Where as discussed previously we interpret solutions to this partial differential equation in the weak sense: b

Transcription

1 Consider the pure initial value problem for a homogeneous system of onservation laws with no soure terms in one spae dimension: Where as disussed previously we interpret solutions to this partial differential equation in the weak sense: b a The Riemann problem is defined as the initial value problem for this system with two valued pieewise onstant initial data. More preisely we have initial data: Hydrodynami Methods The Riemann roblem for Gas Dynamis 1

2 The Riemann problem is a fundamental tool for studying the interation between waves. It has played a entral role both in the theoretial analysis of systems of hyperboli onservation laws and in the development and implementation of pratial numerial solutions of suh systems. Basially the Riemann problem gives the miro-wave struture of the flow. One an think of the propagation of the flow as a set of small sale Riemann problems between adjaent regions followed by the interation between the waves arising from these Riemann problems. This idea was formalized in the fundamental paper of Glimm Solutions in the Large for Nonlinear Hyperboli Systems of Equation that established the first existene theorem for solutions to the initial value problem for nonlinear hyperboli systems of equations as well as numerially by Godunov A Finite Differene Method for the Numerial Computation and Disontinuous Solutions of the Equations of Fluid Dynamis whih forms the basis for many advaned numerial methods. Before proeeding further we need to review briefly the notion of hyperboliity. Hydrodynami Methods The Riemann roblem for Gas Dynamis

3 Smooth solutions of a system of onservation laws are also solutions of a orresponding quasi-linear partial differential equation. Let A = (a ij ) be the Jaobi matrix orresponding to the flux funtion f(u): ij u + fu ( ) = t x i ( ) j u Au ( ) u t x =. Then smooth solutions to also satisfy the equation + This system is said to be hyperboli if the matrix A has all real eigenvalues and a omplete set of eigenvetors: = = = 1 n i i i i i i i i ij Where n is the number of equations in the system. These eigenvetors are alled the harateristi speeds of the hyperboli system. Hydrodynami Methods The Riemann roblem for Gas Dynamis 3

4 For gas dynamis we have the Euler equations: t + v = v + t v v+ = 1 ( v + e) 1 + v( v + e) + v t = For smooth flow we an write this equation in quasi-linear form: D + Dt v = Dv + = Dt De + v = Dt Where is the material derivative of the flow. D Dt = t + v Hydrodynami Methods The Riemann roblem for Gas Dynamis 4

5 The pressure is related to the density and energy by an inomplete equation of state = ( e). If we define the Grüneisen exponent Γ and the sound speed by the formulas: Γ = e ρ = + e = Γ e+ Γ We an write the one dimensional Euler equations in quasi-linear form e ρ u ρ 1 u + Γ u Γ t e u ρ Where u is the x omponent of the veloity vetor. x u e = Hydrodynami Methods The Riemann roblem for Gas Dynamis 5

6 One an easily ompute the eigenvalues and eigenvetors of this system. We have: = u r = l = Γ/ 1 Γ = u r = Γ l = Γ/ = u+ r = l = Γ/ 1 Γ Hydrodynami Methods The Riemann roblem for Gas Dynamis 6

7 The analysis of the harateristis for gas dynamis an be onsiderably simplified by introduing the entropy S (the evaluation of whih requires the omplete equation of state) and using pressure as a fundamental thermodynami variable. By the first law of thermodynamis we have TdS = de+dv and if we write = (S) (reall V=1/ ). Then one an show that: Γ= T S = = S S 1 D + v = Dt Dv + = Dt DS = Dt + u 1 u u = Hydrodynami Methods The Riemann roblem for Gas Dynamis 7

8 In terms of pressure veloity and entropy the eigenvetors beome: 1 1 = u r = 1 l = = u r = l = 1 1 = u+ r = l = Hydrodynami Methods The Riemann roblem for Gas Dynamis 8

9 An important property of gas dynamis and other nonlinear hyperboli systems is the existene of disontinuous solutions. Indeed suh disontinuities are unavoidable and an arise spontaneously from smooth solution through the phenomenon of shok breaking. Suh solutions must be interpreted in the weak sense as desribed in the previous leture. A ritial aspet of disontinuous solutions is that the states on either side of a disontinuity are not arbitrary but must be related by a system of algebrai equations known as the Rankine-Hugoniot equations. The next series of slides will disuss the derivation of these equations and their appliations to gas dynamis. Consider a system of onservation laws u + f = t h in a domain Ω. Suppose that the flow in Ω=is smooth exept aross a smooth moving surfae S(t). We assume that S(t) divides Ω into two regular regions Ω 1 =and=ω. Let φ(xt) be a vetor of C test funtions with support in the interior of Ω. Hydrodynami Methods The Riemann roblem for Gas Dynamis 9

10 Applying the distributional form of the onservation law we have: t+ t dt dx u φ + f φ+ h φ = t Ω t t t+ t t+ t dt dx u φ + f φ+ h φ + dt dx u φ + f φ+ h φ = t Ω () t t Ω () t 1 t Using the fat that u is smooth inside both Ω 1 (t) and Ω (t) we an add the quantity (u t + =f - h) φ== inside the two integrals over these two regions separately to obtain: t+ t t+ t t dt dx ( u φ) + ( fφ) + dt dx ( u φ) + ( fφ) = t Ω () t t Ω () t 1 t Hydrodynami Methods The Riemann roblem for Gas Dynamis 1

11 Applying the divergene theorem to the spae time regions swept out by Ω i (t) we obtain: t+ t t St () Where n(xt) is the spatial unit normal to S(t) and s is the speed of the moving surfae in the diretion n and [] denotes the jump in a quantity aross the surfae S(t) in the diretion n. Sine the test funtion φ and t are arbitrary we have at eah point on the disontinuity surfae the equations: Hydrodynami Methods The Riemann roblem for Gas Dynamis 11

12 Applying the Rankine Hugoniot equations to the gas dynami equations we derive the formulas: ρ ρ s v n = ρ s v n = m 1 1 mv n= mv n 1 1 v n s + n m v v n n = ρ v n s + n m v v n n 1 1 or me e 1 1+ V V 1 =. There are two ases depending on whether the mass flux m is zero or nonzero: m m m = 1 v1 = v + n m s= v n+ V m= v n+ Vm e = e = V V 1+ V V 1 = 1 s = v n= v n. 1 Waves with m are shoks waves with m = ontat disontinuities. Hydrodynami Methods The Riemann roblem for Gas Dynamis 1

13 For perfet gas the Hugoniot equations for shoks an be further simplified. Here we have = / and: m 1 + µ = ρ 1+ µ 1 + µ ρ1 = ρ 1+ µ 1 v1 = v ± 1 + µ n 1 µ s= v n± 1 + µ 1+ µ We see that the flow state behind the shok is ompletely determined by the flow state ahead of the shok the shok normal and the pressure behind the shok. This statement is still true for general equations of state but it may not be possible to expliitly write down the solution as for a perfet gas. More generally given the state ahead of a shok and the shok normal the flow behind the shok is determined by one additional ondition. The variables ommonly used inlude the shok speed the shok Mah number (ratio of the speed of the shok with respet to a fluid at rest and the sound speed) the pressure behind the shok or the density behind the shok. e = e 1 1 = γ 1 µ γ µ 1 + µ 1 Hydrodynami Methods The Riemann roblem for Gas Dynamis 13

14 Homogeneous isotropi onservation laws with no body fores and the Euler equations for gas dynamis (with no gravity) in partiular satisfy the important property of self-similarity the equations are invariant under the transformation t t x x =>=. If u(xt) is a solution then so is u( x t). Sine the initial data for a Riemann problem is invariant under the transformation x αx we onlude that if the solution to the Riemann problem is unique then it must satisfy u(xt) = u( x t) i.e. u(xt) = u(x/t) for t >=. If we let = x/t assume the solution is smooth and insert this into the quasi-linear form of the partial differential equation we obtain: We see that is an eigenvalue of A(u)=df/du and du/d is a orresponding eigenvetor. Suh solutions are alled entered rarefation waves or entered simple waves. Alternatively if u is not smooth then it an onsist of a jump disontinuity onneting two onstant regions and the values of the solution are related by the Rankine-Hugoniot equations. Hydrodynami Methods The Riemann roblem for Gas Dynamis 14

15 Applying the simple wave equations to gas dynamis we obtain the solutions: = ± = ± 1 = 1 1 S = = = 1 1 For the left hand ase the density is found by solving the equation = ( 1 S ) given by the equation of state relation between density pressure and entropy. For a perfet gas this an be solved to obtain: = γ Hydrodynami Methods The Riemann roblem for Gas Dynamis 15

16 The shok wave urve through a given state (indexed by ) is defined as the lous of all states that an be onneted to that state by a shok wave. This urve onsists of two branhes orresponding to a positive or negative mass flux aross the wave. The side of the shok from whih fluid partiles travel into and through the shok is alled the ahead side of the shok. If the given state orresponds to the state ahead of a shok and m is positive we say the wave is forward moving for m negative we all the wave bakward moving. Thermodynamis requires that the entropy produed aross the shok be nonnegative. For most equations of state this orresponds to an inrease in pressure from the ahead to behind side of the shok. Thus only the branh of the wave urve orresponding to inreasing pressure is physial. A derease in pressure aross a wave orresponds to a rarefation wave. The wave urve defined for all is the onatenation of the shok wave urves for > and the rarefation urve for <. It an be shown that this urve is twie ontinuously differentiable. Of partiular interest is the omponent of the wave urve that relates the veloity behind the wave and the pressure behind the wave. The formula for this urve is given on the next slide. Hydrodynami Methods The Riemann roblem for Gas Dynamis 16

17 u= u ± V V V > d S < The funtion V( V ) is found by solving the Hugoniot relation + e= e + V V = ( e V) and = (ev) is the inomplete equation of state. If we define the mass flux aross a general wave m by the formula [] = m[u] we an write the wave urve as: > V V V u= u ± m V = m V < d S Hydrodynami Methods The Riemann roblem for Gas Dynamis 17

18 For a perfet gas Riemann wave urve an be omputed expliitly. γ 1 γ γ 1 γ Hydrodynami Methods The Riemann roblem for Gas Dynamis 18

19 The solution of the Riemann problem for gas dynamis onsistsof onstant regions separated by waves. Moving from left to right these are a bakward shok or rarefation wave a ontat disontinuity and a forward moving shok or rarefation wave. The Riemann data on the left will be the ahead state for the left moving wave while the Riemann data on the right will be the ahead state for the right moving wave. The Rankine-Hugoniot onditions (whih agree with the simple wave onditions for a ontat disontinuity) for the middle wave are the pressure and veloity are ontinuous aross this wave. If we let u m and m denote the ommon values of the pressure and veloity on either side of the ontat and let m l = m( l V l ) and m r = m( r V r ). We have: m = l m l m r = r + ( ) ( ) l m l l r m r r m = l l r + + l r r = + + l l r r l r m. l + r l r Hydrodynami Methods The Riemann roblem for Gas Dynamis 19

20 The set of equations on the previous slide give the mid state pressure as the solution of a nonlinear algebrai equation. One the mid state pressure is determined the mid state veloity is omputed from the orresponding formula. Finally using the mid state pressure the left and right data states and the Hugoniot or rarefation equations as appropriate we evaluate the densities on either side of the ontat disontinuity and the orresponding waves. If a wave is a shok its spae-time position is a line from the origin with slope equal to the shok speed. If it is a rarefation it orresponds to a fan the slope of eah ray of whih is equal to u ± (+ for a forward wave for a bakward wave). Thus the entire struture of the solution to the Riemann problem is determined one the mid state pressure is found. One exeptional ase ours when the wave urves do not interset for positive pressure. In this ase the solution onsists of two rarefation fans eah of whih expands into a vauum. In this ase the mid state veloity is undefined. Hydrodynami Methods The Riemann roblem for Gas Dynamis

21 In onlusion we show that a graphial representation of the Riemann wave urves is useful in understanding the struture of the solution to the Riemann problem. Given the loation of the right state the wave struture is determined by the loation of the left state with respet to the wave urve through the right state. Bakward Shok Wave urve Left Shok Right Shok Forward Shok Wave urve Left Shok Right Rarefation ( r u r ) Left Rarefation Right Shok Forward Rarefation Wave urve Left Rarefation Right Rarefation Bakward Rarefation Wave urve Hydrodynami Methods The Riemann roblem for Gas Dynamis 1

22 1. Derive the quasi-linear form of Euler s equations from the onservation form.. Derive the Hugoniot relations for gas dynamis. 3. Verify by diret alulation that the Riemann wave urve for gas dynamis is C. Hydrodynami Methods The Riemann roblem for Gas Dynamis

23 R. Rihtmyer and K. W. Morton Differene Methods for Initial-Value roblem ( nd ed.) Intersiene ublishers D. Lax Hyperboli Systems of Conservation Laws and the Mathematial Theory of Shok Waves SIAM J. Smoller Shok Waves and Reation-Diffusion Equations Springer-Verlag 198. J. Glimm Solutions in the Large for Nonlinear Hyperboli Systems of Equations Comm. ure Appl. Math Hydrodynami Methods The Riemann roblem for Gas Dynamis 3