Answers to Problem Set Number 04 for MIT (Spring 2008)

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1 Answers to Problem Set Number 04 for MIT (Spring 008) Rodolfo R. Rosales (MIT, Math. Dept., room -337, Cambridge, MA 0139). March 17, 008. Course TA: Timothy Nguyen, MIT, Dept. of Mathematics, Cambridge, MA Contents 1 Problem GaDy03: Rankine Hugoniot conditions for Isentropic Gas Dynamics Statement for GaDy03: Rankine Hugoniot cond. for Isentropic Gas Dynamics Answer for GaDy03: Rankine Hugoniot cond. for Isentropic Gas Dynamics..... Problem GaDy04: Piston problem for Isentropic Gas Dynamics. 4.1 Statement for GaDy04: Piston problem for Isentropic Gas Dynamics Answer for GaDy04: Piston problem for Isentropic Gas Dynamics Problem GaDy03: Rankine Hugoniot conditions for Isentropic Gas Dynamics. 1.1 Statement for GaDy03: Rankine Hugoniot conditions for Isentropic Gas Dynamics. The one dimensional isentropic (constant entropy) Euler equations of Gas Dynamics are given by } ρ t + (ρ u) x = 0, (ρ u) t + (ρ u (1.1) + p) x = 0, where ρ = ρ(x, t), u = u(x, t), and p = p(x, t) are the gas mass density, flow velocity, and pressure, respectively. The first equation implements the conservation of mass, and the second the conservation of momentum. These equations are complemented by an equation of state, relating the pressure to the density. This takes the form p = P(ρ), where P is a function satisfying dp dρ For example, for an ideal gas P = κρ γ, where κ > 0 and 1 < γ < are constants. > 0. (1.) Use conservation to derive the Rankine-Hugoniot jump conditions that shocks for the equations above must satisfy. In other words, consider a solution of the equations of the form } ρ = ρ L and u = u L for x < s t, (1.3) ρ = ρ R and u = u R for x > s t, 1

2 MIT, Spring 008 (Rosales). Answers to PS# 4. Rankine Hugoniot cond.... where ρ L > 0, ρ R > 0, u L, u R, and s are constants. Then find conditions that these constants have to satisfy so that mass and momentum are conserved. Hint 1: In a frame of reference moving with the shock, the fluxes of mass and momentum on each side of the shock must be equal. Hint : Alternatively: use the integral form of the conservation laws across an interval containing the shock; i.e.: a < s t < b. Remark 1.1 One may question whether it makes sense to consider shock wave solutions within the context of the constant entropy assumption. 1 However, it can be shown (though we will not do it here) that the amount of entropy produced within a shock is proportional to the cube of the shock strength for weak shocks. Hence, as long as we use the equations in situations where only weak shocks arise, it does make sense to consider shocks within the context of a constant entropy assumption. 1. Answer for GaDy03: Rankine Hugoniot conditions for Isentropic Gas Dynamics. D1. First derivation. Let U = u s be the flow velocity in the shock frame. In this frame the mass flux is given by ρ U, and the momentum flux by ρ U + p. Hence conservation takes the form ρ L U L = ρ R U R and ρ L U L + p L = ρ R U R + p R. (1.4) Equivalently s = [ρ u] [ρ] = ρ R u R ρ L u L and s = [ρ u + p] ρ R ρ L [ρ u] = ρ R u R + p R ρ L u L p L ρ R u R ρ L u L, (1.5) where the brackets indicate the jumps across the shock of the enclosed quantities. D. Second derivation. Let a and b be fixed points such that a < s t < b for t close to some (arbitrary) t 0. Then the integral form of the equation for the conservation of mass yields d dt b ρ(x, t) dx a }{{} (s t a) ρ L + (b s t) ρ R = ρ L u L ρ R u R. (1.6) From this the first equation in (1.5) follows. Similarly, the integral form of the equation for the conservation of momentum yields d dt b ρ(x, t) u(x, t) dx a }{{} (s t a) ρ L u L + (b s t) ρ R u R From this the second equation in (1.5) follows. = (ρ L u L + p L) (ρ R u R + p R). (1.7) 1 Constant entropy presumes adiabatic conditions, while changes at a shock are anything but slow. Hence shocks must produce entropy.

3 MIT, Spring 008 (Rosales). Answers to PS# 4. Rankine Hugoniot cond Remark 1. Graphical interpretation. Consider the situation where the state ahead of the shock is fixed, and ask the question: What are the states on the back that can be reached by a shock transition, as a function of the mass and momentum fluxes through the shock? We can (graphically) answer this question using equation (1.4) for the shock conditions, which we rewrite in the form ρ U = m and ρ U + p = M, (1.8) where m = ρ R U R is the mass flux through the shock, M = ρ R U R + p R is the momentum flux through the shock, and we have dropped the subscript L from the state on the left. In terms of the specific volume v = 1/ρ, these equations can be re-written as p = M m v, (1.9) where U is recovered from U = m v. This equation has the following simple graphical interpretation in the (p, v) plane: the solutions are the intersections of the curve given by the equation of state p = p(v), and the straight line (called the Rayleigh line) p = M m v. For nice gases, the equation of state p = p(v) is convex, 3 specifically: dp dv = ρ c < 0, and d p > 0, where c > 0 is the sound speed. (1.10) dv Then (1.9) has two solutions, one of them being v = v R we are interested in the other one. The second law of thermodynamics requires shocks to be compressive 4 since the entropy cannot decrease as a fluid particle crosses the shock layer. Hence it must be that v L < v R. Then the fact that p = p(v) is convex implies that its slope at v = v L is less than the slope of the secant line to p = p(v) through v L and v R which from (1.9) is precisely m. Hence ρ L c L < m = ρ L U, so that c L > U. A similar argument shows that c R < U. Thus c R < U < c L. (1.11) The shock is sub-sonic from behind (sound waves can catch up to it) and super-sonic from ahead (sound waves cannot escape from it). The right moving characteristics converge into the shock path. Assume a shock moving to the right: fluid enters the shock from the right, and exits it to the left, so that U > 0. 3 Example: polytropic ideal gases, where p = κ v γ, with κ > 0 and γ > 1 constants. 4 We will not show this here.

4 MIT, Spring 008 (Rosales). Answers to PS# 4. Piston problem... 4 Problem GaDy04: Piston problem for Isentropic Gas Dynamics..1 Statement for GaDy04: Piston problem for Isentropic Gas Dynamics. Imagine a long pipe full of air at rest, with the piston in it. At time t = 0 you start moving the piston at a constant velocity V, which is neither too small, nor too large. 5 What happens? A simple mathematical model for the situation above is as follows: we use the one dimensional isentropic Euler equations of Gas Dynamics (for a polytropic gas) ρ t + (ρ u) x = 0, (ρ u) t + (ρ u + p) x = 0, to model the air, 6 where ρ is the gas density, u is the flow velocity, p = κ ρ γ is the pressure, κ > 0 is a constant, 1 < γ < is a constant, t is time, and x is the length along the pipe, measured from the point where the piston started. The solution to these equations must then be found for t > 0 and x > V t, with the initial and boundary conditions given by Solve this problem. ρ(x, 0) = ρ 0 and u(x, 0) = 0 for x > 0. u(x, t) = V for x = V t > 0. Hint.1 The equations have two sets of characteristics. On the right moving (C + characteristics) given by dx dt = u + c c the quantity r = u + Similarly, on the left moving (C characteristics) given by dx dt = u c the quantity s = u c } } (.1) (.) is constant. (.3) is constant. (.4) Here c = dp/dρ = γ p/ρ > 0 is the sound speed, r is called the right Riemann invariant, and s is called the left Riemann invariant. 5 The neither too small means that we cannot use linearized equations. The nor too large means that we can use simplifying assumptions, such as constant entropy flow. 6 For dry air starting at one atmosphere and 15 degrees Celsius: p = p 0 (ρ/ρ 0 ) γ, where p 0 = dyn/cm, ρ 0 = g/cm 3, and γ =

5 MIT, Spring 008 (Rosales). Answers to PS# 4. Piston problem... 5 You can now reformulate the problem completely in terms of r and s as follows: dx 1. r is constant along = γ + 1 dt 4 dx. s is constant along = 3 γ dt 4 3. r = s = c 0 on x = 0, 4. r + s = V on x = V t, r + 3 γ 4 r + γ s, s, (.5) where c 0 is the sound speed corresponding to ρ 0. Then u and c follow from u = (r + s)/ and c = () (r s)/4. Since c is an increasing function of ρ, ρ follows from knowledge of c. Note that, since c > 0, a realistic physical solution requires r > s everywhere. Note also that the C characteristics that start on x > 0 move backwards through the gas particles (since their velocity u c is less than the gas velocity) and so must eventually approach the gas particles that start next to the piston at time t = 0. Hence it can be argued that Every parcel of gas, at any time, is connected with the initial data (t = 0) on x > 0 via a C characteristic. (.6) Since, along x > 0 for t = 0, s = c 0 /() is identically constant, and since s is constant along the C characteristics, we can use (.6) to argue that The left Riemann invariant s c 0 /() is constant throughout the flow. (.7) HOWEVER, there is a proviso to this conclusion. The argument that s is constant along the C characteristics depends on the solution being smooth. It will not apply across shocks. Thus (.7) holds as long as the flow develops no shocks. At any rate, you can start solving the problem by assuming that (.7) applies. Then the problem reduces to calculating the C + characteristics, along which is r is constant. These are straight lines, since along each of them both r and s are constant, as follows from (.7). But then the problem becomes very simple, and it is very similar to the light turns from red to green and from green to red type of problems in traffic flow. There are two distinct cases that you must consider: A. Case V < 0. In this case no shock waves arise. Start with the assumption that (.7) holds, and then a1. Solve for the C + characteristics that start along x > 0 on t = 0. These carry the value r = c 0 /() why? and all have the same slope (calculate it). a. Solve for the C + characteristics that start along x = V t for t > 0. These carry the value r = V + c 0 /() why? and all have the same slope (calculate it).

6 MIT, Spring 008 (Rosales). Answers to PS# 4. Piston problem... 6 a3. You will find that a1 and a leave an unfilled gap in the C + characteristic field a wedge shaped region in space-time, with its tip at the origin. An expansion fan is needed to fill this region. As long as c 0 /() = V c < V < 0, the process above will yield a complete solution for the problem. Give explicit formulas for u and c as functions of x and t. CHALLENGE QUESTIONS: CQ1. What is special about V = V c? CQ. What is the solution for V < V c? B. Case V > 0. In this case a shock wave arises. Start as in the prior case. When you reach step a3, you will find that instead of a gap in the C + characteristic field, there is region where the characteristics cross: a wedge in space time, with its tip at the origin. Thus a shock is needed to stop the C + characteristics from crossing. Thus, look for a solution where there is a shock along a path x = σ(t), into which the C + characteristics converge (where they end, so the crossing is prevented). In order to proceed further, you will need the jump conditions at the shock. These arise from enforcing conservation of mass and momentum across the shock, which yields two equations. Do not worry about the details of these two equations. The important thing is that, at least in principle (see below for why in principle ), you can think of them in the following way The shock jump conditions for a (right) moving shock with speed speed U have the form: where U = f(r L, r R, s R ) and s L = g(r L, r R, s R ), (.8) f and g are some (known) functions. r L and r R are the values of the right Riemann invariant along the C + characteristics converging into the shock one from the left, and the other from the right, respectively. The condition that these characteristics converge into the shock is equivalent to u L + c L > U > u R + c R, (.9) since the C + characteristics have speed u + c. The C characteristics CROSS the shock path, entering from the right (fluid ahead of the shock) and leaving on the left. They carry the value s R when they enter the shock, and come out on the other side with the value s L. The best way to understand what all this means is to draw a picture in space time, with the shock path, and some typical characteristics: the C + converging into the shock and the C crossing it, right to left as time increases.

7 MIT, Spring 008 (Rosales). Answers to PS# 4. Piston problem... 7 Note: why do we say that the above is in principle? The reason is that no explicit expressions for the functions f and g above exist. These functions are defined implicitly, and (to evaluate them) one must resort to numerical methods. IMPORTANT! Once a shock is introduced, the validity of (.7) must be questioned. When a C characteristic crosses a shock, the argument leading to the conclusion that along it s is constant breaks down since it depends on derivatives existing. In fact, s will have a discontinuity across the shock, so that we can ONLY argue that The left Riemann invariant is constant s c 0 ahead of the shock at x = σ(t). Now you are ready to do the V > 0 case. Proceed as follows. b1. Argue that the solution to (.1.) should be a function of x/t only. Hence: The shock speed dσ = U; should be constant, with the shock path given by x = Ut. dt The the left Riemann invariant should be constant s = s 1 immediately behind the shock, that is: along x = U t 0. Hint: Let u = ũ(x, t) and ρ = ρ(x, t) be the solution to the problem in (.1.). What problem does u = ũ(α x, α t) and ρ = ρ(α x, α t) (where α > 0 is an arbitrary constant) solve? If the solution to (.1.) is unique, what do you conclude? Use this conclusion to show that ũ and ρ are, in fact, functions of x/t only. Now you should be able to argue (do the argument!) that The left Riemann invariant is constant s s 1 behind the shock at x = σ(t), where s 1 is some constant to be determined, different from the constant value ahead of the shock s = s 0 = c 0. b. Solve the problem in the region ahead of the shock x > U t. The solution here is determined by the characteristics (both C + and C ) that start on x > 0 for t = 0. Write formulas for these characteristics. b3. The solution in the region V t < x < U t between the piston and the shock is determined as follows: -1- The C characteristics start on x > 0 at t = 0, and move backwards, carrying the value s = s 0 = c 0 for the left Riemann invariant s. When they reach the shock, they go through it, and the value of s they carry jumps to a new value s = s 1, that should satisfy (.8).

8 MIT, Spring 008 (Rosales). Answers to PS# 4. Piston problem When a C characteristic reaches the piston, the boundary condition there determines a value r = r 1 for the C + characteristic that starts there. -3- The C + characteristic that start at the piston move to the right, and carry a value r = r 1 for the right Riemann invariant. -4- When the C + characteristics starting at the piston reach the shock, they end there. However, through (.8), and the values carried by the other characteristics, they determine the shock speed U, and the value s 1 with which the C characteristics leave the shock (see item -1-). Of course, you do not know the values of U, s 1, and r 1 mentioned above. However, you should be able to write equations (involving the functions f and g in (.8)) that can be used to determine them. WRITE THESE EQUATIONS. You will not be able to solve the equations because you do not have expressions for f and g. This is fine; however, note that it can be shown that the equations have a unique solution that satisfies (.9). Describe what the solution to the problem looks like, in terms of these values. Hint/IMPORTANT: in order to fully understand what the stuff in items -1- through -4- above means, draw a picture in space-time showing typical characteristics and the shock path. C. Finally: what happens in the case V = 0?. Answer for GaDy04: Piston problem for Isentropic Gas Dynamics. Following the process outlined in the hint, we consider the following cases: A. Case V < 0. In this case we expect to find no shock waves, and (.7) to hold. a1. C + characteristics starting on x = ζ 0 at t = 0. These cover the region c 0 t x, for t 0, where they yield... u = 0 and ρ = ρ 0. Proof: along each of these characteristics r is a constant, equal to the value it has at t = 0, where u = 0 and ρ = ρ 0. Thus r = c 0 /(). Because (.7) holds, it turns out that along each of these characteristics both u 0 and ρ ρ 0 are constants. Hence these characteristics have constant slope c 0 and are given by x = c 0 t + ζ. a. C + characteristics starting on x = V τ at t = τ > 0.

9 MIT, Spring 008 (Rosales). Answers to PS# 4. Piston problem... 9 ( These cover the region V t x c 0 + γ + 1 ) V t, for t 0, where they yield 7... u = V and c = c 0 + Proof: along each of these characteristics r is a constant, equal to the value it has at t = τ, where r + s = V, as follows from (.5). Hence, using (.7), it follows that r = V + c 0 /(). Again, as in a1, it then turns out that along each of these characteristics both u and ρ are constants, which we can recover from the values of r and s. Namely: u = V and c = c 0 + slope and are given by x = ( c 0 + γ + 1 V V. V. Hence these characteristics have constant ) (t τ) + V τ. Remark.1 The calculation in this item makes sense only if V < ( c 0 + γ + 1 ) V, namely c 0 = V c < V. We will assume that this holds, for now, and deal with the case V V c later. a3. Region ( c 0 + γ + 1 ) V t < x < c 0 t. Here we have... u = γ + 1 ( x t c 0) and c = γ + 1 c 0 + x γ + 1 t. Proof: This region is an unfilled gap in the C + characteristic field calculated in items a1 and a. An expansion fan is needed to fill this wedge shaped in space-time, with its tip at the origin. In this expansion fan the C + characteristics must all go through the origin, and thus are given by x = (u + c) t, where u and c are constant along each C + characteristic by same argument used in items a1 and a. Hence u + c = x/t, which combined with (.7) yields the result above. c 0 As long as... = V c < V < 0, the process above yields a complete solution for the problem, which is described by the three regions in... a1, a, and a3. Next we concentrate on the CHALLENGE QUESTIONS: Ch1 What is special about V = V c? When V c < V a region forms (in front of the piston) where the gas has accelerated to the piston velocity, and moves together with it the region described in a. Notice that this region grows as more and more gas particles learn of the presence of a moving 7 Note that, since c = c(ρ) is an increasing function of ρ, knowledge of c is equivalent to knowledge of ρ.

10 MIT, Spring 008 (Rosales). Answers to PS# 4. Piston problem piston: the right edge of this region moves (away from the piston) at the C + characteristic ( speed: u + c = V + c 0 + ) V = c 0 + γ + 1 V. As the piston (negative) speed is increased, the C + characteristics find increasingly harder to move away (towards the right) from the piston, and the region described in a becomes smaller. For V = V c this region disappears, as the C + characteristic speed becomes equal to the piston speed. The information about a piston moving at speed V can no longer propagate into the gas. Let us now inspect what happens in the expansion fan region, described in a3. The gas in this region does not directly know of the presence of a piston, since no characteristic starting at the piston reaches this region. What the gas here does it to simply expand into the space left by the block of gas following the piston, and moving away to the left (from the gas at rest on the right) at speed V. In this region the sound speed (hence the density) decreases steadily from a maximum at the right edge c = c 0 to a minimum c = (V V c ) at the left edge. In particular, note that the gas density vanishes at the left edge of the expansion fan region when V = V c. From this we conclude that V = V c is the maximum speed at which the gas can expand, hence if the piston moves faster than V c, the gas cannot follow it. Ch What is the solution for V < V c = c 0? From the prior arguments in Ch1, it should be clear that the solution in this case is For t 0 and x c 0 t... u = 0 and c = c 0. For t 0 and V c t x c 0 t... u = ( x 0) γ + 1 t c = ( x c) γ + 1 t V + V c ( x 0) t c and c = c 0 + γ + 1 = γ + 1 ( x t V c). For t 0 and V t x V c t... vacuum. B. Case V > 0. In this case we expect to need a shock wave, hence we should be careful about the validity of (.7). If we start as in the previous case, and assume that (.7) applies everywhere, we find that the regions described in a1 and a (instead of leaving a gap in the C + characteristic field) overlap, ( and give a multiple valued solution, for c 0 t x c 0 + γ + 1 ) V t and t > 0. Hence a shock is needed to stop the C + characteristics from crossing the shock being along some path x = σ(t), into which the C + characteristics converge, and where they end, so the crossing is prevented.

11 MIT, Spring 008 (Rosales). Answers to PS# 4. Piston problem Of course, once we introduce a shock, we must re-examine the validity of (.7). When the C characteristics cross the shock, the argument leading to the conclusion that along them s is constant breaks down. 8 In fact, s will have a discontinuity across the shock, so that we can now only argue that The left Riemann invariant is constant s c 0 ahead of the shock at x = σ(t). (.10) The big question that we need to answer now is: What happens with the left Riemann invariant behind the shock, in the region x < σ(t)? This is our immediate next task. b1. The solution to (.1.) is a function of x/t only. Hence the shock speed dσ dt = U; is constant, with the shock path given by x = Ut. Furthermore, the values of the variables immediately behind the shock 9 should be constants (independent of time). In particular, the left Riemann invariant has a constant value s = s 1 behind the shock (i.e. the jump in s across the shock is constant). But then the C characteristics propagate this constant value to the whole region behind the shock, and we conclude that The left Riemann invariant is constant s s 1 behind the shock at x = σ(t), (.11) where s 1 is some constant to be determined, different from the constant value ahead of the shock s = s 0 = c 0. Proof: Let u = ũ(x, t) and ρ = ρ(x, t) be the solution to the problem in (.1.). It should be clear that, for any α > 0 constant, u = ũ(α x, α t) and ρ = ρ(α x, α t) is also a solution to the problem in (.1.). Hence, assuming that the solution to (.1.) is unique, we conclude that ũ(x, t) = ũ(α x, α t) and ρ(x, t) = ρ(α x, α t) for any α > 0. Now, take α = 1/t 0, for some t 0 > 0, and evaluate the equalities above at t = t 0. This yields ũ(x, t 0 ) = ũ(x/t 0, 1) and ρ(x, t 0 ) = ρ(x/t 0, 1). Since t 0 is arbitrary, this shows that ũ and ρ are functions of x/t only. Q.E.D. b. Solution in the region ahead of the shock x > U t. The solution here is determined in the same fashion as in item a1 and it is given by.... u = 0 and ρ = ρ 0, with the characteristics given by x = ζ ± c 0 t, with ζ > 0. b3. Solution in the region V t < x < U t between the piston and the shock. The solution here is determined as follows 8 As it depends on taking derivatives. 9 That is, along the line x = U t 0.

12 MIT, Spring 008 (Rosales). Answers to PS# 4. Piston problem The C characteristics start on x > 0 at t = 0, and move backwards, carrying the constant value s = s 0 = c 0 for the left Riemann invariant s. When they reach the shock, they go through it, and the value of s they carry jumps to a new constant value 10 s = s 1, that should satisfy (.8). -- When a C characteristic reaches the piston, the boundary condition there determines a value r = r 1 = V s 1 for the C + characteristic that starts there note that r 1 is a constant, the same for all C + characteristics. -3- The C + characteristics starting at the piston move to the right (till they reach the shock), and carry the constant value r = r 1 for the right Riemann invariant. It follows that In the region between the piston and the shock, the solution is constant, with s s 1 and r r 1 = V s 1. In particular u V. (.1) -4- When the C + characteristics starting at the piston reach the shock, they end there. Since the shock conditions (.8) must apply there, we end up with the equations U = f(r 1, r 0, s 0 ) and s 1 = g(r 1, r 0, s 0 ), (.13) where s 0 = c 0, r 0 = c 0, is the (constant) value of the right Riemann ahead of the shock, and r 1 = V s 1. These are two equations that can be used to determine the two unknowns: the shock speed U, and the value of the left Riemann invariant behind the shock s 1. Note that the condition (.9) takes the form V + c 1 > U > c 0, (.14) where c 1 is the shock speed in the (constant) state behind the shock. This finishes the solution for the case V > 0. Remark. The solution in the case V > 0 consists of two constant states separated by a shock. We can then use the results of the problem GaDy03: Rankine Hugoniot conditions for Isentropic Gas Dynamics to describe the solution to the problem, as follows u = 0, ρ = ρ 0, and p = p 0 = κ ρ γ 0 for x > U t, u = V, ρ = ρ 1, and p = p 1 = κ ρ γ 1 for x < U t, where ρ 1 and U follow from the two equations 10 As follows from (.11). } (.15) U = ρ 1 V = ρ 1 V + p 1 p 0. (.16) ρ 1 ρ 0 ρ 1 V

13 MIT, Spring 008 (Rosales). Answers to PS# 4. Piston problem We now show that these equations have exactly one acceptable solution. Proof: It must be ρ 1 > ρ 0, since this yields U > V > 0, while ρ 1 < ρ 0 yields U < 0, and ρ 1 = ρ 0 yields infinities. Furthermore, the second equality in (.16) is equivalent to (p 1 p 0 ) (ρ 1 ρ 0 ) = ρ 1 ρ 0 V. (.17) }{{} G(ρ 1 ) Clearly G(ρ 0 ) = G (ρ 0 ) = 0, and G (ρ 1 ) > c 0 > 0 for ρ 1 > ρ 0. Hence the curve y = G(ρ 1 ) has exactly one intersection (on ρ 1 > ρ 0 ) with the straight line y = ρ 1 ρ 0 V. Q.E.D. Finally, note that from the first and last term in (.16), it follows that p 1 = p 0 + ρ 1 V (U V ). Use now the first two terms in (.16) to replace ρ 1 V by (ρ 1 ρ 0 ) U. Thus p 1 = p 0 + (ρ 1 ρ 0 ) U (U V ). Interpret this equation as the intersection of the curve y = p(ρ) on the left, with the straight line y = p 0 + (ρ ρ 0 ) U (U V ) on the right assuming U as given. There are two intersections: ρ = ρ 0 and ρ = ρ 1 > ρ 0. Because p = p(ρ) is convex, it immediately follows that c 0 = d p d ρ (ρ 0) < U (U V ) < c 1 = d p d ρ (ρ 1). Thus U > U (U V ) > c 0 and (U V ) < U (U V ) < c 1. Hence (.14) follows. C. Finally: what happens in the case V = 0? Nothing happens! The solution is u = 0 and ρ = ρ 0 for all x, t > 0 THE END.