Partial Differential Equations

Prtil Differentil Equtions If the suject of ordinry differentil equtions is lrge, this is enormous. I m going to exmine only one corner of it, nd will develop only one tool to hndle it: Seprtion of Vriles. Another mjor tool is the method of chrcteristics nd I ll not go eyond mentioning the word. When I develop technique to hndle the het eqution or the potentil eqution, don t think tht it stops there. The sme set of tools will work on the Schroedinger eqution in quntum mechnics nd on the wve eqution in its mny incrntions.. The Het Eqution The flow of het in one dimension is descried y the het conduction eqution P = κa T x (.) where P is the power in the form of het energy flowing towrd positive x through wll nd A is the re of the wll. κ is the wll s therml conductivity. Put this eqution into words nd it sys tht if thin sl of mteril hs temperture on one side different from tht on the other, then het energy will flow through the sl. If the temperture difference is ig or the wll is thin ( T/ x is ig) then there s ig flow. The minus sign sys tht the energy flows from hot towrd cold. When more het comes into region thn leves it, the temperture there will rise. This is descried y the specific het, c. dq = mc dt, or dq dt = mcdt dt (.) Agin in words, the temperture rise in chunk of mteril is proportionl to the mount of het dded to it nd inversely proportionl to its mss. A P (x, t) P (x + x, t) x x + x For sl of re A, thickness x, nd mss density ρ, let the coordintes of the two sides e x nd x + x. dq m = ρa x, nd = P (x, t) P (x + x, t) dt The net power into this volume is the power in from one side minus the power out from the other. Put these three equtions together. dq dt = mcdt dt = ρa x cdt dt = κa T (x, t) x T (x + x, t) + κa x If you let x here, ll you get is =, not very helpful. Insted divide y x first nd then tke the limit. T t = + κa ( ) T (x + x, t) T (x, t) ρca x x x Jmes Nering, University of Mimi

nd in the limit this is Prtil Differentil Equtions T t = κ T cρ x (.3) I ws little cvlier with the nottion in tht I didn t specify the rgument of T on the left side. You could sy tht it ws (x + x/, t), ut in the limit everything is evluted t (x, t) nywy. I lso ssumed tht κ, the therml conductivity, is independent of x. If not, then it stys inside the derivtive, T t = cρ x ( κ T ) x (.4) In Three Dimensions In three dimensions, this ecomes T t = κ cρ T (.5) Roughly speking, the temperture in ox cn chnge ecuse of het flow in ny of three directions. More precisely, the correct three dimensionl eqution tht replces Eq. (.) is H = κ T (.6) where H is the het flow vector. Tht is the power per re in the direction of the energy trnsport. H. d A = dp, the power going cross the re d A. The totl het flowing into volume is ˆn dq dt = dp = H. d A (.7) where the minus sign occurs ecuse this is the het flow in. For smll volume V, you now hve m = ρ V nd mc T = ρ V c T t t = H. da Divide y V nd tke the limit s V. The right hnd side is the divergence, Eq. (9.9). ρc T t = lim V V H. d A =. H = +. κ T = +κ T The lst step gin ssumes tht the therml conductivity, κ, is independent of position.. Seprtion of Vriles How do you solve these equtions? I ll strt with the one-dimensionl cse nd use the method of seprtion of vriles. The trick is to strt y looking for solution to the eqution in the form of product of function of x nd function of t. T (x, t) = f(t)g(x). I do not ssume tht every solution to the eqution will look like this tht s just not true. Wht will hppen is tht I ll e le to express every solution s sum of such fctored forms. Tht this is so is theorem tht I don t pln to prove here. For tht you should go to purely mthemticl text on PDEs.

Prtil Differentil Equtions 3 If you wnt to find out if you hve solution, plug in: T t = κ T cρ x is df dt g = κ cρ f d g dx Denote the constnt y κ/cρ = D nd divide y the product fg. df f dt = D d g g dx (.8) The left side of this eqution is function of t lone, no x. The right side is function of x lone with no t, hence the nme seprtion of vriles. Becuse x nd t cn vry quite independently of ech other, the only wy tht this cn hppen is if the two side re constnt (the sme constnt). df f dt = α nd D d g g dx = α (.9) At this point, the constnt α cn e nything, even complex. For prticulr specified prolem there will e oundry conditions plced on the functions, nd those will constrin the α s. If α is rel nd positive then g(x) = A sinh α/d x + B cosh α/d x nd f(t) = e αt (.) For negtive rel α, the hyperolic functions ecome circulr functions. g(x) = A sin α/d x + B cos α/d x nd f(t) = e αt (.) If α = then g(x) = Ax + B, nd f(t) = constnt (.) For imginry α the f(t) is oscillting nd the g(x) hs oth exponentil nd oscilltory ehvior in spce. This cn relly hppen in very ordinry physicl situtions; see section.3. This nlysis provides solution to the originl eqution (.3) vlid for ny α. A sum of such solutions for different α s is lso solution, for exmple T (x, t) = A e αt sin α /D x + A e αt sin α /D x or ny other liner comintion with vrious α s T (x, t) = f α (t)g α (x) {α s} It is the comined product tht forms solution to the originl prtil differentil eqution, not the seprte fctors. Determining the detils of the sum is jo for Fourier series. Exmple A specific prolem: You hve sl of mteril of thickness L nd t uniform temperture T. Plunge it into ice wter t temperture T = nd find the temperture inside t lter times. The oundry condition here is tht the surfce temperture is zero, T (, t) = T (L, t) =. This constrins the seprted solutions, requiring tht g() = g(l) =. For this to hppen you cn t use the hyperolic

Prtil Differentil Equtions 4 functions of x tht occur when α >, you will need the circulr functions of x, sines nd cosines, implying tht α <. Tht is lso comptile with your expecttion tht the temperture should pproch zero eventully, nd tht needs negtive exponentil in time, Eq. (.). g(x) = A sin kx + B cos kx, with k = α/d nd f(t) = e Dk t g() = implies B =. g(l) = implies sin kl =. The sine vnishes for the vlues nπ where n is ny integer, positive, negtive, or zero. This implies kl = nπ, or k = nπ/l. The corresponding vlues of α re α n = Dn π /L, nd the seprted solution is sin ( nπx/l ) e n π Dt/L (.3) If n = this whole thing vnishes, so it s not much of solution. (Not so fst there! See prolem..) Notice tht the sine is n odd function so when n < this expression just reproduces the positive n solution except for n overll fctor of ( ), nd tht fctor ws ritrry nywy. The negtive n solutions re redundnt, so ignore them. The generl solution is sum of seprted solutions, see prolem.3. T (x, t) = n sin nπx L e n π Dt/L (.4) The prolem now is to determine the coefficients n. This is why Fourier series were invented. (Yes, literlly, the prolem of het conduction is where Fourier series strted.) At time t = you know the temperture distriution is T = T, constnt on < x < L. This generl sum must equl T t time t =. T (x, ) = n sin nπx ( < x < L) L Multiply y sin ( mπx/l ) nd integrte over the domin to isolte the single term, n = m. L dx T sin mπx L = m T [ cos mπ] L mπ = m L L dx sin mπx L This expression for m vnishes for even m, nd when you ssemle the whole series for the temperture you hve T (x, t) = 4 π T mπx π sin Dt/L m L e m (.5) m odd For smll time, this converges, ut very slowly. For lrge time, the convergence is very fst, often needing only one or two terms. As the time pproches infinity, the interior temperture pproches the surfce temperture of zero. The grph shows the temperture profile t sequence of times. T L x

Prtil Differentil Equtions 5 The curves show the temperture dropping very quickly for points ner the surfce (x = or L). It drops more grdully ner the center ut eventully goes to zero. You cn see tht the oundry conditions on the temperture led to these specific oundry conditions on the sines nd cosines. This is exctly wht hppened in the generl development of Fourier series when the fundmentl reltionship, Eq. (5.5), required certin oundry conditions in order to get the orthogonlity of the solutions of the hrmonic oscilltor differentil eqution. Tht the function vnishes t the oundries ws one of the possile wys to insure orthogonlity..3 Oscillting Tempertures Tke very thick sl of mteril nd ssume tht the temperture on one side of it is oscillting. Let the mteril occupy the spce < x < nd t the coordinte x = the temperture is vrying in time s T cos ωt. Is there ny rel sitution in which this hppens? Yes, the surfce temperture of the Erth vries periodiclly from summer to winter (even in Florid). Wht hppens to the temperture underground? The differentil eqution for the temperture is still Eq. (.3), nd ssume tht the temperture inside the mteril pproches T = fr wy from the surfce. Seprtion of vriles is the sme s efore, Eq. (.9), ut this time you know the time dependence t the surfce. It s typicl in cses involving oscilltions tht it is esier to work with complex exponentils thn it is to work with sines nd cosines. For tht reson, specify tht the surfce temperture is T e iωt insted of cosine, understnding tht t the end of the prolem you must tke the rel prt of the result nd throw wy the imginry prt. The imginry prt corresponds to solving the prolem for surfce temperture of sin ωt insted of cosine. It s esier to solve the two prolems together then either one seprtely. (The minus sign in the exponent of e iωt is ritrry; you could use plus insted.) The eqution (.9) sys tht the time dependence tht I expect is The eqution for the x-dependence is then df f dt = α = ( iωe iωt ) e iωt = iω D d g = αg = iωg dx This is gin simple exponentil solution, sy e βx. Sustitute nd you hve Dβ e βx = iωe βx, implying β = ± iω/d (.6) Evlute this s i = (e iπ/) / = e iπ/4 = i Let β = ω/d, then the solution for the x-dependence is g(x) = Ae ( i)β x + Be ( i)β x (.7) Look t the ehvior of these two terms. The first hs fctor tht goes s e +x nd the second goes s e x. The temperture t lrge distnces is supposed to pproch zero, so tht sys tht A =. The solutions for the temperture is now Be iωt e ( i)β x (.8)

Prtil Differentil Equtions 6 The further condition is tht t x = the temperture is T e iωt, so tht tells you tht B = T. T (x, t) = T e iωt e ( i)β x = T e β x e i( ωt+β x) (.9) When you rememer tht I m solving for the rel prt of this solution, the finl result is T ωt = ωt = π/4 ωt = π/ T e β x cos(β x ωt) (.) This hs the ppernce of temperture wve moving into the mteril, leit very strongly dmped one. In hlf wvelength of this wve, β x = π, nd t tht point the mplitude coming from the exponentil fctor out in front is down y fctor of e π =.4. Tht s rely noticele. This is why wine cellrs re cellrs. Also, you cn see tht t distnce where β x > π/ the temperture chnge is reversed from the vlue t the surfce. Some distnce underground, summer nd winter re reversed. This sme sort of eqution comes up with the study of eddy currents in electromgnetism, so the sme sort of results otin..4 Sptil Temperture Distriutions The governing eqution is Eq. (.5). For n exmple of prolem tht flls under this heding, tke cue tht is heted on one side nd cooled on the other five sides. Wht is the temperture distriution within the cue? How does it vry in time? I ll tke simpler version of this prolem to strt with. First, I ll work in two dimensions insted of three; mke it very long rectngulr shped rod, extending in the z-direction. Second, I ll look for the equilirium solution, for which the time derivtive is zero. These restrictions reduce the eqution (.5) to T = T x + T y = (.) I ll specify the temperture T (x, y) on the surfce of the rod to e zero on three fces nd T on the fourth. Plce the coordintes so tht the length of the rod is long the z-xis nd the origin is in one corner of the rectngle. x T (, y) = ( < y < ), T (x, ) = ( < x < ) T (, y) = ( < y < ), T (x, ) = T ( < x < ) (.) y T x O Look t this prolem from severl different ngles, ter it prt, look t lot of specil cses, nd see wht cn go wrong. In the process you ll see different techniques nd especilly lot of pplictions of Fourier series. This single prolem will illustrte mny of the methods used to understnd oundry vlue prolems.

Prtil Differentil Equtions 7 Use the sme method used efore for het flow in one dimension: seprtion of vriles. Assume solution to e the product of function of x nd function of y, then plug into the eqution. T (x, y) = f(x)g(y), then T = d f(x) dx g(y) + f(x) d g(y) dy = Just s in Eq. (.8), when you divide y fg the resulting eqution is seprted into term involving x only nd one involving y only. d f(x) f dx + d g(y) g dy = Becuse x nd y cn e vried independently, these must e constnts dding to zero. d f(x) f dx = α, nd d g(y) g dy = α (.3) As efore, the seprtion constnt cn e ny rel or complex numer until you strt pplying oundry conditions. You recognize tht the solutions to these equtions cn e sines or cosines or exponentils or hyperolic functions or liner functions, depending on wht α is. The oundry conditions stte tht the surfce temperture is held t zero on the surfces x = nd x =. This suggests looking for solutions tht vnish there, nd tht in turn sys you should work with sines of x. In the other direction the surfce temperture vnishes on only one side so you don t need sines in tht cse. The α = cse gives liner functions is x nd in y, nd the fct tht the temperture vnishes on x = nd x = kills these terms. (It does doesn t it?) Pick α to e negtive rel numer: cll it α = k. The ccompnying eqution for g is now d f(x) dx = k f = f(x) = A sin kx + B cos kx d g(y) dy = +k g = g(y) = C sinh ky + D cosh ky (Or exponentils if you prefer.) The comined, seprted solution to T = is (A sin kx + B cos kx)(c sinh ky + D cosh ky) (.4) The generl solution will e sum of these, summed over vrious vlues of k. This is where you hve to pply the oundry conditions to determine the llowed k s. left: T (, y) = = B(C sinh ky + D cosh ky), so B = (This holds for ll y in < y <, so the second fctor cn t vnish unless oth C nd D vnish. If tht is the cse then everything vnishes.) right: T (, y) = = A sin k(c sinh ky + D cosh ky), so sin k = (The fctor with y cn t vnish or everything vnishes. If A = then everything vnishes. All tht s left is sin k =.) ottom: T (x, ) = = A sin kx D, so D =

Prtil Differentil Equtions 8 (If A = everything is zero, so it hs to e D.) You cn now write generl solution tht stisfies three of the four oundry conditions. Comine the coefficients A nd C into one, nd since it will e different for different vlues of k, cll it γ n. T (x, y) = n= γ n sin nπx nπy sinh (.5) The nπ/ ppers ecuse sin k =, nd the limits on n omit the negtive n ecuse they re redundnt. Now to find ll the unknown constnts γ n, nd s efore tht s where Fourier techniques come in. The fourth side, t y =, hs temperture T nd tht implies n= γ n sin nπx nπ sinh = T On this intervl < x < these sine functions re orthogonl, so you tke the sclr product of oth side with the sine. dx sin mπx n= γ n sin nπx nπ sinh = dx sin mπx T γ m sinh mπ [ = T ( ) m ] mπ Just the odd m terms re present, m = l +, so the result is T (x, y) = 4 π T l= l + sinh ( (l + )πy/ ) sinh ( (l + )πx ) sin (l + )π/ (.6) You re not done. Does this mke sense? The dimensions re clerly correct, ut fter tht it tkes some work. There s relly just one prmeter tht you hve to ply round with, nd tht s the rtio /. If it s either very ig or very smll you my e le to check the result. y T x O O If, it looks lmost like one-dimensionl prolem. It is thin sl with temperture T on one side nd zero on the other. There s little vrition long the x-direction, so the equilirium eqution is T = = T x + T y T y

Prtil Differentil Equtions 9 This simply sys tht the second derivtive with respect to y vnishes, so the nswer is the stright line T = A + By, nd with the condition tht you know the temperture t y = nd t y = you esily find T (x, y) T y/ Does the exct nswer look like this? It doesn t seem to, ut look closer. If then ecuse < y < you lso hve y. The hyperolic function fctors in Eq. (.6) will hve very smll rguments, proportionl to /. Recll the power series expnsion of the hyperolic sine: sinh x = x +. These fctors ecome pproximtely The temperture solution is then sinh ( (l + )πy/ ) sinh ( (l + )πy/ ) (l + )π/ (l + )π/ = y T (x, y) 4 π T l= y (l + )πx y sin = T l + Where did tht lst eqution come from? The coefficient of y/ is just the Fourier series of the constnt T in terms of sines on < x <. Wht out the opposite extreme, for which? This is the second picture just ove. Insted of eing short nd wide it is tll nd nrrow. For this cse, look gin t the rguments of the hyperolic sines. Now π/ is lrge nd you cn pproximte the hyperolic functions y going ck to their definition. sinh x = ex + e x ex, for x The denomintors in ll the terms of Eq. (.6) re lrge, e π/ (or lrger still ecuse of the (l + )). This will mke ll the terms in the series extremely smll unless the numertors re correspondingly lrge. This mens tht the temperture stys ner zero unless y is lrge. Tht mkes sense. It s only for y ner the top end tht you re ner to the wll with temperture T. You now hve the cse for which nd y. This mens tht I cn use the pproximte form of the hyperolic function for lrge rguments. sinh ( (l + )πy/ ) sinh ( (l + )π/ ) e(l+)πy/ = e(l+)π(y )/ e (l+)π/ The temperture distriution is now pproximtely T (x, y) 4 π T l= (l + )πx l + e (l+)π( y)/ sin (.7) As compred to the previous pproximtion where, you cn t s esily tell whether this is plusile or not. You cn however lern from it. See lso prolem.3. At the very top, where y = this reduces to the constnt T tht you re supposed to hve t tht position. Recll the Fourier series for constnt on < x <. Move down from y = y the distnce, so tht y =. Tht s distnce from the top equl to the width of the rectngle. It s still rther close to the end, ut look t the series for tht

Prtil Differentil Equtions position. x = x = y = y = T (x, ) 4 π T l= (l + )πx l + e (l+)π sin For l =, the exponentil fctor is e π =.43, nd for l = this fctor is e 3π =.8. This mens tht mesured from the T end, within the very short distnce equl to the width, the temperture hs dropped 95% of the wy down to its limiting vlue of zero. The temperture in the rod is quite uniform until you re very close to the heted end. The Het Flow into the Box All the preceding nlysis nd discussion ws intended to mke this prolem nd its solution sound oh-so-plusile. There s more, nd it isn t pretty. The temperture on one of the four sides ws given s different from the tempertures on the other three sides. Wht will the het flow into the region e? Tht is, wht power must you supply to mintin the temperture T on the single wll? At the eginning of this chpter, Eq. (.), you hve the eqution for the power through n re A, ut tht eqution ssumed tht the temperture grdient T/ x is the sme ll over the re A. If it isn t, you simply turn it into density. P = κ A T x, nd then P A dp da = κ T x (.8) Equivlently, just use the vector form from Eq. (.6), H = κ T. In Eq. (.) the temperture is T long y =, nd the power density (energy / (time. re)) flowing in the +y direction is κ T/ y, so the power density flowing into this re hs the reversed sign, +κ T/ y (.9) The totl power flow is the integrl of this over the re of the top fce. Let L e the length of this long rectngulr rod, its extent in the z-direction. The element of re long the surfce t y = is then da = Ldx, nd the power flow into this fce is L dx κ T y The temperture function is the solution Eq. (.6), so differentite tht eqution with respect to y. L dx κ 4 π T y= [(l + )π/] cosh ( (l + )πy/ ) l + sinh ( (l + )πx ) sin (l + )π/ l= = 4LκT (l + )πx dx sin l= t y =

Prtil Differentil Equtions nd this sum does not converge. I m going to push hed nywy, temporrily pretending tht I didn t notice this minor difficulty with the series. Just go hed nd integrte the series term y term nd hope for the est. = 4LκT = 4LκT π l= l= [ ( ) ] cos (l + )π + π(l + ) l + = This infinite series for the totl power entering the top fce is infinite. The series doesn t converge (use the integrl test). This innocuous-seeming prolem is suddenly pthologicl ecuse it would tke n infinite power source to mintin this temperture difference. Why should tht e? Look t the corners. You re trying to mintin non-zero temperture difference (T ) etween two wlls tht re touching. This cn t hppen, nd the equtions re telling you so! It mens tht the oundry conditions specified in Eq. (.) re impossile to mintin. The temperture on the oundry t y = cn t e constnt ll the wy to the edge. It must drop off to zero s it pproches x = nd x =. This mkes the prolem more difficult, ut then relity is typiclly more complicted thn our simple, idelized models. Does this mke the solution Eq. (.6) vlueless? No, it simply mens tht you cn t push it too hrd. This solution will e good until you get ner the corners, where you cn t possily mintin the constnt-temperture oundry condition. In other regions it will e good pproximtion to the physicl prolem..5 Specified Het Flow In the previous exmples, I specified the temperture on the oundries nd from tht I determined the temperture inside. In the prticulr exmple, the solution ws not physiclly plusile ll the wy to the edge, though the mthemtics were (I hope) enlightening. Insted, I ll reverse the process nd try to specify the size of the het flow, computing the resulting temperture from tht. This time perhps the results will e etter reflection of relity. Eqution (.9) tells you the power density t the surfce, nd I ll exmine the cse for which this is constnt. Cll it F. (There s not conventionl symol, so this will do.) The plus sign occurs ecuse the flow is into the ox. +κ T y (x, ) = F The other three wlls hve the sme zero temperture conditions s Eq. (.). Which forms of the seprted solutions must I use now? The sme ones s efore or different ones? Look gin t the α = solutions to Eqs. (.3). Tht solution is (A + Bx)(C + Dy) In order to hndle the fct tht the temperture is zero t y = nd tht the derivtive with respect to y is given t y =, (A + Bx)(C) = nd (A + Bx)(D) = F /κ, implying C = = B, then AD = F /κ = F κ y (.3) This mtches the oundry conditions t oth y = nd y =. All tht s left is to mke everything work t the other two fces.

y O F Prtil Differentil Equtions x F y/κ O y F y/κ If I cn find solution tht equls F y/κ on the left nd right fces then it will cncel the +F y/κ tht Eq. (.3) provides. But I cn t distur the top nd ottom oundry conditions. The wy to do tht is to find functions tht equl zero t y = nd whose derivtive equls zero t y =. This is fmilir sort of condition tht showed up severl times in chpter five on Fourier series. It is equivlent to sying tht the top surfce is insulted so tht het cn t flow through it. You then use superposition to comine the solution with uniform het flow nd the solution with n insulted oundry. Insted of Eq. (.4), use the opposite sign for α, so the solutions re of the form (A sin ky + B cos ky)(c sinh kx + D cosh kx) I require tht this equls zero t y =, so tht sys ( + B)(C sinh kx + D cosh kx) = so B =. Now require tht the derivtive equls zero t y =, so Ak cos k =, or k = (n + )π for n =,,... The vlue of the temperture is the sme on the left tht it is on the right, so C sinh k + D cosh k = C sinh k + D cosh k = C = D( cosh k)/ sinh k (.3) This is strting to get messy, so it s time to look round nd see if I ve missed nything tht could simplify the clcultion. There s no gurntee tht there is ny simpler wy, ut it is lwys worth looking. The fct tht the system is the sme on the left s on the right mens tht the temperture will e symmetric out the centrl xis of the ox, out x = /. Tht it is even out this point implies tht the hyperolic functions of x should e even out x = /. You cn do this simply y using cosh out tht point. A sin ky ( D cosh k(x )) Put these together nd you hve sum ( (n + n sin )πy ) ( (n + cosh )π(x ) ) n= x (.3) Ech of these terms stisfies Lplce s eqution, stisfies the oundry conditions t y = nd y =, nd is even out the centerline x = /. It is now prolem in Fourier series to mtch the conditions t x =. They re then utomticlly stisfied t x =. ( (n + n sin )πy ) ( (n + cosh )π ) n= = F y κ (.33)

Prtil Differentil Equtions 3 The sines re orthogonl y the theorem Eq. (5.5), so you cn pick out the component n y the orthogonlity of these sis functions. ( (n + u n = sin )πy ), then um, left side = u m, right side ( (m + or, m um, u m cosh )π ) = F um, y κ Write this out; do the integrls, dd the liner term, nd you hve y T (x, y) = F κ 8F ( ) n κπ (n + ) (.34) n= ( (n + sin )πy ) ( (n + cosh )π(x ) ) ( (n + sech )π ) Now nlyze this to see if it mkes sense. I ll look t the sme cses s the lst time: nd. The simpler cse, where the ox is short nd wide, hs. This mkes the rguments of the cosh nd sech lrge, with n / in them. For lrge rgument you cn pproximte the cosh y cosh x e x /, x Now exmine typicl term in the sum (.34), nd I hve to e little more specific nd choose x on the left or right of /. The reson for tht is the preceding eqution requires x lrge nd positive. I ll tke x on the right, s it mkes no difference. The hyperolic functions in (.34) re pproximtely exp ( (n + )π(x )/ ) exp ( (n + )π/ ) = e((n+)π(x )/) As long s x is not ner the end, tht is, not ner x =, the quntity in the exponentil is lrge nd negtive for ll n. The exponentil in turn mkes this extremely smll so tht the entire sum ecomes negligile. The temperture distriution is then the single term T (x, y) F y κ It s essentilly one dimensionl prolem, with the het flow only long the y direction. In the reverse cse for which the ox is tll nd thin,, the rguments of the hyperolic functions re smll. This invites power series expnsion, ut tht pproch doesn t work. The nlysis of this cse is quite tricky, nd I finlly concluded tht it s not worth the troule to write it up. It leds to rther complicted integrl..6 Electrosttics The eqution for the electrosttic potentil in vcuum is exctly the sme s Eq. (.) for the temperture in sttic equilirium, V =, with the electric field E = V. The sme eqution pplies to the grvittionl potentil, Eq. (9.4). Perhps you ve looked into microwve oven. You cn see inside it, ut the microwves ren t supposed to get out. How cn this e? Light is just nother form of electromgnetic rdition, so why

Prtil Differentil Equtions 4 does one EM wve get through while the other one doesn t? I won t solve the whole electromgnetic rdition prolem here, ut I ll look t the sttic nlog to get some generl ide of wht s hppening. z y V L L L x Arrnge set of conducting strips in the x-y plne nd with insultion etween them so tht they don t quite touch ech other. Now pply voltge V on every other one so tht the potentils re lterntely zero nd V. This sets the potentil in the z = plne to e independent of y nd z = : V (x, y) = { V ( < x < L) (L < x < L) V (x + L, y) = V (x, y), ll x, y (.35) Wht is then the potentil ove the plne, z >? Aove the plne V stisfies Lplce s eqution, V = V x + V y + V z = (.36) The potentil is independent of y in the plne, so it will e independent of y everywhere. Seprte vriles in the remining coordintes. V (x, z) = f(x)g(z) = d f dx g + f d g dz = = d f f dx + d g g dz = This is seprted s function of x plus function of y, so the terms re constnts. d f f dx = α, d g g dz = +α (.37) I ve chosen the seprtion constnt in this form ecuse the oundry condition is periodic in x, nd tht implies tht I ll wnt oscillting functions there, not exponentils. The seprted solutions re then f(x) = e iαx nd f(x + L) = f(x) = e Liα =, or Lα = nπ, n =, ±, ±,... The solution for z > is therefore the sum f(x)g(z) = e nπix/l( Ae nπz/l + Be nπz/l) (.38) V (x, z) = n= e nπix/l( A n e nπz/l + B n e nπz/l) (.39) The coefficients A n nd B n re to e determined y Fourier techniques. First however, look t the z-ehvior. As you move wy from the plne towrd positive z, the potentil should not increse

Prtil Differentil Equtions 5 without ound. Terms such s e πz/l however increse with z. This mens tht the coefficients of the terms tht increse exponentilly in z cnnot e there. A n = for n >, nd B n = for n < V (x, z) = A + B + e nπix/l B n e nπz/l + e nπix/l A n e nπz/l (.4) n= n= The comined constnt A + B is relly one constnt; you cn cll it C if you like. Now use the usul Fourier techniques given tht you know the potentil t z =. V (x, ) = C + B n e nπix/l + A n e nπix/l n= n= The sclr product of e mπix/l with this eqution is e mπix/l, V (x, ) LC (m = ) = LB m (m > ) LA m (m < ) Now evlute the integrl on the left side. First, m : e mπix/l, V (x, ) = L L dx e mπix/l { ( L < x < ) V ( < x < L) L = V dx e mπix/l = V L [ = V ( ) m ] mπi L mπi e mπix/l L Then evlute it seprtely for m =, nd you hve, V (x, ) = V L. Now ssemle the result. Before plunging in, look t wht will hppen. The m = term sits y itself. For the other terms, only odd m hve non-zero vlues. V (x, z) = V + V m= +V m= [ ( ) m ] e mπix/l e mπz/l mπi (.4) [ ( ) m ] (.4) e mπix/l e +mπz/l mπi To put this into rel form tht is esier to interpret, chnge vriles, letting m = n in the second sum nd m = n in the first, finlly chnging the sum so tht it is over just the odd terms. V (x, z) = V + V n= +V = V + V = V + π V n= l= [ ( ) n ] e nπix/l e nπz/l nπi [ ( ) n ] e nπix/l e nπz/l +nπi [ ( ) n ] nπ sin(nπx/l)e nπz/l l + sin ( (l + )πx/l ) e (l+)πz/l (.43)

Prtil Differentil Equtions 6 Hving done ll the work to get to the nswer, wht cn I lern from it? Wht does it look like? Are there ny properties of the solution tht re unexpected? Should I hve nticipted the form of the result? Is there n esier wy to get to the result? To see wht it looks like, exmine some vlues of z, the distnce ove the surfce. If z = L, the coefficient for successive terms is l = : π e π =.8 l = : 3π e 3π =.7 5 (.44) The constnt term is the verge potentil, nd the l = term dds only modest ripple, out 5% of the constnt verge vlue. If you move up to z = L the first fctor is. nd tht s little more thn.% ripple. The shrp jumps from +V to zero nd ck dispper rpidly. Tht the oscilltions vnish so quickly with distnce is perhps not wht you would guess until you hve nlyzed such prolem. V (x, L/) V (x, ) The grph shows the potentil function t the surfce, z =, s it oscilltes etween V nd zero. It then shows successive grphs of Eq. (.43) t z = L/, then t z = L, then t z =.5L. The ripple is rely visile t the tht lst distnce. The rdition through the screen of microwve oven is filtered in much the sme wy ecuse the wvelength of the rdition is lrge compred to the size of the holes in the screen. When you write the form of the series for the potentil, Eq. (.4), you cn see this coming if you look for it. The oscillting terms in x re ccompnied y exponentil terms in z, nd the rpid dmping with distnce is lredy pprent: e nπz/l. You don t hve to solve for single coefficient to see tht the oscilltions vnish very rpidly with distnce. The originl potentil on the surfce ws neither even nor odd, ut except for the constnt verge vlue, it is n odd function. z = : V (x, y) = V + { +V / ( < x < L) V / (L < x < L) V (x + L, y) = V (x, y) (.45) Solve the potentil prolem for the constnt V / nd you hve constnt. Solve it for the remining odd function on the oundry nd you should expect n odd function for V (x, z). If you mke these oservtions efore solving the prolem you cn sve yourself some lger, s it will led you to the form of the solution fster. The potentil is periodic on the x-y plne, so periodic oundry conditions re the pproprite ones. You cn express these in more thn one wy, tking s sis for the expnsion either complex exponentils or sines nd cosines. e nπix/l, n =, ±,... or the comintion cos(nπx/l), n =,,... sin(nπx/l), n =,,... (.46) For rndom prolem with no specil symmetry the exponentil choice typiclly leds to esier integrls. In this cse the oundry condition hs some symmetry tht you cn tke dvntge of: it s lmost

Prtil Differentil Equtions 7 odd. The constnt term in Eq. (.3) is the n = element of the cosine set, nd tht s necessrily orthogonl to ll the sines. For the rest, you do the expnsion { +V / ( < x < L) V / (L < x < L) = n sin(nπx/l) The odd term in the oundry condition (.45) is necessrily sum of sines, with no cosines. The cosines re orthogonl to n odd function. See prolem.. More Electrosttic Exmples Specify the electric potentil in the x-y plne to e n rry, periodic in oth the x nd the y-directions. V (x, y, z = ) is V on the rectngle ( < x <, < y < ) s well s in the drkened oxes in the picture; it is zero in the white oxes. Wht is the potentil for z >? z y x The eqution is still Eq. (.36), ut now you hve to do the seprtion of vriles long ll three coordintes, V (x, y, z) = f(x)g(y)h(z). Sustitute into the Lplce eqution nd divide y fgh. d f f dx + d g g dy + d h h dz = These terms re functions of the single vriles x, y, nd z respectively, so the only wy this cn work is if they re seprtely constnt. d f f dx = k, d g g dy = k, d h h dz = k + k = k3 I mde the choice of the signs for these constnts ecuse the oundry function is periodic in x nd in y, so I expect sines nd cosines long those directions. The seprted solution is (A sin k x + B cos k x)(c sin k y + D cos k y)(ee k 3z + F e k 3z ) (.47) Wht out the cse for seprtion constnts of zero? Yes, tht s needed too; the verge vlue of the potentil on the surfce is V /, so just s with the exmple leding to Eq. (.43) this will hve constnt term of tht vlue. The periodicity in x is nd in y it is, so this determines n π k = nπ/, k = mπ/ then k 3 = + m π, n, m =,,... where n nd m re independent integers. Use the experience tht led to Eq. (.45) to write V on the surfce s sum of the constnt V / nd function tht is odd in oth x nd in y. As there, the odd function in x will e represented y sum of sines in x, nd the sme sttement will hold for the y coordinte. This leds to the form of the sum V (x, y, z) = V + n= m= α nm sin ( nπx ) sin ( mπy ) e knmz

Prtil Differentil Equtions 8 where k nm is the k 3 of the preceding eqution. Wht hppened to the other term in z, the one with the positive exponent? Did I sy tht I m looking for solutions in the domin z >? At z = this must mtch the oundry conditions stted, nd s efore, the orthogonlity of the sines on the two domins llows you to determine the coefficients. You simply hve to do two integrls insted of one. See prolem.9. V (x, y, z > ) = V + 8V π odd n odd m ( nπx ) nm sin sin ( mπy ) e knmz (.48).7 Cylindricl Coordintes Rectngulr coordintes re not lwys the right choice. Cylindricl, sphericl, nd other choices re often needed. For cylindricl coordintes, the grdient nd divergence re, from Eqs. (9.4) nd (9.5) V = ˆr V r + ˆφ V r φ + ẑ V z nd. v = r (rv r ) r + r v φ φ + v z z Comine these to get the Lplcin in cylindricl coordintes.. V = V = r ( r V ) + V r r r φ + V z (.49) For electrosttics, the eqution remins V =, nd you cn pproch it the sme wy s efore, using seprtion of vriles. I ll strt with the specil cse for which everything is independent of z. Assume solution of the form V = f(r)g(φ), then V = g r ( r Multiply this y r nd divide y f(r)g(φ) to get r f ( r r f r r f r ) + f r g φ = ) + g g φ = This is seprted. The first terms depends on r lone, nd the second term on φ lone. For this to hold the terms must e constnts. ( r d r df ) = α nd f dr dr d g = α (.5) g dφ The second eqution, in φ, is fmilir. If α is positive this is hrmonic oscilltor, nd tht is the most common wy this solution is pplied. I ll then look t the cse for α, for which the sustitution α = n mkes sense. α = : d g = = g(φ) = A + Bφ dφ r d ( r df ) = = f(r) = C + D ln r dr dr

Prtil Differentil Equtions 9 α = n d > g dφ = n g = g(φ) = A cos nφ + B sin nφ r d f dr + r df dr n f = = f(r) = Cr n + Dr n There s not yet restriction tht n is n integer, though tht s often the cse. Verifying the lst solution for f is esy. A generl solution tht is the sum of ll these terms is V (r, φ) = (C + D ln r)(a + B φ) + n (C n r n + D n r n )(A n cos nφ + B n sin nφ) (.5) Some of these terms should e fmilir: C A is just constnt potentil. D A ln r is the potentil of uniform line chrge; d ln r/dr = /r, nd tht is the wy tht the electric field drops off with distnce from the xis. C A r cos φ is the potentil of uniform field (s is the r sin φ term). Write this in the form C A r cos φ = C A x, nd the grdient of this is C A ˆx. The sine gives ŷ. See prolem.4. Exmple A conducting wire, rdius R, is plced in uniform electric field E, nd perpendiculr to it. Put the wire long the z-xis nd cll the positive x-xis the direction tht the field points. Tht s φ =. In the sence of the wire, the potentil for the uniform field is V = E x = E r cos φ, ecuse V = E ˆx. The totl solution will e in the form of Eq. (.5). Now turn the generl form of the solution into prticulr one for this prolem. The entire rnge of φ from to π ppers here; you cn go ll the wy round the origin nd come ck to where you strted. The potentil is function, mening tht it s single vlued, nd tht elimintes B φ. It lso implies tht ll the n re integers. The pplied field hs potentil tht is even in φ. Tht mens tht you should expect the solution to e even in φ too. Does it relly? You lso hve to note tht the physicl system nd its ttendnt oundry conditions re even in φ it s cylinder. Then too, the first few times tht you do this sort of prolem you should see wht hppens to the odd terms; wht mkes them vnish? I won t eliminte the sin φ terms from the strt, ut I ll clculte them nd show tht they do vnish. V (r, φ) = (C + D ln r)b + (C n r n + D n r n )(A n cos nφ + B n sin nφ) n= Crrying long ll these products of (still unknown) fctors such s D n A n is wkwrd. It mkes it look neter nd it is esier to follow if I comine nd renme some of them. V (r, φ) = C + D ln r + (C n r n + D n r n ) cos nφ + (C nr n + D nr n ) sin nφ (.5) n= As r, the potentil looks like E r cos φ. Tht implies tht C n = for n >, nd tht C n = for ll n, nd tht C = E. Now use the oundry conditions on the cylinder. It is conductor, so in this sttic cse the potentil is constnt ll through it, in prticulr on the surfce. I my s well tke tht constnt to e zero, though it doesn t relly mtter. V (R, φ) = = C + D ln R + (C n R n + D n R n ) cos nφ + (C nr n + D nr n ) sin nφ n= n= n=

Prtil Differentil Equtions Multiply y sin mφ nd integrte over φ. The trigonometric functions re orthogonl, so ll tht survives is = (C mr m + D mr m )π ll m Tht gets rid of ll the rest of the sine terms s promised: D m = for ll m ecuse C m = for ll m. Now repet for cos mφ. = C + D ln R (m = ) nd = (C m R m + D m R m )π (m > ) All of the C m = for m >, so this sys tht the sme pplies to D m. determines D in terms of C nd then E. The m = eqution D = C R = +E R Only the C nd D terms re left, nd tht requires nother oundry condition. When specifying the prolem initilly, I didn t sy whether or not there is ny chrge on the wire. In such cse you could nturlly ssume tht it is zero, ut you hve to sy so explicitly ecuse tht ffects the finl result. Mke it zero. Tht kills the ln r term. The reson for tht goes ck to the interprettion of this term. Its negtive grdient is the electric field, nd tht would e /r, the field of uniform line chrge. If I ssume there isn t one, then D = nd so the sme for C. Put this ll together nd R V (r, φ) = E r cos φ + E cos φ (.53) r The electric field tht this presents is, from Eq. (9.4) E = V = E (ˆr cos φ ˆφ sin φ ) E R ( ˆr r cos φ ˆφ r sin φ ) = E ˆx + E R r (ˆr cos φ + ˆφ sin φ ) As check to see wht this looks like, wht is the electric field t the surfce of the cylinder? E(R, φ) = E (ˆr cos φ ˆφ sin φ ) E R ( ˆr R cos φ ˆφ R sin φ ) = E ˆr cos φ It s perpendiculr to the surfce, s it should e. At the left nd right, φ =, π, it is twice s lrge s the uniform field lone would e t those points.

Prtil Differentil Equtions Prolems. The specific het of prticulr type of stinless steel (CF8M) is 5 J/kg.K. Its therml conductivity is 6. W/m.K nd its density is 775 kg/m 3. A sl of this steel. cm thick is t temperture C nd it is plced into ice wter. Assume the simplest oundry condition tht its surfce temperture stys t zero, nd find the internl temperture t lter times. When is the nd term in the series, Eq. (.5), only 5% of the st? Sketch the temperture distriution then, indicting the scle correctly.. In Eq. (.3) I eliminted the n = solution y fllcious rgument. Wht is α in this cse? This gives one more term in the sum, Eq. (.4). Show tht with the oundry conditions stted, this extr term is zero nywy (this time)..3 In Eq. (.4) you hve the sum of mny terms. Does it still stisfy the originl differentil eqution, Eq. (.3)?.4 In the exmple Eq. (.5) the finl temperture ws zero. Wht if the finl temperture is T? Or wht if I use the Kelvin scle, so tht the finl temperture is 73? Add the pproprite extr term, mking sure tht you still hve solution to the originl differentil eqution nd tht the oundry conditions re stisfied..5 In the exmple Eq. (.5) the finl temperture ws zero on oth sides. Wht if it s zero on just the side t x = while the side t x = L stys t T? Wht is the solution now? Ans: T x/l + (T /π) (/n) sin(nπx/l)e n π Dt/L.6 You hve sl of mteril of thickness L nd t uniform temperture T. The side t x = L is insulted so tht het cn t flow in or out of tht surfce. By Eq. (.), this tells you tht T/ x = t tht surfce. Plunge the other side into ice wter t temperture T = nd find the temperture inside t lter time. The oundry condition on the x = surfce is the sme s in the exmple in the text, T (, t) =. () Seprte vriles nd find the pproprite seprted solutions for these oundry conditions. Are the seprted solutions orthogonl? Use the techniques of Eq. (5.5). () When the lowest order term hs dropped to where its contriution to the temperture t x = L is T /, how ig is the next term in the series? Sketch the temperture distriution in the sl t tht time. Ans: (4T /π) ( n+ ) sin [ (n + )πx/l] e (n+/) π Dt/L, 9.43 5 T.7 In the nlysis leding to Eq. (.6) the temperture t y = ws set to T. If insted, you hve the temperture t x = set to T with ll the other sides t zero, write down the nswer for the temperture within the rod. Now use the fct tht Eq. (.) is liner to write down the solution if oth the sides t y = nd x = re set to T..8 In leding up to Eq. (.5) I didn t exmine the third possiility for the seprtion constnt, tht it s zero. Do so..9 Look t the oundry condition of Eq. (.) gin. Another wy to solve this prolem is to use the solution for which the seprtion constnt is zero, nd to use it to stisfy the conditions t y = nd y =. You will then hve one term in the seprted solution tht is T y/, nd tht mens tht in Eq. (.3) you will hve to choose the seprtion vrile to e positive insted of negtive. Why? Becuse now ll the rest of the terms in the sum over seprted solutions must vnish t y = nd y =. You ve lredy stisfied the oundry conditions on those surfces y using the T y/ term. Now you hve to stisfy the oundry conditions on x = nd x = ecuse the totl temperture

Prtil Differentil Equtions there must e zero. Tht in turn mens tht the sum over ll the rest of the seprted terms must dd to T y/ t x = nd x =. When you nlyze this solution in the sme spirit s the nlysis of Eq. (.6), compre the convergence properties of tht solution to your new one. In prticulr, look t nd to see which version converges etter in ech cse. Ans: T y/ + (T /π) [ ( ) n /n ] sin(nπy/) cosh ( nπ(x /)/ )/ cosh(nπ/). Finish the re-nlysis of the electrosttic oundry vlue prolem Eq. (.45) strting from Eq. (.46). This will get the potentil for z with perhps less work thn efore.. Exmine the solution Eq. (.4) t z = in the light of prolem 5... A thick sl of mteril is lterntely heted nd cooled t its surfce so the its surfce temperture oscilltes s T (, t) = { T ( < t < t ) T (t < t < t ) T (, t + t ) = T (, t) Tht is, the period of the oscilltion is t. Find the temperture inside the mteril, for x >. How does this ehvior differ from the solution in Eq. (.)? Ans: 4T k= ( ) π /(k + ) e β k x sin ( (k + )ωt β k x) ) ; ω = π/t.3 Fill in the missing steps in finding the solution, Eq. (.34)..4 A vrition on the prolem of the lternting potentil strips in section.6. Plce grounded conducting sheet prllel to the x-y plne t height z = d ove it. The potentil there is then V (x, y, z = d) =. Solve for the potentil in the gp etween z = nd z = d. A suggestion: you my find it esier to turn the coordinte system over so tht the grounded sheet is t z = nd the lternting strips re t z = d. This switch of coordintes is in no wy essentil, ut it is it esier. Also, I wnt to point out tht you will need to consider the cse for which the seprtion constnt in Eq. (.37) is zero..5 Strting from Eq. (.5) nd repet the exmple there, ut ssume tht the conducting wire is in n externl electric field E ŷ insted of E ˆx. Repet the clcultion for the potentil nd for the electric field, filling in the detils of the missing steps..6 A very long conducting cylindricl shell of rdius R is split in two long lines prllel to its xis. The two hlves re wired to circuit tht plces one hlf t potentil V nd the other hlf t potentil V. () Wht is the potentil everywhere inside the cylinder? Use the results of section.7 nd ssume solution of the form V V (r, φ) = r n( n cos nφ + n sin nφ ) V Mtch the oundry condition tht V (R, φ) = { V ( < φ < π) V (π < φ < π)

Prtil Differentil Equtions 3 I picked the xis for φ = pointing towrd the split etween the cylinders. No prticulr reson, ut you hve to mke choice. I mke the pproximtion tht the cylinder is infinitely long so tht z dependence doesn t enter. Also, the two hlves of the cylinder lmost touch so I m neglecting the distnce etween them. () Wht is the electric field, V on the centrl xis? Is this nswer more or less wht you would estimte efore solving the prolem? Ans: () E = 4V /πr..7 Solve the preceding prolem outside the cylinder. The integer n cn e either positive or negtive, nd this time you ll need the negtive vlues. (And why must n e n integer?) Ans: (4V /π) n odd (/n)(r/r)n sin nφ.8 In the split cylinder of prolem.6, insert coxil wire of rdius R < R. It is t zero potentil. Now wht is the potentil in the domin R < r < R? You will need oth the positive nd negtive n vlues, (A n r n + B n r n ) sin nφ Ans: (4V /π) m odd sin mφ[ R m r m + R m r m ] / m[r m R m R m R m ].9 Fill in the missing steps in deriving Eq. (.48).. Anlyze how rpidly the solution Eq. (.48) pproches constnt s z increses from zero. Compre Eq. (.44).. A rod clss of second order liner homogeneous differentil equtions cn, with some mnipultion, e put into the form (Sturm-Liouville) (p(x)u ) + q(x)u = λw(x)u Assume tht the functions p, q, nd w re rel, nd use mnipultions much like those tht led to the identity Eq. (5.5). Derive the nlogous identity for this new differentil eqution. When you use seprtion of vriles on equtions involving the Lplcin you will commonly come to n ordinry differentil eqution of exctly this form. The precise detils will depend on the coordinte system you re using s well s other spects of the PDE.. Crry on from Eq. (.3) nd deduce the seprted solution tht stisfies these oundry condition. Show tht it is equivlent to Eq. (.3)..3 The Lplcin in cylindricl coordintes is Eq. (.49). Seprte vriles for the eqution V = nd you will see tht the equtions in z nd φ re fmilir. The eqution in the r vrile is less so, ut you ve seen it (lmost) in Eqs. (4.8) nd (4.). Mke chnge of vriles in the r-differentil eqution, r = kr, nd turn it into exctly the form descried there..4 In the preceding prolem suppose tht there s no z-dependence. Look t the cse where the seprtion constnt is zero for oth the r nd φ functions, finlly ssemling the product of the two for nother solution of the whole eqution. These results provide four different solutions, constnt, function of r lone, function of φ lone, nd function of oth. In ech of these cses, ssume tht these functions re potentils V nd tht E = V is the electric field from ech potentil. Sketch equipotentils for ech cse, then sketch the corresponding vector fields tht they produce ( lot of rrows).

Prtil Differentil Equtions 4.5 Do prolem 8.3 nd now solve it, finding ll solutions to the wve eqution. Ans: f(x vt) + g(x + vt).6 Use the results of prolem.4 to find the potentil in the corner etween two very lrge metl pltes set t right ngles. One t potentil zero, the other t potentil V. Compute the electric field, V nd drw the results. Ans: V ˆφ/πr.7 A thin metl sheet hs stright edge for one of its oundries. Another thin metl sheet is cut the sme wy. The two stright oundries re plced in the sme plne nd lmost, ut not quite touching. Now pply potentil difference etween them, putting one t voltge V nd the other t V. In the region of spce ner to the lmost touching oundry, wht is the electric potentil? From tht, compute nd drw the electric field..8 A sl of het conducting mteril lies etween coordintes x = L nd x = +L, which re t tempertures T nd T respectively. In the stedy stte ( T/ t ), wht is the temperture distriution inside? Now express the result in cylindricl coordintes round the z-xis nd show how it mtches the sum of cylindricl coordinte solutions of T = from prolem.5. Wht if the surfces of the sl hd een specified t y = L nd y = +L insted?.9 The result of prolem.6 hs series of terms tht look like (x n /n) sin nφ (odd n). You cn use complex exponentils, do little rerrnging nd fctoring, nd sum this series. Along the wy you will hve to figure out wht the sum z + z 3 /3 + z 5 /5 + is. Refer to section.7. Finlly of course, the nswer is rel, nd if you look hrd you my find simple interprettion for the result. Be sure you ve done prolem.4 efore trying this lst step. Ans: V (θ + θ )/π. You still hve to decipher wht θ nd θ re..3 Sum the series Eq. (.7) to get closed-form nlytic expression for the temperture distriution. You will find the techniques of section 5.7 useful, ut there re still lot of steps. Recll lso ln(r e iθ ) = ln r + iθ. Ans: (T /π) tn [ sin(πx/) / sinh ( π( y)/ )].3 A generliztion of the prolem specified in Eq. (.). Now the four sides hve tempertures given respectively to e the constnts T, T, T 3, T 4. Note: with little it of foresight, you won t hve to work very hrd t ll to solve this..3 Use the electrosttic equtions from prolem 9. nd ssume tht the electric chrge density is given y ρ = ρ /r, where this is in cylindricl coordintes. () Wht cylindriclly symmetric electric field comes from this chrge distriution? () From E = V wht potentil function V do you get?.33 Repet the preceding prolem, ut now interpret r s referring to sphericl coordintes. Wht is V?.34 The Lplcin in sphericl coordintes is Eq. (9.43). The electrosttic potentil eqution is V = just s efore, ut now tke the specil cse of zimuthl symmetry so tht the potentil function is independent of φ. Apply the method of seprtion of vriles to find solutions of the form f(r)g(θ). You will get two ordinry differentil equtions for f nd g. The second of these equtions is much simpler if you mke the chnge of independent vrile x = cos θ. Use the chin rule couple of times to do so, showing tht the two differentil equtions re ( x ) d g dg x dx dx + Cg = nd d f df r + r dr dr Cf =