1 General boundary conditions in diffusion

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1 Gnral boundary conditions in diffusion Πρόβλημα 4.8 : Δίνεται μονοδιάτατη πλάκα πάχους, που το ένα άκρο της κρατιέται ε θερμοκραία T t και το άλλο ε θερμοκραία T 2 t. Αν η αρχική θερμοκραία της πλάκας ε κάποια δεδομένη χρονική τιγμή δίνεται από την υνάρτηη f για,, να βρείτε την κατανομή της θερμοκραίας για όλες τις μελλοντικές χρονικές τιγμές. Solution: Th tmpratur u,t must satisfy th on-dimnsional diffusion quation with boundary conditions and initial condition u t = 2 u. t 2 u, t = T t, u, t = T 2 t.2 u, = f. In ordr for th initial and boundary conditions to b compatibl w must hav f = T, f = T 2.4 In this cas as th boundary tmpraturs vary, w do not pct any particular quilibrium distribution. W can howvr try a linar solution as in th constant tmpratur cas plus a solution that vanishs at th boundaris. Th part of th solution that vanish at th boundaris will b as usual pandd in th ignfunctions W ar thrfor ld to try a solution of th form u, t = T t + T 2 t T t + c n t Not that this solution automatically satisfis th boundary conditions in.2. It rmains to ask that it satisfis th quation. and thn vntually impos th initial condition.. Substituting.5 into. w obtain.5 t + T 2 t T t + ċ n t = 2 cn t.6 whr ovrdots stand for drivativs with rspct to tim..6 can b rwrittn as 2 ċ n t + cn t = T t T 2 t T t.7

2 Th lft-hand sid of.7 is a -Fourir sris and its cofficints must b qual to th cofficints of th right-hand sid pandd in a -Fourir sris. W hav that T t T 2 t T t = A n with as usual A n t = 2 T t + T 2 t T t d.9 = n 2 T 2 T + 2 n = 2 T + 2 T 2 n Thrfor quation.7 boils down to th following infinit st of quations 2 ċ n t + cn t = A n t, n =, 2,,. Ths quations ar linar first-ordr diffrntial quations and can b thrfor solvd asily as c n t = C n 2t + 2 t.8 2u A n udu. Our solution contains now th infinit sts of constants C n that w can fi by th initial condition: u, = T + T 2 T + c n = f.2 Sinc c n = C n from. w obtain from which w obtain = 2 C n = 2 C n = f T T 2 T f = 2. f T T 2 T d =.4 d + 2T f n + n 2T T 2 d 2T + n 2T 2 This complts th solution of th original quation. t us paus at this stag and s if this solution rducs to th known solution of th systm whn th boundary tmpraturs ar constant, T t = T and T t = T 2. In this cas from.9 w find that A n = for all n. From. and.4 w find that c n t = C n 2t =.5 2

3 2 = fu Finally th full solution in this cas rads +2 2 t u du + 2T + n+ 2T 2 2 t u, t = T + T 2 T +.6 fu u du + T + T 2 n+ which is th corrct solution to th constant boundary tmpratur problm.. A diffrnt form for th solution W will now simplify th solution. somwhat by intgration by parts 2 2 t 2 u 2u A n udu = 2t.7 2T u + n 2T 2u t 2u 2T u + n 2T 2 u du = = 2 T t + n T 2 t 2 2 t T + n T t Combining.7 with. and.4 w obtain 2u T u n T 2 u du c n t = 2 2 t f d 2 T t n T 2 t t Ug.8 th solution.5 now rads u, t = T t+t 2 t T t T t n T 2 t t 2u T u n T 2 u du 2 t fu u du+.9 2u T u n T 2 u du W obsrv howvr that 2 T t n T 2 t = T t T 2 t T t.2

4 by comparing with.8 and.9 or by consulting.6 and.7. Thrfor, u, t = T t + T 2 t T t T t T 2 t T t t = t 2 t fu u du+ 2u T u n T 2 u du 2 2 t fu 2u T u n T 2 u du u du+.22 Not that in this form, it is not at all clar that th solution satisfis th boundary conditions. On would think that u, t = u, t = as vry trm in th sris vanishs in this cas. This is du howvr to th fact that th sris as it stands is badly divrgnt..2 Chcks on th solution t s chck what happns by chcking this form of th solution in a known cas, th on w chckd bfor whr T, T 2 ar constants. Thn w can do th intgral in th last part of t n 2 π 2 = 2 Ug this, th solution.22 bcoms = 2 = 2 u, t = 2 2 u T + n+ T 2 du =.2 T + n+ T 2 2 t T + n+ T 2 2 t 2 t fu T + n+ T 2 t 2 = fu 2 t 2 2 t = u du.24 u du + T + n+ T T + n+ T 2 fu 4 u du + T + n+ T

5 +T + T 2 T whr in th last stp w usd.2. Th solution w found in.26 matchs actly th corrct solution in.6.. inarly varying boundary conditions t us now chck a lss trivial ampl, namly T t = T + T t, T 2 t = T 2 + T 2 t.27 In this cas from. and.4 w can comput c n t = f n + 2T + n+ 2T 2 22 T + n+ 2 = f n + 2T + n+ 2T whr 22 f n = 2 and th full solution in.5 bcoms 2t.28 2t = T n 2 2t.29 T + n+ T 2 fu u, t = T + T t+t 2 T + T 2 T t u du. 2 2 T + n f n + 2T + n+ 2T T + n+ 2 2t Aftr sufficintly long tim th trms in th last lin di ponntially, and th solution rmains th on in th first lin. W would lik now to rsum plicitly th scond part of th sris abov 2 2 T + n+ T 2 = 2 2 T = T 2 6 whr to obtain th last lin w usd th rsults of th appndi. Thrfor th stady part of th solution dropping th ponntially dying trms is n =.2. u s, t = T + T t + T 2 T + T 2 T t T 2 6 Not that it satisfis th boundary conditions as wll as th diffusion quation! 5

6 .4 Gnral drivn solutions Th discussion of th prvious subsction suggsts that in ordr to find th long trm solution w must try a solution of th following ansatz u s, t = T t + T 2 t T t + m a m t.5 m Such an ansatz satisfis trivially th boundary conditions.2. Equation. implis t + T 2 t T t + m ȧ m t = m 2 mm a m 2 m.6 m 2 which can b rarrangd as t + T 2 t T t ȧ m t + and lads to th following quations W st a m+2 = 2 t T t ȧ m t m m 2 m m + m + 2a m+2 m= m =.7 2 a 2 2 =.8 ȧ m t =! a m + m + 2 ȧm, m = 2,, 4,.4 Zt = and solv th quations.8-.4 abov: a 2n = 2n! 2 ȧ m t.4 a 2 = 2 T 2!, a = 2! T 2 t T t Z.42 2 n T n, a 2n+ = 2n +! n T n 2 T n Z n.4 whr f n dnots th n-th drivativ of th function f. Whn T = T 2 = all ths quations giv a m 2 =. Whn T = T 2 =, thn a m 4 = and a 2 = 2 2! T, a = 2! T 2 t T t ȧ 2 ȧ.44 But ȧ 2 = and up to an ponntially dying trm a = 2 T 6 2 T in accordanc with th prvious subsction. Th solution rf9,.4 is implicit. From th dfinition of Z and th formula abov w hav Z = ȧ m t = ȧ 2m + ȧ 2m+ =.45 6

7 = 2 m T m+ 2m! + T m+ 2 T m+ 2m +! 2 m Z m Thrfor Z is dtrmind from th infinit ordr diff. quation Z + 2 m Z m 2m +! = 2 m T m+ 2m! + 2m +! T m+ 2 T m+ 2m +! This quation bing linar with constant cofficints can b solvd by Fourir transform. W dfin Zt W also dfin for convninc 2 τ From which w obtain = τ Zk = τ ikτ m+ 2m! dk Zk ikt, T i t = τ. Equation.47 bcoms dk ik + T k + ikτ m+ 2m! dk T i k ikt.48 ikτ m Zk.49 2m +! ikτ m+ 2m +! T 2 k T k = T k + ikτ m+ T 2m+! 2 k T k + ikτ m 2m+! ikτ cosh ikτ ikτ ikτ T + ikτ ikτ ikτ T2 ikτ ikτ =.5 whr w usd f cosh = n 2n!, f 2 = n 2n +!.5 W may now comput th Fourir transform ã 2m+ k from ã 2n+ k = 2n +! ikτn ik T 2 T Z = ikτn+ 2 2n +!τ and via th invrs Fourir transform a 2m+ t = 2π T2 cosh ikτ T dk ikτn+ 2 2n +!τ ikτ T2 cosh ikτ T ikτ.52 ikt.5 7

8 W now chang variabls to a 2m+ t = u = ikτ, dk = 2 udu.54 iτ 2n +!iπτ 2 C du u 2n+2 T2 coshu T u t τ u2.55 whr C is a path on th compl plan that passs through and is at angl π with th ral 4 ais. Th intgrant has simpl pols at u = i with n a non-zro intgr which corrspond to valus of k = i n2 π 2. Th rsidus ar τ R n = 2n+2 2n +!iπτ 2 T 2 i n T i n2 π2 t τ.56 Clog th contour from blow - or abov + w obtain th sam rsult a 2m+ t = 2 2n+2 2n +!τ 2 T 2 i n2 π 2 n T i n2 π 2 2 Anothr approach to gnral drivn solutions W will mak an ansatz similar to th prvious sction u, t = Th diffusion quation implis as bfor that τ τ n2 π 2 t τ.57 a n t n n 2.58 a m+2 = τȧ m m + m Whil th boundary conditions ar a t = T t, T 2 t = Th solution to th rcursion rlation in 2.59 is a n t 2.6 n= a 2n = τ n a n 2n!, a 2n+ = τ n a n 2n +! 2.6 whr f n t n f. Th first condition in 2.6 implis that a t n = T whil th scond implis that τ n T n T 2 = a 2n + a 2n+ = + a 2n n! so that n= a 2n+ = n= n= n= n= τ n a n 2n +! = T 2 8 n= n= τ n T n 2n! 2.6

9 So th last stp is to solv th linar diffrntial quation for a with constant cofficints: n= τ n a n 2n +! = T 2 n= τ n T n 2n! 2.64 Th solutions of th homognous quation 2.64 ar ponntials and can b found from th charactristic function obtaind by acting on ρt : τ n d n τ n ρ n 2n +! dt n ρt = 2n +! ρt = ρτ ρt 2.65 ρτ Th zros ar at n= n= ρτ = i, n Z {} 2.66 from whr w conclud that th gnral homognous solution is C n n2 t τ 2.67 and is ponntially dying. It is thrfor nough to find a spcial solution of As bfor dfining th Fourir transforms of th various functions as w obtain W also hav ã = a t = T 2 k ikτ n n= 2n! ikτ n n= 2n+! ã 2n k = ikτn 2n! dkã k ikt, ã k = 2π T k = ikτ ikτ dta t ikt 2.68 T2 k cosh ikτ T k 2.69 T k, ã 2n+ k = ikτn 2n +!ãk 2.7 Th Fourir transform with rspct to tim of th original solution 2.58 can b writtn as ikτ ũ, k = cosh ikτ ã k + ã k = 2.7 ikτ ikτ = cosh ikτ T k + ikτ T2 k cosh ikτ T k = 2.72 ikτ ikτ = ikτ T k + ikτ T2 k 2.7 so that finally u, t = ikτ dk ikt ikτ ikτ T k + ikτ T2 k

10 All pols of ikτ ar on th imaginary ais and thrfor th intgral abov is wll dfind and finit. If th boundary conditions ar localizd in tim: T t = Aδt, T 2 t = Bδt t, T k = A 2π, T2 = B 2π ikt 2.75 and th solution bcoms u, t = 2π dk ikτ ikτ ikt ikτ A + ikt t ikτ B 2.76 Appndi W collct hr th cofficints of som -Fourrir sris. For any function f dfind for, th -Fourir sris cofficints ar W tabulat blow som functions cf n = 2 f d. f =, c n = 2 n.2 f =, c n = 2 n+. f = 2 2, c n = 4 n 2 n.4 f =, c n = 2 n 2 n.5 By taking linar combinations w can invrt th transform as follows c n = f = 2 c n = n f = 2 c n = c n = n

The Matrix Exponential

The Matrix Exponential Th Matrix Exponntial (with xrciss) by D. Klain Vrsion 207.0.05 Corrctions and commnts ar wlcom. Th Matrix Exponntial For ach n n complx matrix A, dfin th xponntial of A to b th matrix A A k I + A + k!