THESIS wcmom s ATE UN 1111/ 1/11/1'I/1/1fli/"3'" 1 /f/l/I/llf/Will/W 3 1293 01050 1735 If This is to certify that the dissertation entitled Global Existence and Blow-up of Solutions to Nonlinear Nave Equations presented by Hengli-Jiao has been accepted towards fulfillment of the requirements for Ph . D. degree in Mathematics 1%3Qr— V \Major pro essorv Date 8/ 23/96 MS U is an Affirmative Action/Equal Opportunity Institution 0-12771 LIBRARY ‘ .MiCMQan State University PLACE IN RETURN BOX to remove We checkout from your record. TO AVOID FINES return on or betore date due. —__——————_F DATE DUE DATE DUE DATE DUE *3- JL__ 1 1 GLOBAL EXISTENCE AND BLOW-UP OF SOLUTIONS FOR NONLINEAR WAVE EQUATIONS By HENGLI J IAO A DISSERTATION Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mathematics 1996 Itlr 1,.( r.h.|..I| I‘ .1, ABSTRACT GLOBAL EXISTENCE AND BLOW-UP OF SOLUTIONS FOR NONLINEAR WAVE EQUATIONS By HENGLI J IAO This dissertation studies the global existence and finite-time blow-up of solutions to the nonlinear wave equation u“ — Au = C(u, Du, Dzu) in high space dimensions. For G(u, Du, Dzu) = Iul”, we establish the global existence of solutions with small initial data when p > mfifi for n = 5, p > gigfl for n > 7 respectively; we also prove that most solutions of the equation blow up when l < p < "fl+2"(::‘:)10"—7 for n 2 3. For G(u,Du,Dzu) = lurl”, lutl”, lurrlp 0r luttlpa we prove that most solutions of the equations blow up when n 2 2 and 1 < p S E—i; we also estimate the life-span of the solutions when n 2 2 and p = fl. To: my wife and our parents iii ACKNOWLEDGMENTS My deepest thanks go to my advisor, Dr. Zhengfang Zhou, for his guidance and constant encouragement. His patience and thoughtfulness have made my stay much easier. I also thank my thesis committee members, Drs. T. Y. Li, David Yen, Michael Frazier and Dennis Dunninger, for their time and interest. I appreciate the support of the Department of Mathematics in the last five years when I experienced growth in research and teaching . Never forgotten will be the numerous discussions with many faculty members. iv Contents Introduction 1 1 Finite-Time Blow-Up for Solutions of Semilinear Wave Equations 4 1.1 Preliminaries ............................... 4 1.2 Proof of the Theorems .......................... 8 2 Global Existence for Semilinear Wave Equations in Five Space Di- mensions l3 3 Global Existence for Semilinear Wave Equations in Seven and Nine Space Dimensions 33 4 Finite-time Blow-up and Life Span of Quasilinear Wave Equations 45 4.1 Blow-Up of the Solutions ......................... 45 4.2 Life-span of the solutions ......................... 53 Bibliography 62 Introduction The existence of global solutions of initial value problems for nonlinear partial differ- ential equations has been studied extensively. There are many partial but only few general results. In this dissertation, we are concerned with equations of the following type u“ — Au = G(u, Du, D2u), u(113,0) = f($), Mano) = 9(3), where f,g 6 08°(Rn). We concentrate here on the following topics: (a) Global existence of the solution, i.e., existence of solution u(x, t) for all (x, t) 6 IR" x IR with smooth initial data. (b) Finite-time blow-up of solutions, i.e., u or the derivatives of u of certain order tend to infinity as t tends to some point T0 with To < 00. (c) Life span of a solution, i.e., a solution that exists for all x up to a maximal time To, in the sense that u(x, t) can be defined at most on interval [0, To). Let G(u, Du, D2u) = Iul” with p > 1. (0.1) becomes semilinear wave equation u“ — Au = M", ‘ u(x,0) = f(:r), u(x,0) = g(:r). Equation (0.2) was first studied by John [6] who proved that for n = 3 and 1 < p < (0.2) 1 + J2, a global solution does not exist for any smooth non-trival data with compact support; and for p > 1+ \/2, a global solution exists provided that the initial data are sufficiently small by the contraction mapping theorem. Later, Glassey [3], by using 1 differential inequalities, has showed that for n = 2 and 1 < p < gig the solutions must blow up in finite time. Glassey [4] also proved that for n = 2, p > gig a global solution exists for small data by estimating the solution and using contraction mapping theorem. Moreover, Glassey [4] conjectured that for n 2 2, there exists a critical value po(n) such that most solutions blow up in finite time when 1 < p S po(n), and a small global solutions exist for p > p0(n) in weighted Loo space with weight function gamut) = (t + 1x1 + 2k)"—3—‘(t — m + 2k)? and q = W, where p0(n) is the positive root of quadratic equation (n—1)p2—(n+l)p—2=0. (0.3) n+1+x/n2+10n—7 2(n—1) ' Schaeffer [15] proved that for n = 2 and p = po(n) 2 flag, n = 3 and p:p0(3) _—_ P001) = 1 + J2, the solution must blow up in finite time. To obtain all the results mentioned above, the positivity of the fundamental solution for n S 3 was essential. By using some properties of special functions, Sideris [18] proved that for n 2 4 and 1 < p < p0(n), the solution blows up in finite time provided that some restrictions to the initial data are satisfied. In Chapter 1, we are concerned with the blow-up of solutions to the semilinear wave equation for n > 3 in terms of p. Difficulty arises because the fundamental solution for the wave equation is no longer positive in higher dimensions (n > 3). Our approach to overcome the non-positivity of the fundamental solutions is to per- form analysis near the wave front where the fundamental solution is positive. We will prove that for n 2 2 and 1 < p < po(n), the solutions blow up provided that certain conditions on initial data are satisfied. In comparison with Sideris’s proof, our proof is much simpler. Moreover, we will show that for n 2 3 and p = po(n), the solutions blow up in finite time provided that the initial data are large. The question of existence of the small global solutions is open when n > 3 and p > po(n). In Chapter 2 and 3, by modifying the method of Sideris [17] and Schaeffer [14] for quasilinear wave equations and using some technical estimates, we prove that when p > po(5) = 5910—7 for n = 5, p > Zflafl for n > 7 respectively, there exists a global symmetric solution in weighted L00 space provided the initial data are small. However, the weight function is different from the one that Glassey conjectured. Let G(u, Du, D2u) = [url”, lutlp, luttl” or Iu,,.|” in equation (0.1). Klainerman and Ponce [11] have shown that the equation has a small global solution if the initial data are small enough and p satisfies (n — 1)(p— l)2 — 2p > 0. However, it was conjectured that the critical value is p1(n) = 251% in this case. Sideries [17] studied the case when G(u,Du,D2u) = [Dulp and proved global existence of small radially symmetric solutions when p > 2 and finite-time blow-up of solutions for u}. John [7] and Schaeffer [16] have shown finite—time blow-up of solutions with [utl3 and |u,.|3 for n = 2. By using the same method of [17], Schaeffer [14] has shown p1(5) = %. By performing analysis on wave front, Rammaha [13] has shown finite time blow-up to the nonlinearity lurl” and [utlp provided that p = 24% when n is odd and 1 < p < 241% when n is even. In Chapter 4, by modifying Rammaha’s method and using more technical esti- mates, we prove finite—time blow-up to the nonlinearity |u,.[JD and [utlp with p = if] for any n 2 2 and extend the results to the nonlinearity [um]? and luul” with p = fig for any n 2 2. By carefullyestimating the solution and using blow-up results from ODE, we will also obtain the life-spans of the solutions. Let G(u,Du, Dzu) = lulaut, or lutlau“. John [8] has showed that for n = 3 and a = 1, the solutions blow up in finite time with some restrictions on the initial data. John [7] has also proved that the finite—time blow-up of solutions to the nonlinearity F [(ut)utt with n = 2 and 3 for certain initial data. We also extend this kind of blow-up results to higher space dimensions in chapter 4. Chapter 1 Finite-Time Blow-Up for Solutions of Semilinear Wave Equations 1 .1 Preliminaries Consider the semilinear wave equation utt—Au=]u|p, TERn,t€R, u(x,0) = 0, ut(:r,0) = g(:r), where 9 satisfies the following assumptions through this chapter (a) g e 0m"), gm 2 o and suppu} c {x = M < k}, (b) there exist positive constants kg and k1 with 0 < k0 < k1 < k such that (i) g(:c) ;‘é 0 on {at : k0 < [:13] < k1} (ii) [to is sufficiently close to k such that $3 > 20, where 20 satisfies that for all z 6 [20,1), Pn(z) Z % and Tn(z) Z %, Pn and Tn are the Legendre and Tschebyscheff polynomials of degree n , respectively. We will prove the following blow-up theorems: Theorem 1.1 Let u E 02(R" x [0,T)) with n 2 2 be a solution of (1.1), where T > 0 is the life-span ofu andl < p < po(n) = "*1“ ” +10n—7. Then T < +00. That 2(n—1) is, it blows up in finite time. Theorem 1.2 Consider the equation u“ — Au = IuIP, ((r,t) 6 IR" X [0,T), n 2 2 (1.2) u(z,0) = 0, ut(:z:,0) = i—g(a:), where g satisfies (a) and (b) in Theorem 1.1, p = p0(n) and e is small. Then T < 00 Given u E C2(IR" X [0, T) define a to be the spherical mean of u, i.e, 1 — t : — t d0) u(r, ) wn lel=1 u(rw, ) for all (r,t) 6 IR x [0, T). Then by Daboux’s identity we have (21)“ — Au = (utt —— Au) 2 71F :2 i[ I [u(rw,t)|pdw Z lulp. wn w=l Note that n — 1_ Au(r,t) = a" — ur . We are led to the following radially symmetric equation u“ — "flu, — u" = C(u), (r,t) 6 IR x (0, +00), u(r,0) = 0, ut(r,0) = g(r), where G(u) 2 [u]? (we used u instead of a to simplify notation). It is well-known from Duhamel’s principle that Lemma 1.1 [13] The solution of equation {1.3) has the following form ifn is odd, u(r,t) = u°(r,t) + C'rl—Zrl fat/l r—t+‘r] r+t—T )\2+r2—(t—r)2 2r/\ ) G(U(/\, 7))dAd‘r, where 1-n '“H n—l /\2 + 1‘2 — t2 u (r,t) - Cr / A P 3 ( 2r). )g(A)dA, Ir—t] 2 and C = C(n) > 0. Ifn is even, 1 u(r,t) = u°(r,t) + Griz? jot jot—Tut — T)? — p2)_5p l. A? [(A2 — (r - mm + p)? — A2)]' r—p] L 2 2 2 _ _ 2 T.-. ’\ J” (t T) G(u(/\,r))d/\dpdr, 2 27‘x\ where r+ “0“”):ch Dig—”2W I.-. A.” [(A2_(,._,,)2)((.+,,)2_,\2)]"5 A2+r2_ 2 T.._( m p ) g(A)dAdp. In both cases, u0 is the solution to linear wave equation — Au = 0, u(r,0) = 0, ut(r,0) = g(r). Lemma 1.2 Suppose that u and u0 are same as in Lemma 1.1, then r+t- r > 2 u,(r t)_ u0 (,)rt +Cr— jot—fr”:An_2’ u(A ,1'))dAd1', rt+ ur,0( t)>Cr‘— 3 [A -‘-‘g(,\)d,\ (1.4) r—t fort+ko((r+p)2—A2]’i Ir-pl 2rA Since p E [0,t], A E (r —p,r+p) ('1 [0,10] and t +ko < r < t + k1, imply that A2 2 _ 2 T_2_( J” p ) g(A)dAdp. 2 2_ 2 _ 2 2_ 2 _ _ A+r p>(r p)+r p>r p>r t>.k_0, 2rA _ 2rA — A T k "k A2 + 1.2 _ p2 n—2 > TT ( 2rA ) — r+p we have NIH and 2100,02 0— [0‘02 — pzrip A? [(A2 - (r — mm + p)? — m] lr-pl Changing order of the integral leads to u°(r,t)20r2-?/r:3\%go)dx\ ‘ [(A2_(.~_,,)2)((.~+p)2_A2)]-%p(12_p2)-%dp. IA-tl Note that under the range of the integration (A? — (r — p)2)((r +102 — A2) = (p2 - (r — A)?)(r + A — p)(r p)+r p>r p>r t+r>r+ko>fi 2rA T 2rA T A T A T T+k T k then A2+t2—'r2 1 11—? > '— TT( 2Ar ) T 2 therefore 00,0 2 C- [07 [“’((1_T)2_p2)—%p/l A? [(A2 — (r — mm + p)? — Anti . r—p] G(u(A, r))dAdpdT Changing the order of the integration leads to 2_,_ t r+t—T n t-r 2 l 00,0201? [0 ctr/H“ A?G(u(A,r))dA IMIRA?—(w~—p)?)((r+p)?—A 1‘2 p dp. JU-TV-pz A similar argument to the estimate of u0 yields t-T 2_ T— 2 r 2_ 2_% p _%r_;_ [Mia ( p))(( +p) A] \/(t_7)2_p2deC/\ , therefore r+t— T (,r t) > Crl— 2 "j /_ T‘(G(u A ,.r))dAd1' r— t+r 1.2 Proof of the Theorems Proof of Theorem 1.1. Define F(t) = f . u(r,t)dx, The idea is to show that F(t) blows up in finite time. It is easy to see that Pu) = / ut(r,t)d$, If 2 Tidtiz/R" u(r,t)da: = fun utt(r,t)cl:z: :2 Ln(Au(r,t) + G(u))d:r, by the divergence theorem and support property of u / Au(r,t)d:z: = 0. Therefore for any t > 0, F(t) = /RG(u(r,t)da: Z /Rn |u|p(r,t)da‘ j" u(r, t)d:c /|x| 0 such that g(r) 75 0 for r 6 (r0 — 6,ro + 6). Hence I—n r+t n—l l—n u°(r,t) 2 CrT / A'2_g(A)dA 2 CrT2T / ATg(A)dA 2 C(g)rT'2T, r—t where ro+5 11-] 0(9) = C [W A-rg(A)dA > 0. From Lemma 1.2, we have u(r,t) Z u0(r, t) > 0 for t + ho < r < t + kl. Therefore |u(r,t)|p da' 2 / d2: p / |u°(r,t)| t+ko 0 and a To such that for t > To F0) 2 Aowucwu + own-1. (1.6) After two integrations, we have F(t) 2 A1(C(g )) (t + miller“ + Bt + E (1.7) for t large enough. Since 1 < po(n)< n+1—" —p0 > n+1— ”—21—;29—1- =1. Therefore if t is large enough, the linear term in (1.7) can be absorbed into first term. That is, there exists a T1 > To and an A2 such that for t > T1 Fm 2 means + k)“‘-— (1.8) Since F(t) > 0 for all t and F(0) > 0, hence F(t) > 0 for all t. Combining (1.3) and (1.8) yields mm) 2 mm)...“ + Arr-“P, F (013’ (t) Z F(t)F(t)[wn(t + k)”]"’°A2(C(g))”(t + k)‘"+“"7Tp"p‘l’- Therefore iiim) 2 A30 + k) " 410’???“ 2F< )F(t)- Letd=—(— ——2—1p 2+"—'2lip—1)=(13—1p2—1‘12L—1p—1)+2,then0-<"-”P-2- C(t + k)("_1)("2_31_"p) —> 00 as t —> 00, There exists T2 > T1 such that for t 2 T2, 15%) 2 iA3F2(t)(t + k)-5. Therefore . F0) 1 —5 l _ > _ 2 which implies that lnF(t) — lnF(T2) > [(t + 101-? — (T2 + 151-?) . There exists a T3 > T2, such that for t > T3 lnF(t) > A4(t + Icy-i, F(t) > eAdukV‘i, where A4 = fiAéI—i—E and 1—5 > 0. Hence, there exists a T; > T3 such that for t > T4 F(t) > (t + Icy". (1.9) Then by (1.5) for t > T4 F0) 2 cat-P0 + k)-" 0 for any t > 0, we have F(t)F(t) 2 w;‘PF(t)FE2‘l, for t > T4. Integrating both sides once, we have 1 - - Luz?) +1 +1 50%) — F211;» 2 fl +‘1 (Fe—+10) — F"2—+‘(T4)) 2 Therefore there exists a T5 > T4 such that for t > T5 1-10 - wn_ +1 F2(t) > jfiF”2—+l(t). 2 Hence for t 2 T5, 2L 2 F(t) >( “3‘3 )iFhflm, + t—* i.e., which implies that 1 1 1 I“? 5 +3 (11;: T r:_1 )>(+1n-l ) (tTT5l' ?.—-1 F‘(Ts) F40) ‘11-“ If t ——) 00, we have a contradiction. i.e. T must be finite. Cl Proof of Theorem 1.2. We use the same argument in the proof of theorem 1.1 and note that 6 = —(—-"—2‘—l-p2 + 9—3—1 +1— 2) = 2 in this case, then for t > T2 1 — 2 A§(t + k)"1 which implies that mm) — lnF(T2) 2 A§(zn(t + k) — mar2 + k)). MIH 12 So there exists a T; > T2 such that t > T3; lnF(t) 2 iAélna + k) = ln(t + kfiAi i.e., F(t) 2 (t + k)“; Let 1 A4 = iAé‘ = 4i[w,1,"’A2C'p(g)]% = 4i[w,1,‘pA2§/::6 A?3-‘g(,\)d,\] 2 . If e is sufficiently small, then A4 > 2n, consequently, (1.10) holds. The rest of the proof can be carried out using the exact argument in the proof of theorem 1.1. CI Chapter 2 Global Existence for Semilinear Wave Equations in Five Space Dimensions In this chapter, we will study the existence of global solution for n = 5. We will prove that the global existence of solutions with small initial radially symmetric data for 3 l7 p>_+4£. Theorem 2.1 Consider the radially symmetric wave equation with n = 5 4 U“ — _Ur — arr : C(U), T 2 0, t2 0, 7‘ u(r,0) = f(7‘), ut(r10) -.: g(r), (2'1) where fund 9 are even functions, f,g E C3(IR), and supp{f,g} C {r : [r] S k}. If G satisfies 0G(u) Du |G(u)| _<_ Alul”, | |S BlulpTl, A,B > 0, then the equation has a unique global solution in the space 0,9,). for provided that p > 110(5) = litifl and 23:0(]]f(i)]]oo + llgmlloo) is sufiiciently small, where 13 14 u e 00(1): x 10.00)) | (m). 6 00(1)? x 10.00)), u is even and u(r,t) = 0 if|r] > t+ k, llull = sup(l¢ul + |¢(ru)l + 1410204) < +oo, 43(r, t) = (r + t + 2k)(t — r + 2k)2""‘3 We will prove the theorem by several lemmas. For u 6 SIM, define the operator with r > 0 Tu(r,t) = u°(r,t) + %Tou(r, t), (2.2) where u0(r,t) = Cor’3 (1:1 A(r2 + A2 — t2)g(A)dA, 1% [Cor -3 (1:1 A(r2 + A? —12)f(A)dA (2.3) Tou( (r, t) —/ /I:+I( A( r + A2— (t — r)2)G(u(A,1-))dAd1-. (2.4) By Duhamel Principle, solving (2.1) is equivalent to finding a fixed point for the operator T in QM. The goal in this chapter is to prove T is a contraction map. Therefore We need to prove the following: (i) Tu(r,t) is even in r; (ii) Tu(r,t) = 0 for Ir] 2 t + r; (iii) Tu,rTu and (rTu), are continuous; (iv) ||Tu|| < 00. Since A(r2 + A2 —— (t — T)2)C(u(A, T)) is a odd function in A, therefore Tou(7‘,t) : / Arjtt+;A(2 r + A2— (t _ T)2)G(u(A,T))d), = jo‘ 1:ij (r2 + ,\2 _ (t _ T)2)G(u(A,r))dA, Note that g(A) and f(A) are even functions, A(A2 + r2 — t2)f(A) and A(A2 + r2 — t2)g(A) are odd functions of A, we have u°(—r,t) = u°(r,t). Similarly ngou(r,t) = (:17FT0U(-7‘,t)- Therefore Tu(r, t) = Tu(—r,t). 15 From the expressions of u0 and Tou, we can obtain Tu(r, t) = 0 for [r] _>_ t+ k. It remains to prove (iii) and (iv). For u E 911.1: and 0 < r < t + k, let z(r) = min{1,;1;}. Then we have Ilull IIUH Hence 1110.0) 3 ”$ng (2 5) In order to prove (iii) and (iv), we will prove the following Lemmas. Lemma 2.1 Forp E (“B-g3), 0 < r < t+ k, t1 = maa:{0,%(t — r — 19)} and t2 = ma:r{0, %(t + r — k)}, there exists a constant C1 such that t AZP(A) Cl _ _ ,d < , t. ¢v(A,T)(-"‘ ‘+ ' T - (t — r +2k)2r-2 . AzP(A) C. . _ < . t2 ¢P(A, T) ATrH—TdT T (t - r + 2k)2p-2 Proof of Lemma 2.1. Without loss of generality, we assume t—r Z 0. If r—t+7' > 0, i.e., T > t — r, then ¢(T—t+T,T) = (r—t+r+r+2k)(t—r+2k)2P—3 = (27+r_t+2k)(t_r+2k)2p—3 t AzP(A) _ r AzP(A) t. ¢P(A, T)[=”“+"dT T /0 (2A + t — r + 2k)P(t — r + 21c)1><21»—3)dA - l A 00 A1"? < t — 21: We?) / dA / dA -( “L ) [0(2A+t—r+2k)P +1 (2A+t—r+2k)P g (t — r + 2k)T”(2pT3) (t — r + 21:)”0 + (t —- r + 2hr; /00 AngpdA] . 1 g(t—r+2k)-P-?[(t—r+2k)-?+ 1 ] 1 s (Is—5+ 3 t — r + 2k ‘P(2P‘3)‘5. 5P—2)( ) 16 Forp> fl1f1fl,wehave —p(2p—3)—§<—2p+2and notethatt—r+2k>k then (t — r + 2h)T”(2p‘3)‘iE : (t _ r + 2k)-2P+2(t _ r + 2k)-p(2p-3)-§+2p—2 S k—P(2P-3l-§+2P-’2(t _ 7. + 2k)-2p+2. Ifr—t+1' <0, i.e.,r < t—r, then ¢(t—r—r,r) = (t—T+2k)(27’—(t—r)+2k)2p‘3 t r Az”(A) dr _ ft—r t1 AzP(A) d), :1 ¢P(A, 1') T4” _ o (t — r + 2k)P(t — r — 2A + 2k)P(2P-3) Ift—r S k, then t1 = ma:r{0,-.:;(t—r—k)} =0 ft ' “ Az”(A) 0 (t — 7‘ + 2k)7’(t — r — 2A + 2k)p(2p-3) dA t r Az”(A) < - /0 (t — r + 2k)P(t — r — 2(t - r) + 2k)P(2P-3°d’\ _ k Az”(/\) - k A S (t T r + 2k) 10/0 (-(t — r) + 2k)p(2p—3) 5 (t T r + 2k) 10/0 MAW-31‘“ S (t — r + 2k)—Pk2—P(2P-3). Since p < 2, we have (t — r + 2k)2”’ Z 192“”. Therefore “" Azp(A) ‘1 4W)» 7') Ift—r>k, thent1= %(t-r- k)>() /‘ " ‘1 Az”(A) 0 (t — r + 2k)P(t — r — 2A)P(2P—3) [Azt—r—rd‘l' S (t — r + 2k)‘2P+2k-p(2p—3)+p. dA < t 1‘ t1 Al—p dA T/o (t—r+2k)P t(—r—2A)P(2P 3) <(t—r+2k) PMg +/—T2+ s (t — r + 2k)-P(t — r + k)"’(2p‘3) [i Al-PdA 0 Al-v i] (t _ r _ 2,\ + 2k)p(21A-3)d)‘ t— 7‘ + 2k _ (2 _3) t—r+2k 1— +(t—r—2(—T—)+2k)pp / A Pd). t— 21: iii +(__".+_)—p+1/ 2 4 t-riIZk 2 (t — r — 2A + 2k)-P(2P-3>dA 17 Since ’25 1 k Al‘pd = —— — 2"” f0 A p—1(2) t—ri2k _ _ f 4 A1_pd/\ : (Er-p t r+2k _ E S (Er—pt r+2k g 2 4 2 2 4 t-r k t—gik _ 2-3 2 — 2—3 fl_r42k(t—r-—2A+2k)”(p )g/m (t—r—2A+2k)P

-‘—;L+9,'s>#t—rwkwk,—p(2p—3)<—p+2, and —p(2p — 3) + 1 < -—p + 2, therefore there exists a constant, say Mo, such that for t — r S k /t r in mu) 0 (t — r + 2k)P(t — r — 2A + 2k)p(2p—s) dA 3 M00 — r + 2110-210+2 . Combine all above, there exists a constant M1 such that ‘ AzP(A) d < M1 t1 ¢P(A,T) :lt—HT' T " (t — r + 2k:)21"'2 Now we prove the second inequality. First we consider t+r > k, then t2 = %(t+r—k) > 0 and ¢(t + r — 7, T) = (t + 7- + 2k)(27- _ (t + 1.) + 2k)2p—3 t A210“) r+t t2 AZPO‘) _ ,_,d = (M t2 MA, 'r)l*“+ T f, (t + r + 2k)P(t + r —— 2A + 2k)P(2P-3) r+t t2 Al—p < — /r: (t + r + 2k)P(t + r — 2,\ + 2k)p(2p-3) tiri2k Al—p H(t+7‘+2k till/£2:- +1- HIM?" JJ—(t+r_2)‘+2k)p(2p_3)d)\ dA A similar argument used in the proof of first inequality, there exists a constant Mt; such that for t + r > k, t A2109) M3 Mg, _ < < . t2 (MA, r)l“t+r"dT — (t + r + 2k)210-2 — (t — r + 2k)??-2 18 Ift+r$k,thent2=0and t AzP(/\) t Az"(,\) t ,\l-p t2 ¢p(A3T)l\=t+r-1d7— S A Wth+r'—Td7- S A Wk=t+r—7d’r _ [t (t + r — 7)"? — o (t + r + 2k)P(2T — r — t + 2k)P(210-3) t s (t + r + 2k)-Pk-P<2P-3> / (r + t — T)l_pdT 0 (17' 1 = 13— 2k ‘pk‘p(2”‘3) —— 2‘10 t 2-p ( r+ ) [2_p7‘ +,_p(r+) S (t + r + 2k)’"’k""(2""3)—2 1 (r + t)2"’ _ P < k—P(2P-3) 2k 2p+2 __ t - . — 2_p (+r+ ) Therefore there exists a constant M; such that for O < r < t + k t p ' I [\z (A) :r+t-'r 7' < Ml S M1 - t2 MA, T) “ (t + r + 2k)2v—2 (t — 7' + 2k)2P-2 CI Lemma 2.2 Forp 6 (9434123), there exists a constant Cg such that forO < r < 15+]: 1 )‘2-p 02 _— _ _ T < , t1 ¢p(A, T) A—lr t+ ldT __ (t _ r + 2k)2p-2 t ,\2-p 02 , _ < . t2 ¢p(A, T) A—T‘l’t-Td’r _ (t _ r + 2k)2p—2 Proof. Note p E (Ii—fig, 2), then there exists a number 50 > 0 such that _3+%+¢@+%P+8 _ 4 . P Forr—t+7'>0,i.e.,7'>t—r, t —————’\2-p d t 2k -P<2P-3> ' AH dA /¢P(A,r)l‘="“+’ T‘( “H l / (2/\+(t—r+2k)P l g (t — r + 2k)-P<2P-3)[(t — r + 2k)"’ / AZ‘PdA 0 oo A2-p +f1 (2) + t — r + 2k)(3-p—61p)+p—(3—p—51p) d/\], 19 where 0 < 61 < min{6o,2 — 3}, it is easy to see p — (3 — p + 61p) > 0 and W ./1 (2A + t — 1‘ + 2k)(3-P+51p)+p—(3—p+5,p)d’\ s (t — r + 2k)‘2”+3+5lp [00 A-l-de l — _1- _ -2p+3+61p _ [[351] (t r+2k) . Therefore t A2-p 1-, —l S (t — r + 2k)"’”("”"3’ [(t — r + 2k)"’ + %(t — r + 2k)”2”+3+p5‘] P1 = (t — 7‘ + 2k)‘p(2P—3)—2P+3+P51 [(t _ 1‘ + 2k)p—3—p61 + 13%] l S ‘ kp-3-p51 + _L] (t _ r + 2k)-p(29-3)-2p+3+p51. P51 Since p = 3+60+ " (3+6°)2+8, we have —p(2p — 3) — 2p + 3 + p61 < —2p + 2, i.e., 4 2p2 —- (3 + 51)]? — 1 > O, which requires p > 3+5l+ " i3+51)2+8. Therefore ,\=r—t+rd7’ S k3’P‘P51 + —H (t — r + 2k)'2”+2 . t AZ-p ft—r (WK/MT) p 1 Forr—t+T<0,i.e.,Tk,thent1=%(t—r—k)>0 hum s kl—p(2p_3)(t _ r + 2k)_2p+2 t—r Az—P d t 1' t1 A2—p d)‘ t1 (MA, r)l*=’"“+’ T _/o (t — r + 2k)P(t — r — 2A + 2k)v<2v—3) 5 S (t — r + 2k)"’[(t — r + k)"’(2”‘3) / " [9-de 0 —r4 2k +(t _ 7. _ 2(:T%:_2k)+ +-2k) p(2p- -3)/t—‘+_ A2-pd)‘ _- t—rik +(.t__gi]:)-P+2 [‘42 2 (t _ 7. _ 2)‘ + 2k)- p-(2p- ’3)d)\] Note that t—ri2k _- fl 4 A2_ pd/\< 1 (t T+2k)3_p 5 3— p 4 and —p(2p — 3) — p + 3 S —p + 2, therefor there exists a constant No such that H AH dA < N (t + 210-210+2 t1 45P(A,T) \zt—r—T _ 0 7‘ for t — r > k. Combining all estimates gives t AZ-p _2 +2 in (ya T)A=Ir-t+rld)‘ g 02(t — r + 2k) p . Now let us turn our attention to the second inequality in the lemma If t + r _<_ k, the proof is similar to that of t p AZ (A) l _,+,_ ,dT < Cl(t + r + 2k) 2?“ t2 ¢p(’\ a7.)|/\ and is omitted. Ift+r > k then t2— — %(t+7‘+k) > 0 and ¢(t+r— 7', T): (t+r+2k)(27- —(t+r)+2k)2p‘3 t AZ-p :2 amt-W" r+¢ t2 ‘X2—p (1A »/r (t + 1‘ + 2k)P(t + r — 2) + 2k)P(2p-3) ’ 21 r2 A2_p [pr/51“ 1,) 5 g (t + r + 2k)-P[(t + r + k)"’(2”‘3l / 2 Az‘PdA 0 t 2k 13% +( + r + )—P(2P—3)fl )‘2-pd)‘ 5 2 t k Lj-Sfl (——+—’"2—+-—)2 P “a + r —— 2A + 2k) PW 3ldA] "P(2P-3)+3—p 3—P S (t + 7‘ + 2k)’p[(t + r + k)-p(2P-3) (g) + (13+ T2+ 2k) + Isl-“21"?” t+r+k 2'? p(2p-3)-1( 2 ) 1' Note that —p — p(2p — 3) < —2p + 2 and —p — (2p — 3) + 3 — p < —2p + 2 (which requires p > giifl)’ then we conclude there exists a constant N; such that for t+r > k t A2“? t2 ¢P(A, T) This concludes the proof of the lemma. Cl )3.“da < N,’(t + r + 2k)-2P+2 < N;(t — r + 210-2?” . Lemma 2.3 There exists a constant C3 such that for p E (gfiflfl) r+t- -1' Calla“? < . [0 [Ir- ”TIA T))d)\d'r _ (t _ r + 2k)2P—2 Proof. By assumptions on G and (2.5), we have AA2”(»\)|Iull" < AlluIIPAl‘P MAJ) - war) ’ .+. T u(/\ dAd A P m T AM d d < ./()t ./Ir- H—TIA 7-)) T “u“ /0t ['7'— t'+""l ¢p()\ T) T AG(u(/\,T)) g /\A|u(/\,‘r)|” S 22 LetT+/\=,BandT—/\=a,then r+t— —T [0 / ,))d,\dT Ir— —t+‘r|/\ 1- -p < AlluH”/_ G(/_:( T) ldflda t a+2k)M%M3U%+%fi2 _. _p 2% H, (9:2)“P1 =Anun _ < /(————) t— —r( r ,B-_a 1-P + f? a-P+2k) WP 3Mar/H + —((p2+ik)v2d dfi] (2-7) Now we estimate the following integral for 1 — p — l < —2. Case 1: ift—r—as2, /:+r (LEG-{)1 ‘P 1 Sty—0 Al—p (fl+2k)’2 01:3 —";° (2A+a+2k)’dA tir-a Al—p 2, L 3-0. ”‘10 1 0° -—-—-—— < — APP-Id). fl-r-a (2). + k)’dl\ ‘“ 2’ jug—2 2 _ 1 (t — 7' — a)2"'p _ 2’(l+p—2) 2 ° Sincet—r— a>2, wehave— —(t—r—a)>k. Therefore(1+’2c )(t—r—a)>t—r—a+k i.e.,t—r—a>k—+3(t-r—a+k), then t 2.!) Al—p (Ezfi)p+I—2 k p 1+2 __ < _ _ - — . fhya (2/\+k)’d/\_ 22‘P(l+p—2)(t r (1+ ) 23 From Cases 1 and 2, there exists a constant N3 such that t+r a l—p / ( 2 ) ldfl S N3(t _ 7‘ _ a + k)2_p_l . t —r (5 + 2k)’ 2 Now we can estimate first integral in (2.7) and get -0, 1-p M ‘+' (%) 1 2 . -p(2p-3) .______ L. (“2") d“ (H (fi+2k)v2d t—r /: (a + 2k)'p(2p'3)N3(t — r — a + k)2'p"pda t—r t - 7‘ 2-2p 2 S N3 ( 2 + k) f (a + 2k)'p(2p—3)da —k 2—2p 00 S N3 (LIZLZJE) / (a + 2k)-p(2p-3) —k Nakl-pOP-fi) 2_2 : (p(2p-3)_1)22_2p(t—T+2k) p . The second integral in (2.7) can be estimated as follows fi—a t—r t+r (_)1—P1 2k ‘P(2P‘3)d / ——2 —d (:2 (0+ ) a 1-.. (fl+2k)P2 fl 2 t—r 3(2 + 2k -p(2p—3) t'r -p+1 1+,» (%)1_p 1 ) 1-, (t -— r + 2k) do: t (1,8 _. (p + 2k)‘ '2’ t—r S( 2 t—r + 2k)'p(2p'3)(t — r + 2k)_”+lN3 [_r (t — r — a + 102-1"de t—r S 2”(2p'3)N3(t — r + 2k)-p(2p-3)—p+’/ (t — r — a + k)2-l"pda —O|0 2p(2P-3)N3k3-l-p l+p-3 Where 1 = 3—p+p61 and —p(2p—3) —p+l< —2p+2. (t _ 7. + 2k)-p(2p-3)-p+l , From the estimates of first integral and second integral of (2.7), the result followsD Lemma 2.4 Let u E QM, there exists a constant C; such that for O < 7' < t+ k, Ol=1am12, ‘ HUN” _ 0‘ _ < jo (r +t 7') |G(u(r +t T,T))IdT _ C4(t _ r + 2k)2P‘2’ 24 Hull” a—r+2m%4’ Hull” (t — r + 2lc)21"‘2 ’ Hull” a—r+2mw4' t [0 |r — t + T|a|G(u(r — t + 7',T))|d7' S C; /t(7‘ + t - T)2I6—G-(u(r + t — T,T))|dT S C4 0 0r ‘ 2 00 /(7' — t + T) |—(u(r — t + T,T))|dT S C4 0 07‘ Proof. Note that X’z” )l A°IG(u(/\,T))l s AAaIuww s Allztllp¢p(A(T; we have jot“ + t — T)0|G(u(r + t — 7', T)|dT = /t: A°|G(u(,\, 7))IL=r+t_TdT ‘ /\°’z”()\ ‘A"z”()\) < A p r —Td _<_/4 p 27‘ -Td _[2 Hull WA- +1 T Ilu II waal‘ .. r By Lemma 2.1 and Lemma 2.2, we get the first inequality. Similarly, we can get the second inequality in the lemma. Note that 0G Bu 5r 3r A2 g(ufl, T))|— — A2— S A2A|u|p 1Iu,.| z()||u|| .1 2l|u|| Ave-Ionian? 3’“ (w )) AMT—ms won) All I? W < u ——_—, - ' u(x,-r) By Lemma 2.2, the third and fourth inequalities follow from lemma 2.2. Lemma 2.5 Let u E Qkp and Co r+t— 7' T1u(r.t)— — —T0u( (r, t) #59] -/I | /\(/\2 + r2 — (t — T)2)G(u(/\,’r))d)\d7'. r— t+r Then Tlu, rTlu and (rTlu),. are continuous, and the following inequalities hold Hull" < Inwnm—CMMM’ |_I_UI|” 5,¢(7‘ _—t) _I_IUI|” ¢(__7’_ 15) er1u(r, t)| S C5 |(rT1u),. (r, t)| S Cs 25 Proof. Since A|G(u(A,7))| S A||u||p———Ll::a:)a and AM”) is integrable in A. Also £002 + ’\2 " (t “ 7)2)G(u(A, T)) = 27AG(u(A, 7)) IS integrable in A. Therefore (Tou)r(r,t) = f;/;:: 27AG(u(A,7))dAd7 t +/0 A(1‘2 + A2 — (t — 7)2)G(u(A, 7))L\_:+f+:d7 is continuous Note that Mr” + 12 — (t — r)2)G(u(A, 7))B::i§;: = 22420010, 7))K::::;:dr , then (Tou),(r, t) = 27Ju(r, t) where r+t- 7)‘ (r ,t) 2/0 / u(A ,7))dAd7 +f A2G G,u((A 7))|:\(_:+:+:d7. r— t+7 Therefore, Ju is continuous, and by Lemmas 2.3, 2.4 there exists a constant M such that JIUII” (t — r + 2k)??? Differentiate J u with respect to r, we have IJu(r,t)| S M (2.8) (Ju),(r,t) = fot[3(r + t — 7)G(u(r +t — 7, 7)) — 3(7 — t + 7)G(u(r — t + 7, 7)) +(r + t — 7)2—a—(G(u(7 + t — 7, 7)) — (r — t + 7)2§(G(u(7 — t + 7, 7))]. 07 By lemma 2.4, we can conclude that for 0 < 7' < t + k IIUII” < . . 1(Ju)r(r1t)| — 804 (t _ 7‘ + 2k)2p—2 (2 9) Since Ju(0,t) = 0 and Tou(0,t) = 0 , we have, using (2.8) and (2.9) = r < r lJu(r,t)l I /0 (Ju)r(s,t)ds| _ f0 |(Ju).~(s.t)lds r < p 2.10 _8C4Huii (t—T+2k)2p—2 ( ) Tou(r,t) = f(Tou),(s,t)ds 2 fr 23Ju(s,t)ds. (2.11) o o 26 Hence the following inequalities is true T1u(r, t)_ — f—3T0u(r, t) :00] 2sJu( (s, t)d 322—3—00/ 23/0( Ju) (A t ),dAds which implies that Tlu,rT1u are continuous. Ht—r+2k> §u+r+2m 200 200 S < __ P 72/ |Ju(,s t )lds_ 1.2/08C4||u( n _S+2k)2p_2ds 16COC4Hu|Ip r2 Hull” < . Ift—r+2kS -(t+r+2k), i.e.,rZEi-gfl, then 2_C_o 200 (”Ila”)D <— _/" lJu( )|_ds 72 —2-/0 (t _ r + 2k)2P—2 1 Hall" < 800M IIUH” < . 200M r(t—r+2k)2P‘2_ k (t+r+2k)(t—r+2k)2P-3 (2.12) Now we estimate rTlu Co Co |7T1u(r,t)| = r—2|T0u(r,t)| S 7.22/0 2s|Ju(s, t)|ds_ < ——/0 [Ju( (s ,t)|ds (2.13) Ift—r+2k> %,(t+r+2k) then 200 2C'0 MIIUHP <— —f. .J...)|.._ . —/. Mam—28 < 2C M Hull” < 4C M Hull? " ° U—r+2M%4" ° u+r+2ma—r+2mu4‘ Ht—r+2kS%U+r+2M {Ci ' MIIuHP s = 200M HUN" _ HUIIp o (t — s + 2k)"”"“2 (2p — 3)r (t — r + 2k)29-3 (t + 2k)ZP-3 200M ”all? < 800M Hull” "@p—MrU—r+2M%4*1u—3a+m+2ma—r+2mu4' For (rTlu),, we have (rTlu),(r,t)= —§%Tou(r, t) + g(Tou)r(r,t) 27 We have proved that the first term is continuous and satisfies the estimate. The secod term is also continuous, since Co _ Co _ Co r g(rou),(r,t) _ 7mm) _ 7 f0 (Ju),.(s,t)ds. If t — r + 2k > at + r + 2k), then by (2.10) 2C0 200 804HUHP" Hull” t < < . r ”u(,” )l " r (t — r + 2k)2P-2 — 32000“ (t + r + 2k)(t — r + 2k)2P-3 If t — r + 2k 3 %(t + r + 21:), then by (2.8) p p Z—SEIJUU'JHS 200 MIIUII < 800M Hall 7' (t — r + 21021”-2 _ k (t + r + 2k)(t — r + 2k)2P'3° Lemma 2.6 Let h E C2(R) be an even function with h(r) = 0 for Ir] 2 k. Define r+t u(r,t) = / A(r2 +.\2 -—t2)h(A)d/\ -t and Hull.)o = sup{|v(r,t)|;r E R,t Z 0} for v E C0(R x [0,00)). Then there exists a constant C6 such that llwr'svlloo + llwr"svt||oo + llwr'zvlloo + llwr_2vt||oo + llwr’2vrlloo + llwr‘zvnlloo S Cs(||hl|oo + llh'lloo + llh”||oo), where w(r, t) = t + r + 21:. Proof. Without loss of generality, we assume 1" > 0, t > 0 and note that v(r,t) 7t 0 only whent—k mfg, (2.14) implies r+t+2k 7‘2 vr(rat) Ift—r+2k>%(t+r+2k), by —kk, r+t |v(r,t)| g] |A(r2+A2—t2)h(A)|d/\ r-t k s llhlloo/kIAK/cz + lr— tl - lr+t|)d/\ S 2k2||h||°o(k2 + (9(27‘ + k)) S 8k2||h|loo- So |r‘1v(r,t)| < 8k3||h||oo, therefore lr‘2v(r,t)| S Ilfir‘lv(r,t)| S 8k2||h||oo. If r S k, |r_2v(r,t)| S k 1'-3v(r,t)|S12k2||h||00 + 4k3||h'||00 Therefore lr'zvhxt)! S 12k2|lh||oo + 4k3llh'lloo- |r’lv(r,t)| S klr'2v(r,t) S 12k3||h||00 + 4k3llh'lloo. Ift-r+2k> %(t+r+2k) and by using It—rl r—‘Lfilfl, then 4 (7.3322030) s |;v(r,t)| s «Wu/2”... + 4mm...) Therefore |-‘%v(r,t)| g man/2H0, + 24k3||h’||00. 1. Using similar arguments, we have 5 72k3|lh||oo + 24k3llh'lloo. w 5””, t) To estimate |wr’2v,.t|, we need to estimate |r"lv,t| and Ir‘zvrtl. U, = /::t —2Ath(A)dA + 2r(r + t)2h(r + t) + 2r(r — t)2h(r — t), v” = 6r(r + t)h(r + t) + 6r(r — t)h(r — t) +2r(r + t)2h’(r + t) + 2r(r — t)2h'(r — t). 30 Since h is even and vanises outside (——k, k), we have lr'lvrt(rat)l S 12k||h||oo + 4k2|lh'lloo, lirré r’lvm = O. r-) |—887(r-1v,.t(r, t))| = I6h(r + t) + 6h(r — t) + 10(7‘ + t)h’(r + t) +10(r — t)h'(r — t) + 2(r + t)2h"(r + t) + 2(1‘ — t)2h”(r — t)| S 12|ih|ioo + 20k|lh'||oo + 2k2|lh”||oo, r 6 Ir‘1vrt(r,t)l s [0 IE-(z‘lvrdatfldz g (12||hlloo + 20k||h’||oo + 4k2||h”||oo)r. Therefore |r_2v,.t(7‘, t)| S 12||h||oo + 20k||h’||oo + 4k2||h”||oo. (2.16) A similar argument as before yields lwr‘zvrt(r,t)l S 6k(12llh|loo + 20kIIh”||oo + 4k2IIh”lloo)- Since vt(0,t) = 0, and by (2.16) we have Ivt(r,t)l s [0" lvrt(s,t)lds s (12th)... + 201w“... + 4k2llh”||oo)r3, i.e., Ir"3vt(r, t)l S 12||h||oo + 2Okllh'lloo + 4k2||h”|loo On the other hand r+t Iv.) g / |2sth(s)|ds + 2m + t)2|h(r + t)| + 2m — t)2|h(r — t)| r—t If r > k, then k Iv,(r,t)| g / k2t|s|h(s)|ds + 4rk2||h||oo S 4tk2|lh||oo + 4Tk2|ihlioo S 4(27‘ + k)k2|ihiioo S 12rk2||hlloo 31 i.e., for r > k Ir‘lvt(r,t)| S 12k2||h||oo, Ir-zvt(r,t)| S 12k||h||oo. If r S k, It’lth‘, t)| S k2(12||h||oo + 2Okllh'lloo + 4k2||h"||oo), IT'thU‘a t)| S k(12l|h||oo + 2Okllh’lloo + 4k2||h”||oo). By a similar argument as before, there exists a constant C (k) such taht Iwr"2vt| S C(k)(!|h||oo + Ilh'lloo + llh”||oo), lwr’3vtl S C(k)(||h|loo + llh'lloo + Ilh"||oo)- Lemma 2.7 There exists a constant C: such that lluoll S C7(llflrloo + llf'lloo + ||f"||oo + llglloo + Milk» + ||9"||oo)- Proof. Recall “no” 2 sup{|¢u0i + |¢(ru°)| + |¢(ru0),.|} and note t — r + 2k S 3k on the support of no, then |¢(r,t)u0(r,t)| S (3k:)2”_3 |w(r,t)r'3 (v9(r, t) + vtf(r,t))| |¢(r, t)ru°(r. t)| s ewe—3 |w(r, or” (v90, t) + w. a)! |¢(r,t)(ru°)r(r,t)l S (3k)2”’3|w(r,t)(-2r'3v9(r,t) +r‘2vf(r, t) - 2r‘3v{(r, t) + r'2v11(r,t))l, where r+t v9 = f A(r2 + A2 — t2)g(A)dA, r-t r+t vf = / A(r2 + A2 —t2)f(A)dA. r—t Therefore, The lemma follows from lemma 2.6. 32 Proof of Theorem 2.1. By Lemma 2.5 we have IITUII S lluoll + IITIUII S Huoll +3Csllull” Let 5 = (6C5)—;1_T, hence C5 = $69”. Let 9;. = {u e an Hull 3 a}. 2 Choose z:(||f(‘)||oo + ||g(i)||oo) small enough such that ||u°|| < :35, then .=0 e 1 ”Tun s ||u°|| +30.“qu s ,- + 3 - gel-Pep = 5 Therefore T maps “ink to itself. Exactly same estimate as before but replace C by A(|u|" — M”) and note that l IUI” — lvlpl S (IUI + IUDHIU — vl, then HT“ - Tvll S 305(II’UII + HUID'HHU - U“ S 305(2€)p—1||u — v” = 2P‘2||u - vll- Since p < 2, hence 2""2 < 1, and T is a contraction map on Qg’k, therefore T is a contraction map on flak. By contraction mapping theorem, there exists a unigue fixed point it of 9:1,]: such that Ta = u, i.e., the global solution of the nonlinear wave equation exists. C1 Theorem 2.2 For n = 5 and p Z 2, the global solution of the nonlinear wave equa- tion exists in Qch) where QM defined as before except qb(r, t) = (15+r+2k)(t—r+2/c)p‘l Proof The proof is similar to the proof of Theorem 2.1,and is omitted here. Chapter 3 Global Existence for Semilinear Wave Equations in Seven and Nine Space Dimensions Consider the radially wave equation with n = 7 and 9 n—l Uu— r u,—u,,=F(u), r20,t20, (3.1) u(r,0) = f(r), ut(r,0) = g(r), (3.2) where fand g are even functions, f,g E C3(IR), and supp{f,g} C {r : |r| S k}. Assume |F(u)| S A|u|p, lagginl S Blulp'l, A,B > 0. Let i —k + gate)? ifr 3 —§, 2(7“) = t r if |r| < 129, k — §e(%—">% ifr 2 %. Define u E C°(R X [0,00)) l rzihiu,(rzn—;‘5u)r E C0(R X [0,00)), u is even and u(r,t) = 0 if |r| > t + k, Qp,k,n : ”U” = SUP(|¢(T,t)rznT—7ul + l(r,t)(rz"T_5‘u),.|) < +oo, ¢(T‘,t) : (7‘ + t + 2k)?“ — 7‘ + 2k)2_§lp—E1i-l 33 34 If u E QM, then there exists a constant C such that C||u|| Cllull — < —. |U(T,t) — rzn;7¢(r,t), lily-(7', t) — rzn;5 ¢(r,t) By using the similar method to the case n = 5 and more technical estimates, we obtain the following Theorem 3.1 The equation with n = 7 has a unique global solution in QM provided that p E (M, 2) and Z,=0(Ilfilloo + Ilgilloo) is sufficiently small. Here we note that p1(7) = y%@ is greater than po(7) = 24135. In order to prove the theorem, we need the following lemmas which can be proved by the similar argument to the case n = 5. We omit the detailed proofs of most lemmas. Lemma 3.1 For p E (p0(n),2), there exist constant C1(n,k,p) and C2(n,k,p) such that t Agi—p Cl(n3k’p) / fl |A=l’""’t+‘"| S p_—_1_ _ 3 a tlz 2 p¢P(A,T) (t—r+2k) 2 P t A54." 01(71 ’9 Pl) / "_“7. |A=r+t-T — _l _1______1—2 —3 ) z 2 p¢P(A,r) (t—r+2k)2 p" . ,.__._,. 0102.210) / fl +1 LEV-”Tl S n_-1 -121 2 t1z2p ¢P(A,r) (t—r+2k)2p 2 . H 01(n,k,p) / 1:1 |A=r+t—T S "-1 t2 2 2 P+1¢P(A,r) (t — r + 2k) 2 Forp E(P1(7),2), t A2-p 02(n,k,P) —— _ < t1 ¢p(A, T) lA—Ir“t+TldT _ (t — r + 2k)3p__2 a t XI- 12 d 02 .2¢P(/\, film“ " T< (t—r+2k)3P- -2’ /t(’\)A3-(p1_)| ClT< C2(n,k,p) t1 Z( \2 Ir- t+T| —(t _r+2k)3p_2, t ):3_: 02(72, k3p) _ < . £2”) 2( T)lt\—r+t—TdT _ (t _ r + 2k)3p’2 35 Lemma 3.2 Let u 6 “M and n = 7,9, there exists a constant C3(n,k,p) such that for p E (10002), 2) r+t— —T , f0 / (u(A, mam < C3(”’p’k)l'_'f“ . (3.3) lr- t+T|/\ (t — r + 2l~::)T“"3 For p E (po(9),2) ’ C(92), k>IIuI_I The first inequality can be proved similar to the case n = 5. For the second inequality, we will prove it later. Lemma 3.3 Let u E QM, there exists a constant C4(n,k,p) such that for p 6 (120(7),), 2) C4(n,k,p)I|qu t 3 r_t+T"—§‘Fur—t+r,r (17$ "-1 7: [HI I I (( ))l (,_,+,k)T._T 0402 k ,tleullps (t — r + 2k)'n2_lp’—' 04(nakap)ilull: (t — r +2k)"—5-’-P-”§—" C4(n,k,p)||ullp . (t — r + 2k)-"-‘a'-‘v-“—;-‘ [(Ht— T)n2;3IF(u (r+t— r r))|dr_ < 3 t n—l F / |r—t+TI'2_|a——(u(r—t+r,r))|drS t1 0r /t(r + t — T)nT—l|—aa—§-(u(r + t — r,r))|dr S t2 For 10 6 (121(7). 2) 04(79k7pliluilp (t — r + 2k)32-2’ t / Ir — t + TI3|Q£(u(r — t + r, r))|dT S t1 8?" ‘ _ 3Q]: _ 04(7,k,P)||U|lp f(r+t r) |8T(u(r+t r,r))|dr S (t—r+2k)3P'2' t2 Lemma 3.4 Define r+t-1An_—_-_l n_—3 2 2— — 2 T1u(r,t)= n_ 02/ [I 2 r T P..___3 (A H (t T) )|F(u(A,T))|dAdT, r- -t-+-'r|A 2 2Ar there exists a constant C5(n,k,p) such that for p E (p0(n),2) (1 + (t — r + 2k)"2 ‘5)||u||P ¢(r,t) ’ Irzl2-‘IT1u(r,t)| S C5(n, p, k) 36 (1 + (t — r + 2k)"T‘5)||u||2 about) |rzni+5T1u(r, t)| S C5(n, p, k) Forp E (p1(7),2) H H u p ¢(r.t) ’ IIUII” ¢(T,t) i Lemma 3.5 Let h E C2(R) be an even function with h(r) = 0 for |r| Z k. Define r+t u ,,__3_ A2+r2—t2 v(r,t) _ [MA r P.,_.( W )h(A)dA, IrT1u(r, t)| S C5(7, k,p) Irz(r)T1u(r, t)| S C5(7, k, p) there exists a constant C6(n,k) such that ._t_1 ._._+_1_ __—_I __11 ll'wr "2 vlloo+||wr "2 vtlloo+l|7~vr "2 vrlloo+||wr "2 vmlloo S Ce(n,k)(llhlloo + llh'lloo + Ilh”llo.) when r < g, and -_ti __+_1 __—_l __+_1 ”W "2 vlloo+ ”W n2 vtlloo+llwr n2 vrlloo+llwr n2 vatlloo S 06(n.k)(||hlloo + llh'lloo + llh”||oo) when r 2 g, where w(r,t) = (t + r + 219)??? Lemma 3.6 Let u0 be the solution of the equation n—1 utt— u,—u,.,.=0, r20,t20, r “(7.20) = f(r), “to" 0) : g(r), (3'5) where fand g are even functions, f,g E C3(IR), and supp{f,g} C {r : |r| S k}, then there exists a constant C7(n, k) such that lluoll S 07(n,k)(2(||f‘|loo + llg‘lloo)- i=0 37 Proof of Inequality (3.4). Ift S 12r, then r+t- -1’ f [I_ IA 2(t—T )IF(u (A, T)IdAdT_ < 12r/0 fly” |A2|F(u(A,r)|dAdr r-t+1' r— t+r by (3.3), we are done. Ift > 12r f/Irth T 2(t—T )IF(u (A,r)|dAdr r— t+r| t— 2r r+t- -T r+t— —1' :1) [I 2(t—T )IF(u (A, T) )IdAdr+/t [I 2(t—T )IF(u (A,T)IdAdT 2r r— t+‘rIA r— —t+T|/\ r+t— 1' t- 2r < 2 _ 2r l. 2.. [If t+:|A( IF(u (A ,r)IdAdr+ / j 2(t T)IF(u (A,r)|dAdr Ir— t+r|/\ t- 27' r+t— 1' < 2r/0 /..+._ A2IF(u (A T) )|dAdr+/ / 2(t—T )IF(u (A,T)IdAdT. Ir- “’7' Ir- -t+‘r|/\ Now we only need to estimate the second integral. Note Hall” Apz”(’\)¢p()\fl)’ andt—T—r>t—(t—2r)—r=r>0, hence Ir—t+TI=t—r—rand t— 2r r+t— TA [0 /|i t(—)r )IF(u (A,r)IdAdr r— t-I-‘rIA A2(t — r)IF(u(A,r))I S AA2(t — r) . IIuIIW—r) <./o (HS/1M —TlAp.~.p(A)¢p(A,T)dMT ’7') < Allull” I]: 1:”? T r APzP(( 345p —(T)\,T)d/\dT t—2r t—‘r-I-r —T) dAd 36 L.-. APzP(A (bp(A,T) 71' (') We call the first integral and second integral [1 and [2 respectively in (3.6). [Flu]: ._ mpg? ) ,)‘WT t2 r+k A2—p(t _ T) dAd ft 7 .- APzP A)(r + A + 2k)3P(T _ A + 2k)p(4p-5) T ’ 38 ButA>t—r—r>t—t2—r>t—r—%(t+r-k)=5132Lfli>§,bythedefinition of 2, we have z(A) > %. Note t—r t—(t—2r) ——< t—r—r t—(t—2r)—r =2, i.e., t — 7' < 2(t — r — r), then ‘2 A2 41”t(-— 7') Il_:( )p./t1 ./t:+k T— r ((T—A+2k)p(4p_ SldAdT 2 p :2 2 1+1: ,\2-4p dAd < _ _ _ “(k) hi (t 7' r)._ts(¢-—A-t2kywre) T g M(t2 — t1)(t — t2 — 703—42, where M = 2 (%)p (p(4p — 5) — 1)'1k'p(4p'5)+1. Since t2 — t1 : max{0, g(t + r — k)—} max{0, —(t — r — k)} §=I( +._ )+(Hr—kn—§I—t— % -a—— 2 2r+ 2k__4( r+ )+4 2r _iU—r+2M 11 g 44p‘3Mr(t — r + k)3‘4” Let us estimate t—2r t-r-I-r )‘2()t_1-) [2:]; ji— r—- r /\pr(/\ (15”,(A T)dAdT . 39 Since 2(A) is increasing function /\ > t — 7" — r, z(/\) > 2(t — 7' — r) > 0. Then we have "2' (t — 7' — r)2"‘D “7+” A I < j / 2 _ t2 Zp(t — T — r) t—T—r ¢P(A, T)dAdT “2" (t — 7' — r)2"’ “7+" /\ / dAdT t2 2% - “r — r)(2r — t — r + 2k)P<4P-5) “1-, (T + ,\ + 2103:» LetL=%(t+r)—-’25andt2>0. Thent—2r>3(t+r)>L>%(t+r—k). L (t — T — r)2’P t—r+r ,\ I < / dAd 2 — t2 zP(t — 7' - r)(27' — t — 1‘ + 2k)p(4p’5) t—T—r (7' + A + 2k)3p T F2" (t — T — T)2—p t-T+r ,\ dAd +./L zP(t — T — r)(2*r — t — r + 2k)P(4P—5) t_T_,. (7- + A + 2k)3p T = J1 + .12. L (t _ 7- _ 7-)2—13 t—1'+r A : dAd . J1 ft; zP(t — T — r)(27' — t — r + 2k)P(4P—5) (4-, (T + A + 2k)3p T Sincet—r—TZt—r—L=%+§Z§,wehavez(t—T—r)2§and (t — T — r)2’p L J P(p(4p — 5) — 1)-1k-v<4v-5>+1r(t _ L _ )3—4p, but t-L-T = t-§i-(t+7‘)+%-r = (t—7r+2k) 2 i(t—r+2k)+3—‘fl > 11—6(t—r+2k), l 4 16 16 we have J1 S 2( )P(p(4p _ 5) —1)-lk_p(4p—5)+1164p_37‘(t—- 7‘ + 2k)3—4p . 40 .12 = /t—2r (t — T — 7‘)2-p t—T+r A L 2p“ — T — r)(2'r — t — r + 2k)p(4p-5) t_,_,. (7- + A + 2k)3p dAdT dAd'r ‘ 2’" (t — 7' — r)2“D t—T+r A _ __ 2 p(-—4p 5) f / _(2L t r + k)_p (Zp(t _ 7- _ r) t_T_,, (T + ,\ + 2k)3p 1 S (§)-p(4p—5)(t_ —-r+2k)_ p(—4p 5)] t— 27‘ (t __ 7— _ 7-)2-1’ t—T+r A (2"(t -— T - 7“) t—‘r-r (T + A + 2k)3p dAdT Ifr< %,thent—2r>t—r—§and t-2r (t _ 7- _ r)2_Pd t—T-l-r A A zm—T—r) / (HM/asp r—-—— (t—T—T‘)2_ P t—T+r A =/L d'r/ 21’( t — T — r) t—‘T—r (7‘ + A + 2k)3P 1'27 — — 2") t—'r+r (t 7' 7‘) dr/ ,\ A t—r—g zP(t — 1' — r) t-T—r (7- + A + 2k)3p + :11 +12. Inll,t—T-r>§,thenz(t—T—r)>§and l < 2 p t—"-% t 2_Pd t-'r+r A 1_(;)/L (—T—r) r/M—r (7+,\+2k)3p 14-? t-T+r < (3)” f (t — T — 792-22047 / (r + A + 2k)"2pd,\ t —T—r k t—r— 5 < (-)p f (t — T — r)2'2”(t — r + 2k)1"2p2rd'r _r_g — <(%)p (t_7'+2k)1 2p] 2(t—7'—7')2_2pd'r —OO )3—2p(r _ t + 2k)l—2p. 41 In12,t—T—rg§,z(t—T—r)=t—T—r.Hence [ t—2r t 2—2Pd t—T+r A dA 2 _ [14-4 _ T _ T) 7/14-,- (T + A + 2k)3P % fi+2r A : 2—2p frfl dfl/fi (t—r—fl+A+2k)3pd’\ S 2(t — r + 2k)"3pr fr; 32—2p(fi + r)dfi 4 k )4-2Pr(t — r + 2k)“3p. S (— 4—2p 2 By the estimates of II and [2, note that -—-p(4p — 5) _ 2p + 1 < —4p + 3 and -p(4p _ 5) — 3P < —4p + 3, then there exists a constant N1 such that for r < £2:- J2 S N1T(t — T‘ + 2k)_4p+3. k k k k Ifr Z §,thent—2r < t—r—E andt—T—r > t—(t—2r)——r= r 2 5, 2(t—T—T‘) Z 5 and t—27' ._ _ 2'13 t—‘r+r / (t T r) A dAdT L zP(t — 'r — r)(27' — t — r + 2k)P(4P-5) t_,_r (7- + ,\ + 2k)3p k t-2r t—’r+r /\ < — P 2L — t — 2k -P<4P-5) / t— — H / dAd _(2)( 7‘+ ) L ( T 7‘) t_T_r (T+/\+2k)3P T k - (4 -5) r 2- g (—)p(2L—t—r+2k) P P 3 11-45 2 %(t—7r+2k) B+2r 1 dA jfi (t—r-fi+z\+2k)3p %(t—7r+2k) g (§)”(2L — t — r + 2k)‘P(4p‘5)i(t — 7r + 2k) / flz‘Pdfl r fi+2r A dA [a (t—r—fl+A+2k)3p 42 k i—(t—7r+2k) S ('2‘)p(2L — t — r + 2k)””(4”‘5)§:(t — 77' + 2k)/r fl‘l‘P5fi3+P5‘P [3+2r 1 ffi (t-r—fi+x\+2k)3P k §(t—7r+2k) _<. (§)p(2L — t — r + 2k)_p(4p-5)§(t — 77' + 2k)/r fl-l-IMS [3+2r / (t - r — B + A + 2k)’4p+3+P5dA H k 00 S (5)190 — 71‘ + 2k)(2L — t — r + 2k)‘P(4P-5)r(t _ 7. + 2k)—4p+3+p6[:5 fl—l—padfi S (Slip—p613? (t _ 77' + 2k)(2L — t — r + 2k)—P P4P'5) r(t — r + 2k)-4P+3+PP, Where 6 is small enough such that —p(4p — 5) + 1 + p6 < 0 which is possible by p€(po(9),2)- Note that 2L—t—r+2k=-— ~(t+r+2k) > %(t—r+2k) and t—7r+2k < t—r+2k, then there exists a constant N; such that for r 2 g J2 g N2r(t — r + k)-P(4P-5)+1-4P+3+PS g N2r(t — r + 21\~,)-‘*P+3 Combining the estimates to J1 and J2, there a constant N3 such that for t2 > 0 [2 2 J1 + J2 S N3r(t — r + 2I~c)""”+3 , Combining the estimates to 11 and [2, there exists a constant N; such that r+t— 1' ft ‘/|_ |)\2(t — T) )UIF( (A,T)ldAdT S N4r(t _ 7‘ + 2k)-4p+3. r— t+T for t; > 0. If t; = 0, i.e., t + r S g, then 11 = 0. Therefore we only need to estimate 12. Ifr< §,thent—2r>t-—r—§and -r—§ "7+7 2 _ 2r t—T r 2 _ 12: f j: A? T) dAdr+ t /_ + Agt T) dAdr. t qSP(/\ ,7") t 'r—Aer( t—r—— 'r—er( $1)(A,T) 43 The first integral can be estimated using similar method to ll. The second is 12. Ifr> g, thent—r—r>t—(t—2r)—r=r> %,wehavez(t-r—r) > %. Since A kand t- 2r t— -r+r A2( t _ 7-) I =/ / 2 t 1-, zP(A)¢P( (A T)d’\ t- 27' t— r+r A2-P dAd “kPW’ _—5t)/o ft—T—r zP(t—r-r)(r +A+2k)3P T [< t- 2r t r+r A2-P dAd 2 ‘ kP(4t-P 5)/’;(—1)_P/:_T_r (r + A + 21031) T t _ _ S _kp(4v-5)(t — r + 2k) 3P(t + r)2 Pr(t — 2r) < k k2‘Pk t 2k 3 _ kP(4p-5) r( — r + )_ p S k-4p2+3p+4r(t _ r + 2k)—3p S k—4p2+3p+4r(t _ 7, + 2k)—4p+3 . Combining all estimates, we conclude that there exists a constant C3(9,p,k) such that for p E (po(9),2) r+t- 03(9,p,k)uun (t— F A dAd < . f/l-t+:|)‘2( T): (and): T_(t_r+2k)4p_3 Proof of 3.1: Similar to the proof of theorem 2.1 Theorem 3.2 (almost global existence) Let S(r,t,c) = {(r,t) 6 [WI r — t + 2k S c and r < t+ k,r > 0}. For small n = 7,9 and 6 > 0, the equation {3.1) with 2 Cauchy data (3.2) has a unique solution in 9p,k,9(S(r,t, (m — 1);?) provide that p > po(n) and 23:0(I|f(i)|| + “9(0”) < #17:), where C5(n,p,k) and C7(n,k) as in Lemma 3.4 and Lemma 3.6 respectively. Proof of theorem 3.2: Let 2 1 E — — 5 . ”pH—kn {u E QP.k.9(S(rat1 (4C5(n’p,k)6p—l 1) )l “u” < } 44 Define Tu = u0 + Tlu then by Lemma 3.4, we have ”mu 3 llu°|| + 205(n,p,k)(1+|t- r + 2kl12‘imuup 2 . . 1 < ('l ('l P . _Cv(n,k)(Zi=O(llf n+ug u)+20mm),k)4cs(n,p,k)6p-16 <6 Now the theorem follows from the standard contraction mapping theorem. Chapter 4 Finite-time Blow-up and Life Span of Quasilinear Wave Equations 4.1 Blow-Up of the Solutions Consider the quasilinear wave equation n—l u, — u". = G(Du, Dzu) , (4.1) u(r,0) = 0, u¢(r,0) = g(r). (4.2) We will prove that for n 2 2, the solution of the equation blow up in finite-time for several classes of nonlinearities., we also obtain the life—span of the solutions. Theorem 4.1 Assume (0)9 E 08°(R), g(P) Z 0 and suppw} C {r = M < k}, (b) there exist positive constants kg and k1 with 0 < k0 < k1 < k such that (i) g(x) 16 0 on {cc : k0 < |r| < k1} (ii) k0 is sufiiciently close to k such that 5,3 > 20, where 20 satisfies that for all 2 € [20,1), Pn(z) 2 % and Tu(z) Z %, Pu and Tn are Legendre and Tschebyschefl polynomials of degree n , respectively. Let u E 02(R2 x [0,T)) be a solution of radially wave equation (4.1) with Cauchy data (4.2) and G(Du,Dzu) = lutl”, lurlp, luttlp or |u,.,.|p. T > O is the life span of u and 1 < p S po(n) = %. Then T < +00. That is, it blows up in finite time. 45 46 Lemma 4.1 Let n 2 2 and u E C2(R >< [0,T)) be a solution of the Equation (4.1). Then there exists a constant C > 0 such that fort + k0 < r < t+ k1, t—6 u(r,t) Z C’tl—ZJl [5k 7%lu(r - t + T,T)|pd7‘ (4.3) 2 fort > -(k— k0)— — 36, where 6— — Proof: By Lemma 1.2, we have r+t— T An_1 u,(r t) > u0 (r, t) )+ Crlg— fit/O / ATG(Du(A,T),D2u(A,r))dr r— t+T l—n r+t n— u0(r,t) Z CrT/ ATlg(A)dA > 0 r—t Therefore (,rt )> Cr—l / [W T G,(Du(A T), DPd(A r))dAdr (4.4) r- t+r Note that t > g—(k k0) implies r > k, we can change the order of the integration (4.4), and obtain that A— —r+tG u(r,t) )> Cr— 2 f A_ 2 1k] G(Du,Dzu)deA. (4.5) A— If G(Du,Dzu) = )utlp, then ,_n r "_1 A—r+t u(r.t) > CrT/ AT/ lutlpdrdA. k A—k By Holder’s inequality, A—r+t A4 u,(A, T)d7' P A-r+t 1‘p (/ dr) dA A—k l—n r n—l A-r+t P 2Cr"2—(t—r+k)1""/kA‘T/A k u,(A,T)dT dA u(r,t) > CWT" jkA— > Cr“T"(k — Igor-Pf A2§l|u(/\, A — r + t) — u(A, A — k)|PdA. I: By the support property of u, u(A, A — k) = 0, therefore u(r,t) > 079—3209 — toy-P l1: ABE—Hug, A — r + t)|PdA. (4.6) 47 If G(Du, Dzu) = Iuttlp, l—n r n—l A-r+t u(r,t) > on— / A—z- / luttlpdrdA. k A—k By Holder’s inequality P A—r+t / k (A — r + t — 7')utt(A, T)d7' < A-r+t p A—r+t q i1: _ {/_k lutt| dT}{/A—k (A—r+t—T)} 1 1__ where;+;—l. Then l—n r 11—! A—r'i't P u(r,t) > Cr—T/k AT A k (A — r + t — T)utt(/\,T)d7' A—r+t 5 {f 1. (A—r+t—*r)"d1'}dA _>_C l:2"[(t—r+(k)"+l]lmp :/r)\1;_ — r + t - T)ut(A, T))|’\;\ ”it k+ u(A, 7') A _'+t]p dA 20 —2-"-k—( ko)(:+”“ 10/ A”?— |u(A, A—r+t))PdA (4.7) Combining (4.6) and (4.7) leads to such that u,(r t) > C(t+k)‘T "/ A"— 2 (u(A, A—r+t)|pdA. (4.8) > Ctl—Z‘Q/ A"2 |u(A,A — r + t)|PdA (4.9) k for t > %—-(k k0) Let A— — 7' + r — t then (4. 9) becomes _n t n— u(r,t) 2 CtLT/ (7'+r—t)—2_1|u('r+r—t,7')|pd7' t~r+k 2 Ctl—Z‘f TTl|u(T+r—t,'r)|pd'r t—r-Hc -n t n- 2 Gil—5'] 7—2—1|u(7+r—t,'r)|pd'r 26 48 wherewe usedt—r+k< k—koz26. Note that u,(rt)>Ct1—2_‘/t:+t_1/ G,(DuD2u)dAdr r-t+1' _1—_ r—+t -T 2 Ct—f/ (r —— t + T)— 2 [7 C(Du, Dzu)dAdT 0 r—t+‘r l—n t-6 "_l r+t-‘r 2 CtT / TT G(Du, Dzu)dAdT o r—H-‘r Sincer+t—T>r+t—(t—6)=r+6=r+k‘—2k‘l>t—szfl‘>t—6+k>'r+k,we have 1_n t—6 "_1 1+]: u(r,t) 2 CtT / 7T G(Du,D2u)dAdr (4.10) 0 r-t+T If G(Du, Dzu) 2 lurl”, By Holder’s inequality T+k p T+k l—p / urdA {/ dA} . r-H-T r-t+r _n 1-5 n- ZCtl—2_(t—r+k)l-p/ TTl|u(r—t+7‘,'r)|pd'r o l l—n t-a n- u(r,t) > CtT/ TT 0 -6 n—l > CtIT (—k kg)1 ‘0] TT|u(r—t+7’,7‘)|pd'r. (4.11) 26 If G Du,Dzu 2 u" p, by Holder’s inequality and a similar ar ument to the case g luttlp, we have p 71—! 1—n t—6 T+k u(r,t) > CtT/ TT / (T + k — A)u,.,.(A, T)dA 0 r —t+‘r T+k l—p {/ (1+k—A)qd/\} d7 r—t+r _n t-5 n- 2 CtlT(t — r + k)(1_p)(q+1)/ 7—2—1|u(r — t + T, 7')|pd7' 0 t_6 n—l > CtLE— (13— k0 )(l"p)(q+l) TT|u(r — t + T,T)|pd7' 26 Hence there exists a constant C such that for t + k0 < r < t + In and t > 36, the following holds l—n t_6 n—l u(r,t) > CtT f6 TT|u(T + r — t,'r)|pdr. 2 Lemma 4.2 For anyn Z2, ift+ko 36, then Ctlinlnt or — 4L1 , (r t) > 1-" f p— n l Ctl‘T for 1 Ct‘T‘f 26 Since 1' + k0 < r — t + T < T + k1, lemma 1.2 implies u(r—t+T,T) 2 u0(r—t+T,T) > Cal-Tn therefore 1—n 1‘5 l—n p u(r,t) _>_ CtT/ r"-12(CtT) dr 26 1—_n “5 u m 2 Ct 2 / T 2 " 2 ppdT (4.12) 26 Note that "—51 — "zip = —1 when p = 2%, and —1 < "—;1- — "—gl-p < 0 when 1 < p < fig. (4.12) implies Ctl'Z—nlnt for = 731, u(r,t) 2 p "-1 Ctl_T for 1< p < 2%. Lemma 4.3 [13] Let p = iii—i for any n 2 2, there exists a constant B such that for Tm =Bme1,t+koTm u(r,t) 2 (IKE—"(17102“ . Lemma 4.4 Let 1 < p < it}: for any n 2 2. Then there exists a constant C and Sm such that, fort+ko < r Sm u(r,t) 2 Ct‘E—"thz‘pw, _ nil _ n-l wherea— 2 —2 1).. Proof: If m = 1, it is true because for t + k0 < 7‘ < t + k1 u(r,t) Z u0(r,t) 2 (RI—2"- . Now assume that it is true for m > 1. WE will prove it is true for m + 1 by induction. Let t + k0 < r < t + 1:1 and t > Sm, then by Lemma 4.1 _n t-5 n- u(r,t) 2 Ct‘T/ TT‘|u(r — t + T,T)|PdT 26 50 where t > Sm + 6. By the induction assumption, we have _,, t-6 n_ _n m_ .- p u(r,t) 2 0131—7] T'Tl [CTITTMZmOIM dr m —n t-S n— n- m—l j 2 Ctlz / 7- 21— 21p+°(21=o p)pd1' 1— i1 n-l —1 j > Ct 2n7-n2 — 2 p+a(z.‘=o pug—5 _ m : Ctlzi'T-MZZO p‘)|g—6 _ Therefore there exists a Sm+1 > Sm + 6 such that for t > Sm“ (t _ (5)4220?) _ 5.7.2210” > $442201») . We have u(r,t) 2 Ct‘T'ltC'EIZE‘P') . Lemma 4.5 Let n 2 2 and 1 < p _<_ 13%. Then for any I > 0, there exist a constant C) and an integer m such that ift + k0 < r < t+ [<71 and t > max{Tm, Sm}, u(r,t) > Cltl . Proof: In [13], the case P = % was proved for odd integer n. By using Lemmas 4.1, 4.2, 4.3 and 4.4, we can follow the simliar method in [13] to prove that the inequality holds for any integer. So we omit the proof. For 1 < p < %, note that for any I > 0 there exists an m such that Hence u(r, t) > Ctl for t 2 Sm Proof of theorem 4.1. Ramaha [13] proved that for n odd and G(Du,Dzu) Z |u,|p or lutl”, the solutions must blow up when p = $3}. By the results of lemma 4.1, 4.2, 4.3, 4.4, and 4.5, we can unify both odd and even dimen- sions to show the blow up. The details are omitted. If G(Du, Dzu) Z luttlp or lurrl”, we define t+k F(t) 2/ u(r,t)da: 2/ wnrn‘1u(r,t)dr " o 51 then 1+1: F(t)=/1R utt(r,t)da: 2 can] rn—lutt(r,t)dr " 0 Case 1: C(Du, Dzu) 2 Iuttlp F(t) — F(O) 2 [Qt F(s)ds = fat fray: ut,(r, T)d:rd‘r t 1+]: 1 2 can] / rn" utt(T‘,T)d7‘dT o o 1 1 t 1+1: '5 r+t 3 wn/ (/ r"_1|utt(r,t)|pdr) (/ r"_1dr) dr 0 o o l t u 1 n l- n g a: / (F(t))3(t + 1:)sz g a: F%(t)(t + k)?“ 0 U\ therefore we have 144(1) 2 w;13(t+ k)-‘%+”(F(t) — F(t)», i.e. 2 q F(t) 2 w; (t + k)'(%+”2(12(t) — F(0))P. (4.13) On the other hand F(t) — F(O) = jot/Hi?" Iuttlpdatdr t t+k1 1 2 1.12”] r”_ Iutt|pdrdr 0 t+ko t+k1 t 2 wn/ (/ IuttIPd‘r) r"_1dr t+ko t—k t+k1 t t l-p 2 can] rn’II/ k(t — T)utt(7', T)dr|p (/ (t — T)da) dr t t- t +ko -k t+k1 2 can] r"_1|u(r,t)|p(kq+l)l—pdr t+ko t+k1 Z wn(t + k0)”'1k(q+l)(l’p) |u(r, t)|pdr . (4.14) t+ko Combining this with lemma 4.5, for l = l (n + 95:71), there exists a constant C) and p Tm“) such that for t+ k0 < r < t + k1 and t > Tm“) u(r,t) 2 Cit, 52 Combining (4.14) this with leads to F(t) .— F(O) 2 (,,.,n].c(¢1+1)(1—1v)(k1 _ ko)C'1(t + k0)(n—l)+n+§ETl- 2 Ct2n+p35 (4.15) From (4.13) and (4.15), we have F(t) 2 cu + k>"-<"+”P(F(t) — F(o>)"?-’(F(t) — F(onT‘ . Therefore there exists a T' such that for t > T’ F(t) 2 cr-(nrrtmrtfiTXFU) — Nona? 2 C(Fm — F(onT 2 0,4,1“) Case 2: C(Du, Dzu) Z Iurrlp. By the support property of u and the Divergence Theorem, we have F(t) = / utt(r,t)dx=/ (Au+|u,,.|p)d:r R" |z| Tm’(l) ko+k1 2 u(t + ,t) 2 opt" (4.18) Combining (4.18 and (4.17) leads to F(t) 2 Ct2("+29)‘2 . Then there exists a T" > Tm’(l’) such that t > T" F(t) 2 Ct2("+2‘7). (4.19) From (4.16), there exists To such that for t > To the following holds .. 1 —l n 1 F(t) 2 Ec.;,1,-2t<"+2q><1-P>F"T(MT(7:) (4.20) > OFF—“5" therefore F (t) blows up. D 4.2 Life-span of the solutions Theorem 4.2 Let T be the life span of classic solution of the equation {4.1) with C(Du,D2u) Z |ut|P or luttlp and g(r) replaced by eg(r) for 0 < e < 1. Then there exists a constant C such that T S exp(Cel‘p) ifpzfi—iforn22. Lemma 4.6 Let!) = {(r,t) E R” ( t+ko < r < t+lc1, t 2 26} and u, v 6 C(9) satisfy the following l—n u(r,t) > arl—Tn fkr An—2_l|u(A, A — r + t)|pdA + br 2 (4.21) u(r,t) S arl—2n’ j; Al2i|v(A, A — r + t)|pdA + b7“? (4.22) where a and b are positive constants. Then u(r,t) > u(r,t) for (r,t) E Q (4.23) 54 Proof: we will prove this by contradiction. If this is not true, then 521 = {(r,t) E Q | u(r,t) S u(r,t)} is not empty and closed. Let 02 = {(r0,t) E 91 I such that (ro,t) nearest to r = k }. Since u(k,t) > brl-Tn Z v(k,t) (k,t) ¢ ()1, hence (k,t) ¢ {22. So r0 > 10. Let r E (k,r0), then (r,t) Q (21. Therefore u(r,t) > u(r, t) for k < r < r0, then we have the following u(ro,t) > MT [km A'T‘|u(A, A — r + t)|PdA + br‘T" (4.24) v(r0,t) 3 MT f: AT|U(A, A — r + t)|PdA + br‘T" (4.25) which contradicts to (r0,t) E (11. Therefore (21 must be empty, for any (r, t) E Q u(r,t) > u(r,t) for any (r,t) E Q (4.26) Lemma 4.7 Let S E C2([O,T + k)) and satisfy Saba/0 (3:14 x+k) |S( )lpds+b' Then h(z) = S(x) satisfies the following equations h"(2) + h’(2) = alh(2)l", h(O) = b, 77(0) = 0 , (4.27) where z = 112(1)?) . Proof: Differentiating the equation h(z) = S(x), we have the following S’(x) = h'(z)x i 10 77(2) = (a: + @544) therefore I! h (z) = (x + k)S’(x) + (x + k)2S"(x) . On the other hand, S'(x) = (4+1: [0 |S(s)|pds S =( f——§:——)3/ |S(s)|”ds+(—— +4) 15041 ., —2 , p (:0) = WSW )+ |5($)|, S H (x :14)2 55 Hence —2 , a p (x + k)5 (m) + (x + k)? 'SW) h"(z) + h’(z) = 2(4 + k)S’(x) + (.4 + 14)2 ( = a|S(:v)l" = alh(z)|‘” with h(O) = 5(0) = b and h’(0) = kS’(0) = 0. Lemma 4.8 [1.9] Let 7'1 be the life span of the solution h"(2) + h’(2) = a|h(2)l’”, h(O) = b, 72(0) = 0. (4.28) Then 1)h"(z) > OforO S 2 < 7'1. 2) r1 < +00. 3) h(z) —> +00 as 2 —-> 7'1. Lemma 4.9 [19] Let 2(7)) be the life span of the solution to the following equation hfia+huo=auam h(0) = b7], h’(0) = 0. (4.29) Then 2(17) S 71771”? for 0 < 17 < 1, where 7'1 is defined in Lemma 4.8. Proof of theorem 4.2. Let u(r,t) = S'(r—k)rl—3£ for (r,t) E it. Since A—r+t+ko < A < A—r+t+k1 and "2:— "zip: —1,wehave a]: AT|v(A,A — r + t)|pdA + b = a]; AT-TP(S(A — k)IPdA + b. r-k —l =a/ (4+m mawaww 0 . 7* (4 _ _1_) W. .1. o y+k (r—k)+k =30—k» IV therefore 1—n l-n "k 71—] l—r u(r, t) = S(r — k)r‘T S arT/ AT|v(A, A — r + t)|pdA + brT (4.30) 0 Note that there exists a. constant Co(g) such that l—n u°(r, t) Z Co(g).€rT 56 From (4.6) and (4.7) in the proof of Lemma 4.1, there exists a constant C1 such that u(r, t) Z Clrl—Trn /kr A121|u(A, A — r + t)|pdA + Co(g)erl_2_n Let a = C1 and b = Co(g)e in (4.29). Then 8(4) = 0. f: (“4, —- x i 7.) 15(4)!de + Game, (431) u(r,t) = CWT"- f7. A"T‘|v(A, A — r + t)|PdA + co(g)er‘T”- and by Lemma 4.6 we have u(r,t) > u(r,t) for (r,t) E 0. Let h(z) = h(ln(£fl)) = S(x), then by Lemma 4.7 h"(2) + h’(2) = Cllh(2)|” h(O) .—. Co(g)e, 12(0) = 0. From Lemma 4.8, the life span 2(5) of h satisfies 2(5) S 7151—” where T1 is the life span of the solution of equation (4.27) with a = C1 and b = C0(g). Therefore ln($ Z k) S 7151‘”, and x S kexp(7'151"’) — k. Letx=r—k>t+ko—k,for5smallwehave 1 t < Ekexp (7151'?) . [:1 Theorem 4.3 Let T be the life span of classic solution of the equation (4.1) with C(Du, D2u) Z lurl” or lurrl” and g(r) is replace by eg(r) for O < e < 1. Then there exists a constant C such that T S exp(C€p(l—p)) ifp=§§forn24 57 Proof Let t)=/O/:o t(-—)r )r 2 u(r, 7')drd7' 4+1: fl ”flap/Mr wand u(r, t) = 200,2) + 62242.2) For t + kg S r S t + k1, u°(r,t) 2 Carri—n and r+t- TA Qu(r,) t) > Cr? ”/0 / C(,Du Dzu )dAdr r— t+r Integrating u(r, t) over t + kg S r S t + k yields t-Hc __3 uo / r T (2, t)dr 2 Ce(t+ k)-1 t+ko exchanging the order of integration, we have for t 2 k2 = %(k — k0), t+k r+t-— -1). I: r T" Qu(r, t :dm/Hk '1]: / G(Du D227 )dAdrdr H-ko t+ko lr- t+TI/\ t- k2 _. —/ [MA 7. G(Du ,Dzu)[/::T r’ldr]dAd7' 7+ 0 2t— T-I-ko An—-——l +]_ f T G( Du, D227) )[ftom r‘ldr]dAdr t ’21 T+ko T+k A+t—1' +/: G(Du Dzu)[/r‘,1dr]dAd7' t k, 21— T+koA A—H—T where k2 = (k — 180) First estimate the inner integrals l 2 A+t-1' [+0 rldr2(A+t—r)‘1(A—r—ko)2 (t+k)‘ (A—T—ko) 2(t +1?) 2(t—T)()\—T"ko)2 where we used A — 7' S k and (t + k)‘1(t— 7') S 1. A-H-r A r'ldr Z (A + t — 7')"12(t — 7') Z 2(t + k)'l(t — 7') —t+7' 2 k;1(t + k)-1(t — 2)(A — 2 — 120) 2 7212;1(2 + k)-2(t _ 2)(A — 2 — 720), where we used A — 7' -— kg S 2192 and h(t + k)‘1 S 1. Combining the above estimate, we have I>C(t+k) NA]: (At—7')( —r—ko)A"—2‘ G(Du D22. )dAdr. 58 If C(Du, Dzu) = lurlp, then by Holder’s inequality t-Hc "-1 j j t — 7') — 2 — ko)AT|u,.|PdA42 2 I] j” (t — 7)(A — 7' — ko)A"T3u.dAdTIP 0 T+ko t tau+k "_l [j j (t — T)(A — T — k0)AT“’dAd7-]1‘P 0 r+ko Sincep= 2i} andq= 92:1, EZl—qz —1. A]: t—7')( —A2-—kO)A"T ' 2’9dAd7' < C(t+k)/0(r+ko)"1d7' t_+k). 10 Note u(r,t) = 0 for r 2 t + k, we have < C(t + k)log( t+l€ _ |// ()(t—T A—r—ko)A—2‘u.dAd7-| T+ko( = — t— AT A dAd dAd | AL? 2) u(,.) r) 2 1+]: 71—5 [w(t — 7')A 2 (A — 7' — ko)udAd7'| + t 7+k n_ 2 If f (t —7)A'2_3u(A,7)dA,d7)dAdr| = F(t). 0 r+ko Combining above estimates concludes that for t 2 k2 F”(t) 2 CFP(t)((t + k)log( :k))1"’(t+ 7.)—2 Since H t+k _—_n3 F (t) 2] r T u°(r.t)dr 2 Ge(t+k)-1 t+l€o Integrating twice leads to t k F(t) 2 02(2 + k)log( j; ); t 2 kg From (4.32) and (4.33), we have ., k F (t) 2 052(7 + l<:)—1log(t+ ), t > k2 [8 Integrating two more times concludes the following F(t) 2 022(7 + k)[lo g(t—+£)]2; t 2 723 = 2k. (4.32) (4.33) (4.34) 59 From (4.32) and (4.35), we have the following t+k F"(t) 2 Cep(1"’)(t + k)-2zogP-1( )F(t); t > k3. Since F’(t) > 0, multiply (4.36) by F’(t) and integrate from 184 2 k3 (F(t))2 2 (12024))? + t+l€ k4+k 0220-2)“:+k)-2zogv-1(——k )F2—(k4+k)-2zogP-l( k )1. Since for t2 > t1 2 0 , t2) — F(tll < F’(t2) ’ F( F't < “L t2—t1 — F(O) = 0, So F(k4) S k4F’(k4) Choose 18.; such that k3+k kjcePU-Wa + k)-21092-1( k )21, that is, 104 = O(exp(Cep(1"’))) as e —> 0. By (4.36), F ’(t) 2 F ’(k4)2 + (1234.26))"[92,p(t)F2(t) — 92,2(k4)F 2(124)]- 2 k221i?» F2 (t) Since $0???) 2 C(t + k)_zlogp’l(t—iI;—IE), by (4.37), we have F’(t) > C(t + k)-lzogT(3:—k)r(t) Integrating once leads to log F(t) > czogT(ki4+Tl%), t 2 72., F(k4) — Ift > 185 = Clef, then (4.36) (4.37) (4.38) (4.39) (4.40) (4.41) 60 By (4.35) and (4.41), we have F(t) 2 052(7 + k)22—‘3t, t 2 725 . (4.42) From (4.32) we have F"(t) 2 C(t + 12)-1-uog1-2(t : k )F 2“ (t)FT(7) Combining this with (4.42) leads to F”(t) 2 GeflTurT‘a), t 2 75 . Multiplying F ’(t) to both sides and integrating once 12(4)2 2 GET—”2" (FTa) — FT” (125)). Note that F(t )> F’(k5)(t — 185) > F(k5 )(t—kfi), ift > (96— — 5k5, we have Therefore F’(t) 2 Cam—l2" FPTSU), t 2 k6. One more integration gives 1 20:11 ——_1—-— 2 CE 4 t . 4.43 FTUCS) ( ) Hence the life span is finite. Now we need to prove that the life span T satisfies T S exp(Cep(1‘p)). If T S k6 = C102, by k4 = 0(exp(Cep(1‘P)), we are done. If T 2 k6, by (4.43) and (4.42), we have 1 Gag—15 (_ 211k?” 205m _Ike, which leads to C53 4 MICE” S 1. This 18 a contradiction as e —-> 0. Therefore the life span T must satisfies T g 726 g exp(C5p(l_p)) when 5 is small. El 61 Theorem 4.4 Let u E C2(R >< [0,T)) be a solution of u“ — Au 2 F’(u)ut,orF’(ut)utt (r,t) E R x [0,T) u(r,0) = ef(r), u,(r,0) = eg(r) , where F satisfies F(x) 2 lxl”, 1 < p S 24% and f,g E C8°(R) satisfy the same (4.44) conditions as g of Theorem 4.1, in addition, g- F(g) Z 0. Then T < +00, especially, T < exp(Cel’p) for p = git-i. Proof: We only prove the case F’ (u)ut Define u(r,t) = [0149,5212 , then 7) satisfies $1132 — 1202)] = o. v(r,0) = 0, v,(r,0) = f(r). Therefore Clv — F(vt) = w(r,t) , v(r,0) = 0, v,(r,0) = f(r) , where w(r, t) = F(vt) + g(r) — F(g(r)). Note that v(r, t) = v0(r, t) + Q(2~, t) By lemma 1.2, fort-’rko C(f)2rT , 1_n r+t— TA (422,2) > ch 2 [0 f (A ,T)dAdT r- t+'r r+t— TA > CrT / [4+7 T(g( A)— F9 (2994242 —In r+t TA +Cr2 [cu/HA 2 F()vt d)Ad7' —1n r+t— 7')‘ > CT_ 2 ft/_n2——1F()Ut)dAdT r— t+r l_n r+t—7' >GrT [if T|v(A,A—-—r+ t)|PdA. t+k 62 Hence r+t-‘r’\ 1_n u(,)rt >orT" //_ T|v(,AA—r+t)|PdA+o(f)erT t+k Now following the method of Theorem 4.1, we can prove T < +00 for 1 < p S 2—3. Following the method of Theorem 4.2, we can obtain T < 6$p(061_p) for p = 13%. Bibliography [‘2] [3] l4] [5] [6] [8] Baez, J. C., Segal, I. and Zhengfang Zhou, The global Goursat problem and scattering for nonlinear wave equations, J. of Functional Analysis 93 (2) (1990), 239-269. Glassey, R.T., Blow-up theorem for nonlinear wave equations , Math. Z. 132(1973), 183-203. Glassey, R.T., F inite—time blow—up for solutions of nonlinear wave equations, Math. Z. 177(1981), 323-340. Glassey, R.T., Existence in the large for Du = F (u) in two space dimensions, Math. z. 178(1981), 233-261. Grillakis, M. G., Regularity for the wave equations with a critical nonlinearity, Comm. pure Appl. Math. Vol. XLV, (1992), 749-774. John, F., Blow-up of noninear wave equations in three space dimensions, Man. Math. 28(1979), 235-268. John, F., Blow-up of quasilinear wave equations in three space dimensions, Comm. Pure Appl. Math. 54(1981), 29-51. John, F., Nonexistence of global solutions of Bu 2 g(F(ut)) in two and three space dimensions, Mathematical Research Center, Univ. of Wisconsin, Technical Summary Report No. 2715. 63 64 [9] John, F ., Nonlinear wave equations, Formation of singularities, Pitcher Lecture in the Mathematical Sciences, Lehigh Univ., University Lecture Series, AMS Providence, 1990. [10] Kato, F ., Blow-up of solutions of some nonlinear hyperbolic equations, Comm. Partial Differential Equations 15(1990), 757-762. [11] Klainerman, S. and Ponce, G., Small amplitude solutions to nonlinear evolution equations, Comm. Pure Appl. Math 36(1983), 133-141. [12] Lindblad, H., Blow-up for solutions of Du = [ul’D with small initial data, Comm. Partial Differential Equations 15(6)(1990) 751-821. [13] Rammaha, M.A., Finite-time blow—up for nionlinear wave equations in high di- mensions, Comm. Partial Differential Equations 12(6)(1987), 677-700. [14] Schaeffer, J ., Wave equations with positive nonlinearities, Ph.D. thesis, Indiana University, (1983). [15] Schaeffer, J ., The equation Du = [u]? for the critical value of p, Proc. Roy. Soc. Edinburgh 101A (1985), 31-44. [16] Schaeffer, J ., Finite-time blow-up for u“ — Au 2 H (u,, at) in two space dimen- sions, Comm. Partial Differential Equations 11(5)(1986), 513—543. [17] Sideris, T., Global behavior of solutions to nonlinear wave equations in three space dimensions, Comm. Partial Differential Equations 8(12)(1983), 1291-1323. [18] Sideris, T., Nonexistence of global solutions to semilinear wave equations in high dimensions, J. Differential Equations 52(1984), 378-406. [19] Zhou, Y., Life span of classic solutions to Du = |u|1+° in three space dimensions, J. Partial Differential Equations 5( 1992), 21-32. [20] Zhou, Y., Life span of classical solutions to Du = lulp in two space dimensions, Chinese Ann. Math. 143 (1993), 225-236.