Class #25 Wednesday, April 19, 2018

Transcription

1 Cla # Wedesday, April 9, 8 PDE: More Heat Equatio with Derivative Boudary Coditios Let s do aother heat equatio problem similar to the previous oe. For this oe, I ll use a square plate (N = ), but I m goig to use differet boudary oditios. y T = T old I the previous problem, the bottom was kept hot, ad the other three edges were old. I that ase, eergy moved (i the form of heat) from the hot edge, ito the plate, ad the out of the plate ito the old stuff. L Q = Q = This time, the top boudary is still old, ad the bottom T = T hot boudary is still hot. However, i this problem, the side edges are isulated. That meas that there is o heat L flowig through these edges. This requires us to remember a little bit of physis: sie the heat flow Q is proportioal to temperature gradiet, the this T boudary oditio is the same thig as sayig that = alog both of these edges. Heat a move vertially alog these edges (i.e., i the y diretio, from the bottom to the top of the plate), but it a t ro out of the plate i the x diretio. Additioally, if we had isulated the top edge, T the we would have = there. But that s a differet problem x As before, we will o-dimesioalize this whole problem before startig it: Θ Θ = = = = = Θ ( y = ) = ( x ) ( x ) Θ y = N = = Muh of the early work is the same as before, so we ll steal results from ourselves: ky X = Aos( k Bsi( k Y = Ce De ky ky ( Aos( k Bsi( k )( Ce De ) Θ = = Later, we ll also eed the x derivatives of Θ, so: ( ) ky kasi( k kb os( k Y ( y) Page of 8

2 Cla # Wedesday, April 9, 8 Usig BC #: ( x = ) = = kasi( k) kb os( k) Y ( y) = kb Well, either k or B ould be zero. If we hoose k =, we ed up with othig at all, beause k is iside both the sie ad osie parts of the aswer. So, I gue this is tellig us that B =. Usig BC #: ( x = ) = = ka si( k ) Y ( y ) k = Well, that s a lot like what we had before. So far, the, we have this: = ky ky Θ = A os( Ce De Note that this ifiite series starts at zero, sie we have osie terms, but i the problem from the previous la, we oly started at =, sie we had oly sie terms there. BC #3: I wat to steal some more from ourselves as we evaluate the y-boudary oditios. After some effort the other day, we foud that the Y part of the solutio ame dow to a hyperboli sie futio; we have the same y boudaries here today, so we ll just borrow that aswer. So, our result so far is: Θ = = A os( sih( ( y)) Now, here s a really hard part. I otie that today, starts outig at zero, as just metioed. However, otie that whe =, the k =, too. Reall that whe we started our Separatio of Variables, we were solvig two ODE s: X ( = k X ( Y y Y ( y) = k We proeeded to solve them, resultig i the Θ above. But alog the way, without eve otiig it, we impliitly aumed that k. However, if k =, the these two equatios beome somewhat differet: X ( = Y ( y) = The solutio to these equatios is NOT sie ad osie (or eve expoetials), as we aumed i our aswer. Istead, whe k =, the aswers to these two ODE s are simply: Page of 8

3 Cla # Wedesday, April 9, 8 X k = = C x C, ad Y k = = C 3 y C 4. I other words, the total solutio is atually the thig you see below, where I ve moved the = part out of the summatio ad iluded it as its ow thig: = Θ = A os( sih( ( y)) C x C C y C 3 4 So THIS summatio oly starts at, but we have a extra part stuk oto the ed for =. Beause we are addig this hage to the total solutio, it makes us wish we had see this problem omig before we applied the boudary oditios. But we did t eve kow about this problem util we started doig the boudary oditios! I ay ase, for the side boudary oditios, we a just fix up this ew part of the solutio ad plug it right i there. k = = Redoig BC #: ( x = ) = = ( C )( C3 y C4 ), resultig i C =. Θ k = = C ( C3 y C4 ) Here, I ll ombie C C 3 ito a ew ostat C, ad the same with C 6 = C C 4 : Θ k = C y C = 6 This brigs our total result so far to: Θ = C y C A os( sih( ( y)) 6 = Redoig BC#3: ( y ) Θ = = = C C6 = A os( sih( ( )) So we lear that C = C 6 C6 = = C A os( Our equatio so far: Fially, the last BC: Θ = C y A os( sih( ( y)) = Bottom edge: BC #4: Θ ( y = ) = = C A os( sih( ( )) = = C = A os( sih( ) Page 3 of 8

4 Cla # Wedesday, April 9, 8 We ve see this kid of thig before. Today, I ll ivet a ew parameter (I seem to love doig that ) alled a : a = A sih( ) = = C a os( We eed to solve this for a. This is a Fourier series problem agai: ( C ) os( dx { Mathematia} a = = = for all values of!!!! C = Well, that s uexpeted good ews. So, this boudary oditio redues to: C = Ad the total solutio redues to: Θ = y Well, that s uts. That was way too muh work for suh a simple aswer. This aswer says that the temperature is ot really a futio of x after all, ad that the table gets ooler as you go from the bottom to the top. Θ Θ = = hek! I otie that: = everywhere; hek! Θ ( y = ) = hek! Θ ( y = ) = hek! No-Uiform Temperature alog Bottom Edge Beause this solutio proved too borig, let s do a differet, but similar problem. For this problem, the lower edge temperature is Θ bottom = x, (as opposed to equals hot = ). I other words, the bottom is ool o the left, hot o the right, ad risig steadily i betwee. All we re hagig is BC #4, so we oly eed to go bak to just before we applied BC#4: Θ = C y A os( sih( ( y)) = = x = C A os( sih( ( )) = x C = a os( Page 4 of 8

5 Cla # Wedesday, April 9, 8 ( ) a = x C os x dx a = for eve a 4 = for odd This suggests that we wat to out =, 3,, 7. But i fat, we are outig m =,, 3, 4. 4 So, to isolate the odd values, we use the same old trik as always: a m =. m Θ = Our solutio so far: C ( y ) ( ) = I gue there s oly oe more thig to disover, ad that is C. Reusig BC#, I otie that Θ(x =, y = ) = : = C = C = C = C C = ( ) = = = = ( ) ( ) ( ) sih sih 4 os(( ) sih(( ) ( y)) sih(( ) ) os[ ( ) ] sih[ ( ) ( ) ] sih[ ( ) ] [( ) ] [( ) ] (ifiite sum doe i Mathematia) y 4 os(( ) sih(( ) ( y)) Θ = sih(( ) ) This is plotted i Mathematia olie. Notie the lak of slope alog the side edges, ad the liear profile alog the bottom edge, i agreemet with our boudary oditios. Page of 8

6 Cla # Wedesday, April 9, 8 Usteady Heat Equatio T The heat equatio is: T = where α is a material property related to the odutivity. α t We will o-dimesioalize this as before. The oly ew feature this time is that we eed to α ivet a dimesiole time t: t = t. Makig this substitutio, ad beig too lazy to write L the symbol after every x, y, ad t, the heat equatio beomes: Θ =. Sie is really a t futio of all 3 spatial dimesios x, y, ad z, this problem is way too hard for us. So to simplify it, we ll look at a thi stik, whih has oly oe spatial dimesio: x. So, this is still a D problem, with idepedet variables x ad t. Θ = t I this problem, I ll eed three boudary oditios: two relatig to x, ad oly oe about t. Here s the problem we ll solve today: the boudary oditios are that the left edge (x = ) is held at Θ = ( old ), ad the right edge (x = ) is held at Θ = ( hot ). The third iitial oditio is that Θ = ( old ) everywhere alog the bar whe we start. I other words, it starts off all old, but it heats up over time as heat legth =, of ourse leaks ito the bar from the right had edge. Θ = We ve leared some triks about suh problems from the last two examples. We will aume that the aswer is omprised of two futios X ad T that are multiplied together (Separatio of Variables, or Eigefutio Deompositio ), ad we will add o a extra bit like we did i the last problem for = to aout for the steady state solutio: Θ = X ( T ( t) Θ x Θ = Addig the steady state solutio is a lot like what we did with o-homogeous ODE problems, where we aumed that y omplete = y C y P. So, Θ = X ( T ( t), ad Θ = Θ Θ. If a steady state temperature distributio is reahed, the by defiitio, the temperature is doe hagig. Let s deal with this part first. So, as t, = as t. Therefore, from the heat equatio, t Θ as t. Itegratig twie results i: = Θ = Cx C We a use the BC for the steady state solutio: Θ (x = ) = C =. Page 6 of 8

7 Cla # Wedesday, April 9, 8 Θ (x =) = C =. So, we kow this part already: Θ = x (i the log term, temperature rages from old to hot ) Now, let s do the Separatio of Variables: Θ = X ( T ( t) Θ = X T = XT t = X T The heat equatio therefore beomes: X T XT = X T = X T As before, this is oly poible if eah group is a ostat: X T = k = k X T X k X = T k T = From our hapter o ODE s, the aswers to these two equatios are: X = Aos( k B si( k T = Ce k t Sie I ll be multiplyig these expreios, the ostat C is redudat, ad we ll work it ito A ad B as we ve doe with all the other examples so far: k t [ Aos( k Bsi( k ] e Θ = where agai Θ = Θ Θ Now, we apply the boudary oditios: BC #: Θ(x=) = = Aos() Bsi() e k t (this last bit is Θ ) A = BC #: Θ(x=) = Our aswer so far: k t = Bsi( k) e (this last bit is Θ ) k t = Bsi( k) e Sie B, it must be true that: k = Well, we've see that before! = t Θ = B si( e x Page 7 of 8

8 Cla # Wedesday, April 9, 8 Fially, we apply the iitial oditio: BC #: Θ(t = ) = Note that starts outig at, ot, beause this is a sie series. This is a regular old Fourier series that we ve gotte good at: Mathematia says: B = x si( dx B = = = B si( e x x = B si( (or equivaletly, = ( ) =. B ( si( dx ) So, we re just about doe: = ( ) si t Θ = x e x Some plots are show i Mathematia olie. Eah separate lie represets a time, ad the vertial axis is temperature. Page 8 of 8