Question

2. Solve the heat problem: (Trench: Sec 12.1, 17) 9Uxx = ut, 0 < x < 4, t > 0 u(0, t) = 0, u(4,t) = 0, t> 0 u(3,0) = x2, 0 < x < 4

Answer 1

2. Ansor Griven quyusut, 02244,tro ucot) = 0,4(4,4)=0.00 U2,0) = 2,0x224. Let us suppose the solution invariable seperable G) The soom an, Ucut) = X) Tit) P.D.E becomes, 9x"(al) T(t) = x( 2Tit) => X". T! = constant a x 9T which O.D.E'S dx=0 I zacsay, gives two o. x4 2 and t'-qNT=0 13

cane I - to From @ we have wa X(5) 2 Ge + Cette, 2610,47 where q, Q2 are arbitrary constante now, uco,t)=0 => X10 T(t)=0 X(0)70 [=17442707 ... From A G+C₂20 Again, ucastl=0 2) XC4)=0 [::T(0)407 ... From ANA ge tee IO =407 -477 ANA since -e 70 letra i 1 = et 61 ::G-C2-0 Then X(2)=0. surat) 2x62) T(t)=0 is a trivial solution it so, we neglect core II - d= 0. from ③ we have, x02)=C33+4, 3610,4) -- €3.64 are arbitrary constants As ucout=0 =>x10-0[TD707 from 6 C3.0 te=0 => C4-0. :- X(21) = C33 © Again ucast) = 0 => x64) PO ()#07

:: From @ 4@3=0 => C3=0 :.x (2) 20 Then ucat)=X(X)T(t) = 0 is a it. trivial solution so we neglect it. es cos 7 are come-III:- do. Let, daru? :: From , x"'+pex 20. Therefore, x(31)3 (x) te sin(Ma) :ALFLA) egie arbitrary constants. As uco,t)=0 =>X(0)20. Then from Ô ega = 0 .: X22) - CG sin Cux), LE10,4] Again ucast120 => X(4)=0. Then from @ sin (AM) CG 20. Fon non m trivial solution 70 • Sa (41)=0 = sin(xn), 2-ty-t2 - => Ar=na 4 2212, => MERA + ² = 272 Ana-Ma-man . 2=1,2,3. 16 4 and connesponding eigen functions are X₂ (2) = An sin (2:2), 2=1,2,3,- Thén ☺ becomes, -9dIn=0. q na Tom 1 => TntT6 0.

Therefone, marr Toni 16 Ton => d Tn 9925 2 9. marat. 16 Ton Integrating. we have. In (Tn) = 16 where Inca) q nt+ln(e) arbitrary is a .', 16 Constant. Talt) = 27 ве where on is arbitrary constant Applying principle of superposition, ulaut) = ² ugeait) 22 - ¿ xx Cu Toits. - 26 277 =Ê Ansin (24 SL) Bye 31 4 2=1 a - 12ht = I age sin(4x) 22=1 where an AnBn. Now, ul2,0) =.x then we have, x= an sin (272), which is a pourier @ 221 sine series, 4 where an = 2 2 sin (ntada. 4 n=1,2,3,- o wow, gizarsin (mga) de 3 (2274 a)]] +2.4 CTS (9.2.1 2 X (-2/cos 2x (- 16 ma 4. on sining 0

( 128 64. 22373 =161-molcosnt-1)) + 12973 (Cognit-i). no (cosna-1). *. Qy = 3 (528 2373 eos 27-1 2.(64 B) (eos2n-1) 64 NT 32 2373 32 mn (orar 1) (Cosna-1) ..ana O son 2222,4,6,8, 64 nr (1- v 2 222 :), fonn21,3,5,7,- Requined solution of the heat problem is 9 nr. T6 ucatia I ane sin ( 2x) 22=1 where Ana S O , noeven i.e na2,4,6,8, 64. 11-mar), naodd i'e n=1,3,5,7,--