VI. The Heat Equation. A. The Diffusion Equation Conservation in Action

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1 VI. The Het Eqution c 214, Phiip D. Loewen A. The Diffusion Eqution Conservtion in Action The simpest PDE describing diffusion (of het, perfume, poutnt) ong the x-xis is u t = α 2 u xx. ( ) Theme: Concvity drives Rte of Chnge. (Concvity is u xx ; rte of chnge is u t.) Here α > is given constnt; t is time; x is position; u = u(x, t) is some kind of concentrtion t (x, t). Often u is the concentrtion of het, i.e., temperture: this ppiction eds to the nicknme het eqution for ( ). Derivtion. Consider ong stright eve pipe of sty wter ong the x-xis. Let the st concentrtion t oction x nd time t be u(x, t), in units of kg/m 3. If the cross-section of wter in the pipe is A (with units of m 2 ), the tot mss of st between points x = nd x = b is m [,b] (t) = u(x, t)a dx. Fick s Lw expins how concentrtion tends to equize: t ny test oction, ike x =, the mss fow rte (kg/s) is proportion to the concentrtion grdient nd the re invoved. Thus st diffuses from eft to right t x = t rte of κau x (, t). The sme thing hppens t x = b. Now focus on m [,b] (t), the mss of st in [, b]. If diffusion is the ony process ctive here, then conservtion of mss requires dm [,b] dt = (rte in) (rte out) [ ] = κau x (, t) κau x (b, t) But, proceeding directy, ṁ [,b] (t) = i.e., = κa u xx (x, t) dx. u t (x, t) dx = u t (x, t) dx. So for every re interv [, b], κau xx (x, t) dx [u t (x, t) κau xx (x, t)] dx =. Since the interv [, b] is rbitrry here, the function in brckets cn never hve positive vue. (If it did, we coud zoom in on tiny region where tht function is Fie 214notes, version of 24 Juy 214, pge 1. Typeset t 16:32 Juy 24, 214.

2 2 PHILIP D. LOEWEN wys positive, nd get positive resut for the integr.) Likewise it cn never be negtive. The ony terntive is tht the brcketed expression is zero for x, nd this is the diffusion eqution: u t = α 2 u xx [α 2 = κa]. This ine of resoning ppies to the concentrtion of nything tht stisfies Fick s Lw. Temperture (pproximtey the concentrtion of het per unit mss) is fmous exmpe. Units. Rec the nottion [z] for the units of z. Suppose [t] =s, [x] =m. Then [ ] [ ] u = α 2 2 u t x 2 [u] ( ) s = [u] m 2 [α]2 m 2 [α]2 =. s B. Boundry Vue Probems A we-formed probem for diffusion in < x < hs 3 ingredients: (PDE) concvity drives rte of chnge Fick s Lw pus conservtion: [More gener forms possibe here.] u t = α 2 u xx < x <, t >. (BC) concentrtion nd/or fow-rte info t ech end of the physic interv. In het probems, choices incude (i) u(, t) = = u(, t)... both ends fixed t temperture, (ii) u x (, t) = = u x (, t)... both ends insuted (fow rte is ), (iii) u x (, t) + u(, t) = = u x (, t) bu(, t)... het ost or gined through rdition, (iv) u(, t) = L(t), u(, t) = R(t)... end tempertures decreed by some given fcns of time [e.g., diy temp of ground surfce], not both identicy zero. Cdx (i) (iii) re homogeneous; (iv) is not. Aow different conditions t different ends. For sty wter in stright pipe occupying < x <, Fick s Lw supports option (ii). (IC) initi conditions. One is enough: u(x, ) = f(x), < x <... initi concentrtion or temperture. Function f must be given in probem sttement. Story Probems. A typic physic interprettion of the boundry-vue probem bove invoves temperture in met br tht hs been wrpped in insution so tht het cn fow ong its ength (the x-xis) but not through its sides. The insution my or my not cover the ends of the br: this depends on the boundry condition. The BC u(, t) =, t >, indictes tht the eft end of the br is connected to some Fie 214notes, version of 22 Juy 214, pge 2. Typeset t 16:32 Juy 24, 214.

3 VI. The Het Eqution 3 constnt-temperture energy reservoir ike bucket of ice-wter. If the br is hot to strt with, het wi fow into the bucket nd met some of the ice, but we ssume there is enough ice in there (or somebody shoveing more in t good rte) so tht the temperture does not rise. If the br is very cod t first, het wi fow from the bucket into the br: more wter wi freeze. But we ssume tht there is enough iquid wter (or more running in from hose) tht the reservoir never freezes soid nd strts to chnge temperture. Trnsting Engish text into boundry conditions is ski students shoud be prepred to demonstrte on homework nd tests. C. Four Stndrd Homogeneous Probems The simpest het probem in < x < ooks ike this. (PDE) u t = α 2 u xx, < x <,t >, (BC) x = : Choose frozen (u = ) or insuted (u x = ), t >, x = : Choose frozen (u = ) or insuted (u x = ), t >, (IC) u(x, ) = f(x), < x <. Guessing u(x, t) = X(x)T(t) in (PDE)/(BC) eds s usu to pir of ODE probems, inked by seprtion constnt λ: (i) X (x) + λx(x) =, < x < ; some (BC) inherited from (PDE), (ii) T (t) + α 2 λt(t) =, t >. Depending on origin (BC), (i) wi be n eigenvue probem we recognize, giving sequence of eigenfunctions X n (x) nd eigenvues λ n. E.g., for n insuted end t x = nd frozen (zero-temp) end t x =, we hve u x (, t) = = u(, t) in the origin probem, so X () = = X() in (i), so HPCS eigenfunctions: ( 2n 1 X n (x) = cos 2 ) ( πx 2n 1, λ n = 2 ) 2 π, n = 1, 2, 3,.... Use eigenfunctions from (i) to postute soution in form u(x, t) = [ 1 2 T (t)+ ] Tn (t)x n (x), where eigenfcns X n re known nd time-vrying coeffs T n (t) remin to be found. (Tke T (t) = in FSS, HPSS, HPCS cses.) Initiize T n. IC: f(x) = u(x, ) = [ 1 2 T ()+ ] Tn ()X n (x) gives Propgte T n forwrd in time. T n () = 2 f(x)x n (x) dx. Fie 214notes, version of 21 Juy 214, pge 3. Typeset t 16:32 Juy 24, 214.

4 4 PHILIP D. LOEWEN = u t α 2 u xx = [ 1 2 T (t)+] ( T n (t)x n (x) α 2 T n (t)x n (x)) = [ 1 2 T (t)+] ( T n (t)x n (x) α 2 T n (t)( λ n ) ) X n (x). (Here we used X n = λ nx n. The vues of λ n come out utomticy when we ( ) 2 nπ differentite X n for known choice. We hve λ n = in FSS nd FCS cses, nd λ n = ((2n 1)π/(2)) 2 in the HPSS nd HPCS cses. Notice tht λ n > whenever n 1; the number λ = is n eigenvue ony for the FCS cse, which we wi discuss seprtey beow.) This gives, for ech n, T n(t) + α 2 λ n T n (t) =, t >. The gener soution (for n 1) is T n (t) = A n e α2 λ n t. Note tht A n = T n () = (integr formu bove). [For cse n =, T (t) = A constnt, nd sme integr formu works.] We concude tht in four cses, u(x, t) = [ 1 2 A + ] An e α2 λ n t X n (x), A n = 2 f(x)x n (x) dx. Both Ends Frozen. If u(, t) = = u(, t) in the origin BVP, then X() = = X() in (i), nd we hve eigenfunctions ssocited with FSS. Soution u(x, t) = ( ) b n e α2 n 2 π 2 t/ 2 nπx sin, b n = 2 ( ) nπx f(x) sin dx. Left End Frozen, Right End Insuted. If u(, t) = = u x (, t) in the origin BVP, then X() = = X () in (i), nd we hve eigenfunctions ssocited with HPSS. Soution u(x, t) = ( ) p n e α2 (2n 1) 2 π 2 t/4 2 (2n 1)πx sin, 2 p n = 2 ( ) (2n 1)πx f(x) sin dx. 2 Left End Insuted, Right End Frozen. (Try t home.) u(x, t) = ( ) q n e α2 (2n 1) 2 π 2 t/4 2 (2n 1)πx cos, 2 q n = 2 ( ) (2n 1)πx f(x) cos dx. 2 Fie 214notes, version of 21 Juy 214, pge 4. Typeset t 16:32 Juy 24, 214.

5 VI. The Het Eqution 5 Both Ends Insuted. See Section E beow for some more discussion of this cse. u(x, t) = 2 + ) n e α 2 n 2 π 2 t/ cos( 2 nπx, n = 2 ( ) nπx f(x) cos dx. Consider cse = π to sve gebr. D. Modes For both ends frozen, our series soution is combintion of bsic products u n (x, t) = e α2 n 2t sin(nx). Mode shpes X n (x) = sin(nx) sme s for guitr string, but now they don t oscite s t dvnces. They just decy exponentiy. Lrger n more curvture, fster decy rte. Even for sm t >, decy of high modes is so fst tht 1 3 terms in the series soutions cn give resonbe nswers. Simir remrks ppy to other BC s chosen from either frozen or insuted. E. Long-Term Behviour (1) In cses FSS, HPSS, HPCS, every term in every sum hs the form e α2 λ n t X n (x) for some λ n >. The time-dependent exponenti fctor therefore represents exponenti decy: ech term converges to s t. It foows tht u(x, t) s t. This mkes physic sense, becuse the boundry conditions for these three cses invove t est one end of the br hed t temperture forever. Het wi fow out this end if the br is wrmer thn, nd wi fow in this end if the br is coder. Hence the br wi eventuy come into therm equiibrium t temperture. The series soution found bove ets us ccute excty how the pproch to equiibrium tkes pce. (2) In cse FCS, the story is itte different, becuse when X n (x) = cos(nπx/), subscript n = is owed. Here u(x, t) = e α2 n 2 π 2 t/ 2 cos(nπx/). As t, terms except the first tend to zero. Hence u(x, t) 1 2. This, too, mkes physic sense: the FCS soution comes up when both ends of the br re insuted. Conservtion of het energy impies tht the tot het content of the br cn never chnge. And by the Fourier coefficient formu, the imiting vue cn be expressed s 1 2 = f(x) dx = 1 f(x) dx. Fie 214notes, version of 21 Juy 214, pge 5. Typeset t 16:32 Juy 24, 214.

6 6 PHILIP D. LOEWEN This is precisey the verge vue of the initi temperture u(x, ) = f(x). As expected, het diffuses unti therm equiibrium is reched. (3) Cse FCS is so nice one for conservtion of het energy. Any function u(x, t) stisfying both (PDE) nd (BC) in cse FCS mkes the foowing function constnt in the interv t : F(t) = u(x, t) dx. To see this, ccuute F (t): ccording to the PDE nd BC s, F (t) = for ech t >. See homework probem hetcte for detis. (4) The spti frequency is inked directy to the tempor decy rte. Higher spti modes hve higher temperture grdients, hence higher het fow rtes, nd hence fster neutriztion of bumpy temperture profies. (5) The exponenti decy ssocited with higher modes is so rpid tht keeping ony sm number of ow modes is enough to get resonby ccurte pproximte soutions. We iustrted this on HPSS exmpe in css (Mpe V nimtions re vibe on the Web site). F. Nonhomogeneity: Spitting or Fncy ODE s Probem 1 [mid]. Seprtion of vribes does not work on nonhomogeneous PDE s. E.g., consider the effect of the nonhomogeneous term f(x, t) in the generic form A(t)u tt + B(t)u t + C(t)u = α(x)u xx + β(x)u x + γ(x)u + f(x, t). Trying for u(x, t) = X(x)T(t) eds to A(t)T (t) + B(t)T (t) + C(t)T(t) T(t) = α(x)x (x) + β(x)x (x) + γ(x)x(x) X(x) + f(x, t) X(x)T(t). For nonzero function f, no nice rerrngement wi seprte the vribes t nd x. (Of course, if f(x, t) =, then the PDE is homogeneous nd both sides bove must equ some constnt.) Remedy 1. Use ony the homogeneous prt of whtever PDE is given when buiding n eigenvue probem nd postuting soution form. But use the fu origin PDE for propgtion. The term f(x, t) wi infuence the ODE s for the coefficients T n (t). We hve redy seen some exmpes. Probem 2 [severe]. Seprtion of vribes does not work on nonhomogeneous BC s. E.g., suppose the interv of interest is < x < π, with u(x, ) = 257, u(x, π) = 316. Fie 214notes, version of 24 Juy 214, pge 6. Typeset t 16:32 Juy 24, 214.

7 VI. The Het Eqution 7 If functions u 1 nd u 2 both obey these, the combintion u 1 + u 2 wi not... so superposition of bsic pieces cn t work. Just ignoring the nonhomogeneity nd hoping to compenste ter might suggest FSS postute, T n (t) sin(nx), but this evutes to t x = nd t x = π no mtter wht we do with T n (t). A more fundment cure wi be required. Remedy 2. Revisit the big-picture frmework used for inhomogeneous ODE s: 1. Find one soution. Invent v(x, t) to stisfy BC s. Mny choices exist: it s nice if v cn so stisfy PDE. 2. Buid the rest. Using v(x, t) from bove, spit u(x, t) = v(x, t) + w(x, t) ( ) nd now pursue w. Recommendtion: Write down expicity the fu BVP for w. 3. Seect the nswer of interest. Recipe steps 3 5 for w, then reporting (step 6) for u = v + w. Physic Interprettion. In diffusion probems with Big Four eigenfunctions, the w-probem wi hve soutions converging to constnt (often ) s t. Thus u(x, t) v(x, t) s t, nd w(x, t) = u(x, t) v(x, t) precisey quntifies the discrepncy from the ong-term behviour. Especiy when v = v(x) is independent of t, spitting u = v + w presents the desired soution into two components: the stedy-stte v nd some trnsient w. Usuy the stedystte is pretty esy to find: see the Trench textbook, Section 12.1, for further for discussion, exmpes, nd prctice probems. (Aterntivey, ook into the cssic text on Differenti Equtions nd Boundry Vue Probems by Boyce nd DiPrim. There section ced Seprtion of Vribes; Het Conduction in Rod outines the homogeneous cse, nd ter section ced Other Het Conduction Probems extends this to nonhomogeneous situtions.) Exmpe 1. This one is competey routine: (PDE) u t = α 2 u xx, < x < π, t >, (BC) u(, t) =, u(π, t) = 1, t >, (IC) u(x, ) =, < x < π. Here the BC s re not homogeneous, so spit u(x, t) = v(x, t) + w(x, t), where v is chosen to stisfy (BC): v(, t) =, v(π, t) = 1. ( ) One fesibe (but unwise) choice is v(x, t) = sin(x/2)+x(x π)e t, nd there re mny more. For guidnce bout how to choose v wisey, pug the spit-form u = v +w into (PDE): Fie 214notes, version of 24 Juy 214, pge 7. Typeset t 16:32 Juy 24, 214.

8 8 PHILIP D. LOEWEN (PDE) w t = α 2 w xx + [ α 2 v xx v t ], < x < π, t >. The new PDE for w wi be simpe nd stndrd if the brcketed term invoving v goes wy. Try to rrnge tht s we s ( ): ook for v = v(x) independent of t, so PDE wi hod iff v (x) =. So v(x) = mx + b for some m, b nd the choices b =, m = 1/π give cndidte v(x) = x/π. With this choice, we ook for soution in the form u(x, t) = x + w(x, t). π Pugging this into (PDE)/(BC)/(IC) bove produces some usefu cncetions, nd reduces the given sitution to new probem for the function w: (PDE) w t = α 2 w xx, < x < π, t >, (BC) w(, t) =, w(π, t) =, t >, (IC) w(x, ) = x/π, < x < π. The usu methods bove give the resut for w, so our fin report for u is u(x, t) = x π + w(x, t) = x π + b n e n2 α 2t sin(nx), b n = 2 π π [ x ] sin(nx) dx = 2 π nπ ( 1)n. Exmpe 2. Here the PDE is modified by n extr term on the right-hnd side. The expression x 2 πx is negtive when < x < π, so this represents some kind of het-extrctor operting t different rtes t different points ong the met br in our stndrd interprettion. (PDE) u t = u xx + x 2 πx, < x < π, t >, (BC) u(, t) = 1π, u(π, t) = π4 12, t >, (IC) u(x, ) = 2π x x 3, < x < π. 12 Agin spitting is required, but this time there is choice. Choice 1: Choosing v = v(x) to obey PDE/BC is ide: v (x) = πx x 2 = v (x) = π 2 x2 1 3 x3 + C = v(x) = π 6 x x4 + Cx + E. Now 1π = v() = E nd we get C = 1 from the other end, eving v(x) = π 6 x x4 1x + 1π. ( ) Pugging u(x, t) = v(x) + w(x, t) into the origin probem produces new BVP for w tht is quite mngebe. (This refects crefu dvnce pnning by the probem designer... something sedom vibe in bortory situtions!) (PDE) w t = w xx, < x < π, t >, (BC) w(, t) =, w(π, t) =, t >, Fie 214notes, version of 24 Juy 214, pge 8. Typeset t 16:32 Juy 24, 214.

9 VI. The Het Eqution 9 (IC) w(x, ) = 1(x π), < x < π. The soution wi be u(x, t) = v(x) + w(x, t) with v s in ( ) nd w(x, t) = b n e n2t sin(nx), b n = 2 π π 1[x π] sin(nx) dx = 2 n. Choice 2: There re infinitey mny functions v = v(x) tht mtch the given BC s, nd iner one wi so do the job. Ccution gives ( ) v = 1(π x) + π3 x π 3 12 = 12 1 x + 1π. Spitting u(x, t) = v(x) + w(x, t) for this choice of v eves ess-ttrctive probem for w. (The PDE remins nonhomogeneous, nd the IC s sty ugy.) Substitution produces (PDE) w t = w xx + x 2 πx, < x < π, t >, (BC) w(, t) =, w(π, t) =, t >, (IC) w(x, ) = 2π x x 3 1(π x) π3 x 12 12, < x < π. Stndrd methods wi work on this one; detis in person on request. //// Exmpe 3. Sti ess routine re cses where it s not so esy to find v tht obeys the PDE in ddition to the BC s. Here is sitution tht fits this description: (PDE) u t = u xx + e t sin(4x), < x < π, t >, (BC) u(, t) =, u(π, t) = π, t >, (IC) u(x, ) = x, < x < π. It s cer tht no choice of v = v(x) independent of t cn possiby stisfy the PDE. An inspired guess t some v depending on both t nd x my be possibe (the soution beow suggests one), but sometimes inspirtion is scrce. At such times, remember this: it is impertive to stisfy the BC s. With this in mind, et s work with the simpe choice v(x) = x. This does stisfy the BC s, nd the function u(x, t) = x + w(x, t) soves the stted probem if nd ony if the new unknown w stisfies (PDE) w t = w xx + e t sin(4x), < x < π, t >, (BC) w(, t) =, w(π, t) =, t >, (IC) w(x, ) =, < x < π. This is not so pretty, but not terrifying either. The homogeneous prt of the PDE is w t = w xx, which eds by seprtion of vribes to the eigenvue probem X (x) + λx(x) =, < x < π; X() = = X(π) nd the FSS eigenfunctions. So we postute soution of the form w(x, t) = T n (t) sin(nx). ( ) Fie 214notes, version of 24 Juy 214, pge 9. Typeset t 16:32 Juy 24, 214.

10 1 PHILIP D. LOEWEN The initi condition sys = w(x, ) = T n () sin(nx), so T n () = for ech n by FSS coefficient-extrction. Menwhie, pugging the series ( ) into the fu nonhomogeneous PDE for w gives e t [ sin(4x) = w t w xx = T n (t) + n 2 T n (t) ] sin(nx). How convenient: both sides invove FSS with respect to x hving t-dependent coefficients. Mtching gives { T n e (t) + n2 T n (t) = t, if n = 4,, if n 4. When n 4, this gives T n (t) = T n ()e n2t. But T n () = so in fct T n (t) = for every n 4. When n = 4, we must sove T 4 (t) + 16T 4(t) = e t, T 4 () =. One prticur soution of the ODE here wi hve the form T p 4 (t) = Ce t for some constnt C: substitution gives C + 16C = 1, so C = 1/15. The gener soution of the homogeneous eqution y +16y = is y = Ae 16t, A R, so the gener soution for T 4 is given by T 4 (t) = (1/15)e t + Ae 16t, A R. From this infinite set of possibiities, we seect the function for which = T 4 () = (1/15) + A, i.e., A = 1/15. Now we know bout T 4, nd the other T n, we see tht the sum ( ) hs ony one nonzero term: w(x, t) = T 4 (t) sin(4x) = 1 15 [ e t e 16t] sin(4x). Report: The soution to the origin probem concerning u is u(x, t) = x + w(x, t) = x + 1 [ e t e 16t] sin(4x). 15 //// Fie 214notes, version of 24 Juy 214, pge 1. Typeset t 16:32 Juy 24, 214.