Communications in Mathematical Physics Manuscript-Nr.(will be inserted by hand later)Scalar Conservation Laws with DiscontinuousFlux Function: I. The Viscous Pro�le ConditionS. DiehlDepartment of Mathematics, Lund Institute of Technology, P.O. Box 118, S-221 00 Lund,Sweden, E-mail: diehl@maths.lth.seReceived: / Accepted:Abstract. The equation @u@t + @@x �H(x)f(u) + �1�H(x)�g(u)� = 0, where His Heaviside's step function, appears for example in continuous sedimentation ofsolid particles in a liquid, in two-phase ow, in tra�c-ow analysis and in ionetching. The discontinuity of the ux function at x = 0 causes a discontinuity of asolution, which is not uniquely determined by the initial data. The equation canbe written as a triangular 2� 2 non-strictly hyperbolic system. This augmenta-tion is non-unique and a natural de�nition is given by means of viscous pro�les.By a viscous pro�le we mean a stationary solution of ut + (F �)x = "uxx, whereF � is a smooth approximation of the discontinuous ux, i.e., H is smoothed. Interms of the 2� 2 system, the discontinuity at x = 0 is either a regular Lax, anunder- or overcompressive, a marginal under- or overcompressive or a degener-ate shock wave. In some cases, depending on f and g, there is a unique viscouspro�le (e.g. undercompressive and regular Lax waves) and in some cases thereare in�nitely many (e.g. overcompressive waves). The main purpose of the paperis to show the equivalence between a previously introduced uniqueness conditionfor the discontinuity of the solution at x = 0 and the viscous pro�le condition.1. IntroductionThe scalar conservation law with discontinuous ux function@u(x; t)@t + @@x�F 0�u(x; t); x�� = 0; where F 0(u; x) = (f(u); x > 0g(u); x < 0 (1.1)arises in several applications, for example in continuous sedimentation of solidparticles in a liquid, see Diehl [3] and Chancelier et al. [1], in two-phase ow inporous media, see Gimse and Risebro [6], in tra�c-ow analysis, see Mochon [11],

2 S. Diehland in ion etching in the fabrication of semiconductor devices, see Ross [13]. Inthese applications (except perhaps in tra�c ow) the ux functions f and gare non-convex. The Cauchy problem for a more general equation than (1.1),including a point source s(t) at x = 0, has been analysed by the author in [2].Generally, a solution of (1.1) contains a discontinuity along the t-axis and curvesof discontinuity that go into and emanate from it. For given initial data u(x; 0)there is usually a whole set of pairs of boundary values along x = 0 that canappear in functions u(x; t) satisfying (1.1). In [2] was introduced Condition �,which implies that only one pair of this set is used. Under some regularity as-sumptions, existence and uniqueness results are obtained in [2]. Condition � isa generalization of the classical entropy condition by Oleinik [12] and it is, in[2], justi�ed by studying an extension of Godunov's [7] numerical method. InSection 2 we review some notation and results from [2].In Subsection 3.1 we de�ne a viscous pro�le (corresponding to the discon-tinuity at x = 0 of (1.1)) as a stationary solution of ut + (F �)x = "uxx, whereF �(u; �) (with � > 0) is a smooth approximation of F 0(u; �).Equation (1.1) can be augmented to a 2�2 triangular non-strictly hyperbolicsystem, with ux function independent of x, by adding a scalar conservation lawtaking care of the discontinuity of F 0(u; �), cf. Gimse and Risebro [5]. The uxfunction of this added scalar equation is not uniquely determined. Requiring thatthe viscous pro�le of the 2� 2 system should be the same as the one describedin the preceding paragraph, an explicit formula for the added ux function interms of the smoothing of F 0(u; �) is given in Subsection 3.2. In this context,the Riemann problem is also treated by Gimse and Risebro [5] and, for n � ntriangular systems, by Holden and H�egh-Krohn [8].Depending on f and g, there are three qualitatively di�erent cases of shockwave and viscous wave. In Subsection 3.3 we relate this natural division to theover-, undercompressive waves etc. of the 2� 2 system described above.Section 4 is devoted to showing that Condition � is equivalent to the viscouspro�le condition in the sense that the concentration to the left (u�) and to theright (u+) of the t-axis obtained by Condition � are the only ones such that thereexists a viscous pro�le between u� and u+. This \equivalence" can be seen asa generalization of the \equivalence" between Oleinik's entropy condition andthe viscous pro�le condition for the scalar equation ut + f(u)x = 0. As is wellknown, there exist shock waves, satisfying Oleinik's entropy condition, which areunstable when including di�usion, see Example 2. In the same way, there existshocks along x = 0 in the case f 6= g, satisfying Condition �, which are unstablewhen including di�usion, see Examples 3 and 4.We could have let � = 0 (and " > 0) when examining the uniqueness condi-tions, see Subsection 4.4. However, in the applications referred to above � > 0may provide better models. Compare with that " > 0 usually provides bettermodels than " = 0, that is, there is usually some viscosity or di�usion. For ex-ample in continuous sedimentation, the discontinuous ux function appears dueto a feed inlet to a sedimentation tank. Taking � = 0 corresponds to a pointsource and this is clearly not the case in reality, where the inlet on one hand hasa positive cross-sectional area, on the other causes mixing near x = 0. Further-more, we want to present the qualitatively di�erent viscous pro�les that appearwhen � > 0, since some stability results of these are presented in the subsequentpaper [4]. It is interesting to note that some viscous pro�les are non-monotone.

Scalar Conservation Laws with Discontinuous Flux I 3This occurs mainly for overcompressive shocks, but may occur even for a regularLax shock wave, see Fig. 12.2. Scalar Conservation Laws with Discontinuous Flux FunctionFor solutions of the Cauchy problem, with h 2 C2,ut + h(u)x = 0; x 2 R; t > 0u(x; 0) = u0(x); x 2 R (2.2)the jump condition x0(t) = S(ux+; ux�)and the entropy conditionS(v; ux�) � S(ux+; ux�); 8v between ux� and ux+ (2.3)by Oleinik [12] are assumed to hold, where x = x(t) is a curve of discontinuity,S(�; �) = h(�)�h(�)��� for � 6= � and ux� = u�x(t)� 0; t�.When f 6= g in (1.1) the new feature is the discontinuity that appears atx = 0. We assume that f; g 2 C2 have at most a �nite number of stationarypoints. For u piecewise smooth we introduceu�(t) = lim�&0u(��; t)u�(t) = lim"&0u�(t+ ") :The order of the limit processes is signi�cant for example when a discontinuityemerges from the t-axis. The Cauchy problem of (1.1) can be writtenut + f(u)x = 0; x > 0; t > 0ut + g(u)x = 0; x < 0; t > 0f�u+(t)� = g�u�(t)�; t > 0u(x; 0) = u0(x); x 2 R : (2.4)This problem has been analysed in [2] and we refer to that paper for a fulldescription including proofs and examples. To be able to construct a solutionby the method of characteristics, one needs to �nd boundary functions on eitherside of the t-axis. These functions should satisfy the third equation of (2.4), i.e.,the conservation law over the t-axis. It turns out that these boundary functionsare not only di�cult to obtain but they are also non-unique. Even if f and gare linear functions, the boundary functions are not uniquely determined by thethird equation of (2.4) as the following example shows.Example 1. Let f(u) = u, g(u) = �u and the initial data u(x; 0) = (ul; x < 0ur; x > 0 ,see Fig. 1. Independently of the values u�, with f(u+) = g(u�), the character-istics always emanate from the t-axis. Hence there exist in�nitely many possi-bilities to construct a solution. Two possible choices of u+ and u� are shown in

4 S. Diehlug(u)f(u)ul u� �u u+ ur u ul u+ u� urFig. 1. (Example 1) There are in�nitely many choices of u� and u+.Fig. 1, where the jump from u� to u+ corresponds to a horizontal dotted linefrom the graph of g to the graph of f . As is motivated in [2], of these solutionsthe one with u� = u+ = �u turns out to be the physically correct one. Anothermotivation will be given by the viscous pro�le analysis in Section 4.In the region x > 0, t > 0, the concentration is determined by the equationut+f(u)x = 0. Consider the constant initial datum u(x; 0) = u+, x > 0. For whatboundary values � is it possible to construct a solution satisfying u+(t) � �?Any such solution is obtained by taking the restriction to the region x > 0,t > 0, of the solution of the Riemann problemut + f(u)x = 0; x 2 R; t > 0u(x; 0) = (�; x < 0u+; x > 0 : (2.5)The general solution of (2.5) (with the entropy condition (2.3)) is of the formu(x=t) and can be described by a cone in the x-t-plane, to the left and rightof which the solution is identically equal to � and u+, respectively. Inside thecone the solution is a monotone function of x=t, built up of rarefaction wavesseparated by discontinuities. If the cone lies to the right of the t-axis, we clearlyobtain u+(t) � �. Generally, we de�ne the set of possible � asP (f ;u+) = �� : the solution of (2.5) satis�es u+(0) = � :This set is closely related to the functionf̂ (u;u+) =8 u+see Fig. 2. The function f̂ (�;u+) is non-decreasing and its graph consists of in-creasing parts separated by plateaus, where the function is constant. At theincreasing parts, i.e. when u 2 P , holds f̂ (u;u+) = f(u). Analogously we intro-duce the set N (g;u�) of possible boundary values on the left side of the t-axisand the corresponding non-increasing function �g(�;u�). In Example 1 aboveholds f̂ = f , �g = g and P = N = R, hence all values are possible for u+ and

Scalar Conservation Laws with Discontinuous Flux I 5u+ f̂ f uu f̂f P1u+P�1P1P0P�1Fig. 2. The graphs of f (solid) and the two main possibilities for f̂(�;u+) (dashed). The setP = Si Pi, where P0 = fu+g in the right-hand plot. Note that some intervals are half-openby the de�nition of P .u�. As was mentioned above, the choice u+ = u� = �u, which is the intersectionof the graphs of f̂ and �g, is physically signi�cant.We shall now try to motivate a uniqueness condition for (2.4) by generalizingthe entropy condition (2.3) for problem (2.2), which we can consider as a specialcase of (2.4) with f = g = h. Consider initial data of (2.2) which producea discontinuity with zero speed between u� and u+. According to the jumpcondition and the entropy condition (2.3) the graph of h lies above the horizontalline (the shock speed is zero) through the points �u�; h(u�)� and �u+; h(u+)�,see Fig. 3 (left). This is equivalent touu+u�ĥ(u;u+) h �h(u;u�) u0 ĥ �h uFig. 3. The entropy condition (2.3) can be described by the intersection of ĥ and �h.h(u+) = h(u�) = minu��v�u+ h(v):By de�nition ĥ(u�;u+) = minu��v�u+ h(v) = �h(u+;u�) � , which is the uxlevel of the intersection of ĥ and �h. The entropy condition (2.3) can thus bewritten (for zero-speed shocks)h(u+) = h(u�) = : (2.6)Now suppose the initial datum of (2.2) is constant u0. Then the uniqueentropy solution is this constant u0 = u+ = u�. Again, h(u0) is the ux valueof the intersection of ĥ and �h, see Fig. 3 (right).

6 S. DiehlWith the above discussion in mind we shall need the following notation inorder to formulate Condition �, which involves the intersection of f̂ and �g andwhich can handle (2.4) as well. Let u+; u� 2 R be given. The projection on theu-axis of the intersection of the graphs of f̂ and �g and its ux value, are denotedby �U = �U (u+; u�) = �u 2 R : f̂ (u;u+) = �g(u;u�); = f̂ ( �U ;u+);see Fig. 4. Since f̂ is non-decreasing and �g is non-increasing, �U is an interval.When the set �U only consists of one point, this point is denoted by �u. Furtherintroduce the set of pairs�(u+; u�) = �(�; �) 2 R2 : f(�) = g(�) = :f u g(u) f̂�Uu� �1 �1 �2 �3 u+ = �2 �3�g(u;u�)Fig. 4. An example of the set �U . The dashed line from �1 to �3 is a plateau of �g(�;u�), andfrom �1 to �3 a plateau of f̂(�;u+). Note that � = �(�i; �j) : i; j = 1; 2;3.Condition �. For given u+; u� 2 R holds (u+; u�) 2 �(u+; u�).Note that Condition � reduces to (2.6) when f = g = h, u+ = u+ andu� = u�. As can be seen from the example in Fig. 4 the set � may consist ofmore than one pair. It can be shown, see [2], that for given initial data only onepair in � can be used to start the construction of a solution. The correct choicedepends on the behaviour of the solution in a neighbourhood of the t-axis. Thisalso implies that, generally, boundary functions on either side of the t-axis cannot be given beforehand globally in t. However, with the Riemann initial datau0(x) = (u�; x < 0u+; x > 0 in (2.4), with u� and u+ constant, boundary functions� and � can be given uniquely beforehand as a function of u� and u+, see [2].In fact, with the Riemann initial data, Condition � reduces to the explicit form(�; �) = c(u+; u�), see Theorem 2.1 below, where the function c is de�ned inthe following proposition. Denote the set of intersecting ranges byI(u+; u�) = f̂ (R;u+) \ �g(R;u�):

Scalar Conservation Laws with Discontinuous Flux I 7Proposition 2.1. Let u+; u� 2 R be given. If I(u+; u�) 6= ;, then the set�P (f ;u+) � N (g;u�)� \ �(u+; u�) consists of exactly one pair and hence afunction c is well de�ned byc(u+; u�) = (u+; u�) 2 (P �N ) \ � :Proof. I 6= ; implies �U 6= ;. Since the restrictions f jP and gjN are injective,(u+; u�) 2 (P �N ) \ � () f jP (u+) = gjN (u�) = uniquely determines u+ and u�.Depending on the set �U (u+; u�), the function c yields that there are threequalitatively di�erent types of discontinuity. This is reected in the three di�er-ent types of intersection of the graphs of f̂ and �g, depending on whether none,one or two plateaus are involved.Case 1. �u 2 N \ P , see Fig. 5 and Example 1. The function c yields u+ =u� = �u. f̂ ff f̂�gg g = �g�u u� u+ NP PNu+ u+�uFig. 5. Examples of Case 1 (left) and Case 2a (right). The sets N and P are indicated abovethe u-axis.Case 2a. �u 2 N n P , see Fig. 5.Case 2b. �u 2 P nN (symmetric to Case 2a).Case 3. �U is in�nite or �u 2 {(P [N ), see Fig. 6.Theorem 2.1. Given ul and ur with I(ur ; ul) 6= ;. Then the Riemann problemwith discontinuous ux functionut + f(u)x = 0; x > 0; t > 0ut + g(u)x = 0; x < 0; t > 0f�u+(t)� = g�u�(t)�; t > 0u(x; 0) = (ul; x < 0ur; x > 0 (2.7)has a unique solution which satis�es Condition � for t � 0, is piecewise smoothand for which u�(t) are piecewise monotone. The solution is of the form u(x=t)and the constant boundary values are given by (u+; u�) = c(ur; ul).

8 S. Diehlff g NPNP u+�u u+u�u�u+�Uu�u� gFig. 6. Examples of Case 3: The set �U is in�nite (left) and �u 2 {(P [N) (right). In the left�gure holds u+ = u+.The proof can be found in [2]. Writing (2.7) as a 2� 2 system, see Subsec-tion 3.2, the result on triangular n� n systems by Holden and H�egh-Krohn [8]yields that there exists a unique solution of (2.7) of the form u(x=t), whichsatis�es the viscous pro�le condition. However, there are situations when Condi-tion � yields a (unique) solution, whereas no viscous pro�le exists between anyof the possible states connecting a discontinuity, see Examples 3 and 4 below.The situation is analogous to the one involving Oleinik's entropy condition forthe scalar equation ut + f(u)x = 0, see Example 2 below.3. The Viscous Pro�les3.1. De�nitionWe shall now examine stationary wave solutions of (1.1) by smoothing the dis-continuity in the ux function and by adding a viscosity term. Let H denoteHeaviside's step function and let h be a scalar smooth function of one variablewith h0(x) � 0 for jxj � 1 and h(x) strictly increasing from 0 to 1 in jxj < 1.Then h(x=�)! H(x) as � & 0 in distribution sense andF �(u; x) � h(x=�)f(u) + �1� h(x=�)�g(u)! F 0(u; x); � & 0 :We consider solutions of (1.1) as limit solutions of the parabolic equationut + �F �(u; x)�x = "uxx (3.8)as � and " & 0 with the same speed. Let � and " be �xed and set � = x=".Assume that there is a smooth stationary solution u(x; t) = v(�) of (3.8), i.e.,�F �(v; "�)�� = v00. Of course v also depends on � but we do not write this outexplicitly. We get v0(�) = F ��v(�); "�� + C :Assuming v(�)! u� and v0(�)! 0 as � !�1 we can determine the constantC = �f(u+) = �g(u�) ;

Scalar Conservation Laws with Discontinuous Flux I 9which is in accordance with the third equation of (2.4). The dynamic equationbecomes v0(�) = h("�=�)f�v(�)� + �1� h("�=�)�g�v(�)� � g(u�) : (3.9)We call a solution v of (3.9) a viscous pro�le for a discontinuity between u� andu+ of a solution of (1.1) if v(�)! u� as � !�1.3.2. The Associated Non-Strictly Hyperbolic SystemEquation (1.1) can be written as a 2 � 2 triangular system by adding a scalarequation having H(x) as stationary solution, cf. Gimse and Risebro [5, 6]. Letk(a) be a scalar function with k(0) = k(1) = 0, k(a) � 0 for a 2 [0; 1] andsmooth in (0; 1). Then the Riemann problemat + k(a)x = 0; x 2 R; t > 0a(x; 0) = H(x)has the stationary solution a(x; t) � H(x). Equation (1.1) may then be writtenas a system�ua�t +�af(u) + (1 � a)g(u)k(a) �x = �00� ; x 2 R; t > 0 : (3.10)The eigenvalues of the Jacobian of the ux matrix are�1(u; a) = af 0(u) + (1� a)g0(u) ;�2(u; a) = k0(a) ;so generally (3.10) is a non-strictly hyperbolic system. The sense in which (1.1)and (3.10) are equivalent depends on the function k. Below we shall give a naturalde�nition of k by claiming that the viscous pro�le of (3.9) should be the sameas the corresponding viscous pro�le of (3.10). The latter is usually de�ned byadding a small viscosity matrix times @xx of the state vector on the right-handside of (3.10). It is common to let the viscosity matrix be diagonal and in ourcase it would be unnatural to assume any coupling of viscosity e�ects betweenu and a with the following motivation. The second equation of (3.10) describesa given physical con�guration, which does not depend on u at all. Thus we addonly the term �axx to the second equation. The solution a (with or without theterm �axx) should a�ect u in the �rst equation only via the ux function, bythe original formulation of the problem. Thus we add only the term "uxx to the�rst equation, leading to the parabolic system�ua�t +�af(u) + (1 � a)g(u)k(a) �x = �"u�a�xx ; x 2 R; t > 0 : (3.11)Assume that this system admits stationary solutions u(x; t) = u(�) = u(x=")and a(x; t) = a(�) = a(x=�), with u(�) ! u�, u0(�) ! 0 as � ! �1 anda(�) ! (10 , a0(�) ! 0 as �!�1. Then we obtain the dynamic system, with~a(�) = a("�=�),

10 S. Diehl( _u = ~af(u) + (1� ~a)g(u) � g(u�)_~a = k(~a) ; where _� dd� : (3.12)Note that (u�; 0) and (u+; 1) are �xed points of (3.12).Requiring that (3.9) and (3.12) should describe the same viscous pro�le yields~a(�) = h("�=�) 2 [0; 1]; � 2 R :It is natural to use the same " and � of (3.9) and (3.12), so that a � h. Forjxj < 1, h(x) is strictly increasing. Hence, for j�j < �=" , 0 < a = ~a = h < 1,the second equation of (3.12) uniquely de�nesk(a) = "� h0�h�1(a)�; a 2 (0; 1) : (3.13)Note that k(0+) = k(1�) = 0 and k(a) > 0, a 2 (0; 1), hold as expected and wecan extend k to be continuous in a neighbourhood of [0; 1]. Furthermore,k0(a)!�1; a! (0+1� ; (3.14)(provided the limits exist). This can be seen as follows. The equation _a = k(a)gives formally � + �" = �Z��=" dy = aZ0 d�k(�) (3.15)for small � + �=" > 0 , small a > 0. If k0(a) � C < 1 for small a > 0, thenk(a) = k(0) + k0(�)a � Ca for small a > 0, and the last integral in (3.15) wouldbe divergent, which contradicts the fact that the left-hand side is �nite.3.3. Classi�cation of WavesConsider the system (3.10) with k(a) de�ned by (3.13) and a shock wave (atx = 0) between (u�; 0) and (u+; 1). Depending on f and g and the initial datathere are di�erent types of shock according to Cases 1, 2 and 3 after the proofof Proposition 2.1. In the next section we shall see that this is also a naturaldivision with respect to the corresponding viscous pro�les. Let us compare withthe notion of undercompressive waves etc. of (3.10). The �rst eigenvalue of (3.10)satis�es �1(u�; 0) = g0(u�), �1(u+; 1) = f 0(u+) and the second �2(u�; 0+) =k0(0+) > 0, �2(u+; 1�) = k0(1�) < 0. We say that the wave isundercompressive if g0(u�) < 0; f 0(u+) > 0,marginal undercompressive if g0(u�) = 0; f 0(u+) > 0 or g0(u�) < 0; f 0(u+) = 0,overcompressive if g0(u�) > 0; f 0(u+) < 0,marginal overcompressive if g0(u�) = 0; f 0(u+) < 0 or g0(u�) > 0; f 0(u+) = 0,a regular Lax wave if g0(u�) < 0; f 0(u+) < 0 or g0(u�) > 0; f 0(u+) > 0,degenerate if g0(u�) = f 0(u+) = 0.

Scalar Conservation Laws with Discontinuous Flux I 11The three main types of shock as well as types of viscous pro�les are the follow-ing, referring to the three cases after the proof of Proposition 2.1.Case 1. Here u+ = u� = �u, so there is only a discontinuity in a. Generally,the wave is undercompressive, but it may also be marginal undercompressive ordegenerate. v(�) � �u is the unique viscous pro�le of (3.9), see Section 4.Case 2. Generally, the wave is a regular Lax shock, but it may also bemarginal under or overcompressive or degenerate. The viscous pro�le of (3.9) isunique and satis�es v(�) � u� for � < ��=". If the ratio �=" is small enough, thepro�le is increasing for � > ��=". For larger values of �=" it are non-monotonein j�j < �=", see Section 4.Case 3. Generally, the wave is overcompressive, but it may also be marginalovercompressive or degenerate. There are in�nitely many viscous pro�les of (3.9),of which some can be non-monotone, see Section 4. The smallness assumptionon the ratio �=" is needed in some cases.4. The Equivalence between Condition � and the Viscous Pro�leConditionA piecewise smooth solution u satis�es (u+; u�) 2 P (u+) � N (u�) for all buta �nite number of t's, see [2]. Those events when (u+; u�) =2 P � N occur onlywhen moving plateaus of f̂ and �g meet, i.e., when they have the same functionvalue. The movements of the plateaus depend on the initial data. Condition �resolves this, but the viscous pro�le condition only considers stationary shocks.The situation is completely analogous to the relation between Oleinik's entropycondition (S(u; u�) � the shock speed for all u between u� and u+) and theviscous pro�le condition for the scalar equation ut + f(u)x = 0. Condition � isa generalization of Oleinik's entropy condition, see [2]. Generally, if there existsa viscous pro�le between two states, then the corresponding discontinuity of thehyperbolic equation ful�ls Oleinik's entropy condition in the case f = g and,as we shall prove below, Condition � in the case of f 6= g. However, there aresituations when Oleinik's entropy condition/Condition � admits a discontinuityof the corresponding hyperbolic equation, which does not admit a viscous pro�le.Example 2. The problem ut + (u2 � u4)x = "uxxu(x; 0) = 2H(x)� 1 (4.16)has, with " = 0, the stationary entropy solution u(x; t) = 2H(x) � 1, t � 0,but there is no corresponding viscous pro�le. This is because _v = f(v) � 0for jvj < 1, with equality for v = 0. Thus v(�) � 0 is a solution, and by theuniqueness theorem for ordinary di�erential equations there does not exist anypro�le from �1 to 1, since the graph would then cross the line v = 0. Hencethe solution u(x; t) = 2H(x)� 1 is unstable when including di�usion. Computersimulations indicate that the solution of (4.16) with " > 0 tends to zero pointwiseas t ! 1. Such situations are avoided by Matsumura and Nishihara [10] andJones, Gardner and Kapitula [9], by assuming that S(u; u�) is strictly largerthan the shock speed for all u strictly between u� and u+.

12 S. DiehlAnalogous situations occur in the case of a discontinuous ux function andCondition �, see Examples 3 and 4 at the end of Subsection 4.2.Consider the Riemann problem with discontinuous ux function (2.7). Theset � = �(�; �) 2 P (ur)� N (ul) : f(�) = g(�)contains precisely those pairs for which there exists a piecewise smooth solutionof (2.7) with (u+(t); u�(t)� � (�; �). Theorem 2.1 implies that Condition �picks out the pair (u+; u�) = c(ur; ul) of �. We shall now show that the viscouspro�le condition uniquely picks out the same pair. Note that I(ur ; ul) 6= ; implies� 6= ;.Theorem 4.2. Given ul and ur with I(ur ; ul) 6= ;. Let (u+; u�) = c(ur; ul) andassume that there exists a viscous pro�le between u� and u+. Then Condition �is equivalent to the viscous pro�le condition in the following sense: If the ratio�=" is small enough (see the lemmas below), (u+; u�) is the only pair of � thatadmits a viscous pro�le of (3.9).The theorem will be proved by considering di�erent cases of f and g in thelemmas below. For given ur, ul, f and g, the stationary viscous pro�le obtainedbelow will thus, as �; " & 0 with �=" bounded, describe the discontinuity alongthe t-axis of the solution of (2.7) satisfying Condition �. From now on it willbe convenient to let (up; um) = c(ur ; ul) denote the speci�c pair in � chosen byCondition �. By symmetry we only consider discontinuities with um � up, andwhen we write Case 2 in the sequel we mean Case 2a. In Case 1 holds up = um.By a monotonicity property of the function c (cf. [2]), ul < ur is equivalent toum < up (Cases 2 and 3).It is convenient to write the non-autonomous Eq. (3.9) as an autonomoussystem and to introduce the function V( _v = h("�=�)�f(v) � g(v)� + g(v) � g(u�) � V (v; �)_� = 1 : (4.17)Then the viscous pro�le is obtained directly in the �-v plane.4.1. Case 1The general behaviour is captured by the following lemma. Note that in thiscase up = um = �u.Lemma 4.1. Assume that there is a unique intersection of the graphs of f andg at �u with g(u) ? g(�u) = f(u) 7 f(�u) = u 7 �u ;cf. Fig. 5. Independently of ur and ul this is a case-1 intersection of f̂ and �g with�u 2 int(N \ P ). Then Condition � is equivalent to the viscous pro�le conditionand there is a unique viscous pro�le v(�) � �u.

Scalar Conservation Laws with Discontinuous Flux I 13Proof. Firstly, let u+ = u� = �u according to Condition �. Then V (�u; �) = 0,8�, and V (v; �) = (g(v) � 7 0; � < ��="f(v) � ? 0; � > �=" ; v ? �u ; (4.18)which implies the phase portrait (of (4.17)) in Fig. 7. Hence v(�) � �u is the��u v��=" �="Fig. 7. (Lemma 4.1) Phase plane diagram in the case u+ = u� = �u.unique solution of (4.17) satisfying v(�) ! u� = �u, � !�1.Secondly, let (u+; u�) 2 � with u� < �u. Then f(u+) = g(u�) impliesu� < �u < u+. Further V (�u; �) = g(�u)� g(u�) < 0, 8�, so no orbit can reach u+starting from u� at � = �1.The case u� > �u is analogous.If ur, ul, f and g are such that (4.18) only holds in a neighbourhood of �u,i.e., �u 2 int(N \ P ), the lemma still holds and the modi�cations of the proofare obvious. If �u 2 @(N \P ), e.g. f(u) = �u3 and g(u) = u2, the result are thesame and we omit the details.4.2. Case 2We shall see that there exists a unique viscous pro�le of (4.17), which satis�esv(�) � um for � < ��=".Lemma 4.2. Consider a case-2 intersection at �u with ul such that �g(u;ul) =g(u) is decreasing for u in a neighbourhood of �u; f̂ has a plateau from u0 to u1;u0 < �u < u1 � ur; f(u) > f(u0) = f(u1) = g(�u) � , u 2 (u0; u1); g(u) < ,u 2 (�u; u1); f(u) � for u > u1; cf. Fig. 8. Then Condition � is equivalent tothe viscous pro�le condition. The unique viscous pro�le is non-decreasing.Proof. Assume �rst that g(u) is decreasing for all u. Let (u+; u�) = (u1; �u) =(up; um) according to Condition �. We obtain

14 S. Diehlg f f̂(�;ur)u1 ur�uu0 Fig. 8. The ux functions in Lemma 4.2.V (�u; �) = h("�=�)�f(�u)� � (= 0; � � ��="> 0; � > ��="V (u1; �) = �1� h("�=�)��g(u1) � � (< 0; � < �="= 0; � � �="V (v; �) = g(v) � ? 0; � � ��="; v 7 �u (4.19)V (v; �) = f(v) � > 0; � � �="; v 2 (�u; u1) ;which gives the phase portrait in Fig. 9. Hence there exists a unique viscous��uu1 v��=" �="Fig. 9. (Lemma 4.2) Phase plane diagram and the viscous pro�le in the case when (u+; u�) =(u1; �u).pro�le with v(�) = �u for � � ��=" and v(�) % u1 as � !1.If u� < �u, then (u+; u�) 2 � implies u+ > u1. ThenV (u1; �) = �1� h("�=�)��g(u1)� � + � g(u�) < 0; 8� ;so no curve v(�) can cross the constant line u1 from below. Hence there does notexist any viscous pro�le from u� to u+.

Scalar Conservation Laws with Discontinuous Flux I 15If u� > �u, then (u+; u�) 2 � implies u+ < u0. Then there is a û 2 (u+; u�)such that f(û) > g(u�), g(û) > g(u�), which implies V (û; �) > 0, 8�. Thusthere exists no orbit crossing the constant line û from above.If g(u) is decreasing only in a neighbourhood of [�u; u1], the same conclusionas above can be made.The unique viscous pro�le v(�), see Fig. 9, is non-decreasing, because@V@� = "� h0�"�� ��f(v) � g(v)� (= 0; j�j � �="> 0; j�j < �=" ; v 2 (�u; u1)implies that once the curve v(�) has passed a point (�̂; v̂) with _v(�̂) � 0, it holdsV (v̂; �) > 0, 8� > �̂, so the curve can not cross the horizontal line with value v̂from above.Note that the wave of Lemma 4.2 is marginal undercompressive if g0(um) =g0(�u) < 0 and degenerate if g0(um) = 0.Depending on f , g, ur and ul there are several qualitatively di�erent possi-bilities to obtain a case-2 intersection. However, (4.19) always holds for v closeto u� = �u, so that the pro�le will always be unique. We shall now proceedto other case-2 intersections without going into every detail, since the proofsonly consist in simple phase plane analyses. The phase plane diagrams shownbelow are obtained by computer simulations, where we have used �=" = 1 andh(x) = �1 + sin(x�=2)�=2 for jxj < 1.In Lemma 4.2, the values of f(u) for u > u1 only inuence the phase portraitfor u > u1. If ur = u1 we may assume that f(u) < for u > u1 and the proofabove will yield the following lemma.Lemma 4.3. Given the prerequisites of Lemma 4.2, except that ur = u1 and fis decreasing in a neighbourhood of ur . Then Condition � is equivalent to theviscous pro�le condition. The viscous pro�le is non-decreasing.Since f̂ (u;ur) = and �g(u;ul) = g(u) for u in a neighbourhood of [�u; ur],we have (u+; u�) = (up; um) = (ur; �u). The phase portrait will be qualitativelyas in Fig. 10.−1 −0.5 0 0.5 1 1.5 2

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1.5g = �g f f̂Fig. 10. (Lemma 4.3) Phase portrait (right) when g(u) = �u, f(u) = 1� u, u� = um = 0and u+ = up = 1.

16 S. DiehlLemma 4.4. Given the prerequisites in Lemma 4.2 or Lemma 4.3, but relax theassumption g(u) < , u 2 (�u; u1) = (um; up). If the ratio �=" is so small thatmaxjxj�1;um�u�up ��h(x)�f(u) � g(u)�+ g(u)� �� < "� (up � um) ; (4.20)then Condition � is equivalent to the viscous pro�le condition.Remark. (4.20) is only needed if both g(u) > and f(u) > for smallu� up > 0.Proof. Assume that g(u) > for some u 2 (�u; u1), otherwise one of the twoprevious lemmas applies. Let �rst (u+; u�) = (up; um) according to Condition �.If f(u) < for small u � up > 0, the interesting part of the phase plane isqualitatively as in Fig. 11 or 12. Hence there is a unique pro�le from u� = um−1 −0.5 0 0.5 1 1.5 2 2.5 3

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3f f̂g�gFig. 11. (Lemma 4.4) Phase plane diagram (right) when g(u) = u(u � 1), f(u) = 2 � u,u� = um = 0 and u+ = up = 2.−1 −0.5 0 0.5 1 1.5 2 2.5 3

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2.5f̂fg�gFig. 12. (Lemma 4.4) Phase plane diagram (right) when g(u) = 2u(u � 1), f(u) = 6(2 �u), u� = um = 0 and u+ = up = 2. Note that the viscous pro�le is non-monotone. Thecorresponding shock is a regular Lax wave.to u+ = up independently of the ratio �=". If f(u) > for u > up, thenV (v; �) = f(v) � > 0 for v > up, � > �=", hence the orbits with over-shootin Fig. 12 tend to in�nity. However, invoking (4.20), _v is so small that the orbit

Scalar Conservation Laws with Discontinuous Flux I 17satisfying v(�) = u� for � < ��=" will satisfy v(�=") < u+, hence v(�)% u+ as� !1 (as in Fig. 11).Now consider the other pairs of �. The case u� > um is treated in theproof of Lemma 4.2. Assume that u� < um and f(u) > for some u > up(otherwise this is an empty case). Then (u+; u�) 2 � implies that u+ > up.Requiring (4.20), then for the orbit satisfying v1(�) � u� for � < ��=" holds_v1(�) < "=�(up � um) < "=�(u+ � u�). Hence v1(�=") < u+ and it is easy to seethat all orbits which are < u+ at � = �=" tend to up 6= u+ as � !1.Before ending this subsection, we give examples of unstable discontinuities.They should be compared to Example 2.Example 3. Let the initial data be u0(x) = 2H(x) and the ux functions g(u) =�u and f(u) = (u � 1)2(u � 2)2, see Fig. 13. Here � = �(�; �) : � � 0; � �−1 −0.5 0 0.5 1 1.5 2 2.5 3

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2.5f g = �g f̂Fig. 13. Phase plane diagram (right) in Example 3.0; f(�) = g(�). u(x; t) � 2H(x) is a solution satisfying Condition �. Thereexists no viscous pro�le between 0 and 2, since f(1) = 0 implies V (1; �) =h("�=�) � 1 � 0. The exclusions of the existence of viscous pro�les for all theother pairs of � can be made as in the proof of Lemma 4.2.Example 4. A simpler example than the previous one is obtained by lettingf(u) < for u > 2, e.g. f(u) = (u�1)2(2�u). Then f̂(�; 2) � 2 and � = f(0; 2)g.Hence Condition � is not needed since u(x; t) � 2H(x) is the only possiblesolution. With the same argument as above we conclude that there exists noviscous pro�le between 0 and 2.Situations as in the examples above may occur when f̂ (or �g) has a plateauat the level and there is a û 2 (um; up) with f(û) = . The correspondingdiscontinuities are thus unstable when including di�usion.4.3. Case 3In this case one plateau of �g and one of f̂ lie on the same ux value level = f̂ ( �U ;ur). Recall that (up; um) = c(ur ; ul) is the pair chosen by Condition �.In Case 2 precisely one plateau was involved in the intersection. We chose one off̂ in the previous subsection (corresponding to Case 2a) and saw that in�nitely

18 S. Diehlmany orbits converged to up as � !1. The behaviour of the pro�les for � > �="will thus now also hold for � < ��=". Hence there will be in�nitely many pro�lesfrom um to up. Because of this symmetry between +1 and �1, we shall startby letting g be as simple as possible and consider di�erent f 's, such that a case-3intersection is produced. In some cases we need the smallness condition (4.20)on the ratio �=", which guarantees the existence of monotone pro�les from umto up.Lemma 4.5. Assume that both f(u) > and g(u) > for u 2 (um; up). If(4.20) holds, then Condition � is equivalent to the viscous pro�le condition.Remark. (4.20) is only needed if g(u) > for small um�u > 0 and f(u) > for small u� up > 0.Proof. Let �rst (u+; u�) = (up; um) according to Condition �. The qualitativebehaviour of the interesting part of the phase plane, when g(u) < for smallum � u > 0, is shown in Figs 14 and 15.−1 −0.5 0 0.5 1 1.5 2

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1.5f ĝf = �gFig. 14. (Lemma 4.5) Phase plane diagram (right) when g(u) = u, f(u) = 1 � u, u� =um = 0 and u+ = up = 1. All orbits are viscous pro�les. The corresponding shock wave isovercompressive.−1 −0.5 0 0.5 1 1.5 2

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1.5gf f̂�gFig. 15. (Lemma 4.5) Phase plane diagram (right) when g(u) = u, f(u) = (1� u)2, u� =um = 0 and u+ = up = 1. The corresponding shock wave is marginal overcompressive.If u� < um, then the requirement (u+; u�) 2 � implies u+ > up. ThenV (ui; �) � 0 where ui 2 (um; up) is an intersection of f and g. Hence thereexists no pro�le from u� to u+. The case u� > um is analogous.

Scalar Conservation Laws with Discontinuous Flux I 19If the conditions in the remark hold, (4.20) guarantees that the pro�le satis-fying v(�) = um for � < ��=" will increase to up.Now, we proceed by relaxing the restriction in Lemma 4.5 that the graph off should lie strictly above the level on the whole interval (um; up).Lemma 4.6. Assume that g(u) > for u 2 (um; up), f(u0) = 0 for someu0 2 (um; up) and f(u) > for u 2 (u0; up). If (4.20) holds, then Condition �is equivalent to the viscous pro�le condition.Remark. (4.20) is only needed if g(u) > for small um�u > 0 and f(u) > for small u� up > 0.Proof. Let �rst (u+; u�) = (up; um) according to Condition �. The qualitativebehaviour when f(u) < for smallu�up > 0 and g(u) < for smallum�u > 0 isgiven by the two phase planes in Figs 16 and 17. If both holds (see the remark),(4.20) is required. If instead f(u) > for small u � up > 0, then the orbits−1 −0.5 0 0.5 1 1.5 2 2.5 3

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2.5f g f̂ = �gFig. 16. (Lemma 4.6) Phase plane diagram (right) when g(u) = u, f(u) = (u � 1)2(2� u),u� = um = 0 and u+ = up = 2.−1 −0.5 0 0.5 1 1.5 2 2.5 3

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2.5f�g f̂ gFig. 17. (Lemma 4.6) Phase plane diagram (right) when g(u) = u, f(u) = (1� u)(u � 2),u� = 0 and u+ = 2.in � > �=" just above v = up would increase. Similarly, if g(u) > for smallum�u > 0, this would only a�ect the phase portrait below v = um and � < ��=".

20 S. DiehlIn any case, it follows that all orbits v(�) passing through the points � = �=",u0 < v(�) � up are viscous pro�les.Now assume that u� < um. Then the requirement (u+; u�) 2 � implies thatu+ > up. If the graph of f intersects g at some ui 2 (u�; u+), then we have thesame case as in the previous proof. Otherwise holds f(um) < g(um) < g(u�),which implies V (um; �) < 0 8�, so no orbit can pass the horizontal line v = umfrom below.Finally, assume that u� > um, then u+ < up < u�, which implies V (up; �) >0 8�. Hence no orbit can pass the horizontal line v = up from above.We shall now relax the condition on f further and assume that the graph liesbelow the constant line in a left neighbourhood of up. In order to have a case-3intersection, it follows that f must lie below also in a right neighbourhood ofup.Lemma 4.7. Assume that g(u) > for u 2 (um; up) and that f has a strict localmaximum at up. Then Condition � is equivalent to the viscous pro�le condition.There are precisely one monotone and in�nitely many non-monotone pro�les.Proof. The assumptions imply that up = ur. Phase plane analysis yields that,for (u+; u�) = (ur ; um), there are in�nitely many pro�les & u� as � ! �1and & u+ as � !1, cf. Fig. 18. The pro�le satisfying v(�) = up for � � �=" is−1 −0.5 0 0.5 1 1.5 2 2.5 3

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2.5�g ff̂ gFig. 18. (Lemma 4.7) Phase plane diagram (right) when g(u) = u, f(u) = �(u� 2)2, u� =um = 0 and u+ = up = 2.monotone. Exclusion of other pairs of � can be done exactly as in the proof ofLemma 4.5.Now we shall relax the condition that g lies strictly above the level in theinterval (um; up). Let um < u1 < u2 : : : < uk < up. Assume that g(uj) = 0 forj = 1; : : : ; k. These points must be local minima, otherwise there is no case-3 intersection. If f is as in the previous lemma, then V (uj; �) � 0 for all �.Hence there exists no viscous pro�le between um and up, cf. Example 3. In fact,V (u1; �) � 0 holds for every f satisfying f(u1) � . The only way to obtain acase-3 intersection which admits a viscous pro�le from um to up is to let f(u) > at least for u 2 (u1; up). With this in mind we give the last lemma.Lemma 4.8. Let um < u1 < u2 : : : < uk < up. Assume that g(u) � foru 2 (um; uk) with g(uj) = 0 for j = 1; : : : ; k. Assume further that f(u) > ,

Scalar Conservation Laws with Discontinuous Flux I 21u 2 (u0; up) for some u0 < u1. If (4.20) holds, then Condition � is equivalent tothe viscous pro�le condition.Proof. Assume that u0 > um, otherwise a symmetric case of Lemma 4.6 ap-plies, cf. Fig. 16. Let (u+; u�) = (up; um). The qualitative behaviour is given byFig. 19. Note that the values of g(u) for u > uk do not a�ect the interesting−1 −0.5 0 0.5 1 1.5 2 2.5 3

−1

−0.5

0

0.5

1

1.5

2

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-0.5

0

0.5

1

1.5

2

2.5�g f̂fgFig. 19. (Lemma 4.8) Phase plane diagram (right) when g(u) = u(u � 1)2, f(u) = (0:5 �u)(u� 2), u� = um = 0, u+ = up = 2, u0 = 0:5 and u1 = 1.orbits. Since the graphs of both f and g lie above the line v = in the interval(u0; u1), it follows that V (u; �) > 0 in the rectangle j�j < �=", u0 � u � u1. Weclaim that each orbit passing through a given point in this rectangle is a vis-cous pro�le, which is monotone. Firstly, V (u; �) > 0 in the both strips � < ��=",um < u < u1 (called left strip below) and � > �=" u0 < u < up (called right stripbelow). Secondly, (4.20) ensures that the orbit, say v1(�), that coincide with theupper horizontal boundary of the left strip, i.e. v1(�) = u1 for � � ��=" (hencepassing through the upper left corner of the rectangle), goes into the right strip.Analogously, the orbit with v2(�) = u0 for � � �=" comes from the left strip.Consequently, every orbit passing through the rectangle must lie in both thesestrips, since it cannot cross neither v1(�) nor v2(�). Note that if u0 = u1, the rect-angle deteriorates, but there are still in�nitely many viscous pro�les. Exclusionof other pairs of � can be made as in the previous lemmas.If f also exhibits the analogue local minima as g in the lemma, it shouldnow be clear that the technique above can be used to establish the desiredequivalence.4.4. A Remark when � = 0, " > 0As mentioned in the introduction, � > 0 may provide better models than � = 0in the applications. Above we have let � > 0, so that the viscous pro�les becomesmooth. We have also seen that in many cases there is a boundedness conditionon the ratio �=". The analysis of the equivalence between the uniqueness con-ditions could have been made in the special case � = 0 and " > 0. Then theregion between j�j � �=" in the �gures above disappears and the viscous pro�lesbecome continuous with a jump in their derivative at x = 0. This can be seen

22 S. Diehlas follows. Let ul and ur denote the states on either side of a discontinuity atx = 0. De�ning u as a weak solution of the equationut + �H(x)f(u) + �1�H(x)�g(u)�x = "uxxit can be shown (by a standard procedure taking a test function with supportnear the t-axis) that ut + g(u)x = "uxx x < 0ut + f(u)x = "uxx x < 0"(u+ � u�) = 0f(u+) � "u+x = g(u�)� "u�x :Suppose u(x; t) = v(x=") is a viscous pro�le with v(�) ! ul (ur) as � ! �1(1) and v0(�)! 0 as � !�1 . Then it follows that_v = g(v) � g(ul) x < 0_v = f(v) � f(ur) x > 0v+ = v�f(v+) � _v+ = g(v�)� _v� ; (4.21)where the last equation can be written f(ur) = g(ul). Hence (4.21) is equivalentto (3.9) with � & 0.References1. J.-Ph. Chancelier, M. Cohen de Lara, and F. Pacard. Analysis of a conservation pde withdiscontinuous ux: A model of settler. SIAM J. Appl. Math., 54(4):954{995, 1994.2. S. Diehl. On scalar conservation laws with point source and discontinuous ux function.to appear in SIAM J. Math. Anal., 26(6), 1995.3. S. Diehl. A conservation law with point source and discontinuous ux function modellingcontinuous sedimentation. to appear in SIAM J. Appl. Math., 56(1), 1996.4. S. Diehl and N.-O. Wallin. Scalar conservation laws with discontinuous ux function: II.On the stability of the viscous pro�les. to appear in Comm. Math. Phys.5. T. Gimse and N. H. Risebro. Riemann problems with a discontinuous ux function. InB. Engquist and B. Gustavsson, editors, Third International Conference on HyperbolicProblems, Theory, Numerical Methods and Applications, volume I, pages 488{502, 1990.6. T. Gimse and N. H. Risebro. Solution of the Cauchy problem for a conservation law witha discontinuous ux function. SIAM J. Math. Anal., 23(3):635{648, 1992.7. S. K. Godunov. A �nite di�erence method for the numerical computations of discontin-uous solutions of the equations of uid dynamics. Mat. Sb., pages 271{290, 1959.8. L. Holden and R. H�egh-Krohn. A class of N nonlinear hyperbolic conservation laws. J.Di�erential Equations, 84:73{99, 1990.9. C. K. R. Jones, R. Gardner, and T. Kapitula. Stability of travelling waves for non-convexscalar viscous conservation laws. Comm. Pure Appl. Math., 46:505{526, 1993.10. A. Matsumura and K. Nishihara. Asymptotic stability of travelling waves for scalarconservation laws with non-convex nonlinearity. Comm. Math. Phys., 165:83{96, 1994.11. S. Mochon. An analysis of the tra�c on highways with changing surface conditions.Math.Model., 9(1):1{11, 1987.

Scalar Conservation Laws with Discontinuous Flux I 2312. O. A. Oleinik. Uniqueness and stability of the generalized solution of the Cauchy problemfor a quasi-linear equation. Uspekhi Mat. Nauk, 14:165{170, 1959. Amer. Math. Soc.Transl. Ser. 2 33, (1964), 285-290.13. D. S. Ross. Two new moving boundary problems for scalar conservation laws. Comm.Pure Appl. Math., 41:725{737, 1988.This article was processed by the author using the LaTEX style �le cljour1 from Springer-Verlag.