Asymptotic Behavior of Solutions of a Conservation Law without Convexity Conditions
KUO-SHLJNG CHENG*
Department of Applied Mathematics, National Chiao Tung Vnicersity, Hsinchu, Taiwan 300, Republic of China
Received August 28, 1980
1. INTRODUCTION
In this paper we study the asymptotic behavior for the hyperbolic conser- vation law
u, +.I-@), = 0, t > 0, --oo<x<co, (1.1)
u(x, 0) = uO(x), -al<x<co, (1.2)
with Riemann-like data for (x] large. The function f is a smooth nonlinear function of U. In general, Eq. (1.1) does not have a continuous solution for all time. Shock curves appear after finite time. We will consider a piecewise continuous weak solution of (1.1) 19, lo]. It is well known that across a discontinuity line x =x(t), the solution satisfies the Rankine-Hugoniot condifion (R-H) and the entropy condition (E) ] 111,
(R-H) x’(t) = a@-, u,),
W a(u..,uJ<a(u-,u) for all u between u _ and u I ,
where U- = u(x(t) k 0, t) and a(ui, u2) is the shock speed defined as
a(u, 1 u2) _ .f-(Ul) -J-w u,--2 .
We will consider the solution of (1.1). (1.2) when the initial data u”(x) are Riemann-like data for 1x1 large, or more specifically, when u”(x) satisfies
u”(x) = u, for x < 4,
= u, for x > S, (I.31
‘i: Works partially supported by the National Scmce Council of the Republic of China.
343
0022.039618 l/060343-34%02.OO/O Copyright & 1981 by Academic Press, Inc. All rights of reproduction in any form reserved.
for some constants uI, U, and S, S > 0. For the case when f is convex (or concave) and u, = u,, the asymptotic behavior was discussed by Lax [8], Keylitz [7], DiPerna [4] and Dafermos [3]. For the case when f is convex (or concave) and uI z u,., the asymptotic behavior with decay rates was recently obtained by Liu [ 91. For more generally smooth f; the asymptotic behavior of solutions of (1.1~(1.3) without decay rates was partially answered by Liu [IO]. It is interesting to note that for the case when the initial data are periodic, for generalf, the asymptotic behavior of solutions of (l.l), (1.2) with decay rates was also only partially answered by Greenberg and Tong [6] and Conlon 121. In this paper we will investigate the asymptotic behavior of solutions of (l.l), (1.2) with initial conditions of the form (1.3). We assume that f”(u) vanishes at a finite number of points and changes sign at these points. The main result which we obtain is that the solution approaches that of the corresponding Riemann problem at algebraic rates (we need the assumption that f’“‘(u) f 0 for some n < CO at points f”(u) = 0). In Section 2, we will consider the case whenf”(u) has only one zero. This case will illustrate the nature of the difficulties involved in the general one and will also be needed for the general case. In Section 3, we will consider the general case.
2. THE CASE WHEN~“(U) HAS ONE ZERO
Without loss of generality we assume that the smooth function f satisfies
f “(24) 2 0 for u $0. (2.1)
We need some definitions and notation. The readers are referred to Ballou [l] for details.
DEFINITION 2.1. Let n ( 0 be given and define q* = r*(y) by
Let q > 0 be given and define ra = q*(v) by
Let r < 0 be given and define q** as the unique number that satisfies i;l= (r**)*. Similarly we can define r** for q > 0 as the unique number that satisfies q = (q**)*. Note that r* = +eo and v* = -co are possible.
DEFINIIIOZ 2.2. Under assumption (2.1). the solution of Eq. (1.1) with
the following initial condition
24,(x, 0) - u”,(x) = U[ for x< -s, =CY for -s<x<s. =u r for s < x, where a is a constant, is called uu(x, t).
The following lemmas on properties of special solutions are proved by direct constructions of these solutions.
LEMMA 2.1. If u1 < u, < u) or uI > u, > u,*, then there exists t, and x,, ta > 0, such that for all t > t,,
UJX, t> = u/ for x-at<x,,
= u, for x-oat >x,, where u = cr(u,, u,) and
x n =& [?-.I (2s).
Proof. Since the case u, > u, > uI* can be considered similarly, we prove the case uI < u, < u;” only. We divide the proof into several cases:
A. u,<u,.<O
(i) a ,< u, ; then the solution u,(x, t) is u,(x, t) = u/ for x <x,(f), for x,(t) < x < x2(t), o< t<t,, for x2(t) < x < x3(t), for x,(t) < x, for x <xl(t), for x,(t) < x <x4(f), t, < t < tz, for x4(t) < x, for x < q(t), for x,(t) < x, t, < t,
where h, is the inverse function off’ restricted in u < 0, x,(t) = -s tS’(u,)t. x*(t) = 4 t./-‘(a)& x,(t) = s + a(a, u,) 1. x4(t) satisfies
x;(t)=u (h, (x4(f;+s),u+
X4(t,)=x*(t,)=X3(t,),fZ>t~f*, and -~4(f2) = X,(t,>> x,(t) = x, + du,, 4) 4 X5@*) = x,(t*) =x40*);t, is the time x3(t) meets x2(t) and t, is the time x4(f) meets x,(t). It is easy to calculate t,; in fact,
t, = 2S/(f’(a) -- u(a, 24,)).
Note that f’(a) - ~(a, u,) > 0 is the consequence of entropy condition (E) and assumption (2.1). To see that x4(f) will meet x,(t) at finite time t,, we recall that
f’(q)
<xi(t)
<f’ (h,
(“‘(“t’“))
=:
-y+s ,
which is condition (E). (The strict inequalities are due to assumption (2.1).) We can calculate x:(t),
xC(r)
= _
h’,(b,W+ www -
(X4(9+ W)’
@ I((X4W+ w> - ur)
G
_ minkWx4W
+ WWW
-
(x4(f)+ W>‘l
Cur
- 4
+o.
(2.2)Thus for sufficiently large t, x;(t) <f’(u)). This would ensure that x,(t) meets x,(t) at finite time t,. To find x,, we proceed as in Liu (91. We take A sufficiently large so that u,(x, t) = U, for x = A + at. It is easy to see from (1.1) and (R-H) that
is time invariant. For t > t, = t,, q(t) = (u, - u,)(A - x,). SO V(I)=1(0)= jA (u,(x,O)-u,)dx . u
.s
= 1 (a-u,)dx+ r (u,-U,)dK.
7
-s
Thus 1 x =--- a u, + u/ ur-- u/ [ ___- 2 a (2s). I We prove the lemma in this subcase.(ii) U, < a < u,, then the solution u,(xI t) is u,(x, t) = u/ for X <x,(t), a for x,(t) < x <x*(t), O<t<t,, =U , for x*(t) < x, u,b-9 0 = u/ for x < x3(t), t, (1: UT for x3(f) < x, where
x,(t) = -4 + a@,,
a)t,
x,(t) = s + a(a,
u,)t,
x,(t) = x, + 4% u,)t, t > t,,
XlOl>
= Xl@,)
= x3(1,).
It is easy to see that t, = r, exists and x, can be obtained as in A(i). This proves the lemma for this subcase.
(iii) u,. < a < OI then the solution u,(x, t) is
UJX, 1) = u/ for x <x,(t), for x,(f) < x < x1(t),
x-s
c i
O<t<t,,
=h,
-y-
for x*(t) < x < x.Jt)? = 11, for x1(t) < x,u,(x, t) = u, for x Q x4(t), x-s =h, - ( t i for x,(t) < x <xx,(r), t, c t<t,, = 24, u,(x, t) = uj = 24, where for x3(t) < x, for x < x,(t), for x,(t) < x, x1(t) = -s + u(u,, a) t, x1(t) = s +$‘(a)t, x3(t) = s +f’(u,>t, t, < t, x4(1) satisfies
x;(t)=o
(,,A, (“‘yS)),
x4@,)
= x,(t,) = h@,),
t>t,,
and
x,(t) =x, + u(u,, u,)t = x, + ut, x5(h) = x,(t*) = x,(t*)*
Using arguments similar to those in case A(i), we can prove that t, is finite. Thus we can choose t, = t, in this case. x, can be similarly determined.
(iv> u,** > UT, 0 < Q ,< uT, then the solution u,(x, t) is U,(& t) = u/ for x,<x,(t) =a for x,(t) < x <x*(t), O<t<t,, for x2(t) < x < x3(t), = u, for x3(t) < x, U,(& t> = u/ for x <x4(t),
x-s
= h -j-
(
)
for x4(t) < x(x&), t, < t<t,, for x3(t) < x, for x < x,(t), for x5(t) < x, t, < Lwhere
x,(t) satisfies
xl(t) = 7s + U(U,? a)t,
x2(t) = S t a(a, a,)t = S tf’(a*)t,
X3(f)
= s l tf’(U,)C
xl,(t>=o
(u,,h,
(“‘y S)))
x4@,) = Xl@,) =-a,), t> t,, x,(t) = x, + at, x&) = .df*) = x,(t,).
Similar arguments as in A(i,iii) can be used to complete the proof of the lemma in this subcase.
(v) uT* > 247, ul* < a < u T*, then the solution u~(x, t) is
a = UF(X, t) x-s h, - ( t 1 = u, u,(x, t) = u/ = u,(x, t) x-s = h, - ( t 1 = 24, 505:40:7 4 for x < x1(t), for x,(t) < x <x,(t), for x,(t) < x s q(t), O<lSl,, for x3(t) < x < x4(f), for x,(f) < x, for x ,< x,(t), for x,(t) < x <x,(t), for x5(f) < x <x3(t), I, ctst,, for x3(t) < x S x,(t), for x4(l) < x, for x < x6(t), for x6(t) < x < x3(l), t, < 1s t,: for -q(t) < x S x4(f), for x4(t) < x,
for x < x,(r), X-S
=h, 7
(
1
for x,(f) < x < x4(t), t, <t<t,, = u, for x4(t) < x, &(X, 0 = u/ for x <x,(t), = 24, for x,(t) < x, wherex,(r) = -s + a(u,, ul*)t = -6s +f’(ul*)t, x2(r) = -s +./-‘(a)&
x,(t) = s + u(a, a*)t = s +f’(a*)t, x,(t) = s +f’(q)4 x5(f) satisfies
X;(f)=u (h* (x5(l)r+s) ,h, (x~“‘l+s),)
with x6(f) satisfies with x,(f) satisfies with and finallyx5@,)
=x*0,) =x3@,),
t>t,,
X:(t) = 4% ~&h(f), 0)
x,(b) =x*(b) = X5@,)~ t>t,,
x;(t)=o (u,,h, (“ys))
X7(b)
= X6(4) = x&h
tat,,
The curve x&) has the property that x;(t) < 0. The function u~(x, t) can be constructed as follows: First, we draw a line through (x, t) which is tangent to curve x,(t). This line is unique and contacts with x,(t) at one point, let it be (2, ?). Then we let
.f+S
q.(x,
t) = h, 7
(
)
*
*
The existence of the curve x,,(f) and the solution u,(x, t) can be found in Ballou [ I]. The fact that times f, , 2, t t, and t, exist and are finite can be established similarly. This completes the proof in this subcase.
(4 u,** > uI*, u;* -C a, then the solution u,(x, t) is
= u, u,(x, f) = u/ = u,(x, f) = u, u,(x, t) = u/ = f+(x, I) = u, u,(x, t) = u/ = u, for for for for for for for for for for for for for for for for x,(t) < x < x&>, q(t) < x < X4(f), -Q(f) < x, X,(f) < x < q(t), q(t) < x1 x <x,(0, x,(t) < x <-Q(f), X6(f) < x < x2(t), x*(t) < & x < -q(t)r x,(f) < x <x*(t), x2(t) < x, x < x,(t). X,(f) < x. O<f<f,, t, c f < t,, t, (t<t,, t, ‘c t < t,, t, < f,
where x5(t) satisfies with x6(l) satisfies with x,(f) satisfies with and
x,(t) = -s + u(u,,
ufy = 7s -tf’(U~)f
X*(f) = -s +S’(u,*“)t, x3(t) = -s +f’(a)t, Xq(f) = s + u(a, 24,) f, X;(f)=u (h: (+(f)l+S) ,&)
X5((,)
=-a,> =x&,)9 (>,f,,
X6(t,) = x,(f,) = x,(t,), (>b,X,(f3) = x,(t,) =
x,(f,),t>,t,,
x*(f) = x, -t ut, 44) =x&> = al).
The function u,(x, t) is connected to the curve x6(t) as in A(v). It is straightforward to verify that I, is finite; however, we do not go through this here.
(vii) u,** < u,*, 0 < a < UT*, then the solution u,(x, t) is u,(x, 0 = u/ for x <xl(t), =Ct for x,(t) < x S x*(t), x-s o<tst,, =h, - ( f 1 for x,(f) < x < x3(f), = 24, for x3(f) < x,
u,(x, f) = u/ for x < x4(t), x - s =h, - ( t 1 for X4(f) < x < x,(t), t, <*<t,, = u, for x3(t) < x, u,(x, *> = u/ for x < x,(t), = u, for x,(t) < x, where x,(t) = 4 + a(u,, a)t, x*(t) = s +./-‘(a*)*, x3(*) = s +f’(U,>t, tz < 1, x4(t) satisfies
Xi(f)
= (u,,
h, (“y- ” ))
with
and
x40,) =x1(*,) =-a*,), *a*,,
X!(f) =x, + fft,
with I = x4(t,) = x3(t,). This completes the proof for this subcase. (viii) lp < UT, u;** < u < u:, then the solution u,(x, t) is
U,(& t) = Uf for x < x,(t),
=cI for x,(t) < x < x2(t), O<t<t,,
= u, for x,(t) < x, U,(& *) = u/ for x < x&>,
*, < *, = u, for x3(t) < x, where Xl(f) = 4 + a(u,, a)*, x2(t) = S + a(a, u,)t, x,(t) = x, + at
0x1
u,** < u:, UT < a, then the solution u(x, t) is u&9 t) = u/ for x < x,(t), for x,(t) < x < X?(f), 0 < t<t,, for x#) < x < X3(I), for x&) < x, for x < x,(t), for xl(t) < x < x4(C), t, <t,<t,, =U I for x4(f) < x, u,(x, t) = u/ for x < x5(t), t, ‘c & =24 r for x,(t) < x, where x,(t) = -s + u(u,, z$)t= -s +f’(zq)t, x,(t) = -S -i-f’(a)& x,(t) = s + u(a, u,)t, x,(t) satisfies ~#)=a (h2 (““‘,+“) ,ur) with x4@1) =x*@,) = x30,), t>t,, and finally x,(t) =x, + utwith x5(&) = x,(1,) =x4(1,). This complete the proof for subcase A.
B. u,< O<u, < u:
(i) a < u,, this case is similar to A(i).
(ii,) U, < a < u,, this case is similar to A(ii,iv,v).
(iii) U, < a < ~7, this case is also similar to A(ii) (two shocks). (iv) UT < a, then the solution U(X, t) is
24,(x. t) = u, for x < X,(f), for X2(f) < x < x,(t), 0 < t<f,, =a u, u,(x, I) = u/ for x2(t) < x <x,(t), for x,(t) < x, for x < x,(t).
XfS
ZZ
h2 -
(
t
1
for xl(t) < x < x4(t), 1, < t <t,, = 24, u,(x, t) = u/ = u, for x,(t) < x, for x < x,(t), for x,(t) < x, t, < t, where x*(t) =-s + u(u,, ugt, X2(f) = -S +f’(a)t, X3(t) = S + u(a, u,) t, x,(t) satisfies x;(t)=a (h2 (x4(f)l+s) !z+) with and X4@,)=-M,) = x,(t,h
x5(t) = x, + (32with x,(t,) = x4(t,) = x,(f,). This completes the proof for this subcase B. Q.E.D.
LEMMA 2.2. If0 < u, < u,, then there exists t,, X,(a) and X,(a), t, > 0, such that
for x < x,(a),
for x,(a) < x < x,(a),
Remark. We have a similar theorem for I(, < u, ( 0. ProoJ We divide the range of a into several cases:
(i) 0 <a, this case is trivial.
(ii) u,* < a < 0, then the solution u,(x, 1) is u,(x. t) = u, =a for x <x,(f), for x,(t) < x < x,(t), o<t<t,, for x2(r) < x < x4(f), for x,(t) < x, for x < x,(t), for x,(t) < x <x,(l), I, <t<t,, = u, for x4(t) < x, where x,(t) = -S + u(u!, a)t,
x,(t) = S + a(a, a*)t = S +J’(a*)t, X3(f) = s +J’((u,.>*)t,
x,(t) = s +f’(u,)t, x,(r) satisfies
x;(t)=o (u[,h, (“‘“‘t-“1) with
x&,) = X,(fl> =-%(I*), tat,,
and finally xs(t,) = x3(t,). It is obvious that we can choose t = t,, X,(a) = xs(t,), X,(a) = x4(t,) to complete the proof for this subcase.
(iii) (u,)** < a < u,*, then the solution u,(x, tj is =Cf
x-s
=h, -j--
(
1
= 24, U,(X, 1) = u,x+s
=h, 7
(
1
= u,(x, t) where = 24, for x <x,(t), for x,(t) < x ,< x2(t),for X2(I) < x ,< x3(t), O<t<t,,
for x3(t) < x <x4(t), for x4(t) < x, for x < x,(t), for x,(t) < x < x5(t), for x5(t) < x < x3(t), 1, < t<t,, for x3(t) < x < x,(r), for x,(t) < x, x,(t) satisfies
x,(t) =-s + u(u, up)1 = -s -tf’(u,.)t,
xl(t) = -S + f ‘(a)&
x?(t) = S + a(a, a*)t = S +f’(a*)f$ x,(t) .= s +f’(u,)t,
x;(t)=o (h, (x5(t)I+s).h, (x’(t’(+s),)
with x,(t,> = x+,) = x,(t,), t 2 t,, and t, is the time when x5(t,) =x,(t,).
The curve x,(t) and function u,(x, t) is similar to the case A(v). Choose t, = I,, X,(a) = x5(t,), and X,(a) = x,(12) to complete the proof.
(iv) a < (u,),,, then the solution u,(x, t) is u,(x, t) = u, for x < xl(t),
for x,(t) < x < x.&h
o<r<r,, a for x3(t) < x <x?(t),
where x5(t) satisfies with x6(l) satisfies with for x < x, (1), for x,(t) < x < x,(l), 1, < r B f,, for x,(t) < x, for x < x,(t),
for x,(t) < x < X6(l),
for x,(t) < x <x2(t), for x,(t) < x,x,(t) = -s + a(u,,
ur*)
f,
X*(f)
= 4 +Y(Qt = 4 t o(u,)**, u,)t,
x3(t)
= -s +f’(a)t,
x,(l) = s + u(a, u,) t,
x;(t)=a (h2 (x5ct;+s) ,ur)x5@,)
=x3@*)
=x&J
t> t,,
x;(t)=a (h2 (X6(f)1SS).h2 (“6(‘)1+s)*) x6&)= X,(f,> = X,(t,),
t>t,,
and finally x6(t,) = x,(I,). Choose I, = t,, X,(a) = x,(r,), and X,(a) = x,(t,
to complete the proof, Q.E.D
LEMMA 2.3. ifu, < UT <U,, let
then there exists x,, t, > 0 such that
(i) x;(t) > cs(u,, u?) for all t > 0 except at Jnite points oft, (ii) x,(t) < x, + u(u,, uf) t,
(iii) Cm,-, [xn + a(u,, uf)t -x,(t)] = 0. (iv) u,(x, 1) >, 247 for x > x, + a(u,, u;“) t, t > t,.
Remark. We have a similar lemma for the case uI > u,* > u,. Proof: (i) a < (q)**, then the solution u,(x, t) is
ll,(XI t) = u, = l+.(x, t) = u, u,(x, t) = u, = 2+(x, t) = u, where for x < xl(t), for x,(t) < x < x,(t), for x3(t) < x <x4(t), for x4(t) < x, for x ,< x,(t), for x,(t) < x < x,(t), for x5(t) < x, for x <x,(t), for x,(t) < x < x,(t), for x6(t) < x < x,(t), for x,(t) < x, for x < x,(t), for x,(t) < x,<x,(t), for x9(t) < x, 0 < t<t,, t, <t<t,, tj < I, x,(t) = 4 +&f’(q) t, XI(l) = -7s +f’(u,**)t, x3(t) = 4 +f’(a) t, x,(t) = s + a(a, 24,) t,
x,(t) satisfies
x#)=a (h, (X5”‘,+“) ,q.)
with x5QJ = 4,) =-at,>> t>t,, x6(t) satisfies with andx&J =x2(4) = x,(U tat,,
X,(f)
= x,(f,) + c(u,,
uf>(t
- 13)
= xs(t,)
+f’($w - &), tat,,
x,(t) = x&J + o(u,**,
u,)(t
- tz)
= Q,) +f’(u,)O
- [*I*
Obviously,
Xl@> = Xl @I, 0 < t,<t,, = x,(f): xj < t;
choose x, = x,(t,) -f’(uS)t3, t, = t, to complete the proof for this subcase.
(ii) u,** < a < ul, then the solution u,(x, r) is
u,(x, q = uj for x < x,(t), for x,(t) < x <x2(t),
=Ct for x2(t) < x < X3(f), O<t<r,,
for x3(f) < x < x,(t),
u,(x, t) = U( = = &(X, r) = = u, K,(X, t) = u/ = UJX, t) u, where for x < x1(t), for x,(t) < x < x,(t), for x5(t) < x < x&)9 for x3(r) < x 6 x,(t), for x,(t) < x. for x <x,(t), for x,(t) < x <x3(f), for x3(r) < x < x4(t), for x4(t) < x, t,<t<t2, I, < t, x,(t) satisfies x,(t) = 4 +f’(uJ t, x*(t) = -s +f’(a) t, x3(t) = s +f’(a”)t, -Q(f) = s +f’(u,) t, x;(t)=u (h, (“‘(‘)its),h, (x5(t)r+s),) with and X50,) =x&J = x,(t,), t>t, X6(f) = x,Uz) +S’WN - t2) =x5(b) +S’(m(t - a* It is obvious that we have
x,(t) = x,(0, O<f,<l,, =x&)7 t, < t.
(iii) uI < a ( 0, then the solution u,(x, t) is &(X9 t) = U[ for x < x,(t), =C? for x,(t) < x < X2(f), X-S =h, y-- ( ) O<t<t,, for x,(t) < x ( X.$(f), for x,(t) < x <x4(f), t, <t, =.U r where x,(t) satisfies with It is obvious that for x4(t) < x, x,(t) = 4 + u(u,, a) t,
x2(t) = S + a(a, a*) = S +f’(a”)t, x,(t) = s +f’(uT)t, x,(4 = s +f’(U,>t,
x;(t)=u (u,,h,
( x”“l-s))
x50,)
=x,@,>
= x*(tA t> t,.
x,(f)
=x,(t)
for O<t,<t,, = x5 (4 for t, < t. Choose x, = S, t, = 0 to complete the proof.(iv) 0 <a < UT, this case is similar to the above case. (v) u: < a, this case is trivial.
Before we present our main theorems, we state an ordering principle due originally to Douglis IS]. See also Wu [ 13 1, Ballou [ 1) and Keyfrtz 171.
ORDERING PRINCIPLE. Let the function f be smooth function, and let u(x, t) and u(x, t) be piecewise smooth weak solutions satisfying condition (E) to the Cauchy problem (l.l), (l-2), where the initial data are u(x, 0) and u(x, 0), respectively. Then u(x, 0) < zi(x, 0) Vx E (-co, co) implies u(x, t) < u(x, t) vt 2 0, vx E (-co, 00).
DEFINITION 2.3. For solutions of (1. l), (1.2), (1.3), u(x, t), let x,(t) = sup(x: u(x’, 1) = u, Vx’ <x),
x,(t) = inf(x: u(x’, t) = u, Vx’ > x). Now we state our main theorems.
THEOREM 2.4. If u, < u, c ~7, then there exists t, and x,, t, > 0, such that for all t > I,,
u(x, t) = U[ for x < x0 + ut, = 11, for x > x0 + or,
where u = u(u,, u,) and .
1
X” = ___ ___- 4 -t Ul
u, - u/ - 2 Iv> s.
Remark. We have a similar theorem for u, > u, > u[*. ProoJ Let
M=sup(uO(x):--S<x<S}, (2.3)
m = influ’( -S <x ,< S}. (2.4) From Lemma 2.1, let the solutions corresponding to a = M and a = m be respectively uw(x, t) and u,(x, t) with corresponding x,~, x,, cM, t,,. It is easy to see that x,~ < x,,. If x,,, =x,, which is the case M = m, then we are done. So assume x,~ < x,. Using the Ordering Principle, we have
4&, q < u(x, t) < u‘&, t) vx, vt >, 0. Thus for all t > T= max(t,, tm}, we have
u(x, t) = u, for x < x,+r t at, = 11, for x > x,, -t ot
and
u/ = u,(x, t) < u(x, 1) < u,&, t) = u, for x,~ + at < x < x, + ut.
From Definition 2.3, for t > E it is easy to see that x, + ut < x,(t) < x,(t) < x, + ut.
Furthermore x,(t) and x,.(t) are Lipschitz continuous curves with slopes x;(t) = u(u,, u@,(t) + 0, t)),
and
x:(t) = @, , u(x# - 0, t)),
respectively, and with bounded second derivatives. Since u(x,(t) + 0, t) and
u(x,(t) - 0, t) are both between uI and u, for t > < we have from (E) xi(t) > u(u,, 24,) = u > xi(t).
Thus g(t) - u 20 and x;(t) -a< 0 or (x,(t)-at) is nondecreasing and
x,(t) - ut is nonincreasing for t > i But x,(t) - at <x,(t) - ut for t > i Thus if there exists t, > F such that x,(t,) = x,.(t,,), then we have x,(t) = x,(t) and
x;(t) = u = xi(t) for all t > t,. If this is the case, then we are done. Now suppose the opposite, that is, x,(t) < x,(t) for all t > E then x,(t) - at is nondecreasing and bounded and x,(t) - ut is nonincreasing and bounded for all t > l Hence
‘,iz (x,(t) - ut) = X,, fiz x;(t) = u, * (2.5 >
‘,‘“, (x,(t) - ut) = X,) fim, x:(t) = u, (2.6)
+ 1
with X, <X,. From the entropy condition (E), f’W > +4,%J >f’(u,), we can choose sufficiently small 6 such that
f’(u) > u >f’(u) for all u E (u,, u, + S), u E (u, - 6, 24,). (2.7) From (2.5) and (2.6), we can choose sufficiently large t,, such that
u(x,(t) + 0, t) E (u, - 6, u,.) and u(x,(t) - 0, t) E (u,, u, + 6) for all t > t,.
Now for t> t,, through (x,(t) + 0, t) and (x,(t) - 0, t) we draw charac- teristics backward in time. They would intersect along a discontinuity line whose slope is approximately u due to (2.7) and (R-H). (Note that they
cannot terminate to a contact discontinuity before they meet.) But it is obvious that this discontinuity line violates (E). Q.E.D.
THEOREM 2.5. If0 < u,< u,, let
p,(t) = m;tn j” (4YT t> - u,) dY3 --co .a: q,(r) = my 1 My, 4 - u,) 44 ‘X then 0) A(l) 2 0, q;(t) ,< 0,
(ii) there exists I,, t, > 0, such that fur all t > c,, p,(t) =p,(tJ = Pl(Wh q,(t) = s,(hJ = q,(a)*
Proof: (i) follows from Liu 19, Theorem l(i)]. Let M and m be as defined in (2.3), (2.4). From Lemma 2.2 and the Ordering Principle, we know that
%(X7 q > 4x3 0 > u,(x, a t > 0, --co<x<m, and for t > t, = max(t,+,, t,,,),
u,& 4 > (u, *I*, &(X, q > @I*)** Hence U(X, t) > (u,*)* for all t > t, and
u(x, 4)) = u/ for x < X,(M), = u, for x > X,(m).
Thus for t > I,, U(X, t) are restricted in the region f”(u(x, t) > 0. Hence the theorem follows from Liu 19, Theorem l(ii)]. Q.E.D.
Remark. We have a similar theorem for the case U, < U, < 0.
THEOREM 2.6. If 0 <u, < ur, let p,(r) and q,(r) be as defined in Theorem 2.5. Define the generalized N-waces as
Nz(x, t> = u/ for x -f’(u,)l< - -@,(oo)f”(u,)t,
= u, for x -S’h,>t 2 d2q,(~lfN(u,)4 t > 0.
then we have
(i) the edges of N, and u .have finite distance for all time, i.e.,
+ Ix,(t) -f’w
- d2mwYur)tl
= O(S)9
(ii) (u(x, t) - Nz(x,
t)l
<A; ‘O(S) t- ’ for any x fhat lies between max(x,(t),f ‘(u,)t - +2P,Wf “(u,)t) and min(x,(t),f’(u,)t + dzz )f”(ur)t),(iii) 1 u(x, t) - N,(x, t)l = O(S) t- ‘I2 for x between x,(f) and (f’(u,)t - J--2p,(co)f”(q) or between x,.(t) and (f’(u,)t + d2q,(oo)f”(u,)t),
(iv) 4x, t) =N,( , > P x t I x 1 ies outside the regions of (ii) and (iii), where xl(t) and x,(t) are defined in DeJinition 2.3, A, = minB>rr>(u ,.,. f “(u), and B is a bound for u”(x).
Remark. We have a similar theorem for the case u,. < uI < 0.
ProoJ This theorem is an easy consequence of Lemma 2.2, Theorem 2.5, and Theorem 4 of Liu 191. We omit the proof.
THEOREM 2.7. If u, < UT < u,, then there exist x0, t, > 0, such that 0) x;(t) > 4q, 43
(ii) x,(t) ,< x0 + @q, 46
(iii) lim,_, [x, + @I’ uT”)t -x,(t)] = 0, (iv) ~(4 t) > ((IL:)*)* for x > x,(t), t > to.
Remark. We have a similar theorem for the case I(, > u,* > u,.
ProoJ (i) is obvious. From (i), (xl(t) - o(u,, uT)t) is nondecreasing. But x,(t) < x,,, + u(u(, uf) t, where x, is the x, when a = m in Lemma 2.3 and m is the number defined in (2.4). Thus (x,(f) - a(~,, uT)f) is nondecreasing and bounded. Hence, lim,.,,(x,(t) - u(u[, u:)t) exists; let it be x0. We already proved (ii) and (iii). Now it is easy to see that we can find a time t, sufficiently large, such that u(x, t,) > ((UT)*)* for x,(t,) < x < x, + a(~,, UT) t, and u(x, + o(u,, UT) t, + 0, t,) = UT. Furthermore, the line segment x0 + a(~,, u;“)t, t > t, : is the characteristic line passing through the point (x0 + a(~,, ur) t, + 0, t,) and t, > t,. Thus we have u(xo + ~(a,, uT)t + 0, t) = UT and U(X, t) > UT for all t 2 t,, x > x, + u(u,, uT)t. Hence we can choose t, suf’ficiently large, such that u(x, t) > UT for ail t > t,, x > x0 -t- u(u,, uT)t. Choose to = t, to complete the proof for (iv).
THEOREM 2.8. Under the assumptions of Theorem 2.7, let
4(t) = mfx
jm
(u(Y, t) - u,) dy; xthen
(i> 4’(t) < 0,
(ii) there exists t,, such that q(t) = q(t,)for aZZ t > t,.
Proof: (i) follows from Liu [9, Theorem l(i)]. To prove (ii), take the t, of Theorem 2.7 as the to we want. Assume that the maximum point in the definition of q(t) is taken place at X*(C). We want to prove (1) u(x*(t), t) = u, and u(x, t) is continuous at x*(t), and (2) q’(t) - 0 for all t > to. If x*(t) is a discontinuity, then since x*(t) > x,(t), we must have u(x*(t) - 0, t) > u(x*(t) + 0, t). But in this case, x*(t) is not the maximum point. Hence U(X, t) must be continuous at x*(t). Now if u(x*(t), t) # u,, then x*(f) cannot be the maximum point too. This proves (1). To prove (2), we know that from (l), ak*(t)/dt exists and is equal tof’(u,). From the definition of
q(t), we have
s(t) = jy; -0 (u(Y, t) - u,) dy.
Hence
4’(t) = [W(t) - 0, t) - 11,
] X;(t) +-
J
.X,(l) 0
u, 4
X’(l)
= luW4 - 03
t> - %I X:0) -f(u(x,(t> - 0)) -f(u,)
=o
(R-H).This proves (ii). Q.E.D.
THEOREM 2.9. Under the assumptions of Theorems 2.1 and 2.8, let x0 and q(c0) = q(t,) be the respective constants in Theorems 2.7 and 2.8. Define the following one-sided generalized N-wave
N(x, t) = u, for x < x0 + o(u,, ul*> t,
for x > f’(u,)t -!- dw(u,)t + x0,
Then we haue
(i) there exists t, > 0, such fhat xl(t) = x, + u(u,, $)t for all t > t,,
(ii) Ju(x, t) -N(x, t)l <A-‘O(S) t-’ for x between x, + u(u,, $)t
and min(x,(t),S(u,) t + d2q( ao)f”(u,) t + x0 j.
(iii) [xr(t) -f’(u,)t -
&qb >f%,)t i = O(S),
(iv) ] 24(x, t) - N(x, t)j < O(S) t-l’* for x between x,(t) and f’(u,)t +
d2q(oo)S”(u,)t +x0, where A = min((ui),).(,~RS'(u).
Remark. We have a similar theorem for the case u, < uI* < u,.
Proof: Let -X(t) =x,(t) - x0 -f’(uF) t, then
-X’(l) = xi(t) -f’(@) = u(u,, 24(X,(f) -t 0, t) - u(u,, 24:).
From Theorem 2.7, X(t) -P 0, X(I)-, 0 as t -+ co. Hence we can expand u(u,, U(X)(C) + 0, t) to obtain
u(up u(x[(t) + 0, t) z qu,, uI*) - 2;;y;&
(u - fq)!
But
u: = hQ’(uT)) = h, x, +f’(uf)t - x,
t 9
hence for t large
u(x,(t) + 0, t) r h, tx’@)l- xo 1 z h, -X(t) +f’(ul*) t t Hence or x(t) Z -A 9 with A = 2$“t,, (h;df’(@)))‘, and as
t-+cO.
But as we know X(t) + 0 as t + co. Hence c = co, which means that for some finite t,, X(t,) = 0. This proves (i). Parts (ii), (iii) and (iv) are an easy consequence of Theorem 2.7, Theorem 2.8 and Liu 19: Theorem 41.
LEMMA 2.10. If uI = u, = 0 and f(u) zf’(O)u + Au”’ ’ ‘, n > 1, A > 0, for / ui small, then we have
Iu,(x, t)l = O(S) t- lit*” + I). Proof: If a < 0, then the solution u,(x, t) is
u,(x, t) = 0 for x < x1(t), for x,(t) < x <x2(t), O<t<t,, =(Y for x2(t) < x < x3(t), =o for x,(t) < x, z&(x, t) = 0 for x <x,(t), x + s = h, - ( t ) for x,(t) < x <x4(t), t, <t,
=o
for x,(t) < x, wherex,(t) = -s t.r(0> t,
x2(t)
= -s
tf’(a)t,
x,(t) = s + a(a,
0) t,
x4(t) satisfies x;(t)=o (h, (x4”‘t+S),0)=f(h, (x4(r~ts))/h,(x4(t)ifS), with X4@,) =x&J = x&)9 t> t,.hence
But it is obvious that
x4(t)I+
s >f’(O);
hence (x4(f) + S)/t -f’(O) as t -+ 00. And thus h,((x4(f) + S)/t)+ 0 as t-+ 00. Now for t large, we can have
xi(t) g(O) + Ah:”
(x4(t:+s)-
But on the other hand, from the definition of h,,
(2n + 1) Ah;”
cx4(‘)I+ s 1 +f’(tqzf (h, (x4Q)l+S)) = X’y+s )
Letting X(t) = x4(t) + S, we get
X’(t)
rf’(0) + &
(+ -f’(O))
for t large. HenceX(f) zf’(O)t + O(S) f”(2nA ‘) and
h, (qq [ (2n j l)A (+-‘(o)i]“2n = O(S) t-I/m+ 1).
Similarly we can consider the case a > 0. This completes the proof. Q.E.D. THEOREM 2.11. Zf u, = u, = 0 andf(u) zf’(O)u + AU*“+ ‘, n > 1, A > 0,
for IuI small, then we have
) u(x, t)l = O(S) t - “(2n + I).
ProoJ This theorem is a consequence of Lemma 2.10 and the Ordering Principle.
THEOREM 2.12. /fu,=O <u, andf(u)zf’(O)u +Au2”“. n 2 I, A > 0. for ( u 1 small, define
p(f) = mjn I4 U(Y, 0 dx m
g(t) = m,ax t3; (u(Y, I) - 11,) d?; -x
then
(i) ifp(0) = 0, then p(f) = Ofor all f,
(ii) VP(O) < 0, then p’(f) > Ofor all t arzd p(f) + 0 as f -+ co,
(iii) lyp(f) = min, 1-Y ~ U(U, t) dy = pz’ z&v, f) dy, (hen -u,(t) z-f’(O) t + O(S) fl’” - s for t large, where2nr2”-<2” ‘--~~-~-l=Oo,-l <<<O,
(iv) q’(f)<Ofor all (20,
(v) there exists r0 > 0, such that for all t > t,,
q(t) = q(to).
Prooj From Liu 19, Theorem l(i)], p’(t) > 0 and p(t) < 0; thus if p(O) = 0, then p(t) = 0 for all t which proves (i). Ifp(0) < 0, then we want to prove that the maximum point x,(t) as defined in (iii) is a shock curve with z&,(t) - 0. t) < 0 and 0 < z+,(t) + 0, t) < u(x,(t) - 0, f)*. If u(x, r) is continuous at x,(t), then u(x,(t). t) = 0 and u(x,(t) - E, t) < 0:
u(x,(f) -t- E, t) > 0 for sufficiently small E > 0. But this is impossible, because
the characteristics from the immediate left-hand side of x,(t) will intersect the characteristics from x,(t) immediately. Thus x,(t) must be a shock curve. It is then obvious that u(x,(t) - O1 t) ( 0 and 0 < ~(xJt> + 0, t) < u(x,(t) - 0, t)“. Thus
P’(O = u(xpU) - 0, t> $A’> - VW,(t) - 090) -f(O)1
2.4(x,(t)
0, t) B W,(t) - 0, l)) -f(~(xp(t) + 03
0)
=
-
u(x,(O - 0, t) - u(x,(t) + 0, f) ~fwp(o - 03 t)> -fP> u(x,(t) - 0, f) - 01
> o;this proves (ii). Now we would like to estimate the order of x,(r). From (R-H) and the above arguments, we have
x, @) JW~) - O,O) -fWpU) + 030) R Il(XJf) - 0, f) - 24(x,(t) + 0, t) <f,(u(x P @) _ 0, t)) . .
For t large, ~(x~(f> - 0, I) z h,((x,(f) + S)/t); hence
f (x”(f)I+s)<O and xp(r)t+s -f’(O) as t+ao. Thus for t large, we have u(x,(r) + 0, t) = u(x,(t) - 0, t)* and
x,(f) z fM(-w + SF>) -f(MX,(~) + S)/t)*)
P
hl((Xp(f) + V/f) - h,((x,(t) + S)/t>* =f’(O) + A [I$ + /q--‘/q + . . . + &J”]
=f’(O) + Ah:” (y+S). [1+~+<2+**~+r2”I, where r satisfies
2,@” - y2n - ’ - y2n - 2 - ( . . . - c - 1 = 0, -1 <<<O.
ThUS XL(f) zf’(0) + A (2n + 1) h:” (x,y+s) . y2n rf’(0) + cyn ( xpcf)l+ s -f’(O)) . Hence xp(f) + s
gyo)f + cqsj
f?This proves (iii); (iv) follows from Liu [9, Theorem l(i)]. To prove (v), first, we use the solution u,Jx, f) to prove that u(x, t) > u,(x, t) > (u,)** for all t > t, , where m is the intimum of U(X, 0) and f, is some constant greater than zero. This is by direct construction of the solution u,,,(x, f). We do not want to repeat it here. Assume that the maximum point in the definition of q(f) is taken place at X(t). Since U(X, f) > (u,)**, for all f > t,, X(f) cannot be a shock curve. Hence u(x, t) is continuous at the point X(f) and u(X(f), f) = u,. Direct calculation of q’(f) will prove (v). Q.E.D.
THEOREM 2.13. Under the same assumptions of Theorem 2.12, define the one-sided &-N-wave N,(x, f) as
N,(x, f) = u,
for x >f’W f + ~2q(co)f”(u,) f,
then
(i) Ix,(t) -f’(q) t - \/2q(oo)f”(u,) t ) = O(S)for all t,
(ii) !N,(x, t) - u(x, t)l <A -I(E) O(S) t---I for x between f’(E)t and min(x,(l).f’(u,)t +Jh(03)f”(u,)l),
(iii) (N,(x, t) - u(x, t)l < O(S) t-- li2 for x between x,(t) and f’(u,)t i- ,,/2q( m)f”(u,) t, where E is a small fixed number with 0 < E < u, and q( co) is the constant q(t,J in Theorem 2.12(v) and A(s) = min,,,,,f”(u).
Remark. We can have a similar theorem for the case uI = 0 > u,. Proof: From Theorem 2.12(iii), we find a time t, > 0, such that f’(s) t > x,(t) for t > t,. Thus for t > t,, u(x, t) > E for all x >f’(e)t. Then (i), (ii) and (iii) follows from Theorem 4 of Liu [9\. Q.E.D.
~.THE CASE WHEN~” VANISHES AT n POINTS
AND CHANGES SIGN AT THESE POINTS
Without loss of generality, we assume that f” vanishes at a,, a,,...? a,., where a, < u2 < ... < a,V, and f”(u) < 0 for u < a,, f”(u) > 0 for a, < u < a2 ,..., etc. We also adopt the definitions of u,(x, t), M, m, x,(t) and x,(t) of Section 2. For convenience, we put a, = -co and a,,, , = +co. In this section, we use u(x, t) to denote the solution of (l.l), (1.2) with initial condition (1.3), where f is under the assumption of this section.
We may need direct construction of solution u(x, t) in the proof of the following lemmas and theorems. We will give only some indications and omit the details. These constructions are similar to the constructions in Section 2.
LEMMA 3.1. ff U, E (aim,, a,), U, E (Uj.-,? aj), where 1 < i <j < N + 1% then there exists t, > 0, such that for all t > t,, u,(x, t) E (Al, A,), where A,, A r are two fixed constants with A, E (a,-, , u,) and A,. E (ur, Uj).
LEMMA 3.2. Under the assumptions of Lemma 3.1, there exists t, > 0, such that for all t >, t,, u(x. t) E (A,, A,.), where A,, A, are as in Lemma 3.1. Proof of Lemmas 3.1 and 3.2. Using the Ordering Principle, we can easily establish Lemma 3.2 from Lemma 3.1 if u(x, 0) E (a, aj) for all X. TO
prove Lemma 3.1, we use induction. If a E (a,- i, aj), then Lemma 3.1 is obviously true. Now assume that when a E (uimk, Ui-k+,), Lemma 3.1 is true, and hence Lemma 3.2 is also true when u(x, 0) E (a, a,). We would like
to establish that when a E (ai-.- ,, ai-&, Lemma 3.1 is true. The solutions for the Riemann problems (u[, a) and (a, u,) are combinations of shock waves and rarefaction waves. Let us denote these simple wave resolutions of (G a) and (a, 4 by (u,, u,), (ul, Q,..., (u,, a) and (a, wJ, (wl, WA..., (w,, u,). It is easy to see that at least one of the simple waves (t.,,,, a) and
(a, w,) must be a shock wave. It is this simple consequence of entropy condition (E) that causes the cancellation of waves. Now it is easy to see that shock wave (u,,,, a) or (a, wi) will kill the a-states in a finite time. After that, the remaining rarefaction wave, (urn, a) or (a, w,), will be killed by a combination of type I and type II shocks in a finite time. Thus there exists t, > 0, such that for t > t,, u,(x, t) E (aimk, a,). Using induction hypotheses, we prove that when a E (ai_ k.. i, a,- ,J, Lemma 3.1 is true. Similarly we can consider the case a > a,. This completes the proof of Lemmas 3.1 and 3.2.
Q.E.D.
Remark. We have two similar lemmas when 1 <j < i ( N + 1.
DEFINITION. If the solution of the Riemann problem (u,, u,) consists of a
simple shock wave with f’(u,) > u(u,. u,) >f’(u,) and a(u,, u) > u(u,, u,) > u(u, u,) for all u between u, and u,, then we call (u,, ur) a strict shock.
THEOREM 3.3. If (u,, u,) is a strict shock, then there exists x,, and t,, t, > 0, such that for all t > t, ,
u(x, t) = u/ for x < x0 + o(u,, ur> t, = u, for ~>x~+(u,,u~)~, where
u, + UI ~-
2 N>S.
ProoJ From Lemma 3.2, if u, E [a,.. 1, ail, u, E [a,-, , uj] and i > j (note that the case i < j can be similarly considered), then we can choose A, E (ai.., , u,), A, E (ur, uj) such that (A,, u,) and (u,, A,) are all strict shocks. It is easy to construct the solutions uA,(x, I) and uA,(x, t) directly and find
t,,, tAr and xA,, xA,, such that u&, t) = u/
zz u r u&3 f> = u/ = u,
for x < xA, + u(u,, u,)t,
for x > xa, + u(u,, u,>t, t > t,,, for x < x,, + a(~,, u,)t,
t > t.4; for x > xA, + u(u,, u,)t,
Thus for 1 sufficiently large, say t > t’,, we have
u(x, t) = u, for x < x,, + 0(u,, u,) t,
= 24, for x > xA, + a(u,, ur)t,
and u, = u,,(x, t) < u(x, t) < u,,,(x, t) = u, for xA, + u(u,, u,)t < x <x,, + a(u,, u,.)t. Using the strick shock properties of (u,, ur), we can prove this theorem by using the same arguments as in the proof of Theorem 2.4.
Q.E.D.
DEFINITION. If the solution of the Riemann problem (u,, u,) is a simple rarefaction wave and if u, # a, # u, for all i, then we call (u,, ur) a strict rarefaction wave.
Remark. From this definition, a strict rarefaction wave can have two possibilities only. Either a, ._ , < uI < u, < cli and f”(u) > 0 for all u E (ai. ,.a)) or Uj_, < u,.<u, < Uj andf”(u) <O for all u E (aj-,,a,).
THEOREM 3.4. If (u,, u,) is u strict rurefuction wave, with the proper definitions of p(t) and q(t) in Theorem 2.5 and the definition of N(x, t) in Theorem 2.6: where we have to replace h, by some proper h, and hi is the inverse function of f’(u) restn’cted in (ui _ , , ui), then the proper statements of Theorems 2.5 and 2.6 hold.
Proof: In view of the Ordering Principle and Lemmas 3.1 and 3.2, we can push the solution u(x, t) at a finite time into the interval (aim,, a,) which contains u, and u,. Then the whole story of Liu [ 91 goes and the theorem is
proved. Q.E.D.
For nonstrict shocks and nonstrict rarefaction waves, they can be treated as in Theorems 2.9, 2.11, 2.12, and 2.13. We do not treat them here. Similarly we can treat the case of the combination of shocks and rarefaction waves. For example, if the resolutions of (u,, u,.) to simple waves are (u,, t’,), (u,. cl). (t.*, uI), where (u,, tli) is a shock with f ‘(u,) > o(uI, v,) =f’(c,),
(c,, L.*) is a strict rarefaction wave, (t’*, c~) is a shock with f’(v2) = u(t’*, u,) >f’(u,.), then we can prove that after a finite time, xc(t) = X, -I-
u(u,, u ,) t, x,(t) = X, + a(~, , u,) t and between these two shocks is rarefaction wave (L., . c~). The proof is similar to the proof of Theorem 2.9. For u, = u, = ui, the treatment is almost identical to the treatment of Theorem 2.11. Although we did not consider the case f”(ai) = 0 and f" does not change sign at a,, it is obvious that we can apply our technique to this case as well.
ACKNOWLEDGMENT
I would like to express my thanks to Tai-Ping Liu for introducing me to this area of research and useful suggestions and discusions. I am solely responsible for all possible errors in this paper.
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