Non-linear Scalar Equations

Transcription

1 Non-linear Scalar Equations Professor Dr. E F Toro Laboratory of Applied Mathematics University of Trento, Italy eleuterio.toro@unitn.it August 24, / 44

2 Overview Here we study nonlinear scalar first-order partial differential equations in one space dimension Non-linear hyperbolic equations develop discontinuities, called shocks, in finite time, even if the initial condition is smooth The equations are formulated in integral form, to account for an extended set of solutions that includes those with discontinuities. The enlarged set of solutions is too large, as uniqueness is lost To resolve this difficulty it is necessary to introduce a selection criterion to rule out the spurious solutions. One selection criterion is the Lax entropy condition 2 / 44

3 Definitions and examples Consider the first-order PDE for the unknown function q(x, t) t q + x f(q) = 0. (1) This equation is called a conservation law q is the conserved variable f(q) is the flux function or physical flux, a prescribed function of q The equation is said to be written in differential, conservative form One may express (1) in quasi-linear form as t q + λ(q) x q = 0, λ(q) = d dq f(q) f (q). (2) λ(q) is called characteristic speed 3 / 44

4 Equations of the type (1) may be characterised by the behaviour of the flux f(q) and its derivative, the characteristic speed λ(q) = f (q). There are three cases: Convex flux: λ(q) is a monotone increasing function of q, that is d dq λ(q) = λ (q) = f (q) > 0, q. (3) Concave flux: λ(q) is a monotone decreasing function of q, that is d dq λ(q) = λ (q) = f (q) < 0, q. (4) Non-convex, non-concave flux: λ(q) vanishes for some q, that is d dq λ(q) = λ (q) = f (q) = 0, for some q. (5) 4 / 44

5 Example: The Inviscid Burgers equation t q + x f(q) = 0, f(q) = 1 2 q2, λ(q) = f (q) = q, λ (q) = f (q) = 1 > 0, q. (6) The flux is convex The monotone increasing behaviour of λ(q) means that larger values of q propagate faster than smaller values of q This leads to wave distortion and shock formation We note that the true Burgers equation is viscous, namely t q + x f(q) = α (2) x q, f(q) = 1 2 q2 where α is a viscosity (or diffusion) coefficient. 5 / 44

6 Example: A traffic flow equation t q + x f(q) = 0, f(q) = u max (1 q/q max )q, λ(q) = f (q) = u max (1 2q/q max ), λ (q) = f (q) = 2u max /q max < 0, q. (7) u max 0 and q max > 0 are two constants, with 0 < q q max The flux is concave Larger values of q will propagate more slowly than smaller values of q, the opposite behaviour to that of Burgers equation 6 / 44

7 Example: The Cubic-flux Equation Another example of a conservation law (1) is the cubic-flux equation t q + x f(q) = 0, f(q) = q 3, λ(q) = f (q) = 3q 2, λ (q) = f (q) = 6q < 0 if q < 0, = 0 if q = 0, (8) > 0 if q > 0. This equation is nonconvex/nonconcave for values of q in the set of real numbers and λ (q) = 0 for q = 0. Solutions of this type of equations exhibit a peculiar behaviour and are not studied here. 7 / 44

8 Solution along characteristics Consider the Initial-Value Problem (or Cauchy problem) PDE: } t q + x f(q) = 0, IC: q(x, 0) = h(x). We look for solutions along characteristic curves x = x(t) defined by } dx ODE: = λ(q(x, t)), dt IC: x(0) = x 0. (9) (10) Consider the total derivative of q along one of these curves x(t) dq dt = q dt t dt + q dx x dt = tq + λ(q) x q = 0. (11) Then q(x, t) is constant along the characteristic satisfying IVP (10) Therefore λ(q) is also constant along the characteristic Hence characteristic are straight lines, as for linear advection case 8 / 44

9 Then q(x, t) is found by evaluating the initial data at x(0) = x 0 : the foot of the characteristic, namely q(x, t) = h(x 0 ). (12) The characteristic speed λ(q) may then be evaluated at x 0 so that the solution curves of IVP (10) are Fig. 1 depicts the situation. x = x 0 + λ(h(x 0 ))t. (13) t x = x 0 + λ(h(x 0 ))t x 0 Fig. 1. Characteristic curve x(t) = x 0 + λ(h(x 0 )) t emanating from x 0 : foot of the characteristic. x 9 / 44

10 Relations (12) and (13) give the analytical solution of IVP (9), implicitly The point x 0 depends on (x, t), that is x 0 = x 0 (x, t) Given a point (x, t) the solution is found by evaluating h(x) at x 0 By substituting x 0 from (13) into (12) we obtain q(x, t) = h(x 0 ) = h(x λ(h(x 0 ))t). (14) This, implicit, solution is identical in form to that of the linear advection equation 10 / 44

11 Next we verify that relations (12) and (13) actually define the solution. From (12) we first obtain the t and x partial derivatives of q(x, t), that is t q = h (x 0 ) x 0 t, xq = h (x 0 ) x 0 x. (15) Then we find the t and x partial derivatives of x 0 = x 0 (x, t) from (13). x t = x 0 t + tλ (h)h (x 0 ) x 0 t + λ(h(x 0)) (16) But x and t are independent variables, so x t = 0. Analogously, by differentiating (13) we respect to x we obtain 1 = x 0 x + tλ (h)h (x 0 ) x 0 x. (17) After rearranging equations (16) and (17) we obtain λ(h(x 0 )) + [1 + λ (h(x 0 )h (x 0 )t] x 0 t [1 + λ (h(x 0 ))h (x 0 )t] x 0 x = 0, = 1. (18) 11 / 44

12 From (18) we have x 0 t = λ(h(x 0 )) 1 + λ (h(x 0 )h (x 0 )t, x 0 x = λ (h(x 0 ))h (x 0 )t. (19) By substituting the partial derivatives (19) into (15) we verify that the partial derivatives t q and x q satisfy the quasi-linear form of the PDE in (2). Thus the function (14) is solution of the quasi-linear equation in (2) So solutions to non-linear equations may also be obtained by following characteristics The solution is implicit, see (14), and its computation will generally require an iterative procedure 12 / 44

13 Remarks on the solution along characteristics The denominators in (19) may vanish, rendering the differential form of the equation invalid The method breaks down when two characteristic cross t Double-valued solution here ( ) h x (1) 0 ( ( )) x = x (2) 0 + λ h x (2) 0 t ( ) h x (2) 0 x = x (1) 0 + λ ( ( h x (1) 0 )) t x (1) 0 x (2) 0 Fig. 2. Characteristics from x (1) 0 and x (2) h(x (1) 0 ) and h(x(2) 0 0 carry different initial values ), leading to multi-valued solutions. x 13 / 44

14 Rarefaction wave solution from the Riemann problem Consider the Riemann problem for a general conservation law t q + x f(q) = 0 {, ql if x < 0, q(x, 0) = h(x) = q R if x > 0. (20) We look for similarity solution q(x, t) = g(x/t), with g(ψ) smooth. Then t q = x t 2 g (x/t), x q = 1 t g (x/t). (21) Substitution of these partial derivatives into the PDE t q + λ(q) x q gives g (x/t)[ x t t λ(g(x/t))] = 0, leading to the relation λ(g(x/t)) = x t. (22) 14 / 44

15 Rarefaction wave solution for Burgers equation Specialising the Riemann problem (20) to Burgers equation, for which f(q) = 1 2 q2, λ(q) = f (q) = q. (23) Assuming q L q R (24) we obtain the following rarefaction wave solution q L if x/t < q L, (x, t) R 0, q(x, t) = x/t if q L x/t q R, (x, t) R 1, (25) q R if x/t > q R, (x, t) R / 44

16 q(x, t k ) t = t 1 = 0 t = t 2 q R (c) t = t 3 q L x = 0 x t Tail: x = 0 + λ(q L )t Head: x = 0 + λ(q R )t (b) R 0 R 1 R 2 x = 0 x q(x, 0) q R (a) q L Fig. 4. Rarefaction solution from Riemann problem for Burgers equation. (a): initial condition q(x, 0), (b): picture of characteristics dividing the x-t space into three regions, (c): solution profiles at various times. 16 / 44 x = 0 x

17 Discontinuous solutions: formation of shocks For linear equations with constant coefficients, discontinuities in the solution are present if present at the initial time. Singularities propagate along characteristics For non-linear equations discontinuities may arise from smooth initial conditions, in finite time For smooth solutions of non-linear problems one may still use the method of characteristics to find the solution via the implicit formula (14). Beware however, that characteristics may intersect in finite time. We now study the formation of a shock through the following IVP PDE : t q + x f(q) = 0, f(q) = 1 2 q2, q L = 1 if x < 0, IC : q(x, 0) = h(x) = q M (x) = 1 x if 0 x 1, q R = 0 if x > 1. (26) 17 / 44

18 t (b) x = x q(x, 0) (a) x x = Fig. 5. Example (26). (a): Initial condition at time t = 0, (b): Picture of characteristics. 18 / 44

19 Characteristics emanating from the interval [0, 1] cross at time t = t b = 1. See Fig. 5. For times t, with t < t b, the method of characteristics gives the solution via the implicit formula (14). Given the point (x, t ), the task is to find the solution q(x, t ) at that point. Using (14) q(x, t ) = h(x q(x, t )t ), Recalling that for Burgers equation λ(q) = q and defining by q = q(x, t ) the above equation becomes q = h(x q t ). If 0 x q t 1, then from the IC in (26) q = h(x q t ) = 1 (x q t ) which yields q = 1 x 1 t. 19 / 44

20 From (26), if x q t < 0 then q(x, t ) = 1 and if x q t > 1 the solution is q(x, t ) = 0. Thus the general analytical solution at a point (x, t) for times t < 1 is q L = 1 if x < t, q(x, t) = q M (x) = 1 x 1 t if t x 1, q R = 0 if x > 1. (27) At time t = 1 characteristics cross, a shock wave forms and the method of characteristics is no longer applicable to obtain admissible solutions We shall nevertheless attempt to evolve the solution for times greater than unity, still using the method of characteristics 20 / 44

21 Shock formation: analytical example Fig. 6 shows five solution profiles at times: t = 0 (initial condition), t = 1/2, t = 1, t = 3/2 and t = 2. These profiles have been obtained by evolving the five points on the initial profile denoted as P 1, P 2, P 3, P 4, P 5. q(x, t k ) t = 0 t = 1 2 t = 1 t = 3 2 t = 2 P P 2 P 3 P 4 A 1 A 2 Fitted shock x = P x 2 4 Fig. 6. Analytical solution of the initial-value problem (26) using the method of characteristics. See Table / 44

22 Each of the points P k has an associated characteristic speed given by the initial condition at the appropriate position, recalling again that λ(q) = q. In Table 1, second column, we give these speeds. For example P 1 moves with speed 1 and P 5 moves with speed zero (stationary). The remaining columns of Table 1 give the position of the various points at various times. The reader is encouraged to verify these numbers. Point Speed t = 0 t = 1/2 t = 1 t = 3/2 t = 2 P /2 1 3/2 2 P 2 3/4 1/4 5/8 1 11/8 7/4 P 3 1/2 1/2 3/4 1 5/4 3/2 P 4 1/4 3/4 7/8 1 9/8 5/4 P Table 1. Time variation of the spatial position of the five points on the initial profile at time t = 0, Fig. 5(a). See also Fig / 44

23 Fig. 6 shows profiles at times: t = 0, t = 1/2, t = 1, t = 3/2 and t = 2 Note steepening of wave at time t = 1/2, with respect to IC at t = 0 At time t = 1 a shock forms. Vertical segment (shock) accumulates all initial values 0 q(x, 0) 1. The multivalued segment is regarded as a singularity, with no value assigned to the discontinuity For t = 3/2 we note that the solution profile has folded over becoming triple-valued in the interval 1 < x < 3/2 For example at x = 5/4 the solution takes on the three values q(5/4, 3/2) = 0, q(5/4, 3/2) = 1/2 and q(5/4, 3/2) = 1 At time t = 2 the triple-valued interval has become larger Multivalued solutions are not admitted in the classical theory. Instead, a shock wave is fitted However, there could ambiguity as to where the shock is to be placed, for t > 1 Based on conservation, the equal area rule is imposed (Whitham, 1974). For example, in Fig. 6 A 1 = A 2 gives a shock solution with the discontinuity positioned at x = 3/2 at t = / 44

24 Shock formation: numerical example 1 q(x, t) 0.5 Initial condition Shock Shock x Fig. 7. Shock wave formation from smooth initial condition at time t = 0. Burgers equation solved numerically with the first-order Godunov method on a very fine mesh. The phenomenon of shock formation in non-linear equations calls for the extension of the definition of solution To this end the equations are reformulated in terms of integral relations that no longer require continuity of the solution 24 / 44

25 Integral forms of the equation Consider the general case written in differential conservative form t q(x, t) + x f(q(x, t)) = s(q(x, t)). (28) This equation includes a source term and is thus called a balance law. If s(q(x, t)) = 0 then the equation is a conservation law. We study integral forms, to accommodate discontinuous solutions We shall also derive a condition to be satisfied at discontinuities We consider a control volume V in the x-t plane, Fig. 8 V = [x L, x R ] [t 1, t 2 ], (29) t t2 V t1 xl xr x Fig. 8. Control volume V in x-t space. 25 / 44

26 We integrate equation (28) in space and time in the control volume V xr t2 x L t 1 [ t q(x, t) + x f(q(x, t))] dxdt = xr t2 x L t 1 s(q(x, t)) dxdt. (30) On rearranging the space and time integrals we obtain xr [ t2 ] t2 [ xr ] t q(x, t)dt dx = x f(q(x, t))dx dt x L t 1 t 1 x L xr t2 + s(q(x, t))dxdt. x L t 1 (31) Exact space-time integration gives the integral form of the balance law (28) xr x L q(x, t 2 )dx = xr x L [ t2 t2 ] q(x, t 1 )dx f(q(x R, t))dt f(q(x L, t))dt t 1 t 1 xr t2 + s(q(x, t))dxdt. x L t 1 (32) 26 / 44

27 In the absence of the source term, the integral form states that the amount of q(x, t) in the interval [x L, x R ] at time t = t 2 is equal to the amount of q(x, t) in the interval [x L, x R ] at time t = t 1 plus a difference of time integrals of the fluxes at the extreme points In the presence of a source term this statement is modified appropriately It is also convenient to obtain an averaged version of (32), namely 1 xr q(x, t 2 )dx = x x L 1 xr q(x, t 1 )dx x x L t [ 1 t2 f(q(x R, t))dt 1 t2 ] f(q(x L, t))dt x t t 1 t t 1 + t xr t2 s(q(x, t))dxdt. x t x L t 1 Finite volume methods are constructed from this integral equation. (33) 27 / 44

28 Finite volume formula The integral expression (33) can be written as a finite volume formula used in numerical methods, namely where q new = q old t x [f right f left ] + t s vol (34) q new = 1 xr x x L q(x, t 2 )dx q old = 1 xr x x L q(x, t 1 )dx f right = 1 t f left = 1 t t2 t 1 t2 t 1 f(q(x R, t))dt f(q(x L, t))dt (35) 1 xr t2 s vol = x t x L t 1 s(q(x, t))dxdt 28 / 44

29 An alternative version of the integral form of equation (28) is d dt xr x L q(x, t)dx = f(q(x L, t)) f(q(x R, t)) + xr x L s(q(x, t))dx. (36) This relation says that, in the absence of the source term, the time-rate of change of the total amount of q(x, t) in the interval [x L, x R ] is purely due to the difference of fluxes at the extreme points of the interval [x L, x R ] Integrating in time this equation, reproduces again equation (32) x L For most of this chapter we shall use the homogeneous version (no source) of (32), that is xr t2 t2 t 1 q(x, t 2 )dx = q(x, t 1 )dx+ f(q(x L, t))dt f(q(x R, t))dt. xr x L t 1 (37) 29 / 44

30 Generalised solutions and Rankine-Hugoniot conditions Definition: A generalised (or weak) solution of the conservation law (28) is a function q(x, t) that satisfies the integral form (32), or (36). Weak solutions admit discontinuities (shocks), which satisfy the Rankine-Hugoniot jump condition. Proposition: Rankine-Hugoniot Condition. A discontinuity of a weak solution of the conservation law (28) satisfies the Rankine-Hugoniot jump condition across it, namely f(q(s R, t)) f(q(s L, t)) = [q(s R, t) q(s R, t)] s, (38) where q(s L, t) and q(s R, t) are limiting values from left and right of the discontinuity; f(q(s R, t)) and f(q(s L, t)) are the corresponding flux values and s is the speed of the discontinuity. 30 / 44

31 Proof: Consider the configuration shown in Fig. 9 in the x-t plane, where the line S = S(t) represents a discontinuity of q(x, t). Select two points x L and x R such that x L < S(t) < x R. Assume that q(x, t) and f(q(x, t)) and their derivatives are continuous in x L x < S(t) and in S(t) < x x R. Define q(s L, t) = lim x S(t) q(x, t), q(s R, t) = lim x S(t)+ q(x, t). t S(t) ˆt sl sr xl S(ˆt) xr x Fig. 9. Illustration of the Rankine-Hugoniot condition across a shock. 31 / 44

32 Enforcing the, homogeneous, conservation law in integral form (36) on [x L, x R ] leads to f(q(x L, t)) f(q(x R, t)) = d dt Using Leibniz s formula d dα ξ2 (α) ξ 1 (α) one obtains g(ξ, α) dξ = ξ2 (α) ξ 1 (α) S(t) x L q(x, t) dx + d dt xr S(t) q(x, t) dx. (39) α g(ξ, α) dξ + g(ξ 2, α) dξ 2 dα g(ξ 1, α) dξ 1 dα (40) f(q(x L, t)) f(q(x R, t)) = [q(s L, t) q(s R, t)] s S(t) xr + t q(x, t) dx + t q(x, t) dx. x L S(t) (41) 32 / 44

33 s = d S(t): shock speed (42) dt As the partial derivative t q(x, t) is bounded the relevant integrals in (41) vanish identically as S(t) is approached from left and right and we obtain f(q(s R, t)) f(q(s L, t)) = [q(s R, t) q(s L, t)] s (43) and the claimed result is proved. Defining the jumps f = f(q(s R, t)) f(q(s L, t)) ; q = q(s R, t) q(s L, t) (44) we can relate the speed s of the discontinuity to the jumps as f = s q. (45) This relation at the shock is called the Rankine-Hugoniot condition. For the scalar case considered here one can solve for the speed s as s = f q. (46) 33 / 44

34 Summarising: In order to admit discontinuous solutions one needs to formulate the equations in integral form One may formulate the complete problem in terms of PDEs in differential form for smooth parts and the Rankine-Hugoniot condition across discontinuities Example: Burgers s equation. Assume a shock wave of speed s with states q L and q R. The Rankine-Hugoniot condition gives f(q R ) f(q L ) = 1 2 q2 R 1 2 q2 L = s(q R q L ), from which the shock speed is given by s = 1 2 (q L + q R ). (47) This is a very special case. The shock speed is a simple arithmetic average of the characteristic speeds either side of the shock. 34 / 44

35 Non-uniqueness. Example The enlarged set of solutions of the integral formulation includes smooth (classical) and discontinuous solutions. However, Now the set is too large, it contains spurious, non-physical solutions This requires an admissibility criterion to discard unphysical shocks To illustrate the question of non-uniqueness we consider the example: PDE : t q + x f(q) = 0, f(q) = 1 2 q2, { ql = 0 if x < 0, IC : q(x, 0) = h(x) = q R = 1 if x > 0. (48) 35 / 44

36 Solution 1: rarefaction wave One solution of the problem is the rarefaction wave (smooth) q L = 0 if x/t < 0, q(x, t) = x/t if 0 x/t 1, q R = 1 if x/t > 1. (49) Fig. 10 illustrates solution and the corresponding picture of characteristics. Tail t Rarefaction fan { }} { x = 0 + 1t (Head) x = 0 Fig. 10. Rarefaction solution to problem (48). x 36 / 44

37 Solution 2: shock wave Another, discontinuous, solution (shock) is given as { 0 if x/t < s = 1/2, q(x, t) = 1 if x/t > s = 1/2. (50) t Shock solution x t = 1 2 x t = 1 x = 0 Fig. 11. Shock solution to problem (48). x 37 / 44

38 We now verify that (50) is a solution, that is, it satisfies the integral form xr x xr x L t2 t 1 t2 t L q(x, t 2 )dx = q(x, t 1 )dx+ f(q(x L, t))dt 1 f(q(x R, t))dt. (51) Consider the control volume V = [x L, x R ] [t 1, t 2 ] shown in Fig. 12. Then xr x L q(x, t 2 )dx = xr x L q(x, t 1 )dx = t2 t 1 x2 x L q(x, t 2 )dx + x1 x L q(x, t 1 )dx + f(q(x L, t))dt t2 xr x 2 q(x, t 2 )dx = 1 (x R x 2 ), xr x 1 q(x, t 1 )dx = 1 (x R x 1 ), t 1 f(q(x R, t))dt = 1 2 (t 2 t 1 ). Enforcing the conservation law (51) gives the shock speed x t = 1 2. See Fig. 12. Hence (50) is a weak solution of the equation. 38 / 44

39 t x t = 1 2 t 2 t 1 x L 0 x 1 x 2 x R Fig. 12. Control volume on which to verify that the integral form of the conservation law is satisfied. x 39 / 44

40 Admissible shocks: the Lax entropy condition The proposed solution (50) is not accepted as a physical solution. Rarefaction shocks are excluded. Admissible discontinuities are those arising from compression. This compressibility condition is ensured by the Lax entropy condition. λ(q L ) > s > λ(q R ). (52) s: shock speed, λ(q L ) and λ(q R ) are characteristic speeds. Characteristics run into the shock, compressed by the characteristics. t dx dt = λ(ql) dx dt = S ql dx dt = λ(qr) qr 0 Fig. 13. Picture of characteristics for an entropy-satisfying shock. x 40 / 44

41 The Riemann problem for Burgers s equation PDE : t q + x f(q) = 0, f(q) = 1 2 q2, { ql if x < 0, IC : q(x, 0) = q R if x > 0. q L if x st < 0 q(x, t) = q R if x st > 0 if q L > q R, s = 1 2 (q L + q R ) q L x q(x, t) = t q R if x t q L if q L < x t < q R if x t q R if q L q R. (53) (54) 41 / 44

42 t dx dt = λ(ql) q(x, t) = x t {}}{ dx dt = λ(qr) (b) ql qr ql qr x = 0 x t dx dt = s = 1 (ql + qr) 2 ql qr (a) ql > qr x = 0 Fig. 14. Solution of the Riemann problem for the Burgers equation. Frame (a): shock wave if q L > q R. Frame (b): rarefaction wave if q L q R. x 42 / 44

43 Summary and conclusions We have studied some mathematical aspects of nonlinear scalar first-order partial differential equations in one space dimension We have formulated the equations in integral form so that the resulting family of solutions include those with discontinuities The enlarged set of solutions is too large, uniqueness is lost, for which it is necessary to introduce a selection criterion to rule out the spurious solutions One selection criterion is the Lax entropy condition 43 / 44

44 Exercises for hyperbolic equations. The scalar case 44 / 44

45 Problem 1. Consider the following initial-value problem for the Burgers equation PDE: t q + x f(q) = 0, f(q) = 1 2 q2, 1 if x < 0, IC: q(x, 0) = h(x) = 1 if x > 0. (55) 1 Draw the picture of characteristics associated with the initial condition in the x-t plane. 2 Find the entropy-satisfying solution q(x, t) and draw the corresponding picture of characteristics in the x-t plane. 3 Verify that the above solution satisfies the integral form of the conservation law for an arbitrary quadrilateral control volume. 4 Draw solution profiles at times t 0 = 0, t 1 = 1 and t 2 = 2. Problem 2. Consider the initial value problem (55) for the Burgers equation 44 / 44

46 1 Show that the function 1 if x < 0, q(x, t) = 1 if x > 0 (56) is a solution (shock) of the integral form of the conservation law. 2 Find the shock speed of the above solution. 3 By drawing the picture characteristics in the x-t plane show that the above solution is entropy violating. Problem 3. Consider the Riemann problem for the Burgers equation PDE: t q + x f(q) = 0, f(q) = 1 2 q2, ˆq if x < 0, IC: q(x, 0) = h(x) = ˆq if x > 0, (57) where ˆq > / 44

47 1 Show that the function ˆq if x < 0, q(x, t) = ˆq if x > 0 (58) is a weak solution of the problem. Hint: select an arbitrary control volume in the x-t plane and show that the integral form of the conservation law is satisfied. 2 Verify that the shock solution has speed zero (stationary shock) by applying the Rankine-Hugoniot condition. 3 Plot the solution on the x-t plane, including the picture of characteristics on either side of the discontinuity. 4 Verify that the solution satisfies the Lax entropy condition. Problem 4. Consider the following initial-value problem for the Burgers 44 / 44

48 equation PDE: t q + x f(q) = 0, f(q) = 1 2 q2, 0 if x < 0, IC: q(x, 0) = h(x) = 1 if 0 x 1/2, 0 if x > 1/2. (59) 1 Draw the picture of characteristics and the structure of the solution in the x-t plane. Hint: note that the solution initially contains two distinct waves and there is a critical time ˆt when the left wave catches up with the right wave. 2 Calculate ˆt. 3 Plot the solution profile at time ˆt/2 and at ˆt. 4 What happens after ˆt? Represent the wave structure and the picture of characteristics in the x-t plane up to times much longer than ˆt. Problem 5. Consider the following initial-value problem for the traffic flow 44 / 44

49 equation PDE: t q + x f(q) = 0, f(q) = f(q) = u max (1 q/q max )q, q L if x < 0, IC: q(x, 0) = h(x) = q R if x > 0. (60) 1 Find the characteristic speed λ(q). 2 Assume the initial data q L, q R to be connected by a shock wave. Apply the Rankine-Hugoniot conditions and find an expression for the shock speed. 3 Find conditions on the initial data q L, q R for the shock to be entropy satisfying. Compare your conclusions with the Burgers equation case. 4 What are the conditions on the initial data q L, q R so as to generate a rarefation (entropy satisfying) solution. Problem 6. Consider the inhomogeneous equation t q(x, t) + x f(q(x, t)) = s(q(x, t)). (61) 44 / 44

50 By imitating the procedure to obtain the Rankine-Hugoniot conditions for the homogneous case attempt to obtain the condition for the case that includes a source term s(q). Problem 7: weak solutions and classical solutions. Starting from the integral form d dt xr x L q(x, t)dx = f(q(x L, t)) f(q(x R, t)) + xr x L s(q(x, t))dx (62) and using the fundamental theorem of calculus show that a smooth weak solution also satisfies the differential form of the conservation law. 44 / 44