1/2 and t = 8
6
6
5
5
4
4
3
3
2
2
1
1
0
0
0
1
2
3
4
5
6
0
1
2
3
4
5
6
Figure: Thick Shear Layer flow problem.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
79 / 95
Example 2: Lid Driven Cavity
x ∈ [0, 1]2 and the usual boundary conditions with a drift u = (ū, 0)
parallel to the boundary at the top of the square and u = 0 at the
other sides.
ū = 1.0 and = 10−6 . We use a third-order relaxation method with
128 × 128 grid cells. Stream-functions for Re = 1000 at t = 100 and
Re = 10000, t = 1000.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
80 / 95
Example 2: Lid Driven Cavity
τ = 10−2, t = 100
τ = 10−3, t = 100
1
1
0.9
0.9
0.8
0.8
0.7
0.7
0.6
0.6
0.5
0.5
0.4
0.4
0.3
0.3
0.2
0.2
0.1
0.1
0
0
0.1
0.2
0.3
0.4
0.5
Mapundi K. Banda (Maties)
0.6
0.7
0.8
0.9
1
0
0
0.1
CIMPA/MPE 2013
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Jul 22 - Aug 3
0.9
1
81 / 95
Extensions
Shallow Water Flow in Complex Domains
1
Apply lattice Boltzmann Method.
2
Modified Equilibrium distribution function (Salmon, Dellar).
Turbulence
1 Relaxation time, τ = τ (x) ⇒ ν = ν + ν .
0
t
2
3
LargeREddy Simulation: ψ = ψ + ψ 0 where
ψ 0 = R ψ(y, t)G∆ (x − y) dy G∆ is a filter.
Hence we obtain a Reynold’s subgrid scale tensor:
T (u) ≈ −νt (u)S[u]
Smagorinski.
where
νt (u) = (cs ∆)2 kS[u]k,
Mapundi K. Banda (Maties)
1
kS[u]k = (S[u] : S[u]) 2
CIMPA/MPE 2013
Jul 22 - Aug 3
82 / 95
Extensions
Incompressible Flow with Radiative Heat Transfer
1
Incompressible Navier-Stokes equations modified to include buoyancy
terms.
div u = 0,
∂t u + div (u ⊗ u) + ∇p =
2
τ
∆u + G;
3
Temperature equation with a radiative source term.
∂t gi + ci · ∇gi = J(g ) ⇒ ∂t T + div (uT ) = κ∆T − ∇ · QR .
3
Radiative Transfer solved by Discrete-Ordinate Method.
Z
σa
ω · ∇I + (σa + σs )I =
I dω + σs B(T )
4π S 2
Mapundi K. Banda (Maties)
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Jul 22 - Aug 3
83 / 95
Test Problem 2: Turbulence Mixing Layer Problem
1
Domain: [−1, 1] × [−1, 1] with periodic boundary conditions in
x-direction and free-slip conditions at the top and bottom boundary.
2
Re = 10, 000 = δu∞ /ν0 , based on initial vorticity thickness δ and
free-stream velocity u∞ .
3
Initial Conditions:
y
u(y ) = u∞ tanh( ),
δ
4
δ=
1
14
Superpose two divergence-free functions of the form:
u0 =
5
u∞ = 1,
2y 2
∂ψ 0
∂ψ
, v = − , ψ = 0.001u∞ (cos(8πx) + cos(20πx))e −( δ )
∂y
∂x
Grid: 256 × 256; Smagorinski constant: cs = 0.0316.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
84 / 95
Test Problem 2: Turbulence Mixing Layer Problem
t = 1.3
t = 2.5
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0
−0.2
−0.2
−0.4
−0.4
−0.6
−0.6
−0.8
−0.8
−1
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
−1
−1
−0.8
−0.6
−0.4
−0.2
t = 5
0
0.2
0.4
0.6
0.8
1
0.2
0.4
0.6
0.8
1
t = 10
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0
−0.2
−0.2
−0.4
−0.4
−0.6
−0.6
−0.8
−0.8
−1
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
−1
−1
−0.8
−0.6
−0.4
−0.2
0
Figure: Vorticity obtained by the fifth-order scheme at four different times.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
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Test Problem 4: Incompressible flow with Radiation
11111111111111111111111
00000000000000000000000
adiabatic
L
TC
adiabatic
L
TH
11
00
00
11
00
11
00
11
00
11
00
11
00
11
00
11
00
11
00
11
00
11
00
11
00
11
00
11
00
11
00
11
00
11
00
11
00
11
00
11
00
11
00
11
00
11
L
TC
1
0
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
adiabatic
TH
adiabatic
11111111111111111111111
00000000000000000000000
00000000000000000000000
11111111111111111111111
L
Figure: Geometry: natural (left) and forced (right) convection-radiation.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
86 / 95
Test Problem 4: Natural Convection-Radiation
Ra = 104
Ra = 105
Ra = 106
Ra = 107
1
1
1
1
0.9
0.9
0.9
0.9
0.8
0.8
0.8
0.8
0.7
0.7
0.7
0.7
0.6
0.6
0.6
0.6
0.5
0.5
0.5
0.5
0.4
0.4
0.4
0.4
0.3
0.3
0.3
0.3
0.2
0.2
0.2
0.2
0.1
0.1
0.1
0
0
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
0.1
0.2
0.3
0.4
4
0.5
0.6
0.7
0.8
0.9
1
0.1
0
0
0.1
0.2
0.3
0.4
5
Ra = 10
0.5
0.6
0.7
0.8
0.9
1
0
1
0.9
0.9
0.9
0.9
0.8
0.8
0.8
0.8
0.7
0.7
0.7
0.7
0.6
0.6
0.6
0.6
0.5
0.5
0.5
0.5
0.4
0.4
0.4
0.4
0.3
0.3
0.3
0.3
0.2
0.2
0.2
0.2
0.1
0.1
0.1
0
0
0
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
0.1
0.2
0.3
0.4
0.5
0.4
0.6
0.7
0.8
0.9
1
0.5
0.6
0.7
0.8
0.9
1
0.6
0.7
0.8
0.9
1
7
1
0.2
0.3
Ra = 10
1
0.1
0.2
Ra = 10
1
0
0.1
6
Ra = 10
0.1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
0.1
0.2
0.3
0.4
0.5
Figure: Isotherms (top row) and streamlines (bottom row). From left to right:
Ra = 104 , 105 , 106 and 107 .
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
87 / 95
Test Problem 4: Natural Convection - no Radiation
Ra = 104
Ra = 105
Ra = 106
Ra = 107
1
1
1
1
0.9
0.9
0.9
0.9
0.8
0.8
0.8
0.8
0.7
0.7
0.7
0.7
0.6
0.6
0.6
0.6
0.5
0.5
0.5
0.5
0.4
0.4
0.4
0.4
0.3
0.3
0.3
0.3
0.2
0.2
0.2
0.2
0.1
0.1
0.1
0
0
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
0.1
0.2
0.3
0.4
4
0.5
0.6
0.7
0.8
0.9
1
0.1
0
0
0.1
0.2
0.3
0.4
5
Ra = 10
0.5
0.6
0.7
0.8
0.9
1
0
1
0.9
0.9
0.9
0.9
0.8
0.8
0.8
0.8
0.7
0.7
0.7
0.7
0.6
0.6
0.6
0.6
0.5
0.5
0.5
0.5
0.4
0.4
0.4
0.4
0.3
0.3
0.3
0.3
0.2
0.2
0.2
0.2
0.1
0.1
0.1
0
0
0
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
0.1
0.2
0.3
0.4
0.5
0.4
0.6
0.7
0.8
0.9
1
0.5
0.6
0.7
0.8
0.9
1
0.6
0.7
0.8
0.9
1
7
1
0.2
0.3
Ra = 10
1
0.1
0.2
Ra = 10
1
0
0.1
6
Ra = 10
0.1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
0.1
0.2
0.3
0.4
0.5
Figure: Isotherms (top row) and streamlines (bottom row). From left to right:
Ra = 104 , 105 , 106 and 107 .
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
88 / 95
Dankie, Thank you, E Nkosi
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
89 / 95
References
E. Godlewski, P.-A. Raviart, Numerical Approximation of Hyperbolic
systems of Conservation Laws, Springer, Berlin, 1996
A. Klar, Numerical methods of partial differential equations II:
Hyperbolic conservation laws, Lecture Notes, Kaiserslautern.
C. Hirsch, Numerical Computation of Internal and External Flows I,
Wiley, New York, 2001.
C. Hirsch, Numerical Computation of Internal and External Flows II,
Wiley, New York, 2000.
B. Perthame, Transport Equations in Biology, Birkhäuser, Berlin,
2000.
P. D. Lax, Hyperbolic Partial Differential Equations, AMS, New York,
2006.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
90 / 95
References
M.K. Banda, M. Herty, J.-M. T. Ngnotchouye, Towards a
mathematical analysis for drift-flux multiphase flow models in
networks, SIAM J. Sci. Comput., 31(6), 4633 – 4653 (2010),
DOI:10.1137/080722138.
M.K. Banda, M. Seaid, G. Thoemmes, Lattice Boltzman Simulation
of Dispersion in Two-Dimensional Tidal Flows, Int. J. Numer. Meth.
Engng, 77, 878–900 (2009); DOI: 10.1002/nme.2435.
M. K. Banda, M. Herty, Multiscale modelling for gas flow in pipe
networks, Math. Meth. Appl. Sc., 31(8), 915–936 (2008); DOI:
10.1002/mma.948.
M.K. Banda, M. Seaid, I. Teleaga, Large-Eddy Simulation of Thermal
Flows based on Discrete-Velocity Models, SIAM J. Sci. Comput.,
30(4), 1756–1777 (2008); DOI: 10.1137/070682174.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
91 / 95
References
M.K. Banda, A. Klar, L. Pareschi, M. Seaid, Lattice-Boltzman type
relaxation systems and high order relaxation schemes for
incompressible Navier-Stokes equations, Math. of Comp., 77, 943–965
(2008); DOI: 10.1090/S0025 - 5718 - 07 - 02034 - 0.
M.K. Banda, A. Klar, M. Seaid, A Lattice-Boltzman Relaxation
Scheme for Coupled Convection-Radiation Systems, J. Comput. Phys.,
226, 1408–1431 (2007); DOI: 10.1016/j.jcp.2007.05.030.
M. K. Banda, M. Seaid, Relaxation WENO Schemes for
Multi-Dimensional Hyperbolic Systems of Conservation Laws, Num.
Meth. for PDEs., 23, 1211 – 1334, (2007); DOI: 10.1002/num.20218.
G. Thoemmes M. Seaid, M.K. Banda, Lattice Boltzman Methods for
Shallow Water Flow Applications, Int. J. for Num. Meth. in Fluids,
55(7), 673–692 (2007); DOI 10.1002/fld.1489.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
92 / 95
References
M.K. Banda, M. Herty, A. Klar, Coupling conditions for gas networks
governed by the isothermal Euler equations, NHM, 1(2), 295-314
(2006).
M. K. Banda, W-A. Yong, A. Klar, A Stability Notion of Lattice
Boltzmann Equations, SIAM J. Sci. Comput., 27(6), 2098-2111
(2006).
M. K. Banda, M. Herty, A. Klar, Gas Flow in Pipeline Networks,
NHM, 1(1), 41-56 (2006).
M.K. Banda, M. Herty, Numerical Discretization of Stabilization
Problems with Boundary Controls for Systems of Hyperbolic
Conservation laws, Mathematical Control and Related Fields, 3(2),
121 - 142 (2013).
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
93 / 95
References
M.K. Banda, M. Seaid, Discrete-Velocity models and Lattice
Boltzmann Methods for Convection-Radiation Problems, Novel Trends
in Lattice Boltzmann Methods, Reactive Flow, Physicochemical
Transport and Fluid-Structure-Interaction; Matthias Ehrhardt (ed.);
e-Book series Progress in Computational Physics (PiCP), Vol. 3,
Bentham Science Publishers, pp: 53-90 (2013), Chapter DOI:
10.2174/9781608057160113030006, DOI:
10.2174/97816080571601130301, eISBN: 978-1-60805-716-0, ISBN:
978-1-60805-717-7, ISSN :1879-4461.
Mapundi K. Banda and Mohammed Seaid, Lattice Boltzmann
Simulation for Shallow Water Flow Applications, Hydrodynamics Theory and Model, Dr. Jin - Hai Zheng (Ed.), pp. 255–286 (2012).
ISBN: 978-953-51-0130-7, InTech, Available from:
http://www.intechopen.com/books/hydrodynamics-theory-andmodel/lattice-boltzmannsimulation-for-shallow-water-flow-applications
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
94 / 95
Acknowledgements
CIMPA, IMU, LMS-AMMSI, ICTP, AIMS
Prof. J. Banasiak, College of Mathematics, Computer Science and
Statistics, University of KwaZulu-Natal, Durban.
National Research Foundation, South Africa.
University of KwaZulu-Natal, Witwatersrand and Stellenbosch.
Mapundi K. Banda (Maties)
CIMPA/MPE 2013
Jul 22 - Aug 3
95 / 95