Question

129 similarity variables XER 10 (5.102) 20.Silarity variables for the heat equation: the purpose of this exercise is to derive important canonical solution for the heat equation and to introduce the method of Consider the heat equation Show that w is a solution of (5.102) if and only if () is a solution of the ODE (b) Integrate the ODE and show that the function 5.7 Exercises MY - =0 2.11 +21' = 0, where' = d/d.. Set (x,1)=(x, t)), where 2(x,1)= lution. << >0. (X, 1) = erf 00. is a solution of (5.102), where erf(s) is the error function defined by 2 erf(s) : 1 (c) The complementary error function is defined by 2 erfc(s) : Scd = 1 -erfis). Show that (X.) erfc - the problem L. >0. is a solution of (5.102). (d) Differentiating erf(x/2/7), show that L. K(x,1) = exp 4t 41 is a solution of (5.102). K is called the hear kernel. We shall consider heat kernels in detail in Chapter 8. +y)]. (5,101) which does not uniqueness for

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Answer 1

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2 » cohere a = 3 au and su ut uxx=0 XEIR ,4670 Let u eu,+) = (11n, 1) = 918). RVE au(n,t) I'2 tri at at au (u,t) art 11 1 Zulast) 4 t 2 2x2 put the equation ③ we get, 'Co 4 t 4 t rt = $(dom, 4), '(a). into at an2 L & (0) T 키 Ø" D'a+ 3+ $CA) = "Cd+ 2 * $'(1)=0 Saad facton - е dz e 2² and + ع جے Integrating multiplying both side Ø" (0) za e & '(&) - af den of '(d) = Integrating we get edo (1) constant &'(d) - c [e integrating C e del

Let initia initialy pedo) = to 10 Oca- Sol os Integrating we get ed 2 =0 then 8 (&)- & .= ch é dad do 6 lino Let dos we can find @ bu T er (d) = este dedi +60 ar putting Let enf(s) = 2 e s dn=f(s) n d=0 S 2 VT esa as 2 2 VIT TAT z'e. - 25 VI | asa 2 142s af 25. urt ds2 2 45 S +45 e VIT enf(s) is the solution o e cenfels) - Seram enfis ) = 1-f68) d (1-f(s)) -df (3) S Osoiture CH ds as d² Gel-f(s)): d²f ds2 as2

+28 d sale (1-46) 2 (1-fls) a 52 df dis? E:-(0) = 0 Ôn. 1-f(s) is ako a solution. - 25 df ds tion of o

k into NATE 22 ak(nt 2x 4t Vant 2 2 2 re e at 2²K (net) atram + 2 2+ 2t & 4t 2, 2x2 es VATE x e at + 22 4 t Vitt 8 than 2. 2 ak(mt) X e at 1 2 x 942 Natt 2 tt at Natt 2. Home 2 ge e at t at + 4 tot 8 ENTE 2K(n, t) 2 kin, at "K(n,t) is a solution 2n2 of equation