Total Variation, Compactness, Etc

Appendix A Total Variation, Compactness, etc.

I hate T.V. I hate it as much as peanuts. But I can’t stop eating peanuts. Orson Welles, The New York Herald Tribune (1956)

A key concept in the theory of conservation laws is the notion of total variation,T.V. (u), of a function u of one variable. We deﬁne

T.V. (u):=sup u (xi) u (xi 1) . (A.1) | | i X We will also use the notation u := T.V. (u). The supremum in (A.1)is | |BV taken over all ﬁnite partitions xi such that xi 1 interval I if (we think that) this is not important, or if it is clear which interval we are referring to. For any ﬁnite partition x we can write { i} u (x ) u (x ) = max (u (x ) u (x ) , 0) | i+1 i | i+1 i i i X X min (u (x ) u (x ) , 0) i+1 i i X =: p + n.

Then the total variation of u can be written

T.V. (u)=P + N := sup p +supn. (A.2)

We call P the positive, and N the negative variation, of u. If for the moment we consider the ﬁnite interval I =[a, x], and partitions with a = x < < 1 ··· xn = x, we have that px nx = u(x) u(a), a a

451 452 A Total Variation, Compactness, etc.

x x where we write pa and na to indicate which interval we are considering. Hence

px N x + u(x) u(a). a  a Taking the supremum on the left-hand side we obtain

P x N x u(x) u(a). a a  Similarly, we have that N x P x u(a) u(x), and consequently a a  u(x)=P x N x + u(a). (A.3) a a In other words, any function u(x)inBV can be written as a di↵erence between two increasing functions,1

u(x)=u+(x) u (x), (A.4) x x where u+(x)=u(a)+Pa and u (x)=Na . Let ⇠j denote the points where u is discontinuous. Then we have that

u(⇠ +) u(⇠ ) T.V. (u) < , | j j | 1 j X and hence we see that there can be at most a countable set of points where u(⇠+) = u(⇠ ). Observe6 that functions with ﬁnite total variation are bounded, as

u(x) u(a) + u(a) u(x) u(a) +T.V. (u) . | || | | || | Equation (A.3) has the very useful consequence that if a function u in BV is also di↵erentiable, then

u0(x) dx =T.V. (u) . (A.5) | | Z This equation holds, since

d x d x u0(x) dx = P + N dx = P + N =T.V. (u) . | | dx a dx a Z Z ⇣ ⌘ We can also relate the total variation with the shifted L1-norm. Deﬁne

(u, ")= u(x + ") u(x) dx. (A.6) | | Z If (u, ") is a (nonnegative) continuous function in " with (u, 0) = 0, we say that it is a modulus of continuity for u. More generally, we will use the

1 This decomposition is often called the Jordan decomposition of u. ATotalVariation,Compactness,etc. 453 name modulus of continuity for any continuous function (u, ") vanishing at 2 p " =0 such that (u, ") u( + ") u p,where p is the L -norm. We will need a convenient characterizationk · ofk total variationk·k (in one variable), which is described in the following lemma.

Lemma A.1. Let u be a function in L1(R).If(u, ")/ " is bounded as a function of ", then u is in BV and | |

(u, ") T.V. (u)=lim . (A.7) " 0 " ! | | Conversely, if u is in BV , then (u, ")/ " is bounded, and thus (A.7) holds. In particular, we shall frequently use | |

(u, ") " T.V. (u) (A.8) | | if u is in BV .

Proof. Assume ﬁrst that u is a smooth function. Let xi be a partition of the interval in question. Then { }

xi xi u(x + ") u(x) u (xi) u (xi 1) = u0(x) dx lim dx. " 0 " | | xi 1  ! xi 1 Z Z Summing this over i we get (u, ") T.V. (u) lim inf  " 0 " ! | | for di↵erentiable functions u(x). Let u be an arbitrary bounded function in 1 L , and uk be a sequence of smooth functions such that uk(x) u(x) for almost all x, and u u 0. The triangle inequality shows that! k k k1 !

(u ,") (u, ") 2 u u 1 0. | k | k k kL ! Let x be a partition of the interval. We can now choose u such that { i} k uk (xi)=u (xi) for all i.Then

2 This is not an exponent, but a footnote! Clearly, (u, ")isamodulusofcontinuity if and only if (u, ")=o (1) as " 0. ! 454 A Total Variation, Compactness, etc.

(u, ") (u, ") (u, ") T.V. (u) lim inf lim sup T.V. (u) , (A.9) "   " 0 "  " 0 "  | | ! | | ! | | which imply the lemma. ut Observe that we trivially have

˜(u, "):= sup (u, ) " T.V. (u) . (A.10) " | | | || | For functions in Lp care has to be taken as to which points are used in the supremum, since these functions in general are not deﬁned pointwise. The right choice here is to consider only points xi that are points of approximate continuity 3 of u. Lemma A.1 remains valid. We measure the variation in the case of a function u of two variables u = u(x, y) as follows

T.V.x,y (u)= T.V.x (u)(y) dy + T.V.y (u)(x) dx. (A.11) Z Z The extension to functions of n variables is obvious. We include a useful characterization of total variation. Definition A.2. Let ⌦ Rn be an open subset. We define the set of all functions with finite total✓ variation with respect to ⌦ as follows:

1 BV (⌦)= u Lloc(⌦) sup u(x)div(x) dx < . { 2 | 1 n 1} C0 (⌦;R ), 1 ⌦ 2 k k1 Z 3 Afunctionu is said to be approximately continuous at x if there exists a measurable set A such that limr 0 [x r, x + r] A / [x r, x + r] =1(here B denotes the measure of the set B!), and| uis continuous\ | at| xrelative to| A.(EveryLebesguepoint| | is a point of approximate continuity.) The supremum (A.1)isthencalledtheessential variation of the function. However, in the theory of conservation laws it is customary to use the name total variation in this case, too, and we will follow this custom here. ATotalVariation,Compactness,etc. 455

For u BV (⌦)wewrite 2

Du =sup u(x)div(x) dx, k k 1 n C0 (⌦;R ), 1 ⌦ 2 k k1 Z and u BV (⌦) L1(⌦)wedeﬁne 2 \

u = u 1 + Du . k kBV k kL (⌦) k k Remark A.3. If u is integrable with weak derivatives that are integrable functions, we clearly have Du = u(x) dx. k k |r | Z In one space dimension there is a simple relation between Du and T.V. (u) as the next theorem shows. k k Theorem A.4. Let u be a function in L1(I) where I is an interval. Then

T.V. (u)= Du . (A.12) k k Proof. Assume that u has ﬁnite total variation on I. Let ! be a nonnegative function bounded by unity with support in [ 1, 1] and unit integral. Deﬁne 1 x ! (x)= ! , " " " ⇣ ⌘ and u" = ! u. (A.13) " ⇤ Consider points x

" " u (x)0(x) dx = (u )0(x)(x) dx Z Z " (u )0(x) dx  | | Z T.V. (u) ,  which proves the ﬁrst part of the theorem. Now let u be such that

Du := sup u(x)x(x) dx < . k k C1 1 0 Z 2 1 | | First we infer that

" " (u )0(x)(x) dx = u (x)0(x) dx Z Z = (! u)(x)0(x) dx " ⇤ Z = u(x)(! )0(x) dx " ⇤ Z Du . k k Using that (see Exercise A.1)

f L1(I) =sup f(x)(x) dx, k k C1(I), 0 Z 2 1 | | we conclude that " (u )0(x) dx Du . (A.15) | | k k Z Next we show that u L1. Choose a sequence uj BV C1 such that (see, e.g., [62, p. 172]) 2 2 \

u u a.e., u u 0,j , (A.16) j ! k j kL1 ! !1 and

u0 (x) dx Du ,j . (A.17) j !k k !1 Z For any y, z we have z uj(z)=uj(y)+ uj0 (x) dx. Zy Averaging over some bounded interval J I we obtain ✓ ATotalVariation,Compactness,etc. 457 1 u u (y) dy + u0 (x) dx, (A.18) | j| J | j | j ZJ ZI | | which shows that the u are uniformly bounded, and hence u L1.Thus j 2 u"(x) u(x) ! as " 0 at each point of approximate continuity of u. Using points of approximate! continuity x

ut The next result shows that the generalization (A.11) of the total variation to higher dimension yields a (semi)norm that is equivalent to the one coming from bounded variation.

1 n Theorem A.5. Let u L (K) with K =[a1,b1] [an,bn] R . Then 2 ⇥···⇥ ⇢ Du T.V. (u) n Du . k k  k k Proof. Assume first that u BV (K) L1(K), and define the mollifier of u, viz. 2 \ u" = ! u. " ⇤ Then u" u in L1(K) and lim sup Du" Du < ,see[190, ! " k kk k 1 Thm. 5.3.1]. Let uk denote the function where all but the kth variable remain fixed, namely uk(x0,x)=u(x1,...,xk 1,x,xk,...,xn),x0 =(x1,...,xk 1,xk,...,xn) K0. 2 " 1 Then also uk uk in L ([ak,bk]), which implies, using the lower semiconti- nuity of the bounded! variation [190, Thm. 5.2.1] and Theorem A.4,

" T.V.[ak,bk] (uk) lim inf T.V.[ak,bk] (uk) .  " Fatou’s lemma and Theorem A.4 then imply

" T.V.[ak,bk] (uk) dx0 lim inf T.V.[ak,bk] (uk) dx0  " ZK0 ZK0 " =liminf Dku dx " | | ZK 458 A Total Variation, Compactness, etc.

lim sup D u" dx  | k | " ZK Du < . k k 1 This implies that T.V. (u) n Du .  k k

Assume now that K T.V.[ak,bk] (uk) dx0 < for all k =1,...,n. From 0 1 n 1 Theorem A.4 we have for C0 (⌦; R ), 1, that R 2 k k1  @ u dx0 T.V.[ak,bk] (uk) dx0, @xk  ZK0 ZK0 from which it follows that Du T.V. (u). k k ut

Total variation is used to obtain compactness. The appropriate compactness statement is Kolmogorov’s compactness theorem. We say that a subset M of a complete metric space X is (strongly) compact if any infinite subset of it contains a (strongly) convergent sequence. A set is relatively compact if its closure is compact. A subset of a metric space is called totally bounded if it is contained in a finite union of balls of radius " for any ">0 (we call this finite union an "-net). Our starting theorem is the following result. Theorem A.6. A subset M of a complete metric space X is relatively compact if and only if it is totally bounded.

Proof. Consider first the case where M is relatively compact. Assume that there exists an " for which there is no finite " -net. For any element u M 0 0 1 2 there exists an element u2 M such that u1 u2 "0.Sincetheset u ,u is not an " -net, there2 has to be an u k M suchk that u u " { 1 2} 0 3 2 k 1 3k 0 and u2 u3 "0. Continuing inductively construct a sequence uj such that k k { } u u " ,j= k, k j kk 0 6 which clearly cannot have a convergent subsequence, which yields a contradiction. Hence we conclude that there has to exist an "-net for every ". Assume now that we can find a finite "-net for M for every ">0, and let M1 be an arbitrary infinite subset of M. Construct an "-net for M1 with 1 (1) (1) (j) " = ,say u ,...,u . Now let M be the set of those u M1 such 2 1 N1 1 2 that u u(1) 1 . At least one of M (1),...,M(N1) has to be infinite, since k j k 4 1 1 M1 is infinite. Denote (one of) this by M2 and the corresponding element 1 u2. On this set we construct an "-net with " = 4 . Continuing inductively we construct a nested sequence of subsets Mk+1 Mk for k N such that Mk ⇢ 2 has an "-net with " =1/2k,say u(k),...,u(k) . For arbitrary elements u, v 1 Nk of M we have u v u u + u v 1/2k 1.Thesequence u k k k k with u M isk convergent, kk sincek k k { } k 2 k ATotalVariation,Compactness,etc. 459 1 uk+m uk k 1 , k k2 proving that M contains a convergent sequence. 1 ut A result that simplifies our argument is the following. Lemma A.7. Let M be a subset of a metric space X. Assume that for each ">0, there is a totally bounded set A such that dist(f,A) <"for each f M. Then M is totally bounded. 2 Proof. Let A be such that dist(f,A) <"for each f M.SinceA is totally bounded, there exist points x ,...,x in X such that2A n (x ), where 1 n ✓[j=1B" j (y)= z X z y " . B" { 2 |k k } For any f M there exists by assumption some a A such that a f <". Furthermore,2 a x <"for some j.Thus f 2x < 2",whichprovesk k k jk k jk n M (x ). ✓ B2" j j=1 [ Hence M is totally bounded. ut We can state and prove Kolmogorov’s compactness theorem. Theorem A.8 (Kolmogorov’s compactness theorem). Let M be a subset of Lp(⌦), p [1, ), for some open set ⌦ Rn. Then M is relatively compact if and only2 if1 the following three conditions✓ are fulfilled: (i) M is bounded in Lp(⌦), i.e.,

sup u Lp < . u M k k 1 2 (ii) We have

u( + ") u p ( " ) k · kL  | | for a modulus of continuity that is independent of u M (we let u equal zero outside ⌦). 2 (iii) lim u(x) p dx =0uniformly for u M. ↵ x ⌦ x ↵ | | 2 !1 Z{ 2 || | } Remark A.9. In the case ⌦ is bounded, condition (iii) is clearly superﬂuous.

Proof. We start by proving that conditions (i)–(iii) are sucient to show that M is relatively compact. Let ' be a nonnegative and continuous function such that ' 1, '(x) = 1 on x 1, and '(x)=0whenever x 2. Write  | | | | ' (x)='(x/r). From condition (iii) we see that ' u u p 0 as r . r k r kL ! !1 460 A Total Variation, Compactness, etc.

Using Lemma A.7 we see that it suces to show that M = ' u u M is r { r | 2 } totally bounded. Furthermore, we see that Mr satisﬁes (i) and (ii). In other words, we need to prove only that (i) and (ii) together with the existence of some R so that u =0wheneveru M and x R imply that M is totally 2 | | bounded. Let !" be a molliﬁer, that is, 1 x ! C1, 0 ! 1, !dx=1,!(x)= ! . 2 0   " "n " Z ⇣ ⌘ Then

n " = "(0) = z R z " . B B { 2 |k k } Thus n 1 1/p u !" u Lp " ! Lq 1 max ( z ), (A.20) k ⇤ k  k k |B | z " | | | | which together with (ii) proves uniform convergence as " 0 for u M. ! 2 Using Lemma A.7 we see that it suces to show that N" = u !" u M is totally bounded for any ">0. { ⇤ | 2 } H¨older’s inequality yields

u ! (x) u p ! q , | ⇤ " |k kL k "kL so by (i), functions in N" are uniformly bounded. Another application of H¨older’s inequality implies

u ! (x) u ! (y) = (u(x z) u(y z))! (z) dz | ⇤ " ⇤ " | " Z u( + x y) u Lp !" Lq , k · k k k ATotalVariation,Compactness,etc. 461 which together with (ii) proves that N" is equicontinuous. The Arzel`a–Ascoli theorem implies that N" is relatively compact, and hence totally bounded in p n C( R+r). Since the natural embedding of C( R+r)intoL (R ) is bounded, B p n B it follows that N" totally bounded in L (R ) as well. Thus we have proved that conditions (i)–(iii) imply that M is relatively compact. To prove the converse, we assume that M is relatively compact. Condition (i) is clear. Now let ">0. Since M is relatively compact, we can ﬁnd functions p n u1,...,um in L (R ) such that

m M (u ). ✓ B" j j=1 [ n p n Furthermore, since C0(R )isdenseinL (R ), we may as well assume that n uj C0(R ). Clearly, uj( + y) uj p 0 as y 0, and so there is some 2 k · kL ! ! >0suchthat uj( + y) uj Lp " whenever y <.Ifu M and y <, then pickk some· j such thatk u u <", and| | obtain 2 | | k jkLp

u( + z) u p u( + z) u ( + z) k · kL k · j · kLp + u ( + z) u + u u k j · jkLp k j kLp =2 u u + u ( + z) u k j kLp k j · jkLp 3",  proving (ii).

When r is large enough, r uj = uj for all j, and then, with the same choice of j as above, we obtainB

u u p (u uj) p + u uj p 2 u uj p 2", k Br kL k Br kL k kL  k kL  which proves (iii). ut Helly’s theorem is a simple corollary of Kolmogorov’s compactness theorem. Corollary A.10 (Helly’s theorem). Let h be a sequence of functions deﬁned on an interval [a, b], and assume that this sequence satisﬁes T.V. h

The application of Kolmogorov’s theorem in the context of conservation laws relies on the following result.

n Theorem A.11. Let u⌘ : R [0, ) R be a family of functions such that for each positive T , ⇥ 1 !

n u⌘(x, t) CT , (x, t) R [0,T] | | 2 ⇥ for a constant CT independent of ⌘. Assume in addition for all compact B Rn and for t [0,T] that ⇢ 2

sup u⌘(x + ⇠,t) u⌘(x, t) dx ⌫B,T ( ⇢ ), ⇠ ⇢ B | |  | | | || | Z for a modulus of continuity ⌫. Furthermore, assume for s and t in [0,T] that

u (x, t) u (x, s) dx ! ( t s ) as ⌘ 0, | ⌘ ⌘ |  B,T | | ! ZB for some modulus of continuity ! . Then there exists a sequence ⌘ 0 such T j ! that for each t [0,T] the function u⌘j (t) converges to a function u(t) in 1 n 2 { 1} n Lloc(R ). The convergence is in C([0,T]; Lloc(R )).

Remark A.12. If the spatial total variation of u⌘ is uniformly bounded, then u⌘ has a spatial modulus of continuity.

Proof. Kolmogorov’s theorem implies that for each ﬁxed t [0,T] and for 2 any sequence ⌘j 0 there exists a subsequence (still denoted by ⌘j) ⌘j 0 ! 1 n ! such that u⌘j (t) converges to a function u(t)inLloc(R ). Consider{ now a} dense countable subset E of the interval [0,T]. By possibly taking a further subsequence (which we still denote by u ) we ﬁnd that { ⌘j }

u (x, t) u(x, t) dx 0 as ⌘ 0, for t E. ⌘j ! j ! 2 ZB Now let ">0 be given. Then there exists a positive such that ! (˜) " B,T  for all ˜ .Fixt [0,T]. We can ﬁnd a t E with t t .Thus  2 k 2 | k |

u (x, t) u (x, t ) dx ! ( t t ) " for⌘ ˜ ⌘ | ⌘˜ ⌘˜ k |  B,T | k|   ZB and

u⌘ (x, tk) u⌘ (x, tk) dx " for ⌘j ,⌘j ⌘ and tk E. j1 j2  1 2  2 ZB The triangle inequality yields A.1 Notes 463

u⌘ (x, t) u⌘ (x, t) dx j1 j2 ZB u⌘ (x, t) u⌘ (x, tk) dx + u⌘ (x, tk) u⌘ (x, tk) dx  j1 j1 j1 j2 ZB ZB + u⌘ (x, tk) u⌘ (x, t) dx j2 j2 ZB 3",  1 n proving that for each t [0,T] we have that u⌘(t) u(t)inLloc(R ). The bounded convergence theorem2 then shows that !

sup u⌘(x, t) u(x, t) dx 0 as ⌘ 0, t [0,T ] B | | ! ! 2 Z thereby proving the theorem. ut

A.1 Notes

Extensive discussion about total variation can be found, e.g., in [62], [190], and [6]. The proof of Theorem A.6 is taken from Sobolev [168, pp. 28 ↵]. An alternative proof can be found in Yosida [188, p. 13]. The proof of Theo- rem A.4 is from [62, Thm. 1, p. 217]. The proof of Theorem A.5 follows [62, Thm. 2, p. 220] and [190, Thm. 5.3.5]. Kolmogorov’s compactness theorem, Theorem A.8, was ﬁrst proved by Kolmogorov in 1931 [112] in the case where ⌦ is bounded, p>1, and the translation u(x + ") of u(x) is replaced by the spherical mean of u over a ball of radius " in condition (ii). It was extended to the unbounded case by Tamarkin [171] in 1932 and ﬁnally extended to the case with p =1by Tula jkov [181] in 1933. M. Riesz [155] proved the theorem with translations. See also [64]. For a survey, see [79]. For other proofs of Kolmogorov’s theorem, see, e.g., [168, pp. 28 ↵], [37, pp. 69 f], [188, pp. 275 f], and [186, pp. 201 f].

Exercises

A.1 Show that for any f L1(I)wehave 2

f L1(I) =sup f(x)(x) dx. k k C1(I) 0 Z 2 1 | | 464 A Total Variation, Compactness, etc.

A.2 Show that in Helly’s theorem, Corollary A.10, one can ﬁnd a subsequence hn that converges for all x to some function h of bounded variation. Appendix B The Method of Vanishing Viscosity

Details are the only things that interest. Oscar Wilde, Lord Arthur Savile’s Crime (1891)

In this appendix we will give an alternative proof of existence of solutions of scalar multidimensional conservation laws based on the viscous regularization

m @ uµ + f (uµ)=µuµ,uµ = u , (B.1) t @x j |t=0 0 j=1 j X where as usual u denotes the Laplacian j uxj xj . Our starting point will be the following theorem: P 1 m m 2 m Theorem B.1. Let u0 L (R ) L1(R ) C (R ) with bounded deriva- 1 2 \ \ tives and fj C (R) with bounded derivative. Then the Cauchy problem (B.1) has a classical2 solution, denoted by uµ, that satisﬁes1

µ 2 m m u C (R (0, )) C(R [0, )). (B.2) 2 ⇥ 1 \ ⇥ 1 Furthermore, the solution satisﬁes the maximum principle

µ u (t) m u0 m . (B.3) k kL1(R ) k kL1(R ) µ Let v be another solution with initial data v0 satisfying the same properties as u0.Assumeinadditionthatbothu0 and v0 have ﬁnite total variation and are integrable. Then

u( ,t) v( ,t) 1 m u v 1 m , (B.4) k · · kL (R ) k 0 0kL (R ) for all t 0.

1 The existence and regularity result (B.2)isvalidforsystemsofequationsinone spatial dimension as well.

465 466 B The Method of Vanishing Viscosity

Proof. We present the proof in the one-dimensional case only, that is, with m = 1. Let K denote the heat kernel, that is,

1 x2 K(x, t)= exp . (B.5) p4µ⇡t 4µt ⇣ ⌘ n 1 n Deﬁne functions u recursively as follows: Let u = 0, and deﬁne u to be the solution of

n n 1 n n u + f(u ) = µu ,u = u ,n=0, 1, 2,.... (B.6) t x xx |t=0 0 n Then u (t) C1(R) for t positive. Applying Duhamel’s principle we obtain 2

un(x, t)= K(x y, t)u (y) dy 0 Z t n 1 K(x y, t s)f(u (y, s))y ds dy 0 ZZ t 0 @ n 1 = u (x, t) K(x y, t s)f(u (y, s)) ds dy. (B.7) @x ZZ0 n n n 1 Deﬁne v = u u .Then t n+1 @ n n 1 v (x, t)= K(x y, t s) f(u (y, s)) f(u (y, s)) ds dy. @x ZZ0 Using Lipschitz continuity we obtain

t n+1 n @ v (t) f Lip v (s) L ( ) K(x, t s) dx ds L1(R) k k k k 1 R @x Z0 Z t f Lip 1/2 n k k (t s) v (s) ds.  p⇡µ k kL1(R) Z0 Assume that u M for some constant M. Then we claim that | 0| tn/2 vn(t) M f n , (B.8) L1(R) Lip n/2 n+2 k k  k k µ ( 2 ) where we have introduced the gamma function deﬁned by

1 s p (p)= e s ds. Z0 We shall use the following properties of the gamma function. Let the beta function B(p, q) be deﬁned as BTheMethodofVanishingViscosity 467

1 p 1 q 1 B(p, q)= s (1 s) ds. Z0 Then (p) (q) B(p, q)= . (p + q) 1 After a change of variables the last equality implies that 2 = p⇡. Equa- tion (B.8) is clearly correct for n = 0. Assume it to be correct for n.Then t n+1 n+1 1 1/2 n/2 v (x, t) M f (t s) s ds  k kLip p⇡µ(n+1)/2 ( n+2 ) 2 Z0 (n+1)/2 1 n+1 t 1/2 n/2 = M f (1 s) s ds k kLip p⇡µ(n+1)/2 ( n+2 ) 2 Z0 t(n+1)/2 = M f n+1 . (B.9) Lip (n+1)/2 n+3 k k µ ( 2 )

n Hence we conclude that n v converges uniformly on any bounded strip t [0,T], and that 2 P n u =limun =lim vj n n !1 !1 j=0 X exists. The convergence is uniform on the strip t [0,T]. It remains to show that u is a classical solution of the di↵erential equation.2 We immediately infer that t @ u(x, t)=u0(x, t) K(x y, t s)f(u(y, s)) ds dy. (B.10) @x ZZ0 It remains to show that (B.10) implies that u satisﬁes the di↵erential equation

u + f(u) = µu ,u = u . (B.11) t x xx |t=0 0 Next we want to show that u is di↵erentiable. Deﬁne

n Mn(t)=supmax ux (x, s) . x 0 s t | | 2R   Clearly,

t n 1 1/2 ux (x, t) f Lip (t s) Mn 1(s) ds + M0(t). | |k k p⇡µ Z0 Choose B such that M 0 B/2. Then  M (t) B exp(Ct/µ) (B.12) n  468 B The Method of Vanishing Viscosity if C is chosen such that

1 1 1/2 Cs/µ 1 f Lip s e ds . k k p⇡µ  2 Z0 Inequality (B.12) follows by induction: It clearly holds for n = 0. Assume that it holds for n.Then

t n+1 1 1/2 u (s, x) f Lip (t s) Mn(s) ds + B/2 x k k p⇡µ Z0 t Ct/µ 1 1/2 Cs/µ 1 Be f Lip s e ds +  k k p⇡µ 2 ✓ Z0 ◆ BeCt/µ.  Deﬁne n Nn(t)=supmax uxx(x, s) . x 0 s t | | 2R   Choose B˜ max 2N 0,B2 +1 and C˜ C such that { }

1 1 1/2 2Cs/µ˜ 1 2B˜ f 0 + f 00 s e . k kL1 k kL1 p⇡µ  2 Z0 Then we show inductively that

˜ N (t) Be˜ 2Ct/µ. n  The estimate is valid for n = 0. Assume that it holds for n.Then

un+1(x, t) u0 (x, t) xx  xx t 2 @ + f 00 M (s) + f 0 N (s) K(y, t s) dy ds L1 n L1 n 0 k k k k @x Z Z N  0 t 1 2 1/2 + f 0 + f 00 Mn(s) + Nn(s) (t s) ds k kL1 k kL1 p⇡µ Z0 N  0 t 1 2 2Cs/µ Cs/µ˜ 1/2 + f 0 + f 00 B e + e (t s) ds k kL1 k kL1 p⇡µ Z0 t 2Ct/µ˜ 1 2Cs/µ˜ Be˜ 1+2B˜ f 0 + f 00 e ds  k kL1 k kL1 p⇡µ Z0 ˜ ⇣ ⌘ Be˜ 2Ct/µ.  n n n We have now established that u u uniformly and that ux and uxx both are uniformly bounded (in (x, t) and!n). Lemma B.2 (proved after this theorem) BTheMethodofVanishingViscosity 469 implies that indeed u is di↵erentiable and that ux equals the uniform limit n of ux . Performing an integration by parts in (B.10) we ﬁnd that the limit u satisﬁes

t u(x, t)=u0(x, t) K(x y, t s)f(u(y, s)) ds dy. y ZZ0 Applying Lemma B.3 we conclude that u satisﬁes

u + f(u) = µu ,u = u , t x xx |t=0 0 with the required regularity.2 The proof of (B.3) is nothing but the maximum principle. Consider the auxiliary function

U(x, t)=u(x, t) ⌘(t +(⌘x)2/2). Since U as x , U obtains a maximum on R [0,T], say at the !1 | |!1 ⇥ point (x0,t0). We know that

U(x ,t )=u(x ,t ) ⌘(t +(⌘x )2/2) u (0). 0 0 0 0 0 0 0 Hence ⌘3x2 2u(x ,t ) 2u (0) 2⌘t (1) (B.13) 0  0 0 0 0 O independently of ⌘,sinceu is bounded on R [0,T] by construction. Assume that 0

3/2 3 u (x ,t )+f 0(u(x ,t ))u (x ,t ) µu (x ,t ) ⌘ (1) ⌘ µ⌘ t 0 0 0 0 x 0 0 xx 0 0 O > 0 if ⌘ is suciently small. We have used that f 0(u) is bounded and (B.13). This contradicts the assumption that the maximum was attained for t positive. Thus u(x, t) ⌘(t +(⌘x)2/2) sup U(x, 0)  x 3 2 =sup u0(x) ⌘ x /2 x

sup u 0(x),  x which implies that u sup u . By considering ⌘ negative we ﬁnd that u  0 inf u0, from which we conclude that u u0 . k kL1 k kL1

2 The argument up this equality is valid for one-dimensional systems as well. 470 B The Method of Vanishing Viscosity

Lemma B.6 implies that any solution u satisﬁes the property needed for 1 1 our uniqueness estimate, namely that if u0 is in L ,thenu( ,t)isinL .This is so, since we have that ·

u( ,t) 1 u 1 u( ,t) u 1 Ct. k · kL k 0kL k · 0kL 

Furthermore, since u is of bounded variation (which is the case if u0 is of 1 bounded variation), ux is in L , and thus lim x ux(x, t) = 0. Hence, if u0 is in L1 BV , then we have that | |!1 \ d u(x, t) dx = (f(u) + µu ) dx =0. dt x xx Z Z Hence

u(x, t) dx = u0(x) dx. (B.14) Z Z By the Crandall–Tartar lemma, Lemma 2.13,toprove(B.4)itsucesto show that if u0(x) v0(x), then u(x, t) v(x, t). To this end we ﬁrst add a constant term to the viscous equation. More precisely, let u denote the solution of (for simplicity of notation we we let µ = 1 in this part of the argument) u + f(u) = u , u = u . t x xx |t=0 0 In integral form we may write (cf. (B.10))

u(x, t)= K(x y, t)u (y) dy 0 Z t @ K(x y, t s)f(u(y, s)) ds dy t. @x ZZ0 Furthermore,

u(x, t) u(x, t) t @ K(x y, t s) f(u (y, s)) f(u(y, s)) ds dy + t  @x | | ZZ0 t @ f Lip K(x y, t s) u (y, s) u(y, s) ds dy + t k k @x | | ZZ0 t ds f Lip u(s) u(s) + t k k L1 ⇡(t s) | | Z0 t u(s) u(s) dµ(s)+p t  L1 | | Z0 with the new integrable measure dµ(s)= f Lip/ ⇡(t s). Gronwall’s inequality yields that k k p BTheMethodofVanishingViscosity 471

t pt f u(t) u(t) t exp dµ(s) = t exp 2 k kLip , L1  | | | | p⇡ Z0 ! ⇣ ⌘ which implies that u u in L1 as 0. Thus it suces to prove the monotonicity property! for u and v,where!

v + f(v) = v + , v = v . (B.15) t x xx |t=0 0 Let u0 v0. We want to prove that u v . Assume to the contrary that u(x, t)>v(x, t) for some (x, t), and deﬁne

tˆ=inf t u(x, t) >v(x, t) for some x . { | } Pickx ˆ such that u(ˆx, tˆ)=v(ˆx, tˆ). At this point we have

u (ˆx, tˆ)=v (ˆx, tˆ),u (ˆx, tˆ) v (ˆx, tˆ), and u(ˆx, tˆ) v(ˆx, tˆ). x x xx  xx t t However, this implies the contradiction

= u + f 0(u )u u v + f 0(v )v v at the point (ˆx, tˆ) t x xx t x xx whenever is positive. Hence u(x, t) v(x, t) and the solution operator is monotone, and (B.4) holds.  ut In the above proof we needed the following two results. Lemma B.2. Let C2(I) on the interval I,andassumethat n 2 n ! uniformly. If 0 and 00 are bounded, then is di↵erentiable, and k nkL1 k nkL1

0 0 n ! uniformly as n . !1

Proof. The family 0 is clearly equicontinuous and bounded. The Arzel`a– { n} Ascoli theorem implies that a subsequence n0 k converges uniformly to some function .Then { } x x n = 0 dx dx, k nk ! Z Z from which we conclude that 0 = . We will show that the sequence 0 { n} itself converges to . Assume otherwise. Then we have a subsequence n0 j that does not converge to . The Arzela–Ascoli theorem implies the existence{ } ˜ of a further subsequence n0 that converges to some element ,whichis { j0 } di↵erent from .Butthenwehave

x x ˜ dx=limnk =limn = dx, k j j0 Z !1 0!1 Z 472 B The Method of Vanishing Viscosity which shows that = ˜, which is a contradiction. ut Lemma B.3. Let F (x, t) be a continuous function such that

F (x, t) F (y, t) M x y | | | | uniformly in x, y, t. Deﬁne

t u(x, t)= K(x y, t)u (y) dy + K(x y, t s)F (y, s) ds dy. 0 Z ZZ0 Then u is in C2(Rm (0, )) C(Rm [0, )) and satisﬁes ⇥ 1 \ ⇥ 1 u = u + F (x, t),u = u . t xx |t=0 0

Proof. To simplify the presentation we assume that u0 = 0. First we observe that

t t u(x, t)= F (x, s) ds + K(x y, t s) F (y, s) F (x, s) ds dy. Z0 ZZ0 The natural candidate for the time derivative of u is t @ u (x, t)=F (x, t)+ K(x y, t s) F (y, s) F (x, s) ds dy. (B.16) t @t ZZ0 To show that this is well-deﬁned we ﬁrst observe that @ (1) K(x y, t s) O K(x y, 2(t s)). @t  t s Thus t @ K(x y, t s) F (y, s) F (x, s) ds dy @t | | ZZ0 t 1 M (1) K(x y, 2(t s)) y x dy ds  O t s | | Z0 Z t 1 M (1) ds (1) .  O pt s O Z0 Consider now 1 u(x, t + t) u(x, t) u (x, t) t t t+t 1 F (x, s) ds F (x, t)  t Zt t+t 1 + K(x y, t + t s) F (y, s) F (x, s) ds dy t | | Z Zt BTheMethodofVanishingViscosity 473

t 1 + K(x y, t + t s) K(x y, t s) t ZZ0 @ K(x y, t s) F (y, s) F (x, s) dy ds @t | | t+t 1 F (x, s) ds F (x, t)  t Zt t+t 1 + M K(y, t + t s) y dy ds t | | Zt Z t @ @ + M K(y, t + ✓t s) K(y, t s) y ds dy, @t @t | | ZZ0 for some ✓ [0, 1]. We easily see that the ﬁrst two terms vanish in the limit when t 20. The last term can be estimated as follows (where >0): ! t @ @ K(y, t + ✓t s) K(y, t s) y dy ds @t @t | | ZZ0 t @ @ K(y, t + ✓t s) K(y, t s) y dy ds  @t @t | | Z0 Z t @ @ + K(y, t + ✓t s) + K(y, t s) y dy ds t @t @t | | Z Z ✓ ◆ t @ @ K(y, t + ✓t s) K (y, t s) y dy ds  @t @t | | Z0 Z t 1 + (1) K(y, 2(t + ✓t s)) O t t + ✓t s Z Z ✓ 1 + K(y, 2(t s)) y dy ds. t s | | ◆ Choosing suciently small in the second integral we can make that term less then a prescribed ✏. For this ﬁxed we choose t suciently small to make that integral less than ✏. We conclude that indeed (B.16) holds. By using estimates @ (1) K(x, t) O K(x, 2t), @x  p t @2 (1) K(x, t) O K(x, 2t), @x2  t we conclude that the spatial derivatives are given by t @ u (x, t)= K(x y, t s)F (y, s) ds dy, x @x ZZ0 t @2 u (x, t)= K(x y, t s)F (y, s) ds dy, (B.17) xx @x2 ZZ0 474 B The Method of Vanishing Viscosity from which we conclude that

u (x, t) u (x, t) t xx t @ @2 = F (x, t)+ K(x y, t s) 2 K(x y, t s) F (y, s) ds dy 0 @t @x ZZ ⇣ ⌘ = F (x, t). (B.18)

ut Remark B.4. The lemma is obvious if F is suciently di↵erentiable; see, e.g., [139, Theorem 3, p. 144].

Next, we continue by showing directly that as µ 0, the sequence uµ converges to the unique entropy solution of the conservation! law (B.30).{ We} remark that this convergence was already established in Chapter 3 when we considered error estimates. In order to establish our estimates we shall need the following technical result.

Lemma B.5. Let v : Rm R such that v C1(Rm) and v L1 (Rm). Then ! 2 |r |2 v dx 0 as ⌘ 0. v ⌘ |r | ! ! Z| | Proof. By the inverse function theorem, the set

x v(x)=0, v(x) =0 r 6 n o is a smooth (m 1)-dimensional manifold of Rm.Thus

v dx = v dx. v ⌘ |r | 0< v ⌘ |r | Z| | Z | | The integrand (the norm of the gradient times the characteristic function of the region where v is nonzero and less than ⌘) tends pointwise to zero as ⌘ 0. The lemma| follows| using Lebesgue’s dominated convergence theorem. ! ut The key estimates are contained in the next lemma.

2 m Lemma B.6. Assume that u0 C (R ) with bounded derivatives and ﬁ- nite total variation. Let uµ denote2 the solution of equation (B.1). Then the following estimates hold:

T.V. (uµ(t)) T.V. (uµ(0)) , (B.19) µ µ  u (t) u (s) 1 m C t s . (B.20) k kL (R )  | | BTheMethodofVanishingViscosity 475

0 " i " Proof. We set w = @u /@t and w = @u /@xi for i =1,...,m.Thenweﬁnd that i m @w µ i i + fj0(u )w = µw (B.21) @t xj j=1 X for i =0, 1,...,m. Deﬁne the following continuous approximation to the sign function: 1 for x ⌘, sign (x)= x/⌘ for x <⌘, ⌘ 8 | | <> 1 for x ⌘.  i m Multiply (B.21) by sign⌘(w ) and:> integrate over R [0,T] for some T positive. This yields ⇥

T i m T @w i µ i i sign (w ) dt dx + f 0(u )w sign (w ) dt dx ⌘ j x ⌘ m @t m j R 0 j=1 R 0 Z Z X Z Z T i i = µw sign⌘(w ) dt dx. (B.22) m ZR Z0 The ﬁrst term in (B.22) can be written

T @wi T sign (wi) dt dx = wi sign (wi) dt dx ⌘ ⌘ t m @t m ZR Z0 ZR Z0 T i i i w sign⌘0 (w )wt dt dx. m ZR Z0 Here we have that

T i i i 1 i i w sign⌘0 (w )wt dt dx = w wt dt dx m 0 ⌘ wi ⌘ t T ZR Z Z| | Z  i wt dt dx 0,  wi ⌘ t T ! Z| | Z  using Lemma B.5,whichimplies

T i T @w i @ i sign⌘(w ) dt dx w dt dx m @t ! m @t ZR Z0 ZR Z0 = wi(T ) w i(0) , (B.23) L1 L1 as ⌘ 0. The second term in (B.22) reads ! m T µ i i I := fj0(u )w sign⌘(w ) dt dx m xj j=1 ZR Z0 X 476 B The Method of Vanishing Viscosity

m T i µ i i @w = fj0(u )w sign⌘0 (w ) dt dx m @x j=1 R 0 j X Z Z 1 i µ i = w f 0(u ) w dt dx, ⌘ wi ⌘ ·r Z| t |T  where f 0 =(f10 ,...,fm0 ). This can be estimated as follows:

i I sup f 0(u) w dt dx 0, (B.24) | | u | | wi ⌘ t T r ! Z| | Z  as ⌘ 0. Here the supremum is over u uµ(0) . Finally, ! | |k k1 T T i i i 2 i µ w sign⌘(w ) dt dx = µ w sign⌘0 (w ) dt dx 0. m m r  ZR Z0 ZR Z0 (B.25) Using (B.23), (B.24), and (B.25)in(B.22) we obtain, when ⌘ 0, ! wi(T ) wi(0) 0. (B.26) L1 L1  For i =0thisimplies

t µ µ µ @u u (t) u (s) L1 = dt dx k k m @t ZR Zs t 0 w (t˜) dtdx˜  m ZR Zs t 0 ˜ ˜ = w (t) L1 dt Zs t s w0(0) . | | L1 For i 1weuse(B.26)toprove(B.19). Recalling the results from Ap- pendixA,wedeﬁne

i(u, µ)= u (x + µei) u(x) dx and (u, µ)= i(u, µ). m | | R i=1 Z X Then the inequalities (A.10) hold. We have that BTheMethodofVanishingViscosity 477

" " " i(u ( ,t),µ)= u (x + µei,t) u (x, t) dx · Rm | | Z µ i = w (x + ↵ei,t) d↵ dx m ZR Z0 µ i w (x + ↵ei,t) d↵ dx  m ZR Z0 µ wi( ,t) d↵  · L1 Z0 = µ wi( ,t) | | · L1 i µ w ( , 0) 1 | | · L = µ T.V.xi (u0) dx1 dxi 1 dxi+1 dxm. | | m 1 ··· ··· ZR Thus we ﬁnd that

" " (u ( ,t),µ) T.V. (u ( ,t)) = lim inf · T.V. (u0) , · µ 0 µ  ! | | which proves (B.19). ut From the estimates in Lemma B.6 we may conclude, using Helly’s theorem, Corollary A.10, and Theorem A.11, that there exists a (sub)sequence of uµ 1 m { } that converges uniformly in C([0,T]; Lloc(R )) to a function that we denote by u. It remains to show that u is an entropy solution of the conservation law. Let k be in R.Then

(uµ k) + (f(uµ) f(k)) = µ(uµ k). (B.27) t r· Multiply (B.27) by sign (uµ k) times a nonnegative test function and ⌘ integrate over [0,T] Rm.Weﬁnd,whenwewriteU = uµ k, that ⇥

0= Ut sign⌘(U) ZZ ⇣ + (f(uµ) f(k)) sign (U) µ sign (U)U dx dt r· ⌘ ⌘ ⌘ = Usign (U) ⌘ t ZZ ⇣ µ (f(u ) f(k)) sign (U) + sign0 (U) U dx dt · ⌘ r ⌘ r ⌘ + µ U sign (U) dx dt U sign0 (U)U dxdt r ·r ⌘ ⌘ t ZZ ZZ = Usign (U) + sign (U)(f(uµ) f(k)) dx dt ⌘ t ⌘ ·r ZZ 478 B The Method of Vanishing Viscosity

(U) (U) dx t=0 t=T Z µ sign0 (U)(f(u ) f(k)) Udxdt ⌘ ·r ZZ Usign0 (U)U dx dt ⌘ t ZZ 2 + µ sign (U) U dxdt+ µ U sign0 (U)dxdt. ⌘ r ·r |r | ⌘ ZZ ZZ The third and the fourth integrals tend to zero as ⌘ 0(sincef is Lipschitz ! and x sign (x)⌘0 tends weakly to zero), and the last term is nonpositive. Hence

uµ k + sign (uµ k)(f(uµ) f(k)) dx dt | | t ·r ZZ (B.28) (uµ(0) k) t=T dx µ sign (U) U dxdt. |t=0 r ·r Z ZZ Taking µ 0 we see that the right-hand side tends to zero, and we conclude that !

u k + sign (u k)(f(u) f(k)) dx dt | | t ·r ZZ ⇣ ⌘ (B.29) + (u k) dx (u(T ) k) dx 0, 0 |t=0 |t=T Z Z which is the Kruˇzkov entropy condition. We have proved the following result.

2 m m Theorem B.7. Let u0 C (R ) L1(R ) with bounded derivatives and 2 \1 µ ﬁnite total variation, and let fj C (R) with bounded derivative. Let u be the unique solution of (B.1). Then2 there exists a convergent subsequence of µ 1 m u that converges in C([0,T]; Lloc(R )) to a function u that satisﬁes the Kruˇzkov{ } entropy condition (B.29), and hence is the unique solution of

m @ u + f (u)=0,u = u . (B.30) t @x j |t=0 0 j=1 j X

B.1 Notes

Our proof of Theorem B.1 is taken in part from [96], where a similar result is proved for an equation of the form

ut + j(x, t, u)uxj = µu. j=1 X B.1 Notes 479

We are grateful to H. Hanche-Olsen (private communication) for discussions on the proof of this theorem. We have also used [41]. Other proofs can be found; see, e.g., [138]. The conditions of Theorem B.1 can be weakened con- siderably. Alternative proofs of Theorem B.1 can be obtained using the dimensional splitting construction in Section 4.4. Lemma B.6 is familiar; see, e.g., [138]. Our presentation of Lemma B.6 and Theorem B.7 follows in part Bardos et al. [12]. Recently, Bianchini and Bressan [16, 17, 18] have published results con- cerning the vanishing viscosity method for general systems. More precisely, consider the solution u" of the system

u" + A(u")u" = "u" ,u" = u . t x xx |t=0 0 They prove that u" converges to u, the solution of

u + A(u)u =0,u = u , t x |t=0 0 as " 0. Their assumptions are the following: The matrices A(u) are smooth and! strictly hyperbolic in the neighborhood of a compact set K, the initial data u0 has suciently small total variation, and limx u0(x) K.The "!1 2 proof uses an ingenious decomposition of the function ux in terms of gradients of viscous traveling waves selected by a center manifold technique.

Exercises

B.1 Consider the system of parabolic equations

u + f(u) = µu ,u = u , (B.31) t x xx |t=0 0 n with u0(x) R . 2 a. Show that there exists a solution uµ of (B.31) that satisﬁes the regularity condition (B.2). b. Fix temporarily µ = 1, and consider the equation

u + f(u) = u , u = u , (B.32) t x xx |t=0 0 with solution u . Show that u (t) u(t) 1 0, where u solves (B.31)(withµ = 1). ! c. Assume that the ﬂux functionf satisﬁes

f(u1,...,uj 1,u⇤,uj+1,...,un) = const,ui R,i= j, 2 6

for some j and some u⇤ R. Assume that the jth component u0,j of 2 u satisﬁes u u⇤. Show that the jth component u of the solution 0 0,j  j 480 B The Method of Vanishing Viscosity

u of (B.31) satisﬁes u (x, t) u⇤. j  d. Assume that there are constants u

f(u1,...,uj 1,u ,uj+1,...,un) = const,ui R,i= j, ⇤ 2 6 f(u1,...,uj 1,u⇤,uj+1,...,un) = const,ui R,i= j, 2 6

and that u u0,j u⇤. Show that ⇤  

u uj(x, t) u⇤, ⇤   and hence that the region

n u R u uj u⇤ { 2 | ⇤   } is invariant for the solution of (B.31). Systems with this property appear, e.g., in multiphase ﬂow in porous media and chemical chro- matography. For more on invariant regions, see [93] and [85].