Question

5. Find the solution of the heat conduction problem for each initial condition given: Suxxx = 0<x<211, tu(0,1) = 0, tu(27,1) = 0, t> 0. (a) u(x,0) =(x) = -2sin(3x) - 3sin(4x) + 17sin(9x/2). (b) u(x,0) = f(x) = 8. (Hint: You may skip the integrations by using the result of #2(b).] (c) In each of cases (a) and (b), find the limit of u( 71,1) ast approaches 0. Are they different? Did you expect them to be different? 6. (a) Find the solution of the heat conduction problem: 34.xx = U19 0<x<2, u:(0,1)= 0, 4,(2, 1) = 0, 1>0, u(x,0) = f(x) = 6 -36cos(x) - 81 cos(5/2). (b) What is the limit of u(2.) as 1 approaches ? (c) If the initial condition is, instead, u(x,0) = 4 - x, find the new particular solution. (Hint: the necessary integration has already been done as one of the parts in question #4 above, you just need to apply the numbers from the relevant part of #4 into the general solution.) (d) If the initial condition is, instead, u(x,0) = 21 + 16sin(7x), how would you find the particular solution? (Write down the necessary integral(s), but you do not need to integrate.) (e) For the each of the solutions from (c) and (d), determine the limit of u(2.1) as 1 approaches 00. Are they different than your answer of (b)? Did you expect them to be different?

Answer 1

The given heat of equation is 5 Unx = Ut 5 a 2 Ux! 2 Ulnit) Jaz ət let U(Mt) - X CX TL) so 5o e o XCWTW) = DE X(XTC) => 5 TL) DE XC) = x()& Tik) This means 52X() --C? iw But eso where the negetive sign comes because - boundary conditions on a are periodic and hence . the solution should be oscillatory so x(x)= Acostex) +B sin( ex) and 12TH) =-6² The ot - Tlt)= exp(-0²t) - 1 1 E 1

applying the boundary condition ulo,t)- X(0) T(t) = 0 =) X(0) = 0 so A=0 0 (271,t) = x(272)T(+1=0 => X(217) = 0 so osin ( 2TC C) =0 Cu = usn Hence 0(44)= En Busin ( 25 x) exp(-50%) Now we apply the initial condition ie. @ OC,D)= f(x) = -2894(3x) – 3 bin(4x) + 17 sin () 3-2 sin(3x) -3 sin(4x)+17sin(4x) Clearly ony na 6i8 and a terms will E Bein ( m2) = -2 1. clearly one on reft -3 , Ba = Bq = 17 8o B6=-2, B8 =-3 Hence ucm+)= (-a sin (Ba) exp(-45+) -3 sin(4x) exp(804) +17 Sin ( 93 ) expl-408t) ]

bu W{x,0)= f(x)= 8 Ę Bu sin ( 2x ) = 8 multiplying sin/ mx) on both side and integrating from o to 2R E Bn S." SP ( 22 ) bestand was) dx = lo son (10 ) dx E Bu Te Smm = -82 (605( sang ) Det - Bm = 6 (-2) - 2 mon ex BN: Hence Ulmet) = E() Sim (92) exp ( suht) © ultert)s Ę Bu exp (-59t - 17 expl-toeste) tro for part @ U(tht) lim Ultit) = 0 in part ① ultet): E () (-1) "*" exp(-5.t) for odd n only ům Urt=0 to They are same as expected because after infinite time all the heat leaves the system,