# Second Order Linear Partial Differential Equations. Part III

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1 Secod Order iear Partial Differetial Equatios Part III Oe-dimesioal Heat oductio Equatio revisited; temperature distributio of a bar with isulated eds; ohomogeeous boudary coditios; temperature distributio of a bar with eds kept at arbitrary temperatures; steady-state solutio Previously, we have leared that the geeral solutio of a partial differetial equatio is depedet of boudary coditios The same equatio will have differet geeral solutios uder differet sets of boudary coditios We shall witess this fact, by eamiig additioal eamples of heat coductio problems with ew sets of boudary coditios Keep i mid that, throughout this sectio, we will be solvig the same partial differetial equatio, the homogeeous oe-dimesioal heat coductio equatio: α u = u t where u( is the temperature distributio fuctio of a thi bar, which has legth, ad the positive costat α is the thermo diffusivity costat of the bar The equatio will ow be paired up with ew sets of boudary coditios 008 Zachary S Tseg E-3 -

2 Bar with both eds isulated Now let us cosider the situatio where, istead of them beig kept at costat 0 degree temperature, the two eds of the bar are also sealed with perfect isulatio so that o heat could escape to the outside eviromet (recall that the side of the bar is always perfectly isulated i the oedimesioal assumptio), or vice versa The ew boudary coditios are u (0, = 0 ad u (, = 0 *, reflectig the fact that there will be o heat trasferrig, spatially, across the poits = 0 ad = (Hece, this is a Neuma type problem) The heat coductio problem becomes the iitialboudary value problem below (Heat coductio eq) α u = u t, 0 < <, t > 0, (Boudary coditios) u (0, = 0, ad u (, = 0, (Iitial coditio) u( 0) = f () The first step is the separatio of variables The equatio is the same as before Therefore, it will separate ito the eact same two ordiary differetial equatios as i the first heat coductio problem see earlier The ew boudary coditios separate ito u (0, = 0 X (0)T( = 0 X (0) = 0 or T( = 0 u (, = 0 X ()T( = 0 X () = 0 or T( = 0 As before, we caot choose T( = 0 Else we could oly get the trivial solutio u( = 0, rather tha the geeral solutio Hece, the ew boudary coditios should be X (0) = 0 ad X () = 0 Agai, we ed up with a system of two simultaeous ordiary differetial equatios Plus a set of two boudary coditios that goes with the spatial idepedet variable : * The coditios say that the istataeous rate of chage with respect to the spatial variable (ie, the rate of poit-to-poit heat trasfer), is zero at each ed They do ot suggest that the temperature is costat (that is, there is o chage i temperature through time, which would require u t = 0) at each ed 008 Zachary S Tseg E-3 -

3 X + λx = 0, X (0) = 0 ad X () = 0, T + α λ T = 0 The secod step is to solve the eigevalue problem X + λx = 0, X (0) = 0 ad X () = 0 The result is summarized below ase : If λ < 0: No such λ eists ase : If λ = 0: Zero is a eigevalue Its eigefuctio is the costat fuctio X 0 = (or ay other ozero costa ase 3: If λ > 0: The positive eigevalues λ are π λ =, =,, 3, The correspodig eigefuctios that satisfy the said boudary coditios are π X = cos, =,, 3, The third step is to substitute the positive eigevalues foud above ito the equatio of t ad solve: π T + α T= Zachary S Tseg E-3-3

4 Notice that this is eactly the same equatio as i the first (both eds kept at 0 degree) heat coductio problem, due to the fact that both problems have the same set of eigevalues (but with differet eigefuctios) As a result, the solutios of the secod equatio are just the oes we have gotte the last time T ( α π t / = e, =,, 3, There is this etra eigevalue of λ = 0 that also eeds to be accouted for It has as a eigefuctio the costat X 0 () = Put λ = 0 ito the secod equatio ad we get T = 0, which has oly costat solutios T 0 ( = 0 Thus, we get the (arbitrary) costat fuctio u 0 ( = X 0 ()T 0 ( = 0 as a solutio Therefore, the solutios of the oe-dimesioal heat coductio equatio, with the boudary coditios u (0, = 0 ad u (, = 0, are i the form u 0 ( t ) = 0, α π t / π u ( = X ( T ( = e cos, =,, 3, The geeral solutio is their liear combiatio Hece, for a bar with both eds isulated, the heat coductio problem has geeral solutio: u( = 0 + = e α π t / π cos 008 Zachary S Tseg E-3-4

5 Now set t = 0 ad equate it with the iitial coditio u( 0) = f (): u( 0) = + π cos f ( ) 0 = = We see that the requiremet is that the iitial temperature distributio f () must be a Fourier cosie series That is, it eeds to be a eve periodic fuctio of period If f () is ot already a eve periodic fuctio, the we will eed to epad it ito oe ad use the resultig eve periodic etesio of f () i its place i the above equatio Oce this is doe, the coefficiets s i the particular solutio are just the correspodig Fourier cosie coefficiets of the iitial coditio f () (Ecept for the costat term, where the relatio 0 = a 0 / holds, istead) The eplicit formula for is, therefore, = a = f ( ) cos π d, =,, 3, 0 0 = a 0 / 008 Zachary S Tseg E-3-5

6 Eample: Solve the heat coductio problem 3 u = u t, 0 < < 8, t > 0, u (0, = 0, ad u (8, = 0, u( 0) = 9 3 cos(π/4) 6 cos(π) First ote that α = 3 ad = 8, ad the fact that the boudary coditios idicatig this is a bar with both eds perfectly isulated Substitute them ito the formula we have just derived to obtai the geeral solutio for this problem: u( = 0 + = e 3 π t / 64 π cos 8 heck the iitial coditio f (), ad we see that it is already i the require form of a Fourier cosie series of period 6 Therefore, there is o eed to fid its eve periodic etesio Istead, we just eed to etract the correct Fourier cosie coefficiets from f (): 0 = a 0 / = 9, = a = 3, 6 = a 6 = 6, = a = 0, for all other, 0,, or 6 Note that 0 is actually a 0 /, due to the way we write the costat term of the Fourier series But that should t preset ay more difficulty Sice whe you see a Fourier series, its costat term is already epressed i the form a 0 / Therefore, you could just copy it dow directly to be the 0 term without thikig Fially, the particular solutio is u( = 9 3e 3( ) π t / 64 π cos( ) 6e 4 3(6 ) π t / 64 cos(π ) 008 Zachary S Tseg E-3-6

7 Bar with two eds kept at arbitrary temperatures: A eample of ohomogeeous boudary coditios I both of the heat coductio iitial-boudary value problems we have see, the boudary coditios are homogeeous they are all zeros Now let us look at a eample of heat coductio problem with simple ohomogeeous boudary coditios The geeral set-up is the same as the first eample (where the both eds of the bar were kept at costat 0 degree, but were ot isulated), ecept ow the eds are kept at arbitrary (but costa temperatures of T degrees at the left ed, ad T degrees at the right ed The iitial coditio, as usual, is arbitrary The heat coductio problem is therefore give by the iitial-boudary value problem: α u = u t, 0 < <, t > 0, u(0, = T, ad u(, = T, u( 0) = f () The boudary coditios is ow ohomogeeous (uless T ad T are both 0, the the problem becomes idetical to the earlier eample), because at least oe of the boudary values are ozero The ohomogeeous boudary coditios are rather easy to work with, more so tha we might have reasoably epected First, let us be itroduced to the cocept of the steady-state solutio It is the part of the solutio u( that is idepedet of the time variable t Therefore, it is a fuctio of the spatial variable aloe We ca thusly rewrite the solutio u( as a sum of parts, a time-idepedet part ad a time-depedet part: u( = v() + w( Where v() is the steady-state solutio, which is idepedet of t, ad w( is called the trasiet solutio, which does vary with t 008 Zachary S Tseg E-3-7

8 The Steady-State Solutio The steady-state solutio, v(), of a heat coductio problem is the part of the temperature distributio fuctio that is idepedet of time t It represets the equilibrium temperature distributio To fid it, we ote the fact that it is a fuctio of aloe, yet it has to satisfy the heat coductio equatio Sice v = v ad v t = 0, substitutig them ito the heat coductio equatio we get α v = 0 Divide both sides by α ad itegrate twice with respect to we fid that v() must be i the form of a degree polyomial: v() = A + B The, rewrite the boudary coditios i terms of v: u(0, = v(0) = T, ad u(, = v() = T Apply those coditios to fid that: v(0) = T = A(0) + B = B B = T v() = T = A + B = A + T A = (T T ) / Therefore, T T v ( ) + T = Thig to remember: The steady-state solutio is a time-idepedet fuctio It is obtaied by settig the partial derivative(s) with respect to t i the heat equatio (or, later o, the wave equatio) to costat zero, ad the solvig the equatio for a fuctio that depeds oly o the spatial variable 008 Zachary S Tseg E-3-8

9 ommet: Aother way to uderstad the behavior of v() is to thik from the perspective of separatio of variables You could thik of the steadystate solutio as, durig the separatio of variables, the solutio you would have obtaied if T( =, the costat fuctio Therefore, the solutio is idepedet of time, or time-ivariat Hece, u( = X()T( = X() = v() We ca, i additio, readily see the substitutios required for rewritig the boudary coditios prior to solvig for the steady-state solutio: u(0, = X(0) = v(0) = T, ad u(, = X() = v() = T That is, just reame the fuctio u as v, igore the time variable t, ad put whatever -coordiate specified directly ito v() 008 Zachary S Tseg E-3-9

10 The solutio of bar with two eds kept at arbitrary temperatures Oce the steady-state solutio has bee foud, we ca set it aside for the time beig ad proceed to fid the trasiet part of solutio, w( First we will eed to rewrite the give iitial-boudary value problem slightly Keep i mid that the iitial ad boudary coditios as origially give were meat for the temperature distributio fuctio u( = v() + w( Sice we have already foud v(), we shall ow subtract out the cotributio of v() from the iitial ad boudary values The results will be the coditios that the trasiet solutio w( aloe must satisfy hage i the boudary coditios: u(0, = T = v(0) + w(0, w(0, = T v(0) = 0 u(, = T = v() + w(, w(, = T v() = 0 Note: Recall that u(0, = v(0) = T, ad u(, = v() = T hage i the iitial coditio: u( 0) = f () = v() + w( 0) w( 0) = f () v() osequetly, the trasiet solutio is a fuctio of both ad t that must satisfy the ew iitial-boudary value problem: α w = w t, 0 < <, t > 0, w(0, = 0, ad w(, = 0, w( 0) = f () v() 008 Zachary S Tseg E-3-0

11 008 Zachary S Tseg E-3 - Surprise! Notice that the ew problem just described is precisely the same iitial-boudary value problem associated with the heat coductio of a bar with both eds kept at 0 degree Therefore, the trasiet solutio w( of the curret problem is just the geeral solutio of the previous heat coductio problem (with homogeeous boudary coditios), that of a bar with eds kept costatly at 0 degree: e t w t π π α si ), ( / = = Where the coefficiets are equal to the correspodig Fourier sie coefficiets b of the (ewly rewritte) iitial coditio w( 0) = f () v() (Or those of w( 0) s odd periodic etesio, of period, if it is ot already a odd periodic fuctio of the correct period) Eplicitly, they are give by ( ) = = d v f b 0 si ) ( ) ( π, =,, 3, Fially, combiig the steady-state ad trasiet solutios together, the geeral solutio of the temperature distributio of a bar whose eds are kept at T degrees at the left, ad T degrees at the right, becomes e T T T t w v t u t π π α si ), ( ) ( ), ( / = + + = + =

12 Eample: Solve the heat coductio problem 8 u = u t, 0 < < 5, t > 0, u(0, = 0, u(5, = 90, u( 0) = si(π) 4si(π) + si(6π) First we ote that α = 8 ad = 5 Sice T = 0 ad T = 90, the steady-state solutio is v() = (90 0) / = We the subtract v(0) = 0 from u(0,, v(5) = 90 from u(5,, ad v() = from u( 0) to obtai a ew set of iitial-boudary values that the trasiet solutio w( aloe must satisfy: w(0, = 0, ad w(5, = 0, w( 0) = si(π) 4si(π) + si(6π) Base o α = 8 ad = 5, we write dow the geeral solutio: w( = = e 8 π t / 5 si π The ew iitial coditio, f () v(), is already a odd periodic fuctio of the period T = = 0 Therefore, just etract the correct Fourier sie coefficiets from it: 5 = b 5 =, 0 = b 0 = 4, 30 = b 30 =, = b = 0, for all other, 5, 0, or 30 Add together the steady-state ad trasiet solutios, we have u( = v( ) + w( = e 4e 8(0 ) π t / 5 si(π ) + e 8(30 5 ) π t / 5 8(5 ) π t / 5 si(6π ) si( π ) 008 Zachary S Tseg E-3 -

13 Back to the Steady-State Solutio A thig to remember: the steady-state solutio of the oe-dimesioal homogeeous heat coductio equatio is always i the form v() = A + B Sice it is idepedet of t, the effects of boudary coditios o v() are also simplified as: u( 0, = v( 0 ), ad u ( 0, = v ( 0 ) Nohomogeeous heat coductio equatios (that is, the equatios themselves cotai forcig fuctio terms; ot to be cofused with the homogeeous equatio with accompayig ohomogeeous boudary coditios that we have just see), however, could have differet forms of v() Fact: The steady-state temperature distributio satisfies the property: limu( = v( ) t This relatio is true oly for the solutio of heat coductio equatio (modelig diffusio-like processes that are thermodyamically irreversible) Physically speakig, v() describes the evetual state of maimum etropy as dictated by the secod law of Thermodyamics autio: The above relatio is ot true, i geeral, for solutios of the wave equatio i the et sectio This differece is due to the fact that the wave equatio models wave-like motios which are thermodyamically reversible processes 008 Zachary S Tseg E-3-3

14 Further eamples of steady-state solutios of the heat coductio equatio: Fid v(), give each set of boudary coditios below u(0, = 50, u (6, = 0 We are lookig for a fuctio of the form v() = A + B that satisfies the give boudary coditios Its derivative is the v () = A The two boudary coditios ca be rewritte to be u(0, = v(0) = 50, ad u (6, = v (6) = 0 Hece, v(0) = 50 = A(0) + B = B B = 50 v (6) = 0 = A A = 0 Therefore, v() = = 50 u(0, 4u (0, = 0, u (0, = 5 The two boudary coditios ca be rewritte to be v(0) 4 v (0) = 0, ad v (0) = 5 Hece, v(0) 4 v (0) = 0 = (A(0) + B) 4A = 4A + B v (0) = 5 = A A = 5 Substitute A = 5 ito the first equatio: 0 = 4A + B = 00 + B B = 00 Therefore, v() = Zachary S Tseg E-3-4

15 3 u(0, = 35, u(4, + 3u (4, = 0 Rewritig the boudary coditios: v(0) = 35, ad v(4) + 3 v (4) = 0 Hece, v(0) = 35 = A(0) + B = B B = 35 v(4) + 3 v (4) = 0 = (A(4) + B) + 3A = 7A + B 0 = 7A + 35 A = 5 Therefore, v() = ommet: Notice how i each eample we have see, the steady-state solutio is uiquely determied by the boudary coditios aloe I geeral this is true that the steady-state solutio is almost always idepedet of the iitial coditio The loe eceptio is the isulatededs problem (with boudary coditios u (0, = 0 = u (, ) I this special case, the boudary coditios oly tell us that the steady-state solutio should be a costat, which turs out to be the costat term of the geeral solutio As we have see, that costat is ideed depedet o the iitial coditio it is just the costat term of the iitial coditio, as the latter is epaded ito a Fourier cosie series of period I all other cases the boudary coditios aloe determie the steady-state solutio (therefore, the limitig temperature) of a problem 008 Zachary S Tseg E-3-5

16 Summary: Solvig Secod Order iear Partial Differetial Equatios The Method of Separatio of Variables: 0 (If the boudary coditios are ohomogeeous) Solve for the steady-state solutio, v(), which is a fuctio of oly that satisfies both the PDE ad the boudary coditios Afterwards rewrite the problem s boudary ad iitial coditios to subtract out the cotributio from the steady-state solutio Therefore, the problem is ow trasformed ito oe with homogeeous boudary coditios Separate the PDE ito ODEs of oe idepedet variable each Rewrite the boudary coditios so they associate with oly oe of the variables Oe of the ODEs is a part of a two-poit boudary value problem Solve this problem for its eigevalues ad eigefuctios 3 Solve the other ordiary differetial equatio 4 Multiply the results from steps () ad (3), ad sum up all the products to fid the geeral solutio respect to the give homogeeous boudary coditios Add to it the steady-state solutio (from step 0, if applicable) to fid the overall geeral solutio 5 Epad the iitial coditio ito a suitable (eg a sie or a cosie series, depedig o whichever type the eigefuctios are) Fourier series The compare it agaist u( 0) to fid the coefficiets for the particular solutio 008 Zachary S Tseg E-3-6

17 Summary of Heat oductio Problems Here is a list of heat coductio problems ad their solutios All solutios obey the homogeeous oe-dimesioal heat coductio equatio α u = u t They oly differ i boudary coditios (which are give below for each problem) The iitial coditio is always arbitrary, u( 0) = f () Bar with both eds kept at 0 degree (boudary coditios: u(0, = 0, u(, = 0) u( = = e α π t / π si Epad f () to be a Fourier sie series, the = b Steady-state solutio is v() = 0 Bar with both eds perfectly isulated (boudary coditios: u (0, = 0, u (, = 0) u( = 0 + = e α π t / π cos Epad f () to be a Fourier cosie series, the 0 = a 0 /, ad = a, =,, 3, Steady-state solutio is v() = 0 3 Bar with T degrees at the left ed, ad T degrees at the right ed (boudary coditios: u(0, = T, u(, = T ) T u( = T α π t / + T + e = π si Epad (f () v()) to be a Fourier sie series, the = b T T Steady-state solutio is v ( ) = + T 008 Zachary S Tseg E-3-7

18 Eercises E-3: 7 Fid the steady-state solutio v() of the heat coductio equatio, give each set of boudary coditios below u(0, = 00, u(0, = 00 u(0, = 00, u (0, = 50 3 u (0, = 8, u(0, = 00 4 u (0, = 30, u (0, = 0 5 u(0, + u (0, = 0, u(0, = 00 6 u(0, + u (0, = 0, u(0, u (0, = 00 7 u(0, 0u (0, = 30, u(0, 5u (0, = 0 8 Solve the heat coductio problem of the give iitial coditios u = u t, 0 < < 6, t > 0, u (0, = 0, ad u (6, = 0, (a) u( 0) = π + 3cos(π) 4cos(3π / ) cos(3π), (b) u( 0) = 4, (c) u( 0) =, (d) u( 0) = 0 9 For each particular solutio foud i #8, fid limu( t 0 Solve the heat coductio problem of the give iitial coditios u = u t, 0 < < 6, t > 0, u(0, = 40, ad u(6, = 0, (a) u( 0) = si(π) si(5π / ), (b) u( 0) = 0 For each particular solutio foud i #0, fid limu(4, t 008 Zachary S Tseg E-3-8

19 osider the heat coductio problem 9 u = u t, 0 < < 4, t > 0, u(0, = 3, ad u(4, = 3, u( 0) = f () (a) Fid its geeral solutio (b) Write a eplicit formula to determie the coefficiets (c) Base o (a), fid limu( t 3 osider the heat coductio equatio below, subject to each of the 4 sets of boudary coditios 9 u = u t, 0 < < 0, t > 0, (i) u(0, = 0, ad u(0, = 0, (ii) u (0, = 0, ad u (0, = 0, (iii) u(0, = 0, ad u(0, = 00 (iv) u(0, = 00, ad u(0, = 50 Give the commo iitial coditio u( 0) = 300, determie the temperature at the midpoit of the bar (at = 5) after a very log time has elapsed Which set of boudary coditios will give the highest temperature at that poit? 008 Zachary S Tseg E-3-9

20 Aswers E-3: v() = v() = v() = v() does ot eist 5 v() = 0 6 v() = v() = + 0 π t 9π t / 8π t 8 (a) u( = π+ 3e cos( π) 4e cos(3π / ) e cos(3π ), (b) u( = 4, (d) u( = 0 9 (a) limu( = π ; (b) lim u( = 4 ; (c) lim u( = 5 / 3 ; t (d) lim u( = 0 t t t 8π t 5π t / 0 (a) u( = e si(π) e si(5π / ) (a) ad (b) lim u(4, = 0 (a) t u( = = (b) = ( f ( ) 3) (c) lim u( = 3 t e 9 π t /6 0 4 π si 4 π si d, =,, 3, 3 Boudary coditios (ii) will give the highest temperature At = 5, the temperature is lim u(5, = v(5) = 300 t 008 Zachary S Tseg E-3-0

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