Heat (or Diffusion) equation in 1D*

Transcription

1 Heat (or Diffusio) equatio i D* Derivatio of the D heat equatio Separatio of variables (refresher) Worked eamples *Kreysig, 8 th Ed, Sectios.4b

2 Physical assumptios We cosider temperature i a log thi wire of costat cross sectio ad homogeeous material The wire is perfectly isulated laterally, so heat flows oly alog the wire isulatio heat flow

3 Derivatio of the heat equatio i D Suppose that the thermal coductivity i the wire is K The specific heat is σ The desity of the material is ρ Cross sectioal area is A Deote the temperature at poit at time t by u(, u Heat flow ito bar across face at : KA At the face + δ So the et flow out is : u t u : KA + δ + δ u : KA δ c u, where Coservatio of heat gives : c K σρ u u KA δ σρa δ t

4 oudary ad Iitial Coditios As a first eample, we will assume that the perfectly isulated rod is of fiite legth ad has its eds maitaied at zero temperature. u(, u(, If the iitial temperature distributio i the rod is give by we have the iitial coditio : u(,) f ( ) Evidetly, from the boudary coditios : f () f ( ) f ( ),

5 Solutio by Separatio of Variables. Covert the PDE ito two separate ODEs. Solve the two (well kow) ODEs 3. Compose the solutios to the two ODEs ito a solutio of the origial PDE This agai uses Fourier series

6 Step : PDE ODEs The first step is to assume that the fuctio of two variables has a very special form: the product of two separate fuctios, each of oe variable, that is: where G& Assume that : u (, F( ) G( Differetiatig, we fid: dg dt ad F'' u FG& t u F'' G d F d So that equivaletly F G& G& c G c F '' F'' F G k The two (homogeeous) ODEs are: F'' kf G& kc G

7 Step a: solvig for F* We have to solve: case k µ A + F' ' kf > the geeral solutio is : F( ) Ae Ae ad so A µ + e µ µ + e µ Applyig the boudary coditios It follows that the oly case of case k : F( ) a + b Applyig the boudary coditios : a iterest is k p b case k p < : the solutio is F( ) Acos p + si p Applyig the boudary coditios, we fid F() A F( ) si p, ad so si p p that is : so that p F ( ) si * This aalysis is idetical to the Wave Equatio case we studied earlier

8 Step b: solvig for G c p λ we deote the wave equatio, as for ad,, t e t G G G ) ( λ λ + solutio is : whose geeral have We &

9 Step c: combiig F&G The solutio to the D diffusio equatio is : Iitial coditio : u(,) λ t u(, u(, e si si f ( ) As for the wave equatio, we fid : f ( )si d

10 Aalysig the solutio The solutio to the D diffusio equatio ca be writte as : where () u(, u u (, e The fuctios u The weights (, λ t f ( )si d si This emphasises that the solutio is a weighted sum of fuctios u where () u(, u(, i this case; ad are completely (that is, the costats, c) ad the boudary coditios (, determied by the geeric problem are determied by the iitial coditios, sice We ow take a closer look at the fuctios u

11 Sie term Cosider first the sie term : si Evidetly, the higher the value of the higher the frequecy compoet of f() that is aalysed

12 Epoetial term Cosider et the epoetial term : The λ c. e c Suppose that ad (see slides later) c t.58

13 The epoetial term is e deote t s ad σ Gabor fuctios c t c, the the term is e s σ a Gaussiamultipliedby a sie term.... a Gaussia so if we cosider (, s) isteadof (, we have to work with These are calledgabor fuctiosad are fudametal to sigalprocessigad optics This appears to be a short burst of sie wave i a Gaussia shaped evelope. This is fudametal to AM/FM commuicatios ad to wavelets Deis Gabor (bor udapest 9, died odo 979). eft erli for odo durig the 93s. Evetually professor of electroics at Imperial College. Credited with ivetio of the hologram i 947. Awarded the Nobel prize for physics i 97

14 Recall the solutio The solutio to the D diffusio equatio is : Iitial coditio : u(,) λ t u(, u(, e si si f ( ) As for the wave equatio, we fid : f ( )si d

15 Eample * egth of the bar :8 cms Siusoidal iitial coditios : f ( ) si 8 Copper : K.95 cal/(cm sec o C) (,) u, si f ( ) si 8 ρ 8.9 gm/cm 3 so σ.9 cal/(gm c λ.58 K ρσ u(, e o cm t C).758 si 8 / sec time required for the maimum temperature to reduce to 5 e -.785t 5 t 388s 6.5 mi o : *Kreysig, 8 th Ed, page 63

16 Eample Somewhat more realistically, we assume that the bar is iitially etirely at a costat temperature, the at time t, it is isulated ad queched, that is, the temperatures at its two eds istatly reduce to zero. Iitial coditio : u (,) T oudary coditios : u(, u(,, for all t > u(,, as t

17 Solutio si ), ( ), ( t e t u t u λ the solutio to the D diffusio equatio is : that Recall ) ( si,) ( T f u coditio : Iitial θ θ si si )si ( d T d T d f we fid : the wave equatio, As for

18 Solvig for Evidetly, cos si θ θdθ if m is eve, the this is if (m ) is odd, it equals (m ) ad so, we fid u(, m (m ) T (m ) c t e si(m )

19 Chagig the iitial coditio T T o Suppose that iitially f ( ) T T ( ) if if All the other boudary coditios remai the same: u(, for all t > u(, for all t > t u(, (queched at both eds) (must evetually cool dow to zero)

20 Solutio Recall that the solutio to the D diffusio equatio is : λ t u(, u(, e si Iitial coditio : u(,) si f ( ) We have to solve for the coefficiets usig Fourier series. Istead of orthogoality, we cosult HT

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22 f(φ) A -A / φ which is 8 A f ( φ ) (cosφ + cos 3φ +...) 9 et s shift this by / i the φ directio: θ φ-/ to give: g(θ) A This is ow what we eed betwee < θ < with Α Τ ο -A / θ 8A g( θ ) f ( θ + ) (cos( θ + ) + cos 3( θ + ) +...) 9 8 A (siθ + si 3θ +...) 9 si θ

23 So, ( 8T ) ad the fial solutio for this problem is T (, ( 8T ) si ( ) ep k( ) t Compare this with our previous eample of costat iitial temperature distributio: T (, 4T si ( ) ( ) ep k( ) t

24 Derivatio from electrostatics: the Telegraph Equatio I() Rδ I(+δ) R: Resistace per uit legth C: Capacitace per uit legth V() Cδ δ Ohm s law: V ( + δ) V ( ) IRδ I I V(+δ) V V + δ V + V From I C, we fid : t V I( + δ) I( ) Cδ t ( + δ) I( ) + δ ( ) ( ) δ We fid I V C t V RI V RC V t The diffusio equatio

25 Eample from electrostatics Iitially, a uiform coductor has zero potetial throughout. Oe ed () is the subjected to costat potetial V while the other ed () is held at zero potetial. What is the trasiet potetial distributio? Iitial coditios : u(,), for oudary coditios : u(, V, all t > u(,, all t > We agai use separatio of variables; but we eed to start from scratch because so far we have assumed that the boudary coditios were u(, u(, but this is ot the case here. We ow retrace the steps for the origial solutio to the heat equatio, otig the differeces

26 Step : PDE ODEs The first step is to assume that the fuctio of two variables has a very special form: the product of two separate fuctios, each of oe variable, that is: where G& Assume that : u (, F( ) G( Differetiatig, we fid: dg dt ad F'' u FG& t u F'' G d F d So that equivaletly F G& G& c G c F '' F'' F G k The two (homogeeous) ODEs are: No chage here!! F'' kf G& kc G

27 Step a: solvig for F&G.. k We have to solve: F' ' kf case k so that µ > the geeral solutio is : F( ) u(,) F( ) G() from which A as before Ae + e µ µ case k : so that F( ) a + b Applyig the boudary coditios : F() V b V F( ) a + b, so a F( ) V Sice k, we have G&, so that G is costat (igore)

28 Step b: Solvig for F&G k< case k p < : the solutio is F( ) Acos p + si p Applyig the boudary coditios, we fid F() A F( ) si p, ad so si p p that is : p so that F ( ) si p, ad, as before, we deote λ c This case is as before We have G& + λ G whose geeral solutio is : G ( e λ t

29 Step c: combiig F&G + si ), ( t e V t u λ the solutio to the D diffusio equatio is : I this case, + si ), (. si si si,) ( t e V V t u V d V d V V u θ θ θ λ Solutio is : lis o earlier to last use HT et (or which (by parts) so that coditio : Iitial