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# Citizenship, Community & Cultr SOCD 169

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This 14 page Class Notes was uploaded by Ms. Pauline Hermiston on Thursday October 22, 2015. The Class Notes belongs to SOCD 169 at University of California - San Diego taught by Staff in Fall. Since its upload, it has received 19 views. For similar materials see /class/226818/socd-169-university-of-california-san-diego in Sociology at University of California - San Diego.

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Date Created: 10/22/15

PDE LECTURE NOTES MATH 237A7B 169 12 HEAT EQUATION The heat equation for a function u R1 X R 7gt C is the partial differential equation 121 8 7 A u 0 with u0 ac fac where f is a given function on R By Fourier transforming Eq 121 in the at 7 variables only one nds that 121 implies that 122 a g 5 1205 0 with 12mg g and hence that 1205 e lZZ lnverting the Fourier transform then shows that uh 2 7 F1 et 72flta as 7 f 1equot 5 12fxre A2fx From Example 7 pl 57mm x Mx pinyin and therefore um 7 we 7 gammy This suggests the following theoremR Theorem 121 Let 123 Mg 11277072267171112 be the heat kernel on R Then 1 124 8 7 5A pac 7 y 0 and 1311311741 7 y 67531 where 6 is the 6 7function at at in R More precisely iff is a continuous bounded function on R then uh 2 7 mac 7 mm R7 is a solution to Eq 121 where u0 ac limno ut Proof Direct computations show that 8 7 Ai pac 7 y 0 and an application of Theorem 7 shows limtiopx 7 y 67531 or equivalently that limno fRn pac 7 uniformly on compact subsets of R This shows that limno ut ac uniformly on compact subsets of R Proposition 122 Properties of e AZ 1 For f E L2R dac the function em2f as 7 Pine 7 is smooth in t at fort gt 0 and at E R and is in fact real analytic 2 etA2 acts as a contraction on LPR dac for all p 6 000 andt gt 0 Indeed 3 Moreover pt f 7gt f in LP as t 7gt 0 ewl MWdZI 170 BRUCE K DRIVERl Proof Item 1 is fairly easy to check and is left the reader One just notices that pac 7 y analytically continues to Ret gt 0 and at E C and then shows that it is permissible to differentiate under the integral Item 2 Km mm s n WNW 7 my and hence with the aid of Jensen s inequality we have7 Hm fH ip lfylppnr7ydydr W Rquot Rquot So P is a contraction Vt gt 0 Item 3 It suffices to show7 because of the contractive properties of 17 that p f A f as t l 0 for f 6 CAR Notice that if f has support in the ball of radius R centered at zero7 then lpnfxl lfltyPltx7ydy HfHoo Ptltx7ydy Rquot MSR HfHooCRne Ufl Ry and hence Hp f 7 fH ip 7 m f 7 my HfuoooRnEaWR y SR Therefore p f A f in D as t l 0 Vf 6 CAR l Theorem 123 Forced Heat Equation Suppose g E CbRd and f E Cg 20oo gtlt Rd then t ut7 1p 9691 p f739acd739 0 solves 8 1 8 1 Au f with u0 9 Proof Because of Theorem 1217 we may with out loss of generality assume 9 0 in which case utx 0tpt ft 773d739 Therefore Bu t 8 5t7Ptf0710PT raf TvxldT t 8 p f0ac 70 pT Ef 7 Txd7 and t Queen 7 p 9m 7 7 mm 2 0 2 PDE LECTURE NOTES MATH 237A7B 171 Hence we nd using integration by parts and approximate 6 7 function arguments h t at 8 A l 8 1 a 7 3 um 7 was 70 pi i 7 7 5A H 7mm 7 mm 8 1A m d 7p 090 6111316177 7amp7 7 x T ptf07lgfngfl7Tl a 1 1331 6 E75ApTft77xdr ptf0 17tf0leifgPeft 7xftvx I 121 Extensions of Theorem 121 Proposition 124 Suppose f R 7gt R is a measurable function and there exists constants c C lt 00 such that mm MW Then utac p 96 is smooth for t at E 00 1 X R and for all k E N and all multi7indices a 125 D kutx pa gym fa In particular u satis es the heat equation u Ail2 on 0 0 1 X R Proof The reader may check that m 32495 7 Mame Where q is a polynomial in its variables Let 10 E R and e gt 0 be small then for at E Bac0e and any gt 0 lav 7 ylz W2 7 lellyl M2 2 M2 W2 7 f2 M2 W M2 2 1 732 W 7 572 71 ma a Hence 911 1 sup 67l171l1 S sup l on C some sup own2 1071 at 7 ye717 11w Da 7mm egtgc7eandacEBacoe 10496 7 11 06 egtgc7eandacEBac0e egtgc7eandacEBacoe By choosing close to 0 the reader should check using the above expression that for any 0 lt 6 lt 1t 7 c 2 there is a CMY lt 00 such that gy S CNle lyll In particular 9 E L1 Rn Hence one is justi ed in differentiating past the integrals in pt 96 f and this proves Eq 125 I 172 BRUCE K DRIVERl Lemma 125 There exists a polynomial qnx such that for any gt 0 and 6 gt 07 1 2 1 ylzd lt 671 35 Rn M835 31 7 q 562 6 Proof Making the change of variables y 7 631 and then passing to polar coordinates shows 00 1M25e lylldy6n 1M21e 511ylzdy0Sn71 6 ei zlzrnildr Rquot Rquot 1 Letting A 62 and n 7 fro e MZrndr7 integration by parts shows 00 7M1 00 7M1 1 1 e A n71d 5 7 71 n72d M LI 7 2f 2 4 T A T 1 A n71 7 A 2A6 1 2A ll 2 lterating this equation implies 1 n71 1 n73 A 4 4 A M 2Ae M me 7 2A a 4lt and continuing in this way shows 7 7 n 7 1 H M 7 e we 1 w where 6 is the integer part of n27 i 0 if n is even and i 1 if n is odd and rn is a polynomial Since 00 oo 7 MA War 3 1 Te krldr 6 1 2A it follows that MA e Aqno l for some polynomial qn l Proposition 126 Suppose f E CR R such that S 0617511 then pmf 7gt f uniformly on compact subsets as t l 0 In particular in view of Proposition 124 ut7 at p 96 is a solution to the heat equation with u0x Proof Let M gt 0 be xed and assume S M throughout By uniform continuity of f on compact set7 given 6 gt 0 there exists 6 6t gt 0 such that S e if 117311 3 6 and S M Therefore7 choosing a gt c2 sufficiently sma 117 we 7 mm 7 lmy was 7 y 7 fldy smymwy 7fxdy d C 2 7n2 elmV ew 721Ade Shingle yl m Agile 6 16 y eCNl27Ttin2 e alylzdy M25 So by Lemma 1257 it follows that 1p 7 S e Gawain2 6nqnea51 PDE LECTURE NOTES MATH 237ArB 173 and therefore limsup sup lp 7 S 6 A 0 as e l 0 no lxlltM I Lemma 127 Ifqac is a polynomial on R then t A A PM 7 yqydy 1 Hg nq l Proof Since m x n m 7 mm mm Zaaway dy 2 came ft7 at is a polynomial in at of degree no larger than that ofq Moreover ft7 at solves the heat equation and ft7 ac A qac as t l 0 Since gt7 at 2200 has the same properties of f and A is a bounded operator When acting on polynomials of a xed degree we conclude ft7 at gt7 l Example 128 Suppose qac 1002 1 then t 2 etAZq 951902 90 5A 1112 90 2 I 4A2 1112 35 4 t 2 t2 112 x3 51213 m4 acle as 6th 3t2 Proposition 129 Suppose f E C R and there exists a constant C lt 00 such that 2 7 Z lDo facl g 060 lal2N2 then N tk p f emWW Z gAka 001V as i 10 k0 39 Proof Fix at E R and let 1 fNy1 Z EDaf lya lal 2N1 39 Then by Taylor s theorem With remainder 005 iwa S C y an sup eClxtyV S C y 2N2620lx 1 y 2 S y 2N2620 y 1 te01 and thus A Mamas my 7 A ptyfzvydy S 5 Pt ylyl2N2620lyl1dy R 2 2 1 p1ylyl2N262 allquot dy R t tN1 174 BRUCE K DRIVERl Since f as y and f My agree to order 2N 1 for y near zero7 it follows that N tk N tk N tk Rn mammary 2 Emma 2 re ygtto Z Ewe k0 k0 k0 which completes the proof I 122 Representation Theorem and Regularity In this section suppose that Q is a bounded domain such that Q is a 02 submanifold with 02 boundary and for T gt 0 let QT QT x Q and FT 0T x 012 U x Q C bdQT 0T x 812 U 0 T x Q as in Figure 36 below n Q W FIGURE 36 A cylindrical region QT and the parabolic boundary FT Theorem 1210 Representation Theorem Suppose u 6 02716271 QT QT 1LT X R solves at Au f on QT Then W as mo out was mo am new 9 0Tgtlt 2 apT t 12 lt 6gt W 96 ygtultt y mo 10 dabW 1 2 amen Proof For 1 E C 27110 T X R integration by parts shows fodydt out Aodydt QT 9T 1 tT 1 ot iVoVudydt ou tZO sly ti egalth 9T Q 0Txan 1 T 1 do do ot Aoudydt ou Ody 11 ogt do dt 9T 9 0T gtlt an PDE LECTURE NOTES MATH 237ArB 175 Given 6 gt 0 taking vty pT6ac 7 31 note that v AU 0 and v E 02 10T gtlt implies ft7yPTe7tx 7 ydydt 0 17496 7 yut7 WM 7 pTex 7 yutydy 0TgtltQ Q a 1 8PTe7i y Bu any um imam 7 an amp mad 0T1xan Let 6 L 0 above to complete the proof I Corollary 1211 Suppose f 0 so ut at Aut Then u E C 0T gtlt Proof Extend pac for t S 0 by setting pac 0 ift S 0 It is not to hard to check that this extension is C on R X R 0 Using this notation we may write Eq 126 as 14MB pi 7 yu0ydy a 1 8 Bu 7yuty7pTi71Ey dawn 0oogtlt89 The result follows since now it permissible to differentiate under the integral to show u E C 0T gtlt l Remark 1212 Since at 7gt p x is analytic one may show that as A utx is analytic for all at E 9 123 Weak Max Principles Notation 1213 Let aihbj E C QT satisfy aij aJi and for u 6 029 let n n 127 Lut x Z aiJt myme 2w mam ij1 i1 We say L is elliptic if there exists 9 gt 0 such that Zaijhwmgj 2 am for all 5 e R and m 6 2T Assumption 3 In this section we assume L is elliptic As an example L A is elliptic Lemma 1214 Let L be an elliptic operator as above and suppose u E 02 and 10 E Q is a point where has a local maximum Then Lutxo S 0 for all t e 0T Proof Fix t E 0T and set Bij umfj co Aij aiJt x0 and let ei1 be an orthonormal basis for R such that Aei Aiei Notice that M 2 9 gt 0 for 176 BRUCE K DRIVERl all i By the rst derivative test u1aco 0 for all i and hence Lut 0 ZAijBij ZAJiBij trAB Zei ABei ZAei Bei ZAiei Bei 239 ZAi8 1ut 350 S 0 239 The last inequality if a consequence of the second derivative test which asserts 83utaco S 0 for all u E R l Theorem 1215 Elliptic weak maximum principle Let Q be a bounded domain and L be an elliptic operator as in Eq 127 We now assume that aij and bj are functions ofac alone For each u E C 1029 such that Lu 2 0 on Q ie u is L 7 subharmonic we have 128 maxu 3 max u a bdQ Proof Let us rst assume Lu gt 0 on Q If u and had an interior local maximum at 10 E Q then by Lemma 1214 Luaco S 0 which contradicts the assumption that Luaco gt 0 So if Lu gt 0 on 9 we conclude that Eq 128 holds Now suppose that Lu 2 0 on 9 Let em1 with A gt 0 then Lam A2a11ac 1211 em 2 A A9 1211em By continuity of bac we may choose A suf ciently large so that A9 b1ac gt 0 on 2 in which case ch gt 0 on Q The results in the rst paragraph may now be applied to u6ac eq5ac for any 6 gt 0 to learn lt lt 7 eq5ac u6ac 7 ax u6 7 gaggu eggg cb for all at E Q Letting e l 0 in this expression then implies 3 max u for all at 6 fl bdQ which is equivalent to Eq 128 I Thforem 1216 Parabolic weak maximum principle Assume u E Cl 2 TT T CQT 1 If u 7 Lu 3 0 in QT then 129 maxu maxu 5 FT 2 If u 7 Lu 2 0 in QT then minu minu 5 FT Proof Item 1 follows from Item 2 by replacing u 7gt 7u so it suf ces to prove item 1 We begin by assuming u 7 Lu lt 0on QT and suppose for the sake of contradiction that there exists a point to x0 6 QTT T such that ut0 x0 niax u QT 1 1f t0aco E QT ie 0 lt to lt T then by the rst derivative test Bag to x0 0 and by Lemma 1214 Lut0 350 S 0 Therefore u 7 Lu t0aco 7Lut0 x0 2 0 which contradicts the assumption that u 7 Lu lt 0 in QT PDE LECTURE NOTES MATH 237A7B 177 2 If to7 x0 6 TI T with to T7 then by the rst derivative test7 Egg HT 0 2 0 and by Lemma 1214 Lut07 are S 0 So again u 7 Lu to7 x0 2 0 which contradicts the assumption that u 7 Lu lt 0 in QT Thus we have proved Eq 129 holds if u 7 Lu lt 0 on QT Finally if u 7 Lu 3 0 on QT and e gt 07 the function u tx utac 7 6t satis es u 7 Lu6 3 7e lt 0 Therefore by what we have just proved utac 7 6t 3 maXu6 maxu6 S maXu for all Lac 6 QT ET PT PT Letting e 1 0 in the last equation shows that Eq 129 holds l Corollary 1217 There is at most one solution u E Cl 2 TI T C T to the partial di erential equation 8 8 1Lu withuf onI T Proof If there were another solution 1 then w u 7 u would solve 88 1 Lw with w 0 on PT So by the maximum principle in Theorem 12167 w 0 on QT We now restrict back to L A and we wish to see what can be said when 9 R 7 an unbounded set Theorem 1218 Suppose u E C07T X R N 02 10T gtlt R 7 1 u 7 EAu S 0 on 0T X R and there exists constants Aa lt 00 such that utac S Aealfll for Lac 6 07T X R Then sup utac S K sup u07 txe0TgtltRquot xERquot Proof Recall that solves the heat equation 1 1210 8mm Apir Since both sides of Eq 1210 are analytic as functions in at so7 8p 1 1 8 290 5APMW iiApptaf and therefore for all 739 gt 0 and t lt 739 8p 7 1 8 tix 7pTix iAprn ix 7Similarly since both sides of Eq 1210 are analytic functions in t it follows that 5 1 5174m z EAth 178 BRUCE K DRIVERl That is to say the function 1 7397t 2 2 l emptjlgg fOFOSlltT pm 7 ppm solves the heat equation This can be checked directly as well Let 6739 gt 0 to be chosen later and set vt7 at utac 7 eptac for 0 S t S TQ Since pt7 at is increasing in t7 1 1 1 W2 vtx S Aealil 7 e 7 e7 739 Hence if we require 2 gt a or 739 lt i it will follows that 2 forOStSTQ lim sup vt7 at 700 170le 0992 Therefore we may choose M suf ciently large so that vtx S K supu0z for all 2 M and 0 S t S TQ 2 Since A A 87 v 87 ago we may apply the maximum principle with Q B07 M and T 72 to conclude for Lac 6 QT that ut7 at 7 ept7 at vt7 at S supv0z S K if 0 S t S TQ 2E9 We may now let 6 l 0 in this equation to conclude that 1211 1421 3 K ifO g t g T2 By applying Eq 1211 to ut TQ at we may also conclude utac SKifOStST Repeating this argument then enables us to show ut7 at S K for all 0 S t S T l Corollary 1219 The heat equation u 7 Au 0 on QT X R with u0 E CR has at most one solution in the class offunctions u E C07T X R 02 10T X R which satisfy utac S Aealfll for Lac 6 07T X R for some constants A and a Theorem 1220 Max Principle a la Hamilton Suppose u E 01 2 07T gtlt Rd satis es 1 utac S Aealfll for some Aa for all t S T 2 u0x S 0 for all at 3 88 1 3 Au ie 8 7 Au S 0 Then utac S 0 for all Lac 6 0T gtlt Rd PDE LECTURE NOTES MATH 237A B 179 Proof Special Case Assume 2 lt Au on 0T gtlt Rd u0x lt 0 for all a 6 Rd and there exists M gt 0 such that utx lt 0 if Z M and t E 0TFor the sake of contradiction suppose there is some point Lac 6 0 T X Rd such that utx gt 0 By the intermediate value theorem there exists 739 E 0 t such that u739 a 0 In particular the set u 0 is a non empty closed compact subset of 0 T X B 0 M Let 7r 0T gtlt B0M gt 0T be projection onto the rst factor since u 7E 0 is a compact subset of 0T gtlt B0 M if follows that to 2 mint E 7ru gt 0 Choose a point 330 E B0M such that t0x0 E u 0 ie ut0x0 0 see Figure 37 below Since utx lt 0 for all 0 S t lt to and a 6 Rd ut0x S 0 MltO FIGURE 37 Finding a point t0x0 such that to is as small as possible and ut0 x0 0 for all a 6 Rd with ut0x0 0 This information along with the rst and second derivative tests allows us to conclude mama 0 mama g 0 and g mxo 2 0 This then implies that 0 S g mxo lt Aultt0x0 S 0 which is absurd Hence we conclude that u S 0 on 0 T X Rd General Case Let pt tdlQ e lx be the fundamental solution to the heat equation 825292 2 Apt Let 739 gt 0 to be determined later As in the proof of Theorem 1218 the function 1 T t d2 1 2 gt e4Ttlxl for 0 S t lt 739 pm wax 180 BRUCE K DRIVER1 is still a solution to the heat equation Given 6 gt 0 de ne for t S T2 uet at ut at 7 e 7 6t 7 ept Then 8t7Aue 8t7Au7e 7elt0 u60acu0ac 76 076 3 7elt0 and for t S 72 1 TdZ 6 Hence if we choose 739 such that gt a we will have uet at lt 0 for ml sufficiently large Hence by the special case already proved uet at S 0 for all 0 S t S g and e gt 0 Letting e l 0 implies that ut at S 0 for all 0 S t S T2 As in the proof of Theorem 1218 we may step our way up by applying the previous argument to ut T2ac and then to ut 731 etc to learn utac S 0 for all 0 S t S T l 17611 uet at S Aealill 7 e 7 e 124 NonUniqueness of solutions to the Heat Equation Theorem 1221 See Fritz John 7 For any a gt 1 let 7 e t a tgt 0 1212 gt 7 0 t 0 and de ne 9000ka utacZ 2k k0 Then u E C R2 and 1213 at um and u0 at 0 In particular the heat equation does not have unique solutions Proof We are going to look for a solution to Eq 1213 of the form 00 ut at Zgntxn n0 in which case we have formally that 0 u 7 um 7 gntnn 7 1xn 2 2 l39n n 2N 1gn2il 90 This implies 9n 1214 gn2 To simplify the nal answer we will now assume u0 at 0 ie 91 E 0 in which case Eq 1214 implies 9 E 0 for all n odd We also have with g go i 90 92 21 2 94 43 4 95 6 quot3992 2k PDE LECTURE NOTES MATH 237A7B 181 and hence 00 k 2k 9 096 1215 u t m lt gt lt 7 gt Z W k0 The function utm will solve u um for t7 x E R2 with u07 x 0 provided the convergence in the sum is adequate to justify the above computations Now let gt be given by Eq 1212 and extend g to 7007 0 via 572 8quot where 270 e o bg e o am g for z rem with 7 7T lt 9 lt 7T In order to estimate gkt we will use of the Cauchy estimates on the contour iz 7 ti yt where y is going to be chosen sufficiently close to 0 Now Rez equotquot1M cosa9 izi o cosa9 and hence 87mg e7izi cosae 19Zi From Figure 387 we see FIGURE 38 Here is a picture of the maximum argument 9m that a point z on 83t7 yt may attain Notice that sin 9m ytt y is independent of t and 9m 7 0 as 39y 7gt 0 My min cosa9 77139 lt 9 lt 7T and irew 7 ti 7t is independent oft and 7gt 1 as 39y 7gt 0 Therefore for iz 7 ti yt we have igizii 3 warm 3 yawlwwm 871ih S 87 57 1 prov1ded 7 1s chosen so small that T 2 5 By for w E Btt y7 the Cauchy integral formula and its derivative give 1 92 9W I iz7ti39ytz 7 w dz and k 92 k glt gtw yfi imm dz and in particular 1216 i k k 19 l k if mi 3 f WW idzi e z i ra 27T yt k i ra k 1 iziwi e Wk Wk 182 BRUCE K DRIVERl We now use this to estimate the sum in Eq 1215 as oo gktx2k fifa 00 2k laws W 36 I gmokw l 1 0 1 12 k 77t 7 e I 2 E a 9quotquot k0 Therefore ltilrg utx 0 uniformly for at in compact subsets of R Similarly one may use the estimate in Eq 1216 to show u is smooth and u igwnnemawn k gltkgtltzx2ltk1 7 2k 2k7 1 k1 gk1 0ka ZW quot I 125 The Heat Equation on the Circle and R In this subsection7 let SL L27 z E S 7 be the circle of radius L As usual we will identify functions on SL with 27TL 7 periodic functions on R Given two 27TL periodic functions g let ML 7 WL mama and denote HL 39 ngL to be the 27TL 7 periodic functions f on R such that ffL lt 00 By Fourier s theorem we know that the functions elmL with k E Z form an orthonormal basis for HL and this basis satis es 2 d2 L 7 7 E L 1352 Xk L M Therefore the solution to the heat equation on SL ut law with u07 f 6 HL 2 is given by um Zltfx elt ei L kez 1 WL v 1 k 1 v 2 fye Ldy Mr 71 led WL 7rL pfr7yfydy 77rL where 19596 21L ZanteWL39 71 If f is L periodic then it is nL 7 periodic for all n E N so we also would learn 7rnL ut7 at LpzlLUc 7 for all n E N 77m

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