E = where γ > 1 is a constant spesific to the gas. For air, γ 1.4. Solving for p, we get. 2 ρv2 + (γ 1)E t

Transcription

1 . The Euler equations The Euler equations are often used as a simplification of the Navier-Stokes equations as a model of the flow of a gas. In one space dimension these represent the conservation of mass, momentum and energy, and read ) v E t + v v + p ve + p) Here denotes the density of the gas, v the velocity, p the pressure and E the energy. To close this system, i.e., to reduce the number of unknowns to the number of equations, one can add a constitutive law relating these. Such laws are often called equations of state and are deduced from thermodynamics. For a so called ideal polytropic gas the equation of state takes the form E = p γ + v, where γ > is a constant spesific to the gas. For air, γ.4. Solving for p, we get ) p = γ ) E γ =. v = γ ) E γ q, where the momentum q = v. Inserting this in the Euler equations v v + γ 3 v + γ )E E v =. γe γ t v) In the conserved variables, q and E, this system of conservation laws read q 3) q 3 γ ) q + + γ ) E =. ) E q 3 Set u = Then the Jacobian dfu) reads t γ Eq γ q and fu) = E dfu) = Introducing the enthalpy as the Jacobian can be rewritten as dfu) = γ Eq γ γ 3 3 γ ) q γ Eq q + γ ) E. ) γ q 3 ) q 3 γ) q γ. + γ ) q3 γ E 3 3γ ) q γ q H = E + p = γ E ) γ q = γ ) p + γ v, ) v 3 γ) v γ ). v 3 vh H γ ) v γv γ 3 To find its eigenvalues, we compute the determinant det λi dfu)) = λ [ λ 3 γ)v) λ γv) + γ ) γ ) v H )]

2 + 3 γ v λ γv) + γ ) vh γ ) v 3 = λ [ λ 3vλ + γ3 γ)v + γ ) v + γ )H ] + 3 γ v λ γ + )v3 + γ )Hv [ = λ λ 3vλ + v + ] γ + ) v γ )H γ + )v3 + γ )vh [ = λ λ v)λ v) + ] γ + )v γ )H γ + )v3 + γ )vh [ = λ v) λλ v) + ] γ + )v γ )H [ = λ v) λ v) v )] γ + )v + γ )H [ )] γ = λ v) λ v) H v ). This can be simplified further by introducing the sound speed c, by c = γp. We then calculate H v = γ E γ )v v Therefore Thus the eigenvalues of the Jacobian are = γ E γv ) E = γ v = γ E v ) = γ p γ. detλi dfu)) = λ v) [ λ v) c ]. 4) λ u) = v c, λ u) = v, λ 3 u) = v + c. As for the corresponding eigenvectors, we write these as r i =, y i, z i ), and we see that y i = λ i, and z i = λ i ) γ γ 3)v + λ i γ 3)v. For i = we find that z = v ) γ γ 3)v + vγ 3)v + c cv γ 3)cv ) γ Recall that this is in, q, E) coordinates. = v + c γ cv

3 For i = 3 we similarly calculate ) γp = v + cv γ ) = H cv. z 3 = H + cv, and for i = it is straightforward to see that z = v /. Summing up we have the following eigenvalues and eigenvectors λ u) = v c, r u) = v c, H cv 5) λ u) = v, r u) = v, v λ 3 u) = v + c, r 3 u) = v + c. H + cv At this point it is convenient to introduce the concept of an i Riemann invariant. An i Riemann invariant is a function R = R, q, E) such that R is constant along the integral curves of r i. In other words, an i Riemann invariant satisfies Ru) r i ) =. The usefulness of this is that if we can find for each of the three eigenvectors, two Riemann invariants Ru) and Ru), then we can possibly solve the equations R, q, E) = R l, q l, E l ), R, q, E) = Rl, q l, E l ) to obtain a formula for the rarefaction waves. This is the equivalent to finding an implicit solution of the ordinary differential equation, u = ru), defining the rarefaction curves. It turns out that we have the following Riemann invariants { S, i =, Riemann invariants: v + c γ {, v, i =, Riemann invariants: p, { S, i = 3, Riemann invariants: v c γ, where we have introduced the entropy S S = log ) p γ. Verify these statements! Now we can try to obtain solution formulas for the rarefaction curves, for i =, this curve is given by ) γ p = p l, v = v l + c ) ) γ )/ l. l γ This curve is parameterized by. We must check which half of the curve to use, this will be the part where λ = v c is increasing. On the curve we have v) c) = v l + c ) ) γ )/ ) / l γp) γ l l 3

4 4 = v l + c l γ = v l + c l γ = v l + c l γ ) ) γ )/ ) / γpl l ) ) γ )/ ) γ )/ c l l l γ + l ) ) γ )/. l l ) γ )/ Since γ >, we see that v) c) is decreasing in, and for the -rarefaction wave we must use < l. Since p) is increasing in, this also means that we use the part where p < p l. Therefore we can use p as a parameter in the curve for v and write the -rarefaction curve as v p) = v l + c l γ ) ) γ )/γ) p, p p l. General theory tells us that at least for p close to p l ) this curve can be continued smoothly as a -shock curve. To find the rarefaction curve of the third family, we adopt the viewpoint that u r is fied, and we wish to find u as a function of u r. In the same way as for v this leads to the formula v 3 p) = v r + c r γ p l ) ) γ )/γ) p, p p r. To find how the density varies along the rarefaction curves we can use that the entropy s is constant, leading to ) /γ p =. l p l Now we turn to the computation of the Hugoniot loci. We view the left state u l as fied, and try to find the right state u, recall the notation u = u u l. The Rankine Hugoniot relations for ) are 6) p r s = v s v = v + p s E = ve + p), where s no longer denotes the entropy, but the speed of the discontinuity. Now we introduce new variables by w = v s and m = w. Then the first equation in 6) reads s s l = w + s l w l s l, which implies that m =. Similarly the second equation reads sw + s sw l s l = w + s) l w l + s) + p s m + s = w + w + s l w l l w l s l + p s = w + p + s. Hence mw + p =. Finally the third equation in 6) reads which implies se se l = Ew + Es + pw + ps E l w l E l s p l w l p l s E = E ) l m + pw p l w l + s p l E = E l + p l p ) l m sm w l

5 E + p = m sw c = m γ + w + s) sw c = m γ + w Hence the Rankine Hugoniot conditions are equivalent to 7) m = mw + p = c m γ + w =. We immediately find one solution by setting m =, which implies p = and v =. This is the contact discontinuity. Hence we assume that m to find the other Hugoniot loci. Now we introduce auiliary parameters Using these we have that 8) Then the third equation in 7) reads which can be rearranged as so that 9) π = p p l, and z = l. c c l = π z c l γ + w l = c l and w w l = z. c l π γ z + w l z π ) ) = wl γ z z, wl c l ) = zz π) γ z. Net recall that p = c /γ. Using this, the second equation in 7) reads or which again can be rearranged as z c γ + w = lc l γ ) c z γ + w = c l γ + w l + lw l ) c l π γz + w l z = c l γ + w l. Dividing by c l, we can solve for w l/c l ), viz. ) wl ) = zπ ) γ z. Equating ) and 9) and solving for z yields ) z = βπ + π + β, c l 5

6 6 where Using this epression for z in 9) we get wl c l β = γ + γ. ) = πβ+ π+β γ πβ+ π+β π ) πβ+) π+β) = πβ + ) π ) γ π ) β ) = πβ + γ β. Note that γ > implies β > so that this is always well defined. Since w l = v l s, we can use this to get an epression for the shock speed, βπ + ) s = v l c l γ β, where we use for the first family and + for the third. Net, using 8), v s v l s = z which can be used to epress v as a function of π. βπ + v = v l c l γ β ) ± π + β πβ + c πβ + l γ β ) ) β )π ) = v l c l γ β πβ + ) ) πβ + = v l c l γγ ) π πβ + ) /, where we take the sign for the first family and the + sign for the third. To see how the density varies along the Hugoniot loci, we use that = l z, or 3) = l πβ + π + β, which holds for both the first and the third family. Recall that for the first family we obtained a rarefaction solution if p p l, hence we have a shock solution if p p l. This means that the whole solution curve for waves of the first family is given by ) ) γ )/γ) γ p p l, p p l, 4) v p) = v l + c l ) ) / p γγ ) p l + β p p l, p pl. To find the density along this solution curve we have the formula ) /γ p p l, p pl, 5) p) = l +β p p l β+, p p p l. p l Since the second family is linearly degenerate, we can use the whole integral curve of r. Using the Riemann invariants, this is given simply as 6) v = v l, p = p l.

7 7 For the third family, we take the same point of view as for the shallow water equations; we keep u r fied and look for states u such that the Riemann problem { u <, u, ) = u r >, is solved by a wave shock or rarefaction) of the third family. By much the same calculations as for the first family we end up with ) ) γ )/γ) γ p p r, p p r rarefaction part), 7) v 3 p) = v r c r ) ) / p γγ ) p r + β p p r, p pr shock part). Regarding the density along this curve, it will change according to ) /γ p p r, p pr, 8) 3 p) = r +β p ), p p r. pr β+ p pr Now for any l, the curve v p) is a strictly decreasing function of p for p in the range [, ) taking values in the set, v l + c l /γ )]. Similarly, for any r v p) is a strictly increasing function of p, taking values in the set [v r c r /γ ), ). It follows that these curves will intersect in one point p m, v m ) if v r c r γ v l + c l γ. If this does not hold, then v does not intersect v 3 and we have a solution with vacuum. However if they intersect, then we can find the solution of the Riemann problem by finding p m, v m ).... Sod s shock tube problem. We consider an initial value problem similar to the dam break problem for shallow water. The initial velocity is everywhere zero, but the pressure to the left is higher than the pressure on the right. Specifically we set { <, 9) p, ) = v, ) =,, ) =., We have used γ =.4. In Figure we show the solution to this Riemann problem in the p, v) plane and in the, t) plane. We see that the solution consists of a leftward moving rarefaction wave of the first family, followed by a contact discontinuity and a shock wave of the third family. In Figure we show the 8 Solution in the p, v)plane Solution in the, t) plane v p).8.7 v 4 3 p m,v m ) v p) t p r,v r ) p l,v l ) p 3 Figure. The solution of the Riemann problem 9).

8 8 pressure, velocity, density and the Mach number as functions of /t. The Mach number is defined to be v /c, so that if this is larger than, the flow is called supersonic. The solution found here is actually supersonic between the contact discontinuity and the shock wave. Pressure.4 Velocity. p 8 6 v Density Mach number Figure. Pressure, velocity, density and the Mach number for the solution of 9)... The Euler equations and entropy. We shall show that the physical entropy is in fact also a mathematical entropy for the Euler equations, in the sense that ) S) t + vs), weakly for any weak solution u =, q, E) which is the limit of solutions to the viscous approimation. To this end, it is convenient to introduce the internal specific energy, defined by e = E ) v. Then the Euler equations read ) e + )) v + t t + v) = v) t + v + p ) ) = v + ev + pv =.

9 9 For classical solutions, this is equivalent to the nonconservative form t + v + v = ) v t + vv + p = Verify this! We have that e t + ve + p v =. ) p S = log γ ) γ )e = log γ = γ ) log) loge) logγ ). Thus we see that S = γ > and S e = e <. These inequalities are general, and thermodynamical mumbo-jumbo implies that they hold for any equation of state, not only for polytropic gases. For classical solutions we can compute Therefore S t = S t + S e e t = γ v + v ) + ve + p ) e v = γ ) p ) v γ ) e e ) v e = vs. S t + vs = for smooth solutions to the Euler equations. This states that the entropy of a particle of the gas remains constant as the particle is transported with the velocity v. Furthermore, S) t = t S + S t = v) S vs = vs). Thus for smooth solutions the specific entropy ηu) = Su) is conserved 3) S) t + vs) =. Of course, combining this with ) and viewing the entropy as an independent unknown, we have four equations for three unknowns, so we cannot automatically epect to have a solution. Sometimes one considers models where the energy is not conserved but the entropy is, so called isentropic flow. In models of isentropic flow the third equation in ) is replaced by the conservation of entropy 3). To show that ) holds for viscous limits, we must show that the map u ηu) = u)su), eu)) is conve. We have that η is conve if its Hessian d η is a positive definite matri. For the moment we use the convention that all vectors are column vectors, and for a vector a, a T denotes its transpose. Then we compute d η = d u)su)) = S) T + S ) T + d S.

10 Now we view Su) as Su), eu)) and compute d Su), eu)) = S ) T + S e e ) T + e) T ) + S ee e e) T + S e d e. If we use this in the previous equation, we end up with where C is given by d ηu) = S + S ) ) T + S e e) T + e ) T ) + S ee e e) T S e C, ) C = d e + e) T + e ) T. Trivially we have that =,, ) T. Furthermore so we have E q + q eu) = E, ) T = e + T v, v, ). e = 3, q, The Hessian of e is given by E d 3 q 3 q 4 3 e = q 3 = e v v v. Net, Then e) T + e ) T = e + v v ) + e + v v ) = e + v v v. Now introduce the matri D by C = e v v v e + v v v = v v v. D = v v + e v. We have that D is invertible, and thus d η is positive definite if and only if Dd ηd T is positive definite. Now Dd ηu)d T = S + S ) D D ) T + S e D D e) T + D e D ) T ) + S ee D e D e) T S e DCD T. We compute D =,

11 D e =, DCD T =, and using this Dd ηu)d T = S + S ) + S e + S ee S e S + S S e = S e S e S ee = e. e γ Hence A has three positive eigenvalues, and is positive definite, therefore also d η is positive definite, and η is conve..5 Entropy 5 Specific entropy S η Figure 3. The entropy and specific entropy for the solution of the Riemann problem 9). In Figure 3 we show the entropy and the specific entropy for the solution of Riemann problem 9). The entropy decreases as the shock and the contact discontinuity passes, while it is constant across the rarefaction wave. Analogously to the shallow water equations, we can also check whether ) holds for the solution of the Riemann problem. We know that this will hold if and only if s S + vs. Using the epression giving the shock speed, ), we calculate s S + vs = S s + v ) + l s S + v l S ) βπ + = ± l c l γ β S,

12 where we use the + sign for the first family and the sign for the second. Hence the entropy will decrease if and only if S < for the first family, and S > for the third family. Note in passing that for the contact discontinuity, s = v, and thus s S + vs = v S + v S =. Therefore, as epected, entropy is conserved across a contact discontinuity. We consider shocks of the first family, and view S as a function of π = p/p l. Recall that for these shocks π >. S = S S l p = log To check if hπ) < = h) we differentiate l ) log ) p = γ logz) logπ) ) βπ + = γ log logπ) π + β =: hπ). h β π) = γ π + β)βπ + ) π = γβ )π π + β)βπ + ) ) ππ + β)βπ + ) β = γ )πβ ) π ) ππ + β)βπ + ) β = ππ + β)βπ + ) π π ) β = ππ + β)βπ + ) π + ) <. Thus S is monotonously decreasing along the Hugoniot locus of the first family. We also see that ) holds only if p p l for waves of the first family. For shocks of the third family, an identical computation shows that ) holds only if p p l. Eercise. Show that the La inequalities v l c l > s > v p) c p), if and only if p p l, where v is given by 4), c by 5) and s is the speed of the shock, given by ) using the sign. Similarly we have v 3 p) + c 3 p) > s > v r + c r, where v 3, c 3 and s are given by 7), 8) and ) with the + sign) respectively.