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EXACT SOLUTION AND A TRULY MULTIDIMENSIONAL1

GODUNOV SCHEME FOR THE ACOUSTIC EQUATIONS∗2

WASILIJ BARSUKOW† AND CHRISTIAN KLINGENBERG†3

Abstract. The linearized Euler equations are a valuable system for which the behaviour of4their multi-dimensional solutions can be studied explicitly. Here, the exact solution in three spatial5dimensions is derived. It is then used to obtain a truly multidimensional Godunov finite volume6scheme in two dimensions. Additionally the appearance of logarithmic singularities in the exact7solution of the 4-quadrant Riemann Problem in two dimensions is discussed.8

Key words. Godunov scheme, multidimensional Riemann Problem, acoustic equations, lin-9earized Euler equations10

AMS subject classifications. 35B08, 35B44, 35C05, 35E05, 35E15, 35F40, 35L03, 35Q35,1165M0812

1. Introduction. Though the linearized Euler equations cannot grasp the full13

complexity of inviscid hydrodynamics, they are a paramount example of a hyperbolic14

system of equations in multiple spatial dimensions. Linearizing15

∂tρ+ div (ρv) = 016

∂t(ρv) + div (ρv ⊗ v + p) = 01718

with p(ρ) = Kργ around the state (ρ,v) = (ρ, 0) yields19

∂tρ+ ρdiv v = 0(1)20

∂tv + c2grad ρ

ρ= 0(2)21

22

where one defines c =√p′(ρ). Linearization with respect to a fluid state moving at23

some constant speed U can be easily removed or added via a Galilei transform.24

The same system can be obtained from the Euler equations endowed with an25

energy equation26

∂tρ+ div (ρv) = 027

∂t(ρv) + div (ρv ⊗ v + p) = 028

∂te+ div (v(e+ p)) = 02930

with e = pγ−1 + 1

2ρ|v|2. Linearization around (ρ,v, p) = (ρ, 0, p) yields31

∂tρ+ ρdiv v = 0(3)32

∂tv +grad p

ρ= 0(4)33

∂tp+ ρc2 div v = 0(5)3435

∗Submitted to the editors DATE.†Wurzburg University, Institute for Mathematics, Emil-Fischer-Straße 40, 97074 Wurzburg,

Germany ([email protected], https://www.mathematik.uni-wuerzburg.de/∼barsukow).

1

This manuscript is for review purposes only.

mailto:[email protected]

https://www.mathematik.uni-wuerzburg.de/~barsukow

https://www.mathematik.uni-wuerzburg.de/~barsukow

2 W. BARSUKOW AND C. KLINGENBERG

Equations (4)–(5) are (up to rescaling and renaming) the same as (1)–(2). Both can36

be linearly transformed to the symmetric version37

∂tv + c grad p = 0(6)38

∂tp+ cdiv v = 0(7)3940

This system is the one to be studied in what follows. p will be called pressure and41

v the velocity – just to have names. Due to the different linearizations and the42

symmetrization they are not exactly the physical pressure or velocity any more, but43

still closely related. These equations describe the time evolution of small perturbations44

to a constant state of the fluid. Therefore they will be called the equations of linear45

acoustics. Equations (6)–(7) have been studied among others by [14, 17, 1, 3].46

It should be noted that only the equation for the scalar p is the usual scalar wave47

equation48

∂2t p− c2∆p = 0(8)4950

whereas v fulfills51

∂2t v − c2grad div v = 0(9)5253

The identity ∇× (∇×v) = ∇(∇· v)−∆v links this operator to the vector Laplacian54

in 3-d. By (6)55

∂t(∇× v) = 0(10)5657

but ∇× v needs not be zero initially.58

The system (6)–(7) has a low Mach number limit just as the Euler equations59

(compare [3] and for the Euler case see e.g. [11, 12, 16]). One introduces a small60

parameter ε and inserts the scaling ε−2 in front of the pressure gradient in (4) such61

that the system and its symmetrized version read, respectively,62

∂tv +1

ε2grad p = 0(11)63

∂tp+ c2 div v = 0(12)6465

and66

∂tv +c

εgrad p = 0(13)67

∂tp+c

εdiv v = 0(14)68

69

The transformation which symmetrizes the Jacobian J =

(0 1

ε2

c2 0

)is70

S =

(1

cε

)(15)71

72

such that J = S

(0 c

εcε 0

)S−1. In multiple spatial dimensions the upper left entry73

in S has to be replaced by an appropriate block-identity-matrix. In the following74

preference is given to the symmetric version if nothing else is stated. Regarding75


EXACT SOLUTION AND A GODUNOV SCHEME FOR ACOUSTICS 3

the low Mach number limit the non-symmetrized version is more natural and will76

be reintroduced for studying the low Mach number properties of the scheme. The77

variables of the two systems are given the same names to simplify notation.78

The three-dimensional solution to (6)–(7) is derived in this paper in Section 2 by79

different methods, which are compared. In Section 3 it is applied to a two-dimensional80

Riemann problem, and in Section 4 it is used to obtain a finite volume Godunov81

method. Conclusions are drawn in Section 5.82

2. Exact solution.83

2.1. Review of the one-dimensional case. The 1-d system84

∂tp+ c∂xv = 0(16)85

∂tv + c∂xp = 0(17)8687

is easily solved via characteristics88

∂t(p± v)± c∂x(p± v) = 0(18)8990

The solution at a point x depends only on initial data at points y for which |y−x| = ct.91

This in general is different in multi-dimensional situations: the solution to the scalar92

wave equation (8) at a point x in even spatial dimensions depends on initial data at93

points y for which |x−y| ≤ ct (cf. [4], Section 2.4.1e). In odd spatial dimensions the94

solution depends on data in |x− y| = ct.95

The solution to the vector wave equation (9) will turn out to depend on initial96

data at points y for which |x− y| ≤ ct also in odd spatial dimensions.97

With respect to the numerics of the 1-d system (16)–(17), one can write down the98

exact Godunov scheme by solving the 1-d Riemann problem for this system which is99

the Roe solver [18]. The dimensionally split case applied to multiple dimensions is100

discussed in [8].101

2.2. Multi-dimensional case. In the following, wherever appropriate, the de-102

pendent variables will be named q := (v, p) and the multi-dimensional system written103

as104

∂tq + (J · ∇)q = 0(19)105106

where J is the vector of the Jacobians into x, y and z directions, respectively. For the107

system (7)–(6) in 3-d they are108

J =

0 c0

0c 0

,

0

0 c0

c 0

,

0

00 cc 0

(20)109

110

The three different ways of deriving the solution lead to different, though of course111

equivalent shapes of the solution formula. Certain properties of the solution are more112

immediately seen in one form, and others in the other form. It thus seems appropriate113

to discuss these different approaches.114

2.2.1. Fourier transform and Green’s functions. The very standard pro-115

cedure of finding a solution to any linear equation such as (19) is to decompose the116

initial data q0(x) into Fourier modes117

q0(x) =1

(2π)d/2

∫ddk q0(k) exp(ik · x)(21)118

119



where d is the dimensionality of the space. The coefficients q0(k) of this decomposition120

are called the Fourier transform of q0 and k here characterizes the mode. The time121

evolution of any single mode can be found via the ansatz122

exp(−iω(k)t+ ik · x)(22)123124

where ω(k) is found from the equations by inserting the ansatz. Applying the proce-125

dure to (19) makes ω appear as the eigenvectors ωn (n = 1, . . . d+ 1) of the matrix126

J · k(23)127128

which for the acoustic system yields ωn ∈ 0,±c|k|. Note that this matrix also129

shows up in the study of the Cauchy problem and bicharacteristics (see e.g. [2], VI,130

§3). One then decomposes every Fourier mode with respect to the (orthonormally131

chosen) eigenvectors en (n = 1, . . . , d+ 1) and evolves every part with its own ωn. By132

linearity, the time evolution q(t,x) of the initial data q0(x) is given by adding all the133

time evolutions of the individual modes:134

q(t,x) =1

(2π)d/2

∫ddk

d+1∑n=1

en(en · q0(k)) exp(−iωnt− ik · x)(24)135

136

Given the Fourier transform of any initial data therefore the solution can easily be137

constructed. Due to the fact that the diagonalization is very simple for the acoustic138

system, all the formulae can be worked out in full generality, and for the acoustic139

system one finds140

p(t,x) =1

(2π)d/2

∫ p0(k) + k·v0(k)|k|

2exp(ik · x− ic|k|t)(25)141

+p0(k)− k·v0(k)

|k|

2exp(ik · x + ic|k|t)

ddk(26)142

143

144

v(t,x) =1

(2π)d/2

∫ p0(k) + k·v0(k)|k|

2

k

|k|exp(ik · x− ic|k|t)(27)145

+

k·v0(k)|k| − p0(k)

2

k

|k|exp(ik · x + ic|k|t)(28)146

+

v0(k)− k

|k|k · v0(k)

|k|

exp(ik · x)

)ddk(29)147

148

It is however quite inconvenient to first have to Fourier transform the initial data149

q0(x). A better decomposition is that of δ-distributions, which is trivial1:150

q0(x) =

∫ddx′δ(x− x′)q0(x′)(30)151

152

1The integration sign is meant formally here and denotes the application of the linear functionalδ to the function q0.



Again by linearity, the solution q(t, x) can thus be composed of time evolutions of the153

δ-distributions. These functions deserve a name and are called Green’s functions G;154

the time evolution of δ(x − x′) is G(t,x; x′) and therefore the solution will then be155

written as the convolution156

q(t,x) =

∫ddx′G(t,x; x′)q0(x′)(31)157

158

of the initial data with Green’s function. The Fourier transformation now has to be159

performed only once for the δ-distribution. Taking up the acoustic equations, after160

lengthy evaluations of integrals one finds (defining ξ := x− x′ and ξ := |ξ|)161

Gp(t,x; x′) = −(p+

v · ξct

)1

4π

δ′(ξ − ct)ξ

(32)

162

Gv(t,x; x′) =2

3vδ(ξ)− 1

4π

ξ

ξpδ′(ξ − ct)

ct

(33)

163

− 1

4π

([v − 3

v · ξξ

ξ

ξ

](Θ(ξ − ct)

ξ3+δ(ξ − ct)

ξ2

)+

v · ξξ

ξ

ξ

δ′(ξ − ct)ξ

)(34)164

165

Here δ′ denotes the (distributional) derivative of δ.166

After Green’s function is determined, the solution is obtained as a convolution167

with the initial data. Evaluating the convolution in spherical polar coordinates168

(r, ϑ, ϕ) makes the r-parts simplify, and one is left with only the integrations over169

ϕ and ϑ, which are abbreviated as170

∫dΩ :=

1

4π

π∫0

dϑ sinϑ

2π∫0

dϕ(35)171

172

i.e. this is the average over the sphere. Such spherical means appear already in the173

study of the scalar wave equation ([10, 4]). After calculating the integrals one obtains174

p(t,x) = ∂r

(r

∫dΩ p0

)− 1

r∂r

(r2

∫dΩ n · v0

)(36)175

v(t,x) =2

3v0(x)− 1

r∂r

(r2

∫dΩ p0n

)+ ∂r

(r

∫dΩ (v0 · n)n

)(37)176

−∫

dΩ [v0 − 3(v0 · n)n]−∫ ct

0

dr1

r

∫dΩ [v0 − 3(v0 · n)n](38)177

178

The last equation can be rewritten into179

v(t,x) = v0(x)− 1

r∂r

(r2

∫dΩ p0n

)(39)180

+

∫ ct

0

dr1

r∂r

[1

r∂r

(r3

∫dΩ(v0 · n)n

)− r

∫dΩv0

](40)181

182

where everything (if not stated explictly) is understood to be evaluated at x + rn,183

and wherever it remains, r = ct to be taken at the very end. n is the unit outward184



normal vector in spherical polar coordinates:185

n =

cosϕ sinϑsinϕ sinϑ

cosϑ

(41)186

187

More details on the transition from (38) to (40) are given in Appendix A.188

There is a number of striking differences to the 1-d case that appear in multiple189

spatial dimensions. The appearance of an integration over 1r in (38) or (40) rises the190

question of whether the integration actually gives finite results. It turns out that191

often the square bracket (in both integrals) is zero already, such that the integral is192

harmless. The example in Section 3 below however shows that this is not a general193

feature and that the presence of this term gives rise to singularities of the solution (as194

has been observed already in [1] using a different solution strategy).195

2.2.2. Via the Helmholtz decomposition. Usage of the Helmholtz decom-196

position allows to write down a scalar wave equation for p and for the curl-free part197

of v, whereas the time evolution of the curl is given by ∂t(∇× v) = 0. The solution198

to the scalar wave equation is well-known ([10]) and the solution can then be written199

down as (compare [5])200

p(t,x) = p0(x) +

∫ ct

0

dr r

∫dΩ (div grad p0)− r

∫dΩ div v0(42)201

v(t,x) = v0(x) +

∫ ct

0

dr r

∫dΩ (grad div v0)− r

∫dΩ (grad p0)(43)202

203

Note that the Helmholtz decomposition does not need to be computed explicitly as204

the two time evolutions conveniently reassemble into (43). Again all the quantities205

are to be evaluated at x + rn.206

Equation (43) makes obvious that any change in time of v is a gradient. Indeed,207

the curl must be stationary due to Eq. (10).208

In order to transfer the r-derivatives in (36)–(40) into the derivative operators209

in (42)–(43) one uses the Gauss theorem for the sphere of radius r. For example,210

differentiating211

∂r

(r

∫dΩ p0(x + rn)

)(44)212

213

with respect to r yields214

∂2r

(r

∫dΩ p0(x + rn)

)=

1

r∂r

(r2∂r

∫dΩ p0(x + rn)

)(45)215

216

by elementary manipulations. Differentiation with respect to r can be replaced byn · ∇ inside the spherical mean:217

1

r∂r

(r2∂r

∫dΩ p0(x + rn)

)=

1

r∂r

(r2

∫dΩ n · ∇p0(x + rn)

)(46)218

219

and by Gauss theorem220

=1

r∂r

∫ r

0

dr′(r′2∫

dΩ∇ · ∇p0(x + r′n)

)(47)221

= r

∫dΩ∇ · ∇p0(x + r′n)(48)222

223



Integrating over r, and evaluating at r = ct yields the sought identity224

∂r

(r

∫dΩ p0(x + rn)

)∣∣∣∣r=ct

= p0(x) +

∫ ct

0

dr r

∫dΩ∇ · ∇p0(x + rn)(49)225

226

In a similar but slightly more complicated way the equivalence of the other terms can227

be shown.228

Formula (38) shows that the solution can be reformulated in a way employing only229

first spatial derivatives, whereas the other versions make the impression that second230

derivatives need to be computed. In one-dimensional problems only the values of the231

initial data appear in the solution formulae, not their derivatives. This is different232

in multiple dimensions and can already be observed for the scalar wave equation (as233

dicussed e.g. in [4]).234

Equation (49) shows that although the right-hand side makes you believe that235

the solution depends on initial data at x, and also on initial data inside all of the ball236

around x with radius ct, the left-hand side makes obvious that this is not true. In237

particular the value of the initial data at x has no influence on the solution and the238

explicit appearance of p0(x) on the other side of the equation was artificial, and is239

due to the fact that the integral actually contains −p0(x). The left-hand side shows240

that the solution depends only on the value and the first derivative of p0 along the241

boundary of the aforementioned ball. Also it is not any derivative, but it is the normal242

derivative ∂rp0 = n · ∇p0 which matters.243

Similarly, comparison between Equations (43) and (38) shows that the expressions244

can be simplified. However, now the integral over r is still present in both expressions.245

One might wonder whether it is possible, by some manipulation, to rewrite Equation246

(38) such that it contains only terms involving initial data at x and the boundary of247

the ball of radius ct around x, rather than all of its interior. The following argument248

shows that this is not possible.249

Equation (38) contains an integration over r and further terms. These depend on250

initial data at x and on initial data and its normal derivatives at r = ct. Assume that251

it is possible to rewrite the integral over r such that it also depends only on initial252

data and its normal derivatives at x and at r = ct. Then it would be possible to vary253

any initial data in the region 0 < r < ct without effect on the solution. This however254

is not true, because the solution at x = 0 and time ct = D of the initial data255

p0 = 0(50)256

v0 = ezCx3y2z3(D2 − x2 − y2 − z2)2(51)257258

is259

v(D/c, 0) =CD12

46200ex(52)260

261

and at the same time262

v0(rn) = ezC sin2 ϕ cos3 ϕ sin5 ϑ cos3 ϑ · r8(D2 − r2)2(53)263264

such that265

v0(0) = 0(54)266

v0(rn)|r=ct = 0(55)267

∂rv0(rn)|r=ct = 0(56)268269



The solution depends on the parameter C which only modifies the initial data inside270

0 < r < ct. Therefore, contrary to the scalar wave equation, the acoustic system shows271

dependence on initial data inside the cone x+rn (or in other words, the timelike past)272

even for three spatial dimensions.273

2.2.3. Via exponential map. Using the exponential map, the solution of274

∂t

(vp

)+

(G pDv

)= 0(57)275

276

with G and D (not necessarily commuting) operators is given by277 (uv

)= exp(−tM )

(uv

)0

(58)278

279

with280

M =

(0 GD 0

)(59)281

282

such that283

(vp

)=

∞∑m=0

t2m(G D)m

(2m)!v0

∞∑m=0

t2m(DG )m

(2m)!p0

−

∞∑m=0

t2m+1(G D)m

(2m+ 1)!G p0

∞∑m=0

t2m+1(DG )m

(2m+ 1)!Dv0

(60)284

285

In the case considered here, G = grad and D = div . One can show the equivalence286

to the solution derived above by expanding all qunatities around r = 0, e.g.287

p0(x + rn) =

∞∑m=0

∂mrm!

p0(x)(61)288

289

and using the identity290 ∫dΩninj · · ·nkn`︸︷︷︸

m factors

=1

(m+ 1)!δ(ij · · · δk`)︸︷︷︸m/2 factors

(m even)(62)291

292

where nj denotes the j-th component of n, δij =

1 i = j

0 elsethe Kronecker symbol293

and the round brackets denote a symmetrization, i.e. a sum over all the permutations294

(without any prefactor included). Integration over an odd number of factors yields295

zero by considerations of symmetry. In case of an even number there is no other tensor296

available than the identity δ, and the result has to remain unchanged with respect to297

the exchange of any two indices, which explains the symmetrization. The prefactor298

can be determined by evaluating the integral in some special case that is easy to check299

explicitly, e.g.∫

dΩnznz · · ·nznz = 1m+1 . The approach of writing the solution by300

employing the exponential map is used in [17] as a guideline for numerical solutions,301

but the authors do not derive the solution formulae.302



As an example consider (40) with only the initial data in v0. Here again (upper303

and lower) indices denote the different components304

vi(t,x) = vi0(x) +

∫ ct

0

dr1

r∂r

[1

r∂r

(r3

∫dΩ(vk0nk)ni

)− r

∫dΩvi0

](63)

305

= vi0(x) +

∞∑m=0

∫ ct

0

dr1

r∂r

[1

r∂r

(rm+3

∫dΩ

(nj∂j)m

m!(vk0nk)ni

)(64)306

− rm+1

∫dΩ

(nj∂j)m

m!vi0

](65)307

= vi0(x) +

∞∑m=0

m+ 1

m

rm

m!

[(m+ 3)

∫dΩnkn

i(nj∂j)mvk0 −

∫dΩ(nj∂j)

mvi0

](66)308

= vi0(x) +∞∑m=1

r2m

(2m)!∂i(∂a∂a)m−1∂kv

k0(67)309

310

In the last equality the identity (62) and311

(m+ 3)

∫dΩnkn

i(nj∂j)mvk0 =

1

m+1 (∂a∂a)m/2vi0 + mm+1∂

i(∂a∂a)m2 −1∂kv

k0 m ≥ 2

vi0 m = 0

(68)

312

313

for m even was used. Equation (67) is the one from (60) which completes the example.314

Analogous computations confirm the other parts of the solution.315

3. The two-dimensional Riemann problem. As an example, a particular316

feature of the exact solution (36)–(40), or (42)–(43) of a two-dimensional Riemann317

problem (in the x-z-plane for computational convenience) shall be discussed here. The318

initial velocity shall be v0 = (0, 0, 1)T in the first quadrant and vanish everywhere319

else (see Fig. 1). Also everywhere p0 = 0.320

Fig. 1. Left: Setup of the 2-dimensional Riemann Problem. The only non-vanishing initialdatum is the x-velocity in the first quadrant, indicated by the arrow. As the problem is linear itsmagnitude is of no importance and is chosen to be 1. Right: Sketch of the integration domain.

Computing the x-velocity along x = 0, z > 0, say, results in the following expres-321



sion2:322

vx(t, z) =

∫ ct

0

dr1

r∂r

[1

r∂r

(r3

∫dΩ v0,znznx

)](69)

323

=

∫ ct

0

dr3

r

∫dΩ v0,znznx + r

∫dΩ ∂rv0,znznx

∣∣∣∣r=ct

+ 4

∫dΩ v0,znznx

∣∣∣∣ctr=0

(70)324

325

Here integration by parts has been used twice.326

The initial data v0,z at x+rn have two regions that have to be treated separately327

(see Fig. 1). If ϑ < ϑ0, then v0,z(r, ϑ) = 1 for all ϕ ∈[−π2 ,

π2

]. Otherwise, they are328

only non-zero if r is small enough, i.e. if z + r cosϑ > 0:329

v0,z(r, ϑ) = 1−Θ(r +

z

cosϑ

)=

0 r > − z

cosϑ

1 else(71)330

331

where332

Θ(x− x0) :=

1 x > x0

0 else(72)333

334

Mind for the inequalities that here cosϑ < 0 because ϑ0 >π2 . Additionally the initial335

data vanish again outside ϕ ∈[−π2 ,

π2

], which will be taken care of by the integration336

bounds inside the spherical means.337

The derivative exists only for ϑ > ϑ0 and is338

∂rv0,z = −δ(r +

z

cosϑ

)= −δ(ϑ− ϑ0)

z sinϑcos2 ϑ(73)339

340

with cosϑ0 := − zr .341

With this342

r

∫dΩ ∂rv0,znznx = −r 1

2πzcos3 ϑ0 sinϑ0 =

1

2π

(zr

)2√

1−(zr

)2

(74)343 ∫dΩ v0,znznx =

1

4π

∫ π/2

−π/2dϕ

∫ ϑ0

0

dϑ sin2 ϑ cosϑ cosϕ =1

2π

1

3

(1− z2

r2

) 32

(75)344

345

if r > z and zero otherwise.346

ct∫0

dr3

r

∫dΩ v0,znznx(76)347

=1

2π

z∫0

dr3

r

∫ π

0

dϑ sin2 ϑ cosϑ︸︷︷︸=0

+1

2π

ct∫z

dr3

r

∫ ϑ0

0

dϑ sin2 ϑ cosϑ(77)348

=1

2π

∫ ct

z

dr1

r

(1− z2

r2

) 32

(78)349

=1

2π

1

3

z2

(ct)2

√1−

( zct

)2

− 4

3

√1−

( zct

)2

+ ln1 +

√1−

(zct

)2zct

(79)350

351

2In this paper an index is never meant to indicate a derivative; vx is the x-component of thevelocity v.



On total this gives352

vx(t, z) =1

2πln

1 +

√1−

(zct

)2zct

=1

2πL( zct

)(80)353

354

having defined355

L (s) := ln1 +√

1− s2

s= − ln

s

2− s2

4+O(s4)(81)356

357

One can verify that e−L (s) = tanarcsin s

2.358

Of course the above formulae are valid for z < ct only and vx(t, z) vanishes outside359

|x| ≤ ct by causality.360

Therefore the x-component of the velocity has a logarithmic singularity at z = 0,361

the corner of the initial discontinuity of the z-component. Such a behaviour of the362

solution does not have analoga in one spatial dimension. This has already been363

mentioned in [1] in the context of self-similar solutions to Riemann Problems. Here364

it has been obtained by application of the general formula (36)–(40) which is not365

restricted to self-similar time evolution.366

For Fig. 2–3 the integrals in (36)–(40) have been computed numerically at a finite367

number of points in the x-y-plane. The main difficulty is the appearance of derivatives368

of discontinuous functions, which has been solved by smoothing them out by a small369

amount. The ordinary derivative then approximates the δ-distributions that appear370

in an exact calculation. This explains the smoothing of the jumps and the singularity371

in the plots.372

Fig. 2. Solution of Riemann problem at time ct = 0.25. Left: Pressure. Center: x-velocity.Right: y-velocity. The smoothed out discontinuities are due to finite size sampling of the solution.In green the location of the initial discontinuity.

A vector plot of the flow is shown in Fig. 3.373

4. Godunov finite volume scheme.374

4.1. Procedure. In this section we aim at deriving a two-dimensional finite375

volume scheme, which updates the numerical solution qnij in a Cartesian cell Cij =376

[xi− 12, xi+ 1

2]× [yj− 1

2, yj+ 1

2] at a time tn to a new solution qn+1

ij at time tn+1 = tn+∆t377

using qnij and information from the neighbours of Cij . The grid locations are taken378

equidistant, such that ∆x := xi+ 12− xi− 1

2= const, and ∆y := yj+ 1

2− yj− 1

2= const.379

The knowledge of the exact solution makes it possible to derive a Godunov scheme.380

This is similar in spirit to an idea by Gelfand mentioned in [8, 6] (a translated version381



Fig. 3. The direction of the velocity v(t,x) is indicated by the arrows, color coded is the absolutevalue |v|.

is [7]), but the absence of a(n accessible) published work or details on the procedure382

make it hard to compare it to the approach taken here. From the scarce account383

it seems that an exact solution as presented here was not used and that the present384

approach is more general, in the sense that it might also be used to derive approximate385

solvers.386

This work restricts itself to a derivation of a Godunov scheme via the procedure387

of reconstruction-evolution-averaging ([19, 13]). As it is formulated, the derivation388

encounters technical difficulties, because the solution is needed at all points inside the389

cell. This means an evaluation of the integrals (36)–(40), or (42)–(43), for basically390

all x with no direct simplification. It afterwards undergoes the possibly nontrivial391

task of being integrated over the cell.392

Employing the structure of the conservation law allows to rewrite the volume393

integral into a time integral over the boundary, which is a huge simplification. Now394

the solution is only needed along the boundary of the cell, and one of the components395

of x is zero. Still however one needs to evaluate the solution formulae at a continuous396

set of x values.397

In the following it is shown that for linear systems a Godunov scheme can be398

written down using just one evaluation of the solution formula at a single point in399

space by suitably modifying the initial data.400

Consider a piecewise constant reconstruction of the initial data qij at some point401

in time, which then defines the piecewise constant function q0 with q0(x) = qij if402

x ∈ Cij (Fig. 4). Define the evolution operator Tt such that (Tt q0)(t,x) satisfies (19)403

with (at t = 0) initial data q0(x). Define also the sliding average operator404

(Aq)(x) :=1

∆x∆y

∆x/2∫−∆x/2

dξ

∆y/2∫−∆y/2

dη q(x + s)(82)405

406

where s is the shift vector s = (ξ, η)T. Then we have the following407



Fig. 4. Left: Piecewise constant reconstruction. Right: Application of the sliding average tothe same data amounts to a bilinear interpolation of the values qij interpreted as point values atxij .

Lemma 4.1. The two operators commute:408

AT∆t q0

∣∣∣xij

= T∆tAq0

∣∣∣xij

(83)409410

Proof. By linearity of the equations the time evolution commutes with the sliding411

average (82).412

It thus suffices to find the solution of the problem at time ∆t at xij taking the413

sliding-averaged initial data Aq0. For a 2-d grid this amounts to a bilinear interpola-414

tion of the values qij taken at points xij , see Fig. 4. In practice the four quadrants415

of the integration domain have still to be treated separately.416

In short, for linear systems the last two steps of reconstruction-evolution-averaging417

can be turned around to be reconstruction-averaging-evolution which reduces the418

derivation effort but does not change the result.419

4.2. Notation. In order to cope with the lengthy expressions for the numerical420

scheme, the following notation is used:421

[q]i+ 12

:= qi+1 − qi qi+ 12

:= qi+1 + qi(84)422

[q]i±1 := qi+1 − qi−1(85)423

[[q]]i± 12

:= [q]i+ 12− [q]i− 1

2qi± 1

2:= qi+ 1

2+ qi− 1

2(86)424

425

The only nontrivial identity is426

[q]i± 12

= [q]i+ 12

+ [q]i− 12

= [q]i±1(87)427428

For multiple dimensions the notation is combined, e.g.429

[[q]i±1]j±1 = qi+1,j+1 − qi−1,j+1 − qi+1,j−1 + qi−1,j−1(88)430431

The brackets for different directions commute.432

4.3. Finite volume scheme. It turns out that the solution (36)–(40) can be433

evaluated exactly. This has been performed with Mathematica [20] and one obtains434



the numerical scheme435

un+1 = unij −c∆t

2ε∆x

([p]i±1,j − [[u]]i± 1

2 ,j

)(89)436

− 1

2

(c∆t)2

ε2∆x∆y

(− 1

2π[[[[u]]i± 1

2]]j± 1

2− 1

4[[v]i±1]j±1 +

1

4[[[p]i±1]]j± 1

2

)(90)437

vn+1 = vnij −c∆t

2ε∆y

([p]i,j±1 − [[v]]i,j± 1

2

)(91)438

− 1

2

(c∆t)2

ε2∆x∆y

(− 1

2π[[[[v]]i± 1

2]]j± 1

2− 1

4[[u]i±1]j±1 +

1

4[[[p]]i± 1

2]j±1

)(92)439

pn+1 = pij −c∆t

2ε∆x

([u]i±1,j − [[p]]i± 1

2 ,j

)− c∆t

2ε∆y

([v]i,j±1 − [[p]]i,j± 1

2

)(93)440

− 1

2

(c∆t)2

ε2∆x∆y

(1

4[[[u]i±1]]j± 1

2+

1

4[[[v]]i± 1

2]j±1 − 2 · 1

2π[[[[p]]i± 1

2]]j± 1

2

)(94)441

442

Of course this scheme is conservative and can be written as443

qn+1 = qn − ∆t

∆x

(f

(x)

i+ 12 ,j− f (x)

i− 12 ,j

)− ∆t

∆y

(f

(y)

i,j+ 12

− f (y)

i,j− 12

)(95)444

445

because it is a Godunov scheme. One easily can identify the x-flux through the446

boundary located at xi+ 12:447

f(x)

i+ 12

=1

2

c

ε

pi+ 12 ,j− [u]i+ 1

2 ,j

0ui+ 1

2 ,j− [p]i+ 1

2 ,j

(96)448

+1

2

c∆t

ε∆y· cε

− 12π [[[u]i+ 1

2]]j± 1

2− 1

4 [vi+ 12]j±1 + 1

4 [[pi+ 12]]j± 1

2

014 [[v]i+ 1

2]j±1 − 1

2π [[[p]i+ 12]]j± 1

2

(97)449

450

The corresponding perpendicular flux is its symmetric analogon. The first bracket is451

easily identified as the flux obtained in a 1-d or dimensionally split situation.452

The appearance of prefactors which contain π in schemes derived with the exact453

multi-dimensional evolution operators has already been noticed in [15], but none of454

the schemes mentioned therein matches the one presented here.455

For better analysis of the low Mach number limit the scheme is given in Ap-456

pendix B in the usual variables, i.e. by applying the transformation (15) or, which is457

equivalent, by replacing p 7→ pcε .458

4.4. Numerical results. The scheme (90)–(94) is applied to two test cases. The459

first one is the Riemann Problem considered analytically in Section 3. The second is460

a test of the low Mach number abilities of the scheme.461

4.4.1. Riemann Problem. The initial setup is that of Section 3 (Fig. 1); it is462

solved on a square grid of 101× 101 cells on a domain that is large enough such that463

the disturbance produced by the corner has not reached the boundaries for t = 0.25.464

Here, c = ε = 1. The particular feature of this simulation is that it has been performed465

using a cfl number of 1. This is not possible with dimensionally split schemes, as466

the cfl condition there is ([8], Eq. 8.15, p. 63)467

c∆t <1

1∆x + 1

∆y

(98)468

469



which for square grids gives a maximum cfl number of 0.5. As the present scheme470

is an exact multidimensional Godunov scheme it is stable up to the physical cfl471

number. The results are shown in Fig. 5.472

Fig. 5. Solution of Riemann problem at time ct = 0.25 using scheme (90)–(94). Left: Pressure.Center: x-velocity. Right: y-velocity. Compare the images to Fig. 2. Although the present figurehas been obtained with a finite volume scheme, and Fig. 2 by a sampled evaluation of the exactsolution, the latter, despite larger computational effort, displays more noise. This is due to theparticular choice cfl= 1 (see text).

It is well-known that a cfl number of 1 makes the upwind scheme solve Riemann473

Problems for one-dimensional advection without error. A similar feature is observed in474

this simulation: for y > 0.25 the Riemann Problem in x-direction is yet uninfluenced475

by the corner and the solution is that of the usual one-dimensional Riemann Problem,476

which decays into two waves. These waves don’t show diffusion. However, just as in477

the case of the upwind scheme of linear advection this is an academic artefact. Any478

misalignment of the discontinuity with the grid will destroy this feature and smear out479

shocks just as any other first order scheme. Therefore the described test is not meant480

to stress any particularly good resolution of discontinuities, but it is rather meant to481

show the ability of the scheme to be integrated in a stable way with a cfl number482

which is double the cfl number possible with its dimensionally split counterpart.483

In Fig. 6 the y-component of the velocity obtained with the numerical scheme is484

compared to the analytic solution (80) found in Section 3.485

4.4.2. Low Mach number vortex. The second test shows the properties of486

the scheme in the limit ε → 0. The setup is that of a stationary, divergencefree487

velocity field and constant pressure:488

p0(x) = 1(99)489

v(x) = eϕ

rd r < d

2− rd d ≤ r < 2d

0 else

(100)490

491

The velocity thus has a compact support, which is entirely contained in the com-492

putational domain, discretized by 51 × 51 square cells. Here c = 1 and d = 0.2.493

Zero-gradient boundaries are enforced.494

Fig. 7 shows the error norm at time t = 1 for different cfl numbers; results for495

the scheme (90)–(94) are shown as solid lines. As has been stated earlier, it is stable496

until a cfl number of 1, which is confirmed by a rapid increase of error beyond this497

value. Additionally, the error drops significantly when the cfl number approaches 1498

from below. This drop is more and more abrupt the lower ε is.499



Fig. 6. The y-component of the velocity obtained by the numerical scheme (90)–(94) (cfl= 1)is shown together with the analytic solution (80). Although the latter has only been computed alongone particular axis, the solution is symmetric around the location of the corner in the initial databecause of self-similarity and scale invariance.

The dimensionally split solver is known to display artefacts in the limit ε→ 0 (see500

e.g. [9]). Results obtained with this scheme are shown in the same Figure by dashed501

lines. For small cfl numbers, the flux (97) approaches the dimensionally split case,502

which is confirmed experimentally. For the dimensionally split scheme the error does503

not depend on the cfl number. Also the small stability region cfl< 0.5, as well as504

the growth of the error for decreasing ε are prominent in the Figure.505

Fig. 7. Solid lines show L1 error norms of the numerical solution at time t = 1 for the scheme(90)–(94) as a function of the cfl number and for three choices of ε. Dashed lines, for comparison,display the same for the dimensionally split scheme.

This growth is present also for the case cfl = 1 for the scheme presented here.506

Therefore strictly speaking it is not suitable for the low Mach number regime. At the507

same time, in comparison to the dimensionally split case, the choice cfl = 1 gives a508

considerable improvement, as is shown in Fig. 5 for the case ε = 10−2. The reason509

for this behaviour can be understood by considering the modified equation for the510

scheme (119)–(123) (here it is of advantage to use the scheme in its non-symmetrized511



Fig. 8. Solution of the vortex initial data at time ct = 1 using scheme (90)–(94) for ε = 10−2.The quantity shown in color is the magnitude of the velocity. Left: Exact solution = initial data.Center: cfl = 0.8. Right: cfl = 1.

shape). The modified equation reads512

un+1 − un

∆t+

1

ε2∂xp = ∆x

c

2ε∂x(∂xu+ cfl ∂yv) +O(∆x2)(101)513

vn+1 − vn

∆t+

1

ε2∂yp = ∆x

c

2ε∂y(cfl ∂xu+ ∂yv) +O(∆x2)(102)514

pn+1 − pn

∆t+ c2(∂xu+ ∂yv) = ∆x

c

2ε(∂2xp+ ∂2

yp) +O(∆x2)(103)515516

Here, for simplicity, ∆x = ∆y has been used. One observes that choosing a cfl517

number of 1 makes the velocity diffusion become a gradient of the divergence. In the518

limit ε → 0 the divergence becomes O(ε), and thus the highest order terms in the519

modified equation become O(1). The low Mach number artefacts are then entirely520

due to contributions from the O(∆x2) terms.521

5. Conclusions and outlook. This paper presented an exact solution of the522

linearized Euler equations in three spatial dimensions. It is a paramount example of523

a solution very different to that of advection, with characteristic cones and spherical524

means replacing simple advection along a one-dimensional characteristic. On the525

other hand it also shows differences to the well-understood scalar wave equation, thus526

emphasizing the additional difficulties when dealing with systems of equations. Three527

different ways of derivation have been shown. The advantage of doing so is that they528

make different properties of the solution obvious. Some features are the dependence529

on first derivatives and function values, rather than function values in the initial530

data alone, the dependence on initial data that cannot be reached by traveling with531

one of the characteristic speeds and the appearance of divergent integrals in certain532

situations.533

The multi-dimensional Riemann Problem for the acoustic equations is a prominent534

example for the appearance of such a singularity, which is an intrinsically multi-535

dimensional feature. This paper presented the exact shape of the solution and showed536

that the singularity is logarithmic.537

Furthermore, using the exact solution formulae, a multi-dimensional Godunov538

scheme has been obtained. It has been found that, as expected, its stability region539

extends right up to the maximum allowed physical cfl number, which is twice what540

is possible with a dimensionally split scheme. Additionally, the accuracy increases541

significantly in the vicinity of a cfl number of 1. The method has been shown to542

perform well when applied to Riemann Problems, but not to allow calculations in the543



limit of low Mach numbers in general.544

Appendix A.545

Concentrating on initial data in the velocity only the solution (38) reads546

v(t,x) =2

3v0(x) + ∂r

(r

∫dΩ (v0 · n)n

)−∫

dΩ [v0 − 3(v0 · n)n](104)547

−∫ ct

0

dr1

r

∫dΩ [v0 − 3(v0 · n)n](105)548

549

In order to make the case of stationary solutions v(t,x) = v0(x) appear explicitly in550

the formula one might wonder where the remaining 13v0(x) are hiding. Recall that all551

the functions inside the integrals are to be evaluated at x + rn. One rewrites (using552 ∫dΩn⊗ n = 1

3 id from (62) or by explicit calculation)553

∫dΩ(v0 · n)n

∣∣∣∣r=ct

− 1

3v0(x) =

∫dΩ(v0 · n)n

∣∣∣∣r=ct

−∫

dΩ(v0 · n)n

∣∣∣∣r=0

(106)554

=

∫ ct

0

dr∂r

∫dΩ(v0 · n)n(107)555

556

Now one can regroup the terms. By product rule at the location indicated by an557

arrow558 ∫ ct

0

dr1

r∂r

(r↓· r∂r

∫dΩ(v0 · n)n

)(108)559

=

∫ ct

0

dr∂r

(r∂r

∫dΩ(v0 · n)n

)+

∫ ct

0

dr∂r

∫dΩ(v0 · n)n(109)560

= r∂r

∫dΩ(v0 · n)n

∣∣∣∣r=ct

+

∫ ct

0

dr∂r

∫dΩ(v0 · n)n(110)561

562

which are both terms appearing in the expression for v(t,x). Here563

limr→0

(r∂r

∫dΩ(v0 · n)n

)= 0(111)564

565

was used. Similarly,566

∫ ct

0

dr1

r∂r

(r

∫dΩ[v0 − 3(v0 · n)n]

)(112)567

=

∫ ct

0

dr∂r

∫dΩ[v0 − 3(v0 · n)n] +

∫ ct

0

dr1

r

∫dΩ[v0 − 3(v0 · n)n](113)568

569

By570

limr→0

∫dΩ[v0 − 3(v0 · n)n] = 0(114)571

572



one obtains573

v(t,x) =2

3v0(x) +

1

3v0(x)

(115)

574

+

∫ ct

0

dr1

r∂r

(r2∂r

∫dΩ(v0 · n)n

)−∫ ct

0

dr1

r∂r

(r

∫dΩ[v0 − 3(v0 · n)n]

)(116)575

= v0(x) +

∫ ct

0

dr1

r∂r

[1

r∂r

(r3

∫dΩ(v0 · n)n

)− r

∫dΩv0

](117)576

577

as claimed in (40).578

Appendix B.579

For better comparison to other schemes, here the scheme (90)–(94) is given in the580

variables prior to symmetrization, i.e. such that it is a numerical approximation to581

(11)–(12).582

un+1 = unij −

∆t

2∆x

(1

ε2[p]i±1,j −

c

ε[[u]]i± 1

2,j

)(118)583

− 1

2

∆t

∆x

c∆t

ε∆y

(− 1

2π

c

ε[[[[u]]i± 1

2]]j± 1

2− 1

4

c

ε[[v]i±1]j±1 +

1

4

1

ε2[[[p]i±1]]j± 1

2

)(119)584

vn+1 = vnij −∆t

2∆y

(1

ε2[p]i,j±1 −

c

ε[[v]]i,j± 1

2

)(120)585

− 1

2

∆t

∆y

c∆t

ε∆x

(− 1

2π

c

ε[[[[v]]i± 1

2]]j± 1

2− 1

4

c

ε[[u]i±1]j±1 +

1

4

1

ε2[[[p]]i± 1

2]j±1

)(121)586

pn+1 = pij −∆t

2∆x

(c2[u]i±1,j −

c

ε[[p]]i± 1

2,j

)− ∆t

2∆y

(c2[v]i,j±1 −

c

ε[[p]]i,j± 1

2

)(122)587

− 1

2

c∆t2

ε∆x∆y

(1

4c2[[[u]i±1]]j± 1

2+

1

4c2[[[v]]i± 1

2]j±1 − 2 · 1

2π

c

ε[[[[p]]i± 1

2]]j± 1

2

)(123)588

589

590

Acknowledgements. The authors thank Philip L. Roe and Praveen Chan-591

drasekhar for bringing the solution strategy via Helmholtz decomposition to their592

attention.593

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