Radon measure solutions for scalar. conservation laws. A. Terracina. A.Terracina La Sapienza, Università di Roma 06/09/2017

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1 Radon measure A.Terracina La Sapienza, Università di Roma 06/09/2017

2 Collaboration Michiel Bertsch Flavia Smarrazzo Alberto Tesei

3 Introduction Consider the following Cauchy problem { ut + ϕ(u) x = 0 in R (0, T ] =: Q T u(x, 0) = u 0 in R {0}. (1) where u 0 is a positive Radon measure.

4 Introduction Consider the following Cauchy problem { ut + ϕ(u) x = 0 in R (0, T ] =: Q T u(x, 0) = u 0 in R {0}. (1) where u 0 is a positive Radon measure. T.-P. Liu & M. Pierre, J. Differential Equations (1984) In this paper authors gave a formulation in which the solution becomes L for t > 0, they consider a class of super linear fluxes (in particular fluxes ϕ(u) = u p, p > 1), proving uniqueness and existence.

5 Linear case ϕ(u) = Cu.

6 Linear case ϕ(u) = Cu. The solution is given in a standard way by T Ct (u 0 ). Where for every Radon measure µ, we define in a standard way the translation by T h (µ), ρ R := µ, ρ h R (2) for every ρ C c (R), where ρ h (x) := ρ(x + h) (x R). In this case it is clear that if there is a singular part in the initial data this does not disappear in the solution.

7 Let us consider the following toy case ϕ(u) = 1 1 (1+u), p > 0 p (concave and bounded flux), u 0 = δ 0. Idea: approximate initial data with initial data that are functions { 0 x 1 u 0n = 2n n x < 1 2n.

8 Let us consider the following toy case ϕ(u) = 1 1 (1+u), p > 0 p (concave and bounded flux), u 0 = δ 0. Idea: approximate initial data with initial data that are functions { 0 x 1 u 0n = 2n n x < 1 2n.

9 The sequence of solutions converges to a radon measure such that the singular part persists until time t = 1.

10 The sequence of solutions converges to a radon measure such that the singular part persists until time t = 1. More precisely, fixed T > 1, we have u s (t) = max{1 t, 0}δ 0.

11 The sequence of solutions converges to a radon measure such that the singular part persists until time t = 1. More precisely, fixed T > 1, we have u s (t) = max{1 t, 0}δ 0. The regular part u r is given t] (0, 1) by u r (x, t) := [( p tx 1 ) 1 1 p 1 χ {0<x p t}.

12 The sequence of solutions converges to a radon measure such that the singular part persists until time t = 1. More precisely, fixed T > 1, we have u s (t) = max{1 t, 0}δ 0. The regular part u r is given t] (0, 1) by u r (x, t) := [( p tx 1 ) 1 1 p 1 χ {0<x p t}. For t > 1, u s disappears, moreover u r becomes L (it easily obtained explicitly).

13 The sequence of solutions converges to a radon measure such that the singular part persists until time t = 1. More precisely, fixed T > 1, we have u s (t) = max{1 t, 0}δ 0. The regular part u r is given t] (0, 1) by u r (x, t) := [( p tx 1 ) 1 1 p 1 χ {0<x p t}. For t > 1, u s disappears, moreover u r becomes L (it easily obtained explicitly). It is worth to note: i) There exists a waiting time in (0, + ), first time in which the singular part disappears.

14 The sequence of solutions converges to a radon measure such that the singular part persists until time t = 1. More precisely, fixed T > 1, we have u s (t) = max{1 t, 0}δ 0. The regular part u r is given t] (0, 1) by u r (x, t) := [( p tx 1 ) 1 1 p 1 χ {0<x p t}. For t > 1, u s disappears, moreover u r becomes L (it easily obtained explicitly). It is worth to note: i) There exists a waiting time in (0, + ), first time in which the singular part disappears. ii) For t (0, 1) the support of the singular part is given by the set {0} and we have lim x 0 + u r (x, t) = +.

15 The sequence of solutions converges to a radon measure such that the singular part persists until time t = 1. More precisely, fixed T > 1, we have u s (t) = max{1 t, 0}δ 0. The regular part u r is given t] (0, 1) by u r (x, t) := [( p tx 1 ) 1 1 p 1 χ {0<x p t}. For t > 1, u s disappears, moreover u r becomes L (it easily obtained explicitly). It is worth to note: i) There exists a waiting time in (0, + ), first time in which the singular part disappears. ii) For t (0, 1) the support of the singular part is given by the set {0} and we have lim x 0 + u r (x, t) = +. iii) The singular part is decreasing in time.

16 Similarly, we can study these family of fluxes that are unbounded (sub linear) and concave, ϕ(u) = (1 + u) p 1, 0 < p < 1. Let us consider again initial data u 0 = δ 0. In this case it easy to prove that t n (intersection between shock and rarefaction wave) converges to 0. This suggests that the solution regularizes immediately.

17 Similarly, we can study these family of fluxes that are unbounded (sub linear) and concave, ϕ(u) = (1 + u) p 1, 0 < p < 1. Let us consider again initial data u 0 = δ 0. In this case it easy to prove that t n (intersection between shock and rarefaction wave) converges to 0. This suggests that the solution regularizes immediately. Natural conjectures The existence of a non trivial waiting time is strictly related to the asymptotic behavior of the flux.

18 Similarly, we can study these family of fluxes that are unbounded (sub linear) and concave, ϕ(u) = (1 + u) p 1, 0 < p < 1. Let us consider again initial data u 0 = δ 0. In this case it easy to prove that t n (intersection between shock and rarefaction wave) converges to 0. This suggests that the solution regularizes immediately. Natural conjectures The existence of a non trivial waiting time is strictly related to the asymptotic behavior of the flux. The singular part moves along natural characteristics.

19 Similarly, we can study these family of fluxes that are unbounded (sub linear) and concave, ϕ(u) = (1 + u) p 1, 0 < p < 1. Let us consider again initial data u 0 = δ 0. In this case it easy to prove that t n (intersection between shock and rarefaction wave) converges to 0. This suggests that the solution regularizes immediately. Natural conjectures The existence of a non trivial waiting time is strictly related to the asymptotic behavior of the flux. The singular part moves along natural characteristics. It is necessary to give a formulation for radon measure.

20 Formulation The idea of the formulation is related to that consider for the parabolic problem.

21 Formulation The idea of the formulation is related to that consider for the parabolic problem. F. Smarrazzo & A. Tesei, Arch. Rational Mech. Anal. (2012). M. Porzio, F. Smarrazzo & A. Tesei, Arch. Rational Mech. Anal. (2013). M. Bertsch, F. Smarrazzo & A. Tesei, Analysis & PDE (2013), Nonlinear Anal. (2015), J. Reine Angew. Math. (2016). L.Orsina, M. Porzio & F. Smarrazzo J. Evol. Equ. (2015).

22 Formulation The idea of the formulation is related to that consider for the parabolic problem. F. Smarrazzo & A. Tesei, Arch. Rational Mech. Anal. (2012). M. Porzio, F. Smarrazzo & A. Tesei, Arch. Rational Mech. Anal. (2013). M. Bertsch, F. Smarrazzo & A. Tesei, Analysis & PDE (2013), Nonlinear Anal. (2015), J. Reine Angew. Math. (2016). L.Orsina, M. Porzio & F. Smarrazzo J. Evol. Equ. (2015). Suppose that the initial data is a positive radon measure defined in R, the solution is sought in the space L (0, T ; M + (R)), a subspace of the 2-d radon measure in which there is a good disintegration respect to the time variable.

23 Obviously in the hyperbolic case it is necessary to consider an entropy formulation. Let us assume the following hypothesis for the flux ϕ C([0, )), ϕ(0) = 0, ϕ L (0, ) ; (H 1 ) ϕ(u) there exists lim =: C ϕ. u u

24 Obviously in the hyperbolic case it is necessary to consider an entropy formulation. Let us assume the following hypothesis for the flux ϕ C([0, )), ϕ(0) = 0, ϕ L (0, ) ; (H 1 ) ϕ(u) there exists lim =: C ϕ. u u In particular this assures that the velocity of the characteristics is a priori bounded. The singular part moves along characteristics of the type a + C ϕ t.

25 Obviously in the hyperbolic case it is necessary to consider an entropy formulation. Let us assume the following hypothesis for the flux ϕ C([0, )), ϕ(0) = 0, ϕ L (0, ) ; (H 1 ) ϕ(u) there exists lim =: C ϕ. u u In particular this assures that the velocity of the characteristics is a priori bounded. The singular part moves along characteristics of the type a + C ϕ t. We consider couples (E, F ) of entropy flux entropy such that: E is convex, F = ϕ E, that satisfy { E, F L (0, ), F = E ϕ in (0, ), there exist lim u E(u) u =: C E, lim u F (u) u =: C F. (In particular Kružkov s entropies are admissible if ϕ satisfies H 1 ).

26 A measure u L (0, T ; M + (R)) is a weak solution of the Cauchy problem (1) if for any ζ C 1 ([0, T ]; C 1 c (R)), ζ(, T ) = 0 in R we have Ω T [ ur ζ t + ϕ(u r )ζ x ] dxdt + T 0 u s(, t), ζ t (, t) R dt + C ϕ T 0 u s(, t), ζ x (, t) R dt = u 0, ζ(, 0) R.

27 A measure u L (0, T ; M + (R)) is a weak solution of the Cauchy problem (1) if for any ζ C 1 ([0, T ]; C 1 c (R)), ζ(, T ) = 0 in R we have Ω T [ ur ζ t + ϕ(u r )ζ x ] dxdt + T 0 u s(, t), ζ t (, t) R dt + C ϕ T 0 u s(, t), ζ x (, t) R dt = u 0, ζ(, 0) R. A weak solution of problem (1) is called entropy solution, if for any ζ 0 as above and for any admissible couple (E, F ) we have Ω T [E(u r )ζ t + F (u r )ζ x ] dxdt + C E T 0 u s(, t), ζ t (, t) R dt + +C F T 0 u s(, t), ζ x (, t) R dt R E(u 0r )ζ(x, 0) dx C E u 0s, ζ(, 0) R.

28 Theorem (Bertsch, Smarrazzo, T., Tesei, 2016) Let us assume (H 1 ) problem (1) admits a weak solution. Moreover, if there are not intervals in which the flux ϕ is affine, the solution is also an entropy solution.

29 Theorem (Bertsch, Smarrazzo, T., Tesei, 2016) Let us assume (H 1 ) problem (1) admits a weak solution. Moreover, if there are not intervals in which the flux ϕ is affine, the solution is also an entropy solution. The solution is obtained by using an approximation of the initial data with u 0n L 1 (R) L (R), such that the L 1 is uniformly bounded.

30 Theorem (Bertsch, Smarrazzo, T., Tesei, 2016) Let us assume (H 1 ) problem (1) admits a weak solution. Moreover, if there are not intervals in which the flux ϕ is affine, the solution is also an entropy solution. The solution is obtained by using an approximation of the initial data with u 0n L 1 (R) L (R), such that the L 1 is uniformly bounded. If the flux function ϕ admits affine intervals, it is possible to prove that the solution satisfied a proper measure valued entropy formulation.

31 Remark: we can always assume to be in the case C ϕ = 0. In fact it is enough to consider function ϕ(z) = ϕ(z) C ϕ z, observing that if u is a weak solution (entropy solution) of the problem with flux ϕ then ũ(, t) := T Cϕtu(, t) is a weak solution (entropy solution) of the problem with ϕ.

32 Remark: we can always assume to be in the case C ϕ = 0. In fact it is enough to consider function ϕ(z) = ϕ(z) C ϕ z, observing that if u is a weak solution (entropy solution) of the problem with flux ϕ then ũ(, t) := T Cϕtu(, t) is a weak solution (entropy solution) of the problem with ϕ. Main consequence of the entropy formulation Proposition If u is an entropy solution (C ϕ = 0) u s (t 2 ) u s (t 1 ) for every 0 t 1 t 2.

33 Remark: we can always assume to be in the case C ϕ = 0. In fact it is enough to consider function ϕ(z) = ϕ(z) C ϕ z, observing that if u is a weak solution (entropy solution) of the problem with flux ϕ then ũ(, t) := T Cϕtu(, t) is a weak solution (entropy solution) of the problem with ϕ. Main consequence of the entropy formulation Proposition If u is an entropy solution (C ϕ = 0) u s (t 2 ) u s (t 1 ) for every 0 t 1 t 2. It easy to see that entropy formulation is not enough to assure uniqueness.

34 Remark: we can always assume to be in the case C ϕ = 0. In fact it is enough to consider function ϕ(z) = ϕ(z) C ϕ z, observing that if u is a weak solution (entropy solution) of the problem with flux ϕ then ũ(, t) := T Cϕtu(, t) is a weak solution (entropy solution) of the problem with ϕ. Main consequence of the entropy formulation Proposition If u is an entropy solution (C ϕ = 0) u s (t 2 ) u s (t 1 ) for every 0 t 1 t 2. It easy to see that entropy formulation is not enough to assure uniqueness. It is necessary to give a condition for u r near supp (u s ).

35 Remark: we can always assume to be in the case C ϕ = 0. In fact it is enough to consider function ϕ(z) = ϕ(z) C ϕ z, observing that if u is a weak solution (entropy solution) of the problem with flux ϕ then ũ(, t) := T Cϕtu(, t) is a weak solution (entropy solution) of the problem with ϕ. Main consequence of the entropy formulation Proposition If u is an entropy solution (C ϕ = 0) u s (t 2 ) u s (t 1 ) for every 0 t 1 t 2. It easy to see that entropy formulation is not enough to assure uniqueness. It is necessary to give a condition for u r near supp (u s ). Simple case ϕ bounded, ϕ > 0 (as in the toy case ) (C s ) x 0 supp u s (, t) ess lim u r (x, t) = + q.o. t (0, T ) x x + 0

36 (H 2 ) L 1, K R : ϕ (u)[lϕ(u) + K] [ϕ (u)] 2 < 0

37 (H 2 ) L 1, K R : ϕ (u)[lϕ(u) + K] [ϕ (u)] 2 < 0 Assume (H 2 ) then solution obtained by approximation satisfies (C s ).

38 (H 2 ) L 1, K R : ϕ (u)[lϕ(u) + K] [ϕ (u)] 2 < 0 Assume (H 2 ) then solution obtained by approximation satisfies (C s ). Under condition (H 2 ) we have uniqueness for a class of data in which n the singular part is purely atomic (u 0s = c i δ xi ). i=1

39 (H 2 ) L 1, K R : ϕ (u)[lϕ(u) + K] [ϕ (u)] 2 < 0 Assume (H 2 ) then solution obtained by approximation satisfies (C s ). Under condition (H 2 ) we have uniqueness for a class of data in which n the singular part is purely atomic (u 0s = c i δ xi ). Moreover for general initial data we can prove that the waiting time is strictly positive if and only if ϕ is bounded (Bertsch, Smarrazzo, T., Tesei, 2016). i=1

40 General situation Solve Riemann problem with a delta, more precisely data u 0 is such that { a for x < 0 u 0r = b for x 0, u 0s = δ 0 Solving by approximation as in the toy example we see that if the δ persists for positive time than the values u := u r (0, ) B, u + := u r (0+, ) B +

41 General situation Solve Riemann problem with a delta, more precisely data u 0 is such that { a for x < 0 u 0r = b for x 0, u 0s = δ 0 Solving by approximation as in the toy example we see that if the δ persists for positive time than the values u := u r (0, ) B, u + := u r (0+, ) B + B := {s [0, + ) : ϕ(s) ϕ(k) for every k (s, + )} {+ }, B + := {s [0, + ) : ϕ(s) ϕ(k) for every k (s, + )} {+ }.

42 General situation Solve Riemann problem with a delta, more precisely data u 0 is such that { a for x < 0 u 0r = b for x 0, u 0s = δ 0 Solving by approximation as in the toy example we see that if the δ persists for positive time than the values u := u r (0, ) B, u + := u r (0+, ) B + B := {s [0, + ) : ϕ(s) ϕ(k) for every k (s, + )} {+ }, B + := {s [0, + ) : ϕ(s) ϕ(k) for every k (s, + )} {+ }. In some sense these correspond to boundary conditions + on the left right hand side of the point x = 0 in the sense of Bardos-Le Roux-Nedelec.

43 General situation Solve Riemann problem with a delta, more precisely data u 0 is such that { a for x < 0 u 0r = b for x 0, u 0s = δ 0 Solving by approximation as in the toy example we see that if the δ persists for positive time than the values u := u r (0, ) B, u + := u r (0+, ) B + B := {s [0, + ) : ϕ(s) ϕ(k) for every k (s, + )} {+ }, B + := {s [0, + ) : ϕ(s) ϕ(k) for every k (s, + )} {+ }. In some sense these correspond to boundary conditions + on the left right hand side of the point x = 0 in the sense of Bardos-Le Roux-Nedelec. Heuristically we have uniqueness for initial data for which u 0s is of n the type c i δ xi. i=1

44 In order to prove uniqueness we have to overcome the problem of the existence of the trace.

45 In order to prove uniqueness we have to overcome the problem of the existence of the trace. Suppose that at initial time there is a singular part in x 0 of the type Aδ x0 and that the singular part (in x 0 ) persists until time t. Regarding the condition for the trace of u r near x 0 it is useful to adapt the methods introduced by Otto of the boundary entropy entropy flux pair.

46 In order to prove uniqueness we have to overcome the problem of the existence of the trace. Suppose that at initial time there is a singular part in x 0 of the type Aδ x0 and that the singular part (in x 0 ) persists until time t. Regarding the condition for the trace of u r near x 0 it is useful to adapt the methods introduced by Otto of the boundary entropy entropy flux pair. Considering the boundary condition + along the point x 0, we have the following integral conditions ess lim x x 0 t 0 t sgn (u r (x, t) k)(ϕ(u i r (x, t) ϕ(k))α(t) dt 0 ess lim x x for every k 0, for every α C([0, t]), α 0. sgn (u r (x, t) k)(ϕ(u i r (x, t) ϕ(k))α(t) dt 0

47 Moreover using weak formulation, we must define the evolution of the delta as A(t)δ x0, where t A(t) = A lim ϕ(u(x 0 + δ, t)) ϕ(u(x 0 δ, t)) dt for (0, t]. δ 0 0 A(t) = 0.

48 Moreover using weak formulation, we must define the evolution of the delta as A(t)δ x0, where t A(t) = A lim ϕ(u(x 0 + δ, t)) ϕ(u(x 0 δ, t)) dt for (0, t]. δ 0 0 A(t) = 0. Using previous boundary integral condition it is easy to check that A(t) is nonincreasing.

49 Moreover using weak formulation, we must define the evolution of the delta as A(t)δ x0, where t A(t) = A lim ϕ(u(x 0 + δ, t)) ϕ(u(x 0 δ, t)) dt for (0, t]. δ 0 0 A(t) = 0. Using previous boundary integral condition it is easy to check that A(t) is nonincreasing. We can prove existence and uniqueness in the class of data in which n the singular part is purely atomic, u 0s = c i δ xi, for any bounded flux that satisfies H 1 (Bertsch, Smarrazzo, T., Tesei, 2017). i=1