Laplace s Equation. Laplace s Equation. Given an open set Ω in the (x, y)-plane, Laplace s Equation for u = u(x, y) is 0 = u def

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1 Lplce s Eqution UBC M257/316 Lecture Notes c 2014 by Philip D. Loewen A. The Lplcin For single-vrible function u = u(x, u (x mesures slope nd u (x mesures concvity or curvture. When u = u(x, y depends on two vribles, the grdient ( vector nd the Lplcin ( sclr record the corresponding quntities: u(x, y = (u x (x, y, u y (x, y, u(x, y = u xx (x, y + u yy (x, y. (the grdient (the Lplcin. The Lplcin opertor is importnt enough to deserve n intuitive understnding on its own. However, we ll still think of it s being somehow relted to concvity/curvture. It certinly does the sme job in problems of wve motion or het flow in 2D domins, nd similr extensions hold in 3 or more dimensions. Wve Eqution. For 2D membrne stretched cross wire frme round region Ω in the (x, y-plne (like drum hed, the lterl displcement t point (x, y nd time t is function u = u(x, y, t tht obeys u tt = c 2 (u xx + u yy = c 2 u, (x, y Ω, t > 0. ( Curvture drives ccelertion. Sketch something.] Het Eqution. For 2D metl plte occupying the plne region Ω, sndwiched between insultion slbs on its flt sides so het cn only flow in the (x, y-plne, the temperture t point (x, y nd time t is function u = u(x, y, t tht obeys ( Curvture drives flow rte. u t = 2 (u xx + u yy = 2 u, (x, y Ω, t > 0 Lplce s Eqution. Given n open set Ω in the (x, y-plne, Lplce s Eqution for u = u(x, y is 0 = u def = u xx + u yy, (x, y Ω. ( Applictions: (1 Stedy-stte temperture in 2D region with fixed boundry tempertures. (The cse of 1D region is esy: u (x = 0 implies u(x = mx+c. For higherdimensionl region, however, much greter vriety of solutions is possible. (2 Potentil fields (electrosttic, grvittionl, etc. in regions free from potentil sources (chrges, msses, etc. obey Lplce s eqution. (3 Miniml surfces (e.g., sop films in equilibrium obey the nonliner PDE (1 + u 2 y u xx + 2u x u y u xy + (1 + u 2 x u yy = 0, (x, y Ω. When u is so smll tht terms of second nd higher order in its components re negligible, this is pproximted by Lplce s eqution. File 2014notes, version of 5 August 2014, pge 1. Typeset t 16:38 August 5, 2014.

2 2 PHILIP D. LOEWEN Boundry Conditions. Write Γ for the boundry curve of Ω. (i Dirichlet (prescribed function vlues: function g is given, nd we seek function u obeying ( nd u(x, y = g(x, y for (x, y Γ. (ii Neumnn (prescribed directionl derivtives norml to boundry: function h is given nd we seek u stisfying ( nd u(x, y N(x, y = h(x, y for (x, y Γ. Here N(x, y is the outwrd unit norml to the curve Γ t point (x, y. (iii Mixed (Dirichlet on some segments of Γ, Neumnn on the rest. B. Dirichlet Problem in Rectngulr Box Consider Ω = {(x, y : 0 < x <, 0 < y < b}. Here the boundry curve Γ consists of the four segments on the sides of the rectngle, nd BC s of Dirichlet type cn be drwn right onto the picture: y u(x,b = f 1 (x = g 0 (y u = 0 u(,y=g 1 (y u(x,0=f 0 (x x In detil, functions f 0, f 1, g 0, nd g 1 re given, nd we seek function u obeying both u = 0 in Ω nd u(x, 0 = f 0 (x, 0 < x < (bottom, u(x, b = f 1 (x, 0 < x < (top, u(0, y = g 0 (y, u(, y = g 1 (y, 0 < y < b (left, 0 < y < b (right. Trivil Cse. If f 0 f 1 g 0 g 1 0, wht s u? Physicl intuition nd mthemticl outcome gree: u 0. Simplest Nontrivil Cse. All but one boundry function re zero. Suppose f 1 (x 0, wheres f 0 g 0 g 1 0. Follow our usul 6-step process. 1: Splitting. Not required here. File 2014notes, version of 5 August 2014, pge 2. Typeset t 16:38 August 5, 2014.

3 Lplce s Eqution 3 2: Eigenfunctions. Look for simple nontrivil solutions in product for u(x, y = X(xY (y. Substitute into PDE/BC, remembering tht seprtion of vribles is worse thn futile (it s misleding on nonhomogeneous conditions. So only three BC s give useful info, nmely, Y (0 = 0 (from the bottom, X(0 = 0 (from the left, X( = 0 (from the right. In PDE, substitution leds to 0 = X (xy (y + X(xY (y X (x X(x = Y (y Y (y = λ for some seprtion constnt λ. Since we hve pir of BC s for X nd only one for Y, it is the X-component tht provides well-formed eigenvlue problem: X (x + λx(x = 0, 0 < x < ; X(0 = 0 = X(. The corresponding eigenfunctions re well known (FSS: ( nπx X n (x = sin, n = 1, 2,.... 3: Superposition. Seprtion hs done ll it cn for us. The solution we expect will not hve simple seprted form, but rther combine ll possible product-form solutions found bove like this: ( nπx u(x, y = Y n (y sin. ( Finding those coefficient functions Y n (y will complete the solution. 4: Auxiliry Informtion (formerly Initiliztion. Any series of form ( will stisfy the BC s on the left nd right sides. On the bottom, we need ( nπx 0 = u(x, 0 = Y n (0 sin. Here is FSS expnsion for the zero function, giving Y n (0 = 0, n = 1, 2,.... ( On the top, we need f 1 (x = u(x, b = ( nπx Y n (b sin. This is FSS expnsion for the function f 1. Stndrd coefficient formuls give Y n (b = 2 ( nπx f 1 (x sin dx. ( 0 File 2014notes, version of 5 August 2014, pge 3. Typeset t 16:38 August 5, 2014.

4 4 PHILIP D. LOEWEN 5: Propgtion. Plug series into PDE to get ( ] 2 nπ 0 = u xx + u yy = Y n (y + Y n (y ( nπx sin This is FSS expnsion on 0 < x <, with coefficients independent of x, for the zero function. Agin it requires tht ll coefficients must vnish, i.e., ( 2 nπ Y n (y Y n (y = 0, n = 1, 2,..... Guess Y n = e sy nd plug in to get s = ± nπ, nd hence the generl solution Y n (y = A n e nπy/ + B n e nπy/. Now use ( to get 0 = Y n (0 = A n + B n, so B n = A n, Plug in y = b nd ppel to ( to get A n = Y n (y = A n e nπy/ e nπy/]. Y n (b e nπb/ e = 2 nπb/ e nπb/ e nπb/] 6: Conclusion. The series solution we wnt is u(x, y = A n e nπy/ e nπy/] ( nπx sin, 0 ( nπx f 1 (x sin dx. with constnts A n given in terms of f 1 by the integrl formul bove. This ppernce is typicl: sum of products, with one fctor being n eigenfunction nd the other some kind of exponentil. //// Eigenfunction Setup. Notice tht the key to Step 2 bove ws the presence of homogeneous BC s on the opposite fces (left nd right of the domin Ω. These gve us the FSS eigenproblem encoded in the series solution; the homogeneous BC on the bottom edge helped us only in Step 4, nd nonhomogeneous one could hve been hndled in Step 4 with very little extr work. RTFT. In the textbook by Trench, plese red Section 12.3 nd try problems 5, 9, 19, 31, 34. Symmetry Argument #1. Suppose u = u(x, y obeys Lplce s Eqution in the given rectngle. Define v = u(x, b y. Notice tht v y (x, y = u y (x, b y, v yy (x, y = ( 1 2 u yy (x, b y = u yy x, b y, while v xx (x, y = u xx (x, b y. Now when 0 < y < b, of course 0 < b y < b lso, so we get v xx (x, y + v yy (x, y = u xx (x, b y + u yy (x, b y = 0. File 2014notes, version of 5 August 2014, pge 4. Typeset t 16:38 August 5, 2014.

5 Lplce s Eqution 5 Menwhile, v(x, 0 = u(x, b = f 1 (x, v(x, b = u(x, 0 = f 0 (x = 0, v(0, y = u(0, b y = g 0 (b y = 0, v(, y = u(, b y = g 1 (b y = 0. Now the function u is known, so v is too, nd v solves problem very similr to the u-problem except tht it hs nontrivil temperture on the bottom edge of Ω. Sketch pictoril representtion for the v-problem.] Putting this bck into the nottion of the originl u-problem lets us cover the cse where g 0 f 1 g 1 0 but f 0 0: u(x, y = A n e nπ(b y/ e nπ(b y/] ( nπx sin, 2 ( nπx where A n = e nπb/ e nπb/] f 0 (x sin dx. Symmetry Argument #2. Switching letters x y, b, f g chnges the ppernce but not the vlidity of the solution. (Physiclly it mkes sense too: we re just flipping our metl plte with its stedy temperture distribution long the xis y = x. This swp leves the PDE unchnged, but gives BC where it s g 1 (y tht is nontrivil. So for the cse f 0 g 0 f 1 0, the solution is u(x, y = B n = B n e nπx/b e nπx/b ] sin 2 b e nπ/b e nπ/b] 0 0 ( nπy b ( nπy g 1 (y sin dy. b, (5 Approching this problem directly, we would seprte vribles s shown bove in the PDE, but the seprted BC s would led to the homogeneous conditions Y (0 = 0 = Y (b. So in this sitution, the eigenvlue problem of interest would involve the unknowns Y : Y (y λy (y = 0, 0 < y < b; Y (0 = 0 = Y (b. The nturl series form to postulte would then be u(x, y = nd this is precisely wht we see in line (5 bove. ( nπy X n (x sin, b Prctice. Without lengthy clcultion, find series solution formul for the cse where f 1 g 1 f 0 0 but g 0 0. Splitting nd Superposition. To hndle rbitrry f 0, f 1, g 0, nd g 1, consider four subproblems of the form bove, ech with one nonzero boundry function. Write series solution for ech subproblem, then dd them up. File 2014notes, version of 5 August 2014, pge 5. Typeset t 16:38 August 5, 2014.

6 6 PHILIP D. LOEWEN C. Lplce s Eqution in Polr Coordintes (Pizz Problems In stndrd polr coordintes, where x = r cos θ, y = r sin θ, the Lplcin of given function u = u(r, θ is u = u rr + 1 r u r + 1 r 2u θθ. Proof ide: Define U(x, y = u( x 2 + y 2, tn 1 (y/x, compute U = U xx + U yy with chin rule, express result in terms of r nd θ.] Hence Lplce s eqution is equivlent to 0 = r 2 u = r 2 u rr + ru r + u θθ. Seprtion of this PDE with u(r, θ = R(rΘ(θ leds to r 2 R (r + rr (r R(r = Θ (θ Θ(θ = σ for some seprtion constnt σ, nd this produces two linked ODE problems: (1 r 2 R (r + rr (r σr(r = 0, (2 Θ (θ + σθ(θ = 0. Depending on wht sort of BC s re present, either of these could be the ODE of choice in n eigenvlue problem. Exmple (A. Pizz slice with first bite gone: polr region Ω = {(r, θ : < r < b, 0 < θ < }. u(r,=0 u(b,θ=g(θ u = 0 u(,θ=f(θ u(r,0=0 Here (0, 2π nd > 0, b > re some pressigned constnts. If the BC s on the flt sides re homogeneous, i.e., u(r, 0 = 0, u(r, = 0, < r < b, u(, θ = f(θ, u(b, θ = g(θ, 0 < θ <, then seprtion of vribles in the BC s leds to Θ(0 = 0 = Θ(, File 2014notes, version of 5 August 2014, pge 6. Typeset t 16:38 August 5, 2014.

7 Lplce s Eqution 7 so eqution (2 prticiptes in the eigenvlue problem Θ (θ + σθ(θ = 0, 0 < θ <, Θ(0 = 0 = Θ(. Thus we get FSS eigenfunctions Θ n (θ = sin, nd series-form solution The BC s give f(θ = u(, θ = g(θ = u(b, θ = u(r, θ = R n (r sin. ( nπθ R n ( sin ( nπθ R n (b sin, so R n ( = 2 f(θ sin 0, so R n (b = 2 Plugging the series into the PDE leds to 0 = r 2 u rr + ru r + u θθ = r 2 R n + rr n whence 0 = r 2 R n + rr n ( 2 nπ R n. 0 ( ] 2 nπ R n sin dθ, g(θ sin dθ. ( nπθ This eqution hs Euler type: function R n (r = r s gives solution iff So the generl solution is s(s 1 + s ( 2 nπ = 0, i.e., s = ± nπ. R n (r = A n r nπ/ + B n r nπ/, A n, B n R. When functions f, g re given in detil, the right-hnd sides in the system below re known constnts, nd it is possible to solve for A n, B n : A n nπ/ + B n nπ/ = R n ( = 2 f(θ sin dθ, 0 A n b nπ/ + B n b nπ/ = R n (b = 2 g(θ sin dθ. The series solution is then u(r, θ = A n r nπ/ + B n r nπ/ ] sin Consider sme region with other BC s lter (see (D below. 0 ( nπθ., File 2014notes, version of 5 August 2014, pge 7. Typeset t 16:38 August 5, 2014.

8 8 PHILIP D. LOEWEN Exmple (B. Pizz slice before first bite tken: polr region Ω = {(r, θ : 0 < r < b, 0 < θ < }. u(b,θ=g(θ u(r,=0 u = 0 u(r,0=0 This is the limiting cse 0 + of (1; we use the sme boundry conditions. Mthemticlly, there is no plce for the function f(θ describing u-vlues on the curved inner boundry shown in problem (A, so we don t hve enough informtion to determine both constnts A n nd B n bove. However, problems of physicl interest typiclly hve bounded solutions. The requirement tht u(r, θ behve well ner the boundry (which includes the origin, where r = 0 forces us to choose ll B n = 0, so the solution simplifies to u(r, θ = A n r nπ/ sin, A n = 2 b n g(θ sin dθ. 0 Exmple (B. Infinite pizz slice missing one bite. Prctice: how should the solution shown in prt (A be modified in the limiting cse b +? Exmple (C. Annulus. Solve u = 0 in the polr region Ω = {(r, θ : < r < b}, where > 0 nd b > re given constnts nd u(, θ = f(θ, u(b, θ = g(θ, < θ < π. f(θ u = 0 u(b,θ=g(θ Min Ide: Periodicity implicit in the polr coordinte representtion provides boundry conditions not written explicitly in the problem sttement, nmely, u(r, = u(r, π, u θ (r, = u θ (r, π, < r < b. File 2014notes, version of 5 August 2014, pge 8. Typeset t 16:38 August 5, 2014.

9 Lplce s Eqution 9 Seprting u(r, θ = R(rΘ(θ provides two pieces of boundry informtion bout Θ, so we focus the ODE for Θ in line (2 bove. This produces n eigenvlue problem for Θ: Θ (θ + λθ(θ = 0, < θ < π; Θ( = Θ(π, Θ ( = Θ (π. This is stndrd problem: we know tht its eigenfunctions re precisely the set of ll bsis functions ssocited with the Full Fourier Series on, π]. Hence we postulte solution of the form u(r, θ = 1 2 A 0(r + A n (r cos(nθ + B n (r sin(nθ], for some coefficient functions A n nd B n to be determined. To see how the coefficients evolve with r, plug tht series into Lplce s Eqution: 0 = r 2 u rr + ru r + u θθ = 1 2 r 2 A 0 (r + ra 0 (r] + + r 2 A n (r + ra n (r n2 A n (r ] cos(nθ r 2 B n (r + rb n (r n2 B n (r ] sin(nθ In this full Fourier series for the zero function, ll coefficients must be 0. For cse n 1, this gives pir of identicl Euler-type equtions. Exctly s in problem (A bove, we find the generl solutions A n (r = n r n + b n r n, B n (r = c n r n + d n r n, n, b n, c n, d n R. For cse n = 0, shrewd observtion gives 0 = r 2 A 0(r + ra 0(r! = r d dr (ra 0(r. Hence ra 0(r = b 0 for some b 0 R, nd this implies A 0 (r = b 0 r = A 0 (r = 0 + b 0 ln(r, 0, b 0 R. Thus we my express our series solution s u(r, θ = b 0 ln(r] + n r n + b n r n] cos(nθ + cn r n + d n r n] sin(nθ. Now using stndrd coefficient formuls on the BC s (1 f(θ = u(, θ, g(θ = u(b, θ will give 2 2 systems of liner equtions to solve for ( n, b n nd (c n, d n. Try it!] File 2014notes, version of 5 August 2014, pge 9. Typeset t 16:38 August 5, 2014.

10 10 PHILIP D. LOEWEN Exmple (C. Stedy temperture in disk of rdius b > 0, with given boundry temperture g(θ. This is the limiting cse 0 + of (C bove. Agin we must pply the boundedness requirement imposed by the physicl interprettion (stedy temperture. This tkes the plce of the prescribed temperture f on the inner edge; now boundedness requires b n = 0, d n = 0 for ll n. (Plese think bout b 0 seprtely. The result is u(r, θ = Now the BC u(b, θ = g(θ gives b n n = 1 π b n c n = 1 π r n ( n cos(nθ + c n sin(nθ. g(θ cos(nθ dθ, n = 0,1, 2, 3,..., g(θ sin(nθ dθ, n = 1, 2, 3,.... Averging Property: Midpoint temperture is verge of boundry tempertures. (Proof. Plug in r = 0, recll definition of Fourier Coefficient 0. Importnt Consequence: For ny solution u of Lplce s eqution, in ny 2D domin Ω, there cn be no locl min nd no locl mx for u in interior of Ω. def Reson: Suppose P 0 = (x 0, y 0 is point where locl min occurs. Choose little disk with centre P 0 where ll the boundry vlues re higher thn u(p 0. Then u = 0 in tht disk, nd the verging property just proved is violted. This cn t hppen. Physics: Stedy temperture in 2D region cn t hve isolted locl extrem. This mkes sense hot spots would be unstble. Likewise for sop-film stretched between curved wires simple bumps get pulled down by surfce tension. Poisson Kernel Formul (Optionl: Plug integrl coefficients stright into the series nd interchnge sum nd integrl to get the following: u(r, θ = g(t dt + π = 1 π = 1 π = 1 π n=0 ( n ( r 1 b π ( 1 + π g(t cos(nt dt cos(nθ ] g(t sin(nt dt sin(nθ 1 g(t 2 + ( n r cos(nt cos(nθ + sin(nt sin(nθ]] dt b 1 g(t 2 + ( n r cos(n(t θ] dt b g(t ( n r cos(n(t θ] dt b File 2014notes, version of 5 August 2014, pge 10. Typeset t 16:38 August 5, 2014.

11 Lplce s Eqution 11 For rel x, y with x < 1, geometric series clcultion shows x n cos(ny = n=0 Therefore Re ( x n e iny ( = Re xe iy n n=0 x n cos(ny 1 2 = n=0 n=0 ( xeiy = Re = 1 xeiy 1 + xe iy = x cos(y 1 2x cos(y + x 2. 1 x cos y 1 2x cosy + x 1 1 2x cosy + x x cosy + x 2 1 x 2 1 2x cos y + x 2 Use clc bove with x = r/b nd y = t θ to get fmous nd useful formul: u(r, θ = 1 b 2 r 2 ] 2π b 2 2br cos(t θ + r 2 g(t dt, 0 < r < b, θ R. Defining P( z, t = 1 b 2 r 2 mkes this formul look like mtrixvector 2π b 2 2br cos(t θ + r2 multipliction: u( z = P( z, tg(t dt. Interprettion: Let Y be the set of ll integrble 2π-periodic functions g = g(θ. Let X be the collection of ll well-behved functions u = u(x, y tht stisfy Lplce s Eqution on r < b. Define liner opertor A: X Y like this: Au] = g g(θ = lim u(r, θ. r b E.g., if u(r, θ = r cos θ then g = Au] is the function g(θ = b cos θ, < θ < π. It s esy to find g once u is given, but the relly interesting problem is usully just the opposite, nmely, solve for u in Au] = g. Idelly, we would like to hve n inverse for A, so we could just write u = A 1 g]. Formul ( does exctly this, with P replcing A 1. (Anlogy: For n-dimensionl n vectors, u = Pg iff u k = P kt g t ; for functions, u( z = P( z, tg(t dt. t=1 Froese s notes re good source for this. (See link on course home pge. ( File 2014notes, version of 5 August 2014, pge 11. Typeset t 16:38 August 5, 2014.

12 12 PHILIP D. LOEWEN Exmple (D. Stndrd bitten slice, different boundry conditions. Agin we solve u = 0 in the polr region Ω = {(r, θ : 1 < r < b, 0 < θ < } where = 1, b > 1 nd (0, 2π re given. u(r,=k(r u = 0 u(b,θ=0 u(,θ=0 u(r,0=h(r This time, however, nonhomogeneous boundry dt re given on the flt sides of the domin: u(r, 0 = h(r u(r, = k(r 1 < r < b, u(1, θ = 0 u(b, θ = 0 0 < θ <. Shortcut: If both given functions h nd k re constnt, creful choices of the constnts A nd B together with the substitution u(r, θ = A + Bθ + w(r, θ will reduce this problem to n instnce of (A bove. But if one of these functions is nonconstnt, the method shown below seems inevitble.] A new eigenvlue problem. Seprting u(r, θ = R(rΘ(θ in the homogeneous BC gives R(1Θ(θ = 0 = R(bΘ(θ, i.e., R(1 = 0 = R(b. These homogeneous conditions on R force us to build our eigenvlue problem using the ODE in line (1 of the seprtion-of-vribles result bove. We rrive t this eigenvlue problem for the function R: r 2 R (r + rr (r + λr(r = 0, 1 < r < b; R(1 = 0 = R(b. (11 This is not FSS problem, becuse the ODE hs form different from the one fmilir so fr. To find eigenfunctions will tke grinding cse-by-cse nlysis. Since the ODE hs Euler type, guess R(r = r p nd plug in: r 2 p(p 1r p 2] + r pr p 1] + λr p = 0 p 2 + λ = 0. Cse λ < 0: Write λ = s 2 for some s > 0 nd get p = ±s, so the generl solution is R(r = Ar s + Br s, A, B R. File 2014notes, version of 5 August 2014, pge 12. Typeset t 16:38 August 5, 2014.

13 Lplce s Eqution 13 Now 0 = R(1 = A + B gives B = A, so R(r = A(r s r s, nd 0 = R(b = A(b s b s forces A = 0 (since b s > 1 > b s. Thus only trivil solutions pper when λ < 0. Cse λ = 0: Repeted roots, so R(r = A + B ln(r, A, B R. Now 0 = R(1 = A gives R(r = B ln(r nd 0 = R(b = B ln(b forces B = 0 (since ln(b > 0. Only trivil solutions here too. Cse λ > 0: Write λ = ω 2 for some ω > 0, so p 2 = ω 2 nd p = ±iω. Recll r iω = e iω ln(r = cos(ω ln(r + i sin(ω ln(r, nd tht both rel nd imginry prts give solution. The generl solution is R(r = A cos(ω ln(r + B sin(ω ln(r, A, B R. Now 0 = R(1 = A gives R(r = B sin(ω ln(r, so 0 = R(b = B sin(ω ln(b. This time nonzero vlues of B cn occur, provided ω > 0 stisfies sin(ω ln(b = 0, i.e., ω ln(b = nπ, n = 1, 2, 3,.... Thus we hve sequence of eigenvlues ( 2 nπ λ n = ωn 2 =, n = 1, 2, 3,..., ln(b nd the corresponding eigenfunctions re (multiples of ( nπ R n (r = sin ln(b ln(r, n = 1, 2, 3,.... Expnsion Formuls. Dividing the ODE in (11 by r puts it into Sturm-Liouville form: ( ( = rr + R + λ R = (rr (r + λ R, R(1 = 0 = R(b. r r Hence for ny resonble f = f(r defined for 1 < r < b, we hve f(r = b n R n (r = b n = 2 ln(b b r=1 f(r sin ( nπ b n sin ln(b ln(r, 1 < r < b, ( nπ dr ln(b ln(r, r n = 1, 2, 3,.... ( File 2014notes, version of 5 August 2014, pge 13. Typeset t 16:38 August 5, 2014.

14 14 PHILIP D. LOEWEN Bck in problem (D, we postulte n eigenfunction-series solution: u(r, θ = ( nπ Θ n (θ sin ln(b ln(r for some Θ n to be determined. Plugging in θ = 0 nd θ = will give two opportunities to use (, one with h nd the other with k, nd these will revel the vlues of Θ n (0 nd Θ n (. The evolution of Θ n will be governed by the PDE, i.e., 0 = r 2 u rr + ru r + u θθ = ( Θn (θ r 2 R n (r + rr n (r] + R n (rθ n (θ Now remember the eigenvlue problem: ( 2 nπ r 2 R n (r + rr n (r = R n (r. lnb Hence we hve 0 = Θ (θ ( 2 ( nπ nπ Θ n (θ] sin lnb ln(b ln(r. This is nother chnce to use (, this time with the zero function s the expnsion result. The coefficient formuls give, for ech n, ( 2 nπ 0 = Θ n (θ Θ n (θ, i.e., Θ n (θ = A n e nπθ/ ln b + B n e nπθ/ ln b. lnb Answer: u(r, θ = ] ( nπ A n e nπθ/ln b + B n e nπθ/ ln b sin ln(b ln(r, with A n nd B n determined by using ( with h nd k. File 2014notes, version of 5 August 2014, pge 14. Typeset t 16:38 August 5, 2014.