Fattorini H.O., Kerber A. The Cauchy Problem

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3.5. Symmetric Hyperbolic Systems in the Whole Space

147

functions (where verification of dissipativity merely involves integration by

parts) and then extending it to what essentially is its greatest possible

domain. The fact that the extension d is also dissipative will follow from an

approximation argument showing that d is the closure of do and from

Lemma 3.1.11. Finally, m-dissipativity of & (that is, verification of (3.1.8))

will be a consequence of the maximality of the domain of d and of a duality

argument.

We begin to carry out this program defining do by

dou = Lu. (3.5.1)

The domain of do is by no means critical; we define it as the set 6l ° of all

vector functions u = (u 1, ... , u,) whose components belong to 6D. It follows

from the symmetry of( the Ak that

Dk(Ak(x)U(x), U(x)) = (DAk(x)U(x), U(x))

Hence

+2Re(Ak(x)Dku(x), u(x)).

Re(&ou(x), u(x)) = i E Dk(Ak(x)u(x), u(x))

k=1

(DkAk(x)U(x), U(x))

k=1

+Re(B(x)u(x),u(x)) (x E 18m),

thus

Re(l?0u, u) = Re f (dou(x), u(x)) dx = Re f (K(x)u(x), u(x)) dx,

where

K(x) = B(x)- z E DkAk(x).

(3.5.2)

k=I

The following result is then immediate.

3.5.1

Lemma. The operator do is dissipative in E = L2(12m)P if and

only if K(x) is dissipative in C' for all x, that is

Re(K(x)u,u)<O (xERm,uEC°). (3.5.3)

The formal adjoint of do is defined by

(you=- E Dk(Ak(x)u)+B(x)'u

k=1

Ak(x)DkU+ B(x)'-

DkAk(x)u

k=1

E Ak(x)U+B(x)U

k=1

(3.5.4)

where B' indicates the adjoint of B. The domain of Qo is the same as that of

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Dissipative Operators and Applications

do, and d' is the only operator satisfying the Lagrange identity (dou, v) _

(u, a0' V) or

f (eou(x),v(x))dx= f (u(x),Cov(x))dx (3.5.5)

for all u, v E D(do). If we denote by k(x) the matrix associated with Qo in

the same way K(x) is with do, that is, if

K(x) = B(x)- i E DkAk(x),

k=1

a simple computation yields

K(x) = K(x)' (x E Rm). (3.5.6)

The operator Q is now simply defined as the adjoint of do"

d _ (cT')*; (3.5.7)

in other words, D(d) consists of all u E E such that there exists a v E E

with

(u,d'w) _ (v,w)

(w E D(@0)),

and for those u,

du=v.

Moreover, d is closed (Section 4). In view of the Lagrange identity it is plain

that

Qod')*=Cf.

We show next that

(3.5.8)

(I-&)D(e)=E. (3.5.9)

To this end, denote by X the subspace of E consisting of all elements of the

form

(I-e')u (uED(e'))

Let v be an arbitrary element of E; define a functional

in the subspace 9C

by means of the formula

I((I- &o)w) _ (v,w)

(w E D(&o))

Since d0 is dissipative, the fundamental inequality (3.1.9) implies that 1 is

well defined and bounded (its norm does not surpass IlvII). Extending this

functional to the whole space E with the same norm and applying the Riesz

representation theorem, we see that there exists a u E E such that

(u,w-dow)=(v,w) (wED(¢o)),

which clearly implies that u E D((eo)*) = D(d) and that

(I-Cf)u=v,

(3.5.10)

3.5. Symmetric Hyperbolic Systems in the Whole Space

149

proving (3.5.9). Moreover, it follows from our observation about the norm

of 4) that Ilull < lv11. But we stumble upon a serious difficulty here: in order

to conclude that d is m-dissipative from the information available, we need

to show that u is the only solution of (3.5.10). We obviate this difficulty, as

indicated earlier in this section, by showing that Coo = &; however, a growth

condition on the matrices A1,....Awill have to be postulated.

3.5.2 Theorem.

Assume that

IAk(x)I<CP(Ixl)

(xERm, I<k<m),

(3.5.11)

where p is a continuous positive function increasing so slowly that

(

p(r) =oo. (3.5.12)

Let u E D(d). Then there exists a sequence (un) in D(Cio) such that

un- u,

Cl0un-4Clu

in E; in other words, d = Ro.

We note that the dissipativity assumption (3.5.3) is irrelevant here.

The proof will be carried out by means of an argument involving mollifiers

(see Section 8). Recall that if Q is a function in 6D, which is nonnegative,

vanishes in

I x I > 1, and has integral

1, we define a sequence of

operators in E by

Jnu(x)

= (ft" * u)(x)

=f/n(x - y)u(y)

dy,

where f3n(x) = nmI3(nx). Then each Jn is a bounded operator in E (precisely,

II Jn 11 < 1 for all n). Moreover, Jn is self-adjoint and

Jnu- u (n-*oo)

in the norm of E for any u E E. Finally, Jnu is a 000) function for all u E E.

We shall need some additional information on the interaction of the Jn and

0, which is contained in the following result. It will be necessary to assume

temporarily that the Ak, their partial derivatives, and B are bounded:

lAk(x)1I

ID'Ak(x)I,

IB(x)I<C (xERt, 1< j,k<m). (3.5.13)

3.5.3

Lemma. Assume Ai,...,A n, B satisfy (3.5.13). Then (a)

IIJndu - dJnull < CIIuII

(uED(¢))

(3.5.14)

(in particular, Jnu E D(d)).

(b) If u E D(d),

Il J,,du - WJnuIl - 0

(n -* oo).

(3.5.15)

The proof does not require the dissipativity condition (3.5.3). To

prove (3.5.14) (which only plays an auxiliary role with respect to the more

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Dissipative Operators and Applications

fundamental relation (3.5.15)), we begin with the case u E D(&0). We have

J"60u(x)

= E

f /3 (x

- Y)Ak(Y)Dku(Y) dy + f

/3 (x - y)B(Y)u(y)

k=1

=- k

f pk

x-Y)Ak(Y))u(Y)dY

k=1

+ f

y)B(y)u(y)dy

f (DxO,,(x

y)DkAk(Y))u(Y) dY

k=1

+ f

y)B(y)u(y).

(3.5.16)

On the other hand, it is clear that

E D(60) and

k f

-Y)Ak(x)u(Y) dY

k=1

+ f

y)B(x)u(y) dy. (3.5.17)

Hence

E f

k=1

- k f

$ (x-

Y)DkAk(Y)u(Y)dY

k=1

+ f

y)(B(y)-B(x))u(y) dy. (3.5.18)

We now estimate these integrals. The hypotheses on A1,...,Am clearly imply

that each Ak is uniformly Lipschitz continuous in ll

m. Hence

y)(Ak(Y)- Ak(x))I ,<Clx - YI I D'?3 (x - Y)I

and we also have4

= Ef.k(x -

y), (3.5.19)

IP.(x-Y)DkAk(Y)I

(3.5.20)

Y)(B(Y)- B(x))I < C& (x - y)

(3.5.21)

a Here and in other inequalities C denotes a constant, generally not the same in

different inequalities.

3.5. Symmetric Hyperbolic Systems in the Whole Space

151

for x, y E R m. Then we can write

f E Sn.k(x-Y)+C$ (x-Y) lu(Y)I dy

k=1

E (Sn>k * l ul)(x)+C(I3 * lul)(x)

(x E R

k = I

(3.5.22)

We estimate the right-hand side of (3.5.22) with the help of Young's

theorem (see Section 8). We have J/an dx = 1, whereas

f Sn.k(x) dx = CJ

Ixl IDkan(x)l dx

jxi 6I/n

C Ixl IDka(x)l dx.

Accordingly we obtain

MA U - doJnull < CIIuII

(3.5.23)

for all u E D(Qo).

We prove next the corresponding particular case of (3.5.15). Let

u e D(do), K = supp(u). Integrate by parts the first m integrals on the

right-hand side of (3.5.18) after making the substitution D,kan(x - y) _

- Dy an (x - y) in the integrand. The result is

(JJdou-d0Jnu)(x)=

f $n(x-Y)(Ak(x)-Ak(Y))Dku(Y)dy

k=1

-f

$n(x-Y)(B(x)-B(Y))u(Y)dy. (3.5.24)

Now, if x, y E R',

an(x-Y)IAk(x)-Ak(Y)I <Clx-YINn(x-Y)

<2Cn-lan(x- y),

(3.5.25)

the last inequality resulting from the fact that the support of an is contained

in Ix I < 1/n. Observe next that B must be uniformly continuous in K;

therefore, if r> 0, there exists no such that if x, y are arbitrary elements of

I$n(x-Y)(B(x)-B(Y))l <r/n(x-Y)

(3.5.26)

for n >, no. Making use of (3.5.25) and (3.5.26) combined with Young's

inequality in the estimate obtained taking absolute values in (3.5.24), we

obtain

II Jn9ou - 6oJnuII - 0 (n -+ oo)

(3.5.27)

for any uED(do).

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Dissipative Operators and Applications

To prove (3.5.14), we only have to observe that

JnIL - din=

J.(dO)*-(d0)*Jn C (

i")*-(Jnyy)*

c (e1 JJ - id O,)*

(3.5.28)

(which is a consequence of (4.7) and (4.8)) and use (3.5.23) for do'. Finally,

we show (3.5.15). To this end, we take u E D(d) and a sequence (um) c

D(60) such that u,,, -* u (in the norm of E). Then

IIJ.du - QJnull < IIJnWOu. -

nIlOJnumII+CIIu

- umll

Using (3.5.27) the result follows easily.

In the next result the boundedness assumptions (3.5.13) are entirely

given up in exchange for restrictions on u.

3.5.4

Lemma. Let u E D(d) have compact support. Then (3.5.15)

holds.

Proof.

Let K be the support of u; for a > 0 denote by Ka the set of

all x E R m with dist(x, K) < a. Let

be continuously differentiable

v X v matrix functions in R ,

zero outside of K,, and coinciding with

Al,...,Ar in K; likewise, extend the restriction of B to K to a v X v

continuous matrix function h in Rm vanishing outside K,. Let

MLu= E Ak(x)Dku+B(x)u

(3.5.29)

k=1

and go, 9 the operators defined from L in the same way 610, d were defined

from L. Since supp(J,,u )) c K1, it can be easily verified that

WJnu =nn U.JnU, Jn?u = JJl u,

hence the result follows readily from Lemma 3.5.3.

Proof of Theorem 3.5.2.

We may assume that p is infinitely differen-

tiable. Let q> be another infinitely differentiable function of t with q9(t) = 0

if t < 0, qp(t) =1 for r > 1. Given 0 < r < oo, define

1 (0<s<r)

S (daa

{

<S<Sr)

Jr(S)-

1- f

)

(Sr<S)

where Sr is such that p(s) ' has integral 1 in r < s < Sr. Although fr(s) itself

is merely continuous, it is not difficult to (see \thhat

belongs to 6D, its support being contained in

I X I < Sr; its first partials have

support in r < I X I < Sr and

IDkXr(x)l <C/P(IxI) (xER , 1<k<m). (3.5.30)

3.5. Symmetric Hyperbolic Systems in the Whole Space

153

If u E LP(I8"')v, it is clear that

XrU -U

as r -* oo. On the other hand, if u E D(6 ), a simple computation with

adjoints shows that

1(XrU) = (FAkDkXr)U+Xrll U.

In view of (3.5.11) and (3.5.30), we obtain, taking into account that the

DkXr vanish for

I x I < r,

II(AkDkXr)ull <C f I ul2dx,

jxj >r

which tends to zero as r - oo; accordingly,

d (Xru) - C&u.

(3.5.31)

We combine this result with Lemma 3.5.4: the result is Theorem 3.5.2.

Observing that do is dissipative if 3.5.3 holds, and applying Lemma

3.1.11 we obtain

3.5.5 Theorem. Let do, da be the operators defined by (3.5.1) and

(3.5.4), respectively, and let the matrix function

K(x) = B(x)- z E DkAk(x)

(3.5.32)

k=1

be dissipative for all x E R m. Assume in addition that (3.5.11) holds with a

function p satisfying (3.5.12). Then the operator

LT =Coo=(eo)*

(3.5.33)

is m-dissipative

We note that there are no conditions on the growth of DkAk, B,

other than the unilateral ones implicit in the requirement that the matrix K

be dissipative (and which are of course necessary for dissipativity of do or of

any extension thereof).

On the other hand, although condition (3.5.11)-(3.5.12) is not necessary for

the operator d = (LT')* to be m-dissipative, it is best possible in a certain sense for

the equality d, = d to hold.

*3.5.6

Example. Let p be positive, nondecreasing, and differentiable in r > 0.

Assume that

< cc

(3.5.34)

p(r)

Define p ( r) = p (0) for r < 0. Then the operator

L= p(x2)Dl +p(xI)D2 (3.5.35)

in I8 2 does not satisfy the conclusion of Theorem 3.5.2, that is, Ro * t. (However,

is m-dissipative.) See the author, [1980: 1].

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Dissipative Operators and Applications

The following consequence of Theorem 3.5.5 is immediate consider-

ing the operators CTo - wI and - Coo - wI.

3.5.7 Corollary. (a) Let K satisfy

Re(K(x)u,u)<wIu12 (xERm,uECP)

(3.5.36)

for some w > 0 and assume that (3.5.11)-(3.5.12) holds. Then C = Coo =

(d')*' E (2+(1, W).

(b) If (3.5.36) is strengthened to

I Re(K(x) u, u)I < wI u12

(x E Rm, u E U),

(3.5.37)

then d =Coo=(('o)*EC(1,w).

3.5.8 Remark. In all conditions involving the matrix K(x), we can re-

place it by the "symmetrized" matrix Q(x) defined by

2Q(x) = B(x)+B(x)'- DkAk(x)

(3.5.38)

k=I

since Re(K(x)u, u) = (Q(x)u, u). An obvious advantage is the following: if

Q(x) denotes the matrix corresponding to L', then

Q =Q-

(3.5.39)

3.6.

ISOMETRIC PROPAGATORS AND

CONSERVATIVE OPERATORS. DISSIPATIVE

OPERATORS IN HILBERT SPACE

When the equation

u'(t)=Au(t)

(3.6.1)

models a physical system and the norm I I -I I is the energy of the system (as it

is the case, for instance, with Maxwell's equations in Section 1.6), one may

be led, on physical grounds, to conclude that the energy is constant; if the

model (3.6.1) is to be a reasonable one, we must have

IIS(t)ull =1lull

(u(=-E)

(3.6.2)

in t > 0. We say then that the propagator

is isometric in t > 0. The

characterization of the operators A E (2+ for which the propagator of (3.6.1)

is isometric in t > 0 is rather simple.

The operator A is said to be conservative if

Re(u*, Au) = 0

(u E D(A), u* E O(u))

(3.6.3)

and conservative with respect to 9 (0: D(A) - E* a duality map) if

Re(0(u), Au) = 0

(u E D(A)).

(3.6.4)

3.6. Isometric Propagators and Conservative Operators 155

3.6.1 Theorem.

Let A E (2+(1, 0) and assume the propagator of

(3.6.1) is isometric. Then A is conservative with respect to some duality map.

Conversely, assume that A is densely defined, conservative with respect to

some duality map 0: D(A) - E* and

(XI-A)D(A)=E

(3.6.5)

for some A > 0. Then A E e+(1,0) and the propagator of (3.6.1) is isometric.

Proof.

If A is conservative with respect to 0, it is also dissipative

with respect to 0; in view of (3.6.5) we deduce from Theorem 3.1.8 that

A E (2+(1, 0). Let now S(.) be the propagator of (3.6.1) and u E D(AZ ). If

t, h > 0, we have

S(t + h) u = S(t) u + hAS(t) u +

ft+'(t+h-s)S(s)A2uds

=S(t)u+hAS(t)u+hp(t,h), (3.6.6)

where II p(t, h)II -* 0 as h -* 0, uniformly in t > 0. Then

IIS(t)uIIIIS(t+h)uII>I(0(S(t)u),S(t+h)u)I

Re(6(S(t)u), S(t + h)u)

= II

S(t)ull2 + hRe(0(S(t)u), p(t, h)) (3.6.7)

in view of (3.6.6) and of the fact that A is conservative. Assuming that u

consider the set

e = e(u) = (t >, 0; IIS(t)ull = Ilull).

Since 0 E e, e is nonempty; moreover, e is obviously a closed interval. If

[0, oo), let to be the right endpoint of e and a > 0 so small that IIS(t)uII is

bounded away from zero in to < t < to + a. For any such t we write (3.6.7)

and divide by IIS(t)ull; the result is

IIS(t+h)ull -1IS(t)ull-hp(t,h),

(3.6.8)

where ij is nonnegative and rl(t, h) -* 0 ash -* 0 uniformly in to < t < to + a.

Let now a be a small positive number and S > 0 such that

I q (t, h) I < e for

0<h<Sandto<t<to+a.Letto<tj< < tm= to + a be a partition of

the interval to < t < to + a such that tj - tj

_ I

< S (1 < j < m). Then

0 < IIS(to)ull-IIS(to + a) U11

E (IIS(tj-Dull-lls(t1)ull)

j=1

(tj-tj-I),1(tj,tj-tj_1)<ae.

j=t

Since e is arbitrary, it follows that II S(to + a)ull = IIS(to)ull, which con-

tradicts the fact that to is the right endpoint of e. Hence e = [0, oo) and

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Dissipative Operators and Applications

t - IIS(t)uII is constant in t > 0. Using now the fact that D(A2) is dense in

E, a standard approximation argument takes care of the case u E E.

We prove now the opposite implication. Assume that A E E+(1,0)

and let the solution operator of (3.6.1) be isometric. Take t > 0, u E D(A),

and consider the scalar function

p(s)=Re(u*,S(s)u) (s>0),

where u* E O(S(t)u). Clearly (p is continuously differentiable and has a

relative maximum at s = t. Hence

Re(u*, AS(t)u) = 0.

This proves that (3.6.3) is satisfied for those u E D(A) of the form S(t) u,

u E D(A). This would end the proof if A E (2, but some work remains to be

done in the general case. Let u E D(A), a sequence of positive numbers

tending to zero, u* E By passing if necessary to a subsequence,

we can assume that u* - u* weakly, where u* E O(u) (see the proof of

Lemma 3.1.11) and

(u*, Au) = lim(u*,

u) = 0.

This completes the proof.

The corresponding result for A E e is Theorem 3.1.9 (see (3.1.11) and

surrounding comments).

We have already noted in Section 3.1 that when E is a Hilbert space

an operator A is dissipative if and only if

Re(Au, u) = Re(u, Au) <0 (uED(A))

(3.6.9)

and the notion of dissipativity with respect to a duality map coincides also

with (3.6.9). The Hilbert space theory, as it can be expected, is somewhat

richer than the general theory and connects with some celebrated results of

classical functional analysis. We begin with a characterization of m-dissipa-

tive operators; from now on E = H is assumed to be a (complex) Hilbert

space. A dissipative operator A is maximal dissipative if and only if A has no

proper dissipative extension (if A C B and B is dissipative, then A = B).

3.6.2

Theorem. Let A be m-dissipative. Then A is maximal dissipa-

tive. Conversely, let A be maximal dissipative and closed (or densely defined).

Then A is m-dissipative.

Proof. Let A be m-dissipative, B a proper dissipative extension of

A, and u an element of D(B)\D(A). Write v = (I - A)-'(I - B)u. Then

v e D(A) and (I - A)v = (I - B)v = (I - B)u; since u - v, this shows that

I - B is not one-to-one, contradicting inequality (3.1.9) for B. Then we have

A = B, hence A must be maximal.

Conversely, assume A is maximal; note that if A is densely defined

we may construct using Lemma 3.1.11 a closed dissipative extension of A,