Question

3. (5 points) Find the solution u(x,t) of the equation ut = uxx, subject to the boundary conditions u(0,t) = 1, u(2,t) = 3, and the initial condition u(x,0) = 3x + 1.

Answer 1

THE GIVEN EQUATION IS A HEAT EQUATION WITH NON HOMOGENEOUS BOUNDARY CONDITION;- so, by comparing with general 1-dimensional Heat equation with hon- homogeneous boundary condition : I UA = Uxo boundary condition: with к Boundary J 41014) = C, :- condition ull, A)=Cg Initial condition : 40e,o= f(c) By comparing with given equation we get :- 16=1 (9-3 Ke} 1662) = 3x+1 Now, we Convert the equation 4664) = u Uscx In a system with homogeneous boundary condition transformation: Luba, Al= US, A) & Ait B by the (2) Sol we have & Une = Uxue Ut = Vt By butting in 0 we get: - Mycot = 9 Boundary condition: - Initial conditioni- Vlot) -0 (2,4)=0) VIX,0) = g(x) Now from if :- 26 70 uloit) = Vloit) + Alo) + B I= ot B [ Given :- 410,4)= 1 By @] And 0(0,4)=0 B=1

And from a @ if x=2 we have :- U(0,4)= ula,t) t gA+I 3 ♡ + 2 Ati [Givchi 418,4)=3] 2: 2A B-1 610,4)=0 from A=1 sol julx, t)= U12,A) + x to bom from a (3 Now, if *-o 키 U120) = V12,0) + uti 3x+1= Unio)tato 3x+1-2-1=1140) 2x = 1 no) [Goeiboen - W124,0)=3*+] И sol Unio)=20 (В. solution of the transformed system is Now, the given by : - 0(4,4)= { Dn sin hrxe - Khera t s १२ hol e Where:- l Dn 8819 s glue) .sinhru de ♡ 2 So Du= ooloo Qx. sin hru da 2 [Giveni 1-2 8/26)= an Q From ③] Dn= 2x sinhru dre Q Dn= 20 (singen sinhrdn 2 254 (2) sinn zonder [By product Rule]

2 cos hin da ont fel cas del Eucosh + hr 0 2 Dn= -4 cos hra ] [ + 2 x 2 Tsin hin 2 Dn= - gcoshrt hr 노 sin dyr D kra Dh= 8 hr loshrt y sin hr hara Dn= -8 (-1h+ i sinnro Cashr=ton o] hr Dn - -8 (-1) hr By putting in ③ :- Uln1A)= Bar -8 (-1) h. sin here og A hal hr 2 © Now, By butting Unit) in 2 we geti- ula, A) = § -8 (-1)} sin mrxe + hal ht Q +x+1 [; A=, B=1] v This is the required solution for our given system.