Semilinear Parabolic Equations with Prescribed Energy by Bei Hu and Hong-Ming Yin Department of Mathematics University of Notre Dame Notre Dame, IN 46556, USA. Abstract In this paper we study the following reaction-diusion equation u t =u+ f(u; k(t)) subject to appropriate initial and boundary conditions, where f(u; k(t)) = u p k(t) ork(t)u p with p> and k(t) is an unknown function. An additional energy type condition is imposed in order to nd the solution u(x; t) and k(t). This type of problem is frequently encountered in nuclear reaction process, where the reaction is known to be very strong, but the total energy is controlled. It is shown that the solution blows up in nite time for the rst class of function f for some initial data. For the second class of function f, the solution blows up in nite time if p>n=(n ) while it exists globally in time if <p<n=(n ), no matter how large the initial value is. Partial results are generalized into the case where f(u; k(t)) appears on the boundary. Key Words and Phrases: Inverse Problems, Diusion with Strong Reaction, Global Solvability, Finite time Blowup. AMS(MOS) Subject Classication: 35K57, 35B5. This author is partially supported by National Science Foundation Grant DMS 9-4935
. Introduction Consider a chemical reaction-diusion process, where it is known that the reaction is very strong, say like u p is unknown, say k(t), a function of t. with p>, but the rate with respect to this power On the other hand, the total energy is controlled in the system in order to prevent the blowup phenomenon. That is u(x; t)dx = g(t): This leads to an inverse problem where one needs to nd the solution u(x; t)aswell as the coecient of the strong reaction. Another model arising from the nuclear science is that the growth of temperature is known to be very fast like u p, but some absorption catalystic material is put into the system in such a way that the total mass is conserved. These two models leads us to consider the following parabolic inverse problem: Find u(x; t) and k(t) such that u t u = f(u; k(t)) for (x; t) Q T =(;T]; (.) u (x; t) = for (x; t) S T = [;T]; (.) u(x; ) = u (x) for x ; (.3) where is a bounded domain in R n with a smooth boundary S = and is the outward normal on S. An additional energy condition is prescribed by u(x; t)dx = g(t); t : (:4) The above mathematical problem can also be used to model other phenomena in population dynamics and biological sciences where the total mass is often conserved or known, but the growth of a certain cell is known to be of some form. The research on the well-posedness of a parabolic inverse problem goes back to 6's (see the references in [4]). Since then, considerable progress has been made regarding to various inverse problems for parabolic equations. The reader can nd a large number of references in [4]-[6] and the recent proceedings [9]. The essential dierence between the previous inverse problems and the current one is that the
solution of (.)-(.4) may blow up; this fact is well known when k(t) is given. On the other hand, since our k(t) isgiven in terms of the solution, the problem may have a global solution because of the stabilizing factor k(t). The natural question is whether or not the stabilizing factor is strong enough to prevent the blowup. We shall study this problem in the present work. We shall only discuss two classes of function f(u; k(t)), namely, f(u; k(t)) = u p k(t) or k(t)u p : When the function k(t) > is given, there are a lot of papers (cf. [], [], [4], [7] and the references therein ) dealing with various qualitative properties such as nite time blowup, blowup rate, blowup set. When f(u; k(t)) = u p k(t) or k(t)u p with an unknown k(t), it is not dicult to see that the condition (.4) and the equation (.) imply k(t) = u p dx jj g (t) ; or Hence the reaction term in (.) can be written as f = u p u p dx jj g (t) ; or g (t) u p dx: g (t)u p u p dx: In general, it is not clear which term can dominate the reaction. As there is a nonlocal integral term in the equation (.), the powerful comparison principle is invalid. We prove by applying the energy method that, for the rst class of the function f, the solution will blow up in nite time for a class of initial data. For the second case, one may also believe that the solution will blow up in nite time. Surprisingly, it turns out that the nite time blowup or global existence depends upon the exponent p and the space dimension n. It will be seen in Section 3 that the solution exists globally if p<n=(n ), no matter how large the initial data are. On the other hand, the solution will blow up in nite time if p>n=(n ), provided that initial data satisfy some conditions. This is quite dierent from a regular reaction-diusion equation (cf. [], [], [4], etc.). We shall mention 3
that diusion equations with nonlocal reactions has been considered by a number of authors (cf. [], [7]-[8], [7], etc.). However, none of those deals with similar problems to (.)-(.4), since the energy in these previous problems blows up in nite time. More recently, the authors of [3] studied the problem (.)-(.4) with f(u; k(t)) = u R u dx in one space dimension. Blowup is proved for some special initial value and the blowup rate is also discussed. The argument for proving the blowup property is obviously not suitable for the present situation. The paper is organized in the following way. In Section, we study the problem with f(u; k(t)) = u p where f(u; k(t)) = k(t)u p with p<n=(n for p = n=(n the case with p>n=(n k(t). In Section 3, we show the global existence for the case ). We also prove the global existence ) when the initial value is large enough. In Section 4, we consider ) and prove that the solution blows up in nite time for suitable radially symmetric initial data. Section 5 deals with the case where f(u; k(t)) occurs on the boundary.. Blowup for f(u; k(t)) = u p k(t) Throughout this paper, C denotes generic constants, unless otherwise indicated. Mathematically, we do not require that u(x; t) is nonnegative. Therefore we shall use juj p u instead of u p.for some technical reasons, the argument below isvalid when g(t) = constant, say. This is the case in some applications. The equation (.)-(.4) can now be rewritten as follows (with g(t) = ): where u t =u+juj p u k(t) for x ; t >; u = for x ; t >; (.) u(x; ) = u (x) for x : k(t) = juj p jj udx : The following conditions on the data are assumed throughout this section: H(A) u (x) C 3 () and u (x)dx =: 4
The existence and uniqueness of this system for small t is clear, by the standard theory of parabolic estimates and contraction mapping principle. It is also clear that the solution can be extended in t direction, as long as the L norm of the solution remains nite. Theorem.: The solution of (:) blows up in nite time if p>and is suitably large. jru j dx + ju j p+ dx p + Proof: The proof is based convexity argument (see Levine and Payne [5]). We multiply the equation by u and u t, respectively, and then integrate over, to obtain Let d dt u dx + u t dx + d dt The identity (.3) gives J(t) = jruj dx = jruj dx = d dt J (t) = juj p+ dx p + juj p udx; (.) jj juj p+ dx jruj dx + juj p+ dx: p + : (.3) u tdx : (.4) It follows that J(t) =J() + t u t dxdt: (.5) Introduce a new function I(t) = t u dxdt + A + Bt ; where A and B are two constants to be specied later. Clearly I (t) = u dx +Bt; I (t) = d dt u dx +B: 5
By the identity (.), I (t) = d dt = u dx +B jruj dx + juj p+ dx juj p jj We claim that there exists a constant > such that udx +B: Indeed, the desired the inequality is equivalent to jruj dx + 4( + ) h I (t) 4( + )J(t): (:6) juj p+ dx jj jruj dx + p + juj p udx +B i juj p+ dx : Now Holder's and Young's inequalities imply that for any > juj p dx juj p+ dx + C( ; jj): As p>, we choose and to be small enough such that 4( + ) +p >: Then we have the desired claim if we simply take B C( ; jj): Clearly I (t) = It follows that, for any ">, I (t) 4( + ") t u dx +Bt = u dxdt t t uu t dxdt + u t dxdt + + " Combining the above estimates, we nd that, for >, I (t)i(t) ( + )I (t) 4( + ) h J() + t ( + ) h 4( + ") ( + ) + " u t dxdt ih t h t 6 u dxdt t h u dx +Bt: (:7) u dx +Bt i : (:8) u dxdt + A + Bt i u dx +Bt i : u t dxdt i
Now wechoose " and to be small enough such that +( + )( + "): (:9) If J() 4( + )( + )B (:) " and A is chosen to be large enough, then we have I (t)i(t) ( + )I (t) : It follows (cf. [5]) that exists in (;T ). t u dxdt will blow up in a nite time T, if the solution Remark.: When u (x) = Mis a constant, then u(x; t) = Mis always a solution. In this case, the proof is invalid since in the inequality (.6) the constant B depends on jj, but the condition H(A) implies Mjj =. Consequently, B depends on M, so do the other constants in the proof. Therefore, it is impossible to choose J() to satisfy the desired inequality (.). 3. Global Existence for f(u; k(t)) = k(t)u p For simplicity, we shall assume that g (t) = in Sections 3 and 4. It will be seen that the results can be immediately carried over for a general function g(t) > with g (t) > in this section. In this case the problem (.)-(.4) is equivalent to the following: u t =u+k(t)u p for x ; t >; u = for x ; t>; (3.) u(x; ) = u (x) for x : where k(t) = : u p (x; t)dx 7
We shall assume throughout this section that u (x) is smooth, u (x) > on, and satises the compatibility condition u =on and u (x)dx = m>. By the standard theory of parabolic estimates and contraction mapping principle, the existence and uniqueness of this system for small t is guaranteed. The solution can be extended in t direction, as long as the solution remains nite. Since we assume that the initial data are positive, the solution u(x; t) is positive, by the maximum principle. Theorem 3.: Suppose that <p<n=(n ) for n 3 and <p<when n =or. Then there exists a unique global solution to the system, namely, the solution exists for all t [; ). Proof: The proof here is for n 3. The proof for n = and n = can be obtained with obvious modications. We x a large. Without loss of generality, we assume that jj =. For any <a<, and q + q =,wehave u p+ dx =q u (p+)aq dx u (p+)( =q a)q dx : (3:) There are two free parameters a and q to be determined. We rst let n (p + )aq =(+) n : (3:3) If q = n=(n ) (it is clear that <a< with this choice of q, asp>), then However, the choice q = n=(n (p + )( a)q = n (p ) <p: (3:4) ) will not be good enough in our proof. Since we have a strict inequality in (3.4) when q = n=(n ), we can take q>n=(n ) and q n=(n ) so that (3.4) is still valid for this particular choice of q. Using Holder's inequality again, we obtain, (recalling that jj = ), u p+ dx u (+) n n u (+) n n dx =q dx =q u (p+)( =q a)q dx u p dx (p+)( a)=p : (3:5) 8
Next, multiplying the equation by u and integrating over, we obtain + u+ dx + u jruj dx = t Integrating the equation, we immediately get u(x; t)dx = u (x)dx + t: Hence by Holder's inequality and the fact jj =, u p dt Using (3.4), (3.5), (3.6), (3.7), we obtain + u+ dx + 4 t ( +) If we dene v = u (+)=, then v dx + + t u p+ dx : (3:6) u p dx p p u (x)dx + t u (x)dx = m p > : (3:7) 4 4( +) ru (+)= dx C u (+) n =q jrvj dx C v n n dx : n dx =q : (3:8) Now we use the following (elliptic) Sobolev's embedding theorem (cf. [6]) in the above inequality. Since q>n=(n kwk L r () C krwkl () + kwk L () ), we obtain ; r = n : n v dx + jrvj dx C v dx + : (3:9) t By Gronwall's inequality, we conclude that the right-hand-side of the above inequality is bounded. Therefore sup tt u + dx u + dx t + + u (+) n t u (+) n n n n dx C ; n n n dx dt C : (3:) 9
It follows that u is bounded in L + for any nite. Thus the L p estimates (cf. [3]) implies that u is in W ; (+)=p (Q T) for any. Therefore if we take (+)=p > (n + )=, then by the Sobolev's embedding theorem, u is Holder continuous, for t T, for any T>. It is then easily seen that u C (Q T ). Thus we obtain a global solution. We now turn to the case where p is equal to the critical number n=(n above proof with q = n=(n ) gives (recalling that jj =) ). The u + dx + ru (+)= dx C t C u (+) n u (+) n t + n n n dx =p u p dx n n n dx u (x)dx ; (3:) where at the nal step we have used the inequality (3.7). By Sobolev's embedding, the numerator in the right-hand side of (3.) can be dominated by C u + dx + ru (+)= dx We nowxin (3.) such that ( +)=p > (n +)= and choose u (x)dx to be large enough, then we have the same estimates as (3.). Consequently, we proved Theorem 3.: Suppose that p = n=(n for all t [; ), provided u (x)dx is large enough. ) for n 3. Then the solution exists 4. Blowup of solutions for f(u; k(t)) = k(t)u p In this section we shall construct a solution which blows up in nite time when p>n=(n ). Theorem 4.: Suppose that p>n=(n ). Then there exists a radially symmetric
initial datum u (x) (with =B () ) such that the corresponding solution u(x; t) of (3:) blows up in nite time at x =. Furthermore, x =is the only blowup point. Proof: Let where we choose '(r) = 8 >< >: r for <r; + + r for r ; = p : (4:) It is clear that '(r) isinc [; ] \ C [;)[(;], and ' () = and ' (r). A direct calculation shows that n! n ' p (r)r n dr = = n! n r n p dr + () n! n n p + (); (4:) where () = and! n n! n r n p dr + h n! n + r + i pr n dr = O( n p ); (4:3) is the surface area of unit ball in R n. Thus ()! as!+, since p>n=(n ). From our choice of, we nd that for each (; ), and Since ' is in C,wehave ' rr + n ' r = ( +) (n ) r r + = (n ) ' p for r<; ' rr + n n ' r = = n' p () r + n' p (r) for r : ' rr + n ' r ' p for r<; (4:4) r
in the distribution sense, where Wenow let u (r) ='(r), then = max h i (n ) ;n : (u ) rr + n (u ) r + r ( )( u )p + 4 n n! n n p + () n! n (u ) p r n dr p n! n p (u ) p (u ) p for r<; (u ) p! u p (4.5) provided we take and such that <,<< for some suciently small and. Wenowx=.Thus u (r) will be uniquely determined by the choice of. R Next, we want to give an estimate for the modulus of continuity of the integral B () up (x; t)dx near t = which is uniformly valid for small. Since we want to prove that the solution blows up in nite time, we can always assume for the contrary that it exists for <t<. Note that m = B () h u (x)dx = n! n n + + n (n +) i n : (4:6) n It is clear that m is bounded from above and below uniformly for <. By the constraint condition, we know B () u(x; t)dx = t + m: (4:7) Since u (x) =u (r) is monotone decreasing in r, applying the maximum principle to (r n u r ) gives us u r (r;t). Hence r u(r;t)! n r n =u(r;t) n! n z n dz r u(z; t)n! n z n dz u(z; t)n! n z n dz = t + m;
which implies that, u(r;t) t+m! n r n +m! n r n for <t<: (4:8) Thus u(r; t) is uniformly bounded, say, in[=; ] [; ). Next, by Holder's inequality, one can easily derive k(t) wp n for <t<: (4:9) m p Observe that the function w(r;t)=r n u r satises the following equation where L[] is dened by L[w] = for <r<; <t<; (4:) L[ ]= t rr + n r r pk(t)u p : (4:) Comparing w with a function v satisfying the same equation L[v] =, but equals to on r == and on r =, and equals to r n (u ) r (r) ont=,we easily obtain that u r (3=4;t) c for <t<; (4:) where the constant c is independent ofbecause of the estimates (4.8) and (4.9). We next introduce the auxiliary function as in [], J = w(r;t)+"r n u q ; where we xqsuch that q<p, and p=(q justify this assumption later on) ) <n. We assume that (we shall especially, we x small such that u(3=4;t) u (3=4); (4:3) u(r;t)u(3=4;t) u (3=4) for r 3=4; <t<: 3
A direct calculation shows that L[J] = " (p + q)k(t)r n u p+q nqu q r n u r r n q(q )u q (u r ) (4.4) h = " (p + q)k(t)r n u p+q nqu q J "r n u qi r n q(q )u q (u r ) "nqu q J + "c(r;t); where the coecient of J in the above equation is bounded as long as the solution u(r;t) remains bounded. Since u(r;t), provided c(r;t) = r n u q h r n u q h < ; (p+q)k(t)u p i q +n"q (p + q)k(t) p q +n"q i r n q(q )u q (u r ) "< p q nq k(t)p q : (4:5) We need a lower bound of k(t) for the above tobevalid for small ". Ifwe assume that k(t) k(); (4:6) then (4.5) is valid for small ". On ft =g, for <r, note that p = +, (u ) r +"r(u ) p = r + "rp + r + "rp + h = r + < ; + " p p + + + p p i r + p if " is small enough (independent of). For <r<, (u ) r + "r(u ) p = + "rp + r h = + " p < ; 4 r + r p i
for small ". Since p>qand u>by the assumption (4.3), it follows that (u ) r + "r(u ) q < for r 3=4; for " small enough. Thus J<onft= ; r3=4g. By (4.8) and (4.), we can choose " to be small enough (independent of) so that J <onfr= 3=4;<t<g.Wenowxsuchan".Obviously, J =onfr=; t<g. Thus, the maximum principle implies that J onfr3=4;tg, as long as (4.3) and (4.6) remain valid. Integrating u r "ru q gives us i.e., u q (r;t) q "r + u q (;t) q "r ; u(r;t) h i q (q )" r = h i q =(q ) (q )" r for <r<3=4; where ==(q ) < n=p. It follows that h i p(q ) n p sup n! n u p (r;t)r n dr n! n < (p )" n p : (4:7) Since u(r;t) is uniformly bounded in the domain f r ; t g, the standard parabolic estimates imply that lim sup t! < n! n u p (r;t)r n dr n! n (u ) p r n dr =: (4:8) (4.7) and (4.8) imply that lim sup t! < n! n u p (r;t)r n dr n! n (u ) p r n dr =: (4:9) Therefore, by the continuation argumentint, there exists a (; =) such that (4.3) and (4.6) are valid for <t<, where is independent of. It follows that all of the above estimates are valid for <t<. 5
We now compare u(r;t) with the solution v(r;t) of the following problem where v t =v+ k()vp for r<; t >; v r = c for r =; t >; v = for r =; t >; r v(r;) = u (r) for r ; c = u : r r= By the comparison principle, u(r;t) v(r;t), as long as both solutions exist and t. Let = v t v p, > onfr=; t >g. By using (4.5), we obtain r onfr; t =g. Thus by using the equation for, we can easily obtain that, as long as the solution exists. Hence Especially, v(r;t) v(;t) u p (r) u p () (p )t (p )t =(p ) : =(p ) : Therefore, the solution v(r;t) blows up at a time T <,ifis chosen to be small enough. This proves the theorem. Remark 4.: For a general function g(t), Theorem 4. holds if g(t) g > and g (t) g > for some positive constant g. 5. Nonlinear Boundary Value Problem In this section we shall generalize the above results into the problem with f(u; k(t)) as the boundary function. We rst consider the case where f(u; k(t)) = juj p k(t). 6 u
Consider the following problem: u t =u for x ; t>; u = jujp u k(t) for x ; t; (5.) u(x; ) = u (x) for x : An additional condition is imposed as follows: u(x; t)dx = for t : (5:) We are interested in the case where p>. It is known (cf. [], etc.) that when k(t) =, the solution will blow up in nite time for any nonnegative u (x) which is not identically zero. Will the stabilizing factor k(t) on the boundary be able to prevent the blowup phenomenon? We shall answer this question in this section. We shall assume that u (x) is smooth, say inc 3 (), for convenience. It is clear from the standard theory of parabolic equations that the problem (5.)-(5.) has a local solution. Theorem 5.: The solution of (5.)-(5.) will blow up in nite time if is big enough. J() = jru j dx + ju j p+ ds p + Proof: The proof is similar to that of Theorem.. We multiply the equation (5.) by u and u t, respectively, and integrate over, to get and Let d u dx + jruj dx = juj p+ ds dt u dx + d t dt jruj dx = d dt p + J(t)= I(t) = t jj juj p+ ds ru dx + juj p+ ds p + u dsdt + A + Bt : 7 juj p uds; (5.3) : (5.4)
By a similar calculation, we see that there exists a constant > such that I(t)I (t) ( + )I (t) ; provided that J() is large enough. We shall not give the detail here. Next we consider the case where f(u; k(t)) = k(t)u p on the boundary. In this case, as the interpolation inequality is dierent from the previous situation, we have a dierent inequality about p and n to ensure the existence of a global solution. For simplicity, we again assume that where m = u(x; t)dx = g(t) t + m; u (x)dx: It can be easily seen that the result will be valid for a general bounded smooth function g(t) > ;g (t)>. where Consider the problem: u t =u for x ; t >; u = k(t)up for x ; t>; (5.5) u(x; ) = u (x) for x : k(t) = : u p (x; t)ds We assume that u (x) > and u (x) C 3 (). assumed: u (x) = u (x) u (x)ds for x : The compatibility is also Again the classical theory of parabolic equations implies the problem (5.5) has a unique classical solution locally in time. Moreover, the strong maximum principle implies that u(x; t) > for x as long as it exists. Theorem 5.: For < p <(n )=(n ), if n3, and < p <,if n, the problem (5:5) has a unique solution for all t<. 8
Proof: The proof is based the similar idea to that of Theorem 3.. For any >, we multiply the equation by u to obtain u + dx + t + 4 ru (+)= dx = 4( +) By Holder's inequality, u p+ ds =q u (p+)aq ds u (p+)( u p+ ds : (5.6) u p ds =q a)q ds : We choose, for n 3 (in the case n =orn=wechoose q to be large enough), q> n n ; q n n ; (p+)aq =(+)n n : Then q n, and (p + )( a)q (p )(n ). Since (p )(n ) <p, we get Therefore (p + )( a)q <p: u p+ ds u p ds =q: C u ds (+)q If we dene v(x; t) =u(x; t) (+)=, then by (5.6), v dx + t + 4 jrvj (n ) =q: dx C v n ds (5.7) 4( +) Now we use the following trace inequality (cf. [6]): where r = (n ) n Since q>(n )=(n ), we obtain kwk L r () C krwkl () + kwk L () for n 3; while r is arbitrary for n = or : v dx t + jrvj dx C v dx + : 9 :
Using Gronwall's inequality, we have u + dx t + sup u + (x; t)dx + tt T n u (+) n n n dx C ; u (+) n n dx n n dt C ; (5:8) where the constant C depends only on the known data and T. Wexso that +>n(p ), then by Theorem. in [], sup tt sup u(x; t) C: (5:9) x Thus by Theorems 7.-7. in Chapter V of [3] (it is clear the assumptions (7.4){ (7.6) are satised), we immediately obtain that kuk C; C +;(+)= (Q T ) for some (; ). This estimate implies that the function k(t)u p is uniformly bounded in the space C +;(+)=. Consequently, we can use Schauder's estimate to obtain kuk C: C +;+(=) (Q T ) With the above a priori estimates in hand, we can obtain the existence for t T, for any T>. Remark 5.: Similar to Theorem 3., the above argument works for the critical number p =(n )=(n ) provided inf kuk L ()(t) is large enough, which p tt is the case if we assume that min u (x) to be large enough. Global existence is x guaranteed in this case. Remark 5.: We conjecture that the solution of (5.)-(5.) will blow up in nite time if p>(n )=(n ) and the initial value satises a certain condition. Acknowledgment: The authors would like to thank Professor A. Friedman for some helpful discussion during the preparation of the manuscript. The authors also thank Dr. Henghui hou for the partial motivation of the present research.
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