Question

Solve the heat flow problem: au t> 0, ди (x, t) = 2 (x, t), 0<x< 1, ot дх2 uz(0, t) = uz(1,t) = 0, t>0, u(x,0) = 1- x, 0 < x < 1.

Answer 1

1 Answer It Jxr The one dimentional heat conduction equation is givenby Julxt) - Duille) where as called diffusivity of the substance. Given that Julut) = 2 Juxt), 0<x<1, t> JAN - 0 It و - 1 Subject to to the conditions: ux @ tto x (ot) = 0 uxlit)=0 u(a,0)=1-1 = f(x), OLALI and A are Vering the method of separation of variables, the different solutions of One dimentional heat conduction equation ) u(a,t) = (A P² + Be Px) , PENE Ult) = 4x+3 (B) U(xt) = (A Cospx + 13 Simpx De Pet where 'P' if the reparation constant. In view of the transient nature of the solution, form of the solution of the equation () as u(x,t) = (A CotpX + B Simpx) e pene > day (Apsin px + BP Simpa ) z (=2) the

- (- ApSimpa + BP Cospu Jē pat - AP Smpre 2 - du 2 JK put x=0, weget -pat Ju = (O+Bp.lle Jx eu eleco Isnice Sino=0 Coso = 1 N-0 But (2) Ux(0,4)= 0 (gaven) 20 -pat So we have 0 = Bpe -) eithere p=0 OS) B=0 if P=0, then Ulat) = ato = A This is contradiction the to the obter of eq Hence B=0 il B-0, (xt) = A Cospx @ Put - du Jx put x=1, weget du Ox -Plat - - AP Sinp Bu But ( x - , - UX (1+) = 0 ( giren -pat So we have 0=- AP Simp e - Smp = 0 ( Ato) - P- IT n=1,2,3,...

of the form unt) = 1 / 2 Ao + E An Cos (TX) e nat Thus ulat) as n=1 satisfying the equations 1 and @ Now, we need to bind the obbitrary constants Ag and An Observe that The R.H.S. of 4 is the half-range coline series of fix in (0, 1) by putting t=0 in 6 Here f(x) = u(at) and l=1 Hence Ao=&h fondx A = 'urxo) de 2 (' (1-x)dx -2 (x_)=2(1-3-0+0 => A = 2.5=1 ic. Ao = rd and Ana 2 l s flos cos rex ) deze 2 ( ulxo) cokmx) dx 2 SCO-x) col mes de uso-u u's fretu" SSSo. - 2 こ = 2 2 /0-1) - x) Sinbox) - O-1) -COAKTY) ті

4 >> An = 2 [a-1) Sma Cosmin -(1-0) Sino (only OTT & Coso بک WT) 210 - -ot Gomor MIN : Sino=0 coso = 1 Sinnt=0 Coln=E1 =) An 2 Composer [1-4] The required solution of 1 is given by - NTE Ulr, t) = 1 + § 2 r [-com]carlo x). e n=1