Consequently, the temperature must be the same at each point in the cross section at x. Let:


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1 HW 2 Comments: L13. Derive the het eqution for n inhomogeneous rod where the therml coefficients used in the derivtion of the het eqution for homogeneous rod now become functions of position x in the rod. Solution: The derivtion is virtully the sme s for the homogeneous cse: As in tht cse, the length of rod is L lterl surfce of rod is insulted rod hs constnt therml conductivity k = k (x) (k>0) rod hs constnt specific hetc = c (x) (c>0) rod hs constnt density ρ = ρ (x) (mss/volume) rod hs congruent cross sections perpendiculr to its lterl xis, the xxis; ech cross section hs re A H1. Het flows only prllel to the lterl xis of the rod. Consequently, the temperture must be the sme t ech point in the cross section t x. Let: u = u (x, t) be the temperture t cross section x t time t. (0 x L, t 0) q = q (x, t) be the het flux (the time rte of het trnsfer per unit re) cross the cross section t x t time t. Het flows from left to right when q (x, t) is positive nd flows from right to left when q (x, t) is negtive. e = e (x, t) be the specific internl energy (het energy per unit mss) t cross section x t time t. f = f (x, t), ssumed known, be the time rte per unit volume t which het is generted by processes within the rod. H2. Therml Energy is Conserved: Thetimerteofchngeoftherml energy in ny region V in het conducting solid is equl to the net rte t which het flows into V cross its boundry plus the net rte t which het is produced by sources nd/or sinks in V. Apply H2 to the region V of the rod with x b to obtin: e t ρa dx = for ll <bin [0,L] nd for ll t 0. ( q x + f) Adx (1) 1
2 Lemm 1 If r (x) nd s (x) re continuous functionsonnintervli nd if r (x) dx = then r (x) =s (x) for ll x in I. The Lemm nd (1) imply tht s (x) dx for ll <bin I, e t ρ = q x + f (2) Now, we need to relte the three unknowns e, q, nd u through properties chrcteristic of het flow. H3. (Fourier s Lw of Het Conduction) Het flows from regions of higher temperture to regions of lower temperture nd the rte of flow is greter the more rpidly temperture chnges with respect to distnce. Therefore we ssume tht the rte of het flow is proportionl to the temperture grdient: q (x, t) = ku x with k>0. The therml conductivity k is defined by this reltion. Use of Fourier s lw in (2) give e t ρ =(ku x ) x + f (3) becuse now k is function of x. (You cn expnd this using the product rule but tht turns out not to be n dvntge for mny purposes.) H4. For mny mterils nd over wide temperture rnges the specific internl energy is liner function of the temperture: e = cu + d The specific hetc = c (x) > 0 is defined by this reltion. The units of specific het re [energy]/[mss][deg]. From H4 e t = cu t. So (3) cn be expressed s ρcu t =(ku x ) x + f, the inhomogeneous het eqution (in one sptil dimension). Of course, you cn write this eqution in other equivlent wys. 2
3 L17. Consider homogenous lterlly insulted rod with het flow in one sptil dimension s in clss. Assume tht initilly the temperture in the rod is f (x) for 0 x L, the left end of the rod is lwys held t constnt temperture 0, nd the right hnd end of the rod is insulted. () Formulte n initil boundry vlue problem (IBVP) for the temperture u = u (x, t)in the rod. (b) Find ll nontrivil seprted solutions to the homogeneous equtions in the IBVP in (). Solution: () Since not sources or sinks re specified, ssume there re none. Then the temperture chnge is governed by the bsic het eqution u t = α 2 u xx. The initil temperture in the rod is u (x, 0) which is given s f (x). Sou (x, 0) = f (x). The temperture t the left end of the rod is u (0,t) nd it is mintined t temperture 0; so u (0,t)=0. Therightendoftherodis insulted, which mens there is no het flux there; tht is, q (L, t) =0. By Fourier s lw q (x, t) = ku x (x, t). Set x = L to get u x (L, t) =0t the right end of the rod. The relevnt IBVP is u t = α 2 u xx for 0 <x<l, t>0, (HE) u (x, 0) = f (x) for 0 x L, (IC) u (0,t)=0,u x (L, t) =0 for t 0. (BC) (b) The function u = T (t) X (x) with stisfy (HE) iff (TX) t = α 2 (TX) xx, TX = α 2 TX 00, T α 2 T = X00 X = λ2 where λ 2 is seprtion constnt. (Any letter cn be used in plce of λ 2 ; however,thechoiceof λ 2 is convenient becuse we expect exponentil decy in time.) Thus, T nd X must stisfy T + λ 2 α 2 T =0nd X 00 + λ 2 X =0. The homogeneous boundry conditions will be stisfied if u (0,t) = T (t) X (0) = 0 u x (L, t) = T (t) X 0 (L) =0 If either X (0) 6= 0or X 0 (L) 6= 0, the forgoing equtions would imply T (t) =0 for ll t nd the corresponding seprted solution would be trivil. So, ny nontrivil seprted solutions must stisfy T + λ 2 α 2 T =0nd X 00 + λ 2 X =0 X (0) = 0 X 0 (L) =0. 3
4 You cn solve the differentil eqution for T using n integrting fctor becuse it is first order liner. You cn lso solve it just like you solve second order constnt coefficients by finding one linerly independent solution vi the Euler exponentil guess pproch. Which ever route you follow you find (check this) T (t) =e λ2 α 2 t nd ll its constnt multiples. Likewise if λ 6= 0, the differentil eqution in the BVP for X hs generl solution (check this) X = A cos λx + B sin λx where A nd B re rbitrry constnts. When λ =0, the differentil eqution hs generl solution X = A + Bx nd cn only stisfy the boundry conditions if X (0) = A =0nd X 0 (L) = B =0. Tht is X (x) =0for ll x nd we get the trivil solution for u = TX. So ny nontrivil solutions occur when λ 6= 0. Then X = A cos λx + B sin λx nd this solution to X 00 + λ 2 X =0will stisfy the boundry conditions iff X (0) = A =0, X 0 (L) = Bλcos λl =0. To get n nontrivil solution for X we must tke B 6= 0nd choose λ so tht cos λl =0. The only solutions to this eqution re the odd multiples of π/2 : λl = (2n +1) π for n ny integer, 2 (2n +1)π λ = λ n = for n ny integer. For these vlues of λ the differentil eqution X 00 + λ 2 X =0hs the nontrivil solutions (2n +1)πx X = X n (x) =sinλ n x =sin nd ll its nonzero multiples. Since the T differentil eqution must be solved with the sme λ, the corresponding T fctors re T = T n (t) =e λ2 n α2t = e (2n+1)2 π 2 α 2 t/. Finlly, the seprted solutions to the homogeneous equtions in the IBVP re u = u n (x, t) =T n (t) X n (x) =e (2n+1)2 π 2 α 2 t/ sin nd ll their nonzero multiples. (2n +1)πx 4
5 L21. Let f (s) be twice differentible function of the rel vrible s. In wht follows t is time nd x is the sptil vrible with <x<. Imgine string with infinite extent. () Show tht u = f (x ct) nd v = f (x + ct) re both solutions to the wve eqution in one sptil dimension, which in this cse re the trnsverse deflections of the string. (b) Show tht u = f (x ct) represents trveling wve tht trvels to the right t speed c. This tells you wht c mens in the wve eqution. Hint. At time t =0the wve profile is u = f (x). Drw simple possibility for this profile. Sketch the wve profile t lter time t on the sme set of xes. Now think. (c) Wht is the interprettion of the other solution in ()? Solution: () Let u = f (x ct) nd use the chin rule to obtin Consequently, u t = f 0 (x ct)( c), u tt = f 00 (x ct)( c) 2, u x = f 0 (x ct), u xx = f 00 (x ct). u tt = c 2 f 00 (x ct) =c 2 u xx nd u = f (x ct) is solution of the wve eqution. Likewise for v = f (x + ct). (b) (Remember tht dimensionl nlysis of the wve eqution revels tht the constnt c hs the units of velocity. We wnt to understnd wht it is the velocity of.) At time 0 thewveprofile is u (x, 0) = f (x). This mens tht t time 0 the trnsverse deflection of the string t ny position x is f (x). At time t thewveprofile is u (x, t) =f (x ct). So the trnsverse deflection of the wve t x + ct t time t is u (x + ct, t) =f (x + ct ct) =f (x). Consequently, the trnsverse deflection tht occurred t position x t time 0 occurs t position x + ct t time t. This mens the entire wve is moving to the right with speed c. (c) Likewise, v = f (x + ct) is trvelling wve solution to the wve eqution tht trvels to the left with speed c. 5
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