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Structure preserving schemes for conservation lawsWell-balanced schemes for Euler equations with gravity
Praveen [email protected]
Center for Applicable MathematicsTata Institute of Fundamental Research
Bangalore 560065
International Conference on Advances in Scientific ComputingIIT Madras, 28-30 November, 2016
Supported by Airbus Foundation Chair at TIFR-CAM, Bangalore
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Structure preserving schemes
• Usual considerations: stability and accuracy
• Accuracy is an asymptotic property: ∆x→ 0
• Euler equations
I Density, temperature must be positiveI Smooth solutions: entropy is conservedI Steady solutions: enthalpy is constant along streamlinesI Incompressible: kinetic energy is conservedI With gravity: non-trivial stationary solutions
• Shallow water: energy and potential enstrophy conserved
• Desirable to have numerical solutions satisfy these additionalproperties
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Acknowledgements
• Deep Ray, TIFR-CAM
• Dept. of Mathematics, Univ. of Wurzburg, GermanyI Jonas Berberich (master student)I Juan Pablo Gallego (PhD student)I Christian Klingenberg (professor)I Markus Zenk (PhD student)I Dominik Zoar (master student)
• Heidelberg Institute of Theoretical ScienceI Phillip Edelmann (post-doc)I Fritz Ropke (professor)
• DAADI Research stay programI Passage to India program
• Airbus Foundation Chair at TIFR-CAM
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Euler equations with gravity
Flow propertiesρ = density, u = velocity
p = pressure, E = total energy = e+1
2ρu2
Gravitational potential φ; force per unit volume of fluid
−ρ∇φ
System of conservation laws
∂ρ
∂t+
∂
∂x(ρu) = 0
∂
∂t(ρu) +
∂
∂x(p+ ρu2) = −ρ∂φ
∂x∂E
∂t+
∂
∂x(E + p)u = −ρu∂φ
∂x
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Euler equations with gravity
In compact notation
∂q
∂t+∂f
∂x= −
0ρρu
∂φ∂x
where
q =
ρρuE
, f =
ρup+ ρu2
(E + p)u
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Hydrostatic solutions
• Fluid at rest
ue = 0
• Momentum equationdpedx
= −ρedφ
dx(1)
• Assume ideal gas and some temperature profile Te(x)
pe(x) = ρe(x)RTe(x), R = gas constant
integrate (1) to obtain
pe(x) exp
(−∫ x
x0
φ′(s)RTe(s)
ds
)= p0
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Hydrostatic solutions
• Isothermal hydrostatic state, i.e., Te(x) = Te = const, then
pe(x) exp
(φ(x)
RTe
)= const (2)
Density ρe(x) =pe(x)
RTe• Polytropic hydrostatic state
peρ−νe = const, peT
− νν−1
e = const, ρeT− 1ν−1
e = const (3)
where ν > 1 is some constant. From (1) and (3), we obtain
νRTe(x)
ν − 1+ φ(x) = const
E.g., pressure is
pe(x) = C1 [C2 − φ(x)]ν−1ν
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Well-balanced scheme
• General evolutionary PDE
∂q
∂t= R(q)
Stationary solution qe; scheme is well-balanced if
R(qe) = 0
• We are interested in computing small perturbations
q(x, 0) = qe(x) + εq(x, 0), ε� 1
• Perturbations are governed by linear equation
∂q
∂t= R′(qe)q
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Well-balanced scheme
• Some numerical scheme
∂qh∂t
= Rh(qh)
• qh,e = interpolation of qe onto the mesh
Scheme is well balanced if
Rh(qh,e) = 0
• Suppose scheme is not well-balanced Rh(qh,e) 6= 0. Solution
qh(x, t) = qh,e(x) + εqh(x, t)
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Well-balanced scheme
• Linearize the scheme around qh,e
∂
∂t(qh,e + εqh) = Rh(qh,e + εqh) = Rh(qh,e) + εR′h(qh,e)qh
or∂qh∂t
=1
εRh(qh,e) +R′h(qh,e)qh
• Scheme is consistent of order r: Rh(qh,e) = Chr‖qh,e‖
∂qh∂t
=1
εChr‖qh,e‖+R′h(qh,e)qh
• ε� 1 then first term may dominate the second term; need h� 1and/or r � 1
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Evolution of small perturbations
The initial condition is taken to be the following
φ =1
2x2, ue = 0, ρe(x) = pe(x) = exp(−φ(x))
Add small perturbation to equilibrium pressure
p(x) = exp(−φ(x)) + ε exp(−100(x− 1/2)2), 0 < ε� 1
Non-well-balanced scheme
∂φ
∂x(xi) ≈
φi+1 − φi−12∆x
, reconstruct ρ, u, p
Using exact derivative of potential does not improve results. In practice, φis only available at grid points.
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Evolution of small perturbations
0 0.2 0.4 0.6 0.8 1−2
0
2
4
6
8
10x 10
−4
x
Pre
ssur
e pe
rtur
batio
n
InitialWell−balancedNon well−balanced
0 0.2 0.4 0.6 0.8 1−2
0
2
4
6
8
10x 10
−6
x
Pre
ssur
e pe
rtur
batio
n
InitialWell−balancedNon well−balanced
ε = 10−3, N = 100 cells ε = 10−5, N = 100 cells
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Evolution of small perturbations
0 0.2 0.4 0.6 0.8 1−2
0
2
4
6
8
10x 10
−6
x
Pre
ssur
e pe
rtur
batio
n
InitialWell−balancedNon well−balanced
0 0.2 0.4 0.6 0.8 1−2
0
2
4
6
8
10x 10
−6
xP
ress
ure
pert
urba
tion
InitialWB, cells=100WB, cells=500
ε = 10−5, N = 500 cells ε = 10−5
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1-D finite volume scheme
Define (Xing & Shu, 2013)
ψ(x) = −∫ x
x0
φ′(s)RT (s)
ds, x0 is arbitrary
Euler equations
∂q
∂t+∂f
∂x= −
0ρρu
∂φ∂x
=
0ppu
exp(−ψ(x))
∂
∂xexp(ψ(x))
• Divide domain into N finite volumes each of size ∆x
• i’th cell = (xi− 12, xi+ 1
2)
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1-D finite volume scheme
• semi-discrete finite volume scheme for the i’th cell
dqidt
+fi+ 1
2− fi− 1
2
∆x= e−ψi
(eψi+1
2 − eψi− 1
2
∆x
)
0pipiui
(4)
• ψi, ψi+ 12
etc. are consistent approximations to the function ψ(x)
• consistent numerical flux fi+ 12
= f(qLi+ 1
2
, qRi+ 1
2
)
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1-D finite volume scheme
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1-D finite volume scheme
Def: Property C
The numerical flux f is said to satisfy Property C if for any two states
qL = [ρL, 0, p/(γ − 1)] and qR = [ρR, 0, p/(γ − 1)]
we havef(qL, qR) = [0, p, 0]>
• New set of independent variables
w = w(q, x) :=[ρe−ψ, u, pe−ψ
]>
• We can compute q = q(w, x)
• States for computing numerical flux
qLi+ 1
2
= q(wLi+ 1
2
, xi+ 12), qR
i+ 12
= q(wRi+ 1
2
, xi+ 12)
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1-D finite volume scheme
• First order scheme
wLi+ 1
2
= wi, wRi+ 1
2
= wi+1
• Higher order scheme: Reconstruct w variables, e.g.,
wLi+ 1
2
= wi +1
2m(θ(wi −wi−1), (wi+1 −wi−1)/2, θ(wi+1 −wi))
1. Chandrashekar & Klingenberg, SIAM J. Sci. Comput., Vol. 37, No. 3, 2015
2. Dominik Zoar, Master Thesis, Dept. of Mathematics, Univ. of Wurzburg
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Theorem
The finite volume scheme (4) together with a numerical flux whichsatisfies property C and reconstruction of w variables is well-balanced inthe sense that the initial condition given by
ui = 0, pi exp(−ψi) = const, ∀ i (5)
is preserved by the numerical scheme.
uLi+ 1
2
= uRi+ 1
2
= 0, pLi+ 1
2
= pRi+ 1
2
= pi+ 12
d
dt(ρu)i +
fi+ 12− fi− 1
2
∆x= e−ψi
(eψi+1
2 − eψi− 1
2
∆x
)pi
d
dt(ρu)i +
pi+ 12− pi− 1
2
∆x=pi+ 1
2− pi− 1
2
∆x
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Approximation of source term
• How to approximate ψi, ψi+ 12
, etc. ? Need some quadrature
• To compute the source term in i’th cell, we define
ψ(x) = −∫ x
xi
φ′(s)RT (s)
ds
• piecewise constant temperature
T (x) = Ti+ 12, xi < x < xi+1 (6)
where Ti+ 12
is the logarithmic average given by
Ti+ 12
=Ti+1 − Ti
log Ti+1 − log Ti
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Approximation of source term
• The integrals are evaluated using the approximation of thetemperature given in (6) leading to the following expressions for ψ.
ψi = 0
ψi− 12
= − 1
RTi− 12
∫ xi− 1
2
xi
φ′(s)ds =φi − φi− 1
2
RTi− 12
ψi+ 12
= − 1
RTi+ 12
∫ xi+1
2
xi
φ′(s)ds =φi − φi+ 1
2
RTi+ 12
• Gravitational potential required at faces φi+ 12
φi+ 12
= φ(xi+ 12), φi+ 1
2=
1
2(φi + φi+1)
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Approximation of source term
Theorem
The source term discretization is second order accurate.
Theorem
Any hydrostatic solution which is isothermal or polytropic is exactlypreserved by the finite volume scheme (4).
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Isothermal examples: well-balanced test
Density and pressure are given by
ρe(x) = pe(x) = exp(−φ(x))
N = 100, 1000, final time = 2
Potential 1 Potential 2 Potential 3
φ(x) x 12x
2 sin(2πx)
Table: Potential functions used for well-balanced tests
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Isothermal examples: well-balanced test
Potential Cells Density Velocity Pressure
x 100 8.21676e-15 4.98682e-16 9.19209e-151000 8.00369e-14 1.51719e-14 9.15152e-14
12x
2 100 1.01874e-14 2.49332e-16 1.06837e-141000 1.05202e-13 4.10434e-16 1.11861e-13
sin(2πx) 100 1.12466e-14 5.79978e-16 1.74966e-141000 1.16191e-13 2.93729e-15 1.76361e-13
Table: Error in density, velocity and pressure for isothermal example
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Shock tube under gravitational field
Gravitational field
φ(x) = x
The domain is [0, 1] and the initial conditions are given by
(ρ, u, p) =
{(1, 0, 1) x < 1
2
(0.125, 0, 0.1) x > 12
Solid wall boundary conditions. Final time t = 0.2, N = 100, 2000 cells
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Shock tube under gravitational field
0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
1
1.2
1.4
x
Den
sity
cells = 100cells = 2000
0 0.2 0.4 0.6 0.8 1−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
x
Vel
ocity
cells = 100cells = 2000
0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
1
1.2
1.4
x
Pre
ssur
e
cells = 100cells = 2000
0 0.2 0.4 0.6 0.8 10
0.5
1
1.5
2
2.5
3
3.5
x
Ene
rgy
cells = 100cells = 2000
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2-D Scheme
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Polytropic hydrostatic solution
pressure perturbation with η = 0.1
well-balanced non well-balanced20 equally spaced contours between -0.03 and +0.03
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Polytropic hydrostatic solution
pressure perturbation with η = 0.001
well-balanced non well-balanced20 contours in [-0.00025,+0.00025] 20 contours in [-0.015,+0.0003]
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Rayleigh-Taylor instability
t = 0 t = 2.9
t = 3.8 t = 5.0
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More hydrostatic solutions
We will write the hydrostatic solution as
ρe(x) = ρ0α(x), pe(x) = p0β(x)
From hydrostatic condition
p′e(x) = −ρe(x)φ′(x), i.e., φ′(x) = −p0ρ0
β′(x)
α(x)(7)
Isothermal solution
α(x) = β(x) = exp
(−φ(x)
RT0
), ρ0 =
p0RT0
Polytropic solution
α(x) =
[1 +
ν − 1
sνρν−10
(φ0 − φ(x))
]1/(ν−1), β(x) = (α(x))ν
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More hydrostatic solutions
Constant potential temperatureModeling atmosphere: Exner pressure and potential temperature
π =
(p
p0
) γ−1γ
, θ =T
π
Taking the potential φ(x) = gx, we get
α(x) =
[1− (γ − 1)gx
γRθ0
] 1γ−1
, β(x) = (α(x))γ
Specified Brunt-Vaisala frequency (N)For potential φ(x) = gx and specified frequency N where
N2 = gd
dxln(θ) = const θ = θ0 exp(N2x/g)
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More hydrostatic solutions
The equilibrium functions are given by
α(x) = exp(−N2x/g)
[1 +
(γ − 1)g2
γRθ0N2
{exp(−N2x/g)− 1
}] 1γ−1
β(x) =
[1 +
(γ − 1)g2
γRθ0N2
{exp(−N2x/g)− 1
}] γγ−1
Radiation pressure
p = ρRT +1
3aT 4
Tabulated equation of state(ρi, Ti, pi), i = 1, 2, . . .
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1-D Finite Volume Scheme
See Berberich et al., Int. Conf. on Hyperbolic Problems, Aachen, 1-5 Aug., 2016
dqidt
+fi+ 1
2− fi− 1
2
∆x=
0siuisi
Consistent numerical flux
fi+ 12
= f(qLi+ 1
2
, qRi+ 1
2
)
Gravitational source
s(x, t) =p0ρ0
β′(x)
α(x)ρ(x, t)
Central difference discretization
si =p0ρ0
βi+ 12− βi− 1
2
∆x
ρiαi
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1-D Finite Volume Scheme
Define new set of independent variables
w = w(q, x) := [ρ/α, u, p/β]>
Reconstruct w variables and compute
qLi+ 1
2
= q(wLi+ 1
2
, xi+ 12), qR
i+ 12
= q(wRi+ 1
2
, xi+ 12)
Well-balanced property
The finite volume scheme (4) together with any consistent numerical fluxand reconstruction of w variables is well-balanced in the sense that theinitial condition given by
ui = 0, ρi/αi = const, pi/βi = const, ∀ i
is preserved by the numerical scheme.
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Remarks
• We must know a priori the hydrostatic state (α(x), β(x))
• Well-balanced scheme for any equation of state
• Extended to curvilinear gridsI Seven-League Hydro Code (https://slh-code.org)
• Extended to unstructured gridsI UG3 code (https://bitbucket.org/cpraveen/ug3/wiki/Home)
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Well-balanced DG schemes
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Finite element method
Conservation law with source term stationary solution qe
qt + f(q)x = s(q) f(qe)x = s(qe)
Weak formulation: Find q(t) ∈ V such that
d
dt(q, ϕ) + a(q, ϕ) = (s(q), ϕ), ∀ϕ ∈ V
FEM with quadrature: Find qh(t) ∈ Vh such that
d
dt(qh, ϕh)h + ah(qh, ϕh) = (sh(qh), ϕh)h, ∀ϕh ∈ Vh
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Finite element and well-balanced
Stationary solution is not a polynomial, qe /∈ Vh.
Let
qe,h = Πh(qe), Πh : V → Vh, interpolation or projection
FEM is well-balanced if
ah(qe,h, ϕh) = (sh(qe,h), ϕh)h, ∀ϕh ∈ Vh
How to construct ah, sh to achieve well-balancing ?
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Mesh and basis functions
• Partition domain into disjoint cells
Ci = (xi− 12, xi+ 1
2), ∆xi = xi+ 1
2− xi− 1
2
• Approximate solution inside each cell by a polynomial of degree N
Ii−1 Ii Ii−1
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Discontinuous Galerkin Scheme
Consider the single conservation law with source term
∂q
∂t+∂f
∂x= s(q, x)
Solution inside cell Ci is polynomial of degree N
qh(x, t) =
N∑
j=0
qj(t)ϕj(x), qj(t) = qh(xj , t)
Approximate the flux
f(qh) ≈ fh(x, t) =
N∑
j=0
f(qh(xj , t))ϕj(x) =
N∑
j=0
fj(t)ϕj(x)
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Discontinuous Galerkin Scheme
Gauss-Lobatto-Legendre quadrature
∫
Ci
φ(x)ψ(x)dx ≈ (φ, ψ)h = ∆xi
N∑
q=0
ωqφ(xq)ψ(xq)
Semi-discrete DG: For 0 ≤ j ≤ N
d
dt(qh, ϕj)h + (∂xfh, ϕj)h +[fi+ 1
2− fh(x−
i+ 12
)]ϕj(x−i+ 1
2
)
−[fi− 12− fh(x+
i− 12
)]ϕj(x+i− 1
2
) = (sh, ϕj)h
(1)
where fi+ 12
= f(q−i+ 1
2
, q+i+ 1
2
) is a numerical flux function. This is also
called a DG Spectral Element Method1.
1See arXiv:1511.08739 for more details.5 / 30
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Approximation of source term: isothermal case
Let
Ti = temperature corresponding to the cell average value in cell Ci
Rewrite the source term in the momentum equation as (Xing & Shu)
s = −ρ∂Φ
∂x= ρRTi exp
(Φ
RTi
)∂
∂xexp
(− Φ
RTi
)
Source term approximation: For x ∈ Ci
sh(x) = ρh(x)RTi exp
(Φ(x)
RTi
)∂
∂x
N∑
j=0
exp
(−Φ(xj)
RTi
)ϕj(x) (2)
Source term in the energy equation
1
ρh(ρu)hsh
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Approximation of source term: polytropic case2
Define H(x) inside each cell Ci as
H(x) =ν
ν − 1ln
[ν − 1
ναi(βi − Φ(x))
], x ∈ Ci
αi, βi: constants to be chosen. Rewrite source term
s(x) = −ρ∂Φ
∂x=ν − 1
νρ(βi − Φ(x)) exp(−H(x))
∂
∂xexp(H(x)), x ∈ Ci
Source term approximation: For x ∈ Ci
sh(x) =ν − 1
νρh(x)(βi − Φ(x)) exp(−H(x))
∂
∂x
N∑
j=0
exp(H(xj))ϕj(x)
βi = max0≤j≤N
[ν
ν − 1
pjρj
+ Φ(xj)
], αi = pj∗ρ
−νj∗
2with Markus Zenk 7 / 30
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Well-balanced property
Well-balanced property
Let the initial condition to the DG scheme (1), (2) be obtained byinterpolating the isothermal/polytropic hydrostatic solutioncorresponding to a continuous gravitational potential Φ. Then thescheme (1), (2) preserves the initial condition under any time integrationscheme.
Proof: For continuous hydrostatic solution, qh(0) is continuous. By fluxconsistency
fi+ 12− fh(x−
i+ 12
) = 0, fi− 12− fh(x+
i− 12
) = 0
Above is true even if density is discontinuous, provided flux satisfiescontact property.
=⇒ density and energy equations are well-balanced
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Well-balanced property
Momentum eqn: flux fh has the form
fh(x, t) =
N∑
j=0
pj(t)ϕj(x), pj = pressure at the GLL point xj
Isothermal initial condition, Ti = Te = const. The source term evaluated
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Well-balanced property
at any GLL node xk is given by
sh(xk) = ρh(xk)RTe︸ ︷︷ ︸pk
exp
(Φ(xk)
RTe
) N∑
j=0
exp
(−Φ(xj)
RTe
)∂
∂xϕj(xk)
= pk exp
(Φ(xk)
RTe
) N∑
j=0
exp
(−Φ(xj)
RTe
)∂
∂xϕj(xk)
=N∑
j=0
pk exp
(Φ(xk)
RTe
)exp
(−Φ(xj)
RTe
)∂
∂xϕj(xk)
=N∑
j=0
pj exp
(Φ(xj)
RTe
)exp
(−Φ(xj)
RTe
)∂
∂xϕj(xk)
=N∑
j=0
pj∂
∂xϕj(xk) =
∂
∂xfh(xk)
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Well-balanced property
Since
∂xfh(xk) = sh(xk) at all the GLL nodes xk
we can conclude that
(∂xfh, ϕj)h = (sh, ϕj)h , 0 ≤ j ≤ N
=⇒ scheme is well-balanced for the momentum equation
The proof for polytropic case is similar.
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2-D Euler equations with gravity
∂q
∂t+∂f
∂x+∂g
∂y= s
q = vector of conserved variables, (f , g) = flux vector and s = sourceterm, given by
q =
ρρuρvE
, f =
ρup+ ρu2
ρuv(E + p)u
, g =
ρvρuv
p+ ρv2
(E + p)v
s =
0
−ρ∂Φ∂x
−ρ∂Φ∂y
−(u∂Φ∂x + v ∂Φ
∂y
)
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Mesh and basis functions
A quadrilateral cell K and reference cell K = [0, 1]× [0, 1]
12
3
4
K
x
y
ξ
η
1 2
34
K
TK
1-D GLL points: ξr ∈ [0, 1], 0 ≤ r ≤ NTensor product of GLL points
N = 1 N = 2 N = 3
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Numerical Results
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1-D hydrostatic test: Well-balanced property
Potential
Φ = x, Φ = sin(2πx)
Initial state
u = 0, ρ = p = exp(−Φ(x))
Mesh ρu ρ E
25x25 1.03822e-13 2.72604e-14 9.53913e-1450x50 1.04783e-13 2.67559e-14 9.36725e-14
100x100 1.05019e-13 2.66323e-14 9.34503e-14200x200 1.05088e-13 2.66601e-14 9.33861e-14
Table: Well-balanced test on Cartesian mesh using Q1 and potential Φ = y
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1-D hydrostatic test: Well-balanced property
Mesh ρu ρ E
25x25 1.04518e-13 2.7548e-14 9.64205e-1450x50 1.04983e-13 2.69317e-14 9.43158e-14
100x100 1.05069e-13 2.69998e-14 9.39126e-14200x200 1.05089e-13 2.68828e-14 9.462e-14
Table: Well-balanced test on Cartesian mesh using Q2 and potential Φ = x
Mesh ρu ρ E
25x25 9.23424e-13 2.31405e-13 8.16645e-1350x50 9.36459e-13 2.28315e-13 8.04602e-13
100x100 9.39613e-13 2.28001e-13 8.03005e-13200x200 9.40422e-13 2.2792e-13 8.02653e-13
Table: Well-balanced test on Cartesian mesh using Q1 and Φ = sin(2πx)
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1-D hydrostatic test: Well-balanced property
Mesh ρu ρ E
25x25 9.34536e-13 2.35173e-13 8.30316e-1350x50 9.39556e-13 2.29908e-13 8.10055e-13
100x100 9.40442e-13 2.28538e-13 8.04638e-13200x200 9.40668e-13 2.28051e-13 8.0357e-13
Table: Well-balanced test on Cartesian mesh using Q2 and Φ = sin(2πx)
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1-D hydrostatic test: Evolution of perturbations
Potential
Φ(x) = x
Initial condition
u = 0, ρ = exp(−x), p = exp(−x) + η exp(−100(x− 1/2)2)
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1-D hydrostatic test: Evolution of perturbations
0.0 0.2 0.4 0.6 0.8 1.0
x
−0.002
0.000
0.002
0.004
0.006
0.008
0.010
Pre
ssu
rep
ert
urb
ati
on
Initial
100 cells
1000 cells
0.0 0.2 0.4 0.6 0.8 1.0
x
−0.002
0.000
0.002
0.004
0.006
0.008
0.010
Pre
ssu
rep
ert
urb
ati
on
Initial
50 cells
500 cells
(a) (b)
Figure: Evolution of perturbations for η = 10−2: (a) Q1 (b) Q2
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1-D hydrostatic test: Evolution of perturbations
0.0 0.2 0.4 0.6 0.8 1.0
x
−0.00002
0.00000
0.00002
0.00004
0.00006
0.00008
0.00010
Pre
ssu
rep
ert
urb
ati
on
Initial
100 cells
1000 cells
0.0 0.2 0.4 0.6 0.8 1.0
x
−0.00002
0.00000
0.00002
0.00004
0.00006
0.00008
0.00010
Pre
ssu
rep
ert
urb
ati
on
Initial
50 cells
500 cells
(a) (b)
Figure: Evolution of perturbations for η = 10−4: (a) Q1 (b) Q2
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Shock tube problem
Potential Φ(x) = x and initial condition
(ρ, u, p) =
{(1, 0, 1) x < 1
2
(0.125, 0, 0.1) x > 12
0.0 0.2 0.4 0.6 0.8 1.0
x
0.0
0.2
0.4
0.6
0.8
1.0
1.2
Den
sity
Q1, 100 cells
Q2, 100 cells
FVM, 2000 cells
0.0 0.2 0.4 0.6 0.8 1.0
x
0.0
0.2
0.4
0.6
0.8
1.0
1.2
Den
sity
Q1, 200 cells
Q2, 200 cells
FVM, 2000 cells
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2-D hydrostatic test: Well-balanced test
PotentialΦ = x+ y
Hydrostatic solution is given by
ρ = ρ0 exp
(−ρ0g
p0(x+ y)
), p = p0 exp
(−ρ0g
p0(x+ y)
), u = v = 0
ρu ρv ρ E
Q1, 25× 25 9.85926e-14 9.85855e-14 5.32357e-14 1.55361e-13Q1, 50× 50 9.94493e-14 9.94451e-14 5.37084e-14 1.56669e-13Q1, 100× 100 9.96481e-14 9.96474e-14 5.38404e-14 1.57062e-13
Q2, 25× 25 9.9256e-14 9.92682e-14 5.39863e-14 1.57435e-13Q2, 50× 50 9.961e-14 9.96538e-14 5.41091e-14 1.57521e-13Q2, 100× 100 9.95889e-14 9.97907e-14 5.43145e-14 1.57728e-13
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2-D hydrostatic test: Evolution of perturbation
p = p0 exp
(−ρ0g
p0(x+ y)
)+ η exp
(−100
ρ0g
p0[(x− 0.3)2 + (y − 0.3)2]
)
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2-D hydrostatic test: Evolution of perturbation
x
y
0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
xy
0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
(a) (b)
Figure: Pressure perturbation on 50× 50 mesh at time t = 0.15 (a) Q1 (b) Q2.Showing 20 contours lines between −0.0002 to +0.0002
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2-D hydrostatic test: Evolution of perturbation
x
y
0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
xy
0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
(a) (b)
Figure: Pressure perturbation on 200× 200 mesh at time t = 0.15 (a) Q1 (b) Q2.Showing 20 contours lines between −0.0002 to +0.0002
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2-D hydrostatic test: Discontinuous density
Ligther fluid on top
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2-D hydrostatic test: Discontinuous density
Heavier fluid on top
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Radial Rayleigh-Taylor problem
30× 30 cells 956 cells
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Radial Rayleigh-Taylor problem: PerturbationsCartesian mesh of 240× 240 and Q1 basis
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Well-balanced versus non well-balanced
(a) (b) (c)Well balanced test for radial Rayleigh-Taylor problem on 50× 50 mesh.(a) Initial density, (b) well-balanced scheme, density at t = 1.5, (c) non
well-balanced scheme, density at t = 1.5
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Kinetic energy and entropy consitency
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Kinetic energy
Kinetic energy per unit volume
K =1
2⇢u2
satisfies the following equation
d
dt
Z
⌦
Kdx =
Z
⌦
p@u
@xdx�4
3
Z
⌦
µ
✓@u
@x
◆2
dx Z
⌦
p@u
@xdx
Work done by pressure forces, absent in incompressible flowsIrreversible destruction due to molecular di↵usion
Note: Convection contributes to only flux of KE across @⌦
Praveen. C (TIFR-CAM) KEP/Entropy stable schemes 10 May 2013 10 / 51
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Entropy conditionEntropy-Entropy flux pair: U(u), F (u)
U(u) is strictly convex and U 0(u)f 0(u) = F 0(u)
Then, for hyperbolic problem (Euler equation)
@u
@t+@f
@x= 0 =) U 0(u)
@u
@t+U 0(u)f 0(u)
@u
@x= 0
+@U(u)
@t+@F (u)
@x= 0
For discontinuous solutions, only inequality
@U(u)
@t+@F (u)
@x 0
R⌦
U(u)dx for an isolated system decreases with time
Second law of thermodynamics
Praveen. C (TIFR-CAM) KEP/Entropy stable schemes 10 May 2013 13 / 51
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• Jameson: KE consistent
• Ismail & Roe: Entropy consistent
• KEPEC: Both KE and entropy consistent
• Ch., CICP, vol. 14, no. 5, 2013
• Ray et al., CICP, vol. 19, no. 5, 2016
Euler equations
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FVM for NS equationThe semi-discrete finite volume method for NS equations using thecentered KEP and entropy conservative flux is stable for the kineticenergy and entropy, i.e.,
X
j
�xdKj
dt=
X
j
"�uj+ 1
2
�xpj+ 1
2� 4
3µ
✓�uj+ 1
2
�x
◆2#�x
X
j
�uj+ 1
2
�xpj+ 1
2
��x
and
X
j
�xdUj
dt= �
X
j
"8µ�j+ 1
2
3
✓�uj+ 1
2
�x
◆2
+
RTjTj+1
✓�Tj+ 1
2
�x
◆2#�x 0
• Expect good stability property
• No control of density/pressure
Praveen. C (TIFR-CAM) KEP/Entropy stable schemes 10 May 2013 20 / 51
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DNS of 3-D Taylor-Green vortex: central scheme
fj+ 12
= 12(fj + fj+1)
0
2
4
6
8
10
12
14
16
18
0 2 4 6 8 10 12 14 16 18 20
Enst
roph
y
Time
Enstrophy
spectralcentral avg (256x256x256)
0
0.005
0.01
0.015
0.02
0.025
0.03
0 2 4 6 8 10 12 14 16 18 20
Dis
sipa
tion
Rat
e
Time
Dissipation Rate
spectralcentral avg (256x256x256)
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0 2 4 6 8 10 12 14 16 18 20
KE
Time
Kinetic Energy
spectralcentral avg (256x256x256)
Praveen. C (TIFR-CAM) KEP/Entropy stable schemes 10 May 2013 42 / 1
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DNS of 3-D Taylor-Green vortex
0
1
2
3
4
5
6
7
8
9
10
11
0 5 10 15 20En
stro
phy
Time
Enstrophy
spectralKEPES (128x128x128)KEPES (256x256x256)
Praveen. C (TIFR-CAM) KEP/Entropy stable schemes 10 May 2013 46 / 1
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DNS of 3-D Taylor-Green vortex
0
0.002
0.004
0.006
0.008
0.01
0.012
0.014
0 5 10 15 20
Dis
sipa
tion
Rat
e
Time
Dissipation Rate
spectralKEPES (128x128x128)KEPES (256x256x256)
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0 5 10 15 20KE
Time
Kinetic Energy
spectralKEPES (128x128x128)KEPES (256x256x256)
Praveen. C (TIFR-CAM) KEP/Entropy stable schemes 10 May 2013 47 / 1
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Isentropic vortex advection: Density contours
(KEP, 50 ⇥ 50, T=57) (KEP, 100 ⇥ 100, T=87) (KEP4, 50 ⇥ 50, T=100)
(KEP4, 100 ⇥ 100, T=100) (ROE-EC, 100 ⇥ 100, T=100) (Others configs, T=100)
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Ideal compressible MHD
w =⇥⇢, ⇢u1, ⇢u2, ⇢u3, E, B1, B2, B3
⇤>
f1 =
266666666664
⇢u1
p + ⇢u21 + 1
2|B|2 � B2
1
⇢u1u2 � B1B2
⇢u1u3 � B1B3
(E + p + 12|B|2)u1 � (u · B)B1
0u1B2 � u2B1
u1B3 � u3B1
377777777775
, f2 =
266666666664
⇢u2
⇢u1u2 � B1B2
p + ⇢u22 + 1
2|B|2 � B2
2
⇢u2u3 � B2B3
(E + p + 12|B|2)u2 � (u · B)B2
u2B1 � u1B2
0u2B3 � u3B2
377777777775
Magnetic field must be divergence free: r · B = 0, an intrinsic property
@
@tr · B = 0
Entropy pair
U = � ⇢s
� � 1, F↵ = � ⇢su↵
� � 1, s = ln(p⇢��) (3)
7 / 28Ch. & Klingenberg, SINUM, vol. 54, issue 2, 2016 Yee & Sjogreen, ICOSAHOM, Rio, June, 2016
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Thank You