Question

Find the solution to the heat equation on the infinite domain

∂u∂t=k∂2u∂x2,−∞<x<∞,t>0,u(x,0)={x,0,|x|<1|x|>1.∂u∂t=k∂2u∂x2,−∞<x<∞,t>0,u(x,0)={x,|x|<10,|x|>1.

in terms of the error function.

Q1 (10 points) Find the solution to the heat equation on the infinite domain azu ди at k -00<x<0, t>0, ar2 u(x,0) (X, 1x < 1 10, [] > 1. in terms of the error function. + Drag and drop your files or click to browse...

Answer 1

all 22 given heat equation 24 224 -K -0<x< 29 it>O at U(3,0) = sz, 18121 = f(x) 0, JX171 we know solution to this problem is given by explore than one $(4) rakit u(x, t) = -(4-8) 4 Kt 1-14-3)? 568). expl 4 Kit ja trosern os akt -1 -09 1-01-82 (A) exp limit as t + Note- g breaked -1 I wh (2) : x , = 0, <XCI otherwise 1 - (X-8) 2 1 & expl ).cs +c) * 4 KA Sunkt U(x, t) = 0 + 18.exp CS Scanned with CamScanner

let z = (x-8) rakt We know essor function erf (0) - 96 ex?dx dz = Jakt ? ds = 2- -J4 Ktdz Now z varies from Z= Xt1 Juka as 8- Varles to z- SAKA from af to 1 x-1 | Sukt Ul, t) = 1 JUKKA (X=. S4K4 ) * - Jakt exp(-22)+z HH J4K& 2-1 Jakt 1 1 ( 16 Z 34K+) exp(-22)42 xt! Jakt x-1 Sakt exp(-22)/2-54kt. 14 4kt 2. expl-22d/2 xo X+1 Jakt 2+1 Juke 2 Sol X. I, - sakt• Iz ] we solve II & Ig repectivly. CS Scanned with CamScanner

L kt -22 22 11 e d2 한d t AKE ez2d2 +) o H J4RA 본 JARt e22d2 t Jaktan E ē 22d2 제 해 Jakt NIRt 22 22 JÃ 01 - 현 ē dz 오 xt eff [est()- ] 내 19 Kt 세 22 2. è dz I 11 ut -z2a P dp 十zdz- J4Kt 에 제 Je Jact P -22 e 2 대 RA +! Jakt 2. 2. 어세 외k -() 4Kt e It 은 CS 2. Scanned with CamScanner

from equation K JqK 又 U(X,大) 示 erf J- / 十 eTV 2 (HK, 3 -(H) 體) 11 piecent colore *人 (4) *大 e (x) =(x+y)) UCXjk) - 宗 (行 「K 2 e9好 CS Ans Scanned with CamScanner
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