Prof. Dr. I. Nasser Phys 571, T131 5-Nov-13 Green function.doc. 2 ρ(r)

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1 Gree fuctio.doc Solutio of ihomogeeous ordiary differetial equatios usig Gree's fuctios G(, ') (9.7, 10.5) Historical Itroductio: Gree's fuctios are auiliary fuctios i the solutio of liear partial differetial equatios. Gree's fuctio is amed for the self-taught Eglish mathematicia George Gree ( ), who ivestigated electricity ad magetism i a thoroughly mathematical fashio. I 188 Gree published a privately prited booklet, itroducig what is ow called the Gree fuctio. This was igored util William Thomso (Lord Kelvi) discovered it, recogized its great value ad had it published ie years after Gree's death. Berhard Riema gave it the ame "Gree fuctio" For eample, i electrodyamics, we are cocered with fidig solutios to the Poisso equatio: ρ(r) Φ (r) = (I) εo ad the Laplace equatio: Φ (r) = 0 (II) I fact, the Laplace equatio is the homogeeous versio of the Poisso equatio. The Gree s fuctio allows us to determie the electrostatic potetial from volume ad surface itegrals: (III) This geeral form ca be used i 1,, or 3 dimesios. I geeral, the Gree s fuctio must be costructed to satisfy the appropriate boudary coditios. I some cases, it may be difficult or icoveiet to fid a Gree s fuctio that geerates a solutio with the correct boudary coditios. I these situatios, we ca still use Eq. (III) to obtai a solutio to the Poisso equatio (I) ad the add the appropriate liear combiatios of solutios to the Laplace s equatio (II) to adjust the boudary values. Here, we are goig to discussed oe method of costructig Gree s fuctios that works for oe-dimesioal systems. Net, we discuss aother method that is geeralizable for higher dimesioal systems. 1- Homogeeous Equatio Start with the secod order liear homogeeous differetial equatios (Eige value equatios), which ca be writte as a eigevalue problem of the form: L ˆ( ) y ( ) = E y ( ) (1) where ˆL is a operator ivolvig derivatives w.r.t., E is a eigevalue ad y is a eigefuctio (which satisfies some specified boudary coditios). The geeral case that we are iterested i is called a Sturm-Liouville 1 problem, for which oe ca show that the eigevalues are real, ad the eigefuctios are orthoormal, i.e. b m = y * m( ) y ( ) = δ m () a where a ad b are the upper ad lower limits of the regio where we are solvig the problem, ad we have also ormalized the solutios. 1

2 Gree fuctio.doc 1 The geeral Sturm-Liouville problem, see Arfeke, has a weight fuctio w() multiplyig the eigevalue o the RHS of Eq. (1) ad the same weight fuctio multiplies the itegrad show i the LHS of the orthogoality ad ormalizatio coditio, Eq. (). Furthermore the eigefuctios may be comple, i which case oe must take the comple cojugate of either y or ym i i Eq. (). Here, to keep the otatio simple, we will just cosider eamples with w() = 1 ad real eigefuctios. Eample, Fid the solutio of the differetial equatio: d y( ) 1 + y ( ) = λy ( ) 4 d y( ) 1 = λ y( ) 4 i the iterval 0 π, with the boudary coditios y(0) = y( π ) = 0. Aswer: This correspods to ˆ d L = (4) cos ad This is just the simple harmoic oscillator equatio, ad so the solutios are ( ) si ( ). The boudary coditio y (0) = 0 elimiates ( ) cos ad the coditio y ( π ) = 0 gives a positive iteger. (Note: For = 0 the solutio vaishes ad takig < 0 just gives the same solutio as that for the correspodig positive value of because si ( ) = si ( ) ). Hece we oly eed cosider positive iteger.) The ormalized eigefuctios are therefore: y ( ) = si ( ), π = 1,,3, (5) ad the eigevalues i Eq. (3) are: 1 λ = 4 (6) sice the equatio satisfied by y ( ) is y "( ) + y( ) = 0. (3)

3 Gree fuctio.doc Defiitios Dirac delta fuctio 1. Defiitio as limit. The Dirac delta fuctio ca be thought of as a rectagular pulse that grows arrower ad arrower while simultaeously growig larger ad larger. rect(, b) = y width = b height = 1/b (so total area = 1) δ() = lim(b 0) rect(, b) Note that the itegral of the delta fuctio is the area uder the curve, ad has bee held costat at 1 throughout the limit process. δ ( ) = 1 Shiftig the origi. Just as a parabola ca be shifted away from the origi by writig y = ( 0 ) istead of just y =, ay fuctio ca be shifted by pluggig i 0 i place of its usual argumet. y δ( - 0 ) = 0 Shiftig the positio of the peak does t affect the total area if the itegral is take from to. δ ( 0 ) = 1 Disclaimer: Mathematicias will object that the Dirac delta fuctio defied this way (or ay other way, for that matter) is ot a real fuctio. That is true, but physicists recogize that for all practical purposes you really ca just thik of the delta fuctio as a very large peak. 3

4 Gree fuctio.doc θ ( ). Defiitio as derivative of step fuctio. The step fuctio, also called the Heaviside step fuctio is usually defied like 1 this: 1, if > 0 θ ( ) = 0, if 0 It s a fuctio whose oly feature is a step up from 0 to 1, at the origi: What s the derivative of this fuctio? Well, the slope is zero for < 0 ad the slope is zero for > 0. What about right at the origi? The slope is ifiite! So the derivative of this fuctio is a fuctio which is zero everywhere ecept at the origi, where it s ifiite ad the itegral of the derivative fuctio from to must be 1 because θ ( ) is the ati-derivative ad has a value of 1 at = ad a value of 0 at = (thik Fudametal Theorem of Calculus). dθ Eample: Prove that = δ ( ), where θ ( ) is the step fuctio, see the figure. Aswer: θ ( ) 1, if > 0 = 0, if 0 dθ df f ( ) = f ( ) θ( ) θ( ) df = f ( ) = f ( ) f ( ) f (0) 0 ( ) dθ = f (0) = f ( ) δ( ) = δ ( ) A ifiite peak at the origi whose itegral is 1? Soud familiar? Therefore δ() ca also be defied as the derivative of the step fuctio. Note that oe of the uses of step fuctio is to write r > (the larger of r ad r ') as: r = rθ( r r' ) + > r' θ( r' r). Similarly, r < (the smaller of r ad r ' ) as: r = rθ( r' r) + < r' θ( r r') The ( r> ) = [ rθ( r r' ) + r' θ( r' r)] r r ; = θ( r r' ) + rδ( r r') r' δ( r' r) = θ( r r' ) Where r ' is treated as a costat. Also ( r< ) = [ rθ( r' r) + r' θ( r r')] r r = θ r' r rδ( r r') r' δ( r' r) = θ r' r ( ) ( ) 4

5 Gree fuctio.doc Where r ' is treated as a costat It ca be proved that ( ) g r, r ' = sikr coskr is a solutio of the differetial equatio: < > d d k g r r r r dr r dr = r δ by direct substitutio. (, ') ( ') 3. Defiitio as Fourier trasform. We ve see that takig the Fourier trasform of a fuctio gives you the frequecy compoets of the fuctio. What about takig the Fourier trasform of a pure sie or cosie wave oscillatig at ω 0? There is oly oe frequecy compoet, so the Fourier trasform must be a sigle, very large peak at ω 0 (or possibly two peaks, oe at ω 0 ad oe at ω 0 ). A delta fuctio! 4. Defiitio as desity. What s the desity of a 1 kg poit mass located at the origi? Well, it s a fuctio that must be zero everywhere ecept at the origi ad it must be ifiitely large at the origi because for a mass that truly occupies oly a sigle poit, the mass must have bee ifiitely compressed. How about the itegral? The itegral of the desity must give you the mass, which is 1 kg. A fuctio that is zero everywhere ecept at the origi, ad has a itegral equal to 1? Souds like the delta fuctio agai! More precisely, this would be a three-dimesioal aalog to the regular delta fuctio δ(), because the desity must be itegrated over three dimesios i order to give the mass. This is sometimes writte δ(r) or as δ 3 (r): δ 3 (r) = δ() δ(y) δ(z) Properties 1. Itegral. Oe of the most importat properties of the delta fuctio has already bee metioed: it itegrates to 1.. Siftig property. Whe a delta fuctio δ( 0 ) multiplies aother fuctio f(), the product must be zero everywhere ecept at the locatio of the ifiite peak, 0. At that locatio, the product is ifiite like the delta fuctio, but it might be a larger or smaller ifiity (ow you see why mathematicias do t like physicists), depedig o whether the value of f() at that poit is larger or smaller tha 1. I other words, the area of the product fuctio is ot ecessarily 1 ay more, it is modified by the value of f() at the ifiite peak. This is called the siftig property of the delta fuctio: δ ) f ( ) = f ( ) ( 0 0 Mathematicias may call the delta fuctio a fuctioal, because it is really oly well-defied iside itegrals like this, i terms of what it does to other fuctios. (A fuctioal is somethig that operates o fuctios, rather tha a fuctio which is somethig that operates o variables.) 5

6 Gree fuctio.doc 3. Symmetry. A few other properties ca be readily see from the defiitio of the delta fuctio: a. δ( ) = δ() (Note that δ() behaves as if it were a eve fuctio) b. δ( 0 ) = δ( + 0 ) 4. Liear systems. If a physical system has liear resposes ad if its respose to delta fuctios ( impulses ) is kow, the i theory the output of this system ca be determied for almost ay iput, o matter how comple. This rather amazig property of liear systems is a result of the followig: almost ay arbitrary fuctio ca be decomposed ito (or sampled by ) a liear combiatio of delta fuctios, each weighted appropriately, ad each of which produces its ow impulse respose. Thus, by applicatio of the superpositio priciple, the overall respose to the arbitrary iput ca be foud by addig up all of the idividual impulse resposes. More Properties 1- δ *( ) = δ ( ) it is a real fuctio - δ ( 0) = 1 It is ormalized a+ 3- δ ( a) = 0, a δ ( a) = 1 4- f ( ) δ ( ) = f (0) a 5- f ( ) δ ( a) = f ( a) δ ( a) d f ( ) ( ) f ( ) ( ) f (0) δ = δ = δ( a ) = δ( ) a d 8- = δ ( ) d 1 9- ( δ ( )) = δ ( ) Closure property: Cosider a complete orthoormal set of fuctio ϕ1( ), ϕ( ), the for ay complete fuctio f ( ) = aϕ( ) Where a = f * ( ') ϕ( ') ' 6

7 Gree fuctio.doc For The f f * ( ) = ( ') ϕ( ') ' ϕ( ) = f ( ) = * * ( ') ϕ( ') ϕ( ) ', δ ' ϕ( ') ϕ( ) δ ( ') δ 1 im ϕ ( ) = e ϕ π 1 = π im ( ϕ ϕ ') ( ') e m = 7

8 Gree fuctio.doc - Ihomogeeous Equatios =========================5/11/013=========================== Theorem: the solutio of the ihomogeeous equatio L ˆ( ) y( ) f ( ) f ( ) G(, ') f ( ') ' = ϕ0 + = is give by: where ϕ 0 is the solutio of the equatio: L ˆ( ) ϕ ( ) 0 0 = ad is the solutio of the equatio: L ˆ( ) G(, ') = δ ( ') The defiitio of the above equatio is ot uique. Some books are used δ ( ') ad others are used 4 πδ ( '). Proof: Start with the equatio: Oe fids: f ( ) G(, ') f ( ') ' = ϕ0 + ϕ0 [ ] [ δ ] f Lf ( ) ( ) = L ( ) + LG ( ) (, ') f( ') ' = 0 + (, ') ( ') ' = f ( ) ======================================================= Gree s fuctios, the topic of this hadout, appear whe we cosider the ihomogeeous equatio aalogous to Eq. (1) L ˆ( ) y ( ) = f( ) (7) where L ˆ( ) is a liear, self-adjoit differetial operator, y( ) is the ukow fuctio, ad f() is a kow o-homogeeous term. For a discussio of the cocept of self-adjoit ad o selfadjoit differetial operators please refer, for eample, to the tet by Arfeke. Operatioally, we ca write a solutio to equatio (1) as y = Lˆ f 1 ( ) ( ) where L 1 ( ) is the iverse of the differetial operator L ˆ( ). Sice L ˆ( ) is a differetial operator, it is reasoable to epect its iverse to be a itegral operator. We epect the usual properties of iverses to hold, ˆˆ 1 ˆ 1 ˆ I LL = L L = where I is the idetity operator. More specifically, we defie the iverse operator as 1 L f G f ( ) = (, ') ( ') ' where the kerel G(, ') is the Gree's Fuctio associated with the differetial operator L. Note that G(, ') is a two-poit fuctio which depeds o ad '. To complete the idea of the iverse operator L, we itroduce the Dirac delta fuctio as the idetity operator I. Recall the properties of the Dirac delta fuctio δ ( ) are: 8

9 Gree fuctio.doc f ( ) = δ( ') f ( ') ', δ( ') ' = 1 The Gree's fuctio G(, ') the satisfies L ( G ) (, ') = δ ( ') (8) Importat ote: The defiitio i equatio (8) is ot uique. Some books are used δ ( ') ad others are used 4 πδ ( '). The solutio to equatio (7) ca the be writte directly i terms of the Gree's fuctio as y( ) = G(, ') f ( ') ' (9) To prove that equatio (9) is ideed a solutio to equatio (7), simply substitute as follows: L( ) y( ) = L( ) G(, ') f ( ') ' = L( ) G(, ') f ( ') ' = δ ( ') f ( ') ' = f ( ) Note that we have used the liearity of the differetial ad iverse operators i additio to equatios (4), (5), ad (6) to arrive at the fial aswer. We emphasize that the same Gree s fuctio applies for ay f ( ), ad so it oly has to be calculated oce for a give differetial operator L ad boudary coditios. Commet: The gree s fuctio G(, ') represets the respose of the system to a uit impulse at = '. G(, ') is the field at the observer s poit r caused by a uit source at the source poit r ', the the field at r caused by a source distributio is the itegral of over the whole rage of occupied by the source. 3- A simple eample Eample, Fid the solutio of the differetial equatio: d y( ) 1 + y ( ) = si( ) (A) 4 i the iterval 0 π, with the boudary coditios y(0) = y( π ) = 0. a = 0 b = π Solutio: 1 st method: The geeral solutio of this equatio is: 4 y( ) = Acos( ) + B si( ) si( ) (B) 15 complemetary solutio of the equatio particular solutio of the equatio d y ( ) 1 i + y ( ) = 0 Im( e ) 4 1 D + 4 (10) 9

10 Gree fuctio.doc i Im( e ) si( ) 4 Note: The particular solutio is give as: y p = = = 1 1 si( ) D + ( i) Usig the above boudary coditios, oe fids A = B = 0, ad 4 y( ) = si( ) 15 (C) See the plottig of this solutio H.W. check that (C) sactisfies (A). d method: Closed form epressio for the Gree's fuctio I may useful cases, oe ca obtai a closed form epressio for the Gree's fuctio by startig with the defiig equatio, Eq. (8). We will illustrate this for the eample i the previous sectio for which Eq. (8) is 1 G"(, ') + G(, ') = δ ( ') (11) 4 Remember that ' is fied (ad lies betwee 0 ad π) while is a variable, ad the derivatives are with respect to. We solve this equatio separately i the two regios i- 0 < ', ad ii- ' < π. I each regio separately the equatio is G + (1/4)G = 0, for which the solutios are G(, ') = A( ')cos( ) + B( ')si( ) (1) Where the costats A ad B will deped o '. Sice y (0) = 0, we require G(0, ') = 0 ad so, for the solutio i the regio 0 < ', the cosie is elimiated. Similarly G(π, ) = 0 ad so, for the regio ' < π, the sie is elimiated. Hece the solutio is: GI B( ')si( / ) (0 < ') G(, ') = (13) GII A( ')cos( / ) ( ' < π ) GI GII a = 0 b =π ' 10

11 Gree fuctio.doc Now: How do we determie the two coefficiets A ad B? The aswer will be as follows: I- We ca get oe relatio betwee them by requirig that the solutio is cotiuous at = ', i.e. the limit as ' from below is equal to the limit as ' from above. This gives: B( ')si( '/) = A( ') cos( '/) (14) II- The secod relatio betwee A ad B is obtaied by itegratig Eq. (11) from ' ε to ' + ε, ad takig the limit ε 0, which gives: so Where Hece dg (, ') dg dg lim + 0 = 1 ε 0 ' + ε ' ε ' B( ') GI cos( ) (0 < ') dg (, ') = ' A ( ') GII si( ) ( ' < π ) has a discotiuity of 1 at = ', i.e. ' ' A ' B ' GII GI = 1 si( ) cos( ) = 1 (18) Solvig Eqs. (14) ad (18) gives B( ') = cos( '/) (19) A ( ') = si( '/ ) Substitutig ito Eq. (3) gives ' cos( )si( ) (0 < ') G(, ') = (0) ' si( )cos( ) ( ' < π ) A sketch of the solutio is show i the figure below. The discotiuity i slope at = ' (I took ' = ϕ = 3 π /4) is clearly see. (15) (16) (17) 11

12 Gree fuctio.doc ', ad >, the larger of ad It is istructive to rewrite Eq. (0) i terms of <, the smaller of ad '. Oe has G(, ') = cos( > / )si( < / ) (1) Irrespective of which is larger, which shows that G is symmetric uder iterchage of ad '. We ow apply the closed form epressio for G i Eq. (0) to solve our simple eample, Eq. (A), with the fuctio: y( ) = cos( / ) si( / ) f ( ) d si( / ) cos( / ) f ( ) d 0 with f () = si(), ad usig formulae for sies ad cosies of sums of agles ad itegratig gives: π where we agai used formulae for sums ad differeces of agles. This result is i agreemet with Eq. (C). 4- Summary We have show how to solve liear, ihomogeeous, ordiary differetial equatios by usig Gree s fuctios. These ca be represeted i terms of eigefuctios, see et sectio, ad i may cases ca alteratively be evaluated i closed form, see Sec. 6. The advatage of the Gree s fuctio approach is that the Gree s fuctio oly eeds to be computed oce for a give differetial operator L ad boudary coditios, ad this result ca the be used to solve for ay fuctio f ( ) o the RHS of Eq. (9) by usig Eq. (0). The advatages of Gree s fuctios may ot be readily apparet from the simple eamples preseted here. However, they are used i may advaced applicatios i physics. We ca summarize the properties of the oe-dimesioal Gree s fuctio as follows: 1- G(, ξ ) is symmetric uder iterchage of ad ξ. 1

13 Gree fuctio.doc - Both GI (, ξ ) ad (, ) ξ satisfy the homogeeous equatio LGI (, ξ ) = δ( ξ), a < ξ LGII (, ξ) = δ( ξ), ξ < b 3- GI (, ξ ) satisfies the boudary coditio at = a. Similarly GII (, ξ ) satisfies the boudary coditio at = b. 4- G(, ξ ) is a cotiuous fuctio of, i.e. lim G (, ξ ) = lim G (, ξ ). 5- dg (, ξ ) is a discotiuous ad the discotiuity is give by. dg II dg I = 1 = ξ = ξ 6- Geerates a superpositio priciple for the solutio uder geeral forcig fuctios: y( ) = G(, ') f ( ') ' 3 rd method: Eigefuctios Epasio y ξ I this sectio we will obtai a epressio for the Gree's fuctio i terms of the eigefuctios ( ) of the homogeeous equatio: L( ) y ( ) = λ y ( ). (I) which have the same boudary coditios as y( ). We assume that the solutio y ( ) of the ihomogeeous equatio: L ( ) y( ) λ y( ) = f ( ) (II) To solve (II), we epad y ( ) ad f ( ) i eige fuctios of the operator L i (I), ad oe obtais: y( ) = c y ( ), I ξ f ( ) = d y ( ), d = y f for some choice of the costats c ad d. Substitutig () ito Eq. (I) gives II () c ( λ λ) y ( ) = dy ( ) ( ) To determie the c we multiply Eq. ( ) by oe of the eigefuctios, y m ( ) say, ad itegrate use the orthogoality of the eigefuctios, this gives: d y f c = = ( d = y f ) λ λ λ λ Substitutig for c ito Eq. () gives 13

14 Gree fuctio.doc b f ( ) y f f ( ) y( ) = = y ( ') f ( ') ' λ λ λ λ = b a a G(, ', λ) f ( ') ' where y ( ') y ( ) G(, ', λ) G(, ') = λ λ (3) Notice the symmetric property of the fuctio: G(, ') i the argumet ad '. Furthermore, G(, ') oly depeds o the eigefuctios of the correspodig homogeeous equatio, i.e. o the boudary coditios ad L. It is idepedet of f ( ) ad so ca be computed oce ad for all, ad the applied to ay f ( ) just by doig the itegral i Eq. (9). Eample: A strig of legth is vibratig with frequecy ω. The equatio ad boudary coditios are: du ( ) Determie the gree fuctio. Solutio: + k u( ) = 0, u(0) = u( ) = 0, k = d φ ( ) Start with the eigevalue equatio: ( ), coditios φ(0) = φ( ) = 0, we have: π The eigevalues λ =, = 1,,3, π Normalized eigefuctio: u = si The Gree s fuctio π π ' si si u( ') u( ) G(, ') = = λ λ π k ω c = λφ λ = ad usig the boudary Eample, Fid the solutio of the differetial equatio: d y( ) = f ( ), f ( ) = si( ) (A) i the iterval 0 π, with the boudary coditios y(0) = y( π ) = 0. d φ ( ) Aswer: Start with the eigevalue equatio: = λφ( ), λ = ad usig the boudary coditios φ (0) = φ ( π) = 0, we have: 14

15 Gree fuctio.doc φ ( ) = si ( ), = 1,,3, (B) π ad si( ')si( ) G(, ') = π = 1 λ The: π π si( ')si( ) y( ) = G(, ') f ( ') ' si( ') ' π = π 0 0 = 1 π si( ) π si( ) 1 1 = si( ')si( ') ' = = si( )cos( ) = si( ) π = 1 π 4 0 π δ, This is because of the orthogoality of the si( ) i the iterval from 0 to π oly the = cotributes ad the itegral for this case is π. term i agreemet with Eq. (C). Eample, Fid the solutio of the differetial equatio d y( ) 1 + y ( ) = si( ) 4 Aswer: π π si( ')si( ) 1 y( ) = G(, ') f ( ') ' si( ') ', λ π = π = λ = 1 π si( ) π si( ) 8 4 = si( ')si( ') ' si( )cos( ) si( ) π = = = π 4 4 = 1 0 π δ, Because of the orthogoality of the si( ) i the iterval from 0 to π oly the = term cotributes, ad the itegral for this case is π. H. W. Do eample i Arfke. H. W. Fid the solutio of the differetial equatio: d y( ) 1 + y ( ) = f ( ), f ( ) = 4 15

16 Gree fuctio.doc i the iterval 0 π, with the boudary coditios y(0) = y( π ) = 0. Aswer: a- The differetial equatio recipe gives: y( ) = π si( /) b- Direct ad Gree fuctio gives: y( ) = π si( /) c- Eigefuctio gives: Note: si( ) = 1 ) y( ) = ( 1) ( 4, 16

17 Gree fuctio.doc Eample: For the D.E. y ''( ) + y( ) = f ( ) such that y(0) = y( π / ) = 0, calculate G(, ') ad solve for y( ). Plot both fuctios. Aswer: For the fuctios: A ( ')si( ) (0 < ') G(, ') = B( ')cos( ) ( ' < π / ) dg (, ') A ( ')cos( ) (0 < ') = B( ')si( ) ( ' < π / ) cos( ')si( ) (0 < ') G(, ') = si( ')cos( ) ( ' < π / ) 0 π / y( ) = cos( ) si( ') f ( ') ' si( ) cos( ') f ( ') ' Note that: 1- Usig the Gree s fuctio may seem to be a complicated way to proceed, especially for certai choices of f ( ). However, you should realize that the elemetary derivatio of the solutio may ot be so simple i other cases, ad you should ote that the Gree s fuctio applies to all possible choices of the fuctio o the RHS, f ( ). Furthermore, we will see i the et sectio that oe ca ofte get a closed form epressio for G, rather tha a ifiite series. It is the much easier to fid a closed form epressio for the solutio. - The orthogoal fuctio epasio method ca easily be eteded to two ad three dimesios. For eample if { u ( )} ;{ v ( ) } ad { w ( ) } deote the complete fuctios i the, y, ad z directios respectively, the the three dimesioal Gree s fuctio ca be writte as: π π ' mπy mπy ' πz πz ' si si si si si si a a a a a a G(, ') = 4π λ Where ml,, m,, π mπ π λm,, = α+ βm+ γ = + + { } a b c See Eq i Jackso for a eample. I Jackso, he used ad Eample: A hollow, grouded, coductig cube with sides of legth a has a charge desity 17

18 Gree fuctio.doc π πy πz ρ = ρo si si si a a a placed withi its iterior ( 0 < < a, 0 < y < a, 0 < z < a ). a. Usig the method of eigefuctio epasio, show that the Gree s fuctio π π ' mπy mπy ' πz πz ' si si si si si si 3 a a a a a a G(, ') = πa ml,, + m + b. Fid the electrostatic potetial V (, y, z ) iside the cube. Solutio: a- The eigefuctio is: Ψ ( ) = si ( α ) si ( β y ) si ( γ z ) Ad the eigevalues are: λm,, { α βm γ } Ψ G(, ') = 4π 8 m,, m abc π mπ π = + + = + + a b c ( ) Ψ ( ') m,, m,, λ ml,, m,, π Here, a = b = c a, the λm,, = { α+ βm+ γ } = { + m + } a, ad π π ' mπy mπy ' πz πz ' si si si si si si a 8 a a a a a a G(, ') = 4π π 3 a ml,, + m + ( ) b- Iside the cube: 1 V (, y, z) = ρ( ') G(, ') dτ ' 4πε o V aaa 3 ρo π ' πy ' πz ' V (, y, z) = ' dy ' dz 'si si si 4π aε o a a a 000 π π ' mπy mπy ' πz πz ' si si si si si si a a a a a a + m + ml,, ( ) From orthogoality, the oly o-vaishig itegrals are for = m = = 1 aaa 3ρo 1 π mπy πz π ' πy ' πz ' V (, y, z ) = si si si ' dy ' dz 'si si si 4π aεo 3 a a a 000 a a a 1 ρ π π π = 3 a o m y z si si si π εo a a a ( a /) 3 18