M(t) = u t (x, t) dx.

Transcription

1 2 Hat Equatio 2. Drivatio Rf: Strauss, Sctio.3. Blow w provid two drivatios of th hat quatio, This quatio is also kow as th diffusio quatio. 2.. Diffusio u t ku xx = k >. (2. Cosidr a liquid i which a dy is big diffusd through th liquid. Th dy will mov from highr coctratio to lowr coctratio. Lt u(x, t b th coctratio (mass pr uit lgth of th dy at positio x i th pip at tim t. Th total mass of dy i th pip from x to x at tim t is giv by Thrfor, M(t = dm dt = x x x x u(x, t dx. u t (x, t dx. By Fick s Law, dm = flow i flow out = ku x (x, t ku x (x, t, dt whr k > is a proportioality costat. That is, th flow rat is proportioal to th coctratio gradit. Thrfor, x Now diffrtiatig with rspct to x, w hav Or, This is kow as th diffusio quatio Hat Flow x u t (x, t dx = ku x (x, t ku x (x, t. u t (x, t = ku xx (x, t. u t = ku xx. W ow giv a altrat drivatio of (2. from th study of hat flow. Lt D b a rgio i R. Lt x = [x,..., x ] T b a vctor i R. Lt u(x, t b th tmpratur at poit x,

2 tim t, ad lt H(t b th total amout of hat (i caloris cotaid i D. Lt c b th spcific hat of th matrial ad ρ its dsity (mass pr uit volum. Th H(t = cρu(x, t dx. Thrfor, th chag i hat is giv by dh dt = D D cρu t (x, t dx. Fourir s Law says that hat flows from hot to cold rgios at a rat κ > proportioal to th tmpratur gradit. Th oly way hat will lav D is through th boudary. That is, dh dt = κ u ds. D whr D is th boudary of D, is th outward uit ormal vctor to D ad ds is th surfac masur ovr D. Thrfor, w hav cρu t (x, t dx = κ u ds. D Rcall that for a vctor fild F, th Divrgc Thorm says F ds = F dx. (Rf: S Strauss, Appdix A.3. Thrfor, w hav cρu t (x, t dx = (κ u dx. This lads us to th partial diffrtial quatio D D D D D cρu t = (κ u. If c, ρ ad κ ar costats, w ar ld to th hat quatio whr k = κ/cρ > ad u = i= u x i x i. u t = k u, 2.2 Hat Equatio o a Itrval i R 2.2. Sparatio of Variabls Cosidr th iitial/boudary valu problm o a itrval I i R, u t = ku xx x I, t > u(x, = φ(x x I u satisfis crtai BCs. I practic, th most commo boudary coditios ar th followig: (2.2 2

3 . Dirichlt (I = (, l : u(, t = = u(l, t. 2. Numa (I = (, l : u x (, t = = u x (l, t. 3. Robi (I = (, l : u x (, t a u(, t = ad u x (l, t + a l u(l, t =. 4. Priodic (I = ( l, l: u( l, t = u(l, t ad u x ( l, t = u x (l, t. W will giv spcific xampls blow whr w cosidr som of ths boudary coditios. First, howvr, w prst th tchiqu of sparatio of variabls. This tchiqu ivolvs lookig for a solutio of a particular form. I particular, w look for a solutio of th form u(x, t = X(xT (t for fuctios X, T to b dtrmid. Suppos w ca fid a solutio of (2.2 of this form. Pluggig a fuctio u = XT ito th hat quatio, w arriv at th quatio Dividig this quatio by kxt, w hav XT kx T =. T kt = X X = λ. for som costat λ. Thrfor, if thr xists a solutio u(x, t = X(xT (t of th hat quatio, th T ad X must satisfy th quatios T kt = λ X X = λ for som costat λ. I additio, i ordr for u to satisfy our boudary coditios, w d our fuctio X to satisfy our boudary coditios. That is, w d to fid fuctios X ad scalars λ such that { X (x = λx(x x I (2.3 X satisfis our BCs. This problm is kow as a igvalu problm. I particular, a costat λ which satisfis (2.3 for som fuctio X, ot idtically zro, is calld a igvalu of x 2 for th giv boudary coditios. Th fuctio X is calld a igfuctio with associatd igvalu λ. Thrfor, i ordr to fid a solutio of (2.2 of th form u(x, t = X(xT (t our first goal is to fid all solutios of our igvalu problm (2.3. Lt s look at som xampls blow. Exampl. (Dirichlt Boudary Coditios Fid all solutios to th igvalu problm { X = λx < x < l (2.4 X( = = X(l. 3

4 Ay positiv igvalus? First, w chck if w hav ay positiv igvalus. That is, w chck if thr xists ay λ = β 2 >. Our igvalu problm (2.4 bcoms { X + β 2 X = < x < l Th solutios of this ODE ar giv by X( = = X(l. X(x = C cos(βx + D si(βx. Th boudary coditio Th boudary coditio X( = = C =. X(l = = si(βl = = β = π l Thrfor, w hav a squc of positiv igvalus with corrspodig igfuctios λ = ( π 2 l X (x = D si ( π l x. =, 2,.... Is zro a igvalu? Nxt, w look to s if zro is a igvalu. If zro is a igvalu, our igvalu problm (2.4 bcoms { X = < x < l X( = = X(l. Th gral solutio of th ODE is giv by Th boudary coditio Th boudary coditio X(x = C + Dx. X( = = C =. X(l = = D =. Thrfor, th oly solutio of th igvalu problm for λ = is X(x =. By dfiitio, th zro fuctio is ot a igfuctio. Thrfor, λ = is ot a igvalu. Ay gativ igvalus? Last, w chck for gativ igvalus. That is, w look for a igvalu λ = γ 2. I this cas, our igvalu problm (2.4 bcoms { X γ 2 X = < x < l X( = = X(l. 4

5 Th solutios of this ODE ar giv by X(x = C cosh(γx + D sih(γx. Th boudary coditio Th boudary coditio X( = = C =. X(l = = D =. Thrfor, thr ar o gativ igvalus. Cosqutly, all th solutios of (2.4 ar giv by λ = ( π 2 ( π X (x = D si l l x =, 2,.... Exampl 2. (Priodic Boudary Coditios Fid all solutios to th igvalu problm { X = λx l < x < l X( l = X(l, X ( l = X (2.5 (l. Ay positiv igvalus? First, w chck if w hav ay positiv igvalus. That is, w chck if thr xists ay λ = β 2 >. Our igvalu problm (2.5 bcoms { X + β 2 X = l < x < l X( l = X(l, X ( l = X (l. Th solutios of this ODE ar giv by Th boudary coditio Th boudary coditio X(x = C cos(βx + D si(βx. X( l = X(l = D si(βl = = D = or β = π l. X ( l = X (l = Cβ si(βl = = C = or β = π l. Thrfor, w hav a squc of positiv igvalus with corrspodig igfuctios X (x = C cos λ = ( π 2 l ( π ( π l x + D si l x. 5

6 Is zro a igvalu? Nxt, w look to s if zro is a igvalu. If zro is a igvalu, our igvalu problm (2.5 bcoms { X = l < x < l X( l = X(l, X ( l = X (l. Th gral solutio of th ODE is giv by X(x = C + Dx. Th boudary coditio Th boudary coditio X( l = X(l = D =. X ( l = X (l is automatically satisfid if D =. Thrfor, λ = is a igvalu with corrspodig igfuctio X (x = C. Ay gativ igvalus? Last, w chck for gativ igvalus. That is, w look for a igvalu λ = γ 2. I this cas, our igvalu problm (2.5 bcoms { X γ 2 X = l < x < l X( l = X(l, X ( l = X (l. Th solutios of this ODE ar giv by Th boudary coditio Th boudary coditio X(x = C cosh(γx + D sih(γx. X( l = X(l = D sih(γl = = D =. X ( l = X (l = Cγ sih(γl = = C =. Thrfor, thr ar o gativ igvalus. Cosqutly, all th solutios of (2.5 ar giv by ( π 2 λ = l ( π X (x = C cos l x λ = X (x = C. ( π + D si l x =, 2,... 6

7 Now that w hav do a coupl of xampls of solvig igvalu problms, w rtur to usig th mthod of sparatio of variabls to solv (2.2. Rcall that i ordr for a fuctio of th form u(x, t = X(xT (t to b a solutio of th hat quatio o a itrval I R which satisfis giv boudary coditios, w d X to b a solutio of th igvalu problm, { X = λx x I X satisfis crtai BCs for som scalar λ ad T to b a solutio of th ODE T = kλt. W hav giv som xampls abov of how to solv th igvalu problm. Oc w hav solvd th igvalu problm, w d to solv our quatio for T. I particular, for ay scalar λ, th solutio of th ODE for T is giv by T (t = A kλt for a arbitrary costat A. Thrfor, for ach igfuctio X with corrspodig igvalu λ, w hav a solutio T such that th fuctio u (x, t = T (tx (x is a solutio of th hat quatio o th itrval I which satisfis our boudary coditios. Not that w hav ot yt accoutd for our iitial coditio u(x, = φ(x. W will look at that xt. First, w rmark that if {u } is a squc of solutios of th hat quatio o I which satisfy our boudary coditios, tha ay fiit liar combiatio of ths solutios will also giv us a solutio. That is, u(x, t N u (x, t = will b a solutio of th hat quatio o I which satisfis our boudary coditios, assumig ach u is such a solutio. I fact, o ca show that a ifiit sris of th form u(x, t u (x, t = will also b a solutio of th hat quatio, udr propr covrgc assumptios of this sris. W will omit discussio of this issu hr Satisfyig our Iitial Coditios W rtur to tryig to satisfy our iitial coditios. Assum w hav foud all solutios of our igvalu problm. W lt {X } dot our squc of igfuctios ad {λ } dot our squc of igvalus. Th for ach λ, w hav a solutio T of our quatio for T. Lt u(x, t = X (xt (t = A X (x kλt. 7

8 Our goal is to choos A appropriatly such that our iitial coditio is satisfid. I particular, w d to choos A such that u(x, = A X (x = φ(x. I ordr to fid A satisfyig this coditio, w us th followig orthogoality proprty of igfuctios. First, w mak som dfiitios. For two ral-valud fuctios f ad g dfid o Ω, f, g = f(xg(x dx Ω is dfid as th L 2 ir product of f ad g o Ω. Th L 2 orm of f o Ω is dfid as f 2 L 2 (Ω = f, f = f(x 2 dx. W say fuctios f ad g ar orthogoal o Ω R if f, g = f(xg(x dx =. W say boudary coditios ar symmtric if Ω [f (xg(x f(xg (x] x=b x=a = for all fuctios f ad g satisfyig th boudary coditios. Lmma 3. Cosidr th igvalu problm (2.3 with symmtric boudary coditios. If X, X m ar two igfuctios of (2.3 with distict igvalus, th X ad X m ar orthogoal. Proof. Lt I = [a, b]. b λ X (xx m (x dx = a = b a b = a b X (xx m (x dx Ω X (xx m(x dx X (xx m (x x=b x=a a X (xx m(x dx + [X (xx m(x X (xx m (x] x=b x=a b = λ m X (xx m (x dx, usig th fact that th boudary coditios ar symmtric. Thrfor, a (λ λ m b a X (xx m (x dx =, 8

9 but λ λ m bcaus th igvalus ar assumd to b distict. Thrfor, as claimd. b a X (xx m (x dx =, W ca us this lmma to fid cofficits A such that A X (x = φ(x. I particular, multiplyig both sids of this quatio by X m for a fixd m ad itgratig ovr I, w hav A m X m, X m = X m, φ, which implis A m = X m, φ X m, X m. Exampl 4. (Dirichlt Boudary Coditios I th cas of Dirichlt boudary coditios o th itrval [, l], w showd arlir that our igvalus ad igfuctios ar giv by ( π 2 ( π λ =, X (x = si l l x =, 2,.... Our solutios for T ar giv by Now lt u(x, t = T (t = A kλ t = A k(π/l2t. X (xt (t = = = ( π A si l x k(π/l2t. Usig th fact that Dirichlt boudary coditios ar symmtric (chck this!, our cofficits A m ar giv by A m = X m, φ X m, X m = l si (mπx/l φ(x dx l si2 (mπx/l dx = 2 l l si ( mπ x φ(x dx. l Thrfor, th solutio of (2.2 o th itrval I = [, l] with Dirichlt boudary coditios is giv by ( π u(x, t = A si l x k(π/l2 t whr A = 2 l = l ( π si l x φ(x dx. 9

10 Exampl 5. (Priodic Boudary Coditios I th cas of priodic boudary coditios o th itrval [ l, l], w showd arlir that our igvalus ad igfuctios ar giv by λ = ( π 2, X (x = l λ =, X (x = C. ( π cos si l x ( π =, 2,... l x Thrfor, our solutios for T ar giv by { T (t = A kλt A = k(π/l2 t =, 2,... A =. Now usig th fact that for ay itgr, u (x, t = X (xt (t is a solutio of th hat quatio which satisfis our priodic boudary coditios, w dfi u(x, t = X (xt (t = A + = Now, takig ito accout our iitial coditio, w wat u(x, = A + = [ ( π ( π ] A cos l x + B si l x k(π/l2t. [ ( π ( π ] A cos l x + B si l x = φ(x. It rmais oly to fid cofficits satisfyig this quatio. From our arlir discussio, usig th fact that priodic boudary coditios ar symmtric (chck this!, w kow that igfuctios corrspodig to distict igvalus will b orthogoal. For priodic boudary coditios, howvr, w hav two igfuctios for ach positiv igvalu. Spcifically, cos ( πx ad si ( πx ar both igfuctios corrspodig to th igvalu l l λ = (π/l 2. Could w b so lucky that ths igfuctios would also b orthogoal? By a straightforward calculatio, o ca show that l l ( π ( π cos l x si l x dx =. Thy ar orthogoal! This is ot mrly coicidtal. I fact, for ay igvalu λ of (2.3 with multiplicty m (maig it has m liarly idpdt igfuctios, th igfuctios may always b chos to b orthogoal. This procss is kow as th Gram-Schmidt orthogoalizatio mthod. Th fact that all our igfuctios ar mutually orthogoal will allow us to calculat cofficits A, B so that our iitial coditio is satisfid. Usig th tchiqu dscribd

11 abov, ad lttig f, g dot th L 2 ir product o [ l, l], w s that A = A = B =, φ, = l φ(x dx 2l l cos(πx/l, φ cos(πx/l, cos(πx/l = l si(πx/l, φ si(πx/l, si(πx/l = l l l l l ( π cos si l x φ(x dx ( π l x φ(x dx. Thrfor, th solutio of (2.2 o th itrval I = [ l, l] with priodic boudary coditios is giv by u(x, t = A + = [ ( π ( π ] A cos l x + B si l x k(π/l2 t whr Fourir Sris A = 2l A = l B = l l l l l l l φ(x dx ( π cos l x φ(x dx ( π si l x φ(x dx. I th cas of Dirichlt boudary coditios abov, w lookd for cofficits so that φ(x = = ( π A si l x. W showd that if w could writ our fuctio φ i trms of this ifiit sris, our cofficits would b giv by th formula A = 2 l l ( π si l x φ(x dx. For a giv fuctio φ dfid o (, l th ifiit sris φ = ( π A si l x whr A 2 l l ( π si l x φ(x dx is calld th Fourir si sris of φ. Not: Th otatio just mas th sris associatd with φ. It dos t imply that th sris cssarily covrgs to φ.

12 I th cas of priodic boudary coditios abov, w lookd for cofficits so that φ(x = A + = W showd that i this cas our cofficits must b giv by [ ( π ( π ] A cos l x + B si l x. (2.6 A = 2l A = l B = l l l l l l l φ(x dx For a giv fuctio φ dfid o ( l, l th sris φ A + = ( π cos l x φ(x dx ( π si l x φ(x dx. [ ( π ( π ] A cos l x + B si l x (2.7 whr A, B ar dfid i (2.7 is calld th full Fourir sris of φ. Mor grally, for a squc of igfuctios {X } of (2.3 which satisfy crtai boudary coditios, w dfi th gral Fourir sris of a fuctio φ as φ A X (x whr A X, φ X, X. Rmark. Cosidr th xampl abov whr w lookd to solv th hat quatio o a itrval with Dirichlt boudary coditios. (A similar rmark holds for th cas of priodic or othr boudary coditios. I ordr that our iitial coditio b satisfid, w dd to fid cofficits A such that φ(x = = ( π A si l x. W showd that if φ could b rprstd i trms of this ifiit sris, th our cofficits must b giv by A = 2 l ( π si l l x φ(x dx. Whil this is th cssary form of our cofficits ad dfiig u(x, t = ( π A si l x k(π/l2 t for A dfid abov is th appropriat formal dfiitio of our solutio, i ordr to vrify that this costructio actually satisfis our iitial/boudary valu problm (2.2 with Dirichlt boudary coditios, w would d to vrify th followig. 2

13 . Th Fourir si sris of φ covrgs to φ (i som ss. 2. Th ifiit sris actually satisfis th hat quatio. W will ot discuss ths issus hr, but rfr th radr to covrgc rsults i Strauss as wll as th ots from 22a. Latr, i th cours, w will prov a L 2 covrgc rsult of igfuctios. Complx Form of Full Fourir Sris. It is somtims usful to writ th full Fourir sris i complx form. W do so as follows. Th igfuctios associatd with th full Fourir sris ar giv by { ( π ( π } cos l x, si l x for =,, 2,.... Usig dmoivr s formula, w ca writ ( π cos l x ( π si l x iθ = cos θ + i si θ, = iπx/l + iπx/l 2 = iπx/l iπx/l. 2i Now, of cours, ay liar combiatio of igfuctios is also a igfuctio. Thrfor, w s that ( π ( π cos l x + i si l x = iπx/l ( π ( π cos l x i si l x = iπx/l ar also igfuctios. Thrfor, th igfuctios associatd with th full Fourir sris ca b writt as { iπx/l } =..., 2,,,, 2,.... Now, lt s suppos w ca rprst a giv fuctio φ as a ifiit sris xpasio i trms of ths igfuctios. That is, w wat to fid cofficits C such that φ(x = = C iπx/l. As dscribd arlir, igfuctios corrspodig to distict igvalus will b orthogoal, as priodic boudary coditios ar symmtric. Thrfor, w hav l l iπx/l imπx/l dx = for m. 3

14 For igfuctios corrspodig to th sam igvalu, w d to chck th L 2 ir product. I particular, for th igvalu λ = (π/l 2, w hav two igfuctios: iπx/l ad iπx/l. By a straightforward calculatio, w s that l l l l iπx/l iπx/l dx = for iπx/l iπx/l dx = 2l. Thrfor, our cofficits C would d to b giv by C = 2l l l φ(x iπx/l dx. Cosqutly, th complx form of th full Fourir sris for a fuctio φ dfid o ( l, l is giv by φ C iπx/l whr C = l φ(x iπx/l dx. 2l = 2.3 Fourir Trasforms 2.3. Motivatio Rf: Strauss, Sctio 2.3 W would ow lik to tur to studyig th hat quatio o th whol ral li. Cosidr th iitial-valu problm, { ut = ku xx, < x < (2.8 u(x, = φ(x. I th cas of th hat quatio o a itrval, w foud a solutio u usig Fourir sris. For th cas of th hat quatio o th whol ral li, th Fourir sris will b rplacd by th Fourir trasform. Abov, w discussd th complx form of th full Fourir sris for a giv fuctio φ. I particular, for a fuctio φ dfid o th itrval [ l, l] w dfi its full Fourir sris as φ = C iπx/l whr C = 2l l l l φ(x iπx/l dx. Pluggig th cofficits C ito th ifiit sris, w s that [ l ] φ φ(y iπy/l dy iπx/l. 2l = Now, lttig k = π/l, w ca writ this as [ φ 2π = l l l ] π φ(y i(y xk dy l. 4

15 Th distac btw poits k is giv by k = π/l. As l +, w ca thik of k dk ad th ifiit sum bcomig a itgral. Roughly, w hav φ 2π φ(y i(y xk dy dk. Blow w will us this motivatio to dfi th Fourir trasform of a fuctio φ ad th show how th Fourir trasform ca b usd to solv th hat quatio (amog othrs o th whol ral li, ad mor grally i R Dfiitios ad Proprtis of th Fourir Trasform W say f L (R if f(x dx < +. R For f L (R, w dfi its Fourir trasform at a poit ξ R as f(ξ (2π /2 R ix ξ f(x dx. W dfi its ivrs Fourir trasform at th poit ξ R as ˇf(ξ (2π /2 R ix ξ f(x dx. Rmark. Somtims th costats i frot ar dfid diffrtly. I.. - i som books, th Fourir trasform is dfid with a costat of istad of, ad i accordac th (2π (2π /2 ivrs Fourir trasform is dfid with a costat rplacig th costat abov. (2π /2 Thorm 6. (Plachrl s Thorm If u L (R L 2 (R, th û, ǔ L 2 (R ad û L 2 (R = ǔ L 2 (R = u L 2 (R. I ordr to prov this thorm, w d to prov som prlimiary facts. Claim 7. Lt Th Proof. f(ξ = = f(ξ = f(x = ɛ x 2. (2ɛ /2 ξ 2 /4ɛ. (2π /2 R ix ξ ɛ x 2 dx ( ix ξ ɛx2 (2π /2 dx 5 ( ixξ ɛx2 dx.

16 Thrfor, w just d to look at Compltig th squar, w hav Thrfor, w hav ɛx 2 ixξ = ɛ [ [ = ɛ x + ixξ ɛx2 dx = ixξ ɛx2 dx. ( 2 iξ 2ɛ x 2 + iξx + ɛ ( ] 2 iξ + ɛ 2ɛ ( 2 iξ. 2ɛ ( ] 2 iξ 2ɛ ɛ[x+(iξ/2ɛ]2 ξ2 /4ɛ dx. Now makig th chag of variabls, z = [x + (iξ/2ɛ], w hav ɛ[x+(iξ/2ɛ]2 ξ2 /4ɛ dx = ξ2 /4ɛ ɛz2 dz whr Γ is th li i th complx pla giv by { Γ z C : y = x + iξ } 2ɛ, x R. Without loss of grality, w assum ξ >. A similar aalysis works if ξ <. Now ɛz2 dz = dz Γ lim R + Γ R ɛz2 whr Γ R is th li sgmt i th complx pla giv by { Γ R z C : z = x + iξ } 2ɛ, x R. Now dfi Λ R, Λ2 R ad Λ3 R as show i th pictur blow. That is, Λ R {x R : x R} { Λ 2 R z C : z = x + iy, x, y R, x = R, y ξ } 2ɛ { Λ 3 R z C : z = x + iy; x, y R; x = R, y ξ }. 2ɛ Im ( z Γ 3 Λ R R ξ/2ε Λ R Γ R R 2 Λ R R ( z 6

17 From complx aalysis, w kow that ɛz2 dz = C whr C is th closd curv giv by C = Γ R Λ R Λ2 R Λ3 R travrsd i th coutr-clockwis dirctio. Thrfor, w hav dz = dz Γ R ɛz2 Λ R ɛz2 whr th itgral o th right-had sid is th li itgral giv by Λ R = Λ 3 R Λ R Λ2 R travrsd i th dirctio show. Thrfor, ɛz2 dz = dz. Γ lim R + Λ R ɛz2 But, as R +, Thrfor, Cosqutly, w hav Thrfor, w hav f(ξ = = Λ j R Λ R Γ ɛz2 dz forj = 2, 3 ɛz2 dz ɛz2 dz = ɛx2 dx. ixξ ɛx2 dx = ξ2 /4ɛ = ξ2 /4ɛ = ξ2 /4ɛ ɛx2 dx. Γ = ξ2 /4ɛ ɛ π. ɛz2 dz ɛx2 dx d x x2 ɛ ( ( ξ2 /4ɛ ξ2 /4ɛ π π (2π /2 ɛ ɛ (2ɛ /2 ξ 2 /4ɛ, as claimd. 7

18 Claim 8. Lt w(x = u v(x = u(x yv(y dy L R (R. That is, lt w b th covolutio of u ad v. Th ŵ(ξ = û v(ξ = (2π /2 û(ξ v(ξ. Proof. By dfiitio, ŵ(ξ = (2π /2 R ix ξ w(x dx = (2π /2 R ix ξ u(x yv(y dy dx R [ ] = (2π /2 R R i(x y ξ u(x y dx iy ξ v(y dy = (2π (2π /2 R /2 û(ξ iy ξ v(y dy = û(ξ R iy ξ v(y dy = û(ξ(2π /2 v(ξ = (2π /2 û(ξ v(ξ. Now w us ths two claims to prov Plachrl s Thorm. Proof of Thorm 6. By assumptio, u L (R L 2 (R. Lt v(x u( x w(x u v(x. Thrfor, v L (R L 2 (R ad w L (R L(R. First, w hav v(ξ = (2π /2 R ix ξ v(x dx = (2π /2 R ix ξ u( x dx = (2π /2 R iy ξ u(y dy = (2π /2 R iy ξ u(y dy Thrfor, usig Claim 8, = û(ξ. ŵ(ξ = (2π /2 û(ξ v(ξ = (2π /2 û(ξû(ξ = (2π /2 û 2. 8

19 Nxt, w us th fact that if f, g ar i L (R, th f, ĝ ar i L (R, ad, morovr, f(xĝ(x dx = R f(ξg(ξ dx. R (2.9 This fact ca b s by dirct substitutio, as show blow, [ ] f(xĝ(x dx = f(x R R (2π /2 R ix ξ g(ξ dξ dx [ ] = R (2π /2 R ix ξ f(x dx g(ξ dξ = f(ξg(ξ dξ. R Thrfor, lttig f(x = ɛ x 2 ad lttig g(x = w(x as dfid abov, substitutig f ad g ito (2.9 ad usig Claim 7 to calculat th Fourir trasform of f, w hav R ɛ ξ 2 ŵ(ξ dξ = R /4ɛ (2ɛ /2 x 2 w(x dx. (2. Now w tak th limit of both sids abov as ɛ +. First, lim ɛ + R ɛ ξ 2 ŵ(ξ dξ = ŵ(ξ dξ. R (2. Scod, w claim lim ɛ + (2ɛ /2 R x 2 /4ɛ w(x dx = (2π /2 w(. (2.2 W prov this claim as follows. I particular, w will prov that (4πɛ /2 R x 2 /4ɛ w(x dx w( as ɛ +. First, w ot that (4πɛ /2 R x 2 /4ɛ dx =. (2.3 This follows dirctly from th fact that Thrfor, (4πɛ /2 R x 2 /4ɛ w(x dx w( = z2 dz = π. (4πɛ /2 R x 2 /4ɛ [w(x w(] dx. Now, w will show that for all γ > thr xists a ɛ > such that (4πɛ /2 R x 2 /4ɛ [w(x w(] dx < γ 9

20 for < ɛ < ɛ, thus, provig (2.2. Lt B(, δ b th ball of radius δ about. (W will choos δ sufficitly small blow. Now brak up th itgral abov ito two pics, as follows, (4πɛ /2 R x 2 /4ɛ [w(x w(] dx x 2 /4ɛ [w(x w(] dx (4πɛ /2 B(,δ + x 2 /4ɛ [w(x w(] dx (4πɛ /2 I + J. R B(,δ First, for trm I, w hav x 2 /4ɛ [w(x w(] dx (4πɛ /2 w(x w( L (B(,δ B(,δ (4πɛ /2 R x 2 /4ɛ dx < γ 2, for δ sufficitly small, usig th fact that w C(R ad (2.3. Now for δ fixd small, w look at trm J, (4πɛ /2 R x 2 /4ɛ [w(x w(] dx B(,δ (4πɛ /2 R x 2 /4ɛ w(x dx B(,δ + (4πɛ /2 R x 2 /4ɛ w( dx B(,δ /4ɛ (4πɛ /2 x 2 w(x dx L (R B(,δ R + δ2 /8ɛ 2 /2 w( (8πɛ /2 R x 2 /8ɛ dx B(,δ C /4ɛ (4πɛ /2 δ 2 + /8ɛ C δ2 (8πɛ /2 R x 2 /8ɛ dx C + /8ɛ C δ2 < γ 2 (4πɛ /2 δ 2 /4ɛ for ɛ sufficitly small, usig th fact that for a fixd δ, Thrfor, w coclud that lim /4ɛ ɛ + (4πɛ /2 δ 2 = = lim /8ɛ ɛ + δ2. I + J < γ for ɛ chos sufficitly small, ad, thus, (2.2 is prov. 2

21 Now combiig (2. ad (2.2 with (2., w coclud that ŵ(ξ dξ = (2π /2 w(. R Now ŵ(ξ = (2π /2 û 2 ad w( = u v( = R u(xv( x dx = R u(xu(x dx = R u 2 dx. Thrfor, w coclud that or (2π R /2 û 2 dξ = (2π /2 u R 2 dx, û L 2 = u L 2, as claimd. A similar tchiqu ca b usd to show that ǔ L 2 = u L 2. Dfiig th Fourir Trasform o L 2 (R. For f L (R, that is, f such that R f(x dx < + it is clar that th Fourir trasform is wll-dfid, i.. - th itgral covrgs. If f / L (R, th itgral may ot covrg. Hr w dscrib how w dfi th Fourir trasform of a fuctio f L 2 (R (but which may ot b i L (R. Lt f L 2 (R. Approximat f by a squc of fuctios {f k } such that f k L (R L 2 (R ad f k f L 2 as k +. By Plachrl s thorm, f k f j L 2 = f k f j L 2 = f k f j L 2 as k, j +. Thrfor, { f k } is a Cauchy squc i L 2 (R, ad, thrfor, covrgs to som g L 2 (R. W dfi th Fourir trasform of f by this fuctio g. That is, f g. Othr Proprtis of th Fourir Trasform. Assum u, v L 2 (R. Th (a u(xv(x dx = R û(ξ v(ξ dξ. R (b α i x i u(ξ = (iξ i αiû(ξ. (c û v(ξ = (2π /2 û(ξ v(ξ. (d u = ˇû 2

22 Proof of (a. Lt α C. By Plachrl s Thorm, w hav Now usig th fact that = u + αv L 2 = û + αv L 2 u + αv R 2 dx = û + αv R 2 dξ w hav a + b 2 = (a + b (a + b = a 2 + ab + ab + b 2, u R 2 + u(αv + u(αv + αv 2 dx = û R 2 + û( αv + û( αv + αv 2 dξ, which implis u(αv + u(αv dx = R û( αv + û( αv dξ. R Lttig α =, w hav uv + uv dx = R û v + û v dξ. R (2.4 Lttig α = i, w hav iuv iuv dx = R iû v iû v dξ. R (2.5 Multiplyig (2.5 by i, w hav uv + uv dx = R û v + û v dξ. R (2.6 Addig (2.4 ad (2.6, w hav 2 uv dx = 2 R û v dξ, R ad, thrfor, proprty (a is provd. Proof of (b. W assum u is smooth ad has compact support. W ca us a approximatio argumt to prov th sam quality for ay u such that α i x i u L 2 (R. Usig th dfiitio of Fourir trasform ad itgratig by parts, w hav xi u(ξ = (2π /2 R ix ξ xi u(x dx = iξ (2π /2 R i ix ξ u(x dx = iξ i (2π /2 R ix ξ u(x dx = iξ i û(ξ. 22

23 Proprty (b follows by applyig th sam argumt to highr drivativs. Proof of (c. Proprty (c was alrady provd i Claim 8 i th cas wh u ad v ar i L (R. O ca us a approximatio argumt to prov th rsult i th gral cas. Proof of (d. whr Fix z R. Dfi th fuctio v ɛ (x ix z ɛ x 2. Thrfor, v ɛ (ξ = (2π /2 R ix ξ v ɛ (x dx = (2π /2 R ix ξ ix z ɛ x 2 dx = (2π /2 R ix (ξ z ɛ x 2 dx = f(ξ z f(x ɛ x 2. By Claim 7, w kow that f(ξ = /4ɛ (2ɛ /2 ξ 2. Thrfor, w hav v ɛ (ξ = /4ɛ (2ɛ /2 ξ z 2. Now usig (2.9, with f = u ad g = v ɛ, w hav û(ξv R ɛ (ξ dξ = u(x v R ɛ (x dx = û(ξ R iξ z ɛ ξ 2 dξ = u(x (2ɛ /2 R x z 2 /4ɛ dx Takig th limit as ɛ +, w hav lim û(ξ ɛ R iξ z ɛ ξ 2 dξ = û(ξ R iξ z dξ + ad lim u(x ɛ + (2ɛ /2 R x z 2 /4ɛ dx = (2π /2 u(z, usig (2.2. Thrfor, w coclud that (2π /2 u(z = R iξ z û(ξ dξ, which implis u(z = (2π /2 R iξ z û(ξ dξ = ˇû(z, as dsird. 23

24 2.4 Solvig th Hat Equatio i R 2.4. Usig th Fourir Trasform to Solv th Iitial-Valu Problm for th Hat Equatio i R Cosidr th iitial-valu problm for th hat quatio o R, { ut = ku xx x R, t > u(x, = φ(x. Applyig th Fourir trasform to th hat quatio, w hav û t (ξ, t = kû xx (ξ, t = û t = k(iξ 2 û = kξ 2 û Solvig this ODE ad usig th iitial coditio u(x, = φ(x, w hav û(ξ, t = φ(ξ kξ2t. Thrfor, u(x, t = (2π /2 = 2π = 2π = 2π = 2π R ix ξ û(ξ, t dξ ix ξ kξ φ(ξ 2t dξ [ ] ix ξ iy ξ φ(y dy kξ2t dξ 2π [ ] φ(y i(y x ξ kξ2t dξ dy 2π φ(y f(y x dy whr f(ξ = kξ2t. By Claim 7, f(ξ = ktξ2 implis f(z = 2kt z2 /4kt. Thrfor, f(y x = 2kt (x y2 /4kt, ad cosqutly, th solutio of th iitial-valu problm for th hat quatio o R is giv by u(x, t = φ(y (x y2 /4kt dy (2.7 4kπt for t >. W ca us a similar aalysis to solv th problm i highr dimsios. Cosidr th iitial-valu problm for th hat quatio i R, { ut = k u x R, t > (2.8 u(x, = φ(x. 24

25 Employig th Fourir trasform as i th -D cas, w arriv at th solutio formula for t >. u(x, t = φ(y (4kπt /2 R x y 2 /4kt dy (2.9 Rmark. Abov, w hav show that if thr xists a solutio u of th hat quatio (2.8, th u has th form (2.9. It rmais to vrify that this solutio formula actually satisfis th iitial-valu problm. I particular, lookig at th th formulas (2.7 ad (2.9, w s that th fuctios giv ar ot actually dfid at t =. Thrfor, to say that th solutio formulas actually satisfy th iitial-valu problms, w ma to say that lim t + u(x, t = φ(x. Thorm 9. (Rf: Evas, p. 47 Assum φ C(R L (R, ad dfi u by (2.9. Th. u C (R (,. 2. u t k u = for all x R, t >. 3. lim u(x, t = φ(x. (x,t (x, x,x R,t> x 2 t /2 4kt Proof.. Sic th fuctio is ifiitly diffrtiabl, with uiformly boudd drivativs of all ordrs o R [δ, for all δ >, w ar justifid i passig th drivativs isid th itgral isid ad s that u C (R (,. 2. By a straightforward calculatio, w s that th fuctio H(x, t x y 2 /4kt (4kπt /2 satisfis th hat quatio for all t >. Agai usig th fact that this fuctio is ifiitly diffrtiabl, w ca justify passig th drivativs isid th itgral ad coclud that u t (x, t k u(x, t = [(H R t k x H(x y, t]φ(y dy =. sic H(x, t solvs th hat quatio. 3. Fix a poit x R ad ɛ >. W d to show thr xists a δ > such that u(x, t φ(x < ɛ for (x, t (x, < δ. That is, w d to show (4πkt /2 R x y 2 /4kt φ(y dy φ(x < ɛ 25

26 for (x, t (x, < δ whr δ is chos sufficitly small. Th proof is similar to th proof of (2.2. I particular, usig (2.3, w writ (4πkt /2 R x y 2 /4kt φ(y dy φ(x (2.2 = (4πkt /2 R x y 2 /4kt [φ(y φ(x ] dy. Lt B(x, γ b th ball of radius γ about x. W look at th itgral i (2.2 ovr B(x, γ. Usig th fact that φ is cotiuous, w s that x y 2 /4kt [φ(y φ(x (4πkt /2 ] dy φ(y φ(x L (B(x,γ < ɛ 2 B(x,γ for γ chos sufficitly small. For this choic of γ, w look at th itgral i (2.2 ovr th complmt of B(x, γ. For y R B(x, γ, x x < γ, w hav 2 y x y x + x x < y x + γ 2 < y x + 2 y x. Thrfor, o this pic of th itgral, w hav y x < 2 y x. Thrfor, this pic of th itgral is boudd as follows, x y 2 /4kt [φ(y φ(x (4πkt /2 ] dy C x y 2 /6kt dy. t /2 R B(x,γ R B(x,γ W claim th right-had sid abov ca b mad arbitrarily small by takig t arbitrarily small. W prov this i th cas of =. Th proof for highr dimsios is similar. By makig a chag of variabls, w hav C t /2 R B(x,γ (x y 2 /6kt dy = C = C t /2 γ γ/ t y2 /6kt dỹ z2 dz as t +. I summary, tak γ sufficitly small such that th pic of th itgral i (2.2 ovr B(x, γ is boudd by ɛ/2. Th, choos δ < γ/2 sufficitly small such that Part (3 of th thorm follows. C γ t z2 dz < ɛ Th Fudamtal Solutio Cosidr agai th solutio formula (2.9 for th iitial-valu problm for th hat quatio i R. Dfi th fuctio /4kt H(x, t (4πkt /2 x 2 t > (2.2 t < 26

27 As ca b vrifid dirctly, H is a solutio of th hat quatio for t >. I additio, w ca writ th solutio of (2.8 as u(x, t = [H(t φ](x = H(x y, tφ(y dy. R As H has ths ic proprtis, w call H(x, t th fudamtal solutio of th hat quatio. Lt s look at th fudamtal solutio a littl mor closly. As mtiod abov, H itslf a solutio of th hat quatio. That is, H t k H = x R, t >. What kid of iitial coditios dos H satisfy? W otic that for x, lim t + H(x, t =. Howvr, for x =, lim t + H(x, t =. I additio, usig (2.3, w s that H(x, t dx = R for t >. Thrfor, lim t + R H(x, t dx =. What kid of fuctio satisfis ths proprtis? Wll, actually, o fuctio satisfis ths proprtis. Ituitivly, th ida is that a fuctio satisfyig ths proprtis rprsts a poit mass locatd at th origi. This objct is kow as a dlta fuctio. W mphasiz that th dlta fuctio is ot a fuctio! Istad, it is part of a group of objcts calld distributios which act o fuctios. W mak ths idas mor prcis i th xt sctio, ad th will rtur to discussig th fudamtal solutio of th hat quatio. 2.5 Distributios Rf: Strauss, Sctio Dfiitios ad Exampls W bgi by makig som dfiitios. W say a fuctio φ : R R has compact support if φ outsid a closd, boudd st i R. W say φ is a tst fuctio if φ is a ifiitly diffrtiabl fuctio with compact support. Lt D dot th st of all tst fuctios. W say F : D R is a distributio if F is a cotiuous, liar fuctioal which assigs a ral umbr to vry tst fuctio φ D. W lt (F, φ dot th ral umbr associatd with this distributio. Exampl. Lt g : R R b ay boudd fuctio. W dfi th distributio associatd with g as th map F g : D R which assigs to a tst fuctio φ th ral umbr That is, F g : φ g(xφ(x dx. (F g, φ = 27 g(xφ(x dx.

28 O particular xampl is th Havisid fuctio, dfid as { x H(x = x <. Th, th distributio F H associatd with th Havisid fuctio is th map which assigs to a tst fuctio φ th ral umbr (F H, φ = φ(x dx. That is, F H : φ φ(x dx. Exampl. Th Dlta fuctio δ (which is ot actually a fuctio! is th distributio δ : D R which assigs to a tst fuctio φ th ral umbr φ(. That is, Rmarks. (δ, φ = φ(. (a W somtims writ δ R (xφ(x dx = φ(, howvr, this is rathr iformal ad ot accurat bcaus δ (x is ot a fuctio! It should b thought of as purly otatioal. (b W ca talk about th dlta fuctio ctrd at a poit othr tha x = as follows. For a fixd y R, w dfi δ y : D R to b th distributio which assigs to a tst fuctio φ th ral umbr φ(y. That is, Drivativs of Distributios (δ y, φ = φ(y. W ow dfi drivativs of distributios. Lt F : D R b a distributio. W dfi th drivativ of th distributio F as th distributio G : D R such that (G, φ = (F, φ. for all φ D. W dot th drivativ of F by F. Th F is th distributio such that for all φ D. (F, φ = (F, φ 28

29 Exampl 2. For g a C fuctio, w dfi th distributio associatd with g as F g : D R such that (F g, φ = Thrfor, itgratig by parts, w hav (F g, φ = = g(xφ(x dx. g(xφ (x dx g (xφ(x dx. By dfiitio, th drivativ of F g, dotd F g is th distributio such that (F g, φ = (F g, φ for all φ D. Thrfor, F g is th distributio associatd with th fuctio g. That is, (F g, φ = (F g, φ = g (xφ(x dx. Exampl 3. Lt H b th Havisid fuctio dfid abov. Lt F H : D R b th distributio associatd with H, discussd abov. Th th drivativ of F H, dotd F H must satisfy (F H, φ = (F H, φ for all φ D. Now Thrfor, (F H, φ = = lim b b φ (x dx φ (x dx = lim b φ(x x=b x= = φ(. (F H, φ = (F H, φ = φ(. That is, th drivativ of th distributio associatd with th Havisid fuctio is th dlta fuctio. W will b usig this fact blow Covrgc of Distributios Lt F : D R b a squc of distributios. W say F covrgs wakly to F if for all φ D. (F, φ (F, φ 29

30 2.5.4 Th Fudamtal Solutio of th Hat Equatio Rvisitd I this sctio, w will show that th fudamtal solutio H of th hat quatio (2.2 ca b thought of as a solutio of th followig iitial-valu problm, { Ht k H = x R, t > (2.22 H(x, = δ. Now, first of all, w must b carful i what w ma by sayig that H(x, = δ. Clarly, th fuctio H dfid i (2.2 is ot v dfid at t =. Ad, ow w r askig that it quals a distributio? W d to mak this mor prcis. First of all, wh w writ H(x, = δ, w rally ma = i th ss of distributios. I additio, w rally ma to say that lim t + H(x, t = δ i th ss of distributios. Lt s stat this mor prcisly ow. Lt F H(t b th distributio associatd with H(t dfid by (F H(t, φ = H(x, tφ(x dx. R Now to say that lim t + H(x, t = δ i th ss of distributios, w ma that lim t +(F H(t, φ = (δ, φ = φ(. (2.23 Thrfor, to summariz, by sayig that H is a solutio of th iitial-valu problm (2.22, w rally ma that H is a solutio of th hat quatio for t > ad (2.23 holds. W ow prov that i fact, our fudamtal solutio dfid i (2.2 is a solutio of (2.22 i this ss. Showig that H satisfis th hat quatio for t > is a straightforward calculatio. Thrfor, w oly focus o showig that th iitial coditio is satisfid. I particular, w d to prov that (2.23 holds. W procd as follows. lim t +(F H(t, φ = lim H(x, tφ(x dx t R + = lim t + (4πkt /2 R x 2 /4kt φ(x dx. But, ow i (2.2, w provd that this limit is xactly φ(! Thrfor, w hav prov (2.23. Cosqutly, w ca thik of our fudamtal solutio as a solutio of (2.22. Th bauty of this formulatio is th followig. If H is a solutio of (2.22, th dfi u(x, t [H(t φ](x = H(x y, tφ(y dy. R Th ida is that u should b a solutio of (2.8. W giv th formal argumt blow. u(x, = [H( φ](x = H(x y, φ(y dy = φ(x. R I additio, u t k x u = H R t (x y, tφ(y dy k R x H(x y, tφ(y dy = [H R t (x y, t k x H(x y, t]φ(y dy =. 3

31 Not: Ths calculatios ar formal i th ss that w ar igorig covrgc issus, tc. To vrify that this formulatio actually givs you a solutio, you d to dal with ths issus. Of cours, for ic iitial data, w hav show that this formulatio dos giv us a solutio to th hat quatio. 2.6 Proprtis of th Hat Equatio 2.6. Ivariac Proprtis Cosidr th quatio u t = ku xx, < x <. Th quatio satisfis th followig ivariac proprtis, (a Th traslat u(x y, t of ay solutio u(x, t is aothr solutio for ay fixd y. (b Ay drivativ (u x, u t, u xx, tc. of a solutio is agai a solutio. (c A liar combiatio of solutios is agai a solutio. (d A itgral of a solutio is agai a solutio (assumig propr covrgc. ( If u(x, t is a solutio, so is th dilatd fuctio u( ax, at for ay a >. This ca b provd by th chai rul. Lt v(x, t = u( ax, at. Th v t = au t ad Ad, thrfor, v x = au x. v xx = a au xx = au xx Proprtis of Solutios. Smoothss of Solutios. As ca b s from th abov thorm, solutios of th hat quatio ar ifiitly diffrtiabl. Thrfor, v if thr ar sigularitis i th iitial data, thy ar istatly smoothd out ad th solutio u(x, t C (R (,. 2. Domai of Dpdc. Th valu of th solutio at th poit x, tim t dpds o th valu of th iitial data o th whol ral li. I othr words, thr is a ifiit domai of dpdc for solutios to th hat quatio. This is i cotrast to hyprbolic quatios whr solutios ar kow to hav fiit domais of dpdc. 3

32 2.7 Ihomogous Hat Equatio Rf: Strauss, Sc. 3.3; Evas, Sc c. I this sctio, w cosidr th iitial-valu problm for th ihomogous hat quatio o som domai Ω (ot cssarily boudd i R, { ut k u = f(x, t x Ω, t > (2.24 u(x, = φ(x. W claim that w ca us solutios of th homogous quatio to costruct solutios of th ihomogous quatio Motivatio Cosidr th followig ODE: { ut + au = f(t u( = φ, whr a is a costat. By multiplyig by th itgratig at, w s th solutio is giv by u(t = at φ + t a(t s f(s ds. I othr words, th solutio u(t is th propagator at applid to th iitial data, plus th propagator covolvd with th oliar trm. I othr words, if w lt S(t b th oprator which multiplis fuctios by at, w s that th solutio of th homogous problm, { ut + au = u( = φ is giv by S(tφ = at φ. Furthr, th solutio of th ihomogous problm is giv by u(t = S(tφ + t S(t sf(s ds. W claim that this sam tchiqu will allow us to fid a solutio of th ihomogous hat quatio. Big abl to costruct solutios of th ihomogous problm from solutios of th homogous problm is kow as Duhaml s pricipl. W show blow how this ida works. Suppos w ca solv th homogous problm, { ut k u = x Ω (2.25 u(x, = φ(x. That is, assum th solutio of (2.25 is giv by u h (x, t = S(tφ(x for som solutio oprator S(t. W claim that th solutio of th ihomogous problm (2.24 is giv by u(x, t = S(tφ(x + t 32 S(t sf(x, s ds.

33 At last formally, w s that u t k u = [ t k ]S(tφ(x + [ t k ] = + S(t tf(x, t + = S(f(x, t = f(x, t. t t S(t sf(x, s ds [ t k ]S(t sf(x, s ds Blow, w show how w us this ida to costruct solutios of th hat quatio o R ad o boudd domais Ω R Ihomogous Hat Equatio i R Cosidr th ihomogous problm, { ut k u = f(x, t x R, t > u(x, = φ(x. (2.26 From arlir, w kow th solutio of th corrspodig homogous iitial-valu problm { ut k u = x R, t > (2.27 u(x, = φ(x is giv by u h (x, t = H(x y, tφ(y dy. R That is, w ca thik of th solutio oprator S(t associatd with th hat quatio o R as dfid by S(tφ(x = H(x y, tφ(y dy. R Thrfor, w xpct th solutio of th ihomogous hat quatio to b giv by t u(x, t = S(tφ(x + S(t sf(x, s ds t = H(x y, tφ(y dy + R H(x y, t sf(y, s dy ds. R Now, w hav alrady show that u h (x, t = R H(x y, tφ(y dy satisfis (2.27, (at last for ic fuctios φ. Thrfor, it suffics to show that u p (x, t t R H(x y, t sf(y, s dy ds (2.28 satisfis (2.26 with zro iitial data. If w ca prov this, th u(x, t = u h (x, t + u p (x, t will clarly solv (2.26 with iitial data φ. I th followig thorm, w prov that u p satisfis th ihomogous hat quatio with zro iitial data. 33

34 Thorm 4. Assum f C 2 (R [, (maig f is twic cotiuously diffrtiabl i th spatial variabls ad oc cotiuously diffrtiabl i th tim variabl ad has compact support. Dfi u as i (2.28. Th. u C 2 (R (,. 2. u t (x, t k u(x, t = f(x, t for all x R, t >. 3. lim (x,t (x, x R,t> u(x, t = for all x R. Proof.. Sic H has a sigularity at (,, w caot justify passig th drivativs isid th itgral. Istad, w mak a chag of variabls as follows. I particular, lttig ỹ = x y ad s = t s, w hav t t H(x y, t sf(y, s dy ds = H(ỹ, sf(x ỹ, t s dỹ d s. R R For as of otatio, w drop th otatio. Now by assumptio f C 2 (R [, ad H(y, s is smooth ar s = t >. Thrfor, w hav t t H(y, sf(x y, t s dy ds = R ad t xi x j H(y, sf(x y, t s dy ds = R Thrfor, u C 2 (R (,. t t H(y, s R t f(x y, t s dy ds + H(y, tf(x y, dy R R H(y, s xi x j f(x y, t s dy ds. 2. Now w d to calculat u t k u. Usig th sam chag of variabls as abov, w hav t [ t k x ] H(y, sf(x y, t s dy ds R t = H(y, s[ R t k x ]f(x y, t s dy ds + H(y, tf(x y, dy R t = H(y, s[ R s k y ]f(x y, t s dy ds + H(y, tf(x y, dy. R Now, w would lik to itgrat by parts to put th drivativs o H as w kow H is a solutio of th hat quatio. Howvr, w kow H has a sigularity at s =. 34

35 To avoid this, w brak up th itgral ito th itrvals [, ɛ] ad [ɛ, t]. Thrfor, w writ t [ t k x ]u = H(y, s[ ɛ R s k y ]f(x y, t s dy ds ɛ + H(y, s[ R s k y ]f(x y, t s dy ds + H(y, tf(x y, dy R I ɛ + J ɛ + K. First, for J ɛ, w hav ɛ H(y, s[ R s k y]f(x y, t s dy ds ɛ ( f t L + k f L H(y, s dy ds R ɛc H(y, t dy Cɛ, R usig (2.3. For I ɛ, usig th assumptio that f has compact support, w itgrat by parts as follows, t H(y, s[ ɛ R s k y ]f(x y, t s dy ds t = [ ɛ R s k y ]H(y, sf(x y, t s dy ds s=t H(y, sf(x y, t s dy R s=ɛ = + H(y, ɛf(x y, t ɛ dy H(y, tf(x y, dy R R = H(y, ɛf(x y, t ɛ dy K. R Thrfor, I ɛ + K = H(y, ɛf(x y, t ɛ dy. R Now, u t k u = lim ɛ +[I ɛ + J ɛ + K] = lim H(y, ɛf(x y, t ɛ dy ɛ R + = lim ɛ + (4πkɛ /2 R y 2 /4kɛ f(x y, t ɛ dy = f(x, t, 35

36 usig th sam tchiqu w usd to prov part (3 of Thorm To prov that th limit as t + of our solutio u(x, t is, w us th fact that t u(x, t L (R = H(y, sf(x y, t s dy ds R t f L (R [,t] H(y, s dy ds R t C (4πks /2 R y 2 /4ks dy ds = Ct, usig (2.3. Thrfor, as t +, u(x, t, as claimd Ihomogous Hat Equatio o Boudd Domais I this sctio, w cosidr th iitial/boudary valu problm for th ihomogous hat quatio o a itrval I R, u t ku xx = f(x, t x I R, t > u(x, = φ(x x I u satisfis crtai BCs t >. Usig Duhaml s pricipl, w xpct th solutio to b giv by u(x, t = S(tφ(x + t S(t sf(x, s ds whr u h (x, t = S(tφ(x is th solutio of th homogous quatio ad S(t is th solutio oprator associatd with th homogous problm. As show arlir, th solutio of th homogous problm with symmtric boudary coditios, u t ku xx = x I, t > u(x, = φ(x x I u satisfis symmtric BCs t > is giv by u(x, t = A X (x kλ t = whr X ar th igfuctios ad λ th corrspodig igvalus of th igvalu problm, { X = λx x I ad th cofficits A ar dfid by X satisfis symmtric BCs, A = X, φ X, X, 36

37 whr th ir product is tak ovr I. Thrfor, th solutio oprator associatd with th homogous quatio is giv by whr S(tφ = A X (x kλ t = A = X, φ X, X,. Thrfor, w xpct th solutio of th ihomogous quatio to b giv by whr u(x, t = S(tφ(x + = t A X (x λt + = S(t sf(x, s ds t B (sx (x λ(t s ds = B (s X, f(s X, X. I fact, for ic fuctios φ ad f, this formula givs us a solutio of th ihomogous iitial/boudary-valu problm for th hat quatio. W omit proof of this fact hr. W cosidr a xampl. Exampl 5. Solv th ihomogous iitial/boudary valu problm for th hat quatio o [, l] with Dirichlt boudary coditios, u t ku xx = f(x, t x [, l], t > u(x, = φ(x x [, l] u(, t = = u(l, t t > Th solutio of th homogous problm with iitial data φ, u t ku xx = x [, l], t > u(x, = φ(x x [, l] u(, t = = u(l, t t > is giv by whr u h (x, t = A = 2 l = l ( π A si l x k(π/l2t, ( π si l x φ(x dx. (2.29 Thrfor, th solutio of th ihomogous problm with zro iitial data is giv by u p (x, t = t = ( π B (s si l x k(π/l2 (t s ds 37

38 whr B (s = 2 l l Cosqutly, th solutio of (2.29 is giv by u(x, t = = ( π t A si l x k(π/l2t + ( π si l x f(x, s dx. = ( π B (s si l x k(π/l2 (t s ds whr A ad B (s ar as dfid abov Ihomogous Boudary Data I this sctio w cosidr th cas of th hat quatio o a itrval with ihomogous boudary data. W will us th mthod of shiftig th data to rduc this problm to a ihomogous quatio with homogous boudary data. Cosidr th followig xampl. Exampl 6. Cosidr u t ku xx = u(x, = φ(x u(, t = g(t u(l, t = h(t. W itroduc a w fuctio U(x, t such that < x < l < x < l (2.3 U(x, t = l [(l xg(t + xh(t]. Assum u(x, t is a solutio of (2.3. Th lt v(x, t u(x, t U(x, t. Thrfor, v t kv xx = (u t ku xx (U t ku xx = U t = l [(l xg (t + xh (t]. Furthr, v(x, = u(x, U(x, = φ(x l v(, t = u(, t U(, t = v(l, t = u(l, t U(l, t =. [(l xg( + xh(] 38

39 Thrfor, v is a solutio of th ihomogous hat quatio o th itrval [, l] with homogous boudary data, v t kv xx = [(l l xg (t + xh (t] < x < l v(x, = φ(x [(l xg( + xh(] < x < l l v(, t = = v(l, t. W ca solv this problm for v usig th tchiqu of th prvious sctio. Th w ca solv our origial problm (2.3 usig th fact that u(x, t = v(x, t + U(x, t. 2.8 Maximum Pricipl ad Uiquss of Solutios I this sctio, w prov what is kow as th maximum pricipl for th hat quatio. W will th us this pricipl to prov uiquss of solutios to th iitial-valu problm for th hat quatio Maximum Pricipl for th Hat Equatio First, w prov th maximum pricipl for solutios of th hat quatio o boudd domais. Lt Ω R b a op, boudd st. W dfi th parabolic cylidr as Ω T Ω (, T ]. W dfi th parabolic boudary of Γ T as Γ T Ω T Ω T. t T Ω T Γ T Ω R W ow stat th maximum pricipl for solutios to th hat quatio. Thorm 7. (Maximum Pricipl o Boudd Domais (Rf: Evas, p. 54. Lt Ω b a op, boudd st i R. Lt Ω T ad Γ T b as dfid abov. Assum u is sufficitly smooth, (spcifically, assum u C 2 (Ω T C(Ω T ad u solvs th hat quatio i Ω T. Th,. max Ω T u(x, t = max u(x, t. (Wak maximum pricipl Γ T 39

40 2. If Ω is coctd ad thr xists a poit (x, t Ω T such that u(x, t = max u(x, t, Ω T th u is costat i Ω t. (Strog maximum pricipl Proof. (Rf: Strauss, p. 42 (wak; Evas, Sc (strog Hr, w will oly prov th wak maximum pricipl. Th radr is rfrrd to Evas for a proof of th strog maximum pricipl. W will prov th wak maximum pricipl i th cas = ad Ω = (, l. Lt M = max u(x, t. Γ T W must show that u(x, t M throughout Ω T. Lt ɛ > ad dfi v(x, t = u(x, t + ɛx 2. W claim that v(x, t M + ɛl 2 throughout Ω T. Assumig this, w ca coclud that u(x, t M + ɛ(l 2 x 2 o Ω T ad cosqutly, u(x, t M o Ω T. Thrfor, w oly d to show that v(x, t M + ɛl 2 throughout Ω T. By dfiitio, w kow v(x, t M + ɛl 2 o Γ T (that is, o th lis t =, x = ad x = l. W will prov that v caot attai its maximum i Ω T, ad, thrfor, v(x, t M + ɛl 2 throughout Ω T. W prov this as follows. First, by dfiitio of v, w s v satisfis v t kv xx = u t k(u + ɛx 2 xx = u t ku xx 2ɛk = 2ɛk <. (2.3 Iquality (2.3 is kow as th diffusio iquality. W will us (2.3 to prov that v caot achiv its maximum i Ω T. Suppos v attais its maximum at a poit (x, t Ω T such that < x < l, < t < T. This would imply that v t (x, t = ad v xx (x, t, ad, cosqutly that v t (x, t kv xx (x, t, which cotradicts th diffusio iquality (2.3. Thrfor, v caot attai its maximum at such a poit. Suppos v attais its maximum at a poit (x, T o th top dg of Ω T. Th v x (x, T = ad v xx (x, T, as bfor. Furthrmor, v t (x, T = lim δ + v(x, T v(x, T δ δ Agai, this cotradicts th diffusio iquality. But v must hav a maximum somwhr i th closd rctagl Ω T. Thrfor, this maximum must occur somwhr o th parabolic boudary Γ T. Thrfor, v(x, t M + ɛl 2 throughout Ω T, ad, w ca coclud that u(x, t M throughout Ω T, as dsird.. 4

41 W ow prov a maximum pricipl for solutios to th hat quatio o all of R. Without havig a boudary coditio, w d to impos som growth assumptios o th bhavior of th solutios as x +. Thorm 8. (Maximum Pricipl o R (Rf: Evas, p. 57 Suppos u (sufficitly smooth solvs { ut k u = x R (, T u = φ ad u satisfis th growth stimat u(x, t A a x 2 x R x R, t T for costats A, a >. Th sup u(x, t = sup φ(x, t. R [,T ] R Proof. (For simplicity, w tak k =, but th proof works for arbitrary k >. Assum Thrfor, thr xists ɛ > such that Fix y R ad µ >. Lt v(x, t u(x, t 4aT <. 4a(T + ɛ <. µ (T + ɛ t x y 2 4(T +ɛ t /2. Usig th fact that u is a solutio of th hat quatio, it is straightforward to show that v is a solutio of th hat quatio, v t v =, x R (, T ]. Fix r > ad lt U B(y, r, U T = B(y, r (, T ]. From th maximum pricipl, w kow max U T v(x, t = max v(x, t. Γ T W will ow show that max ΓT v(x, t φ(x. First, look at (x, t o th bas of Γ T. That is, tak (x, t = (x,. Th v(x, = u(x, µ (T + ɛ u(x, = φ(x. x y 2 4(T +ɛ /2 4

42 Now tak (x, t o th sids of Γ T. That is, choos x R such that x y = r ad t such that t T. Th µ v(x, t = u(x, t (T + ɛ t r 2 /2 µ A a x 2 (T + ɛ t r 2 /2 µ A a( y +r2 (T + ɛ r 2 /2 4(T +ɛ t 4(T +ɛ t 4(T +ɛ. By assumptio, 4a(T + ɛ <. Thrfor, /4(T + ɛ = a + γ for som γ >. Thrfor, v(x, t A a( y +r2 µ(4(a + γ /2 (a+γr2 sup φ(x R for r sufficitly larg. Thrfor, w hav show that v(x, t sup R φ(x for (x, t Γ T, ad, thus, by th maximum pricipl max U T v(x, t = max v(x, t, Γ T w hav v(x, t sup R φ(x for all (x, t U T. W ca us this sam argumt for all y R, to coclud that v(y, t sup φ(x R for all y R, t T as log as 4aT <. Th usig th dfiitio of v ad takig th limit as µ +, w coclud that u(y, t sup φ(x R for all y R, t T as log as 4aT <. If 4aT, w divid th itrval [, T ] ito subitrvals [, T ], [T, 2T ], tc., whr T = ad prform th sam calculatios o ach of ths subitrvals. 8a Usig th Maximum Pricipl to Prov Uiquss W ow us th maximum pricipls prov abov to prov uiquss of solutios to th hat quatio. Thorm 9. (Uiquss o boudd domais Lt Ω b a op, boudd st i R ad dfi Ω T ad Γ T as abov. Cosidr th iitial/boudary valu problm, { ut k u = f x Ω T u = g x Γ T (2.32 Assum f ad g ar cotiuous. Th thr xists at most o (smooth solutio of (

43 Proof. Suppos thr xist two smooth solutios u ad v. Lt w u v. Th w is a solutio of { wt k w = x Ω T w = x Γ T. By th wak maximum pricipl, (2.33 max w(x, t = max w(x, t =. Ω T Γ T Thrfor, w(x, t o Ω T, which implis u(x, t v(x, t o Ω T. Nxt, lt w = v u. Thrfor, w is also a solutio of (2.33, ad, by th wak maximum pricipl, max Ω T w(x, t = max Γ T w(x, t =. Cosqutly, w(x, t o Ω T, which implis v(x, t u(x, t o Ω T. Thrfor, w hav u(x, t v(x, t (x, t Ω T v(x, t u(x, t (x, t Ω T. Cosqutly, w coclud that u = v for i Ω T, as dsird. Thorm 2. (Uiquss o R (Rf: Evas, p. 58. Cosidr th iitial-valu problm { ut k u = f x R (, T u(x, = g x R (2.34. Assum f ad g ar cotiuous. Th thr xists at most o (smooth solutio of (2.34 satisfyig th growth stimat u(x, t A a x 2. Proof. Assum u, v ar two solutios. Lt w u v ad lt w = v u. Now apply th maximum pricipl o R to show that u v. Rmark. Thr ar, i fact, ifiitly may solutios of { ut k u = x R (, T u(x, = x R. (Rf: Joh, Chaptr 7. All of thos solutios othr tha u(x, t grow fastr tha x 2 as x +. 43

44 2.8.3 Ergy Mthods to prov Uiquss W ow prst a altrat tchiqu to prov uiquss of solutios to th hat quatio o boudd domais. Thorm 2. Lt Ω b a op, boudd st i R. Lt T >. Lt Ω T, Γ T b th parabolic cylidr ad parabolic boudary dfid arlir. Cosidr th iitial/boudary valu problm, { ut k u = f (x, t Ω T (2.35 u(x, t = g (x, t Γ T. Thr xists at most o (smooth solutio u of (2.35. Proof. Suppos thr xist two solutios u ad v. Lt w = u v. Th w solvs { wt k w = (x, t Ω T w = (x, t Γ T. (2.36 Lt E(t Ω w 2 (x, t dx. Now, E( = Ω w2 (x, dx =. W claim that E (t ad, thrfor, E(t = for t T. By usig th fact that w is a solutio of (2.36 ad itgratig by parts, w s that E (t = 2 ww t dx Ω = 2k w w dx Ω = 2k = 2k Ω Ω w 2 dx + 2k w 2 dx, Ω w w ν ds(x usig th fact that w = o Γ T ad k >. Thrfor, E(t = for t T. Usig th assumptio that w is smooth, this implis w i Ω T, ad, thrfor, u v i Ω T. 44