PHYS 301 HOMEWORK #10 (Optional HW)

PHYS 301 HOMEWORK #10 (Optional HW) 1. Conside the Legende diffeential equation : 1 - x 2 y'' - 2xy' + m m + 1 y = 0 Make the substitution x = cos q and show the Legende equation tansfoms into d 2 y 2 + cot q + m m + 1 y = 0 We will need to tansfom each tem in the equation using the substitution x = cos q. Going though the equation, we fist tansfom the (1 - x 2 ) tem. Since x=cos q, it is easy to show that: 1 - x 2 = 1 - cos 2 q=sin 2 q Now we have to tansfom the deivate tems. Although we fist enounte y'', we fist have to tansfom y'. We do this using the chain ule : dx = dx Since x = cos q, then dx = - sin q fl dx = -1 so that sin q dx = -1 sin q ª u Ou definition of u will become han in just a moment. Now, to tansfom y'', we apply the chain ule again : d 2 y dx = d 2 dx dx = du dx = du dx We use the definition of u to find :

2 phys301-2011hw10s.nb du = d -1 sin q = cos q sin 2 q - 1 sinq d 2 y 2 Using ou expesssion fo /dx fom above, we get : d 2 y dx = du 2 dx = cos q sin 2 q - 1 sinq d 2 y -1 2 sinq And the Legende diffeential equation tansfoms as : 1 - x 2 y'' - 2xy' + m m + 1 y Ø sin 2 q cos q sin 2 q - 1 sinq d 2 y -1 2 sinq - 2 cos q -1 sin q + m m + 1 y = -cos q sin q + d2 y 2 d 2 y 2 + cos q sin q QED + 2 cos q sin q + m m + 1 y = + m m + 1 y = 0 2. Conside fou chages aligned along the x axis : a chage of - 2 q at x = +a a chage of + q at x = +2 a a chage of + 2 q at x = -a a chage of - q at x = -2 a Calculate the electostatic potential of this aay of chages. The following diagam will be useful fo the solution :

phys301-2011hw10s.nb 3 P 1 2 4 3 f q -q 2q -2q q -2a,0 -a,0 a,0 2a,0 We ae asked to find the potential at P due to the fou chages shown.ou fist step is to apply the pinciple of supeposition to this situation, ecognizing that the total potential measued at P will be the sum of the potentials due to the fou chages.thus, we must find a way to expess the potential due to each chage.we know that the potential due to a chage Q a distance d away is : V = kq d whee k is a constant and you might ecall fom electonamics that this constant is given by : k = 1 4 pe 0 Fo now, we will just keep the constant as k. Thus, fo each chage, we need to compute its distance fom P in tems of and q.we can do this using the law of cosines. (1) 1 2 = 2 + a 2-2 a cos qfl 1 = 1 + a 2-2 a cos q 2 2 = 2 + 2a 2-2 2a cos qfl 2 1 + 2a 2-2 2a cos q

4 phys301-2011hw10s.nb 3 2 = 2 + a 2-2 a cos f= 2 + a 2-2 a cos 180 -q = 2 + a 2 + 2 a cosq = 1 + a 2 + 2 a cos q 4 2 = 2 + 2a 2-2 2a cos f= 2 + 2a 2-2 2a cos 180 -q = 2 + 2a 2 + 2 2a cos q= 1 + 2a 2 + 2 2a cos q Note that in the expessions fo 3 and 4, we fist define the distance in tems of the angle f and ecognize that f=180-q (and that cos(180-q) = -cos q). Since potential vaies as the invese of the distance (see eq.(1)), we can wite each of the 1/distance expessions in tems of Legende polynomials.recalling that the geneating function fo Legende polynomials is : 1 =S Pl x h l 1-2hx + h 2 (2) we can wite the potential due to the fist chage as: V 1 = -2qk 1 = -2 qk -2qk = 1-2 a cos q+ a 2 S P l cos q a l (3) Notice that eq. (3) is identical to eq. (2) with h = (a/) and x = cos q. We use the geneating function ( eq. (1)) to wite each of the emaining potentials : V 2 = kq 2 = kq = 1-2 2a cos q+ 2a 2 kq S Pl cos q 2a l (4) Compae the adical expession in eq. (4) to the analogous expession in eq. (2). In eq. (4), x = cos q as befoe, but now h = 2 a/, esulting in the (2 a/) tem in the infinite seies.

phys301-2011hw10s.nb 5 V 3 = 2qk 3 = 2qk 2qk = 1 + 2 a cos q+ a 2 2qk S -1 l P l cos q a l S Pl cos q -a l = In the adical expession above, notice that h = -(a/), thus geneating the -1 l tem. Finally, we have fo V 4 : V 4 =- qk 4 =- - qk qk = 1 + 2 2a cosq+ 2a 2 S Pl cos q -2a l =- qk S -1 l P l cos q 2a l (5) (6) We find the total potential by summing the fou individual potentials : V = V 1 + V 3 + V 2 + V 4 We can add in any ode, of couse, but gouping the potentials in this way allows us to goup accoding to common tems. We can wite these goupings as : V 1 + V 3 =-2 qk -2qk S P l cos q a l + 2qk S 2Pl cos q a l = -4qk l odd S -1 l P l cos q a l = S Pl cos q a l l odd notice that even l tems sum to zeo. Fo the second gouping, we have : V 2 + V 4 = qk S Pl cos q 2a l - qk S -1 l P l cos q 2a l = qk S 2 l P l cos q a l 1 - -1 l = 2qk S 2 l P l cos q a l l=odd Again the even l tems cancel, and notice that we have a 2 l tem. If we sum all fou tems and expand, we find ou potential takes the fom: V = -4qk P 1 cos q a + P 3 cos q a 3 + P 5 cos q a 5 +... +

6 phys301-2011hw10s.nb 2qk 2 1 P 1 cos q a 1 + 2 3 P 3 cos q a 3 + 2 5 P 5 cos q a 5 +... = 12 q k P 3 cos q a 3 + 60 q k P 5 cos q a 5 +... Notice that the fist non - zeo tem is the l = 3 tem; all the even tems ae zeo and the l = 1 tems cancel. Altenate Solution We can also solve this poblem making use of the esults of the electic dipole posted in the classnote: http://www.luc.edu/faculty/dslavsk/couses/phys301/classnotes/legendep.pdf In this wite-up, efe to the section "The Electic Dipole", and note that thee ae chages of ±q at ±a fom the oigin. Notice in the gaph above fo the octupole poblem (the poblem with fou chages), that we have essentially a pai of dipoles: the inne dipole consists of chages of 2q at ± a, and the oute dipole consists of chages ±q at ±2a. Theefoe, we can find the potential due to the inne dipole by taking eq. (33) of the classnote and substituting -2q fo q, giving us: V inne =- 4qk S Pl cos q a l l odd The potential due to the oute dipole can be detemined fom eq. (33) by substituting 2 a fo a, giving us : V oute = 2qk S 2 l P l cos q a l l=odd And we obtain the same expession as befoe when we sum the potentials of the inne and oute dipoles. 3. Boas p. 581 no. 1. We ae asked to expand the function :

phys301-2011hw10s.nb 7-1, -1 < x < 0 f x = 1 0< x < 1 In a Legende seies. Legende seies can be expessed as : f x = S cm P m x m=0 whee the P m (x) ae the Legende polynomials, and the coefficients ae detemined fom: c m = 2m+ 1 1f x P m x dx 2-1 We can save ouselves a little time by noticing that ou f (x) is an odd function; ecalling that Legende polynomials ae odd functions fo odd values of m and even fo even values of m, we can see immediately that all the even coefficients will be zeo since they will be the esult of integating an odd function (f (x) times an even Legende polynomial) fom - 1 to 1. Theefoe, we need only compute the odd coefficients and we find : c 1 = 3 1 2 f x P 1 x dx = 2 ÿ 3 1-1 2 1 ÿ P 1 x dx = 3 0 2 c 3 = 2 ÿ 7 1 11 2 1 ÿ P 3 x dx = 7 0 0 2 5x3-3x dx = -7 8 c 5 = 2 ÿ 11 1 11 2 1 ÿ P 5 x dx = 11 0 0 8 15 x - 70 x3 + 63 x 5 dx = 11 16 so that the fist thee non - zeo tems of the Legende seies is : f x = c 1 P 1 x + c 3 P 3 x + c 5 P 5 x +... = 3 2 x - 7 8 ÿ 1 2 5x3-3x + 11 16 ÿ 1 8 15 x - 70 x3 + 63 x 5 +... We can wite a shot Mathematica pogam to veify :

8 phys301-2011hw10s.nb Clea c, m, f f x_ : Which 1 x 0, 1, 0 x 1, 1 c m_ : c m 2 m 1 2 Integate LegendeP m, x f x, x, 1, 1 Plot Sum c n LegendeP n, t, n, 0, 51, t, 1, 1 1.0 0.5-1.0-0.5 0.5 1.0-0.5-1.0 4. Boas p. 581 no. 4. We compute the Legende seies as above, finding coefficients fom : Do Pint "fo n ", n, " c n ", 2 n 1 2 Integate AcSin x LegendeP n, x, x, 1, 1, n, 0, 6 fo n 0 c n 0 fo n 1 c n 3 8 fo n 2 c n 0 fo n 3 c n 7 128 fo n 4 c n 0 fo n 5 c n 11 512 fo n 6 c n 0 Notice that the even coefficients ae zeo, eminding us that ac sin (x) has an odd powe expansion :

phys301-2011hw10s.nb 9 Seies AcSin x, x, 0, 7 x x3 6 3x5 40 5x7 112 O x 8 So that the Legende seies expansion fo ac sin x is : ac sin x = p 8 3P 1 x + 7 16 P 3 x + 11 64 P 5 x +,,, Veifying with a plot : Clea c, m, f f x_ : AcSin x c m_ : c m 2 m 1 2 Integate f x LegendeP m, x, x, 1, 1 Plot Sum c n LegendeP n, t, n, 0, 51, t, 1, 1 1.5 1.0 0.5-1.0-0.5 0.5 1.0-0.5-1.0-1.5 5. Boas p. 626, no. 2 Following the example in the book and the wok done in class, we know that sepaating vaiables leads to the geneal solution : Applying bounday conditions : T x, y = A cos kx + B sin kx Ce ky + De -ky T x, y Ø 0asyØ 0 fl C = 0 T x, y = 0 when x = 0 fl A = 0 T x, y = 0 when x = 20 cm fl sin 20 k = 0 fl 20 k = n p o k = n p 20 These thee conditions tells us that the geneal solution is of the fom :

10 phys301-2011hw10s.nb -n p y 20 T x, y = B sin n p x 20 e The bounday condition along the bottom edge tells us that : T x, 0 = B sin n p x 20 = 0, 0 < x < 10 100, 10 < x < 20 Now, we can see that thee is no single value of B that will satisfy this equation fo all values of n and x. Howeve, we also know that we can expand this function as a Fouie sine seies and find the values of the coefficients that will allow us to model this function along the lowe edge. Expanding ou function as a Fouie sine seies fom - 20 to 20, we know that the values of the coefficients ae detemined fom : b n = 2 L 0 L f x sin n p x L dx Fo this paticula function, L = 20 and since f (x) = 0 fom 0 to 10, ou integal becomes : b n = 2 20 20 100 sin n p x 20 dx 10 Integate 100 Sin n x 20, x, 10, 20 10 200 Cos n Cos n 2 n Evaluating coefficients : Simplify, Assumptions Mod n, 2 0 200 1 n n Simplify, Assumptions Mod n, 2 1 200 n Theefoe, we can summaize ou coefficients as : b n = 200 n p, n odd -400, n= 2, 6, 10,... n p 0, n = 4, 8, 12,...

phys301-2011hw10s.nb 11 and ou solution becomes : -n p y 20 T x, y = S bn sin n p x 20 e n=1 Constucting a function and plotting, we get : Clea temp temp x_, y_ : 200 Sum Cos n 2 Cos n Sin n x 20 Exp n y 20 n, n, 1, 1000 ContouPlot temp x, y, x, 0, 20, y, 0, 20, Contous 10 6. Boas p. 626 no. 3. In this case, ou geneal solution becomes : T x, y = A cos kx + B sin kx Ce ky + De -ky Application of the bounday conditions gives us again that C = 0 = A, and we find k fom : T x, y = 0 when x =p fl sin k p = n pflk = n

12 phys301-2011hw10s.nb so that ou geneal solution becomes : T x, y = B sin nx e -ny T (x, 0) = B sin (n x) = cos x, so that we again expand T (x, 0) in a Fouie sine seies between - p and p and evaluate the coefficients that will model this bounday condition : b n = 2 L 0 L f x sin nx dx = 2 p p cos x sin nx dx 0 Integate Cos x Sin n x, x, 0, 2 2n 1 Cos n 1 n 2 and we can see that these coefficients ae zeo when n is odd, and when n is even : 4n b n = p n 2-1 so that ou solution is finally : T x, y = 4 p S n sin nx e -ny even n n 2-1 7. Boas p. 627 no. 7 The geneal fom of the solution to this poblem is the same as the pio two, except now the plate is not infinite in length. If the plate is 1 cm long and the tempeatue at y = 1 is zeo, Boas shows (p. 624) that we can model this bounday condition bysetting : Ce ky + De -ky = 1 2 ek 1-y - 1 2 e-k 1-y = sinh k 1 - y Ou geneal solution becomes : so that : T x, y = B sin kx sinh k 1 - y T p, y = 0 fl k = n

phys301-2011hw10s.nb 13 T x, y = B sin nx sinh n 1 - y T x, 0 = B sin nx sinh n = cos x As befoe, we define a function on (-p, p) and find the coefficients of its Fouie sine seies. So we have : T x, 0 = cos x = S Bn sinh n sin nx = S bn sin nx fl B n = n=1 n=1 b n sinh n We compute the b n 's fom: b n Integate Cos x Sin n x, x, 0, 2 2n 1 Cos n 1 n 2 epoducing the esult fom the pevious poblem. Theefoe, the final tempeatue distibution is: T x, y =SB n sin nx sinh n 1 - y = 4 p n sinh n 1 - y sin nx S n=even n 2-1 sinh n 8. Boas p. 627 no. 10 As befoe, the geneal solution is : T x, y = A cos kx + B sin kx Ce ky + De -ky The bounday condition at y = 10 allows us to wite : Ce ky + De -ky = 1 2 ek 10-y - e -k 10-y = sinh k 10 - y The bounday condition that T = 0 when x = 0 fl A = 0, and the bounday condition thatt (10, y) = 0 fl sin 10 k = 0 fl k = 10/n p. The solution then takes the fom : T x, y = B sin n p x 10 sinh n p 10 - y 10

14 phys301-2011hw10s.nb so that T (x, 0) = B sin[n p x/10] sinh[n p] = 100 We can expand this function (f (x) = 100) in a Fouie sine seies T x, 0 = 100 = S Bn sin n p x 10 sinh n p = S bn sin n p x 10 n=1 n=1 and the Fouie coefficients ae detemined fom : b n = B n sinh n p = 2 10 0 1 100 ÿ sin n p x 10 dx = 400 n p odd n 0 even n theefoe, The solution is then : B n = T x, y = 400 p b n sinh n p = 400 fo odd n. n p sinh n p sin n p x 10 sinh n p 10 - y 10 S n odd n sinh n p 9. Boas, p. 650 no. 1. Following the teatment in Boas, we find that the geneal solution to Laplace' s equation in spheical coodinates is : T, q = A l + B -l l+1 P l cos q Since we ae asked to find the tempeatue distibution inside the sphee, we know the coefficient B must be zeo, othewise we have a singulaity at the oigin = 0. Thus the solution becomes : T, q = A l P l cos q We ae given the bounday condition that T (1, q) = 35 cos 4 q which allows us to wite: T 1, q = AP l cos q = 35 cos 4 q Since no single value of A can satisfy this equation fo all values of q, we ealize that a seies of solutions of the fom :

phys301-2011hw10s.nb 15 T 1, q =S Al P l cos q = 35 cos 4 q We can find the A l by ealizing they ae just the coefficients of the appopiate Legende seies, and we can compute them fom: A l = 2l+ 1 135 x 4 P l x dx whee we set x = cos q 2-1 Recalling that Legende polynomials ae even/odd fo even/odd values of l, and that ou function 35 x 4 ) is even, we expect to get non-zeo coefficients only fo even values of l: Do Pint "fo l ", l, " A l ", 2 l 1 2 Integate 35 x^4 LegendeP l, x, x, 1, 1, l, 0, 8, 2 fo l 0 A l 7 fo l 2 A l 20 fo l 4 A l 8 fo l 6 A l 0 fo l 8 A l 0 So ou solution fo the tempeatue distibution inside the sphee is : T, q =S Al l P l cos q = 7P 0 cos q + 20 2 P 2 cos q + 8 4 P 4 cosq 10. Boas, p. 650 no. 2 This poblem is vey simila to the poblem above. We find the geneal solution to Laplace' s equation in spheical coodinates; set B = 0 since we ae asked to find the tempeatue distibution inside the sphee, and find the values of the coefficients A l by expanding the bounday condition as a Legende seies. We have: and at = 1 : T, q =S Al l P l cos q

16 phys301-2011hw10s.nb T 1, q =S Al P l cos q = cos q-cos 3 q In othe wods, the A' s ae the coefficients of the Legende seies of cos q - cos 3 q which we compute fom: Do Pint "fo l ", l, " A l ", 2 l 1 2 Integate x x^3 LegendeP l, x, x, 1, 1, l, 0, 9 Befoe we integate, let' s notice that the integal will be zeo wheneve l is even, so we expect to have only odd non - zeo coefficients : fo l 0 A l 0 fo l 1 A l 2 5 fo l 2 A l 0 fo l 3 A l 2 5 fo l 4 A l 0 fo l 5 A l 0 fo l 6 A l 0 fo l 7 A l 0 fo l 8 A l 0 fo l 9 A l 0 and ou solution is : T, q =S Al l P l cos q = 2 5 P 1 cos q - 2 5 3 P 3 cos q