Modelli Clamfim Equazione del Calore Lezione ottobre 2014

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1 CLAMFIM Bologna Modell Clamfm Equazone del Calore Lezone ottobre 2014 professor Danele Rtell danele.rtell@unbo.t 1/24?

2 Convoluton The convoluton of two functons g(t) and f(t) s the functon (g f)(t) = + g(t x)f(x)dx 2/24?

3 Eserczo Se Dmostrare che 0 se t 0 f(t) = e t se t > 0 f f(t) = te t 3/24?

4 Eserczo Se Dmostrare che Eserczo Se a > 0 Dmostrare che 0 se t 0 f(t) = e t se t > 0 f f(t) = te t e ta se t 0 f(t) = 0 se t < 0 f f(t) = te at 3/24?

5 Eserczo Se a > 0, b > 0, a b e ta se t 0 f(t) = 0 se t < 0 g(t) = e tb se t 0 0 se t < 0 Dmostrare che f g(t) = e at e bt b a 4/24?

6 Eserczo Calcolare, se a > 0 la trasformata d Fourer ˆf(s) d 1 t < a f(t) = 0 t > a 5/24?

7 Eserczo Calcolare, se a > 0 la trasformata d Fourer ˆf(s) d 1 t < a f(t) = 0 t > a Calcolare la trasformata d Fourer ˆf(s) d 1 t 2 t < 1 f(t) = 0 t > 1 5/24?

8 artal dfferental equatons A partal dfferental equaton (DE) s a relaton that nvolves partal dervatves of an unknown functon. Let the unknown functon be u, and x, y, z,... be ndependent varables,.e., u = u(x, y, z,...). Often, one of these varables represents the tme. F (x, y, z, u, u x, u y, u z, u xx, u xy,, u xxx, ) = 0. (pde) 6/24?

9 artal dfferental equatons A partal dfferental equaton (DE) s a relaton that nvolves partal dervatves of an unknown functon. Let the unknown functon be u, and x, y, z,... be ndependent varables,.e., u = u(x, y, z,...). Often, one of these varables represents the tme. F (x, y, z, u, u x, u y, u z, u xx, u xy,, u xxx, ) = 0. (pde) We have used the subscrpt notaton for the partal dfferentaton u x = u x, u xy = 2 u x y, We wll always assume that the unknown functon u s suffcently well behaved so that all necessary partal dervatves exst and correspondng mxed partal dervatves are equal 6/24?

10 As n the case of ordnary dfferental equatons, we defne the order of (pde) to be the hghest order partal dervatve appearng n the equaton. Furthermore, we say that (pde) s lnear f F s lnear as a functon of the varables u, u x, u y, u z, u xx,,.e., F s a lnear combnaton of the unknown functon and ts dervatves. 7/24?

11 As n the case of ordnary dfferental equatons, we defne the order of (pde) to be the hghest order partal dervatve appearng n the equaton. Furthermore, we say that (pde) s lnear f F s lnear as a functon of the varables u, u x, u y, u z, u xx,,.e., F s a lnear combnaton of the unknown functon and ts dervatves. The followng are examples of partal DEs: u x + u y = 3u z 2x 2 5z u xx + u y = x 2 frst order lnear second order lnear 7/24?

12 By a soluton of (pde) we mean a contnuous functon u = u(x, y, z, ), wth contnuous partal dervatves, whch, when substtuted n (pde), reduces equaton (pde) to an dentty. For nstance u(x, t) = xe x t solves u t = u xx 2u x 8/24?

13 By a soluton of (pde) we mean a contnuous functon u = u(x, y, z, ), wth contnuous partal dervatves, whch, when substtuted n (pde), reduces equaton (pde) to an dentty. For nstance u(x, t) = xe x t solves u t = u xx 2u x Example For the frst-order partal DE for the unknown u = u(x, y) u x + u y = 0 t s possble to show that u = φ(x y) where φ s any functon havng contnuous frst-order partal dervatves s a soluton. Indeed, snce u x = φ (x y) and u y = φ (x y) t mmedately follows that u x + u y = φ (x y) φ (x y) = 0 8/24?

14 The heat equaton Besdes ths equaton arses n Mathematcal hyscs t s of nterest n Mathematcal Fnance also. We deal wth the partal dfferental equaton u t (x, t) = u xx (x, t) for t > 0, x R (H) u(x, 0) = f(x) for t = 0, x R we assume u and u x fnte as x, t > 0 and both f(x) and u(x, t) are defned for < x < +. 9/24?

15 The heat equaton Besdes ths equaton arses n Mathematcal hyscs t s of nterest n Mathematcal Fnance also. We deal wth the partal dfferental equaton u t (x, t) = u xx (x, t) for t > 0, x R (H) u(x, 0) = f(x) for t = 0, x R we assume u and u x fnte as x, t > 0 and both f(x) and u(x, t) are defned for < x < +. Knowng the Fourer transform of the Gaussan s essental for the treatment we are about to gve. The dea s to take the Fourer transform of both sdes of the heat equaton, wth respect to x thnkng t as a parameter 9/24?

16 Theorem Soluton to (H) s gven by the formula u(x, t) = 1 + ( (x y) 2) exp f(y)dy 4π t 4t 10/24?

17 Theorem Soluton to (H) s gven by the formula u(x, t) = 1 + ( (x y) 2) exp 4π t 4t or, ntroducng the heat kernel H(x, t) = 1 4π t e x2 /4t f(y)dy by the formula u(x, t) = H(x y, t) f(y)dy 10/24?

18 The heat equaton u t (x, t) = u xx (x, t) for t > 0, x R u(x, 0) = f(x) for t = 0, x R (H) Theorem Soluton to (H) s gven by the formula u(x, t) = 1 + ( (x y) 2) exp f(y)dy 4π t 4t 11/24?

19 or, ntroducng the heat kernel H(x, t) = 1 4π t e x2 /4t by the formula u(x, t) = H(x y, t) f(y)dy 12/24?

20 The Fourer transform of the rght hand sde of the equaton (H), u xx (x, t), s Fu xx (s, t) = (2π s) 2 Fu(s, t) = 4π 2 s 2 Fu(s, t) 13/24?

21 The Fourer transform of the rght hand sde of the equaton (H), u xx (x, t), s Fu xx (s, t) = (2π s) 2 Fu(s, t) = 4π 2 s 2 Fu(s, t) For the left hand sde, u t (x, t), we do somethng dfferent. We have 13/24?

22 The Fourer transform of the rght hand sde of the equaton (H), u xx (x, t), s Fu xx (s, t) = (2π s) 2 Fu(s, t) = 4π 2 s 2 Fu(s, t) For the left hand sde, u t (x, t), we do somethng dfferent. We have Fu t (s, t) = + u t (x, t)e 2π sx dx 13/24?

23 The Fourer transform of the rght hand sde of the equaton (H), u xx (x, t), s Fu xx (s, t) = (2π s) 2 Fu(s, t) = 4π 2 s 2 Fu(s, t) For the left hand sde, u t (x, t), we do somethng dfferent. We have Fu t (s, t) = = t + + u t (x, t)e 2π sx dx u(x, t)e 2π sx dx 13/24?

24 The Fourer transform of the rght hand sde of the equaton (H), u xx (x, t), s Fu xx (s, t) = (2π s) 2 Fu(s, t) = 4π 2 s 2 Fu(s, t) For the left hand sde, u t (x, t), we do somethng dfferent. We have Fu t (s, t) = = t + + = Fu(s, t) t u t (x, t)e 2π sx dx u(x, t)e 2π sx dx 13/24?

25 Thus takng the Fourer transform (wth respect to x) of both sdes of the equaton u t (x, t) = u xx (x, t) leads to t Fu(s, t) = 4π2 s 2 Fu(s, t) 14/24?

26 Thus takng the Fourer transform (wth respect to x) of both sdes of the equaton u t (x, t) = u xx (x, t) leads to t Fu(s, t) = 4π2 s 2 Fu(s, t) Ths s a dfferental equaton n t, an ordnary dfferental equaton, despte the partal dervatve symbol, and we can solve t: Fu(s, t) = Fu(s, 0) e 4π2 s 2 t 14/24?

27 Thus takng the Fourer transform (wth respect to x) of both sdes of the equaton u t (x, t) = u xx (x, t) leads to t Fu(s, t) = 4π2 s 2 Fu(s, t) Ths s a dfferental equaton n t, an ordnary dfferental equaton, despte the partal dervatve symbol, and we can solve t: Fu(s, t) = Fu(s, 0) e 4π2 s 2 t What s the ntal condton Fu(s, 0)? Fu(s, 0) = + u(x, 0)e 2π sx dx = + f(x)e 2π sx dx = Ff(s) 14/24?

28 uttng t all together Fu(s, t) = Ff(s) e 4π2 s 2 t 15/24?

29 uttng t all together Fu(s, t) = Ff(s) e 4π2 s 2 t We recognze that the exponental factor on the rght hand sde s the Fourer transform of the heat (Gaussan) hernel H(x, t) = 1 ( ) x 2 exp 4π t 4t snce for α > 0 f(x) = e αx2 ˆf(s) = π α e π2 α s2 15/24?

30 We then have a product of two Fourer transforms Fu(s, t) = Ff(s) FH(s, t) and we nvert ths to obtan a convoluton n the x doman u(x, t) = H(x, t) f(x) or, wrtten out u(x, t) = 1 4π t + exp ( (x y) 2) f(y)dy 4t 16/24?

31 We then have a product of two Fourer transforms Fu(s, t) = Ff(s) FH(s, t) and we nvert ths to obtan a convoluton n the x doman u(x, t) = H(x, t) f(x) or, wrtten out u(x, t) = 1 4π t + ( (x y) 2) exp f(y)dy 4t The functon H(x, t) s also called Green s functon, or fundamental soluton for the heat equaton. 16/24?

32 Ths technque apples also to the problem u t (x, t) = cu xx (x, t) for t > 0, x R u(x, 0) = f(x) for t = 0, x R (H c ) where u and u x fnte as x, t > 0 17/24?

33 Ths technque apples also to the problem u t (x, t) = cu xx (x, t) for t > 0, x R u(x, 0) = f(x) for t = 0, x R (H c ) where u and u x fnte as x, t > 0 whose soluton s 1 + ( (x y) 2) u(x, t) = exp f(y)dy 4π ct 4ct 17/24?

34 Ths s not surprsng snce, f we assume that u(x, t) solves u t = cu xx, f we defne w(x, t) = u(x, t/c) then w s soluton of w t = w xx. In fact 18/24?

35 Ths s not surprsng snce, f we assume that u(x, t) solves u t = cu xx, f we defne w(x, t) = u(x, t/c) then w s soluton of w t = w xx. In fact w t (x, t) = 1 c u t(x, t/c) 18/24?

36 Ths s not surprsng snce, f we assume that u(x, t) solves u t = cu xx, f we defne w(x, t) = u(x, t/c) then w s soluton of w t = w xx. In fact w t (x, t) = 1 c u t(x, t/c) = 1 c c u xx(x, t/c) 18/24?

37 Ths s not surprsng snce, f we assume that u(x, t) solves u t = cu xx, f we defne w(x, t) = u(x, t/c) then w s soluton of w t = w xx. In fact w t (x, t) = 1 c u t(x, t/c) = 1 c c u xx(x, t/c) = u xx (x, t/c) 18/24?

38 Ths s not surprsng snce, f we assume that u(x, t) solves u t = cu xx, f we defne w(x, t) = u(x, t/c) then w s soluton of w t = w xx. In fact w t (x, t) = 1 c u t(x, t/c) = 1 c c u xx(x, t/c) = u xx (x, t/c) = w xx (x, t) 18/24?

39 Ths s not surprsng snce, f we assume that u(x, t) solves u t = cu xx, f we defne w(x, t) = u(x, t/c) then w s soluton of w t = w xx. In fact w t (x, t) = 1 c u t(x, t/c) = 1 c c u xx(x, t/c) = u xx (x, t/c) = w xx (x, t) That s we can assume wthout loss of generalty c = 1 and concern only wth u t = u xx 18/24?

40 Remark The heat kernel H(x, t) s defned for t > 0 only and s an odd functon of x Fgure 1: Graph of the heat kernel for dfferent values of t 19/24?

41 ropertes of H ) H t (x, t) = H xx (x, t) for each t > 0, x R ) ) lm t 0 + H(x, t)dx = 1 for each t > 0 H(x, t) = 0 x 0 + x = 0 20/24?

42 ropertes of H ) H t (x, t) = H xx (x, t) for each t > 0, x R ) ) lm t 0 + H(x, t)dx = 1 for each t > 0 H(x, t) = 0 x 0 + x = 0 ) stems from a drect calculaton. ) follows from the change of varable q = x = dx = 4t dq whch mples 4t + H(x, t) dx = 1 π + e q2 dq = 1 20/24?

43 Eventually to obtan ) when x 0 we change varable puttng s = 1 t and we use Hosptal rule 21/24?

44 Eventually to obtan ) when x 0 we change varable puttng s = 1 t and we use Hosptal rule 1 lm e x2 /4t t 0 + 4π t 21/24?

45 Eventually to obtan ) when x 0 we change varable puttng s = 1 t and we use Hosptal rule 1 s lm e x2 /4t = lm t 0 + 4π t s + 4π e sx 2 /4 21/24?

46 Eventually to obtan ) when x 0 we change varable puttng s = 1 t and we use Hosptal rule 1 s lm e x2 /4t = lm = lm 1 t 0 + 4π t s + 4π e sx 2 /4 s + x 2 sπ e = 0 sx2 /4 21/24?

47 Eventually to obtan ) when x 0 we change varable puttng s = 1 t and we use Hosptal rule 1 s lm e x2 /4t = lm = lm 1 t 0 + 4π t s + 4π e sx 2 /4 s + x 2 sπ e = 0 sx2 /4 whle for x = 0 ) s obvous 21/24?

48 Remark Usng the change of varable y = x+2s t = dy = 2 t ds we can wrte soluton of (H) as u(x, t) = 1 π e s2 f(x + 2s t)ds (H s ) 22/24?

49 Exercse rove, usng (H s ) that u(x, t) = x 2 + 2t solves u t = u xx x R, t > 0 u(x, 0) = x 2 x R (pb1) 23/24?

50 Exercse rove, usng (H s ) that u(x, t) = x 2 + 2t solves u t = u xx x R, t > 0 u(x, 0) = x 2 x R (pb1) From (H s ) we can wrte u(x, t) = 1 π e s2 ( x + 2s t) 2 ds 23/24?

51 Exercse rove, usng (H s ) that u(x, t) = x 2 + 2t solves u t = u xx x R, t > 0 u(x, 0) = x 2 x R (pb1) From (H s ) we can wrte Now u(x, t) = 1 π e s2 ( x + 2s t) 2 ds ( e s2 x + 2s 2 t) ( = e s 2 x 2 + 4xs ) t + 4s 2 t 23/24?

52 Exercse rove, usng (H s ) that u(x, t) = x 2 + 2t solves u t = u xx x R, t > 0 u(x, 0) = x 2 x R (pb1) From (H s ) we can wrte Now u(x, t) = 1 π e s2 ( x + 2s t) 2 ds ( e s2 x + 2s 2 t) ( = e s 2 x 2 + 4xs ) t + 4s 2 t Observe that s 4xs t s an odd functon of s, so that 23/24?

53 u(x, t) = 1 π e s2 ( x 2 + 4s 2 t ) ds 24/24?

54 u(x, t) = 1 π = x2 π e s2 ( x 2 + 4s 2 t ) ds e s2 ds + 4t π s 2 e s2 ds 24/24?

55 u(x, t) = 1 π = x2 π = x 2 + 4t π c e s2 ( x 2 + 4s 2 t ) ds e s2 ds + 4t π s 2 e s2 ds 24/24?

56 u(x, t) = 1 π = x2 π = x 2 + 4t π c e s2 ( x 2 + 4s 2 t ) ds e s2 ds + 4t π s 2 e s2 ds To fnd the value of the constant c we can mpose that the found functons solves (pb1) 24/24?

57 u(x, t) = 1 π = x2 π = x 2 + 4t π c e s2 ( x 2 + 4s 2 t ) ds e s2 ds + 4t π s 2 e s2 ds To fnd the value of the constant c we can mpose that the found functons solves (pb1) It s u t = 4 π c, u xx = 2 24/24?

58 u(x, t) = 1 π = x2 π = x 2 + 4t π c e s2 ( x 2 + 4s 2 t ) ds e s2 ds + 4t π s 2 e s2 ds To fnd the value of the constant c we can mpose that the found functons solves (pb1) It s u t = 4 π c, u xx = 2 then c = π 2 = s 2 e s2 ds 24/24?

59 u(x, t) = 1 π = x2 π = x 2 + 4t π c e s2 ( x 2 + 4s 2 t ) ds e s2 ds + 4t π 24/24? s 2 e s2 ds To fnd the value of the constant c we can mpose that the found functons solves (pb1) It s u t = 4 π c, u xx = 2 then c = π 2 = s 2 e s2 ds Concluson: soluton to (pb1) s u(x, t) = x 2 + 4t π π 2 = x2 + 2t

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