Heat/Di usion Equation. 2 = 0 k constant w(x; 0) = '(x) initial condition. ( w2 2 ) t (kww x ) x + k(w x ) 2 dx. (w x ) 2 dx 0.
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1 Hat/Di k constant w(x; ) '(x) initial condition w(; t) w(l; t) boundary conditions Enrgy stimat: So w(w t kw xx ) ( w ) t (kww x ) x + k(w x ) or and thrfor E(t) R l Z l Z l ( w ) t (kww x ) x + k(w x ) dx ( w ) tdx kww x j l + k Z d l w dt dx k Z l Z l (w x ) dx (w x ) dx w dx is a non-incrasing function of tim or quivalntly E(t) E For th initial valu roblm if '(x) thn E() and E(t) i.. uniqunss. Stability Considr u t ku xx u(x; ) '(x) v t kv xx v(x; ) (x) Lt w(x; t) u(x; t) v(x; t) w(x; ) u(x; ) v(x; ) '(x) (x)
2 From th nrgy stimat w gt Z l [u(x; t) Z l v(x; t)] dx ['(x) (x; t)] dx So small changs in ' causs small changs in v in th L norm. Th maximum rincil will giv us stability in th maximum norm. Maximum Princil + X u a ij (x; t) + j i b i i + c(x; t)u f Strong vrsion: Assum Lu, M max(u). Assum u M at an intrior oint (x ; t ) and on of th following is tru. c and M arbitrary. c and M 3. M and c arbitrary Thn u M vrywhr in th domain (i.. u is constant). Wak Vrsion: If Lu and c(x; t) thn th maxiumum occurs initially or on th boundary. So if juj < M at t ; x ; x l thn u < M For a minimum rincil Proof (wak vrsion in on dimnsion) Intuitivly at a maximum th rst drivativs ar zro < so w hav a contradiction. Th roblm is that w may hav a saddl oint So instad w suly a rturbation argumnt. D n: v(x; t) u(x; t) + x > Wish: v(x; t) < M + l with M max ju(x; t)j so u(x; t) < M + (l x )
3 Formally: v(x; t) < M + l on t and x and x l Also () v t kv xx u t k(u + x ) xx u t ku xx k < Assum thr is a maximum at (x ; t ) in th intrior. At such a maximum v t and v xx. Hnc, v t kv xx at (x ; t ). This contradicts (*). Hnc, v(x; t) < M + l. Not that low ordr trms dstroy this rorty. Considr u t u xx + au a > u(x; ) sin x u(; t) u(; t) has a solution (a )t sin(x) which grows in tim if a >. Lmma: I Z x dx roof: On roof is by a comlx intgral and Cauchy s Thorm. Anothr roof is I Z Z Z Z Z (x +y ) dxdy r x + y x r cos y sin r rdrd d dr Z r rdr r dr r j 3
4 Prortis of hat u(x y; t) is a solution (shift). Any drivativ of u is a solution 3. Any linar combinations of solutions is a solution 4. Th intgral R S(x y; t)g(y)dy is a solution 5. u( ax; at) a > is a solution 6. uu() 4k x t. Thn u( ax; at) 4k ax at u(x; t). 4
5 @u u(x; ) (x) roof of #4: If S(x,t) is a solution of (*) thn by linarity so is v(x; t) Z S(x y; t)g(y)dy Thn v t v xx v t kv xx Z Z Z S t (x y; t)g(y)dy S xx (x y; t)g(y)dy [S t (x y; t) ks xx (x y; t)] g(y)dy roof of (5) v u( ax; at) v t au t ( ax; at) v xx au xx ( ax; at) v t kv xx a(u t ku xx ) 5
6 D n Q t kq xx ( Q(x; ) x > Q(x; ) x < Assum Q(x; t) g() x thn: So Thrfor Q t x g () t Q x g () Q xx g () Q t kq xx t g () + g () W now tak th limit as t #. D n 8 So g () C Q(x; t) g() C Thn x > Q(x; ) C >< x < Q(x; ) C >: Z g () Z x Z t g () 4 g () d + C d + C C + C d + C C C C + C Thrfor Q(x; t) + x Z d 6
7 Error Function Rmmbr Z d Z d Z d D n xz rf(x) xz d rf c(x) rf(x) rf() rf() rf( x) rf(x) d Z d + xz d + rf(x) Thn + rf Q(x; t) x 7
8 D n Thn S is also a solution and Z S(x y; t)'(y)dy is also a solution. Dos it satisfy th initial condition u(x; ) '(x)? Z Z Z S(x y; t)'(y)dy (x y; (y)dy Q(x y; t)'(y)j Assum ' is zro at in nity thn Z Q(x (y)dy Using ( Q(s; ) s > Q(s; ) s < W hav u(x; ) Z xz Q(x @y (y)dy 'jx '(x) 8
9 Conclusion Z x (x y) '(y)dy Z S(x; t)dx Z lim S(x; t) (x) t d 9
10 Imortanc of Limiting Procss Thn a ossibl solution is For x xd and t! thn @ u(x; ) 4 t x 4t 3 Howvr, for x 4t constant and t! th solution is not boundd!! Hnc, this is not a lgitimat solution
11 Main Thorm Thorm: u(x; ) '(x) j(x)j < Thn Z (x y) '(y)dy uc < x < ; < t < ach drivativ of u(x; t) satis s () lim t# '(x) roof S(z; t) Z S(x y; t)'(y)dy z Z S(z; t)'(x z)dz Lt z kt z kt. Thn 4 xz 4 '(x kt)d Assum j'(x)j M. Thn ju(x; t)j M Z 4 4 d M
12 Di rntiating Z const (x y; t)'(y)dy Z const t M 4 '(x kt)d Z 4 d const t M z kt z '(x z)dz Similarly Rmmbring n n const M t Z Z n 4 S(x; t)dx w gt d const t M u(x; t) '(x) Z 4 Z S(x y; t) ['(y) '(x)] dy 4 W assum that '(x) is continuous. Hnc, j'(y) '(x)j " whn jx yj. So ju(x; t) '(x)j boundd by kt Z h '(x i kt) '(x) d :::d + kt {z } " Z :::d jj> kt {z } " bcaus is small t is small So j'(y) '(x)j "!
13 Backward This givs growth in tim and is not wll osd. Equivalntly w ar trying to solv th hat quation backward in tim, i.. givn distribution of tmratur at tim T nd original tmratur. Considr u n (x; ) sin(nx)! as n! n u n (x; t) n sin(nx)n kt is unboundd Black Schols s V ss + rsv s rv + V t smarkt valu of asst bing otiond ttim constant volatility rconstant intrst rat Givn th nal valu w wish to dtrmin how to ric it initially. So w hav backward roblm! - but w hav additional minus sign ) wll osd 3
14 Exrcis Thn ( '(x; ) '(x; ) jxj < l jxj > l Z Zl l (x y) '(y)dy (x y) dy x l Z x+l q dq q x y 4 x l Z x+l x + l rf q dq 4 x+l Z x l x l rf q dq Not: Solution is di rnt from zro for all x whn t >. 4
15 Lowr Ordr Trms u t ku xx u(x; ) '(x) u Lt v(x; t) t u(x; t) so t v(x; t) v(x; ) '(x) So t v t v t v Hnc, v t kv xx v(x; ) '(x) v(x; t) Z Z t (x y) '(y)dy (x y) '(y)dy 5
16 Convction - Di usion u t + cu x ku xx u(x; ) '(x) Lt v(x; t) u(x + ct; t) or v(x ct; t) v(y; ) {z } y Thn using th c So c @ Thrfor v(x; t) Z (x y) '(y)dy Z (x ct y) '(y)dy For xaml choos Thn 8 < Z : '(x) ( x < jxj > Z (x ct y) '(y)dy + (x ct y) 9 '(y)dy ; 6
17 D n x ct y. Thn y + x ct. 8 >< + x ct Z Z 9 > dy + dy+ >: x ct >; 8 3 >< Z x ct Z Z 6B 4@ + C A >: dy B 4@ + rf( x ct ) x ct Z C A dy 39 > 7 5 >; 7
18 Comarison: Wav Equation Di usion Equation Prorty Wav Di usion sd nit in nit singularity along charactristics disaar wll osd t > X X wll osd t < X X maximum rincil X X Enrgy consrvd dcays information transortd dcays 8
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