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An overdetermined initial and boundary-value problem for a reaction-diffusion equation
Nonlmear Analysis, Theory, Printed in Great Britain.
Methods
& Applications,
Vol.
19, No. 3, pp. 2S9-269,
1992. 0
0362-546X192 $5.00+ .oO 1992 Pergamon Press Ltd
AN OVERDETERMINED INITIAL AND BOUNDARY-VALUE FOR A REACTION-DIFFUSION EQUATIONJ-
PROBLEM
ARTURO DE PABLOand JUAN LUIS VAZQUEZ Departamento de Matemiticas, Universidad Autbnoma, 28049 Madrid, Spain (Received 14 March 1991; received for publication 21 August 1991) Key words and phrases: Reaction-diffusion propagation.
1. INTRODUCTION CONSIDER
equation, overdetermined problem, free boundaries, finite
AND DESCRIPTION
OF RESULTS
the Cauchy problem Ut = (U”),, + UP t 4% 0) = &d-e
when (x, t) E Q = R x (0, 00) for x E R
(1.1)
with m > 1, p < 1 and m + p 2 2. This represents a process of slow diffusion of a substance of density u = U(X,t) accompanied by strong reaction; indeed, the diffusion coefficient, mu”-’ tends to 0 while the reaction coefficient, r/‘-l, tends to 00 as u + 0. We assume that the initial function u,, is nonnegative and locally integrable in the line R and look for nonnegative weak solutions of the above problem. Questions of existence, nonuniqueness of solutions of this problem under the conditipns m > 1 and p < 1, as well as the propagation of supports with finite or infinite velocity have been studied by the authors in the papers [l, 21. Thus, [l] shows existence of (global) solutions for problem (1.1) under a natural growth assumption on the initial data, uO(x) = o( ]x]~‘(*-~)). In general the solutions of the initial-value problem need not be unique due to the non-Lipschitz character of the reaction term u p. Actually, it is shown in [l] that a maximal and a minimal solution exist for every admissible initial data. The analysis of the uniqueness question performed in [2] offers the following results: (i) the maximal solution is always positive everywhere in Q; (ii) if p < 2 - m there exists a unique solution of the initial-value problem unless u0 = 0 (almost uniqueness); (iii) for 2 - m I p < 1 uniqueness holds only when the support of (uO)is the whole line, R. Moreover, the minimal solution has the property of finite speed of propagation and interfaces arise whenever the support of u0 is not the whole line; (iv) in the critical exponent casep = 2 - m the interfaces propagate with a minimum speed c* = 2dm > 0.For p 1 2 - m the minimum speed is 0. Here we are interested in understanding the phenomenon of nonuniqueness. Therefore, we restrict ourselves to the exponent range 2 - m I p < 1. We remind the reader that negative exponents p I 0 are also allowed, with the convention that up = 0 wherever u = 0. We will consider initial data u0 such that problem (1.1) admits more than one solution and we will try to t Work supported by DGICYT Project PB86-0112-CO2 and EEC Contract SCl-0019-C. 259
260
A. DE PABLOand J. L. VAZQUEZ
describe the set of such solutions. Our idea is to prescribe the interface and show that under some assumptions the resulting overdetermined problem is well-posed. This can be expressed as follows: the instability of the level u = 0 due to the part of the equation involving reaction, u, = up, together with the finite propagation property of the diffusion equation u, = (u*)~~, will result in the possibility of infinitely many solutions of the full equation having different interfaces. However, once the interface is determined, so is the solution. In order to clearly formulate our ideas and avoid unnecessary complications we may consider a nonnegative initial function u0 E C(R) fl L”(R) whose support is bounded from above, more specifically such that uO(x) > 0 for x < 0, u&x) = 0 for x 2 0. (1.2) (We could easily replace u0 E L”(R) by the general growth assumption U&Y)= o(]xI~‘(“-~)) in the sequel with only some technical differences. Another, may be more interesting, possibility is to consider compactly supported initial data: this will be briefly commented upon in the final section, but let us now say that no essential novelties appear.) We consider a possible interface x = r(t) and pose the problem: under which conditions on r there exists a (unique) solution to problem (1.1) such that sup(x E R: u(x, t) > 0) = t(t)? With the information
(1.3)
of the interface position, problem (1.1) becomes u, = (u”),
(1.4b)
u(x, 0) = uO(x)
for x E R
u(x, t) > 0
iff x < T(t), t 2 0
1 (1.4c)
+ 2.P
when (x, t) E Q
(1.4a)
(the equation being satisfied in the sense of distributions the following mixed problem:
in Q). This problem is equivalent to
(1Sa)
2.4,= (u”),
(1.5b)
u(x, 0) = u,(x)
for x s 0
(1Sc)
24(x,t) > 0
for x c r(t), t 2 0
(1.5d)
u(<(t), t) = 0
for t 2 0
(u”),(C(t), t) = 0
for t > 0
1 (1.3
+ 2P
(1.4)
when t > 0 and x < l(t)
(1.5)
where both Dirichlet and Neumann conditions are prescribed at the lateral boundary x = r(t), t > 0. Usually, for given initial data u0 in some appropriate class a problem like this has a solution consisting of a function u(x, t) and a so-called free boundary T(t), both are uniquely determined. This is not the case here. We want to prescribe u0 and l and find u which solves (1.4) or (1.5). Clearly, some conditions on c(t) are necessary for the existence of a solution to such a problem. Thus, t(t) must be nonnegative and nondecreasing, since we know that the support of any solution of (1.1) is necessarily expanding in time [ 11. More precise information is obtained by considering the (right-hand) interface s(t) of the minimal solution to problem (1 .l), cf. [l], which is a particular instance of solution of (1.4) or (1.5). An obvious comparison argument shows that c must satisfy for t 2 0. r(t) 1 s(t) (1.6)
Initial and boundary-value problem for a reaction-diffusion
equation
261
As we shall see comparison of free boundaries is not sufficient, but it leads us in the right direction. The actual result implies comparison of the speeds of the interfaces. In order to state our main result we need to introduce the concept of nonincreasing majorant ii, of the initial value uO, defined by ii,(x) = sup(u,(y): y 2 x). (1.7) Let 5(f) be the interface of the minimal solution to (1.1) with initial data ii,. THEOREM1.1. Problem (1.4) admits a solution for every nonnegative continuous u0 satisfying (1.2) and every nonnegative continuous < satisfying for every f2 > C, > 0 <(f*) - r(tl) 2 W
(1.8)
- .%fi).
The result has the disadvantage of depending on comparison with a function $(t), of which not much is known in general, apart from being nonnegative and nondecreasing. However, it clearly shows that problem (1.4) has a solution for every r which has a large enough derivative. Notice also that the solutions corresponding to different values of r are necessarily different. It is interesting to point out a situation where the interface f((t) is explicit. This happens in the critical exponent value p = 2 - m and for initial data satisfying the condition
#y(x)
5
s
1x1
for x < 0.
(1.9)
Clearly, the same condition is satisfied by iiO. Then, as shown in [l, corollary 5.61, the support of the minimal solution ti expands with the minimum speed c* = 2Jm, i.e. s(i) = $(t) = 2Jm t. Condition (1.8) transforms into the following necessary and sufficient criterion. COROLLARY 1.2. Let m + p = 2 and assume u,, satisfies (1.2) and also (1.9). Then problem (1.4) can be solved if and only if (<(t,) - r(t,))/(tz - ti) > 2dm for every t, > t, > 0. More generally, if m + p = 2 equation solution V(x, t, c) given by
(1 .l) admits for every c z c* a travel&g
um-yx,t;C)
=
Zrn -(ctm
l
- x)+
wave (1.10)
where z is related to c by
c=z+E
(1.11)
2
Equation (1.11) gives two values of z for each c > c *. It has been proved in [l] that CTis a minimal solution of (1.1) if we choose the highest of the two values of z resulting from (1.11) when c > c,. Using (1 .lO) we get another criterion. COROLLARY 1.3. Let m + p = 2 and assume that u0 satisfies u,“-‘(x)
I
ZyjJ
t-x),
for some z > dm. Then for every function r such that for every t > 0 D’&t) there exists a solution of problem (1.4).
(1.12) 2 c = z + m/z
262
A. DE PABLOand J. L. VAZQUEZ
In this case, when u,, satisfies (1.12) with z > dm, it is easy to see that the condition on <, though sufficient, is not necessary. The next section is devoted to proving these results. To complete the well-posedness of problem (1.4) we prove in Section 3 the uniqueness of solutions under some additional regularity on the initial data. THEOREM
1.4. Let u0 L 0 be a uniformly continuous function which satisfies (1.2) and is also: (i) monotone decreasing near x = 0 and (ii) uniformly positive away from x = 0. Let 5 2 0 be a nondecreasing continuous function. Then problem (1.4) admits at most one solution u I 0. The above conditions on u0 are in our opinion technical and not essential. In the case p 5 0 we are able to prove uniqueness for general u,, (see theorem 3.2), but unfortunately we have been unable to eliminate the technical conditions when p > 0. It is interesting to compare our results for the overdetermined problem (1.5) with the initial and Dirichlet boundary-value (1.5e) from problem (1.5).
problem (IDP) obtained by eliminating the Neumann condition
THEOREM
1.5. Let u0 be a continuous, nonnegative and bounded real function satisfying (1.2) and let [ be a continuous function in [0, co) with ((0) = 0. Then there exists a continuous weak solution U(X,t) of problem (IDP), which is smooth and positive in 1(x, t): t > 0, x < r(t)). Assume now that u0 and c satisfy the uniqueness requirements of theorem 1.4. Then the solution of the full problem (1.5) is unique. This means that whenever the full problem (1.5) has a solution, for instance if condition (1.8) is satisfied, the solution of problem (IDP) automatically satisfies the Neumann condition (1.5e). On the other hand, when no solutions of the overdetermined problem (1.5) exists, for instance when m + p = 2 and l’(t) < 2dm the solution of (IDP) cannot satisfy the Neumann condition. In the last section we consider the case of compactly supported initial data and study the existence and uniqueness of a solution to problem (1.1) with prescribed (compact) support for each time. The same comment on the mixed problem (1.5) can be done in this situation. 2. CONSTRUCTION
OF SOLUTIONS
In constructing solutions to problem (1.4) or (1.5) we rely, both for m + p = 2 and m + p > 2, on comparison with an important particular solution E(x, t) constructed in [2] with zero initial data, E( *, 0) = 0. This solution is an absolute minimal solution in the sense that for any solution to (1.1) and for any (x0, to) E Q we have 2.4(x0,to) > 0 =a u(x, t) z E(x - x,, t - to>
in Q.
(2.1)
Moreover, E has the self-similar form E(x, t) = t”q((xlt+)
(2.2)
where Q”= l/(1 - p), p = (m - p)/(2(1 - p)) and q: [0, co) -+ [0, 00) is a continuous and decreasing function with compact support. Observe that the support of E is bounded by two interfaces x = &p(t) where p(t) = p. to, /3 2 1. This represents a minimal growth rate for the
Initial and boundary-value problem for a reaction-diffusion
equation
263
interface of every solution to (l.l), i.e. we have the growth condition on the interface r(t) of problem (1.4): for t 2 to L 0. (2.3) r(t) 2 &I) + P& - M When m + p = 2 we have p = 1 and (2.3) implies the already explained minimal speed of propagation of the interface, with p0 = c* > 0. After these preliminaries, we begin the construction of solutions for problem (1.4) with the critical case, which is somewhat easier. Proof of corollary 1.2. Step 1. Associated to problem (1.1) we define the map H which sends each nonnegative function u0 into the minimal solution u = ZZ(u,) to the Cauchy problem (1.1). Let also x = s,(t) be the (right-hand) interface of u. First of all we observe that when u0 satisfies (1.9), the interface of the minimal solution u = ZZ(u,) is exactly s,(t) = 2Jmt, cf. [l]. Now take r > 0. We next construct a solution to (1.4) in the special case t(t) = r + 2dmt. This means that we can obtain solutions to (1.1) whose support jumps instantly from zero to rat t = 0. For that purpose we consider, for fixed E > 0, the minimal solution (2.4)
u, = ~@rJ + EXto,J
where x1 stands for the characteristic function of the set I. Using the absolute minimal solution E, (2.2), to compare from below and the travelling wave with minimal velocity U, = U(x, t; 2dm), see (1. lo), to compare from above, we obtain Z.&(x,t) 2 cot > 0
for 0 I x 5 r, t > 0
uE(x, t) 5 U,(x - r - 6, t) where 6 = (dm/(m
- l)),P’
for x E R, t > 0
(2.5) (2.6)
and c,, = p(O) with Edas in (2.2). Thus
r + 2Jmt 5 s,,(t) 5 r + 6 + 2Jmt
for t > 0.
(2.7)
On the other hand u,, I uEZfor er 2 Q, and therefore there exists the limit v = lim u, E-O
(2.8)
which is a solution to (1.4a), (1.4b) and also (1.4~) holds with c(t) = r + 2dmt. Finally v(x, t) 5 U,(x - r, t) = U,(x - r - 2Jmt, 0).
(2.9)
Step 2. Now let c(t) L 0 be any continuous function such that r(O) = 0 and (r(t2) - r(tl))/(t2 - tr) 2 2Jm for every t2 > t, > 0. We fix t > 0 and n E N and consider the partition (tj = jt/n), 0 5 j I n - 1, of the interval [0, t]. We construct an approximate solution un as follows: we set u”(0) = uo, and then proceed iteratively in each subinterval as in step 1, extending the support of the solution at every mesh time tj up to r(t), 0 5 T I t and solving equation (1.4a) in the time interval [tj, tj+l].
A. DE PABLO and J. L. VAZQUEZ
264
To be precise, and that
assume that un is known 0x9
for 0 s t 5 tj. We define
I r(t) for 0 I t I tj
s,(t)
(2.10)
t) 5 U*(x - r(t), 0)
un in [tj, tj+J by means Un(X,
with 0 I t I tj,l
in [0, tj], that the interface
T +
tj)
=
of the formula
Ei_;H(z&)(X,
T)
(2.11a)
- tj, where (2.1 lb)
Ujl~(x) = Un(X9 tj) + EXIO,g(t,)](x)* We have, as before Zln(X, 5)
2
Cg(T
-
fortj
tj)
(2.12)
and Un(X,
Thanks
5)
to our induction
I
U*(X
-
r(tj)*
assumptions
?T -
for
tj)
tj
<
7 I
on U” and r, this implies
tj+l ,
x
E
R.
(2.13)
for the solution (2.14)
U%, r) 5 V*(x - r(r), 0) if
tj 5
T I
tj+l
and for the interface S@(r) =
in the same time interval. Once we have concluded limit
r(tj)
+
2dm(r
-
tj)
The induction is thus complete. the construction of the approximate
(2.15)
solutions
zP we may pass to the
v = lim 2.4’. n-cc
(2.16)
It is easy to check that v is a solution of (1.4a) and (1.4b). Thanks to the above estimates we can show that s,.(t) converges to r(t), which actually is the interface for v (use (2.12) and (2.13)). We have obtained therefore a solution of problem (1.4). n
Remark solution
2.1. It is easy to check that the solution obtained by the previous method is the minimal of the problem (1:4) if we consider in each iteration, instead of (2.11) the function U”(X,
t +
tj)
=
Ei,;
~~oH([uJJE
-
61+)(X,
(2.17)
T).
Also the proof of corollary 1.3 is immediate using in (2.10) instead of U, the travelling wave U(x, t; c) with velocity c = z + m/z. On the contrary, the condition on { in corollary 1.3 is not sharp as follows from the asymptotic result proved in [l, theorem 5.51. PROPOSITION
2.2.
is asymptotically
If u,, is bounded linear and s(t)/t
then the interface of the minimal + 2dm as t + 00.
solution
to problem
To treat the general case m + p L 2 we first suppose that the initial datum decreasing in (-00, 0), so that the minimal solution to (1.1) is also monotone.
(1.1)
is monotone
Initial and boundary-value
problem
for a reaction-diffusion
265
equation
LEMMA2.3. If u0 is nonincreasing in (-00, 0), then the minimal solution li = H(u,) to (1.1) is nonincreasing in (-00, r(t)) for each t > 0. Proof, Apply the classical strong maximum principle [3] to the equation satisfied by (u,), to get the desired conclusion. n
With this property in mind we can repeat the above construction
of the sequence u”.
LEMMA2.4. Assume u0 is nonincreasing in (-00, 0) and satisfying (1.2), let 2 = H(u,) be the minimal solution with initial data u,, and let g(t) be its interface. Then for every continuous function &j(t) such that [(t2) - T(tl) z s^(t2) - f(tJ for every t, > t, > 0, problem (1.4) admits a solution. Proof. Recall the notation of the proof of corollary 1.2. Observe that if fi(*, t) is nonincreasing we have fi(x, r) I ti(x - x0, r) for every (x, r) E Q, x0 2 0. Applying a comparison between minimal solutions in each iteration we get for r I tj by induction 3((t)
I
s,“(t)
Un(X,
5) 2
E(X
-
r(tj),
U’(X,
t)
ti(X
-
r(tj)
5
‘5 -
(2.18)
tj)
+ s”(tj),
t)
and thus t(tj)
+ Po(t
-
tj)”
5
Sun(t)
I
S(T)
+
<(tj)
-
(2.19)
S*(tj)
for tj< ‘5~ tj+l. We now need the condition on r: [(t2) - r(tJ remaining details follow the proof of corollary 1.2. n
2 3(t2) - .f(tl).
For the
Now let u0 be any nonnegative continuous function satisfying (1.2), ii, be its nonincreasing majorant and ii = H(ii,). Then theorem 1.1 is obtained as a consequence of the above lemma. n 3. UNIQUENESS Here we prove theorem 1.4. The strategy consists of, given any solution to problem (1.4), obtaining an approximation sequence of resealed supersolutions and use comparison to conclude that our solution is the maximal solution to (1.4). Clearly this implies uniqueness. Before doing that we recall that comparison does not hold in general for the solutions of problem (l.l), precisely because uniqueness fails. Nevertheless we have the following partial comparison principle. LEMMA3.1. Let u1 , u2 be a subsolution and a supersolution respectively to equation (1.4a), and let s,(t), s2(t) be their interfaces. Assume that u,(x, 0) <
u,k
s1(t) <
0)
SZ(f)
for x < s,(O) for t 2 0.
(3.1) (3.2)
Then u,(x, t) <
u2(x,
t)
for x < s2(t), t > 0.
(3.3)
A. DE PABLOand J. L. VAZQUEZ
266
By supersolution (resp. subsolution) we understand a continuous and nonnegative function u(x, t) which satisfies U, - (u”), - up I 0 (resp. I 0) in the sense of distributions in Q. In our application such super- and subsolutions are smooth whenever they are positive. Therefore, the proof of lemma 3.1 consists of using the classical maximum principle in the region x < si(t), where both functions are positive and the equation is parabolic nondegenerate. Proof of theorem 1.4. Given a solution u to problem (1.4) we define the function Uk(X, t) = u/&x,
It is easy to check that uk is a supersolution for the interfaces
=
k>
t) = ku(x - p, k”?)
P
+
l,p>O.
(3.4)
to equation (1.4a) for every k > 1, Also we have,
W=‘O
>
t(t)
for t 2 0.
Now we compare the initial values. If u0 is monotone for -2~ < x < 0, we immediately have uk(-&
0)
-
uO(x)
>
for --u < x < 217= r(O).
0
On the other hand, put A = inf u0 > 0, and take E > 0 small such that E + JE < A. From X5-‘I the uniform continuity of u,, we can choose p > 0 small enough such that ukb,
0) - u&d = kuo(x - P) - uoW > (k - l&(x) L
- ke
(k - l)A - kc
> kdc - (E + &) = 0 for x I -v (take k = 1 + JE). Applying lemma 3.1 we get that any solution to problem (1.4) lies below &. Since r& tends to u as k tends to 1 and p tends to 0 (E tends to 0), we conclude that u is the maximal solution n to problem (1.4), and since u is arbitrary this implies that u is the only solution.
Let us remark that when u0 is monotone of uO. We also notice that alternative proofs on z+,, for instance (u,“-‘)“(x) s -co. None when the exponent p 5 0 we prove a clean
we do not need to assume the uniform continuity of uniqueness can be done under other restrictions of these restrictions seems to be essential. Actually, uniqueness result.
THEOREM 3.2. Let m > 1,p I 0 and m + p 2 2 and let uo,
Proof. Let u be the solution to problem (1.4) obtained in theorem 1.1. From remark 2.1 we assume that u is the minimal solution to that problem. Now let u be any other solution to (1.4). We fix T > 0 and put QT = R x (0,T). Since u and u satisfy (1.4a) and (1.4b) in the sense of
Initial and boundary-value
problem
for a reaction-diffusion
equation
267
distributions, we may have that for every p E C”(Qr) with compact support in x for every 0 % t 5 T, the following identity holds: 01
(v - 4(x, t)P(x* t) d-x
=I-t1 iR
[(u - u)cp, + (u” - zP)q,
+ (up - up)(p](x, T) dxdr.
0.R
(3.5)
If u0 is bounded, we have u and u bounded in Qr [2], so equality (3.5) still holds for test functions $!JE Cm(Qr) fl L’(Q*). If we set P(X, s) - zP(X, s) u(x, 4 - u(x, s)
a = a(x, S) =
mzF1(x,
if u(x, s) f u(x, s) (3.6) otherwise
s)
the above equality becomes
=
[(u - U)(Pt+ w,) + (UP- up)p](x, T) dx dz.
(3.7)
+ x2) for some constant k > 0 to be chosen. We We try the test function ~(x, T) = e “‘-“/(l observe that 0 5 4x, 4 5 c(m, p, IIu~/I~, 77 and qt = -@, Pxx 5 V* Then if k is large enough we have pt + a~,, < 0, and since u 2 U, (3.7) is reduced to 01
s
I
R (u
- u)(x, t)G46 t) dx (VP - up)(x, r)p(x, r) dx dr.
(3.8)
Now we observe that u(x, z) > u(x, r) > 0 for x < c(r) and then up(x, r) - up(x, T) 5 0 for x < c(s), while v(x, r) = u(x, r) = 0 for x I r(s) so that vp(x, r) = up(x, T) = 0 for x L T(T). This implies, within (3.8), that v(*, t) = u(-, t) for every 0 5 t I T. n 4. EXISTENCE
FOR
THE
DIRICHLET
PROBLEM
In this section we outline the proof of theorem 1.5. The idea consists of first adding E > 0 to the initial data and taking as boundary condition U(X,t) = E for t 2 0 and x = c(t). In this way we obtain a nondegenerate problem that can be solved by means of the classical quasilinear theory [3], thus obtaining a classical solution U, > E. We then let E + 0 and use the monotonicity of the solution with respect to E to pass to the limit U(X,t) = lim u,(x, t), which E-O is a weak solution of the problem originally proposed.
A. DEPABLOand J. L. VAZQUEZ
268
5. SOLUTIONS We consider
here the case where the initial z+,(x) > 0
Then, problem
WITH COMPACT
given two functions (1.4) by
data have compact uO(x) = 0
for 1x1 < R,
u(x, t) > 0
SUPPORT say
otherwise.
1 R, we substitute
if and only if -r,(t)
support,
(5.1)
the boundary
< x < &(t),
condition
(1.4~) in
t 2 0
(1.4c’)
and study the resulting problem (1.4’) = (1.4a) + (1.4b) + (1.4~‘). The construction of the solutions is exactly the same as for problem (1.4), but here we also have to introduce the nondecreasing majorant of u,-,. Define then the functions &,1(X) = suPluo(Y): Y 2 xl (5.2) %,2(X) = suPl%(Y): and let -si(t), s*(t) be the interfaces of the minimal H(uO,i), H(u&. We have the following theorem. THEOREM 5.1. For every continuous functions & 2 R, i = I,2 satisfying
Y 5 xf solutions
associated
to those initial
values,
function u0 satisfying (5. I), and every pair of continuous for every t2 > t1 > 0
(1.4’) admits
(5.3)
a solution.
For the uniqueness of solutions we follow the proof using a different resealed sequence.
of theorem
1.4 given in Section
3, but
THEOREM 5.2. Let u. be a Lipschitz continuous function which is monotone near 1x1 = R. Then for every pair of continuous functions ri 2 R, problem (1.4’) admits at most one solution. Proof.
Given
a solution
u of the above problem, u,(x,
which is a supersolution
t) = ku(k-“x,
to equation
we define
the function k>
/?-‘-2*t)
1
(1.5a) if 0 s o( I (m - p)/2.
(5.4) Its interfaces
satisfy
for t r 0 ck,i(f) = ka~i(km-1-2at) > [i(t) if 0 < Q! < (m - 1)/2. The comparison between the initial values of u and uk is again easy near the border, using the monotonicity. To compare in the interior assume f+,(x) 2 A for 1x1 5 R - q and let M be the Lipschitz constant of uO. We have u,(x,
0) - uo(x) = ku,(k-“x)
for 1x1 I R - q if 0 < (Y < 6/(MR).
- uO(x)
2 (k -
l)u,(x)
- M(k”
1 (k -
l)A - Ma(k
-
As in the proof of theorem
- l)R 1)R > 0 1.4 this implies uniqueness.
n
Initial
and boundary-value
6. SOME
problem
EXTENSIONS
for a reaction-diffusion
AND
equation
269
COMMENTS
We have assumed for simplicity in the construction of solutions to problem (1.4), that the interface is given by a continuous function r(t). However, it is clear from the proofs of Section 2 that we may allow < to have positive jump discontinuities. In particular, if we take t(O+) = r > 0 in theorem 1.1 we have to deal with the function
and compare its interface with r. Moreover, we may set r(t) = ~0 for I E [to, m), with C, L 0 by simply using the construction of this paper for 0 I t < t,, and then taking u equal to the maximal solution for t 2 t,, . The same ideas apply to problem (1.4’). Besides, in the existence and uniqueness theorems we do not assume that the slope of the initial data is bounded near the interface(s), so we may have also discontinuous initial values. Observe that the minimal solution to problem (1.1) has finite interface(s) even in this case. In all our constructions we have insisted in constructing solutions with connected support for each time t > 0. However, the existence of the special solution E(x, t) shows that we can start a new “piece” of the positivity set of a solution at any instant and at any place in the complement of the support. The interface of this new piece can be chosen subject to the above explained conditions until it merges with previous components. See also [2]. REFERENCES 1. DE PABLO A. & VAZQUEZ J. L., Travelling waves and finite propagation in a reaction-diffusion equation, J. diff. Eqns 93, 19-61 (1991). 2. DE PABLO A. & VAZQUEZ J. L., Balance between strong reaction and slow diffusion, Communs partial diff. Eqns 15(2), 159-183 (1990). 3. LADYZENSKAJA 0. A., SOLONNIKOV V. A. & URALCEVA N. N., Linear and quasilinear equation of parabolic type,
Transl. Math. Monogr.,
Am. math. Sot. Transl. (1968).
Methods
& Applications,
Vol.
19, No. 3, pp. 2S9-269,
1992. 0
0362-546X192 $5.00+ .oO 1992 Pergamon Press Ltd
AN OVERDETERMINED INITIAL AND BOUNDARY-VALUE FOR A REACTION-DIFFUSION EQUATIONJ-
PROBLEM
ARTURO DE PABLOand JUAN LUIS VAZQUEZ Departamento de Matemiticas, Universidad Autbnoma, 28049 Madrid, Spain (Received 14 March 1991; received for publication 21 August 1991) Key words and phrases: Reaction-diffusion propagation.
1. INTRODUCTION CONSIDER
equation, overdetermined problem, free boundaries, finite
AND DESCRIPTION
OF RESULTS
the Cauchy problem Ut = (U”),, + UP t 4% 0) = &d-e
when (x, t) E Q = R x (0, 00) for x E R
(1.1)
with m > 1, p < 1 and m + p 2 2. This represents a process of slow diffusion of a substance of density u = U(X,t) accompanied by strong reaction; indeed, the diffusion coefficient, mu”-’ tends to 0 while the reaction coefficient, r/‘-l, tends to 00 as u + 0. We assume that the initial function u,, is nonnegative and locally integrable in the line R and look for nonnegative weak solutions of the above problem. Questions of existence, nonuniqueness of solutions of this problem under the conditipns m > 1 and p < 1, as well as the propagation of supports with finite or infinite velocity have been studied by the authors in the papers [l, 21. Thus, [l] shows existence of (global) solutions for problem (1.1) under a natural growth assumption on the initial data, uO(x) = o( ]x]~‘(*-~)). In general the solutions of the initial-value problem need not be unique due to the non-Lipschitz character of the reaction term u p. Actually, it is shown in [l] that a maximal and a minimal solution exist for every admissible initial data. The analysis of the uniqueness question performed in [2] offers the following results: (i) the maximal solution is always positive everywhere in Q; (ii) if p < 2 - m there exists a unique solution of the initial-value problem unless u0 = 0 (almost uniqueness); (iii) for 2 - m I p < 1 uniqueness holds only when the support of (uO)is the whole line, R. Moreover, the minimal solution has the property of finite speed of propagation and interfaces arise whenever the support of u0 is not the whole line; (iv) in the critical exponent casep = 2 - m the interfaces propagate with a minimum speed c* = 2dm > 0.For p 1 2 - m the minimum speed is 0. Here we are interested in understanding the phenomenon of nonuniqueness. Therefore, we restrict ourselves to the exponent range 2 - m I p < 1. We remind the reader that negative exponents p I 0 are also allowed, with the convention that up = 0 wherever u = 0. We will consider initial data u0 such that problem (1.1) admits more than one solution and we will try to t Work supported by DGICYT Project PB86-0112-CO2 and EEC Contract SCl-0019-C. 259
260
A. DE PABLOand J. L. VAZQUEZ
describe the set of such solutions. Our idea is to prescribe the interface and show that under some assumptions the resulting overdetermined problem is well-posed. This can be expressed as follows: the instability of the level u = 0 due to the part of the equation involving reaction, u, = up, together with the finite propagation property of the diffusion equation u, = (u*)~~, will result in the possibility of infinitely many solutions of the full equation having different interfaces. However, once the interface is determined, so is the solution. In order to clearly formulate our ideas and avoid unnecessary complications we may consider a nonnegative initial function u0 E C(R) fl L”(R) whose support is bounded from above, more specifically such that uO(x) > 0 for x < 0, u&x) = 0 for x 2 0. (1.2) (We could easily replace u0 E L”(R) by the general growth assumption U&Y)= o(]xI~‘(“-~)) in the sequel with only some technical differences. Another, may be more interesting, possibility is to consider compactly supported initial data: this will be briefly commented upon in the final section, but let us now say that no essential novelties appear.) We consider a possible interface x = r(t) and pose the problem: under which conditions on r there exists a (unique) solution to problem (1.1) such that sup(x E R: u(x, t) > 0) = t(t)? With the information
(1.3)
of the interface position, problem (1.1) becomes u, = (u”),
(1.4b)
u(x, 0) = uO(x)
for x E R
u(x, t) > 0
iff x < T(t), t 2 0
1 (1.4c)
+ 2.P
when (x, t) E Q
(1.4a)
(the equation being satisfied in the sense of distributions the following mixed problem:
in Q). This problem is equivalent to
(1Sa)
2.4,= (u”),
(1.5b)
u(x, 0) = u,(x)
for x s 0
(1Sc)
24(x,t) > 0
for x c r(t), t 2 0
(1.5d)
u(<(t), t) = 0
for t 2 0
(u”),(C(t), t) = 0
for t > 0
1 (1.3
+ 2P
(1.4)
when t > 0 and x < l(t)
(1.5)
where both Dirichlet and Neumann conditions are prescribed at the lateral boundary x = r(t), t > 0. Usually, for given initial data u0 in some appropriate class a problem like this has a solution consisting of a function u(x, t) and a so-called free boundary T(t), both are uniquely determined. This is not the case here. We want to prescribe u0 and l and find u which solves (1.4) or (1.5). Clearly, some conditions on c(t) are necessary for the existence of a solution to such a problem. Thus, t(t) must be nonnegative and nondecreasing, since we know that the support of any solution of (1.1) is necessarily expanding in time [ 11. More precise information is obtained by considering the (right-hand) interface s(t) of the minimal solution to problem (1 .l), cf. [l], which is a particular instance of solution of (1.4) or (1.5). An obvious comparison argument shows that c must satisfy for t 2 0. r(t) 1 s(t) (1.6)
Initial and boundary-value problem for a reaction-diffusion
equation
261
As we shall see comparison of free boundaries is not sufficient, but it leads us in the right direction. The actual result implies comparison of the speeds of the interfaces. In order to state our main result we need to introduce the concept of nonincreasing majorant ii, of the initial value uO, defined by ii,(x) = sup(u,(y): y 2 x). (1.7) Let 5(f) be the interface of the minimal solution to (1.1) with initial data ii,. THEOREM1.1. Problem (1.4) admits a solution for every nonnegative continuous u0 satisfying (1.2) and every nonnegative continuous < satisfying for every f2 > C, > 0 <(f*) - r(tl) 2 W
(1.8)
- .%fi).
The result has the disadvantage of depending on comparison with a function $(t), of which not much is known in general, apart from being nonnegative and nondecreasing. However, it clearly shows that problem (1.4) has a solution for every r which has a large enough derivative. Notice also that the solutions corresponding to different values of r are necessarily different. It is interesting to point out a situation where the interface f((t) is explicit. This happens in the critical exponent value p = 2 - m and for initial data satisfying the condition
#y(x)
5
s
1x1
for x < 0.
(1.9)
Clearly, the same condition is satisfied by iiO. Then, as shown in [l, corollary 5.61, the support of the minimal solution ti expands with the minimum speed c* = 2Jm, i.e. s(i) = $(t) = 2Jm t. Condition (1.8) transforms into the following necessary and sufficient criterion. COROLLARY 1.2. Let m + p = 2 and assume u,, satisfies (1.2) and also (1.9). Then problem (1.4) can be solved if and only if (<(t,) - r(t,))/(tz - ti) > 2dm for every t, > t, > 0. More generally, if m + p = 2 equation solution V(x, t, c) given by
(1 .l) admits for every c z c* a travel&g
um-yx,t;C)
=
Zrn -(ctm
l
- x)+
wave (1.10)
where z is related to c by
c=z+E
(1.11)
2
Equation (1.11) gives two values of z for each c > c *. It has been proved in [l] that CTis a minimal solution of (1.1) if we choose the highest of the two values of z resulting from (1.11) when c > c,. Using (1 .lO) we get another criterion. COROLLARY 1.3. Let m + p = 2 and assume that u0 satisfies u,“-‘(x)
I
ZyjJ
t-x),
for some z > dm. Then for every function r such that for every t > 0 D’&t) there exists a solution of problem (1.4).
(1.12) 2 c = z + m/z
262
A. DE PABLOand J. L. VAZQUEZ
In this case, when u,, satisfies (1.12) with z > dm, it is easy to see that the condition on <, though sufficient, is not necessary. The next section is devoted to proving these results. To complete the well-posedness of problem (1.4) we prove in Section 3 the uniqueness of solutions under some additional regularity on the initial data. THEOREM
1.4. Let u0 L 0 be a uniformly continuous function which satisfies (1.2) and is also: (i) monotone decreasing near x = 0 and (ii) uniformly positive away from x = 0. Let 5 2 0 be a nondecreasing continuous function. Then problem (1.4) admits at most one solution u I 0. The above conditions on u0 are in our opinion technical and not essential. In the case p 5 0 we are able to prove uniqueness for general u,, (see theorem 3.2), but unfortunately we have been unable to eliminate the technical conditions when p > 0. It is interesting to compare our results for the overdetermined problem (1.5) with the initial and Dirichlet boundary-value (1.5e) from problem (1.5).
problem (IDP) obtained by eliminating the Neumann condition
THEOREM
1.5. Let u0 be a continuous, nonnegative and bounded real function satisfying (1.2) and let [ be a continuous function in [0, co) with ((0) = 0. Then there exists a continuous weak solution U(X,t) of problem (IDP), which is smooth and positive in 1(x, t): t > 0, x < r(t)). Assume now that u0 and c satisfy the uniqueness requirements of theorem 1.4. Then the solution of the full problem (1.5) is unique. This means that whenever the full problem (1.5) has a solution, for instance if condition (1.8) is satisfied, the solution of problem (IDP) automatically satisfies the Neumann condition (1.5e). On the other hand, when no solutions of the overdetermined problem (1.5) exists, for instance when m + p = 2 and l’(t) < 2dm the solution of (IDP) cannot satisfy the Neumann condition. In the last section we consider the case of compactly supported initial data and study the existence and uniqueness of a solution to problem (1.1) with prescribed (compact) support for each time. The same comment on the mixed problem (1.5) can be done in this situation. 2. CONSTRUCTION
OF SOLUTIONS
In constructing solutions to problem (1.4) or (1.5) we rely, both for m + p = 2 and m + p > 2, on comparison with an important particular solution E(x, t) constructed in [2] with zero initial data, E( *, 0) = 0. This solution is an absolute minimal solution in the sense that for any solution to (1.1) and for any (x0, to) E Q we have 2.4(x0,to) > 0 =a u(x, t) z E(x - x,, t - to>
in Q.
(2.1)
Moreover, E has the self-similar form E(x, t) = t”q((xlt+)
(2.2)
where Q”= l/(1 - p), p = (m - p)/(2(1 - p)) and q: [0, co) -+ [0, 00) is a continuous and decreasing function with compact support. Observe that the support of E is bounded by two interfaces x = &p(t) where p(t) = p. to, /3 2 1. This represents a minimal growth rate for the
Initial and boundary-value problem for a reaction-diffusion
equation
263
interface of every solution to (l.l), i.e. we have the growth condition on the interface r(t) of problem (1.4): for t 2 to L 0. (2.3) r(t) 2 &I) + P& - M When m + p = 2 we have p = 1 and (2.3) implies the already explained minimal speed of propagation of the interface, with p0 = c* > 0. After these preliminaries, we begin the construction of solutions for problem (1.4) with the critical case, which is somewhat easier. Proof of corollary 1.2. Step 1. Associated to problem (1.1) we define the map H which sends each nonnegative function u0 into the minimal solution u = ZZ(u,) to the Cauchy problem (1.1). Let also x = s,(t) be the (right-hand) interface of u. First of all we observe that when u0 satisfies (1.9), the interface of the minimal solution u = ZZ(u,) is exactly s,(t) = 2Jmt, cf. [l]. Now take r > 0. We next construct a solution to (1.4) in the special case t(t) = r + 2dmt. This means that we can obtain solutions to (1.1) whose support jumps instantly from zero to rat t = 0. For that purpose we consider, for fixed E > 0, the minimal solution (2.4)
u, = ~@rJ + EXto,J
where x1 stands for the characteristic function of the set I. Using the absolute minimal solution E, (2.2), to compare from below and the travelling wave with minimal velocity U, = U(x, t; 2dm), see (1. lo), to compare from above, we obtain Z.&(x,t) 2 cot > 0
for 0 I x 5 r, t > 0
uE(x, t) 5 U,(x - r - 6, t) where 6 = (dm/(m
- l)),P’
for x E R, t > 0
(2.5) (2.6)
and c,, = p(O) with Edas in (2.2). Thus
r + 2Jmt 5 s,,(t) 5 r + 6 + 2Jmt
for t > 0.
(2.7)
On the other hand u,, I uEZfor er 2 Q, and therefore there exists the limit v = lim u, E-O
(2.8)
which is a solution to (1.4a), (1.4b) and also (1.4~) holds with c(t) = r + 2dmt. Finally v(x, t) 5 U,(x - r, t) = U,(x - r - 2Jmt, 0).
(2.9)
Step 2. Now let c(t) L 0 be any continuous function such that r(O) = 0 and (r(t2) - r(tl))/(t2 - tr) 2 2Jm for every t2 > t, > 0. We fix t > 0 and n E N and consider the partition (tj = jt/n), 0 5 j I n - 1, of the interval [0, t]. We construct an approximate solution un as follows: we set u”(0) = uo, and then proceed iteratively in each subinterval as in step 1, extending the support of the solution at every mesh time tj up to r(t), 0 5 T I t and solving equation (1.4a) in the time interval [tj, tj+l].
A. DE PABLO and J. L. VAZQUEZ
264
To be precise, and that
assume that un is known 0x9
for 0 s t 5 tj. We define
I r(t) for 0 I t I tj
s,(t)
(2.10)
t) 5 U*(x - r(t), 0)
un in [tj, tj+J by means Un(X,
with 0 I t I tj,l
in [0, tj], that the interface
T +
tj)
=
of the formula
Ei_;H(z&)(X,
T)
(2.11a)
- tj, where (2.1 lb)
Ujl~(x) = Un(X9 tj) + EXIO,g(t,)](x)* We have, as before Zln(X, 5)
2
Cg(T
-
fortj
tj)
(2.12)
and Un(X,
Thanks
5)
to our induction
I
U*(X
-
r(tj)*
assumptions
?T -
for
tj)
tj
<
7 I
on U” and r, this implies
tj+l ,
x
E
R.
(2.13)
for the solution (2.14)
U%, r) 5 V*(x - r(r), 0) if
tj 5
T I
tj+l
and for the interface S@(r) =
in the same time interval. Once we have concluded limit
r(tj)
+
2dm(r
-
tj)
The induction is thus complete. the construction of the approximate
(2.15)
solutions
zP we may pass to the
v = lim 2.4’. n-cc
(2.16)
It is easy to check that v is a solution of (1.4a) and (1.4b). Thanks to the above estimates we can show that s,.(t) converges to r(t), which actually is the interface for v (use (2.12) and (2.13)). We have obtained therefore a solution of problem (1.4). n
Remark solution
2.1. It is easy to check that the solution obtained by the previous method is the minimal of the problem (1:4) if we consider in each iteration, instead of (2.11) the function U”(X,
t +
tj)
=
Ei,;
~~oH([uJJE
-
61+)(X,
(2.17)
T).
Also the proof of corollary 1.3 is immediate using in (2.10) instead of U, the travelling wave U(x, t; c) with velocity c = z + m/z. On the contrary, the condition on { in corollary 1.3 is not sharp as follows from the asymptotic result proved in [l, theorem 5.51. PROPOSITION
2.2.
is asymptotically
If u,, is bounded linear and s(t)/t
then the interface of the minimal + 2dm as t + 00.
solution
to problem
To treat the general case m + p L 2 we first suppose that the initial datum decreasing in (-00, 0), so that the minimal solution to (1.1) is also monotone.
(1.1)
is monotone
Initial and boundary-value
problem
for a reaction-diffusion
265
equation
LEMMA2.3. If u0 is nonincreasing in (-00, 0), then the minimal solution li = H(u,) to (1.1) is nonincreasing in (-00, r(t)) for each t > 0. Proof, Apply the classical strong maximum principle [3] to the equation satisfied by (u,), to get the desired conclusion. n
With this property in mind we can repeat the above construction
of the sequence u”.
LEMMA2.4. Assume u0 is nonincreasing in (-00, 0) and satisfying (1.2), let 2 = H(u,) be the minimal solution with initial data u,, and let g(t) be its interface. Then for every continuous function &j(t) such that [(t2) - T(tl) z s^(t2) - f(tJ for every t, > t, > 0, problem (1.4) admits a solution. Proof. Recall the notation of the proof of corollary 1.2. Observe that if fi(*, t) is nonincreasing we have fi(x, r) I ti(x - x0, r) for every (x, r) E Q, x0 2 0. Applying a comparison between minimal solutions in each iteration we get for r I tj by induction 3((t)
I
s,“(t)
Un(X,
5) 2
E(X
-
r(tj),
U’(X,
t)
ti(X
-
r(tj)
5
‘5 -
(2.18)
tj)
+ s”(tj),
t)
and thus t(tj)
+ Po(t
-
tj)”
5
Sun(t)
I
S(T)
+
<(tj)
-
(2.19)
S*(tj)
for tj< ‘5~ tj+l. We now need the condition on r: [(t2) - r(tJ remaining details follow the proof of corollary 1.2. n
2 3(t2) - .f(tl).
For the
Now let u0 be any nonnegative continuous function satisfying (1.2), ii, be its nonincreasing majorant and ii = H(ii,). Then theorem 1.1 is obtained as a consequence of the above lemma. n 3. UNIQUENESS Here we prove theorem 1.4. The strategy consists of, given any solution to problem (1.4), obtaining an approximation sequence of resealed supersolutions and use comparison to conclude that our solution is the maximal solution to (1.4). Clearly this implies uniqueness. Before doing that we recall that comparison does not hold in general for the solutions of problem (l.l), precisely because uniqueness fails. Nevertheless we have the following partial comparison principle. LEMMA3.1. Let u1 , u2 be a subsolution and a supersolution respectively to equation (1.4a), and let s,(t), s2(t) be their interfaces. Assume that u,(x, 0) <
u,k
s1(t) <
0)
SZ(f)
for x < s,(O) for t 2 0.
(3.1) (3.2)
Then u,(x, t) <
u2(x,
t)
for x < s2(t), t > 0.
(3.3)
A. DE PABLOand J. L. VAZQUEZ
266
By supersolution (resp. subsolution) we understand a continuous and nonnegative function u(x, t) which satisfies U, - (u”), - up I 0 (resp. I 0) in the sense of distributions in Q. In our application such super- and subsolutions are smooth whenever they are positive. Therefore, the proof of lemma 3.1 consists of using the classical maximum principle in the region x < si(t), where both functions are positive and the equation is parabolic nondegenerate. Proof of theorem 1.4. Given a solution u to problem (1.4) we define the function Uk(X, t) = u/&x,
It is easy to check that uk is a supersolution for the interfaces
=
k>
t) = ku(x - p, k”?)
P
+
l,p>O.
(3.4)
to equation (1.4a) for every k > 1, Also we have,
W=‘O
>
t(t)
for t 2 0.
Now we compare the initial values. If u0 is monotone for -2~ < x < 0, we immediately have uk(-&
0)
-
uO(x)
>
for --u < x < 217= r(O).
0
On the other hand, put A = inf u0 > 0, and take E > 0 small such that E + JE < A. From X5-‘I the uniform continuity of u,, we can choose p > 0 small enough such that ukb,
0) - u&d = kuo(x - P) - uoW > (k - l&(x) L
- ke
(k - l)A - kc
> kdc - (E + &) = 0 for x I -v (take k = 1 + JE). Applying lemma 3.1 we get that any solution to problem (1.4) lies below &. Since r& tends to u as k tends to 1 and p tends to 0 (E tends to 0), we conclude that u is the maximal solution n to problem (1.4), and since u is arbitrary this implies that u is the only solution.
Let us remark that when u0 is monotone of uO. We also notice that alternative proofs on z+,, for instance (u,“-‘)“(x) s -co. None when the exponent p 5 0 we prove a clean
we do not need to assume the uniform continuity of uniqueness can be done under other restrictions of these restrictions seems to be essential. Actually, uniqueness result.
THEOREM 3.2. Let m > 1,p I 0 and m + p 2 2 and let uo,
Proof. Let u be the solution to problem (1.4) obtained in theorem 1.1. From remark 2.1 we assume that u is the minimal solution to that problem. Now let u be any other solution to (1.4). We fix T > 0 and put QT = R x (0,T). Since u and u satisfy (1.4a) and (1.4b) in the sense of
Initial and boundary-value
problem
for a reaction-diffusion
equation
267
distributions, we may have that for every p E C”(Qr) with compact support in x for every 0 % t 5 T, the following identity holds: 01
(v - 4(x, t)P(x* t) d-x
=I-t1 iR
[(u - u)cp, + (u” - zP)q,
+ (up - up)(p](x, T) dxdr.
0.R
(3.5)
If u0 is bounded, we have u and u bounded in Qr [2], so equality (3.5) still holds for test functions $!JE Cm(Qr) fl L’(Q*). If we set P(X, s) - zP(X, s) u(x, 4 - u(x, s)
a = a(x, S) =
mzF1(x,
if u(x, s) f u(x, s) (3.6) otherwise
s)
the above equality becomes
=
[(u - U)(Pt+ w,) + (UP- up)p](x, T) dx dz.
(3.7)
+ x2) for some constant k > 0 to be chosen. We We try the test function ~(x, T) = e “‘-“/(l observe that 0 5 4x, 4 5 c(m, p, IIu~/I~, 77 and qt = -@, Pxx 5 V* Then if k is large enough we have pt + a~,, < 0, and since u 2 U, (3.7) is reduced to 01
s
I
R (u
- u)(x, t)G46 t) dx (VP - up)(x, r)p(x, r) dx dr.
(3.8)
Now we observe that u(x, z) > u(x, r) > 0 for x < c(r) and then up(x, r) - up(x, T) 5 0 for x < c(s), while v(x, r) = u(x, r) = 0 for x I r(s) so that vp(x, r) = up(x, T) = 0 for x L T(T). This implies, within (3.8), that v(*, t) = u(-, t) for every 0 5 t I T. n 4. EXISTENCE
FOR
THE
DIRICHLET
PROBLEM
In this section we outline the proof of theorem 1.5. The idea consists of first adding E > 0 to the initial data and taking as boundary condition U(X,t) = E for t 2 0 and x = c(t). In this way we obtain a nondegenerate problem that can be solved by means of the classical quasilinear theory [3], thus obtaining a classical solution U, > E. We then let E + 0 and use the monotonicity of the solution with respect to E to pass to the limit U(X,t) = lim u,(x, t), which E-O is a weak solution of the problem originally proposed.
A. DEPABLOand J. L. VAZQUEZ
268
5. SOLUTIONS We consider
here the case where the initial z+,(x) > 0
Then, problem
WITH COMPACT
given two functions (1.4) by
data have compact uO(x) = 0
for 1x1 < R,
u(x, t) > 0
SUPPORT say
otherwise.
1 R, we substitute
if and only if -r,(t)
support,
(5.1)
the boundary
< x < &(t),
condition
(1.4~) in
t 2 0
(1.4c’)
and study the resulting problem (1.4’) = (1.4a) + (1.4b) + (1.4~‘). The construction of the solutions is exactly the same as for problem (1.4), but here we also have to introduce the nondecreasing majorant of u,-,. Define then the functions &,1(X) = suPluo(Y): Y 2 xl (5.2) %,2(X) = suPl%(Y): and let -si(t), s*(t) be the interfaces of the minimal H(uO,i), H(u&. We have the following theorem. THEOREM 5.1. For every continuous functions & 2 R, i = I,2 satisfying
Y 5 xf solutions
associated
to those initial
values,
function u0 satisfying (5. I), and every pair of continuous for every t2 > t1 > 0
(1.4’) admits
(5.3)
a solution.
For the uniqueness of solutions we follow the proof using a different resealed sequence.
of theorem
1.4 given in Section
3, but
THEOREM 5.2. Let u. be a Lipschitz continuous function which is monotone near 1x1 = R. Then for every pair of continuous functions ri 2 R, problem (1.4’) admits at most one solution. Proof.
Given
a solution
u of the above problem, u,(x,
which is a supersolution
t) = ku(k-“x,
to equation
we define
the function k>
/?-‘-2*t)
1
(1.5a) if 0 s o( I (m - p)/2.
(5.4) Its interfaces
satisfy
for t r 0 ck,i(f) = ka~i(km-1-2at) > [i(t) if 0 < Q! < (m - 1)/2. The comparison between the initial values of u and uk is again easy near the border, using the monotonicity. To compare in the interior assume f+,(x) 2 A for 1x1 5 R - q and let M be the Lipschitz constant of uO. We have u,(x,
0) - uo(x) = ku,(k-“x)
for 1x1 I R - q if 0 < (Y < 6/(MR).
- uO(x)
2 (k -
l)u,(x)
- M(k”
1 (k -
l)A - Ma(k
-
As in the proof of theorem
- l)R 1)R > 0 1.4 this implies uniqueness.
n
Initial
and boundary-value
6. SOME
problem
EXTENSIONS
for a reaction-diffusion
AND
equation
269
COMMENTS
We have assumed for simplicity in the construction of solutions to problem (1.4), that the interface is given by a continuous function r(t). However, it is clear from the proofs of Section 2 that we may allow < to have positive jump discontinuities. In particular, if we take t(O+) = r > 0 in theorem 1.1 we have to deal with the function
and compare its interface with r. Moreover, we may set r(t) = ~0 for I E [to, m), with C, L 0 by simply using the construction of this paper for 0 I t < t,, and then taking u equal to the maximal solution for t 2 t,, . The same ideas apply to problem (1.4’). Besides, in the existence and uniqueness theorems we do not assume that the slope of the initial data is bounded near the interface(s), so we may have also discontinuous initial values. Observe that the minimal solution to problem (1.1) has finite interface(s) even in this case. In all our constructions we have insisted in constructing solutions with connected support for each time t > 0. However, the existence of the special solution E(x, t) shows that we can start a new “piece” of the positivity set of a solution at any instant and at any place in the complement of the support. The interface of this new piece can be chosen subject to the above explained conditions until it merges with previous components. See also [2]. REFERENCES 1. DE PABLO A. & VAZQUEZ J. L., Travelling waves and finite propagation in a reaction-diffusion equation, J. diff. Eqns 93, 19-61 (1991). 2. DE PABLO A. & VAZQUEZ J. L., Balance between strong reaction and slow diffusion, Communs partial diff. Eqns 15(2), 159-183 (1990). 3. LADYZENSKAJA 0. A., SOLONNIKOV V. A. & URALCEVA N. N., Linear and quasilinear equation of parabolic type,
Transl. Math. Monogr.,
Am. math. Sot. Transl. (1968).