Finer regularity of an entropy solution for 1-d scalar conservation laws with non uniform convex flux

Finer regularity of an entropy solution for 1-d scalar conservation laws with non uniform convex flux

ADIMURTHI(*) - SHYAMSUNDARGHOSHAL(**) - G.D. VEERAPPAGOWDA(***)

ABSTRACT- Consider a scalar conservation law in one space dimension with initial data inL¹:If the fluxf is inC²and locally uniformly convex, then for allt>0, the entropy solution is locally in BV (functions of bounded variation) in space variable.

Inthis case it was shown in [5], that for all most everyt>0, locally, the solution is in SBV (Special functions of bounded variations). Furthermore it was shown with an example that for almost everywhere int>0cannot be removed. This paper deals with the regularity ofthe entropysolutions ofthe strict convexC¹fluxfwhichneed not beinC²and locallyuniformlyconvex. Inthiscase, the entropy solutionneed not be locally in BV in space variable, but the composition with the derivative of the flux function is locally in BV. Here we prove that, this composition is locally is in SBV on all most everyt>0. Furthermore we show that this is optimal.

MATHEMATICSSUBJECTCLASSIFICATION(2010). 35B65, 35L65, 35L67.

KEYWORDS. Hamilton-Jacobi equation, scalar conservation laws, characteristic lines, sbv functions, uniformly convex.

1. Introduction

Let f :R!R be a locally Lipschitz continuous function and u02 L¹(R): Consider the following scalar conservation law in one space di- (*) Indirizzo dell'A.: Center for Applicable Mathematics, Tata Institute of Fundamental Research, PO Bag No. 6503, Bangalore-560065, India.

E-mail: aditi@math.tifrbng.res.in

(**) Indirizzo dell'A.: Laboratoire de MatheÂmatiques, Universite de Franche- ComteÂ, 25030 BesancËon, France.

E-mail: ssghosha@univ-fcomte.fr

The second author would like to thank French ANR project ``CoToCoLa'' for the support.

(***) Indirizzo dell'A.: Center for Applicable Mathematics, Tata Institute of Fundamental Research, PO Bag No. 6503, Bangalore-560065, India.

E-mail: gowda@math.tifrbng.res.in

mension

@u

@t ‡ @

@xf(u)ˆ0 if x2R; t>0:

(1:1)

u(x;0)ˆu0(x) if x2R:

A functionu2L¹(RR‡) is said to be a solution of (1.1) ifuis a weak solution of (1.1) satisfying Kruzkov [14] entropy condition. By using van- ishing viscosity, Kruzkov [14] proved that there exists a unique entropy solution to (1.1).

Under the convexity assumption on the flux, Lax and Oleinik [12] have obtained an explicit formula for the solution. In order to state their result, let us recall some definitions and their properties of convex functions [4].

Let f :R!Rbe a convex function. Then

(i) f is said to be a strictly convex function if for allx6ˆya2(0;1), f(ax‡(1 a)y)<af(x)‡(1 a)f(y):

(ii) f 2C²(R) is said to be locally uniformly convex if for all compact set KR, there exist a constant C(K)>0 such that f⁰⁰(x)C(K) for all x2K. It is said to be uniformly convex inRif there existC>0 such that

f⁰⁰(x)Cfor allx2R.

(iii) f is said to be of super linear growth if

juj!1lim f(u)

juj ˆ 1:

(iv)Associate to a convex function f;letfbe its convex dual defined by f(p)ˆsup

q fpq f(q)g:

Then for anyq2R;

jpj!1lim f(p)

jpj lim

jpj!1 q p jpj

f(q) jpj

ˆ lim

jpj!1q p jpj sinceqis arbitrary and lettingjqj ! 1to obtain

jpj!1lim f(p)

jpj ˆ 1:

(v) Letfbe inC¹(R) and strictly convex. Thenf⁰is a strictly increasing function. Furthermore, it can be shown that if f is of super linear growth, then f ˆf; (f)⁰ˆ(f⁰) ¹. Therefore f is in C¹(R) with super linear

growth and is strictly convex. With these preliminaries, we recall the fol- lowing (see [12]):

THEOREM1.1. (Lax and Olenik) Assume that f is in C¹(R);strictly convex and of super linear growth. Then there exist a function y(x;t)such that x7!y(x;t)is a non decreasing function and for all t>0, a.e. x2R;

the solution u of(1.1)is explicitly given by f⁰(u(x;t))ˆx y(x;t)

t :

(1:2)

As an immediate consequence of this theorem, we have (1)For each t>0;x7!f⁰(u(x;t))2BV_loc(R):

(2)If f 2C²and locally uniformly convex, then for all t>0, x7!u(x;t) is in BVloc(R):

In this context, Alberto Bressan raised the following question.

Problem 1.Assume that for allt>0;x7!u(x;t)2BV_loc(R);then is it in SBV_loc(R)?

Under the suitable condition on f;Ambrosio-De Lellis [5] proved the following

THEOREM 1.2. (Ambrosio-De Lellis) Assume further that f 2C²(R) and locally uniformly convex. Then there exist a countable set S(0;1) such that for all t2=S; x7!u(x;t)2SBV_loc(R):Furthermore, they gave an example for which S6ˆf:

Observe that from the explicit formula, if f^{0 1} is a locally Lipschitz continuous, then the solution is in BVloc(R) in space variable which is guaranteed iffisC²and locally uniformly convex. This is what assumed in [5]. In [7], the condition of locally uniformly convexity has been relaxed to prove the same result as in [5] provided the zero set off⁰⁰is countable andu is in BVloc:In [15] even the convexity condition is relaxed under the hy- pothesis that f isC²and the zero set of f⁰⁰is countable. For further gen- eralization see [6], [8], [9], [10], [16].

Now the natural question is ``What happens if f 2C¹and satisfying the hypothesis of Lax-Olenik Theorem?'' In this case, the problem of Bressan can be reformulated as follows.

Problem 2.

(i) Doesx7!u(x;t)2BV_loc(R) for allt>0.

(ii) Doesx7!f⁰(u(x;t))2SBV_loc(R) for a.e.t>0:

(iii) If (ii) is true, does there exist a u02L¹(R) and a countable set Sˆ ftngsuch that f⁰(u(;tn))62SBVloc(R) for alln:

In view of (1.2), the main result of this paper is to show that the Problem 2 has an affirmative answer and we have the following

THEOREM1.3. Let f satisfies the hypothesis of Theorem1.1 andube the solution of (1.1). Then

(1) There exist a countable set S(0;1) such that for all t62S;x7!f⁰(u(x;t))2SBV_loc(R):

(2) Assume that f(0)ˆf⁰(0)ˆ0 and there exist c>0 and0<g<1 such that for alla0b;f satisfies

f^{0 1}(b) f⁰¹(a)c(b a)^g:

Then there exist a u02L¹(R) such that for all t>0, x7!u(x;t)62 BV_loc(R):

(3)There exist a u₀2L¹(R)and a decreasing sequenceft_ngsuch that x7!f⁰(u(x;t_n))62SBV_loc(R):

For (1), we follow the similar path as in [5]. The idea in the proof of (2) lies in the construction of asymptotically single shock packets as in [1]. The proof of (3) lies in the backward construction (see Appendix) as in [2], [3].

REMARK1.4. There are many fluxes satisfying the hypothesis of (2).

For example, letp>2;then f(u)ˆjuj^p

p is one of such function.

2. Preliminaries

Let f be as in Theorem 1.1. Letu02L¹(R):LetA2Rand define v₀(x)ˆ

Z^x

A

u₀(u)du (2:1)

f(p)ˆmax

q fpq f(q)g (2:2)

the convex dual of f. Then from the hypothesis of f, it follows that f2C¹(R);strictly convex and having super linear growth.

Lett>0;x2R;define v(x;t)ˆmin

y v₀(y)‡tf x y t

n o

: (2:3)

Then we have the following (see [12])

THEOREM 2.1. (Hopf): v2Lip(RR_‡) and is the unique viscosity solution of the following Hamilton Jacobi equation.

@v

@t‡f @v

@x ˆ0 if x2R; t>0

(2:4)

v(x;0)ˆv0(x) if x2R furthermore it satisfies

(i) Dynamic programming principle: Let0s<t:then v(x;t)ˆmin

y2R v(y;s)‡(t s)f x y t s

n o

(2:5)

(ii) There exist a M>0 depending on u₀ and f such that for all minimizers of y of(2:5)

x y t s

M:

(2:6)

Next we define the different notions of characteristics and their properties without proof.

DEFINITION2.2. Let 0s<t;x2R:Then

(i) Characteristic setsch(x;s;t);ch(x;t) are defined by ch(x;s;t) ˆ fy2R; yis a minimizer in 2:5)g ch(x;t) ˆ ch(x;0;t):

(ii) y2ch(x;s;t) is called a characteristic point and the line joining (x;t) and (y;s) is called a characteristic line segment.

(iii) Extreme characteristic points

y‡(x;s;t) ˆ maxfy2ch(x;s;t)g y (x;s;t) ˆ minfy2ch(x;s;t)g y(x;t) ˆ y(x;0;t)

(iv) The characteristic line segment associated toy(x;s;t) are called extreme characteristic line segments.

DEFINITION 2.3. (proper intersection) Let 0s_i<t_i and for iˆ1;2;r_iˆ[s_i;t_i]!Rbe affine maps. Thenr₁andr₂are said to intersect properly if there exist a u₀2( max (s₁;s₂);min (t₁;t₂)) such that r1(u0)ˆr2(u0):

Properties of Characteristics(see [12])

(1) No two distinct characteristic line segments intersect properly.

(2) Let y₁ 2ch(x;s;t) and y₂2ch(y₁;s): Then y₂2ch(x;t) and (x;t), (y₁;s), (y₂;0) are collinear. That is

x y2

t ˆx y1

t s ˆy1 y2

s : (2:7)

(3) x7!y(x;s;t) are non decreasing functions and at points of con- tinuityxofy_‡(;s;t)

y‡(x;s;t)ˆy (x;s;t):

(2:8)

(4) Lax-Oleinik explicit formula: Let u(x;t)ˆ@v

@x(x;t); then u is the unique solution of (1.1) satisfying

f⁰(u(x;t))ˆx y_‡(x;s;t)

t s for a:e:x2R:

(2:9)

DEFINITION2.4 (characteristic curve). Let 0s<tanda2R. Define the characteristic curvej(t;s;a) passing throughaby

j(t;s;a)ˆinffj2R;y (j;s;t)ag:

(2:10)

Then we have the following LEMMA2.5. Let0s<t, then

(1) Let x_n!x₀;y_n2ch(x_n;s;t)such that y_n!y₀. T hen y₀2ch(x₀;s;t).

(2) lim

x"x0y‡(x;s;t)ˆy (x0;s;t)andlim

x#x0y (x;s;t)ˆy‡(x0;s;t):

y‡(x;s;t)<y (j(t;s;a);s;t)if x<j(t;s;a) (2:11)

y (x;s;t)y_‡(j(t;s;a);s;t)if x>j(t;s;a) (2:12)

y (j(t;s;a);s;t)ay‡(j(t;s;a);s;t):

(2:13) (3)

(4) t7!j(t;s;a)is a Lipschitz continuous function with limt!sj(t;s;a)ˆa:

(2:14)

PROOF. From (2.5), we have for ally,

v(x_n;t) ˆ v(y_n;s)‡(t s)f x_n y_n t s

v(y;s)‡(t s)f xn y t s

: Sincevis continuous, hence lettingn! 1to obtain

v(x0;t) ˆ v(y0;s)‡(t s)f x₀ y₀ t s

v(y;s)‡(t s)f x₀ y t s

: Hencey₀2Ch(x₀;x;t):This proves (1).

From the non proper intersection of characteristic line segments it follows that if x<x₀, then y_‡(x;s;t)y (x₀;s;t). Let x_n"x₀ and y‡(xn;s;t)!y0. Then from (1),y02ch(x0;s;t) and hence

y (x0;s;t)y0ˆlimy‡(xn;s;t)y (x0;s;t):

This proves (2).

From (2) and definition ofj(;s;a) gives

y‡(x;s;t)<y (j(t;s;a);s;t) if x<j(t;s;a) y (x;s;t)y‡(j(t;s;a);s;t) if x>j(t;s;a):

Choose a sequence x_n<j(t;s;a) such that x_n"j(t;s;a) and choose a n₀>0 such that for alln>n₀;

y (xn;s;t)ˆy‡(xn;s;t):

Sincey (x_n;s;t)<aand hence from (2) we have y (j(t;s;a);s;t) ˆ lim

xn"j(t;s;a)y‡(xn;s;t)

ˆ lim

xn"j(t;s;a)y (xn;s;t) a:

Similarly other inequality holds and this proves (3).

Letst₁<t₂and letybe the extreme characteristic points andr(u) be the characteristic line segments at (j(t2;s;a);t2). Thengare given by

g(u)ˆj(t2;s;a)‡(u t2)j(t2;s;a) y

t2 s:

(2:15)

From (2.13) and non proper intersection of characteristic line segments, it follows that for alls<t1<t2,

g (t₁)j(t₁;s;a)g_‡(t₁):

(2:16)

Hence from (2.6), (2.15) and (2.16) we have

jj(t₁;s;a) j(t₂;s;a)j 2Mjt₂ t₁j jj(t;s;a) yj Mjt sj

which proves (2.15) and hence (4). This proves the Lemma. p We follow the same notations as in [5]. Fort>0;x2R;u2[0;t] define

m(x;t) ˆ x y(x;t) t

g(u) ˆ x‡m(x;t)(u t) I(x;t) ˆ (y (x;t);y_‡(x;t))

C(x;t) ˆ f(j;u);0<u<t;g (u)<j<g_‡(u)g D(t) ˆ fx2R;y‡(x;t) has a jump atxg

I(t) ˆ S

x2D(t)I(x;t)

c(t) ˆ S

x2D(t)C(x;t) F(t) ˆ jI(t)j:

WherejI(t)jdenotes the Lebesgue measure ofI(t). Then we have the fol- lowing

LEMMA 2.6. Let 0<s<t;j;x₁;x₂2R; x₁6ˆx₂ and (j;s)2C(x;t).

Then

I(x1;t)\I(x2;t)ˆf; C(x1;t)\C(x2;t)ˆf (2:17)

I…j;s† I…x;t†;C…j;s† C…x;t†

(2:18)

I(s)I(t) (2:19)

PROOF. (2.17) and (2.18) follows from the non proper intersection of characteristic line segments. Let x2D(s);j₀ˆj(t;s;x) and yˆ y(j₀;s;t). Then from (2.13), y xy_‡. Suppose y ˆx, then from (2.7), the points (j₀;t);(x;s);(y_‡(x;s);0) lie on a straight line and also the points (j₀;t);(x;s);(y (x;s);0) lie on a straight line. Hencey‡(x;s)ˆy (x;s)

which contradicts the fact thatx2D(s). Hencey <x<y‡and this implies y (j₀;t)y (x;s)<y‡(x;s)y‡(j₀;t). This proves (2.19) and hence the

Lemma. p

LEMMA2.7. Let t>0;x2R. Assume that at(x;t);y(x;t)satisfies the following hypothesis.

(i) y_‡(x;t)ˆy (x;t).

(ii) Leth>0;x h62D(t). Assume one of the following holds lim_h!0

x h62D(t)

y_‡(x;t) y_‡(x h;t)

h ˆ 1; lim_h!0

x‡h62D(t)

y_‡(x‡h;t) y_‡(x;t)

h ˆ 1:

(2:20)

Then for all t<tand z2R,

x62ch(z;t;t):

(2:21)

PROOF. Suppose not, then there exist a z02R;t0>t such that x2ch(z0;t;t0). Sincey‡(x;t)2ch(x;t) and hence from (2.7) we have

x y‡(x;t)

t ˆz0 x t₀ t: (2:22)

Letaˆz₀ x

t₀ t be this common value. Let h(z)ˆ

Z^z

A

u(u;t)du‡(t₀ t)f z0 z t₀ t

: (2:23)

Then from (2.5), we have

h(x)ˆmin

z2R h(z):

(2:24)

Without loss of generality we can assume that lim_n!0

x n62D(t)

y_‡(x;t) y_‡(x h;t)

h ˆ 1

and similar argument hold if the other limit holds. Let^>1‡ t t₀ tand h₀>0 such that for allu2(0;h₀)nD(t)

y‡(x u;t)>^u y‡(x;t) (2:25)

Sincef⁰ ˆ(f⁰) ¹, is a strictly increasing function, hence forh2(0;h₀), and from (2.9), (2.2) and (2.25) we have

h(x) h(x h)ˆ Z^x

A

u(u;t)du Z^{x h}

A

u(u;t)du

‡(t0 t)f z0 x t0 t

(t0 t)f z0 x‡h t0 t

ˆ Z^h

0

u(x u;t)du‡(t0 t) f(a) f a‡ h

t0 t

ˆ Z^h

0

f⁰ x u y‡(x u;t) t

du‡(t0 t) f(a) f a‡ h

t0 t

Z^h

0

f⁰

x y‡(x;t)

t ‡^ 1

t u

du

‡ (t0 t) f

a) fa‡ h t0 t

ˆ t

^ 1

f(a‡^ 1 t

h

f(a)

‡ (t0 t) f(a) f a‡ h

t0 t

:

Let g(s)ˆf(a‡s) f(a), then g(0)ˆ0 and g is a strictly convex function. Let a>b>0, then we have g(b)ˆgb

aa

<

1 b

a g(0)‡ b

ag(a)ˆb

ag(a). Hence g(b) b <g(a)

a . L et h>0;aˆ ^ 1 t

h> n t₀ tˆb.

Therefore fromg(b) b <g(a)

a implies that t

h(^ 1) f a‡ ^ 1 t

h

f(a)

>t₀ t

h f a‡ h t0 t

f(a)

This implies that h(x)>h(x h) contradicting (2.24). This proves the

Lemma. p

COROLLARY2.8. Let(x;t)satisfies the hypothesis of lemma(2.3)and t>t. Letj₀ˆj(t;t;x), then(x;t)2C(j₀;t).

PROOF. From (2.13) and (2.21) it follows that y (j₀;t;t)<x<y‡(j₀;t;t):

From (2.7) it follows that (y (j₀;t;t);y‡(j₀;t;t)) ftg C(j₀;t) and hence (x;t)2C(j₀;t). This proves the corollary. p PROOF OF THEOREM 1.3. In view of the finite speed propagation, without loss of generality, assume thatu0 is as follows:

u₀(x)ˆ

u if x<A u₀(x) if x2(A;B) u‡ if x>B 8<

(3:1) :

whereuare constants andu₀2L¹(A;B). Let mˆinfu0; Mˆsupu0

(3:2)

a…t† ˆA‡tf⁰…m†;b…t† ˆB‡tf⁰…M†

(3:3)

then it follows that

u(x;t)ˆ u if x<a(t) u‡ if x>b(t)

Hence from (2.9)

y_‡(x;t)ˆy (x;t)ˆ x tf⁰(u ) if x<a(t) x tf⁰(u_‡) if x>b(t):

(3:4)

As a consequence of this we have

LEMMA3.9. For t>0;F(t)<1and is a nondecreasing function.

PROOF. From (3.4),D(t)[a(t);b(t)] and hence F(t)ˆ jI(t)j ˆ X

x2D(t)

(y_‡(x;t) y (x;t))b(t) a(t)<1:

From (2.19),Fis a non decreasing function and this proves the Lemma.

Letdy(;t) be the Distributional derivatives ofx7!y(x;t). Thendy(;t) is a non negative measure. Decomposedy(;t) as follows:

dy(;t)ˆftdx‡dnt‡dm_t

where the measures are mutually singular, withft2L¹_loc(R);dntis the jump

part and m_t is the Cantor part (see [5]). From (3.4) it follows that the support ofdm_t;dnt are in [a(t);b(t)] anddy(:;t)ˆdxforx62[a(t);b(t)]. p For a measurable setAR, denote jAj to be its Lebesgue measure.

Then from the Besicovitch differentiation Theorem (see [13]) it follows that there exists a setE_t[a(t);b(t)] with the property:

jE_tj ˆ0; E_t\D_tˆf; m_t(R)ˆm_t(E_t) (3:5)

and form_t a.e.x2E_t

h!0;xhlim2=D(t)

2h‡nt[x h;x‡h]

m_t([x h;x‡h]) ˆ0:

(3:6)

To prove theorem (1.3), we need to replace the symmetric limit by one sided limit in (3.6) and is given by the following

LEMMA3.10. Form_t a:e:x2Et, we have (3:7) 1 ˆ lim

h!0;x h=2D(t)

y‡(x;t) y‡(x h;t)

h ˆ lim

h!0;x‡h2=D(t)

y‡(x‡h;t) y‡(x;t)

h :

To prove this lemma, one uses a covering lemma different from that of Besicovitch covering lemma. For the sake of completeness, proof of this has been given in the appendix.

LEMMA3.11. Let t >0, then there exist a constant C>0such that for allt>t,

F(t) F(t)Cm_t(E_t):

(3:8)

PROOF. Forx2R, denotej₀(x)ˆj(t;t;x). Letx2E_tsatisfying (3.5) and (3.6). Then from (3.6), (3.7), corollary (2.1), chooseh(x) small such that xh(x)62D(t) and

‰x h…x†;x‡h…x†Š ftg C…j₀…x†;t†

(3:9)

nt([x h(x);x‡h(x)])1

2m_t([x h(x);x‡h(x)]) (3:10)

Since

nt([x h(x);x‡h(x)])ˆ X

z2[x h(x);x‡n(x)]\D(t)

jI(z;t)j

therefore from (2.17), (2.18), (3.9) and (3.10), we have jI(j₀(x);t)j P

(j;t)2C(j0(x);t)\(D(t)ftg)jI(j;t)j

(y_‡(x‡h(x);t) y (x h(x);t)) P

z2[x h(x);x‡h(x)]\D(t)ftg)I(z;t) m_t([x‡h(x);x h(x)] nt([x h(x);x‡h(x)])

1

2m_t([x h(x);x‡h(x)]):

(3:11)

Nowf[x h(x);x‡h(x)]g_x2EtcoversEtand hence by Besicovitch covering Lemma, there exists a countable disjoint set of intervalsf[x_j h_j;x_j‡h_j]g which satisfies (3.11) and

X

j

m_j[x_j h_j;x_j‡h_j]Cm_t(E_t) for someC>0:Therefore

F(t) F(t) X

j

jI(j₀(xj));t)j X

j

X

(j;t)2C((j0(xj);t)\(D(t)ftg)

jI(j;t)j

1 2

X

j

m_t[(x_j h_j;x_j‡h_j])

C 2m_t(Et):

This proves (3.8) and hence the Lemma. p

Before going to the next Lemma we need the following:

Let f be as in Theorem (1.3) with f(0)ˆf⁰(0)ˆ0: Since f(u)=u is a strictly increasing function we therefore haveuf⁰(u) f(u)>0 foru6ˆ0:

Since f(0)ˆf⁰(0)ˆ0;hence choosea<0 andb₀>0 such that b₀f⁰(b₀) f(b₀)>af⁰(a) f(a)

and let g(u)ˆuf⁰(u) f(u) af⁰(a)‡f(a): Then g(0)<0 and g(b₀)>0:

Hence there exist a b2[0;b₀) such that g(b)ˆ0: That is there exist a<0<bsuch that

bf⁰(b) af⁰(a)ˆf(b) f(a)>0 (3:12)

Let a<0<b be chosen as above and mˆf(b) f(a)

b a : Let A<B;

g₁(t)ˆA‡tf⁰(b);g₂(t)ˆB‡tf⁰(a) and g₁; and g₂ intersect at

Tˆ B A

f⁰(b) f⁰(a): Clearly if Xˆg₁(T)ˆg₂(T); then A<X<B. L et CˆA‡T(f⁰(b) m)ˆB‡T(f⁰(a) m) and g₃(t)ˆC‡tm then by convexity,A<C<Bandg₃(T)ˆX:Furthermore we have the following identity

b(C A)‡a(B C) ˆ T[b(f⁰(b) m) a(f⁰(a) m)]

ˆ T[bf⁰(b) af⁰(a)‡f(a) f(b)]

ˆ 0 (3:13)

Now defineu_0;A;B by

u_0;A;B(x)ˆ

0 if x62[A;B]

b if A<x<C a if C<x<B 8>

<

>: (3:14)

Then from (3.13), we have Z^B

A

u_0;A;B(x)dxˆb(C A)‡a(B C)ˆ0 (3:15)

Let u_A;B be the solution of (1.1) with initial data u_0;A;B: Then from the choice ofa;b andCfor 0<t<T; u_A;B is given by

u_A;B(x;t)ˆ

0 if x62[A;B]

f^{0 1} x A t

if A<x<g₁(t) b if g₁(t)<x<g₃(t) a if g₃(t)<x<g₂(t) f^{0 1} x B

t

if g₃(t)<x<B 8>

>>

>>

>>

>>

>>

><

>>

>>

>>

>>

>>

>>

: (3:16)

Then we have the following

LEMMA 3.12. Under the notations as above, let j(t)ˆj(t;0;C)be the characteristic curve passing through C and defined as in (2.10) for the solution u_A;B. Then for all t>0:

j(t)ˆg₃(t) if t<T (3:17)

A<j…t†<B if t>T:

(3:18)

For t>T;uA;B is given by

u_A;B(x;t)ˆ

0 if x62(A;B)

f⁰¹ x A t

if A<x<j(t)

f⁰¹ x B t

if j(t)<x<B:

8>

>>

>>

>>

><

>>

>>

>>

>>

: (3:19)

PROOF. (3.17) follows from (3.16) and j(T)ˆX:Suppose (3.18) is not valid, then define

T1ˆsupft>T; A<j(t)<Bg:

(3:20)

Then atT₁, eitherj(T₁)ˆAorj(T₁)ˆB. Without loss of generality, we can assume thatj(T1)ˆB. Then forT<t<T1;uA;Bis given by (3.19) and A;B2ch(j(T1);T1). Therefore we have from (2.3)

v(j(T1);T1)ˆ Z^B

A

u_0;A;B(u)du‡T1f(0)ˆT1f j(T1) A T₁

(3:21)

Hence from (3:15) and (3:21), we havef(0)ˆf j(T1) A T₁

ˆf B A T₁

: Since 0ˆf⁰(f⁰(0))ˆf⁰(0) and therefore 0 is a critical point for f. This implies that B Aˆ0 which is a contradiction and this proves

the Lemma. p

COROLLARY 3.13. Let f , u_0;A;B, u_A;B are as in Lemma 3.4. Further- more assume that there exist c>0 and 0<g<1 such that for all a<0<b;

f⁰¹(b) f^{0 1}(a)c(b a)^g (3:22)

Then for all t>0

TV(u_A;B(:;t))min b a;c B A t

_g

(3:23)

PROOF. Let 0<t<T, then from (3.16)

uA;B(j(t) ;t) uA;B(j(t)‡;t)ˆb a:

Lett>T;then from (3.19) and (3.22)

u_A;B(j(t) ;t) u_A;B(j(t)‡;t) ˆ f^{0 1} j(t) A t

f^{0 1} j(t) B t

c B A t

_g

:

This proves (3.23) and hence the corollary. p

For proving (3) of the theorem (1.3), we need the following back- ward construction which was proved in [1] in obtaining the solution of control problems for conservation laws. Since the conservation law is irreversible, given a target function g2L¹ at time tˆT>0; in general there may not exist a u02L¹(R) such that the corresponding solution u of (1.1) may not coincide with g at tˆT: This is mainly because, according to the Lax-Oleinik formula, g must satisfy the compatibility condition, namely, x7!x Tf⁰(g(x)) is a nondecreasing function. Using the backward construction we can prove that it is also necessary. Basic idea of the proof is by discretization of the function x Tf⁰(g(x)) and using the convexity assumption on f, we build the approximate solution and show that this solution converges and attains the value g(x) at time tˆT:

For the sake of completeness we will give the proof of this result in the appendix.

Letr:[A1;A2]![A1;A2] be a non decreasing function. Lett0>0 and

M ˆ max

x2[A1;A2]f^{0 1x} _t^r(x)₀ g₁(t) ˆ A₁‡f⁰( M)(t t₀) g₂(t) ˆ A₂‡f⁰(M)(t t₀) Then we have the following.

LEMMA3.14 (Backward construction). Let t0>0;r;[A1;A2];g₁andg₂ are defined as above. Assume that f⁰(0)ˆ0;Then we have the following:

There exist a u₀2L¹(A₁;A₂)withku₀k₁depends onr;t₀;(A₂ A₁):With this u₀, define

u_0;A1;A2(x)ˆ 0 if x62(A₁;A₂) u0(x) if x2(A1;A2):

(

Let uA1;A2 be the entropy solution of (1.1)with u0;A1;A2 as its initial data.

Then uA1;A2 satisfies (i) Let0<tt₀;then

u_A1;A2(x;t)ˆ

0 if x62(A1;A2) f^{0 1} x r(x)

t0

if x2(A1;A2); tˆt0

8>

<

>:

(ii) Let t>t₀;then for x62(g₁(t);g₂(t));u_A1;A2(x;t)ˆ0:

PROOF OFTHEOREM1.3. (1) Letu0be as in (3.1) andube the solution of (1.1). LetF(t);m_tandE_tbe as in Lemma 3.11. From Lemma 3.9,t7!F(t) is a non decreasing function and hence there exist a countable set S(0;1) such thatF is continuous on (0;1)nS:Therefore for allt2=S, F(t‡)ˆF(t): Hence from (3.8)m_t(Et)ˆ0;for all t62S: Therefore from Lax-Olenik explicit formula (2.9), x7!f⁰(u(x;t))2SBV(R) for all t62S:

This proves (1).

(2) Letfb_ng 2(0;1) be a sequence such that for n2; 0<b_n<b_{n 1} andb_n <b_{n 1} 1

n^1=g:LetI_nˆ b_{n 1} 1 n^1=g; 1

bn 1

andu_nbe the solution constructed in Lemma with A_n ˆb_{n 1} 1

n^1=g;B_nˆb_{n 1};u_0;nˆu_0;An;Bn: Define

u₀(x)ˆ u_0;n(x) if x2I_n

0 otherwise

(

u(x;t)ˆ u_n(x;t) if x2I_n

0 otherwise

(

In view of Rankine-Hugoniot condition, uis the solution of (1.1) with datau₀:From corollary 3.2, for anyt>0;

TV(u(:;t)) X

n2

min b a;c B_n A_n t

_g

ˆ X

n2

min b a; c t^gn

n o

ˆ 1:

This proves (2).

(3) Let 0<h<1

2andM2Rbe such that(1 h)²

h >f⁰(M)> 2h 1 2h:Let g(u;t;A)ˆA‡(u t)f⁰(M) and defineft_ngand intervals fI_nˆ(A_n;B_n)g

as follows:

t0ˆ1; B0ˆ1 A0ˆ1 h

tn‡1ˆtn h^n‡1; Bn‡1ˆg(tn‡1;tn;An);An‡1ˆBn‡1 h^n‡1 Then asn! 1;we have

tn‡1 ˆ1 h h² h^n‡1!1 h

1 hˆ1 2h 1 h ˆt1:

A_n‡1ˆA₀ (h‡h². . .‡h^n‡1)f⁰(M) h^n‡1!1 h h

1 hf⁰(M)ˆA₁: Sinceh<1

2and(1 h)²

h >f⁰(M), we therefore havet1 >0 andA1 >0:

Let r_n:I_n!I_n be a non decreasing Cantor-Vitali function with r_n(A_n)ˆA_n;r_n(B_n)ˆB_n: Then from Lemma 3.5, there exist a u_0;n2 L¹(R) with support inI_n such that the solutionu_n of (1.1) satisfies

f⁰(u_n(x;t))ˆ

0 if x62In;0<ttn

x r_n(x)

tn if x2I_n;tˆt_n: 8<

(3:24) :

Since f⁰(0)ˆ0, henceun(x;tn)ˆ0 for allx62In and f⁰(un(x;tn))max

x2In

x r_n(x) tn

2h_n t1 2

t1<f⁰(M):

Henceun(x;tn)M. Sinceun(x;tn)ˆ0 forx>Bnand therefore from the method of characteristics, fort>tn;x>g(t;tn;Bn);un further satisfies

un(x;t)ˆ0 (3:25)

Now defineu₀2L¹(R) by

u₀(x)ˆ u_0;n(x) if x2I_n

0 otherwise

(

and letube the solution of (1.1). Then from (3.24) and (3.25),usatisfies

f⁰(u(x;t))ˆ

f⁰(u_n(x;t)) if x2I_n; t2(0;t_n) x r_n(x)

t if x2In; tˆtn: 8>

<

>:

Since r_n62SBV_loc(R), hence f⁰(u(x;tn))62SBV_loc(R) for all n. This proves (3) and hence the theorem.

4. Appendix

PROOF OF LEMMA 3.5. Let C1<C2 and r₀ be any non decreasing function. Then there exists u₀2L¹(R);u2L¹(RR_‡)such that u is the solution of (1.1)with initial data u₀. Moreover u satisfies

f^p(u(x;T))ˆx r₀(x) T

PROOF. Without loss of generality we can assume thatr₀:[C₁;C₂]! [C₁;C₂] is a non decreasing left continuous function with r₀(C_i)ˆC_i for iˆ1;2.

Step 1: Discretization of r₀: L et n1 and C₁ˆy₀<y₁ <

<ynˆC2 be such that jyi yi‡1j 1=n. From the left continuity, define C1ˆx0x1 xnˆC2 by r₀¹[y0;y_i]ˆ[x0;x_i] and

r_n(x)ˆ

x if x2=[C₁;C₂]

y0x_[y0;y1](x)‡ⁿP¹

iˆ1yix_(yi;yi‡1](x) if x2[C1;C2]:

8<

: Then forx2R,

jr_n(x) r₀(x)j 1=n:

Step 2: Letf(_f)ˆmin

x2R f(x) and definea_i;n;b_i;n;s_i;n;a_i;n(t);b_i;n(t);s_i;n(t), (See Figure 1) by

Figure 1.

f⁰(ai;n) ˆ y_i x_i T f⁰(b_i;n) ˆ y_i x_i‡1

T

f⁰(s_i;n) ˆ f(b_i;n) f(a_i;n) b_i;n a_i;n a_i;n(t) ˆy_i;n‡f⁰(a_i;n)(t T) b_i;n(t) ˆy_i;n‡f⁰(b_i;n)(t T) s_i;n(t) ˆy_i;n‡f⁰(s_i;n)(t T):

Now f⁰(ai;n)ˆy_i x_i

T y_i x_i‡1

T ˆf⁰(bi;n) and f⁰(bi;n)ˆy_i x_i‡1

T

y_i‡1 x_i‡1

T ˆf⁰(a_i‡1;n). Henceb_i;nminfa_i;n;a_i‡1;ng and from convexity, a_i;ns_i;n b_i;n. Therefore for all 0tT; a_i;n(t)s_i;n(t)b_i;n(t). Let si;n(0)ˆaiand for (x;t)2R[0;T], define

f⁰(un(x;t))ˆ

f⁰(ai;n) if ai;n(t)<x<si;n(t) f⁰(b_i;n) if s_i;n(t)<x<b_i;n(t) x_i‡1 y

t if b_i;n(t)<x<a_i‡1;n(t):

8>

><

>>

:

Sincef⁰(_f)ˆ0;a_0;nˆa_n;nˆ_f, we have

f⁰(u_n(x;t))ˆ0 if x2=[C₁;C₂]; 0tT:

Then clearlyunsatisfies

u_nt‡f(u_n)_x ˆ 0 R(0;T)

f⁰(u_n(x;T)) ˆ

x r_n(x)

T if C₁<x<C₂ 0 if x2=[C₁;C₂]:

8<

:

Also

un(x;0)ˆ

ai;n if ai;n<x<si;n(0);

b_i;n if s_i;n(0)<x<a_i‡1;n f if x2=[C1;C2]:

8>

<

>: Hence

jf⁰(un(x;0))j Maxfjf⁰(ai;n)j;jf⁰(bi;n)jg jC2 C1j

T

and

TV(f⁰(un(x;0)))X

jf⁰(ai;n) f⁰(bi;n)j ‡X

jf⁰(bi;n) f⁰(ai‡1;n)j

ˆX x_i x_i‡1 T

‡X y_i y_i‡1 T

2(C₂ C₁)

T :

Hence from Helley's theorem, there exist a subsequence still denoted ff⁰(un(;0))g which converges to a function z pointwise. Let u0(x)ˆ (f⁰) ¹(z(x)), hence un(x;0)!u0(x). From super linear growth,fun(;0)g are uniformly bounded, hence u02L¹(R). Further more by construc- tion,u_n satisfies

f⁰(u_n(x;T))ˆx r_n(x)

T :

Letube the solution of (1.1) withu₀as its initial data. From the dominated convergence Theorem, u_n(;0)!u₀ in L¹_loc(R), hence from L¹ con- tractivity,u_n(x;T)!u(x;T) a.e.x2R. Therefore for a.e.x2R,

f⁰(u(x;T))ˆx r₀(x)

T :

This proves the Lemma. p

In order to prove Lemma 3.10, we need the following covering lemma (see [11]).

LEMMA4.1. Let Iˆ(a;b)andm be a non negative finite measure on I. Let EI ande₀>0such thatm(E)e₀, wheremdenotes the outer measure. Let < be a family of open intervals covering E. Then there exists a sub family <1 < which consists of mutually disjoint open intervals such that

X

J2<1

m(J)e0

(4:1) 4

PROOF. L etE [_J2<Jˆ [¹_nˆ1(a_n;b_n), with (a_n;b_n)\(a_m;b_m)ˆffor n6ˆm. L ete_nˆm((a_n;b_n)), then

e₀m(E)X¹

nˆ1

m((a_n;b_n))ˆX¹

nˆ1

e_n:

Choose [an;b_n](an;bn) such thatm([an;b_n])3e_n

4. Since<is a cover of [a_n;b_n] and hence there exists a finite number of open setsfJ_i⁽ⁿ⁾g <

such that [an;b_n] [J⁽ⁿ⁾_i (an;bn). By discarding some of the sets if necessary, we can assume that for eachi,J⁽ⁿ⁾_i is not contained in[_k6ˆiJ⁽ⁿ⁾_k . Letx_i2J_i⁽ⁿ⁾n [_k6ˆiJ⁽ⁿ⁾_k and by reordering of the indicies we can assume that a_n<x₁<x₂< <b_n. Hence G⁽ⁿ⁾_e ˆ fJ⁽ⁿ⁾_i :i is eveng and G⁽ⁿ⁾_o ˆ fJ⁽ⁿ⁾_i :i is oddg consists of disjoint sets and

X

i2G⁽ⁿ⁾e

m(J_i⁽ⁿ⁾)‡ X

i2G⁽ⁿ⁾o

m(J⁽ⁿ⁾_i )m([an;b_n])3en

4 : ChooseG⁽ⁿ⁾to be one of the familyG⁽ⁿ⁾_e orG⁽ⁿ⁾_o such that

X

i2G⁽ⁿ⁾

m(J_i⁽ⁿ⁾)3 en

8 en

4:

Let<₁ˆ [¹_nˆ1G⁽ⁿ⁾, then<₁ <consists of disjoint open intervals and X

J2<1

m(J)X

n

m(G⁽ⁿ⁾)1 4

Xene0

4:

This proves the Lemma. p

LEMMA 4.2. Letm;gbe non negative mutually singular measures on (a;b). Let D(m)and D(g) be the points of discontinuities ofm([a;x])and g([a;x])respectively. Let E(a;b)n(D(m)[D(g))be a measurable set with m(E)>0andg(E)ˆ0. Then forma.e. x2E

limh!0

m((x;x‡h]) g((x;x‡h])ˆlim

h!0

m((x h;x]) g((x h;x])ˆ 1:

(4:2)

PROOF. L etN>0 and define E_Nˆ

x2E : lim

h!0

m((x;x‡h]) g((x;x‡h])<N

:

Claim: m(E_N)ˆ0. Suppose not and let m(E_N)e0>0 for some e0. Let E_NU be an open set. Then for x2E_N, chooseh_n >0 such that [x;x‡h_n]U and m((x;x‡h_n])<Ng((x;x‡h_n]). Since x2=D(m)[D(g) and hence by continuity, choose a(x)<x<b(x) such that I(x)ˆ (a(x);b(x))U and m((a(x);b(x)))Ng((a(x);b(x))). Let < ˆ fI(x)g_x2EN is a cover of E_Nand hence by lemma 4.1 there exists a subfamily <1 of

< such that

e₀

4 X

J2<1

m…J† NX

J2<1

g…J†

ˆ Ng…[J2<1J†

Ng…U†:

Hencee0

4Ninffg(U):E_NUg ˆ0, sinceg(E)ˆ0. This is a contradiction and hence the claim. Sincen

x2E:lim

h!0

m((x;x‡h]) g((x;x‡h])<1o

ˆ [¹_Nˆ1E_Nand hence form a.e.x2E

h!0lim

m((x;x‡h]) g((x;x‡h])ˆ 1:

This proves the Lemma. p

PROOF OFLEMMA3.2:

PROOF. Take mˆm_t;gˆdx;EˆE_t in Lemma 4.2. Since for x2E_t, x h2=D(m); m((x;x h])y_‡(x;t) y_‡(x h;t) and hence the Lemma.

p

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Manoscritto pervenuto in redazione il 13 Novembre 2012.