Physics 246 Electricity nd Mgnetism I, Fll 26, Lecture 22 1 Summry: Method of Seprtion of Vribles 1. Seprtion of Vribles in Crtesin Coordintes 2. Fourier Series Suggested Reding: Griffiths: Chpter 3, Section 3.3, pges 127-145. Wngsness: Chpter 11, Section 11.4 nd 11.5, pges 185-21. Kreyszig: Chpter 1, pges 565-625.
Physics 246 Electricity nd Mgnetism I, Fll 26, Lecture 22 2 Method of Seprtion of Vribles In solving prtil differentil equtions such s Lplce s eqution, 2 V (x, y, z) = it is possible to write the solution V (x, y, z) s the product of three functions, ech of single vrible V (x, y, z) = X(x)Y(y)Z(z) I should point out tht not ll solutions of Lplce s eqution cn be written in this form, however, ll solutions cn be written s (perhps infinite) sum of such seprble terms. Exmple:
Physics 246 Electricity nd Mgnetism I, Fll 26, Lecture 22 3 Find the potentil V (x, y, z) inside the slot shown in the bove drwing. The boundry conditions to be stisfied re: 1) t y = ; V = for < z < ; < x < 2) t y = π; V = for < z < ; < x < 3) for y π; V = V (y) t x = ; < z < Note: this third boundry condition defines the potentil long n infinite strip t x =. Wht hppens t lrge x? Does V? We know tht for n infinite line of chrge V log r which does not pproch zero t lrge r. Recll the problem of point chrge +q ner conducting plne. It induces negtive chrge distribution in the plne ner it hving totl chrge -q fr from +q we would see no net chrge nd s r, V. At the surfce x =, y π, V (, y, ) = V (y) is function of the surfce chrge density there. Thus, the presence of the conducting pltes t y = nd y = π gurntees tht for lrge x (>> π) we will see no net chrge nd,
Physics 246 Electricity nd Mgnetism I, Fll 26, Lecture 22 4 therefore, V s x. This gives us our lst boundry condition: 4) Between y = nd y = π, t lrge x, V. In the region of interest, which lies within the slot, there is no chrge density (ϱ = ) therefore, Lplce s eqution will give us the potentil V. There is no z-dependence to this problem two dimensionl nd 2 V (x, y) = 2 V x 2 + 2 V y 2 = Try the function V (x, y) = X(x)Y(y) [ ] 2 x + 2 X(x)Y(y) = 2 y 2 Y(y) 2 X x 2 + X(x) 2 Y y 2 = If we now divide both sides of this eqution by the product XY we obtin 1 d 2 X(x) + 1 d 2 Y(y) = X(x) dx 2 Y(y) dy 2 f(x) + g(y) =
Physics 246 Electricity nd Mgnetism I, Fll 26, Lecture 22 5 since the first term is only function of x nd the second term is only function of y. Since x nd y re independent vribles, the only wy to stisfy this eqution is if the two functions re seprtely equl (nd opposite) to the sme constnt. Then, 1 d 2 X(x) = C X(x) dx 2 1 1 d 2 Y(y) = C Y(y) dy 2 1 so tht their sum is zero. These two equtions cn now be solved seprtely. The eqution for X is d 2 X dx 2 C 1X = which hs solution X(x) = Ae kx + Be kx, where C 1 = k 2 Tht this is the solution cn be esily verified by direct substitution: dx dx = kaekx kbe kx nd d 2 X dx 2 = k2 Ae kx + k 2 Be kx = k 2 X
Physics 246 Electricity nd Mgnetism I, Fll 26, Lecture 22 6 is Similrly, the solution to the eqution for Y, d 2 Y dy 2 + k2 Y = Y(y) = C sin ky + D cos ky On substituting, we hve dy = kc cos ky kd sin ky dy nd d 2 Y dy = 2 k2 C sin ky k 2 D cos ky = k 2 Y Thus, the solution to the two dimensionl problem is V (x, y) = X(x) Y(y) = (Ae kx + Be kx )(C sin ky + D cos ky) Next we need to use the boundry conditions to determine the five constnts: A, B, C, D nd k. B.C. (4) requires tht V s x. If the constnt k > then the term Ae kx cn only be if A =. On the other hnd, if k <, the term Be kx will only be zero if B =. In either cse we would hve V (x, y) = e kx (C sin ky + D cos ky)
Physics 246 Electricity nd Mgnetism I, Fll 26, Lecture 22 7 where (for the cse k > ) C = BC nd D = BD B.C. (1) requires tht V (x, ) =. Since cos ky t y = we require tht D =. So now, V (x, y) = e kx C sin ky B.C. (2) requires tht V (x, π) =. Except for the trivil cse where C = this will only be true if sin kπ =, i.e., if k = n, n integer. Then V (x, y) = e nx C sin ny Finlly, we hve B.C. (3) which sttes tht t x = we hve V (, y) = V (y). The solution which we hve found will fit this boundry condition if V (y) = C sin ny but not otherwise. No one choice of either C or n will solve the generl problem! Note, however, tht if V 1 nd V 2 re both solutions, then α 1 V 1 + α 2 V 2 is lso solution 2 (α 1 V 1 + α 2 V 2 ) = α 1 2 V 1 + α 2 2 V 2 = For exmple, we might hve V 1 = e n 1x sin n 1 y, nd V 2 = e n 2x sin n 2 y This suggests tht for the generl boundry condition V (y) we should try the expression
Physics 246 Electricity nd Mgnetism I, Fll 26, Lecture 22 8 V (x, y) = C k e kx sin ky, k=1 k integer with V (, y) = C k sin ky = V (y) k=1 The expnsion of V (, y) = V (y) in n infinite series of sine terms is clled the Fourier Sine Series. In generl, one cn expnd ny odd function of y s sine series nd ny even function of y cn be expnded s cosine series. To do this, however, we need to determine ll of the C k. To help us with this we use Fourier s Trick: π sin ky sin ny dy = π 2 δ kn If we tke the expression C k sin ky = V (y) k=1
Physics 246 Electricity nd Mgnetism I, Fll 26, Lecture 22 9 nd multiply both sides by sin ny nd integrte, we hve ( π ) π sin ny C k sin ky dy = V (y) sin ny dy k=1 or π C k sin ny sin ky dy = k=1 } {{} π V (y) sin ny dy π 2 δ kn In the sum over k, the δ kn picks out the one vlue of k which is equl to n, thus C n π 2 = π V (y) sin ny dy so tht finlly our solution to Lplce s eqution in the region between the pltes is with V (x, y) = C n e nx sin ny, n=1 n integer
Physics 246 Electricity nd Mgnetism I, Fll 26, Lecture 22 1 C n = 2 π π V (y) sin ny dy Note: A result which is probbly new to the student is the ide tht essentilly ny function V (y) cn be expnded in terms of Fourier series (either sine or cosine, or in the generl cse, both). The expressions V (y) = C(k) sin ky nd k=1 π C(k) = 2 V (y) sin ny dy π define the Fourier series representtion of the originl function nd the Fourier coefficients. In the generl cse we would replce the discrete summtion by n integrl nd this would define Fourier trnsform pir. For exmple, we might hve 2 F(k) = f(x) cos kx dx π nd f(x) = 2 π F(k) cos kx dk
Physics 246 Electricity nd Mgnetism I, Fll 26, Lecture 22 11 The Fourier series representtions of constnt potentil V = V is shown in the figure below. Here you see the trunction effects observed when the series is limited to 1, 5, 1 or 1 terms.
Physics 246 Electricity nd Mgnetism I, Fll 26, Lecture 22 12 Problem 3.14: Infinitely Long Rectngulr Pipe For the infinitely long rectngulr pipe shown bove we hve the following boundry conditions: 1) V = t x =, y, < z < 2) V = t y =, x b, < z < 3) V = t y =, x b, < z <
Physics 246 Electricity nd Mgnetism I, Fll 26, Lecture 22 13 4) V = V (y) t x =, y, < z < Find the potentil, V (x, y, z), inside the pipe. First we notice tht this problem is ctully independent of z, thus it reduces to two-dimensionl problem for V (x, y). The generl form of such solution cn immeditely be written down, however, we should consider the boundry conditions first: The region of interest is finite in extent in the x nd y directions (nd independent of z). Thus, solutions of the forms e kx nd e kx re llowble. The boundry conditions in the x-direction suggest tht the x-dependence should involve the terms e kx, e kx, cosh kx, or sinh kx while the boundry conditions in the y-direction suggest tht the y-dependence should involve terms like e ikx, e ikx, sin kxor cos kx. Thus, we expect solutions of the form V (x, y) = (Ae kx + Be kx )(C sin ky + D cos ky) Now, more creful exmintion of the boundry conditions shows tht for the y dependence, the fct tht V = t y = requires tht D =. The fct tht V = t y = requires tht ky y= = k = nπ. Finlly, we note tht if x = we require tht V = so tht we need B = A.
Physics 246 Electricity nd Mgnetism I, Fll 26, Lecture 22 14 The form of the solution is then V (x, y) = C sin( nπ y) sinh(nπ y) where we hve collected the constnts into C = 2 A C. To mtch the remining boundry condition (V = V (y) t x = b) we try superposition V (x, y) = C n sinh( nπ y) sin(nπ x) n= Then, t x = b we would hve V (b, y) = V (y) = C n sinh( nπ b) sin(nπ x) n= Now, multiplying both sides by sin( mπ y) nd integrting over y from to, we obtin V (y) sin( mπ y) dy = C n sinh( nπ b) sin( nπ y) sin(mπy) dy n= }{{} 2 δ mn giving C m sinh( mπ b) 2 = V (y) sin( mπ y) dy
Physics 246 Electricity nd Mgnetism I, Fll 26, Lecture 22 15 Solving this eqution for C m gives [ ] 2 C m = sinh( mπ b) V (y) sin( mπ y) dy nd V (x, y) = C m sinh( mπ x) sin(mπ z y) m= If V (y) = V, constnt, then we cn redily evlute ll of the coefficients since { if m is even, sin( mπ y) dy = 2 mπ Then, [ ] 2V C m = sinh( mπ b) 2 mπ, Then, the solution for V (y) = V is V (x, y) = 4V π n=1,3,5,... if m is odd m odd sinh( nπ x) n sinh( nπ b) sin(nπ y) A much better wy to write this summtion over only the odd vlues of n is to let n = 2l + 1 (this wy n is odd for ny integer vlue of l), replce n by 2l + 1 wherever it ppers nd then sum over l. V (x, y) = 4V π l= sinh( (2l+1)π x) (2l + 1) sinh( (2l+1)π + 1)π y) b) sin((2l