ON THE UNIQUENESS OF SOLUTION TO GENERALIZED CHAPLYGIN GAS

The main object of the paper is finding a unique solution to Riemann problem for generalized Chaplygin gas model. That is a model of the dark energy in Universe introduced in the last decade. It permits an infinite mass concentration so one has to consider solutions containing the Dirac delta function. Although it was easy to construct solution to any Riemann problem, the usual admissibility conditions, overcompressiveness, do not exclude unwanted delta-type waves when a classical solution exists. We are using Shadow Wave approach in order to solve that uniqueness problem since they are well adopted for using Lax entropy–entropy flux conditions and there is a rich family of convex entropies.


Introduction.
A generalized Chaplygin gas appears in a number cosmology theories and it is a model for a compressible fluid with a pressure inversely proportional to a gas energy density, p = −C/ρ α , C > 0, 0 < α < 1, see [2] for the first model, and [9] for some more advanced models.It is used as a model for the dark energy in the Universe.(We will use C = 1 in the rest of the paper for simplicity.)The system consists from the mass and momentum conservation laws , where u denotes a velocity of the gas.In this paper we use the momentum variable q = ρu: The physical region for both systems is ρ > 0 and the sound speed of the system tends to zero as ρ → ∞.Note that we do not have vacuum states due to division by zero in the flux.That is, we do not loose a solution by rewriting the original only one state variable contains the Dirac delta function.Let us also mention [17] where one can find a definition for more singular non-classical objects -δ shock waves, and [10] describing something that looks like √ δ (see also [14] that contains some additional properties of such waves).The papers [5] and [13] contain some examples of delta shock formation from classical shocks.Note that all these objects can be substituted by appropriate shadow waves used in this paper.
The paper is organized as follows.Section 2 contains some basic properties of the given system.Section 3 contains the existence proof to the Riemann problem but without uniqueness.In order to gain uniqueness, in Section 4, we use convex entropy -entropy flux pair.Our intention is to employ entropy conditions in order to weed out inadmissible solutions.Actually, our aim was to prove that a simple shadow wave solution (SDW for short) to our problem is admissible for all convex entropy pairs only at the points that can not be connected by two shock solution.In general, there are two entropy conditions that SDW solution has to satisfy.The first one is connected with δ -part, and the second one with δ-part.We proved that first entropy condition is true for all convex entropy pairs we have found.But with the second condition we were only partially successful.In Section 5 we gave some partial uniqueness results.We also proved local theorem which gave us the existence of the points that can be connected with the left-hand state by two shock solution where entropy condition is not satisfied while overcompressibility condition is.Thus, we have proved that the entropy conditions are better than well-known overcompressibility condition in that case.That is the first result of that kind up to our knowledge.

Figure 1. Classical waves
A solution is given as a combination of the rarefaction waves for the points (ρ, q) above the curves R 1 and R 2 .In areas between the curves R 1 and S 2 (S 1 and R 2 , resp.) one can always find a solution in the form R 1 + S 2 (S 1 + R 2 , resp.).Bellow the curves S 1 and S 2 we have a solution consisting from two shocks.But not in the complete area (for more details see [21]): Only if (ρ 1 , q 1 ) is above the curve Γ ss = Γ ss (ρ 0 , q 0 ) : That curve is obtained as a boundary of all possible S 1 + S 2 combinations and is not included in that area -i.e.there are no classical solutions when (ρ, q) ∈ Γ ss .
(See Figure 1.) 3. Shadow waves.In this section we are looking for non-classical (singular) solution below the curve Γ ss .We are using a simple shadow wave type of solution which is defined as robustly as possible in order to improve chances of obtaining some sort of uniqueness.A big advantage of this type of solutions is that it also includes delta and singular shocks as special cases.That was one of the main reasons why we have chosen to look at solution in the form of the simple shadow wave.
Lemma 3.1.There exists a simple shadow wave (SDW for short) written in the form that solves (1, 3) if and only if Proof.Using Lemma 1 from [15] one gets the following formulas for its derivatives Here and bellow the sign "≈" denotes a limit as ε → 0 while [y] := y 1 − y 0 is the standard designation of jump in a variable y across a shock front.The support of delta function δ and its derivative δ is the line x = ct.One immediately sees that the only possibility to avoid a trivial case (when both ρ i,ε and q i,ε , i = 0, 1, are zero) is ρ i,ε , q i,ε ∼ ε −1 , i = 0, 1.So, let us denote and Riemann problem (1, 3) reduces to the system of the following equations Denote by ρ α so called Rankine-Hugoniot deficits for the first and second equation of the system, resp.One immediately gets κ 2 = cκ 1 from the second equation.The third and fourth equation determines The only possible relation between unknowns ξ i , χ i , i = 0, 1, is and it fixes the fourth equation.The first and the third equation in (4) uniquely determines a strength of SDW The variable ρ denotes the density so κ 1 > 0 (the case κ 1 = 0 corresponds to a shock).From the first equation in (4) we have and the positivity of κ 1 implies that one has to take plus sign in (5).A simple computation gives Thus, an SDW solution to (1), (3) exists if and only if i.e. a point (ρ 1 , q 1 ) has to be below the curve or above the curve Remark 1.Note that in the (simple) SDW given by (3.1) we have only used constant mean-states (ρ 0,ε , q 0,ε ), (ρ 1,ε , q 1,ε ) and a constant SDW speed curve x = ct.That is the simplest form of a SDW solution, but in the case of our Riemann problem that is enough since the initial data does not contain a delta function and initial states (ρ 0 , q 0 ), (ρ 1 , q 1 ) in the Riemann initial data are constant.Otherwise, one may use a type of SDW called weighted SDW (for more details see [15]).
The curve given by (7) coincides with (2) and is above Γ ss .Therefore, the region of the data (ρ 1 , q 1 ) situated between this curve and Γ ss corresponds exactly to S 1 + S 2 solution, meaning that a solution to Riemann problem is not unique: For (ρ 1 , q 1 ) between these curves both S1+S2 and SDW solution exists.Also, both solutions exist above the curve (8).One has to exclude SDW or S1+S2 solution.The overcompressibility condition is often used in order to gain a uniqueness of delta shock -type solutions.It means that λ i (ρ 0 , q 0 ) ≥ c ≥ λ i (ρ 1 , q 1 ) should be true for i = 1, 2.
That relation for system (1) is satisfied if Let us denote by x := q 0 ρ 1 − q 1 ρ 0 and note that (9) implies x > 0. Take ρ 1 > ρ 0 first.Note that condition where Note that the condition for SDW existence (6) means that x > x * .So, further on we will look only at x satisfying x > max{x * , z} =: x 0 .
Let us first note that equals zero when ρ 1 = ρ 0 , and its first derivative with respect to ρ 1 is negative for Let ρ 1 < ρ 0 , now.Using the same notation and arguments (with ρ 0 and ρ 1 interchanged) as above, one could see that both conditions in (9) Therefore, one sees that (ρ 1 , q 1 ) can be connected by an overcompressive SDW with (ρ 0 , q 0 ) if and only if it lies bellow the curve Remark 2. As one could see, the curve (8) lies above Γ oc and SDW solution above ( 8) is not overcompresive.If (ρ 1 , q 1 ) lies below Γ oc and above Γ ss a solution to (1, 3) is not unique (see Figure 2): One can construct both S1+S2 and the overcompressive SDW solution to that problem.Our aim is to use a possibility of using convex entropy -entropy flux pair for SDWs.That possibility was one of the major reasons of use SDWs to reconstruct non-classical solution to conservation law systems (see [15] for examples).
The solution concepts used in [20] and [21] share that property.Basically, all three concepts give solutions with the same distributional limit.The authors of these papers simply excluded unwanted delta shocks in the above area without an explanation.We will try to use Lax entropy condition.The first task will be to find as broad as possible a family of convex entropies for system (1).
4. Convex entropies.Suppose that a conservation laws system posses convex entropy -entropy flux pair (called convex entropy pair bellow) (η, Q).According SDW solution exists but is not overcompressive above that line SDW solution exists below that line to the entropy conditions from [15], a SDW solution (ρ, q) to ( 1) is admissible if It is not so hard to find one convex entropy pair.Analogously to the known energy function for other gas dynamic models, we have the following pair of functions η = 1 2 Substitution of these functions in (11) gives a different set of admissible points (ρ 1 , q 1 ) than the overcompressibility condition.But there is still a non-empty intersection of that set with {(ρ 1 , q 1 ) : there exists a S1+S2 solution connecting (ρ 0 , q 0 ) and (ρ 1 , q 1 )}.Even more, the overcompressive and entropic sets of admissible states (ρ 1 , q 1 ) are not comparable as one could see on the Figure 3.Note that a situation is different in the case of Chaplygin gas with α = 1 (see [16]), where use only of the energy η = q 2 +1 ρ as a convex entropy is enough to single out a unique solution to Riemann problem, and the overcompressibility condition gives the same one.
Let us now try to find some more convex entropies.Using the standard procedure (see [6] for example) one can find that an entropy function η satisfies where B = 3+α 2(1+α) .For l ≤ 0 a function is not convex and consequently, a function g nor η can not be convex.Fix l > 0. Then g(v From [18] one gets where I ν (x) denote modified Bessel function of the first kind, while K ν (x) denote modified Bessel function of the second kind.Using the original variables (ρ, q), we have where and λ := √ l > 0.
Proof.It is known that entropy function is convex if its Hessian matrix is positive definite.So, in order to prove that η 1 and η 2 are convex it is enough to prove that the principal minors of a Hessian matrix of η 1/2 are all positive.We use the following relations in the proof below:

Determinant of Hessian matrix is given by
Since 1 1+α > α 1+α , for α ∈ (0, 1), it is clear that D 1 is positive.On the same way as above one gets ≥ 0, Since, ∂ 2 ∂ρ 2 η 2 (ρ, q) > 0 and D 2 > 0, one concludes that η 2 is also a convex function.In the proof of non-convexity of the functions η 3 , η 4 one follows the same arguments and one also uses the following For example, determinant of the Hessian matrix of the function η 3 is D 3 is negative which means that η 3 is not a convex function.Same follows for a function η 4 .
Therefore, using the original variables (ρ, q) one can conclude that all convex η obtained by the separation of variables are linear combination of the functions η 1 and η 2 from (12) for every λ > 0. Appropriate entropy flux functions are given by Remark 3. In order to get more convex entropies one can try to separate variables on the different way.For example, we can separate variables by . As a result we get the following convex entropy function and appropriate entropy flux function given by We choose to represent results obtained by using convex entropy-entropy flux pairs (η 1 , Q 1 ), (η 2 , Q 2 ).We did not make a significant improvement with the above pair (η, Q) so we omit those results.
→ 0 as ε → 0, i = 1, 2 and since the modified Bessel functions of the second kind satisfy we have By virtue of the above relation, the first relation in (11) for the entropy pair (η 1 , Q 1 ) becomes It is now clear that that relation is true for every λ if and only if c(ξ But that relation is always true when SDW is a solution to the system as one could see above.Obviously, the same holds for the second entropy pair (η 2 , Q 2 ).
In order to prove uniqueness of solution one needs to prove that the second relation in ( 11) is always non-positive for (ρ 1 , q 1 ) lying on Γ ss (ρ 0 , q 0 ) and bellow for every λ > 0, while above it is positive at least for some λ > 0. We were not able to complete that process, and we will present partial results about that in the rest of the paper.We left that question open. 5. Partial uniqueness results.In order to prove that the second relation in (11) is satisfied in some cases we will use the following equality for modified Bessel functions of the second kind Also, the following inequalities hold for every x > 0 and 0 < α < 1 (also see Figure 4) Inequality (15) follows from relation proved in [8], since 1 1+α > 1 2 , 0 < α < 1. Inequality ( 16) follows from paper [1], where the function x → x ν e x K ν (x) is proved to be monotone decreasing on (0, ∞) for all ν < 1 2 , while lim Here, we can use this results since 1+α in order to simplify the future notation.Note that 0 < A < 1 for 0 < α < 1.
We start our analysis by looking at the second entropy inequality in (11).By a simple substitution and use of the above relations we get the left-hand side of the second relation in (11) to be in the form for the first entropy pair (η 1 , Q 1 ).There was used that ρ For points (ρ 1 , q 1 ) ∈ Γ ss (ρ 0 , q 0 ) we have E 1 λ = e 2λ q 0 ρ 0 Ê1 λ , where ) .

(17)
In the same way as above, one can determine that the left-hand side of the second relation in (11) for the second entropy pair (η 2 , Q 2 ) equals For (ρ 1 , q 1 ) lying on Γ ss (ρ 0 , q 0 ) we have
The following theorem is very important.We claim that there exist points above a curve Γ ss that satisfy the overcompressibility condition but not the entropy one.So we may avoid non-uniqueness at least at these points by using the entropy admissibility condition with or without overcompressibility.We still do not know whether there are some points where the overcompressibility is stronger condition than the entropy condition.Let us add that we did not find any numerical example for that until now.Theorem 5.1.For every α ∈ (0, 1) and every point (ρ 0 , q 0 ) there exists its neighborhood such that there exist points above the curve Γ ss where overcompressibility condition is satisfied but entropy conditions is not for λ large enough.
A simple computation gives Due to continuity of all functions used in this analysis overcompressibility condition is satisfied on Γ β in a neighborhood of ρ 0 , too.
Let as now check the second entropy condition for the first entropy pair (η 1 , Q 1 ) using the above data.
We have ).

Using the relation
from [11] and the fact that ) .Now, using relation (16) and letting λ → ∞, we have that Since the exponential function decreases to zero at infinity faster than any power of λ, the above is true if has only one root x β in the interval (0, 1) given by (note that x = 1 is one root of the equation h β (x) = 0 for any β ∈ (0, 1)).For each β ∈ (0, 1) we obtain an interval (α β , 1), α β := x 2 β such that entropy condition does not hold i.e. the function h β is positive for all α in the interval (α β , 1). Obviously, With β → 0 we have x β → 0, so the function h β (x) is positive for α ∈ (0, 1) and β small enough.
Since the function h is continuous, one can conclude the following: For any α ∈ (0, 1) there exist some β ∈ (0, 1) (sufficiently small β if α is sufficiently small) such that entropy condition, when λ → ∞, is not satisfied in the neighborhood of point ρ 0 , on the curve Γ β for the first entropy pair.This completes the proof.
Remark 4. The left-hand side of the entropy condition, for the second entropy pair (η 2 , Q 2 ), q 1 given by Γ β and ρ 1 = ρ 0 equals If we compare the entropy conditions for (η 1 , Q 1 ) and (η 2 , Q 2 ), we can see that

Ẽ2
λ,β | ρ0=ρ1 = Ẽ1 λ,β | ρ0=ρ1 holds.So, we can conclude that the second entropy condition for the second entropy pair, (η 2 , Q 2 ), ρ 1 in the neighborhood of ρ 0 and λ sufficiently large is satisfied if and only if it is satisfied for the first entropy pair.
After proving that there are cases when entropy condition is more restrictive than the overcompressibility one, we will present some results that illustrates usefulness of the entropy condition.We start with the one describing asymptotic behavior of the entropy condition as parameter λ tends to zero or infinity.
Proof.Since we have already proved that the first relation in (11) holds true for any λ > 0, we just need to prove that the second relation in (11) holds true for λ sufficiently small and large.Let as check condition for the first entropy pair (η 1 , Q 1 ) and λ → 0. One could easily check that lim λ→0 E i λ = 0, i = 1, 2 follows from ( 17) and (18).Even more, lim λ→0 E i λ = 0, i = 1, 2 holds for any q, without limiting analysis to the Γ ss curve.We want to show that Êi λ are decreasing in λ = 0. Using the formulas .
In order to prove that Ê1 λ is decreasing in λ = 0 one may use the following equalities taken from [22] , where I ν (x) denotes modified Bessel function of the first kind.Using the identity Γ(ν)Γ(1 − ν) = π sin (πν) one gets Replacing the identity (21) into the first derivative with respect to λ given above and arranging the terms in ascending powers of λ one gets the following form of the first derivative 1+α ).
Take now λ to be large enough.Using the notation from ( 17) and ( 18) one can easily conclude that Ê1 λ ≤ 0 using the following: -Each term in ( 17) and ( 18) are close to zero for λ sufficiently large.
-The terms dominate the other three terms in (17) for λ large enough.Same holds for the second entropy pair.
Remark 5. We have performed a lot of tests in order to check the validity of the above proposition for each λ > 0. It seems that Γ ss is always in the entropic region, but we did not succeed to prove it.One can look at Figures 6 and 7.It should be noted that lack of precise enough approximations for Bessel function of   15) and ( 16) are not enough to prove nonpositivity of Ê1 λ the second kind represents a serious setback in this analysis.Even though it looks like that inequalities ( 15) and ( 16) can be quite helpful, they are not enough to prove (global) non-positivity of E i λ , i = 1, 2 (for example see Figure 5).As one can see below, in some special cases those inequalities were very helpful, but only locally.Note that inequalities (15) and (16) give us only one (lower or upper) bound for Bessel functions.In order to get the other bound one can use inequality (20).Using (20) one gets the lower bound for K α 1+α (x) But, one can easily check that not even the inequality given above is enough to prove non-positivity of E i λ , i = 1, 2. Numerical experiments we have done 1 confirmed our assertion.So, one has to look for better approximations of the Bessel functions or to find an alternative way to prove non-positivity of the entropy functions.
If we use the entropy condition as the admissible one that means that we have to prove that Γ ss is entropic in order to have a solution.If the curve Γ ss is optimal, that would imply prove uniqueness.That would be the second open question we left open.
In the rest of this section we shall present some special cases when Γ ss is entropic.
Example 1.The relations in (11) are satisfied if (ρ 1 , q 1 ) is lying on Γ ss and one of the following conditions is satisfied sufficiently small.(ii) α is close enough to 1.

Since e −2λρ
→ 0, it is clear that E 1 λ ≤ 0 holds.So, the second entropy condition holds for (η 1 , Q 1 ).The same holds for the second entropy pair (η 2 , Q 2 ).(ii) Let as take α close enough to 1. Then A will be close to 1 and So, where the relation , α ∈ (0, 1) was used.
(ii) One can conclude that ρ is sufficiently small.Using relations (14), (15) and ( 16) one gets can try to avoid use of derivatives, because of there complexity and focus attention on use of some other mathematical tool.