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16 Strongly Hyperbolic First-Order Systems in One Space Dimension
I6 Strongly Hyperbolic First-Order Systems in One Space Dimension Some of the results in the preceding sections, in particular those concerning the existence and uniqueness of the solutions to the Cauchy problem, can be extended to certain partial differential equations and systems with variable coefficients, generally called hyperbolic. Here we shall limit ourselves to firstorder systems with variable coefficients, when the number of space variables is equal to one. These systems are of the form
a
a
at
ax
L = - - A(x, t ) -
- B(x, t).
Here A(x, t ) and B(x, t ) are m x m matrices whose entries are smooth functions, say C' functions of (x, t ) in R x 1-T, T [ (R c R" open, T > 0); of course, a/& stands for I d / a t where Z is the m x m identity matrix. We shall assume that the system L is strongly hyperbolic (cf. Definition 15.3):
(1 6.1)
For every xo E 0,to E ] - T, T [ , the eigenvalues of A@,, are real and distinct.
to)
LEMMA16.1. Zf(16.1)holds, therearem C'functionsof(x, t ) i n 0 x 1-T, T [ , 11,. . . , A,,,, representing at each point ( x , t ) the eigenvalues of A(x, t ) . Prooj Since the eigenvalues of A(x, t ) are real and distinct, we may'choose their indices so as to have I , < I 2 < * * . < I , everywhere in R x 1 - T, T [ . Let ( x o , t o ) be an arbitrary point in the latter set, I: thejth eigenvalue of A ( x o , to). Let us denote by f(I; x , t ) the characteristic polynomial of A(x, t ) : f ( A ; x,t ) = det(II - A(x, t ) ) . We have
f(2: ;xo
I
fxq; xo , t o ) # 0.
t o ) = 0,
We may therefore apply the implicit function theorem and conclude that there is a C' function A(x, t ) in a neighborhood of ( x o , t o ) satisfying there 132
Sect. 161
133
STRONGLY HYPERBOLIC FIRST-ORDER SYSTEMS
f (A(x, t ) ; x , t ) = 0 and equal to A; when x have l ( x , t ) = Aj(x,t).
= xo , t = t o . We
must necessarily Q.E.D.
Remark 16.1. The lemma would still be true if the eigenvalues ceased t o be real, provided that they remain distinct. In general, if multiple eigenvalues are allowed, the lemma is not valid. A simple example is provided by
W e now set P j ( x , t ) = (2ni)-'
'
( z l - A(x, t))-' dz,
7,
where y j is a circle in the complex plane o n which n o eigenvalue of A(x, t ) lies and whose interior contains one and only one eigenvalue of A ( x , t ) , the j t h one, Aj(x, t ) . It is clear that P j ( x , t ) is a C' function of ( x , t ) ; it is t h e j t h spectralprojector of A(x, 1 ) . The range g j ( x , t ) of the matrix Pj(x,t)actingon C" is one-dimensional: It is the j t h eigenspace of A(x, t ) (which is onedimensional). We have I = P , +.**+P,. This implies that, for each (x, t ) , Y , ( Xr),, . . . , Y J x , t ) span the whole space C". Of course, we have Pi' = Pi for every j .
LEMMA 16.2. Every point ( x o , t o ) of 0 x ] - T, T [ has an open neighborhood Bo in this set where a matrix-valued function T ( x , t ) is dejined, having the following properties : (16.2)
r ( x , t ) is C' in Soand invertible at every point of So;
(16.3)
we have,for every (x, t ) E Bo,
r-l(x,t)A(x, t ) q x , t ) =
i
U x , r)
o
0
0
... 0
: 1.
A,(x, t )
Proof: F o r each j , we arbitrarily select a nonzero vector v: in T j ( x 0 ,t o ) a n d we set v,(x, t )
=Pj(X,
t)V?.
When x = x o , t = t o , we have vj = v?. hence the vj then form a basis of C" and this will remain true in a small neighborhood of ( x o , to), which we take to be Bo . Let us write rn
134
[Chap. I1
THE CAUCHY PROBLEM
where (el, . . ., em) is the canonical basis of C"; i.e., for each k,all the components of ek are zero, except the kth one, which is equal to one. The matrix ($(x, t))ljj,kjm will be our matrix T(x, t ) . We leave to the student the Q.E.D. verification that it possesses the properties (1 6.2) and (16.3). We are now in a position to study the following local Cauchy problem. In what follows, U will be a small open subset of R, 6 a small number such that 0 < 6 < T . The Cauchy problem is stated as follows : (16.4)
Lu=f(x,t),
(16.5)
XEU,
u(x, 0) = u,(x),
It(
x E u.
About the data, we assume thatf(x, t) is a C' function of (x, t) in R x ] - T, T [ while uo(x) is a C' function in Q. We transform the problem by setting
where T(x, t) is the matrix in Lemma 16.2 when we apply it to (xo , t o ) where xo is a point in U and t o = 0. From now on, until the contrary is specified, U and 6 will be assumed to be sufficiently small so that U x ] - 6,6[ is contained in Bo . We have
where we have set (16.7)
B, = B - rI+ A r , .
Problem (16.4)-(16.5) is thus transformed into (16.8) (16.9)
V , - T - ~ A T V , - B , V = ~ ( ~ , ~ ) ,X E U , V(X, 0) = V0(X),
It1
<6,
x E u.
The advantage of the new system lies in the fact that T-'AT is diagonal [cf. (16.3)]. This advantage will be exploited fdly in the solution of the Cauchy problem. It is instructive to consider first the case where m = 1, i.e., L is a first-order scalar linear partial differential operator.
Sect. 161
STRONGLY HYPERBOLIC FIRST-ORDER SYSTEMS
135
The Scalar Case In this case A and B are complex-valued functions. We shall write A(x, t), instead of A(x, t). Hypothesis (1 6.1) demands that 1be real. We notice that the principal part of L,
a
L - - - A(& O - at
t)
a ax -3
can easily be transformed, by a change of variables, into a/&. It should be emphasized, however, that this transformation is only local, that is, valid in a small open set (which we may identify with the set goabove). Indeed, we solve the nonlinear ordinary differential equation dx ds
- = -A(x,
( 16.10)
s),
with initial condition
-
(16.11)
-y,
where y is a point sufficiently near xo . For small s, (16.10) has a unique solution, which we denote by x = x(y, s). Consider then the change of variables ( 16.12)
t = s.
x = x(y, s),
By virtue of (16.1I), the derivative of x with respect to y does not vanish if s is close to zero (for s = 0, it is equal to one). Consequently, (16.12) is a true C' change of variables. We have
a
a
as
at
a ax
1(x, t ) - = Lo
-=--
The operator L is transformed into
a
L" = - - B ( x ( ~s), , as
S)
(rather, one should say that this is the expression of L in the new coordinates y, s). Since the lines s = 0 and t = 0 coincide and (16.1 1) holds, the problem (16.5)-(16.5) is transformed into (1 6.13)
(16.14)
L # u = m y , s), s),
u, Y E u.
x(y, s) E
u(y, 0) = uo(y>,
Is1
< 6,
136
THE CAUCHY PROBLEM
[Chap. I1
But because of the expression of L’, this latter problem can be viewed as a Cauchy problem (depending on the parameter y ) for a first-order linear ordinary differential equation, and we know how to solve it. It suffices to recall the considerations at the beginning of Sec. 11. We have
We have set (16.16)
P(y, s) = fB(x(y, s’), s’) ds’. 0
Formula (16.15) can be interpreted in an interesting geometrical manner. Notice first that y , s are the new coordinates of the point (x, t). In particular, s = t. On the other hand, when s’varies from 0 to s = t , the point (x(y, s’), s’) ranges over an arc of curve joining ( y , 0) to (x, t ) . The curve in question is the integral curve of (16.10)-(16.11). If it passes through (x, t ) , we shall denote it by C(x, t). At the point (x, t ) , Lo is a partial differentiation in the direction of its tangent. This defines an orientation on C(x, t ) , which is the same as the one defined by taking time as a parameter (we go forward if time increases). By the fundamental theorem o n ordinary differential equations, applied to (16.10)-(16.1 I), through every point ( x , t ) there is one and only one arc of integral curve of (16.10)-(16.11). Of course, C ( x , t ) = C(x,, tl) means that ( x , t ) and (xi, tl) lie on one and the same such integral curve. If U and 6 are sufficiently small, U x 1-6, 6[ will be “trivially fibered” by these curves C(x, t ) , in fact by the curves C(x, 0) as x ranges over U. This means that the effect of (16.12) is to “straighten up” the curves C(x, 0) and transform them into segments of vertical lines x = const. Now, in this picture, y is completely determined by the fact that ( y , 0) is the intersection of C(x, t ) with the straight line t = 0. As for P(y, s) it is equal to the integral of 3 along C(x, I), with respect to the measure dt, from ( y , 0) to (x, t). We shall write B,(x, t ) instead of P(y, s). We have
B,(x, t ) =
f
B(x’, t’) dt‘,
C(x, 1 )
keeping in mind that the bounds are 0 and t ; (x’, t ‘ ) is the variable point on C(x, t). As for (16.5), it reads
(16.17)
keeping in mind that y is dejned by the fact that ( y , 0) is the intersection of C(x, t ) with the axis t = 0; here also the integration is performed from t’ = 0 to t‘ = t even when t < 0.
Sect. 161
137
STRONGLY HYPERBOLIC FIRST-ORDER SYSTEMS
Remark 16.2. Suppose that f = 0. Then the value of u at ( x , t ) depends only on the value of uo at y . In a sense, we may say that the “influence” of the initial data propagates along the curves C(x, t ) .
The General Case We return now to the general (but “ diagonalized ”) problem (16.8)-( 16.9). To solve it we shall now combine the considerations in the scalar case with Picards interation method. But first let us simplify (16.8)-(16.9) by substituting u(x, t ) - uo(x) for u(x, t ) . This amounts to the same as assuming that uo = 0 (note that g must then be replaced by g + T-’ATu,, + B , uo). Next we define a sequence of functions d k )(k = 0, 1, 2, . . .) as follows: ( 16.1 8)
$)
-
r- l A r U ( k )= B X
#
“(k-1)
+ g,
dJj
r=o
= 0,
agreeing that = 0. We shall show that, if U and 6 are sufficiently small, the functions d k )converge uniformly in U x ] - 6, 6[ to a C’ function u(x, t) which is a solution of (16.8)-(16.9); grad d k ) will converge uniformly to grad u. We set w(k)
= u(k)
- u(k
k = 0 , 1 , ....
- 1 ),
Then we define u
+m
=
1 dk),
k=O
and we shall show ,that the sequence {d”}is summable. We derive from (16.18), by subtraction (16.19) (16.20)
wjO)
w;k) -
- r - 1 A r wx (0)= g,
d y x , 0 ) = 0,
r-lArwkk) = B ~ ( ~, - ~w ‘’~ ) ( x0 ,) = 0,
k > 0.
Equations (16.19) and (16.20) are easy to solve. They are essentially the conjunction of m scalar equations. Indeed, let us call d k ) j thejth component of wtk),( B , dk))j that of B , d k !and , g’ that of g. Equation (16.19) is the conjunction of the following equations (when j = 1, . . . , m): (16.21)j
~ . ~ ( o=) jj J g,
w y x , 0 ) = 0,
where L j = d / d t - Aj(x, t) d/dx, whereas (16.20) is the conjunction of the equations (16.22),
LJ . W (=~{ )B ~ #~
( ~ - ‘ ) } j ,
w ( ~ ) ~ (0) x= , 0,
k > 0.
138
[Chap. I1
THE CAUCHY PROBLEM
We call Cj(x, t ) the segment of integral curve of the vector field LJ which joins the axis t = 0 to the point (x, t). We apply formula (16.17): (16.23)
dk-')}j(X', t') dt',
w ( ~ ) ~ (tx) ,=
k > 0,
whereas (1 6.24)
W(O)j(X,
t) =
We introduce now the following notation: Notation 16.1. We denote by 9(x, t ) the smallest compact set containing the point ( x , t ) and hauing the following property: If(x', t') E 9 ( x , t ) , every curue segment Cj(x', t') ( j = 1, . . . , m ) is contained in 9(x, t ) .
We then set, for any vector-valued continuous function p(x, t), IIIcp(x, t>lll = SUP 1 4G', t') I . P(x. 1)
We derive at once from (16.23), IIIw(~)(x, t)lll
S CI t I I I I w ( ~ - ~ ) ( x ,
k > 0,
t>lll,
hence, by iteration, IIIw(~)(x, t)III
5 ~ ~ l ~ l ~ k l 01l11,l ~ ~ ko>~ 0.~ ~ ,
Since, on the other hand, (16.24) implies IIIw'o'(x, t)lIl
5 CI t I
Illdx, t)lII,
we finally obtain (1 6.25)
IIIW(~'(X,
t)lll
S
(eltl)k+'Ills(x, t)lIl,
',
1
k = 0, 1, .-..
Consequently, for I t I < C - the series wCk)converges uniformly. An inequality analogous t o (16.25) can be derived for grad d k )instead of d k ) (g then also has to be modified). To see this, one first estimates w$k) using formula (16.23). This is not as difficult as it may seem, if one first straightens up the curve Cj as indicated in the scalar case. To estimate wik) one uses the ) wik)in combination with (16.21) and (16.22) (expressing estimates of d k and wj') in terms of the other quantities). In these latter estimates, one must take advantage of the fact that the eigenvalues Aj(x, t ) are C' functions of (x, t) [this is then true for the curves Cj(x, t ) ] and of the assumption that the original data f(x, t), uo(x) are also once continuously differentiable. We leave the details to the student. Finally we have obtained an expression of u of the form u = ( I - T)-'g
Sect. 161
STRONGLY HYPERBOLIC FIRST-ORDER SYSTEMS
139
where T is an integral operator. This integral operator T is such that, as shown by formula (16.25), the value of u a t the point ( x , t ) depends only on the values of g in the set 9 ( x , t ) (we are assuming here that uo = 0). We encounter once more the " domain of influence" phenomenon (cf. Theorem 14.1) : In the present case, the influence propagates along the curves Cj(x, t ) .
Example 16.1. Consider the Cauchy problem for the wave equation in one space variable : (16.26)
urr- u,,
=f,
~ ( 0X,) = u ~ ( x ) ,
~ ~ (X0) = , u,(x).
Set u1 = u, u2 = u, - u, and call u the vector with components u l , u2, f the one with components 0, J and uo the one with components u 0 , u1 - uox. The problem (16.26) is then transformed into the following Cauchy problem :
+
Here L , = a/at - a / a x , L, = a/at a l a x ; c,(x, t ) is the curve X I = --t' + (x + t ) , and C,(x, t) is the curve x' = t' + ( x - t). Supposing, for instance, that t is greater than zero, the set 9 ( x , t ) is the portion of the (closed) backward light-cone, with vertex at (x, t ) , which lies in the half-plane t 2 0.
Exercises
16.1. Here x = ( x ' , . . . , x"), n 2 1, and (16.28)
a
L
=-at
a 1 Uj(X, t) ad' n
where the coeffcients aj are real-valued C" functions in an open subset 0 of R"+l. Show that every point ( x o , t o ) of Q has an open neighborhood B0c z2 where the system of ordinary differential equations
(16.29) has a unique solution x
(16.30)
= x(y, s)
x'(
s=ro
such that
=yj,
j = 1 , ...,
where y is a point R",close to xo . Prove that
(16.31)
x = x ( y , s),
t = s,
defines a Cmchange of variables near ( x o , to).
140
[Chap. I1
THE CAUCHY PROBLEM
Apply these facts to the solution of the Cauchy problem (16.32) where u,, is a sufficiently regular function of x in some neighborhood of xoin R". Solve the same problem with Lu =f(instead of Lu = O),fbeing a smooth function in an open neighborhood of ( x o , to) in R"". In particular, let ui be the solution of (16.32) when uo(x)= xj. Relate u = ( u ' , . . . , u") to the transformation ( I 6.31). 16.2. Use the notation of Exercise 16.1. We assume now that u j ( x , s) = uj(x), independent of s ( j = 1, . . . ,n). Denote by y(x, t ) the function obtained by inverting with respect to y the function x = x(y, t ) in (16.31). Set (16.33)
y ( x , t)
= T,x.
Prove that, for sufficiently small values of I t 1, small neighborhood of xo , we have (16.34)
To = Z,
the identity;
I t'I
and for x in a suitably
TtT, ,= T,,,. .
Compute T, in the following cases: (16.35) the u j are all constant; (16.36)
L = -a- - a at ax
(n = 1);
(16.37) 16.3. Using the same notation and hypothesis as in Exercise 16.2, give an example in the case where n = 1 in which the operator T , , defined in (16.33), does not exist for all real values of t. 16.4. Use the notation as in Exercise 16.1, with R = R"". We suppose here that there is a constant K > 0 such that
(16.38)
Igrad,ujI
IrC inthewholeofR"",
j = 1, ... ,n.
Prove that the solution of (16.29)-(16.30) exists for all y E R" and all t E R. 16.5. Consider L = I a / d t - A , where I is the identity m x m matrix and (16.39) where the A j ( x ) are rn x rn matrices with entries that are analytic functions of x in some open neighborhood of the origin in R". Let h be an arbitrary
Sect. 161
STRONGLY HYPERBOLIC FIRST-ORDER SYSTEMS
141
function in R",valued in C". which can be extended to C" as an entire function. Prove that the series at the right of
(1 6.40)
2- k5! (Akh)(x)
+m
(erAh)(x)=
.k
k=o
converges, provided that 1 x [ and I t I are sufficiently small, and defines an analytic function of x in a suitably small open neighborhood of the origin. 16.6. We use the same notation as in Exercise 16.5, except we assume now that m = I (thus the coefficients A j are scalar functions). Compute the operator e f A , defined in (16.40), in each one of the cases (16.35), (16.361, (16.37), and also in the case (16.41)
a
L=-dr
"z 2
( n = I).
Compare with the operator T , defined in (16.33). 16.7. Consider L = a/at - I ( x ) iijdx (we are in the case of one space variable, x ; i.e., n = I ) and suppose that I is an analytic function of x in a neighborhood of zero. Give an interpretation of the operator (16.40) in this case as a "motion" in the complex plane (near the origin) and derive from it that the Cauchy problem is well posed for L if and only if I is real-valued.
a
a
at
ax
L = - - A(x, t ) -
- B(x, t).
Here A(x, t ) and B(x, t ) are m x m matrices whose entries are smooth functions, say C' functions of (x, t ) in R x 1-T, T [ (R c R" open, T > 0); of course, a/& stands for I d / a t where Z is the m x m identity matrix. We shall assume that the system L is strongly hyperbolic (cf. Definition 15.3):
(1 6.1)
For every xo E 0,to E ] - T, T [ , the eigenvalues of A@,, are real and distinct.
to)
LEMMA16.1. Zf(16.1)holds, therearem C'functionsof(x, t ) i n 0 x 1-T, T [ , 11,. . . , A,,,, representing at each point ( x , t ) the eigenvalues of A(x, t ) . Prooj Since the eigenvalues of A(x, t ) are real and distinct, we may'choose their indices so as to have I , < I 2 < * * . < I , everywhere in R x 1 - T, T [ . Let ( x o , t o ) be an arbitrary point in the latter set, I: thejth eigenvalue of A ( x o , to). Let us denote by f(I; x , t ) the characteristic polynomial of A(x, t ) : f ( A ; x,t ) = det(II - A(x, t ) ) . We have
f(2: ;xo
I
fxq; xo , t o ) # 0.
t o ) = 0,
We may therefore apply the implicit function theorem and conclude that there is a C' function A(x, t ) in a neighborhood of ( x o , t o ) satisfying there 132
Sect. 161
133
STRONGLY HYPERBOLIC FIRST-ORDER SYSTEMS
f (A(x, t ) ; x , t ) = 0 and equal to A; when x have l ( x , t ) = Aj(x,t).
= xo , t = t o . We
must necessarily Q.E.D.
Remark 16.1. The lemma would still be true if the eigenvalues ceased t o be real, provided that they remain distinct. In general, if multiple eigenvalues are allowed, the lemma is not valid. A simple example is provided by
W e now set P j ( x , t ) = (2ni)-'
'
( z l - A(x, t))-' dz,
7,
where y j is a circle in the complex plane o n which n o eigenvalue of A(x, t ) lies and whose interior contains one and only one eigenvalue of A ( x , t ) , the j t h one, Aj(x, t ) . It is clear that P j ( x , t ) is a C' function of ( x , t ) ; it is t h e j t h spectralprojector of A(x, 1 ) . The range g j ( x , t ) of the matrix Pj(x,t)actingon C" is one-dimensional: It is the j t h eigenspace of A(x, t ) (which is onedimensional). We have I = P , +.**+P,. This implies that, for each (x, t ) , Y , ( Xr),, . . . , Y J x , t ) span the whole space C". Of course, we have Pi' = Pi for every j .
LEMMA 16.2. Every point ( x o , t o ) of 0 x ] - T, T [ has an open neighborhood Bo in this set where a matrix-valued function T ( x , t ) is dejined, having the following properties : (16.2)
r ( x , t ) is C' in Soand invertible at every point of So;
(16.3)
we have,for every (x, t ) E Bo,
r-l(x,t)A(x, t ) q x , t ) =
i
U x , r)
o
0
0
... 0
: 1.
A,(x, t )
Proof: F o r each j , we arbitrarily select a nonzero vector v: in T j ( x 0 ,t o ) a n d we set v,(x, t )
=Pj(X,
t)V?.
When x = x o , t = t o , we have vj = v?. hence the vj then form a basis of C" and this will remain true in a small neighborhood of ( x o , to), which we take to be Bo . Let us write rn
134
[Chap. I1
THE CAUCHY PROBLEM
where (el, . . ., em) is the canonical basis of C"; i.e., for each k,all the components of ek are zero, except the kth one, which is equal to one. The matrix ($(x, t))ljj,kjm will be our matrix T(x, t ) . We leave to the student the Q.E.D. verification that it possesses the properties (1 6.2) and (16.3). We are now in a position to study the following local Cauchy problem. In what follows, U will be a small open subset of R, 6 a small number such that 0 < 6 < T . The Cauchy problem is stated as follows : (16.4)
Lu=f(x,t),
(16.5)
XEU,
u(x, 0) = u,(x),
It(
x E u.
About the data, we assume thatf(x, t) is a C' function of (x, t) in R x ] - T, T [ while uo(x) is a C' function in Q. We transform the problem by setting
where T(x, t) is the matrix in Lemma 16.2 when we apply it to (xo , t o ) where xo is a point in U and t o = 0. From now on, until the contrary is specified, U and 6 will be assumed to be sufficiently small so that U x ] - 6,6[ is contained in Bo . We have
where we have set (16.7)
B, = B - rI+ A r , .
Problem (16.4)-(16.5) is thus transformed into (16.8) (16.9)
V , - T - ~ A T V , - B , V = ~ ( ~ , ~ ) ,X E U , V(X, 0) = V0(X),
It1
<6,
x E u.
The advantage of the new system lies in the fact that T-'AT is diagonal [cf. (16.3)]. This advantage will be exploited fdly in the solution of the Cauchy problem. It is instructive to consider first the case where m = 1, i.e., L is a first-order scalar linear partial differential operator.
Sect. 161
STRONGLY HYPERBOLIC FIRST-ORDER SYSTEMS
135
The Scalar Case In this case A and B are complex-valued functions. We shall write A(x, t), instead of A(x, t). Hypothesis (1 6.1) demands that 1be real. We notice that the principal part of L,
a
L - - - A(& O - at
t)
a ax -3
can easily be transformed, by a change of variables, into a/&. It should be emphasized, however, that this transformation is only local, that is, valid in a small open set (which we may identify with the set goabove). Indeed, we solve the nonlinear ordinary differential equation dx ds
- = -A(x,
( 16.10)
s),
with initial condition
-
(16.11)
-y,
where y is a point sufficiently near xo . For small s, (16.10) has a unique solution, which we denote by x = x(y, s). Consider then the change of variables ( 16.12)
t = s.
x = x(y, s),
By virtue of (16.1I), the derivative of x with respect to y does not vanish if s is close to zero (for s = 0, it is equal to one). Consequently, (16.12) is a true C' change of variables. We have
a
a
as
at
a ax
1(x, t ) - = Lo
-=--
The operator L is transformed into
a
L" = - - B ( x ( ~s), , as
S)
(rather, one should say that this is the expression of L in the new coordinates y, s). Since the lines s = 0 and t = 0 coincide and (16.1 1) holds, the problem (16.5)-(16.5) is transformed into (1 6.13)
(16.14)
L # u = m y , s), s),
u, Y E u.
x(y, s) E
u(y, 0) = uo(y>,
Is1
< 6,
136
THE CAUCHY PROBLEM
[Chap. I1
But because of the expression of L’, this latter problem can be viewed as a Cauchy problem (depending on the parameter y ) for a first-order linear ordinary differential equation, and we know how to solve it. It suffices to recall the considerations at the beginning of Sec. 11. We have
We have set (16.16)
P(y, s) = fB(x(y, s’), s’) ds’. 0
Formula (16.15) can be interpreted in an interesting geometrical manner. Notice first that y , s are the new coordinates of the point (x, t). In particular, s = t. On the other hand, when s’varies from 0 to s = t , the point (x(y, s’), s’) ranges over an arc of curve joining ( y , 0) to (x, t ) . The curve in question is the integral curve of (16.10)-(16.11). If it passes through (x, t ) , we shall denote it by C(x, t). At the point (x, t ) , Lo is a partial differentiation in the direction of its tangent. This defines an orientation on C(x, t ) , which is the same as the one defined by taking time as a parameter (we go forward if time increases). By the fundamental theorem o n ordinary differential equations, applied to (16.10)-(16.1 I), through every point ( x , t ) there is one and only one arc of integral curve of (16.10)-(16.11). Of course, C ( x , t ) = C(x,, tl) means that ( x , t ) and (xi, tl) lie on one and the same such integral curve. If U and 6 are sufficiently small, U x 1-6, 6[ will be “trivially fibered” by these curves C(x, t ) , in fact by the curves C(x, 0) as x ranges over U. This means that the effect of (16.12) is to “straighten up” the curves C(x, 0) and transform them into segments of vertical lines x = const. Now, in this picture, y is completely determined by the fact that ( y , 0) is the intersection of C(x, t ) with the straight line t = 0. As for P(y, s) it is equal to the integral of 3 along C(x, I), with respect to the measure dt, from ( y , 0) to (x, t). We shall write B,(x, t ) instead of P(y, s). We have
B,(x, t ) =
f
B(x’, t’) dt‘,
C(x, 1 )
keeping in mind that the bounds are 0 and t ; (x’, t ‘ ) is the variable point on C(x, t). As for (16.5), it reads
(16.17)
keeping in mind that y is dejned by the fact that ( y , 0) is the intersection of C(x, t ) with the axis t = 0; here also the integration is performed from t’ = 0 to t‘ = t even when t < 0.
Sect. 161
137
STRONGLY HYPERBOLIC FIRST-ORDER SYSTEMS
Remark 16.2. Suppose that f = 0. Then the value of u at ( x , t ) depends only on the value of uo at y . In a sense, we may say that the “influence” of the initial data propagates along the curves C(x, t ) .
The General Case We return now to the general (but “ diagonalized ”) problem (16.8)-( 16.9). To solve it we shall now combine the considerations in the scalar case with Picards interation method. But first let us simplify (16.8)-(16.9) by substituting u(x, t ) - uo(x) for u(x, t ) . This amounts to the same as assuming that uo = 0 (note that g must then be replaced by g + T-’ATu,, + B , uo). Next we define a sequence of functions d k )(k = 0, 1, 2, . . .) as follows: ( 16.1 8)
$)
-
r- l A r U ( k )= B X
#
“(k-1)
+ g,
dJj
r=o
= 0,
agreeing that = 0. We shall show that, if U and 6 are sufficiently small, the functions d k )converge uniformly in U x ] - 6, 6[ to a C’ function u(x, t) which is a solution of (16.8)-(16.9); grad d k ) will converge uniformly to grad u. We set w(k)
= u(k)
- u(k
k = 0 , 1 , ....
- 1 ),
Then we define u
+m
=
1 dk),
k=O
and we shall show ,that the sequence {d”}is summable. We derive from (16.18), by subtraction (16.19) (16.20)
wjO)
w;k) -
- r - 1 A r wx (0)= g,
d y x , 0 ) = 0,
r-lArwkk) = B ~ ( ~, - ~w ‘’~ ) ( x0 ,) = 0,
k > 0.
Equations (16.19) and (16.20) are easy to solve. They are essentially the conjunction of m scalar equations. Indeed, let us call d k ) j thejth component of wtk),( B , dk))j that of B , d k !and , g’ that of g. Equation (16.19) is the conjunction of the following equations (when j = 1, . . . , m): (16.21)j
~ . ~ ( o=) jj J g,
w y x , 0 ) = 0,
where L j = d / d t - Aj(x, t) d/dx, whereas (16.20) is the conjunction of the equations (16.22),
LJ . W (=~{ )B ~ #~
( ~ - ‘ ) } j ,
w ( ~ ) ~ (0) x= , 0,
k > 0.
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[Chap. I1
THE CAUCHY PROBLEM
We call Cj(x, t ) the segment of integral curve of the vector field LJ which joins the axis t = 0 to the point (x, t). We apply formula (16.17): (16.23)
dk-')}j(X', t') dt',
w ( ~ ) ~ (tx) ,=
k > 0,
whereas (1 6.24)
W(O)j(X,
t) =
We introduce now the following notation: Notation 16.1. We denote by 9(x, t ) the smallest compact set containing the point ( x , t ) and hauing the following property: If(x', t') E 9 ( x , t ) , every curue segment Cj(x', t') ( j = 1, . . . , m ) is contained in 9(x, t ) .
We then set, for any vector-valued continuous function p(x, t), IIIcp(x, t>lll = SUP 1 4G', t') I . P(x. 1)
We derive at once from (16.23), IIIw(~)(x, t)lll
S CI t I I I I w ( ~ - ~ ) ( x ,
k > 0,
t>lll,
hence, by iteration, IIIw(~)(x, t)III
5 ~ ~ l ~ l ~ k l 01l11,l ~ ~ ko>~ 0.~ ~ ,
Since, on the other hand, (16.24) implies IIIw'o'(x, t)lIl
5 CI t I
Illdx, t)lII,
we finally obtain (1 6.25)
IIIW(~'(X,
t)lll
S
(eltl)k+'Ills(x, t)lIl,
',
1
k = 0, 1, .-..
Consequently, for I t I < C - the series wCk)converges uniformly. An inequality analogous t o (16.25) can be derived for grad d k )instead of d k ) (g then also has to be modified). To see this, one first estimates w$k) using formula (16.23). This is not as difficult as it may seem, if one first straightens up the curve Cj as indicated in the scalar case. To estimate wik) one uses the ) wik)in combination with (16.21) and (16.22) (expressing estimates of d k and wj') in terms of the other quantities). In these latter estimates, one must take advantage of the fact that the eigenvalues Aj(x, t ) are C' functions of (x, t) [this is then true for the curves Cj(x, t ) ] and of the assumption that the original data f(x, t), uo(x) are also once continuously differentiable. We leave the details to the student. Finally we have obtained an expression of u of the form u = ( I - T)-'g
Sect. 161
STRONGLY HYPERBOLIC FIRST-ORDER SYSTEMS
139
where T is an integral operator. This integral operator T is such that, as shown by formula (16.25), the value of u a t the point ( x , t ) depends only on the values of g in the set 9 ( x , t ) (we are assuming here that uo = 0). We encounter once more the " domain of influence" phenomenon (cf. Theorem 14.1) : In the present case, the influence propagates along the curves Cj(x, t ) .
Example 16.1. Consider the Cauchy problem for the wave equation in one space variable : (16.26)
urr- u,,
=f,
~ ( 0X,) = u ~ ( x ) ,
~ ~ (X0) = , u,(x).
Set u1 = u, u2 = u, - u, and call u the vector with components u l , u2, f the one with components 0, J and uo the one with components u 0 , u1 - uox. The problem (16.26) is then transformed into the following Cauchy problem :
+
Here L , = a/at - a / a x , L, = a/at a l a x ; c,(x, t ) is the curve X I = --t' + (x + t ) , and C,(x, t) is the curve x' = t' + ( x - t). Supposing, for instance, that t is greater than zero, the set 9 ( x , t ) is the portion of the (closed) backward light-cone, with vertex at (x, t ) , which lies in the half-plane t 2 0.
Exercises
16.1. Here x = ( x ' , . . . , x"), n 2 1, and (16.28)
a
L
=-at
a 1 Uj(X, t) ad' n
where the coeffcients aj are real-valued C" functions in an open subset 0 of R"+l. Show that every point ( x o , t o ) of Q has an open neighborhood B0c z2 where the system of ordinary differential equations
(16.29) has a unique solution x
(16.30)
= x(y, s)
x'(
s=ro
such that
=yj,
j = 1 , ...,
where y is a point R",close to xo . Prove that
(16.31)
x = x ( y , s),
t = s,
defines a Cmchange of variables near ( x o , to).
140
[Chap. I1
THE CAUCHY PROBLEM
Apply these facts to the solution of the Cauchy problem (16.32) where u,, is a sufficiently regular function of x in some neighborhood of xoin R". Solve the same problem with Lu =f(instead of Lu = O),fbeing a smooth function in an open neighborhood of ( x o , to) in R"". In particular, let ui be the solution of (16.32) when uo(x)= xj. Relate u = ( u ' , . . . , u") to the transformation ( I 6.31). 16.2. Use the notation of Exercise 16.1. We assume now that u j ( x , s) = uj(x), independent of s ( j = 1, . . . ,n). Denote by y(x, t ) the function obtained by inverting with respect to y the function x = x(y, t ) in (16.31). Set (16.33)
y ( x , t)
= T,x.
Prove that, for sufficiently small values of I t 1, small neighborhood of xo , we have (16.34)
To = Z,
the identity;
I t'I
and for x in a suitably
TtT, ,= T,,,. .
Compute T, in the following cases: (16.35) the u j are all constant; (16.36)
L = -a- - a at ax
(n = 1);
(16.37) 16.3. Using the same notation and hypothesis as in Exercise 16.2, give an example in the case where n = 1 in which the operator T , , defined in (16.33), does not exist for all real values of t. 16.4. Use the notation as in Exercise 16.1, with R = R"". We suppose here that there is a constant K > 0 such that
(16.38)
Igrad,ujI
IrC inthewholeofR"",
j = 1, ... ,n.
Prove that the solution of (16.29)-(16.30) exists for all y E R" and all t E R. 16.5. Consider L = I a / d t - A , where I is the identity m x m matrix and (16.39) where the A j ( x ) are rn x rn matrices with entries that are analytic functions of x in some open neighborhood of the origin in R". Let h be an arbitrary
Sect. 161
STRONGLY HYPERBOLIC FIRST-ORDER SYSTEMS
141
function in R",valued in C". which can be extended to C" as an entire function. Prove that the series at the right of
(1 6.40)
2- k5! (Akh)(x)
+m
(erAh)(x)=
.k
k=o
converges, provided that 1 x [ and I t I are sufficiently small, and defines an analytic function of x in a suitably small open neighborhood of the origin. 16.6. We use the same notation as in Exercise 16.5, except we assume now that m = I (thus the coefficients A j are scalar functions). Compute the operator e f A , defined in (16.40), in each one of the cases (16.35), (16.361, (16.37), and also in the case (16.41)
a
L=-dr
"z 2
( n = I).
Compare with the operator T , defined in (16.33). 16.7. Consider L = a/at - I ( x ) iijdx (we are in the case of one space variable, x ; i.e., n = I ) and suppose that I is an analytic function of x in a neighborhood of zero. Give an interpretation of the operator (16.40) in this case as a "motion" in the complex plane (near the origin) and derive from it that the Cauchy problem is well posed for L if and only if I is real-valued.