Question

Solve the heat flow problem: ди ди - (x, t) = 2 — (x, t), 0<x<1, t> 0, д дх2 и(0, t) = (1,1) = 0, t>0, и(x, 0) = 1 +3 cos(x) – 2 cos(3лх), 0<x<1.

Answer 1

gu.(mt) at tro OLULI Kunn au Auxiliary equation solve the heat flow problem: 2010 (20.t) ,0<x<1, t>o ux(ort) = Ux(1 t) = o. u(,0) = 1 + 3 Cos(H) - 2 cos (310) the heat conduction in a finite rod uf OnLl, to u (0,0) = f) (where e=1) Comparing the given heat equation au at dur I K=2 And Initial condition u(2,0) = f(x) ie f(n) = 1+3 COSCH) - 2 Cos(3*) Let us assume the solution of Equation (1) az ucht) = X(n) T (t) [By separation variable method] x" constant = a (say). X" -ax=0 & T'-20T=0 Once again in we observe that it be negative and therefore and therefore equal to – x² Hence (D+) X=0 mataro m = f(x. Х RT i.e x"+ 2x=0 T' tart=0

m, t20 - 0 = m = - 202 similary (Di+22YT-0 Auxiliary equation AC e faz therefore X(n) = Bsinl&x) cos&x) + and TU Boundary condition Unco.t) = 0, Ux(lt) = 2 (By separation variable method x'(0)-o and x' (4) -- 0 use the condition. x'() = 0 - & A sin(ax) + Bx cos(an) X'(o) -0.A.0 + Ba| - To=B (since ato, cos(o) = 1 sin()= 0) And x' (1) = 0 RA sin(a) + 0.a cosca -O = -&A sin & = 0 x!() = (3 470,-470) sin (a) = 0 sin (2) = sin(at) 10=nM n=1,2,3 - Hence the solution, X(u) = A cos (4x) and Tt) Tt - c.pt (d = 1) in

and en eeriet therefore xn(n)- Ancos (4x) Tn (t) = corresponding eigen function therefore -antz un(*t) cos (NX) therefore ucu,t) = { an cos( nnn) e +2 an e ĕ 201² n=1 where । s (nn) dx an = dh Now ucnol - fx) = 1 + 3 cos (r) - 2 Cos (3x) Suno) cos Wher e=.1, u(no) = f(x) giren therefore (1+3 cos (6) - 2 cos(34%)) cos(ing 2 [ s'cos (nem) elu + 3 s'coscn) cos(he) din s'coscntiu) cos (3 pm) di] 7 an - 2 [- sina A + 3 f'costumg ees (etny dla -2 s'cos (nong cos(3 rug dx 2 [8 s Coshot x) cos(MN) dx-2 scos en ma) conferin, an [ 3 11-212] da of an = an 0 2.

where I = I'cos (nn) cos(rujan Result: s'cos (man) Cosminde m=n 13 man use Result for n=1 then II s'costa) Costajan á for na otherwis I is zero similarly I2 Cos (non) Cos (34x) dx use Result for n = 3 then I f'cos (3r) cos(3014) dn -) 12 = 1 for n=3. otherwise I2 is zero 2 therefore an [3 flo cos (nom) Costingdy - 2 of (nam) cos(Mu) dan For = 1 and n=3 then ay = 2[3.Ž -2.0] - Tai = 3 and 2[3.0-2. z - Taz az 11

therefore the required solution Equation (2) -212 + az los(34nje -207 isn't -18117 ucrit) = a costaje ē ulrit) = 3 cossringe + (-2) Cos(311) (: a = 3, --2) aza otherwise an=o) therefore -ant uchrt) = 3 cos (onje -1802 2 Cos (3x) e (Ans).