On the structure of entropy solutions to the Riemann problem for a degenerate nonlinear parabolic equation

We find an explicit form of entropy solutions to a Riemann problem for a degenerate nonlinear parabolic equation with piecewise constant velocity and diffusion coefficients. It is demonstrated that this solution corresponds to the minimum point of some strictly convex function of a finite number of variables.


Introduction
In a half-plane Π = {(t, x) | t > 0, x ∈ R}, we consider a nonlinear parabolic equation where v(u), a(u) ∈ L ∞ (R), a(u) ≥ 0. Since the diffusion coefficient a(u) may take zero value, equation ( 1) is degenerate.In the case when a(u) ≡ 0 it reduces to a first order conservation law where ϕ (u) = v(u).Similarly, a general equation ( 1) can be written in the conservative form u t + ϕ(u) x − tA(u) xx = 0 with A (u) = a 2 (u), which allows to define weak solutions of this equation.Unfortunately, weak solutions to a Cauchy problem for equation (1) are not unique in general, and some additional entropy conditions are required.We consider the Cauchy problem with initial data u(0, x) = u 0 (x), where u 0 (x) ∈ L ∞ (R).Recall the notion of entropy solution (e.s. for short) in the sense of Carrillo [1].Entropy condition (4) means that for each nonnegative test function (5) In the case of conservation laws (2) the notion of e.s.reduces to the notion of generalized e.s. in the sense of Kruzhkov [2].Taking in (4) k = ±M , M ≥ u ∞ , we derive that that is, an e.s.u of (1), ( 3) is a weak solution of this problem.It is known that an e.s. of ( 1), (3) always exists and is unique.In general multidimensional setting this was demonstrated in [2] for conservation laws and in [1] for the general case.If to be precise, in [1] the case of usual diffusion term A(u) xx was studied but the proofs can be readily adapted to the case of the self-similar diffusion tA(u) xx .
If u = u(t, x) is a piecewise C 2 -smooth e.s. of equation ( 1) then it must satisfy this equation in classic sense in each smoothness domain.Applying relation (1) to a test function f = f (t, x) ∈ C 2 0 (Π) supported in a neighborhood of a discontinuity line x = x(t) and integrating by parts, we then obtain the identity a.e. on the line x = x(t).Here we denote by [w] the jump of a function w = w(t, x) on the line x = x(t) so that [w] = w(t, x(t)+) − w(t, x(t)−), where w(t, x(t)±) = lim y→x(t)± w(t, y).
Since the functions f , f x are arbitrary and independent on the line x = x(t), identity (6) implies the following two relations of Rankine-Hugoniot type Similarly, it follows from entropy relation (5), after integration by parts, that Since the function A(u) increases, it follows from (7) that A(u) = const when u lies between the values u(t, x(t)−) and u(t, x(t)+).This implies that [sign(u − k)(A(u) − A(k))] = 0 and in view of arbitrariness of f ≥ 0 it follows from (9) that In the case when k lies out of the interval with the endpoints u ± .= u(t, x(t)±) relation (10) follows from (8) and fulfils with equality sign.When u − < k < u + this relation reads where A(u) ± x = A(u) x (t, x(t)±).Adding (8) to this relation and dividing the result by 2, we arrive at the following analogue of the famous Oleinik condition (see [3]) known for conservation laws.
In the case u + < u − this condition has the form and can be derived similarly.Geometric interpretation of these conditions is that the graph of the flux function ϕ(u) lies not below (not above) of the segment connecting the points (u − , ϕ(u − ) − tA(u) − x ), (u + , ϕ(u + ) − tA(u) + x ) when u − ≤ u ≤ u + (respectively, when u + ≤ u ≤ u − ), see Figure 1.We take here into account that in view of condition (8) the vector (−x (t), 1) is a normal to the indicated segment.We also notice that it follows from relations (11), (12) with k = u ± and from the Rankine-Hugoniot condition (8) that 2 The case of piecewise constant coefficients.
Below we will assume that the functions v(u), a(u We will study problem (1), (3) with the Riemann data u 0 (x) = α, x < 0, β, x > 0. Since this problem is invariant under the scaling transformations t → λt, x → λx, λ > 0 then, by the uniqueness, the e.s.u = u(t, x) is self-similar: u(t, x) = u(λt, λx).This implies that u = u(x/t).Suppose that a k > 0. Then in a domain where u k < u(ξ) < u k+1 with ξ = x/t equation (1) reduces to the second order ODE is the error function.Therefore, it is natural to seek the e.s. of our problem in the following form where and we agree that a n = 0, F (−∞) = 0, F (+∞) = 1.We also assume that ξ k+1 > ξ k whenever a k > 0. The rays x = ξ k t for finite ξ k are (weak or strong) discontinuity lines of u, they correspond to discontinuity points ξ k of the function u(ξ).Observe that conditions (7), (8) turns into the following relations at points ξ k Here w(ξ k ±) denotes unilateral limits of a function w(ξ) at the point ξ k .Similarly, the Oleinik condition (11) reads Notice that our solution (13) is a nonstrictly increasing function of the self-similar variable ξ and, therefore, Let us firstly analyze the solution (13) in the case In this situation ξ = ξ k is a weak discontinuity point, the function u(ξ) itself is continuous, only its derivative u (ξ) may be discontinuous.Moreover, it follows from (17) that both functions u(ξ) and u (ξ) are continuous at point If a k−1 > a k = 0 then again u(ξ) is continuous at u k and (15) reduces to the relation where we use the fact that ϕ Finally, since the function ϕ(u) is affine on the segment [u k−1 , u k ] and A(u) (ξ k ±) ≥ 0, then entropy relation ( 16) is always satisfied.Now we consider the case when there exists a nontrivial family of mutually equaled values ξ i , ξ i = ξ k for i = k, . . ., l, where l > k.We can assume that this family is maximal, that is, Then a i = 0 for i = k, . . ., l − 1, and the point ξ = c .= ξ k is a discontinuity point of u(ξ) with the unilateral limits Since a(u) = 0 for u(c−) < u < u(c+), we find that A(u(c−)) = A(u(c+)) and condition (14) is satisfied.Further, we notice that Therefore, condition (15) can be written in the form where, as is easy to verify, In the similar way we can write the Oleinik condition (16) as follows We use here the fact the function ϕ(u) is piecewise affine and, therefore, it is enough to verify the Oleinik condition (16) only at the nodal points k = u j .

The entropy function
We introduce the convex cone Ω ⊂ R d consisting of points ξ = (ξ 1 , . . ., ξ d ) with increasing coordinates, Each point ξ ∈ Ω determines a function u(ξ) in correspondence with formula (13).Assume firstly that ξ ∈ Int Ω, that is, the values ξ k are strictly increasing.Then conditions (17), ( 18), ( 19), (20) coincides with the equality ∂ ∂ξ k E( ξ) = 0, where We will call this function the entropy because it depends only on the discontinuities of a solution.Thus, for ξ ∈ Int Ω the e.s.(13) corresponds to a critical point of the entropy.We are going to demonstrate that the entropy is strictly convex and coercive in Ω.Therefore, it has a unique global minimum point in Ω.In the case when this minimum point lies in Int Ω it is a unique critical point.
Proof.If E( ξ) ≤ c then it follows from nonnegativity of all terms in (25) that for all k = 0, . . ., n − 1 Relation (26) implies the estimate where m = min k=0,...,n−1,a k >0 On the other hand, if a 0 = 0 then (u 1 − u 0 )(ξ 1 − v 0 ) 2 ≤ 2c, in view of ( 27) with k = 0, and In any case, To get an upper bound, we remark that in the case a n−1 > 0 it follows from (28) with ), which implies the estimate In both cases Since all coordinates of ξ lie between ξ 1 and ξ d , estimates (29), (30) imply the bound Further, since F (x) = 1 √ 2π e −x 2 /2 < 1, the function F (x) is Lipschitz with constant 1 and it follows from (28) that We find that ξ k+1 − ξ k ≥ a k δ (this also includes the case a k = 0).We conclude tat the set E( ξ) ≤ c lies in the compact set By the continuity of E( ξ) the set E( ξ) ≤ c is a closed subset of K and therefore is compact.
We take c > N .
= inf E( ξ).Then the set E( ξ) ≤ c is not empty.By Proposition 1 this set is compact and therefore the continuous function E( ξ) reaches the minimal value on it, which is evidently equal to N .We proved the existence of global minimum E( ξ0 ) = min E( ξ).The uniqueness of the minimum point is a consequence of strict convexity of the entropy, which is stated in Proposition 2 below.The following lemma plays a key role.Lemma 1.The function P (x, y) = − ln(F (x) − F (y)) is strictly convex in the half-plane x > y.
Proof.The function P (x, y) is infinitely differentiable in the domain x > y.To prove the lemma, we need to establish that the Hessian D 2 P is positive definite at every point.By the direct computation we find We have to prove positive definiteness of the matrix Q = (F (x) − F (y)) 2 D 2 P (x, y) with the components Since F (x) = e −x 2 /2 , then F (x) = −xF (x) and the diagonal elements of this matrix can be written in the form By Cauchy mean value theorem there exists such a value z ∈ (y, x) that Therefore, and it follows that 0, then the matrix Q > 0, as was to be proved.
Proof.Since 1 − F (x) = F (−x), we see that P (+∞, x) = P (−x, −∞), and it is sufficient to prove the strict convexity of the function P (x, −∞) = − ln F (x).By Lemma 1 in the limit as y → −∞ we obtain that this function is convex, moreover, But this contradicts our assumption.We conclude that d 2 dx 2 P (x, −∞) > 0 and the function P (x, −∞) is strictly convex.
In view of (25) the entropy E( ξ) is a linear combination of the functions P k with positive coefficients, and convexity of the entropy readily follows from the statements of Lemma 1 and Corollary 1.To establish the strict convexity, we have to demonstrate that the Hessian matrix D 2 E( ξ) is strictly positive.Assume that for some Since E( ξ) is a linear combination of convex functions P k ( ξ) with positive coefficients, we find that This can be written in the form i,j=k,k+1 In view of Lemma 1 and Corollary 1 the functions P k in above equalities are strictly convex as functions of either two variables (ξ k , ξ k+1 ) or single variable ξ k+1 .Therefore, these equalities imply that in any case ζ k+1 = 0, k = 0, . . ., n − 2, and ζ n = 0 if a n−1 = 0 (when d = n).We conclude that all coordinates ζ i = 0, i = 1, . . ., d.Hence, equality (31) can hold only for ζ = 0 and the matrix D 2 P ( ξ) > 0 for all ξ ∈ Ω.This completes the proof.

The variational formulation
Let ξ0 = (ξ 1 , . . ., ξ d ) ∈ Ω be the unique minimum point of E( ξ).The necessary and sufficient condition for ξ0 to be a minimum point is the following one so that T ( ξ0 ) is the tangent cone to Ω at the point ξ0 .If ξ0 ∈ Int Ω then T ( ξ0 ) = R d and (32) reduces to the requirement ∇E( ξ0 ) = 0.As we have already demonstrated, this requirement coincides with jump conditions (17), ( 18), ( 19), (20) for all k = 1, . . ., d.But these conditions are equivalent to the statement that the function ( 13) is an e.s. of (1), (3).In the general situation when ξ0 can belong to the boundary of Ω, the coordinates of ξ0 may coincides.Let ξ k = • • • = ξ l = c be a maximal family of coinciding coordinates, that is, ξ k−1 < ξ k = ξ l < ξ l+1 (it is possible here that k = l).Then, as is easy to realize, the vector p = (p 1 , . . ., p d ), with arbitrary increasing coordinates p k ≤ • • • ≤ p l and with zero remaining coordinates, belong to the tangent cone T ( ξ0 ).In view of (32) for any such a vector.Using the summation by parts formula, we realize that the above condition is equivalent to the following requirements Recall that a i = 0 for k ≤ i < l.By the direct computation we find where A(u) (c±) are given by ( 22), (23).Putting these expressions into (33), (34), we obtain exactly the jump conditions (21), (24).Therefore, the function (13) corresponding to the point ξ0 is an e.s. of ( 1), (3).Conversely, if ( 13) is an e.s. then relations (33), (34) holds for all groups of coinciding coordinates.As is easy to verify, this is equivalent to the criterion (32).We have proved our main result.
Remark 1. Adding to the entropy (25) the constant we obtain the alternative variant of the entropy If we consider the values v k , a k as a piecewise constant approximation of an arbitrary velocity function v(u) and, respectively, a diffusion function a(u) ≥ 0 then, passing in (35) to the limit as max(u k+1 − u k ) → 0, we find that the entropy Existence and uniqueness of a minimal point in this case is trivial.By Theorem 1 and Remark 1 we obtain new, variational formulation of the entropy solution.