with nonnegative, bounded, continuous initial values and positive numbers

Transcription

1 APPLICATIONES MATHEMATICAE 2722) pp J RENC LAWOWICZWarszawa) GLOBAL EXISTENCE AND BLOW-UP FOR A COMPLETELY COUPLED FUJITA TYPE SYSTEM Abstract The Fujita type global existence and blow-up theorems are proved for a reaction-diffusion system of m equations m > 1) in the form u it = u i +u p i i+1 i = 1m 1 u mt = u m +u p m 1 x R N t > with nonnegative bounded continuous initial values and positive numbers p i The dependence on p i of the length of existence time its finiteness or infinitude) is established 1 Introduction The main concern of this paper is to examine the following system of reaction-diffusion equations: { uit u 11) i = u p i i+1 i = 1m 1 u mt u m = u p m 1 for x R N t > p i > ; and u i x) = u i x) i = 1m x R N where u i are nonnegative continuous bounded functions We consider the behaviour of classical nonnegative solutions for 11) as regards their maximal existence time We describe the cases of global existence and finite blow-up time for every solution of 11) in the second situation a nontrivial solution) in terms of p i and N This work generalizes results of Fujita [F1] [F2] for the scalar Cauchy problem) Escobedo and Herrero [EH] for the system 11) with m = 2) and the author [R] for the system 11) with m = 3) We want to mention that the method used here is strictly connected with the special form of the 2 Mathematics Subject Classification: 35B3 35B5 35K55 35K57 Key words and phrases: blow-up global existence reaction-diffusion system Research supported by KBN grant no 2PO3A [23]

2 24 J Renc l awowicz system 11) In the case of weakly coupled systems proofs are much more complicated we refer to [EL] for the case of two equations) We introduce some notations to formulate the main theorems Let A m be a matrix of the form p 1 p 2 12) A m = pm 1 p m We denote by α = α 1 α m ) the unique solution of 13) A m I)α t = 11) t Let 14) δ = deta m I) = 1) m+1 m Whenever δ we have 15) p j α 1 = 1+p 1 +p 1 p 2 ++ m 1 m p j 1 ) p j 1 α i = 1+p i +p i p i+1 ++ j i 1 p j m p i = 2m j 1 Theorem 11 Suppose deta m I) = Then every solution of 11) is global Theorem 12 Suppose deta m I) and max i=1m α i < Then all the solutions of 11) are global Theorem 13 Suppose deta m I) and max i=1m α i N/2 Then every nontrivial solution of 11) blows up in a finite time The plan of this paper is the following: we prove some auxiliary lemmas in Section 2 global existence theorems is the content of the last section whereas the global nonexistence theorem is proved in Section 3 2 Preliminaries Let St) be the semigroup operator for the heat equation ie ) St)ξ x) = 4πt) N/2 x y 2 ξ y)dy 4t R N

3 Coupled Fujita type system 25 A classical solution of 11) satisfies the following variation of constants formulas: 21) It follows that u i t) = St)u i + u m t) = St)u m + St s)u p i i+1 s)ds i = 1m 1 u i τ) = Sτ t i )u i t i )+ Sτ t i )u i t i ) St s)u p m 1 s)ds τ t i τ t m u m τ) = Sτ t m )u m t m )+ Sτ t m )u m t m ) Sτ t i s)u p i i+1 s)ds i = 1m 1 Sτ t m s)u p m 1 s)ds We remark that if x i t i ) are such that u i x i t i ) > i = 1m then by positivity of St) u i τ) > for τ > t i Taking t = max i=1m t i we have u i xτ) > for x R N τ > t i = 1m Lemma 21 Let u = u 1 u m ) with u = u 1 u m ) be a solution of 11) Then we can choose τ = τu 1 u m ) and some constants c > a > such that min i u i τ)) ce a x 2 Proof We can assume that for instance u 1 Then there exists R > satisfying ν 1 = inf{u 1 x) : x < R} > By formula 21) we have ) x u 1 t) St)u 1 ν 1 4πt) N/2 2 ) y 2 dy 4t 4t We put Hence y R y R u 1 t) = u 1 t+τ ) for some τ > a 1 = 1 ) y c 1 = ν 1 4πτ ) N/2 2 dy 4τ 4τ u 1 ) = u 1 τ ) > c 1 a 1 x 2 ) To obtain the assertion we use an inductive argument By 21) u i τ ) τ Sτ s)ss)u i+1) ) p i ds

4 26 J Renc l awowicz If p i 1 then Otherwise u i τ) τ u i τ) Sτ s)ss)u i+1) ) p i ds = τsτ)u i+1) ) p i τ Sτ s)ss)u i+1) ) p i ds = τsτ)u p i i+1) It is clear that if we assume that u i+1) satisfies u i+1) c a x 2 ) for some constants c and a then an analogous estimate holds true for u i τ) Replacing u i+1) by u i+1 τ ) for some τ if necessary we get the conclusion To fix ideas we assume henceforth that m maxα i = α 1 if p j > 1 i Lemma 22 Let u 1 t)u m t)) be a bounded solution of 11) in some strip [T) R N < T Assume that m p j > 1 Let n be an integer such that 22) p i 1 for 1 i n p i < 1 for n < i m Then there exists a positive constant C depending on p i only such that 23) t m j=n+1 p jα 1 St)u m j=n+1 p j 1 C if n < m t α 1 St)u 1 C if n = m t [T) Proof Using 21) for i = 1 in 21) for i = m we get u m t) St s)ss)u 1 ) p m ds Consider first n = m Then by the Jensen inequality for p m 1 we have 24) u m t) St)u 1 ) p m ds = tst)u 1 ) p m Using 24) in 21) for i = m 1 and the Jensen inequality for p m p m 1 1 we get u m 1 t) St s)sss)u 1 ) p m ) p m 1 ds s p m 1 St)u 1 ) p mp m 1 ds 1 p m 1 +1 St)u 1) p mp m 1 t p m 1+1

5 We now reason by induction Assume that Coupled Fujita type system 27 25) u m j t) c j St)u 1 ) π j t r j where 26) π j = p m j π j 1 π = p m r j = p m j r j 1 +1 r = 1 c j = 1 r j c p m j j 1 c = 1 Substituting 25) in 21) for i = m j 1 by the Jensen inequality we obtain u m j+1) t) whence by 26) St s)[c j Ss)u 1 ) π j s r j ] p m j+1) ds c p m j+1) j and for j = m 1 25) gives s r jp m j 1 St)u 1 ) p m j 1π j ds c p m j 1 j r j p m j 1 +1 St)u 1) p m j 1π j t r jp m j 1+1 u m j 1 t) c j+1 St)u 1 ) π j+1 t r j+1 27) u 1 t) c m 1 St)u 1 ) π m 1 t r m 1 where 28) π m 1 = m p j r m 1 = p m 1 +1)p m 2 +1)p m 3 +1))p 1 +1 m ) = α 1 p j 1 c 1 m 1 = p m 1 +1) p 1p m 2 p m 2 p m 1 +1)+1) p 1p m 3 [p m 1 +1)p m 2 +1))p 1 +1] Iteratingthisschemeandsettingp = π m 1 r = r m 1 weobtainanestimate for u 1 t) from below: m 29) u 1 t) A k i St)u 1) pk t r1+p+p2 ++p k 1) where 21) i=1 1/A k 1 = rpk 1 [rp+1)] pk 2 [rp 2 +p+1)] pk 3 [rp k 1 ++p 2 +p+1)]

6 28 J Renc l awowicz 21) [cont] 1/A k m = p m r +1) pk 1 /p m [p m rp+1)+1] pk 2 /p m [p m rp 2 +p+1)+1] pk 3 /p m [p m rp k 2 ++p 2 +p+1)+1] p/p m 1/A k m 1 = p m 1 +1) pk /p m 1 p m ) p m 1 p m r+p m 1 +1) pk 1 /p m 1 p m ) [p m 1 p m rp+1)+p m 1 +1] pk 2 /p m 1 p m ) [p m 1 p m rp k 2 ++p 2 +p+1)+p m 1 +1] p/p m 1p m ) 1/A k m i = [p m ip m 1 +1)++1)] pk /p m i p m ) We can rewrite 1/A k 1 [p m i p m i+1 p m r +1) ++1)+1)+1] pk 1 /p m i p m ) [p m i p m rp+1)+1)++1)+1] pk 2 /p m i p m ) [p m i p m rp k 2 ++p+1)+1) in 21) as 211) 1/A k 1 = r pk 1)/) k 1 p j )+1] p/p m ip m ) ) p k j 1 By assumptionandp > 1 wehavemaxα i = α 1 sothefollowing inequalities remain true for j : r1+p++p j ) = α 1 )1+p++p j ) > 1 r1+p++p j ) > 1+p m 1 r1+p++p j ) > 1+p m 2 p m 1 +p m 2 r1+p++p j ) > 1+p m i p m 1 +p m i p m 2 ++p m i for i = 1m 2 Using this in 21) for A k 2 to A k m we obtain 212) 1/A k m [rp m +1)] pk 1 /p m [rp+1)p m +1)] pk 2 /p m [rp k 2 ++p 2 +p+1)p m +1)] p/p m [ k 2 = [rp m +1)] 1 p k p pm p j+1 ) p k j 1] 1/pm 1 1/A k m 1 [rp m 1 p m +1)] pk 1 /p m 1 p m ) [rp+1)p m 1 p m +1)] pk 2 /p m 1 p m ) [rp k 2 ++p 2 +p+1) p m 1 p m +1)] p/p m 1p m ) r pk /p m 1 p m )

7 Coupled Fujita type system 29 = r pk /p m 1 p m ) [rp m 1 p m +1)] [ k 2 p j ) A k m i ) i j= p m j [ i j= 1 p m 1 pm p k p ) p k j 1] 1/pm 1 p m ) ) ] p k 1[ i p m j +1 r rp+1) [ rp k 2 ++p+1) = r pk[ i j= )]p k p p m j +1 i j= k 2 j= )] p k 2 p m j +1 )] pr p p m j +1 k p j+1 1 ) p k j 1 for i = 1m 2 Substituting 211) 212) and 213) for i = 1m 2 into 29) we infer 214) St)u 1 ) pk t r pk p k 1 1 [1+ p r m 2 i= [ k 1 [ 1+ i j= p j+1 1 pm + p p m 1 pm ++ p m 2 j= p m j k )p p p i j= p m j) 1] m j ut) ) p k j 1] 1+ 1 pm + 1 p m 1 pm ++ m 2 j= p m j) 1 Using 15) and 22) we observe that 1+ p p ++ = 1+p 1 +p 1 p 2 ++p 1 p m 1 = r p m p m p = 1+ r 1 p m p m p 2 p i m 1 p m j p m j = p for all i = 1m 2 j= j= whence by 214) we obtain 215) St)u 1 ) pk t r pk 1 r pk 1 r 1+p) pk p r 1 p [ k 1 p j+1 1 ] ) k j 1 ] 1+ r 1 p ut)

8 21 J Renc l awowicz Thus 216) Now we have to estimate B k = lnb k = k 1 p j+1 1 ) k j 1 = k k p k j [lnp j 1) ln)] p j ) k j 1 p ) plnp pk 1 1 ) 2 ln)pk = pk 1 p ln [ p )p k 1 ] p = ln = ln[b pk 1 ] where b = p p/) /) > Employing this in 215) we finally have St)u 1 t r p k 1 p k 1)r p k p r 1 p k ) r p k ) 1+p) p k ) p b pk 1 r 1 p k 1+ ) p ) ut) 1/pk Using ut) < and letting k we get 217) t r/) St)u 1 c < where c = cprb) = cp i ) i = 1m Noting r = α 1 ) we get the assertion Now we assume that n < m ie p i < 1 for n < i m Then instead of 24) we get using the Jensen inequality for p m < 1 u m t) St)u p m 1 ds = tst)u p m 1 and analogously for p m j < 1 ie for m j > n) we can prove inductively exactly as 25) 218) u m j t) c j St)u π j 1 tr j where c j π j r j are given by 26) Using218) forj = m n 1in 21) fori = n andthejenseninequality for p n 1 we get u n t) c m n St)u π m n 1 1 ) p n t r m n where π m n 1 = m j=n+1 Again arguing by induction as in the proof of formula 25) we conclude u m j t) c j St)u π m n 1 1 ) p np m j t r j p j

9 for j m n and Coupled Fujita type system 211 u 1 t) c m 1 St)u π m n 1 1 ) p np 1 t r m 1 219) = c m 1 St)u π m n 1 1 ) π m 1/π m n 1 t r m 1 where π m n 1 = m i=n+1 p i by 26) and π m 1 r m 1 c m 1 are given by 28) Let us remark that π m 1 /π m n 1 = n i=1 p i 1 and since π m 1 > 1 also l i=1 p i π m 1 > 1 for any n l m It follows that we apply henceforth the Jensen inequality with an onent greater than 1 Therefore repeating the above considerations we get instead of 29) 22) u 1 t) m A k ist)u π m n 1 i=1 1 ) pk /π m n 1 t r p k 1 where the A k i are defined by 21) To estimate m i=1 Ak i we proceed in the same way as before because we have not used n there) whence St)u π m n 1 1 t π m n 1α 1 c < after letting k Thus the proof is complete Lemma 23 Let the assumptions of Lemma 22 be satisfied Then we can find a constant c > c = cp i ) i = 1m such that for t > 221) St)u π m n 1 1 t π m n 1α 1 c < Proof For τt we have by 21) for i = 1 u 1 t+τ) = St+τ)u 1 + = St)u 1 τ)+ t+τ St+τ s)u p 1 2 s)ds St s)u 2 τ +s)) p 1 ds It now follows that we can replace u 1 by u 1 τ) in 23) whence we have the conclusion by Lemma 21 and setting t = τ 3 Proof of Theorem 13 We prove Theorem 13 by contradiction Assuming that 22) holds and maxα i N/2 we see by 15) that α 1 N/2 and p = m p j > 1 We shall derive a lower bound for a solution which contradicts the bounds obtained in Lemmas 22 and 23 By Lemma 21 we have for nontrivial initial data of 11) u 1 x) c e a x 2 for some constants c > a > Employing this in 21) for i = 1 since ) a x St)e a x 2 = 1+4at) N/ at

10 212 J Renc l awowicz we get ) a x 31) u 1 t) St)u 1 c 1+4at) N/ at Using this bound in 21) for i = m we have 32) u m t) c p m = c p m St s)u 1 s)) p m ds St s)1+4as) Npm/2 apm x 2 ) ds 1+4as 1+4as) Np m 1)/2 1+4as+4ap m t s)) N/2 ap m x 2 1+4as+4ap m t s) ) ds Recall the definition of the number n in Lemma 22 By induction we prove that the following estimate holds for m j > n and some constant C: 33) u m j t) c1+4at) N/2 4at j) β aπj j x 2 ) 1+4aπ j t where j = 2j 1 for j > = > 1 34) m π j = p m j π j 1 = p j π = p m k=m j β j = p m j β j 1 + N 2 1 p m j)+1 β 1 = and t > j/4a) Let n < m Then p m < 1 p n+1 < 1 and p n 1 by definition and for the function f m s) = 1+4as+4ap m t s) we have f s) = 4a1 p m ) > Thus f m ) f m s) f m t) for s [t] Using this we estimate 32) as follows: u m t) c p m 1+4at) N/2 apm x 2 ) t 1+4as) Npm 1)/2 ds 1+4ap m t c1+4at) N/2 4at) 1 Npm 1)/2 apm x 2 1+4ap m t so we have 33) for j = )

11 Coupled Fujita type system 213 Substituting 33) in 21) for i = m j 1 we have u m j+1) t) c = c j/4a) j/4a) St s)u m j s)) p m j 1 ds St s)1+4as) Np m j 1/2 4as j) p m j 1β j apm j 1 π j x 2 ) ds 1+4aπ j s j/4a) 1+4as) Np m j 1/2 1+4aπ j s) N/2 1+4aπ j s+4aπ j p m j 1 t s)) N/2 ap m j 1 π j x 2 ) 4as j) p m j 1β j ds 1+4aπ j s+4aπ j p m j 1 t s) Considerf m j 1 s) = 1+4aπ j s+4aπ j+1 t s) Fromf m j 1 s) = 4aπ j1 p m j 1 ) > we can conclude u m j+1) t) c p m j 1 j 1+4aπ j t) N/2 aπj+1 x 2 ) 1+4aπ j+1 t j/4a) 1+4as) Np m j 1/2 4as j) p m j 1β j 1+4aπ j s) N/2 ds To estimate the integral notice that 35) 36) and 1+4aπ j t) N/2 π N/2 j 1+4at) N/2 since π j < 1 1+4aπ j t) N/2 1+4aπ m 1 t) N/2 π N/2 m 1 1+4at) N/2 j 1 37) 4at j c 1+4at) for t > 2j/4a) with c = j 2j +1 Thus integrating we get u m j 1 t) c1+4at) N/2 aπj+1 x 2 ) 1+4aπ j+1 t 4at 2j )1+N1 p m j 1)/2+p m j 1 β j

12 214 J Renc l awowicz for t > 2j /4a) This in view of 34) gives 33) with j replaced by j +1 Analogously for m j n we have the following estimate for any t > j/4a): 38) u m j t) c1+4at) N/2 4at j) β j aπ j x 2 1+4aπ m n 1 t with j β j π j defined in 34) We proceed in a similar way Assuming 38) for some j m n and using 21) for i = m j +1) we obtain u m j 1 t) c j/4a) 1+4as) Np m j 1/2 4as j) p m j 1β j 1+4aπ m n 1 s) N/2 1+4aπ m n 1 s+4aπ j p m j 1 t s)) N/2 ap m j 1 π j x 2 ) ds 1+4aπ m n 1 s+4aπ j p m j 1 t s) Therefore we have g m j 1 s) = 1+4aπ m n 1 s+π j p m j 1 t s)) so g m j 1s) = 4aπ m n 1 1 π ) n ) j+1 = 4aπ m n 1 1 p k π m n 1 and since p k 1 for k n we get g m j 1 s) Thus u m j 1 t) c1+4aπ j+1 t) N/2 aπj+1 x 2 ) 1+4aπ m n 1 t j/4a) k=m j 1 1+4as) Np m j 1/2 4as j) p m j 1β j 1+4aπ m n 1 s) N/2 ds Using35) withπ j replaced by π m n 1 < 1 36) withπ j+1 andintegrating we obtain 38) with j replaced by j +1 and t > j+1 /4a) We need a lower bound for u 1 t) By assumption m p j > 1 whence by 22) p 1 > 1 but we can have p 2 < 1 or p 2 1 In the former case we have m j = 2 > n = 1 therefore 33) gives 39) u 2 t) c1+4at) N/2 4at m 2 ) β aπm 2 x 2 m 2 1+4aπ m 2 t Otherwise from m j = 2 n and 38) we get 31) u 2 t) c1+4at) N/2 4at m 2 ) β aπm 2 x 2 ) m 2 1+4aπ m n 1 t ) )

13 Coupled Fujita type system 215 Defining M = π m n 1 with π m 2 = M for n = 1) we conclude 311) u 2 t) c1+4at) N/2 4at m 2 ) β aπm 2 x 2 ) m 2 1+4aMt Using 21) for i = 1 we obtain as p 1 > 1 312) u 1 t) c m 2 /4a) m 2 /4a) St s)u 2 s)) p 1 ds 1+4as) Np 1/2 1+4aMs) N/2 1+4aMs+4aπ m 1 t s)) N/2 aπ m 1 x 2 ) 4as m 2 ) p 1β m 2 ds 1+4aMs+4aπ m 1 t s) c1+4aπ m 1 t) N/2 aπm 1 x 2 ) 1+4aMt m 2 /4a) 4as m 2 ) p 1β m 2 1+4aMs) N/2 1+4as) Np 1/2 ds c1+4at) N/2 aπm 1 x 2 1+4aMt ) m 1 /4a) 1+4as) β m 1 1 ds By recursive definition 34) and a routine inductive calculation we get β j = 1+p m j +p m j p m j 1 + +p m j p m 1 + N 2 1 p m j p m ) therefore using the definition of p r α 1 see 15) 28)) we obtain β m 1 = 1+p 1 +p 1 p 2 ++p 1 p m 1 + N 2 1 p 1p m ) = r + N 2 1 p) Since α 1 N/2 p > 1 and r = α 1 p 1) we see that β m 1 so 312) gives ) ) ap x u 1 t) 1+4at) N/ at log 1+4aMt 1+ m 1

14 216 J Renc l awowicz for t > m 1 /4a) It follows that [ )] M ) 1+4at apm x St)u 1 t)) M 1+4at) NM/2 2 log St) 1+ m 1 1+4aMt [ )] M 1+4at = c1+4at) NM/2 log 1+ m 1 1+4aMp+1)t) N/2 1+4aMt) N/2 apm x 2 ) 1+4aMp+1)t [ ] N/2 c1+4at) NM/2 1+4aMt apm x 2 ) 1+4aMt)p+1) 1+4aMp+1)t [ )] M 1+4at log 1+ m 1 Putting x = in the last estimate we have [ )] M 1+4at 1+4at) NM/2 St)u 1 t)) M c log 1+ m 1 Therefore for t > max{1 m 1 /4a)} we obtain using α 1 N/2 and M = π m n 1 [ )] πm n 1 313) t π m n 1α 1 St)u 1 t)) π m n 1 1+4at c log 1+ m 1 It is clear that for t large enough 313) is incompatible with the bound 221) This implies that u 1 t) blows up in a finite time which concludes the proof of the theorem in this case If n = m ie p i 1i = 1m then instead of 38) we obtain 314) u m j t) c1+4at) N/2 4at j) β aπj j x 2 ) 1+4at for t > j/4a) and with jβ j π j defined as above Then we can repeat our considerations concerning the lower bound for u 1 t) in the case n < m simply using 314) instead of 33) and 38) and starting from 311) replacing everywhere M = π m n 1 by 1 Finally we conclude that for t > max{1 m 1 /4a)} ) 315) t α 1 1+4at St)u 1 t) clog 1+ m 1 This contradicts 221) in the case n = m and the proof is complete

15 4 Proofs of Theorems 11 and 12 Coupled Fujita type system 217 Proof of Theorem 11 By assumption m p j = 1 We are looking for a global supersolution to 11) of the form 41) u 1 u m A 1 e θ 1t = A m e θ mt We choose A i i = 1m so large that u i L A i Then 41) is a supersolution to 11) if for all t 42) This is equivalent to 43) u it u i u p i i+1 i = 1m 1 u mt u m u p m 1 θ i > A 1 i A p i i+1 p iθ i+1 θ i )t) i = 1m 1 θ m > A 1 m A p m 1 p m θ 1 θ m )t) If we take θ i+1 = θ i /p i i = 1m 1 then from m p j = 1 we get θ m = θ 1 / m 1 p j = θ 1 p m so 43) holds for θ 1 large enough Thus the solution of 11) is global Proof of Theorem 12 In this case from p j 14) 15) we have m p j < 1 We assert that u i ) m i=1 of the form 44) u 1 u m = A 1 t+t ) θ 1 A m t+t ) θ m is a global supersolution to 11) for some positive constants A i θ i We have to choose t such that u i x) u i for x R N The inequalities 42) imply that 45) θ i p i θ i+1 1 θ m p m θ 1 1 i = 1m 1 Noting that 45) has the form A I) θ) 1 where θ = θ 1 θ m ) 1 = 11) A is given by 12) we assume θ i = α i so θ i > as maxα i < Then 44) satisfies 42) provided that 46) A i θ i A p i i+1 i = 1m 1 A m θ m A p m 1 and t is large enough Thus the proof is complete

16 218 J Renc l awowicz References [EH] MEscobedoandMAHerreroBoundednessandblowupforasemilinear reaction-diffusion system J Differential Equations ) [EL] MEscobedoandHALevineCriticalblowupandglobalexistencenumbers for a weakly coupled system of reaction-diffusion equations Arch Rational Mech Anal ) 47 1 [F1] HFujitaOntheblowingupofsolutionsoftheCauchyproblemforu t = u+ u 1+α JFacSciUnivTokyoSectIAMath131966) [F2] On some nonexistence and nonuniqueness theorems for nonlinear parabolic equations in: Proc Sympos Pure Math 18 Amer Math Soc [R] JRenc lawowiczglobalexistenceandblowupofsolutionsforacompletely coupled Fujita type system of reaction-diffusion equations Appl MathWarsaw) ) Joanna Renc lawowicz Institute of Mathematics Polish Academy of Sciences Śniadeckich 8-95 Warszawa Poland jr@impangovpl Received on