Journal of Computational Physics 258 (2014) 118–136
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Journal of Computational Physics
www.elsevier.com/locate/jcp
A symmetry preserving dissipative artificial viscosity in an r–zstaggered Lagrangian discretization
Pavel Váchal a,∗, Burton Wendroff b
a Faculty of Nuclear Sciences and Physical Engineering, Czech Technical University in Prague, Brehová 7, 115 19 Praha 1, Czech Republicb Theoretical Division, Group T-5, Los Alamos National Laboratory, MS B284, Los Alamos, NM 87544, USA
a r t i c l e i n f o a b s t r a c t
Article history:Received 26 February 2013Received in revised form 20 August 2013Accepted 20 October 2013Available online 26 October 2013
We present an artificial viscous force for two-dimensional axi-symmetric r–z geometry andlogically rectangular grids that is dissipative, conserves the z-component of momentumand preserves spherical symmetry on an equi-angular polar grid. The method turns outto be robust and performs well for spherically symmetric problems on various grid types,without any need for problem- or grid-dependent parameters.
The basic structure of Lagrangian staggered grid schemes involves geometric objects, namely, cells and points. Each cellc has a time-independent mass mc , a volume V c , and an internal specific energy εc . Each point p has a time-independentmass mp , a position Xp , and a velocity Up = (up, v p).
Cells and points influence each other through certain matrices. In particular, cells produce forces on points. We expressthis as Newton’s law
mpdUp
dt=
∑c(p)
Fpc, (1)
where by∑
c(p) we mean the sum over cells that have p as a vertex.Now, (1) holds in all coordinate systems and all cell geometries and all dimensions. However, in this note we are specif-
ically concerned with the two-dimensional axi-symmetric geometry and with logically rectangular grids. So
Xp = (rp, zp), (2)
and cells are (possibly degenerate) quadrilaterals. Except that the cells are not really quadrilaterals but are solid cylindricalfigures obtained by rotating the cells around the z-axis. The masses and forces in (1) are then defined on this basis.
The grid being logically rectangular means that we can identify p with the pair (i, j) where i and j range over sets ofconsecutive integers.
P. Váchal, B. Wendroff / Journal of Computational Physics 258 (2014) 118–136 119
One test of any r–z scheme is whether or not it will preserve spherically symmetric data centered anywhere on thez-axis. This can be achieved with quadrilateral cells if the grid is symmetric, that is, it is polar with rays spaced by auniform angle. Even then, any scheme of the type described above may fail to preserve symmetry. A popular way to getaround this is to use the area-weighted scheme [3], defining the Cartesian force matrix FD
pc in which the (r, z) coordinatesare treated as if they were (x, y) Cartesian coordinates, but then writing
mpdUp
dt= rp
∑c(p)
FDpc . (3)
Why this approach preserves symmetry is well explained in the original reference.Taking the cell masses as fundamental, there are several ways to define the cylindrical nodal mass in such a way that
mp
rp(4)
is both independent of angle and time invariant for a symmetric grid. Our particular choice of nodal mass for the cal-culations presented in this paper is given in Section 6.4.1. It is not hard to show that with this, spherical symmetry ispreserved.
The area-weighted scheme is attractive because it easily converts any Cartesian scheme into an axi-symmetric one. Onedrawback addressed in this paper is that the artificial viscosity so constructed may not be dissipative and may not conservez-momentum.
That is, if now FDpc is just the Cartesian pressure force, and FAD
pc is the Cartesian artificial viscous force, and then the total
force is Fpc = FDpc + FAD
pc so that
mpdUp
dt= rp
∑c(p)
(FD
pc + FADpc
). (5)
In a typical staggered grid approach to energy conservation the internal energy change is based on the equation
mcd εc
dt= −
∑p(c)
rp(FD
pc + FADpc
) · Up. (6)
The problem is that the contribution to the internal energy from the viscous term is not dissipative, that is it does notnecessarily increase the internal energy, since the inequality∑
p(c)
rpFADpc · Up � 0 (7)
is not necessarily true. This is the case for the tensor viscosity of [2].We construct below a dissipative artificial viscous force defined on arbitrary logically rectangular grids that preserves
spherical symmetry if the grid is equi-angular, polar, and symmetric centered anywhere on the z-axis. Furthermore, thisforce conserves the z-component of momentum on internal cells.
Our viscous force is based on the Laplacian of the velocity in r–z geometry. We then present an extension of this viscositythat is applicable when spherical symmetry might not exist.
2. An edge viscous force
Cells and points have associated edges: four edges for each full quadrilateral cell, and for each interior point p four edgesthat have p as an endpoint. Edges are indexed by e, and the two edges of c that have p as an endpoint have the indexese(p, c).
We are going to define a cylindrical artificial viscous force for cell c at node p, FApc that is not area-weighted, so that the
equation for velocity becomes
mpdUp
dt=
∑c(p)
(rpFD
pc + FApc
). (8)
This force will be a sum of edge forces, namely,
FApc =
∑e(p,c)
fpe. (9)
These edge forces will have the following form: for the radial component of the force
120 P. Váchal, B. Wendroff / Journal of Computational Physics 258 (2014) 118–136
(fpe)r = σce
(re �upe − ae
ue
re
), (10)
while the z-component is
(fpe)z = σcere �vpe, (11)
where re is the arithmetic average r of the edge, ue is the average of the radial components of the endpoint velocities,σce � 0 supplies the proper dimensions (density times velocity times length).
The exact form of the dimension-carrying coefficient σce critically affects the behavior of our method, but it plays no role in thesymmetry and conservation properties of the scheme. The exact definition of σce will be given in Section 6.4.1.
In Section 3.1 the constant ae � 0 will be constructed so that if the grid is equi-angular polar then symmetry will be preserved. ae
will depend only on a few neighbors of p.1
As for the velocity difference �Upe , if the edge e(p, c) connects point p of c to point q of c then
�Upe = (�upe,�vpe) = Uq − Up . (12)
Then clearly z-momentum is conserved, since the z-component of the force at p is the negative of it at q. It is easy to seethat this viscous force is dissipative. Since
Until a later section on general grids, we will assume that we are given an equi-angular polar grid, that is
ri, j = R j sin(iγ ), (15a)
zi, j = R j cos(iγ ), (15b)
so that both the spherical radii R j and the angular interval γ are known.
3.1. Acceleration
We assume symmetric data, including the σ coefficients. There is no loss of generality of our results if we set σce = 1for all cells. In Fig. 1 we show a part of an equi-angular polar grid. While Q can be anywhere on the z-axis, with no lossof generality we suppose in (15) and in most of the following text that it is located at (0,0). The points 1, 0, and 3 lie onthe circle of radius R0, while the points 2, 0, and 4 lie on the ray with angle θ . Consider first the viscous force contributionfrom the edges (0,2) and (0,4). For all such edges forming a straight line we set ae = 0.
Now, if the velocity field is symmetric then U2, U0, and U4 are directed radially (all inward or all outward) with mag-nitudes independent of angle. That is, each of those velocities is of the form Uk = ±‖Uk‖(sin θ, cos θ) for k ∈ {0,2,4}, with‖Uk‖ independent of θ . Since each rk = Rk sin θ it follows that the viscous force at point 0 from these edges has the form
f0,rays = h sin θ(sin θ, cos θ), h independent of angle, (16)
that is, it is directed radially and the components have sin θ as a common factor.For the edges (0,1) and (0,3), from symmetry we can safely assume that the velocity vectors at 1, 0, and 3 are radial
unit vectors. Thus the components of velocity U = (u, v) are
u1 = sin(θ − γ ), v1 = cos(θ − γ ), (17a)
u0 = sin θ, v0 = cos θ, (17b)
u3 = sin(θ + γ ), v3 = cos(θ + γ ). (17c)
For the r averages we have
1 This form is motivated by the Laplacian of a vector in r–z geometry. For example from [1, p. 182], it follows that the Laplacian is
L(U)r = 1
r
[∂
∂r
(r∂u
∂r
)+ ∂
∂z
(r∂u
∂z
)− u
r
], L(U)z = 1
r
[∂
∂r
(r∂v
∂r
)+ ∂
∂z
(r∂v
∂z
)].
P. Váchal, B. Wendroff / Journal of Computational Physics 258 (2014) 118–136 121
Fig. 1. Grid.
1
2(r1 + r0) = 1
2R0
(sin θ + sin(θ − γ )
), (18a)
1
2(r3 + r0) = 1
2R0
(sin θ + sin(θ + γ )
). (18b)
Consider first the z-component of the viscous force at 0, which is
(f0,circ)z = 1
2R0
[(cos(θ − γ ) − cos θ
)(sin θ + sin(θ − γ )
) + (cos(θ + γ ) − cos θ
)(sin θ + sin(θ + γ )
)].
This reduces to
(f0,circ)z = −2R0 sin2 γ sin θ cos θ. (19)
The radial component of the force is
(f0,circ)r = 1
2R0
[(sin(θ − γ ) − sin θ
)(sin(θ − γ ) + sin θ
) + (sin(θ + γ ) − sin θ
)(sin θ + sin(θ + γ )
)] − 2a/R0.
(For these edges ue/re = 1/R0 and there are two contributions to that term.) This reduces to
(f0,circ)r = −2R0 sin2 γ sin θ sin θ + R0 sin2 γ − 2a/R0. (20)
Therefore, if
a = R20 sin2 γ
2, (21)
this force is radial and because of the common sin θ factor multiplying the components the acceleration will be radial andindependent of angle. Thus
ae ={0 on rays,
12 R2
j sin2 γ on a circle of radius R j .(22)
3.2. Internal energy
Instead of (6), the internal energy equation using our edge viscosity will now be
mcdεc
dt= −
∑p(c)
(rpFD
pc + FApc
) · Up, (23)
and we have to verify that symmetry is maintained, that is, we have to show that
δεc ≡ 1
mc
∑p(c)
FApc · Up (24)
is a function only of R .
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So consider in Fig. 1 the cell with vertexes (1,5,2,0) and its cylindrical volume V . Let �1 be the planar area of thetriangle with vertexes (1, Q ,0) and let �2 be the planar area of the triangle with vertexes (5, Q ,2). Note that the meshorigin Q is on the z-axis, but not necessarily at the coordinate origin. Then
V = 1
3
[(r1 + r0)�1 − (r5 + r2)�2
]. (25)
Note that �1 and �2 are independent of angle.We show in Appendix A that for this cell,∑
p(c)
FApc · Up = A(r1 + r0) + B(r5 + r2), (26)
where A and B are independent of angle. Then
1
3ρcδεc = A(r1 + r0) + B(r5 + r2)
(r1 + r0)�1 − (r5 + r2)�2= A + B r5+r2
r1+r0
�1 − r5+r2r1+r0
�2, (27)
but
r5 + r2
r1 + r0(28)
is independent of angle, and therefore so is δεc .
4. Boundary conditions
4.1. Velocity
Clearly, points on the z-axis (r = 0) require special treatment. In the case of full symmetry, the viscous force we havedefined for points at the same spherical radius R has the form
h sin θp(sin θp, cos θp), (29)
where h is independent of angle. We have assumed that the interior nodal masses can be written
mp = β sin θp, (30)
where β is independent of angle. It follows that the contribution to the viscous acceleration is
dδUp
dt≡ h
β(sin θp, cos θp) (31)
for all interior points at the same R .Then, in order to preserve symmetry at the boundary point r = 0 with this R it seems that we have no choice but to
take
dδu
dt= 0, (32a)
dδv
dt= h
β. (32b)
The fact of the matter is, that there is a simple generally applicable boundary condition that preserves symmetry, which we show inSection 5.1. However, it is interesting at this point to see what the factor h is. We have to return here to rectangular polarindexing, so the grid points are (R j sin θi, R j cos θi), and the velocities can be written (ui, j, vi, j) = g j(sin θi, cos θi). Then thecontribution to coefficient h from (16) is
1
2
[(R j+1 + R j)(g j+1 − g j) + (R j + R j−1)(g j−1 − g j)
]. (33)
From (20) the contribution is
−2g j R j sin2 γ . (34)
So,
h = 1 [(R j+1 + R j)(g j+1 − g j) + (R j + R j−1)(g j−1 − g j)
] − 2g j R j sin2 γ (35)
2
P. Váchal, B. Wendroff / Journal of Computational Physics 258 (2014) 118–136 123
and thus
dδv
dt= h
β= 1
β
(1
2
[(R j+1 + R j)(g j+1 − g j) + (R j + R j−1)(g j−1 − g j)
] − 2g j R j sin2 γ
). (36)
Note that the momentum equation β dδvdt = h on the z-axis is consistent with the area weighted momentum equations.
4.2. Total energy
We have shown in Section 3.2 that our viscous force preserves internal energy symmetry for all cells, including cellswith an edge on the z-axis. However, above we have obtained the acceleration by defining the ratio of force to mass forpoints on that boundary – neither force nor mass has been chosen. In particular, the viscous force defined in (10), (11) isnot what defines that acceleration. This has an effect on total energy conservation, so let us review that concept.
To start, generalize the definitions of mp and mc by introducing a mass matrix with components mpc such that
mc =∑p(c)
mpc (37)
and
mp =∑c(p)
mpc (38)
and
dmpc
dt= 0. (39)
Define a cell kinetic energy Kc by
Kc =∑p(c)
mpc1
2Up · Up (40)
and a total cell energy as
Ec = mcεc + Kc . (41)
Next, we claim and prove in Appendix A that for points not on the z-axis there is a nodal flux function Gpc , that is,∑c(p)
Gpc = 0 (42)
and for cells that do not have an edge on the z-axis
dEc
dt=
∑p(c)
Gpc (43)
so that for those cells∑c
dEc
dt=
∑c
∑p(c)
Gpc = 0. (44)
The problem for cells with an edge on the z-axis is that the proof of (44) involves a division by mp at points for whichrp = 0. However, for pure area weighted forces, that is excluding our viscous force, using mpc from (4), both mpc and Fpc arezero if p is a point on the z-axis. Then as long as the acceleration is finite those points are not present in the proof of (44)in Appendix A, so cells with an edge on the z-axis can be included in the conservation of total energy.
This is not the case for our viscous force. Forces at the nodes of a cell, such as our viscous force, are defined at all nodesof all cells, including nodes on the z-axis. In order to guarantee non-negative and symmetric energy change in the z-axisboundary cells, the “work” of the z-axis viscous force on the boundary nodes must be included in the internal energy changeof those cells. However, the contribution to the acceleration of the z-axis nodes from the viscous force is not obtained fromthis force divided by a nodal mass – as it would have to be for energy conservation up to the boundary – but by theconstraint of symmetry preservation.
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5. Viscosity on general grids
Up to this point we have assumed that we know the complete structure of the polar grid, in particular that initially thedata and the grid are spherically symmetric. We consider here the possibility of treating a wider range of applications.
In fact, we suppose only one restriction on the logically rectangular grid: the z-axis must be one of the grid curves, thatis either i = const (i-curve) or j = const ( j-curve). We know of no other scheme where this is not the case. Suppose it is ani-curve. Then if there are circles and rays the rays must be i-curves. Thus, we take
ae = 0 on all i-curves. (45)
Now we know that only the j-curves are possibly circles. Returning to the polar case, if (15) holds then we prove in Ap-pendix A that
and therefore the acceleration from (48) can be written as
dδvi, j
dt= ri, j+1 + ri, j
2mi, j(vi, j+1 − vi, j) + ri, j−1 + ri, j
2mi, j(vi, j−1 − vi, j)
+ ri−1, j + 2ri, j + ri+1, j
2mi, j(vi−1, j − 2vi, j + vi+1, j). (49)
Note that
ri, j+1 + ri, j
mi, j,
ri, j−1 + ri, j
mi, j,
as well as
ri−1, j + 2ri, j + ri+1, j
mi, j= ri−1, j + ri, j
mi, j+ ri, j + ri+1, j
mi, j
are independent of i.Now suppose that the z-axis is the i = 0 ray. We apply the reflection condition
v−1, j = v1, j
and define the z-component of acceleration at (0, j) as
dδv0, j
dt= r1, j+1 + r1, j
2m(v0, j+1 − v0, j) + r1, j−1 + r1, j
2m(v0, j−1 − v0, j) + 2
0 + 2r1, j + r2, j
2m(v1, j − v0, j), (50)
1, j 1, j 1, j
P. Váchal, B. Wendroff / Journal of Computational Physics 258 (2014) 118–136 125
Fig. 2. z-Axis acceleration. Weights w are average edge r divided by mp . Acceleration at (0, j) is w1(v0, j+1 − v0, j)+ w3(v0, j−1 − v0, j)+ 2(w2 + w4)(v1, j −v0, j).
where we used the “weights” (i.e., the coefficients of the velocity differences in (49)) from the first off-axis node (1, j).Those weights are defined and independent of i for ri, j > 0, and will maintain symmetry at the z-axis.
This situation is shown in Fig. 2, where we denoted
Such choice of acceleration of nodes on z-axis also makes sense for a general grid.
6. Numerical results
We have defined for any logically rectangular grid an edge type artificial viscosity that we now call LapEdge, and wehave shown that it is dissipative and preserves spherical symmetry on a symmetric (polar equi-angular) grid. In order toperform actual numerical tests of the validity of this viscosity we need a code that also computes the pressure forces. Wehave such a code which does area-weighted (AW) pressure forces. Our code does this in the same way as the code ALE Inc[11], that is, the r–z force at a node p is the Cartesian force at p multiplied by rp (3). We also use the same nodal massesas ALE Inc for both pressure force and LapEdge. We have the option of using subcell pressures [4] and we have decided tohave them fully on as a default.
All LapEdge tests ran to completion and negative contributions by the viscous forces to the energies were never created.Also, there are potentially adjustable parameters in the factor σce , but these were set at their fixed default values for alltests, that is, there is no tuning of these parameters.
As a benchmark for comparison we also perform all tests with the tensor viscosity of Campbell and Shashkov [2] CS(for different formulations see also [17,8]). These are also done as area-weighted just as implemented in the code ALE Inc.The CS results are presented in a separate section since some of those tests did not run to completion, usually because ofthe existence of negative energy. Although one could blame our interpretation of CS for this, CS is known not to be alwaysdissipative. To ensure that viscous CS forces are dissipative, we always set negative contributions to the internal energyto zero. Some tuning was necessary to get the best results with CS, including the possible use of limiters (see [2] for adefinition), and, following [2], taking a very small CFL number for polar grids. We do not give the details of all the tuningand tricks we needed to run the CS method. Still, we do show the results (the best ones obtained for each test), because asfar as we understand CS is a standard and thus, as mentioned above, we are showing it here as a benchmark for the readerto have as a comparison.
At the end of the numerical part, we give in Section 6.4 some details of the implementation, including the definition ofσce and the complete algorithm of one time step.
We first present two classic problems, Noh and Sedov, on three grids – symmetric, randomly perturbed symmetric, andrectangular. Then we comment on the Saltzman piston problem.
126 P. Váchal, B. Wendroff / Journal of Computational Physics 258 (2014) 118–136
The Noh implosion test problem [13] is a spherical infinite strength shock propagating out from the origin. Initially,density is set to 1 and pressure to 0 everywhere, the velocity field is given by vectors of norm 1 pointing toward theorigin, that is, Up = −Xp/R with R = ‖Xp‖, and the ideal polytropic gas equation of state P = (Γ − 1)ρε with adiabaticindex Γ = 5/3 is used. The exact solution at time t inside the shocked region (R < t/3) is ρ = 64, P = 64/3, U = 0, andoutside the shock (R > t/3) we have ρ = (1+ t/R)2, P = 0, Up = −Xp/R . In our simulations we approximate the zero initialpressure by P = 10−6. We will show results at the final time t = 0.6.
The 3D version of the Sedov blast wave test problem [16] describes the evolution of a blast wave in a point-symmetricexplosion. Initially, density is 1 everywhere and the total energy Etotal = 0.851072 is stored at the origin in the form ofinternal energy. Again, the ideal polytropic gas equation of state is used, this time with adiabatic index Γ = 7/5. At thefinal time tfinal = 1.0 the exact solution is a spherically symmetric diverging shock whose front is at radius R = 1 and has adensity peak ρ = 6.0. The exact solution shown in the plots as a reference was obtained by code [6].
6.1.1. Symmetric polar gridThe grid here is an equi-angular polar grid with uniform radial spacing, covering one quadrant of a circle with radius
Rmax. The initial positions of its nodes are prescribed as
(r, z)i, j = R j (sinϕi, cosϕi), 0 � i � M, 0 � j � N
with
ϕi = i�ϕ, �ϕ = π
2/M,
R j = j�R, �R = Rmax/N,
where M is the number of cells in the angular direction and N the number of cells in radial direction.The radius of the computational domain is Rmax = 1 for the Noh test and Rmax = 1.2 for the Sedov test. For the Sedov
test, all the initial energy is stored in the cells connected to the origin.Radial profiles of density for calculations with increasing grid resolution for both Noh and Sedov are shown in Fig. 3.
These are actually interconnected scatter plots, showing values in all cells, which demonstrates the perfect symmetry.Note: The ubiquitous Lagrangian wall-heating problem [14] is clearly evident here.
6.1.2. Perturbed polar gridHere we show results obtained on polar grids which have been initially perturbed as follows. Suppose we have an
equi-angular and radially equidistant polar grid with spacing �R × �ϕ and perturb it so that each interior node is movedoff its original position by (c1α�R, c2α�ϕ), where −1 � c1 � 1 and −1 � c2 � 1 are random numbers and α is a parameterwhich controls the level of distortion.
Thus, in this section each initial grid has M cells in the angular direction and N cells in the radial direction and coveringone quadrant of a circle with radius Rmax. The nodes are defined by
(r, z)i, j = Ri, j(sinϕi, j, cosϕi, j), 0 � i � M, 0 � j � N
P. Váchal, B. Wendroff / Journal of Computational Physics 258 (2014) 118–136 127
Fig. 4. Example of a perturbed polar initial grid with 20 × 20 cells, Rmax = 1 and α = 0.05.
Fig. 5. Noh on perturbed polar grids with α = 0.05, LapEdge. Top: on a 20 × 20 grid. Bottom: on a 40 × 40 grid. For initial 20 × 20 grid, see Fig. 4.
with
ϕi, j = i�ϕ + αc1i, j�ϕ, �ϕ = π
2/M,
Ri, j = j�R + αc2i, j�R, �R = Rmax/N,
where c1 and c2 are randomly generated numbers between −1 and 1. To keep the domain boundary unperturbed, we setthe corresponding c’s to zero, more precisely
128 P. Váchal, B. Wendroff / Journal of Computational Physics 258 (2014) 118–136
Fig. 6. Sedov on perturbed polar grids with α = 0.075, LapEdge. Top: on a 20 × 20 grid. Bottom: on a 40 × 40 grid.
• for Noh problem, set c1 = c2 = 0 if i ∈ {0, M} or j ∈ {0, N},• for Sedov problem, set c1 = c2 = 0 if i ∈ {0, M} or j ∈ {0,1, N}.
The reason for the latter, i.e. for keeping the cells connected to the origin regular, is to clearly define the initial ball (sym-metric volume) where high energy is deposited. Note that for |α| � 0.5 the initial grid is never tangled, and for the Nohproblem the initial velocity field is symmetric, only the initial nodal positions are perturbed. We show in Fig. 4 an ex-ample of the initial perturbed grid with 20 × 20 cells, Rmax = 1 and α = 0.05 together with a zoom in of a portionof it.
See Fig. 5 for Noh and Fig. 6 for Sedov.
6.1.3. Rectangular gridIn this section we run the Noh problem on the computational domain (r, z) ∈ [0,1] × [0,1] and the Sedov problem on
(r, z) ∈ [0,1.2] × [0,1.2]. This domain is covered by a grid of N × N square cells.In the Sedov test, all initial energy Etotal is stored in one cell at the origin.Refer to Fig. 7 for Noh and to Fig. 8 for Sedov. Each test is shown on three different mesh resolutions, namely N = 20,
40 and 80.
6.2. Comparison with CS
We show here the results obtained running the tensor viscosity CS [2].Figs. 9(a), 10 and 11 contain Noh on all the grids but not all the resolutions. Note: For finer resolutions, we had to take
a very small CFL number.Fig. 9(b) shows Sedov on the symmetric polar grid for all the resolutions, whereas on the perturbed polar and rectangular
grids this test failed.
P. Váchal, B. Wendroff / Journal of Computational Physics 258 (2014) 118–136 129
A well-known difficult test is the Saltzman piston problem [15,5], which we run here in the r–z coordinate system. Thecomputational domain is (r, z) ∈ [0.0.1,0.1], covered by a 100 × 10 grid with nodal positions prescribed by
ri, j = i�r, 0 � i � 10,
zi, j = j�z + (10 − i) sin( j�rπ)�z, 0 � j � 100,
with
�r = 0.1, �z = 1
.
10 100
130 P. Váchal, B. Wendroff / Journal of Computational Physics 258 (2014) 118–136
Initially, the domain is filled with gas at rest, with ρ = 1 and ε = 10−6 everywhere. The ideal polytropic gas equation ofstate with Γ = 5/3 is used. At time t = 0, a piston located at z = 0 starts to move inward with constant velocity U = (0,1),triggering an infinite-strength shock moving in the same direction with constant speed 4/3. Therefore the shock is reflectedfrom the opposite wall (z = 1) at time 0.75.
We show the outcome at time t = 0.7 before the first shock reflection in Fig. 12. Although LapEdge in the default setupruns past the first reflection time to t = 0.925, the results at both times are poor; clearly more work is required on themethod for this case.
For results of the Saltzman test by CS in Cartesian coordinates see [2], for r–z results using an improved version see[10].
P. Váchal, B. Wendroff / Journal of Computational Physics 258 (2014) 118–136 131
Fig. 10. Noh by CS [2] on perturbed polar grids with α = 0.05. Top: on a 20 × 20 grid. Bottom: on a 40 × 40 grid. For initial 20 × 20 grid, see Fig. 4.
6.4. Implementation details
6.4.1. Nodal mass and dimensional factor σThe mass of a node is computed as
mp =∑
mpc,
c(p)
132 P. Váchal, B. Wendroff / Journal of Computational Physics 258 (2014) 118–136
Fig. 12. Saltzman piston problem t = 0.7 by LapEdge.
P. Váchal, B. Wendroff / Journal of Computational Physics 258 (2014) 118–136 133
where the mass matrix component mpc is now the mass of a quadrilateral subcell, given by node p, the center of cell c, andmidpoints of the two edges emerging from p. Refer e.g. to [7] for an analysis of possible nodal mass definitions.
To define the dimensional factor σce , we use the linear as well as nonlinear part of artificial viscosity module originallyproposed by Kuropatenko [9]. In particular, we take
σce = Sc�charec κc
(having the dimension velocity-times-length-times-density), where
• Sc is the switch to turn off viscosity in expanding cells, which is detected by the sign of velocity divergence.• κc is the viscosity module
κc = ρc
(k2
Γc + 1
4�Uc +
√(k2
Γc + 1
4�Uc
)2
+ (k1sc)2
), (51)
where Γc is the adiabatic gas constant, ρc resp. sc the cell’s density resp. speed of sound and
�Uc = maxe(c)
‖�Ue‖.
The constants k1 and k2 are used to control the amount of linear and nonlinear of artificial viscosity used. We usek1 = k2 = 1 in all LapEdge calculations.
• �charec is the characteristic length
�charec = Ac
4�e,
where Ac is the area (Cartesian volume) of cell c and �e is the length of edge e. This characteristic length was used inall LapEdge numerical results.
6.4.2. The solution algorithmAlgorithm 1 describes step by step the process of advancing the solution from time tn to time tn+1 = tn + �t .
Algorithm 1: Advancing in time.Input : Values at time tn
Iterate with index 〈ν〉: (inner iter. for velocity convergence)• Compute corner forces (cell and subcell pressure, viscous)
F〈ν〉 = F ({X,U,ρ}n+ 12 ,〈ν−1〉,Un,Pn+ 1
2 ,(k−1),m)
• Compute acceleration d δUdt and apply boundary conditions
• Update velocity Un+1,〈ν〉 = Un + �t d δUdt
• Compute velocity at half-time Un+ 12 ,〈ν〉 = 1
2 (Un + Un+1,〈ν〉)• Get node positions at half-time Xn+ 1
2 ,〈ν〉 = Xn + 12 �t Un+ 1
2 ,〈ν〉
• Update density ρn+ 12 ,〈ν〉 from Xn+ 1
2 ,〈ν〉 and mass m
• Set {X,U,ρ}n+ 12 ,(k) = {X,U,ρ}n+ 1
2 ,〈ν〉 and F(k) = F〈ν〉
• Update energy: εn+1,(k) = εn − �t∑
p(c) F(k) · Un+ 12 ,(k)
Check that viscous forces are dissipative (their work � 0)
• Move the grid Xn,(k) = Xn + �t Un+ 12 ,(k)
• Update density ρn+1,(k) from Xn+1,(k) and mass m• Compute pressure from EOS and its time-centered value
Pn+1,(k) = P({ρ,ε}n+1,(k)), Pn+ 12 ,(k) = 1
2 (Pn +Pn+1,(k))
• Set {X,ρ, ε,P}n+1 = {X, ε,ρ,P}n+1,(k) and Un+1 = Un+1,〈ν〉
Inner and outer iterations are used to maintain dissipation of energy. More specifically, it has been proved in Section 2that our artificial viscosity is dissipative in continuous time. To ensure dissipation also in the discrete case, inner iterations
on the velocity update are carried out so that velocity converges, and thus velocity Un+ 12 ,(k) which is used in the energy
update (in scalar product with the forces) is the same which was used to compute the forces. (Note that this inner-outeriteration on velocity was also needed in [12].) Further note that energy is conserved locally, since the forces F(k) in thevelocity update are exactly the same as those used to update energy.
134 P. Váchal, B. Wendroff / Journal of Computational Physics 258 (2014) 118–136
In practice we only perform one inner iteration (on 〈ν〉) and two outer ones (on (k)), which corresponds to one predictorand one corrector, but we check the sign of viscous work after the corrector to ensure that this is sufficient for dissipation.So far in all the test runs this was the case. Let us remark here that the tensor viscosity [2] in r–z is not dissipative byconstruction, so even iterating velocity until convergence does not ensure dissipation and we have to cut off negative worka posteriori when it appears, so that energy is not perfectly conserved.
It remains to describe the computation of forces, denoted as F(·) in the Algorithm 1, and boundary conditions foracceleration. For interior nodes the process is as follows. In each corner pc of cell c, we compute the Cartesian (cell andsubcell) pressure force FD
pc as in [3,4], multiply it by the r-coordinate of node p (area weighting) and then add our essentially
r–z viscous force FApc given by (9)–(11), as in (8). These forces are stored corner by corner, so that we can sum over cells
sharing node p in the velocity update and over the nodes of cell c in the energy update. The acceleration of the node isnow obtained by simply dividing the total force acting on the node by its mass.
For a node p on the z-axis (rp = 0), the acceleration due to pressure forces is given by rqFDp /( 1
2 mq), where FDp is the total
Cartesian pressure force at p and q is the interior node neighboring to p (i.e. nearest neighbor on the same circle/ j-curve:p = (0, j), q = (1, j) if the grid covers one quadrant only). The acceleration of z-axis nodes due to our viscous force iscovered in Section 5.1.
The time step length �t is provided by the user for the first step (typically �t1 = 10−6 for our tests), and for all othersit is calculated as
�tn = min(CCFL�t, β�tn−1), �t = min
c
mine(c) �e
sc,
where �e is the edge length, sc is the cell’s speed of sound, CCFL ensures satisfaction of the CFL condition and β prevents�t from growing too fast between two consecutive time steps. By default we set CCFL = 0.25, β = 1.2.
Optionally, we can use the smaller median instead of shortest edge, i.e. replace the numerator in definition of �t by theshortest distance between midpoints of opposite edges in the cell. Similarly as the other parameters, this choice does notaffect the result significantly.
7. Summary
A key component of staggered grid Lagrangian codes is the ability to decrease kinetic energy and increase internal energyduring shock compression. This is relatively straightforward in one space dimension and in two Cartesian dimensions. Inthe latter case the tensor viscosity of [2], among others, is known to effectively achieve this. In cylindrical axi-symmetric(r–z) geometry there also may be the need to exactly preserve spherical symmetry if at all possible; indeed, for a logicallyrectangular grid composed of straight edge quadrilaterals this is only possible if the grid is symmetric, that is, equi-angularpolar. In that case the area-weighted scheme [3] readily converts Cartesian forces into symmetry preserving r–z forces. Thisincludes the tensor viscous force. However, that r–z force is not guaranteed to increase internal energy. In this paper wehave proposed an r–z viscous force that always increases internal energy if it is turned on. It is modeled after the cylindricalgeometry Laplacian of the velocity. We proved that with it spherical symmetry is preserved on an equi-angular polar grid.It is also formulated to be applicable to any logically rectangular grid.
We have computed the classic Noh and Sedov test problems, first on a sequence of successively more finely resolvedsymmetric grids. Symmetry was always preserved. In order to obtain a different perspective, were also able to run in ourcode the tensor viscosity on those grids. In this case our viscosity seems to be slightly more accurate.
Next, in order to test the generality of our code we did Noh and Sedov on a small randomly perturbed symmetric grid,and on three successively finer resolutions of an initially square grid. All ran to completion without any limiters or tinkeringwith CFL, Kuropatenko factors k1, k2 or subcell pressures. The results seem not to be strongly dependent on the grid beingsymmetric. We also computed the difficult Saltzman problem with a less than satisfactory result.
Concerning Noh and Sedov, after some tuning, CS tensor viscosity produced comparable results on the symmetric grid,while on the perturbed and square grids some test results were comparable but some did not run to completion.
Acknowledgements
The authors gratefully acknowledge the following technical contributions to this work. Preliminary calculations by MilanKucharík and Richard Liska that supported the correctness of the theory. Before we had our own code we received invaluablehelp from Raphaël Loubère with the code ALE Inc [11]. Conversations with Pierre-Henri Maire about Lagrangian methods.Misha Shashkov has generously shared with us his considerable knowledge and experience with staggered grid Lagrangianmethods. Burton Wendroff [18] has an early version of the theory restricted to equi-angular polar grids. Careful reviews andcomments from both anonymous reviewers are highly appreciated.
This work was performed under the auspices of the National Nuclear Security Administration of the US Department ofEnergy at Los Alamos National Laboratory, under Contract W-7405-ENG-36 and Contract DE-AC52-06NA25396. The authorsacknowledge the partial support of the DOE Advanced Simulation and Computing (ASC) Program and the DOE Office ofScience ASCR Program.
Pavel Váchal has been partly supported by the Czech Science Foundation project P201/12/P554 and RVO: 68407700.
P. Váchal, B. Wendroff / Journal of Computational Physics 258 (2014) 118–136 135
Appendix A. Proofs
Proof of (20).(sin(θ − γ ) − sin θ
)(sin(θ − γ ) + sin θ
) = sin2(θ − γ ) − sin2 θ
= sin2 θ cos2 γ − 2 sin θ cosγ cos θ sinγ + cos2 θ sin2 γ − sin2 θ,(sin(θ + γ ) − sin θ
136 P. Váchal, B. Wendroff / Journal of Computational Physics 258 (2014) 118–136
Proof of (46).
R−2j ‖Xi+1, j − Xi−1, j‖2 = [
sin(θ + γ ) − sin(θ − γ )]2 + [
cos(θ + γ ) − cos(θ − γ )]2
= [sin θ cosγ + cos θ sinγ − sin θ cosγ + cos θ sinγ ]2
+ [cos θ cosγ − sin θ sinγ − cos θ cosγ − sin θ sinγ ]2
= (2 cos θ sinγ )2 + (2 sin θ sinγ )2. �Proof of (44). Using the full force as in (1)∑
c
dEc
dt=
∑c
(dKc
dt+ mc
dεc
dt
)=
∑c
(∑p(c)
mpc1
2
d(Up · Up)
dt−
∑p(c)
Fpc · Up
)
=∑
c
(∑p(c)
mpcUp · dUp
dt−
∑p(c)
Fpc · Up
)
=∑
c
(∑p(c)
mpcUp ·∑c′(p)
Fpc′
mp−
∑p(c)
Fpc · Up
).
Interchanging the sums over c and c′ and using (38) we see that∑c
dEc
dt= 0,
and therefore that
Gpc = mpcUp ·∑c′(p)
Fpc′
mp− Fpc · Up (A.1)
or, also,
mp Gpc = Up ·∑c′(p)
(mpcFpc′ − mpc′Fpc). � (A.2)
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