UNIFORM FORMULA FOR THE RIEMANN SOLUTIONS OF A SCALAR COMBUSTION MODEL∗
2016-11-24ChunlongYANGGaoweiCAOXiaozhouYANGWuhanInstituteofPhysicsandMathematicsChineseAcademyofSciencesWuhan430071Chinamailgwcaowipmaccn
Chunlong YANGGaowei CAOXiaozhou YANGWuhan Institute of Physics and Mathematics,Chinese Academy of Sciences,Wuhan 430071,China E-mail:gwcao@wipm.ac.cn
UNIFORM FORMULA FOR THE RIEMANN SOLUTIONS OF A SCALAR COMBUSTION MODEL∗
In this paper,we construct a uniform formula for the Riemann solutions of the simplified Chapman-Jouguet model.Firstly,we define a new functional,and then,we obtain that the Riemann solutions can be expressed by the maximum value point of this functional, while Riemann solutions may contain some of strong detonation waves,Chapman-Jouguet detonation waves and contact discontinuities.Finally,Chapman-Jouguet deflagration waves are also discussed.
Riemann solution;Chapman-Jouguet model
2010 MR Subject Classification35L65
1 Introduction
Combustion phenomenon are investigated by the conservation laws and the internal mechanism in reaction zones.The two kinds of combustion waves,detonation waves and deflagration waves,were observed through experiments by French chemists in 1881.Chapman and Jouguet established a model of hyperbolic equations for combustion theory.The research history began from the classical Euler equations for gas dynamics and assumed that the reaction rate was infinitely large.Zeldovich,von Neumann,and Doring established another model which assumed the reaction rate was finite.However,these two models are difficult and complex.Then Madja[4]developed a simplified model,which was a 2×2 system,where a lumped variable was introduced to represent density,velocity or temperature.This model decribles the phenomena of detonation waves,deflagration waves and exhibited instability.From Majda model, one obtained simplified ZND model and simplified Chapman-Jouguet model.
Consider the following simplified Chapman-Jouguet model for combustion

where u is a lumped variable,representing density,velocity or temperature,z∈[o,1]is a variable representing the fraction of unburnt gas,q>o are constants representing the binding energy and Uiis the ignition temperature.We require f′′≥α>o.
A global existence theorem for this model was proved by Ying and Teng in[9].To study its properties,Liu and Zhang[2]discussed its Riemann problem,and introduced an admissibility criterion through the study of elementary waves.Then the uniqueness is ensured by requiring the solution to have minimum instability,or equivalently,by replacing combustion waves by nonreacting waves whenever possible.For the Riemann problem,the combustion waves contain one of the following wavessuch as strong detonation waves,Chapman-Jouguet detonation waves, Chapman-Jouguet deflagration waves or contact discontinuities.These different waves have different expressions and properties.For more relevant works on model(1.1)–(1.3),see[1,3,5–8,1o].
In this paper,we construct a functional F,then the Riemann solutions can be expressed by the the maximum value point of F under some assumptions.In this way,we obtian a uniform formula of the Riemann solution which contains either wave of the following strong detonation waves(denoted by SDT),Chapman-Jouguet detonation waves(denoted by CJDT)or contact discontinuities(denoted by J).The case of Chapman-Jouguet deflagration waves(denoted by CJDF)is also discussed(as usual,shock waves and rarefaction waves are denoted by S and R, respectively).
We consider the Riemann problem of model(1.1)–(1.3)with initial data

where u−>Ui>u+and u−>u++q,or

where u−
For any point(x,t)∈ℝ×ℝ+in the x−t plane,there exists a=a(x,t)∈ℝ uniquely such that f′(a)=x/t.Define a functional of u:

Then,for any(x,t)∈ℝ×ℝ+,let

We state the main results of this paper as the following theorem.
Theorem 1.1For the Riemann initial data(1.4)or(1.5),u(x,t)defined by(1.7)combining with z(x,t),determined by u(x,t)according to(1.2),is the solution to the problem (1.1)–(1.3).
If f′(Ui) 2.1For the Riemann Initial Data(1.4) We consider the case of Riemann initial data(1.4),first. For any given(x,t)∈ℝ×ℝ+,with x/t=f′(a).If uo;by(1.4),we have uo(x−tf′(u))=u+and zo(x−tf′(u))=1,then ¯z(u)=zo(x).From(1.6),we have,if u Similarly,if u>a,we have f′(u)>f′(a)=x/t,that is,x−tf′(u) so the maximum value point of F(·;x,t)only can be u−,u+or a. Notice that F(a−o;x,t)=−qzo(x)x/t≤o,and F(a+o;x,t)=o,we can calculate the maximum value point of F(·;x,t)as follows. (I)If a≤u+,then u(x,t)=u−(see Fig.1). Fig.1(I)a≤u+ (II)If u+a and u+ Fig.2(II)u+ Since u−>u++q≥u++qzo(x),from(2.4)(2.5),we have F(u−;x,t)≥F(u+;x,t)holds if and only if (III)If a≥u−,then u(x,t)=u+for F(u+;x,t)>o,and u(x,t)=a for F(u+;x,t)≤o (see Fig.3).Since u−≤a,from(2.1),we have Fig.3(III)u+ Since a≥u−>u++q≥u++qzo(x),from(2.7),we have F(u+;x,t)>o holds if and only if We can classify the solutions into four cases as follows. Case 1f(u++q) Fig.4f(u++q) In this case,it has Recall that x/t=f′(a),and f′is strictly increasing. (2.1)can be rewritten as:if x/t≤f′(u+),then u(x,t)=u−. For(2.1),if x So(2.1)can be rewritten as:ifx/t For(2.1),(2.8)always holds for a≥u−.So(2.1)can be rewritten as:if x/t≥f′(u−), then u(x,t)=u+. Therefore,from(1.2)and formula(1.7),it has(see Fig.5) Fig.5SDT Case 2f(u++q) Fig.6f(u++q) In this case,it has Recall that x/t=f′(a),and f′is strictly increasing. (2.1)can be rewritten as:if x/t≤f′(u+),then u(x,t)=u−. For(2.1),if x and if x>o,then f′(a)>o and(2.6)is equivalent to which is impossible. So(2.1)can be rewritten as:then u(x,t)=u−;if For(2.1),if x which always holds since a≥u−>u+;if x>o,then(2.8)is equivalent to which also holds for a≥u−>u+.So(2.1)can be rewritten as:if x/t≥f′(u−),then u(x,t)=u+. Therefore,from(1.2)and formula(1.7),it has(see Fig.7) Fig.7S and J Case 3f(u++q)≥f(u+),f(u−)>f(uc)(see Fig.8) Here,uc≥u++q satisfies following equality Fig.8Left:f′(u+)<0;Right:f′(u+)>0 In this case,it has Recall that x/t=f′(a),and f′is strictly increasing. (2.1)can be rewritten as:if x/t≤f′(u+),then u(x,t)=u−. For(2.1),if x So(2.1)can be rewritten as:if f′(u+) For(2.1),(2.8)always holds for a≥u−.So(2.1)can be rewritten as:if x/t≥f′(u−), then u(x,t)=u+. Therefore,from(1.2)and formula(1.7),it has(see Fig.9) Fig.9SDT Case 4f(u++q)≥f(u+),f(u−)≤f(uc)(see Fig.1o) Fig.10Left:f′(u+)<0;Right:f′(u+)>0 Here,ucis determined by(2.18). In this case,it has Recall that x/t=f′(a),and f′is strictly increasing. (2.1)can be rewritten as:if x/t≤f′(u+),then u(x,t)=u−. For(2.1),if x which always holds for a For(2.1),(2.8)holds if and only if a>uc.So(2.1)can be rewritten as:if f′(u−)≤x/t≤f′(uc),then u(x,t)=a;if x/t>f′(uc),then u(x,t)=u+. Therefore,from(1.2)and formula(1.7),it has(see Fig.11) Fig.11CJDT 2.2For the Riemann Initial Data(1.5) Next,we consider the Riemann initial data(1.5).Similar to(1.4),for any given(x,t)∈ℝ×ℝ+,with x/t=f′(a),we have Thus,for any given(x,t)∈ℝ×ℝ+,the derivative of F(u;x,t)about u is so the maximum value point of F(·;x,t)only can be u−,u+or a. Notice that u− (I)If a≤u−,then u(x,t)=u−(see Fig.12). Fig.12a≤u_2 Proof of the Theorem





































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