Question

(1 point) Solve the heat problem with non-homogeneous boundary conditions

∂u∂t(x,t)=∂2u∂x2(x,t),  0<x<3,  t>0∂u∂t(x,t)=∂2u∂x2(x,t),  0<x<3,  t>0

u(0,t)=0,  u(3,t)=2,  t>0,u(0,t)=0,  u(3,t)=2,  t>0,

u(x,0)=23x,  0<x<3.u(x,0)=23x,  0<x<3.

Recall that we find h(x)h(x), set v(x,t)=u(x,t)−h(x)v(x,t)=u(x,t)−h(x), solve a heat problem for v(x,t)v(x,t) and write u(x,t)=v(x,t)+h(x)u(x,t)=v(x,t)+h(x).

Find h(x)h(x)

  h(x)=h(x)=

The solution u(x,t)u(x,t) can be written as

u(x,t)=h(x)+v(x,t),u(x,t)=h(x)+v(x,t),

where

v(x,t)=∑n=1∞aneλntϕn(x)v(x,t)=∑n=1∞aneλntϕn(x)

  v(x,t)=∑n=1∞v(x,t)=∑n=1∞

Finally, find

  limt→∞u(x,t)=limt→∞u(x,t)=

(1 point) Solve the heat problem with non-homogeneous boundary conditions au ди (x, t) at (2, t), 0<x<3, t> 0 ar2 u(0,t) = 0, u(3, t) = 2, t>0, u(t,0) = 0 <3 <3 Recall that we find h(2), set v(x, t) = u(x, t) – h(2), solve a heat problem for v(x, t) and write u(x, t) = v(x, t) +h(2). Find h(c) h() = 2/3x The solution u(x, t) can be written as u(x, t) = h(x) +v(x, t), where vla,t) = Żanet+8a() v(2,t) = n=1 Finally, find lim u(, t) =

Please show all work.

Answer 1

Sely Ju (x,t) It OLUL 3 3 Given heat egh with non-homogeneous boundary condition is ju - 0 Jell2(4,4), 0281 < 3", to ulo,t) = 0 4/3, t) too @) (boundary bondiston) U(no) 을거 (3) (Initial Enduction We relve it wring reperation of variable method let ų (u,t) = X(1) T(t) (4) Put this in eqh I then get XTI X'T (variable reperated Take are X" Х -d (ray) T constant (5) aa where A is reperable X" td x = 0 ch (0) & T+AT = 0 4(e,t) - 0 - X60) = & 4(3, t) X(3) = 2 wing d , da X" tox = 0 + X(1 A + Bu 0 = A + B. o A = & X(3) = 2 2 - 0 +B(3) — 38= 2 -> 18- } / X (o)

dlo H = X" & 2 X X = A + Bean A & B ary where are arbitr constants. X (0) =0 o = Aelo + Be-20 → = A (1) + BCU A+B=0 0 B = -A → x(x) = A[ exhaeth رسم م -Y3 > X (3) = 2 A[ets = A[e3x- elr? ee e34 e34 A = 2 e 3x ㅋ el? 34 - A of eur 0972 2. do daq? all && 2x = 0 IX (H) A Casant & B Sinan X (o) A Carto + B Sinto A (C) + B (0) A 11

X (3) = 2 B Sina 3 = 2 7 B Sin 32 =2 B = 2 C -Sin 32 #2) U 2 Sina H uring di ,t=0 TE =) T= A. Tlo) T dlo, q2 T' - Q²T (10-2²).T ( T'E T's & → m-22) m=q2 Tapet 다. (A.E.) 2 d> ta . 2 T't «2 Т 22T T' dT तर - 27 dT T 22 dt T bg (T) q²t + logo 11

- leg t o log (I) e-p? T → e-&?t C - 2²t ce - → Tn (t) Ge-q2t Un (uit) Xn (1) Tn (t) = 2 Sinan Gedat but 26 an = An Sineu eht = u(n, t) an Sinan & Lt hal But here given u (s, t) h(H) + 4 (s, t) - h (1) u(at) - u (nt)

HR BW - h(x) = 8 an sincs er? anedut (1) ha n=1 C Mest) = { ane On (1) (x, t) anedut In (21) hal lim u (nt) lim 1 an Sinar ex?t tas t700 n=1 e non ancor (one (con an Sinan nel an Sinar (1 hal il an Sinan 6 0 n=1 CA (0) an sinase hal lim ust)