Solution of inhomogeneous ordinary differential equations using Green's functions G( x, x ')

Transcription

1 Solutio of ihomogeeous ordiary differetial equatios usig Gree's fuctios G(, ') 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 (793 84), who ivestigated electricity ad magetism i a thoroughly mathematical fashio. I 88 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,, 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. - 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: Lˆ( ) y ( ) = Ey ( ) () 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 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. The geeral Sturm-Liouville problem, see Arfeke, has a weight fuctio w() multiplyig the eigevalue o the RHS of Eq. () ad the same weight fuctio multiplies the itegrad show i the LHS of the orthogoality ad

2 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() = ad real eigefuctios. Eample, Fid the solutio of the differetial equatio: d y( ) + y ( ) = λy ( ) 4 d y( ) = 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) This is just the simple harmoic oscillator equatio, ad so the solutios are cos( ) ad 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 =,,3, (5) ad the eigevalues i Eq. (3) are: λ = 4 (6) sice the equatio satisfied by y ( ) is y "( ) + y( ) = 0. (3)

3 Defiitios Dirac delta fuctio. 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 = /b (so total area = ) 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 throughout the limit process. δ ( ) = 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 ) = 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 θ ( ). Defiitio as derivative of step fuctio. The step fuctio, also called the Heaviside step fuctio is usually defied like this:, if > 0 θ ( ) = 0, if 0 It s a fuctio whose oly feature is a step up from 0 to, 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 because θ ( ) is the ati-derivative ad has a value of 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 ( ), 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? 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 Where r ' is treated as a costat It ca be proved that ( ) g r, r ' = si kr cos kr is a solutio of the differetial equatio: < > d d + + k g rr 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 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 kg. A fuctio that is zero everywhere ecept at the origi, ad has a itegral equal to? 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. Itegral. Oe of the most importat properties of the delta fuctio has already bee metioed: it itegrates to.. 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. I other words, the area of the product fuctio is ot ecessarily 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 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 - δ *( ) = δ( ) it is a real fuctio - δ ( 0) = It is ormalized a+ 3- δ ( a) = 0, a δ ( a) = a 4- f ( ) δ ( ) = f (0 ) 5- f ( ) δ( a) = f ( a) δ( a) d f ( ) ( ) f ( ) ( ) f (0 δ = δ = 0 0 δ( a ) = δ( ) a d 8- = δ ( ) d 9- ( δ ( )) = δ ( ) Closure property: Cosider a complete orthoormal set of fuctio ϕ( ), ϕ( ), the for ay complete fuctio f ( ) = aϕ( ) Where a = f * ( ') ϕ( ') ' 6

7 For The f f ϕ d ϕ * ( ) = ( ') ( ') ' ( ) = f ϕ ϕ d δ( ) = ϕ ϕ * * ( ') ( ') ( ) ', ' ( ') ( ) δ ( ') δ im ϕ ( ) = e ϕ π = π im ( ϕ ϕ') ( ') e m = 7

8 - Ihomogeeous Equatios =========================8//03=========================== Theorem: the solutio of the ihomogeeous equatio L ˆ( ) y ( ) = f( ) is give by: f ( ) = ϕ0 + G(, ') f ( ') d ' 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: = ϕ0 + f ( ) G(, ') f ( ') d ' 0 [ δ ] [ ] Lf ( ) ( ) = L ( ) ϕ + LG ( ) (, ') f( ') ' = 0 + (, ') f ( ') ' = f ( ) ======================================================= Gree s fuctios, the topic of this hadout, appear whe we cosider the ihomogeeous equatio aalogous to Eq. () 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 () as y = Lˆ f ( ) ( ) where L ( ) 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 ˆ = L L ˆ = I LL where I is the idetity operator. More specifically, we defie the iverse operator as L f ( ) G(, ') f ( ') d ' = 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 f ( ) = δ( ') f ( ') ', δ( ') ' = The Gree's fuctio G(, ') the satisfies LG ( ) (, ') = 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 ( ') d ' (9) To prove that equatio (9) is ideed a solutio to equatio (7), simply substitute as follows: L ( ) y ( ) = L ( ) G (, ') f( ') ' = LG ( ) (, ') f( ') ' = δ ( ') f ( ') d ' = 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( ) + y ( ) = si( ) (A) 4 i the iterval 0 π, with the boudary coditios y(0) = y( π ) = 0. a = 0 b = π Solutio: st method: The geeral solutio of this equatio is: 4 y( ) = Acos( ) + B si( ) si( ) 5 complemetary solutio of the equatio particular solutio of the equatio d y ( ) i + y ( ) = 0 Im( e ) 4 D + 4 (0) (B) 9

10 Note: The particular solutio is give as: y p i Im( e ) si( ) 4 si( ) D + ( i) = = = Usig the above boudary coditios, oe fids A = B = 0, ad 4 y( ) = si( ) (C) 5 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 G"(, ') + G(, ') = δ ( ') () 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 + (/4)G = 0, for which the solutios are G (, ') = A ( ')cos( ) + B ( ')si( ) () 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(, ') = (3) GII A ( ')cos( / ) ( ' < π ) GI GII a = 0 b = π ' 0

11 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( '/ ) (4) II- The secod relatio betwee A ad B is obtaied by itegratig Eq. () from ' ε to ' + ε, ad takig the limit ε 0, which gives: so Where Hece dg (, ') dg dg lim + 0 = ε 0 ' + ε ' ε ' B( ') GI cos( ) (0 < ') dg (, ') = ' A ( ') GII si( ) ( ' < π ) has a discotiuity of at = ', i.e. ' ' A ' B ' GII GI = si( ) cos( ) = (8) Solvig Eqs. (4) ad (8) gives B( ') = cos( '/ ) (9) 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. (5) (6) (7)

12 It is istructive to rewrite Eq. (0) i terms of <, the smaller of ad ', ad >, the larger of ad '. Oe has G(, ') = cos( > / )si( < / ) () 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:

13 - G(, ξ ) is symmetric uder iterchage of ad ξ. - Both GI (, ξ ) ad G (, ) ξ satisfy the homogeeous equatio LG (, ξ) = δ( ξ), a < ξ I 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- ξ I ξ dg (, ξ ) is a discotiuous ad the discotiuity is give by. dg II dg I = = ξ = ξ 6- Geerates a superpositio priciple for the solutio uder geeral forcig fuctios: y( ) = G(, ') f ( ') d ' II 3

14 Gree's Fuctios for self-adjoit operator, Sturm-Liouville equatio Arfke, 0.5, page 663 H.W. Problems 0.5.,,3,0, For the self-adjoit operator, Sturm-Liouville equatio is defied as: d d L= p ( ) + q ( ), () Defie the Gree s equatio by the equatio: d d p ( ) + q ( ) G (, ξ) = δ( ξ) () where p ( ), q ( ) ad G(, ξ ) are cotiuous fuctios i the iterval [a,b]. Itegratig both sides of equatio () fromξ ε to ξ + ε, oe obtais: ξ+ ε ξ+ ε d dg (, ξ ) p( ) + q( ) G (, ξ ) = (3) ξ ε ξ ε ξ+ ε If we ow let ε 0, the itegral q ( ) G (, ξ ) vaishes because q ( ) ad G(, ξ ) are ξ ε cotiuous fuctios of. The equatio (3) reduces to: or dg (, ξ ) p ( ) = 4 ξ+ ε ξ ε dg (, ξ) dg (, ξ) = (5) p( ξ ) ξ+ ε ξ ε Where it is implied thatε 0. Sice there is a discotiuity i the derivative of the Gree s fuctio at = ξ, it is coveiet to cosider the two itervals a < ξ adξ < b separately ad write the Gree s fuctio i the followig form: cu ( ), a < ξ G(, ξ ) = (6) cv ( ), ξ < b We ca summarize the followig properties of the oe-dimesioal Gree s fuctio. - Both G (, ξ ) ad G (, ) ξ satisfy the homogeeous equatio LG (, ξ), a < ξ LG(, ξ), ξ < b - G (, ) ξ satisfies the boudary coditio at = a. Similarly G (, ) ξ satisfies the boudary coditio at = b. 3- G(, ξ ) is a cotiuous fuctio of, i.e. lim G(, ξ) = lim G(, ξ). ξ ξ dg (, ξ ) 4- is a discotiuous ad the discotiuity is give by. dg dg = (7) p( ) = ξ = ξ ξ (4)

15 These properties ca be used for the eplicit costructio of the Gree s fuctio. Let u( ) be a solutio of Lu( ) = 0 satisfyig the boudary coditio at = a. Similarly, Let v( ) be a solutio of Lv ( ) = 0 satisfyig the boudary coditio at = b. Followig the defiitio of the Gree s fuctio it follows that: cu ( ), a < ξ G(, ξ ) = (8) cv ( ), ξ < b Sice G(, ξ ) is a cotiuous fuctio of = ξ, we have cv ( ξ) cu ( ξ) = 0 (9) ad equatio (7) implies cv '( ξ) cu '( ξ) = (0) p( ξ ) Equatios (9) ad (0) ca be solved to obtai c ad c. A o-trivial solutio eists oly if the determiat v ( ξ) u( ξ) 0 () v '( ξ) u' ( ξ) The left had side of () is the Wroskai of v ( ξ ) ad u ( ξ ), W( ξ ) = uv' u' v. W ( ξ ) will be o-zero if v ( ξ ) ad u ( ξ ) are liearly idepedet. If this coditio is satisfied, from (9) ad (0), we get: v ( ) c = ξ p( ξ) W ( ξ) () Similarly, u ( ) c = ξ p( ξ) W ( ξ) (3) Usig: A = W ( ξ) p( ξ), we have v ( ξ ) c =, (4) A u ( ξ ) c = (5) A Ad (8) will be v( ξ ) u( ), a < ξ G(, ξ ) = A u( ξ ) v( ) (6), ξ < b A

16 H.W. : From equatio (), d d Ly = p( ) + q( ) y = 0 d d Lu = p( ) + q( ) u = 0 p'( ) q ( ) pu ( ) " + up ' '( ) + qu ( ) = 0 u" + u' + u= 0, p ( ) p ( ) d d Lv = p( ) + q( ) v = 0 p'( ) q ( ) pv ( ) " + v' p'( ) + qv ( ) = 0 v" + v' + v= 0 p ( ) p ( ), check the followig: - W ( ) = uv ' u ' v dw ( ) p '( ) = uv '' u '' v = W p( ) - W ( ) = u v' u' v uv ' u ' v d v W ( ) = u u = u u H.W. - For the D.E. y ''( ) y( ) = f ( ) such that y(0) = y( ) = 0, calculate G(, '). - Costruct the Gree fuctio for the differetial equatio : y ( ) + 3 y ( ) = F( ), Subject to the boudary coditios: y(0) = y() = For the boudary value problem y "( ) + ω y( ) = f ( ), where f ( ) is a kow fuctio ad y satisfies the boudary coditios y(0) = y( ) = 0, calculate the Gree s fuctio. 4- For the D.E. y ''( ) y( ) = such that y(0) = y( ) = 0. Calculate G(, '). 6

17 Gree s fuctio step by step with eample Eample: For the o-homogeeous differetial equatio. d a y( ) = e L( ) such that y(0) = y '() = 0. a- Calculate G(, '). b- Solve for y( ) usig part b. (A) directly. (A) c- Check your aswer by solvig Aswer: To fid the solutio, usig the Gree s fuctio, of the o-homogeeous differetial equatio: L ( ) y ( ) = f( ), () with the give boudary coditios, we have to do the followig: - It is the most importat step to have the geeral solutio of the homogeeous equatio: L ( ) y ( ) = 0. For eample: the solutio of homogeeous equatio of () d y( ) = 0 y ( ) = c + d () - Use the geeral solutio i () as a geeral solutio of LGξ ( ) (, ) = 0, with the costats defied i the two regios. For eample: GI (, ξ) = c + d, a < ξ G(, ξ ) = (3) GII (, ξ) = c + d, ξ < b 3- Defie the o-homogeeous equatio LG ( ) (, ξ) = δ( ξ) 4- Use the i- boudary coditios give i the equatio, y(0) = 0 ad y '() = 0, to have d = 0, ad c = 0, the c (0 < ξ ) G(, ξ ) = d ( ξ < ) (4) ii- Use the cotiuity of the Gree s fuctios GI (, ξ) = GII (, ξ) to have cξ = d, ad iiidg dg Use the discotiuity = to have c 0= c =. From step (ii) oe has d = ξ = ξ = ξ 7

18 iv- Fially, we have the target Gree s fuctio i the for: GI (, ξ) = (0 < ξ) G(, ξ ) = GII (, ξ) = ξ ( ξ < ) Fially, calculate the itegral b b y ( ) = G (, t ) f () t dt = G (, t ) f () t dt + G (, t ) f () t dt to have, usig f ( ) = e a, the I a a a a at at e e a a a 0 0 y ( ) = G (, t ) f () t dt = te dt + e dt = + II (5) c- With simple itegratio, oe fids: a a a e e y ''( ) = e y '( ) = + b y ( ) = + b + c, a a a e Use the boudary coditios y(0) = y '() = 0, oe fids c =, b =, The a a a a e e y( ) = + a a a 8

19 Eample: For the D.E. Arfeke method y ''( ) + e a = 0 such that y(0) = y '() = 0. a- Solve for y( ). b- Calculate G(, '). c- Solve for y( ) usig part b. a a a e e a- y ''( ) = e y '( ) = + b y ( ) = + b + c, a a a e Use the boudary coditios y(0) = y '() = 0, oe fids c =, b =, The a a a a e e y( ) = + a a a b- For the equatio y ''( ) = 0 y ( ) = c + d the G ''( ) = 0 G ( ) = c + d c + d (0 < ') G(, ') = c + d ( ' < ) With the B.C. y(0 = )0 d = 0, y '() = 0 c = 0, the, c = cu ( ) u( ) = (0 < ') G(, ') =, d = dv( ) v( ) = ( ' < ) The Wroskai at ay coveiet poit gives: W ( ) = uv ' vu ' = 0 () = v( ') u( ') p ( ) = A= W( ') p ( ') =, c = =, d = = ' ad the Gree s A A fuctio i the form: (0 < ') G(, ') = ' ( ' < ) Usig f ( ) = e a, the a a at at e e y ( ) = G (, ') f ( ') ' = te dt + e dt = + a a a 0 0 9

20 Gree s fuctio for Bessel fuctio (Afke 0.5.8, 0.5.3) For readig Fid the Gree fuctio for the equatio: d y dy + + ( k ) y = f ( ) with the boudary coditios y(0) = fiite, y( a) = 0. Buttig the above equatio i Sturm-Liouville i the form: d dy f ( ) + ( k ) y = where P ( ) =.The Gree fuctio G (, ξ ) i the form: d dg + ( k ) G = δ( ξ) With the geeral solutio c J( k ) + c N ( k ). Hece: dj ( k ) + dn ( k ) (0 < ξ ) G(, ') = cj ( k ) + cn ( k ) ( ξ < a) With the boudary coditios y (0) = fiite d = 0 ad u( ) = J ( k ). With y( a ) = 0, we J ( ka) have cj( ka) + cn ( ka) = 0 c = c N ( ka) ad J( k ) N ( ka) J( ka) N ( k ) v( ) =. N ( ka) With kowig the u( ) ad v( ), we have to evaluate the Wroskia W( ξ ) = uv' u' v. To do so, we ca choose ay coveiet poit such as 0. k (! ) lim J( k ), lim N ( k ) 0! 0 π k H.W. Prove that J ( ) ' ' ( ) ( [ ( ), ( )]) ( ) ka J ka W ξ = u v u v = W J k N k = N ( ka) N ( ka) π J ( ka) p( ξ ) ξ A W ( ξ ) p( ξ = = ) = N ( ka) πξ ξ Hece π J( kξ) N ( ka) J( ka) N ( kξ) J ( k ) (0 < ξ ) N ( ka) G(, ') = π J( k ) N ( ka) J( ka) N ( k ) J ( k ξ) ( ξ < a ) N ( ka) 0

21 Gree s Fuctio i Free Space Eample: Solve G ( rr, ') = δ ( r r ') usig the Fourier s trasformatio. Aswer: Start with the Fourier trasformatio i oe dimesio: ik ik ( ) = ( ) ( ) ( ) ( ) π = π G e G k d k G ik e G k d k I three dimesios: i kr 3 3 i kr G() r = e G( k) d k G() r = ( ik) e G( k) d k ( π) ( π) For a cotiuous fuctio i three dimesios, oe fids: i δ ( r r' ) = e 3 π ( ) k. ' 3 ( r r ) Thus G ( rr, ') = δ ( r r ') will be: i kr i k. ( r r' ) 3 G ( r) + δ ( r r') = ( ik ) e G ( ) d + e d k = ( π) k k ( π) This implies: i kr' i kr 3 3 {( k ) G( ) + e } e d k = 0 ( π ) k Cosequetly, i i k r' e k G( k) + e = 0 G( k ) = k Thus: k r' ( π) ( π) d k i k r' i kr. i kr. e G ( r r' ) = e G( ) d = e d 3 k k 3 k k i kr e = d *, 3 ( π ) k k H.W. Do the itegratio with the otatios: d k= k dkd Ω= k dk siθdθdφ ad k R= kr cosθ, oe fids: Aswer: si( kr) G ( r r' ) = dk = π R = k 4πR 4π r - r' 0 π / br ± i q r e 4π [*Help: prove the stadard itegral I = dτ = ] r b + q R = r - r'

22 ± ikr e H.W. Prove that the Gree s fuctio G () r = is the solutio of the scalar wave equatio: r Proof: We have two cases: Case I: 0 ( ) r, we have to prove that ( k ) ± k G() r = 4 πδ () r () ikr e + = 0 r ikr ikr e d d e k ikr Start with = r ( ) e, = r r dr dr r r Case II: 0 r =, we have to prove that ( ) ikr e ( + k ) = 0, r k G() r d r 4 π δ() r d r 4π = = I this case let us costruct the solutio of a scalar wave equatio i ay volume V of free space havig a arbitrary small radius ξ ad havig a arbitrary source ρ() r, the a) b) sphere of radius ξ G() r d r G() r d r G() r d s ( ) 3 3 = = usig the Gree's idetity ikr d e ik ξ ik ξ = rˆ rˆr d Ω= 4π ik ξe e dr r ξ ikr ikr 3 e ξ e ikr k G () r d r = k r drd Ω= 4π e dr r ik ik 0 0 ik ξ ik ξ = 4π ik ξe + e From the fial results of equatios A ad B, oe fids ( ) 3 + k G() r d r = 4π (A) (B)

23 Applicatios of Gree s fuctio The Bor Approimatio (9.7., ) Eercise: 9.7.8: Itegral scatterig equatio for statioary states The time idepedet Schrodiger equatio i the form: k + V () r Ψ () r = EΨ (), r E = > 0 µ µ could be chaged to the o-homogeeous equatio: µ + k Ψ ( r) = U( r) Ψ ( r), U( r) = V ( r) We claim that the solutios may be writte i the form: 3 ' Ψ ( r) = ϕ i ( r) + G( r, r ') U( r') Ψ( r' ) d r, where ϕ ( r ) is a solutio of the homogeeous equatio: i Eercise: 9.7.6: The solutios of ( ) + k ϕ i ( r) = 0,, ad G(r) is a solutio of 3 + k G( r, r ') = δ (r - r '),. H. W. Proof the claim: ± ik r r ' ± e + k G( rr, ') = δ ( rr, ') are G ( rr, ') =. 4 π r r' These are two liearly idepedet solutios of a secod order differetial equatio. Oe is ikr represetig the outgoig wave, e + ikr, ad the other the icomig wave, e. We will be iterest i outgoig wave. The Bor approimatio 3 ' Ψ ( r ) = ϕi( r) + G( r, r ') U( r') d r ϕi( r') + G(, ) U( r") i( r'') r' r" ϕ. 3 with d r = r drd Ω= r dr siθdθdφ. This procedure ca be repeated ad yields the Bor epasio. The Bor approimatio is the first term i the Bor epasio. ± ik r r ' 3 ' ikr e 3 ' Ψ( r) ϕi( r) + G( rr, ') U( r') ϕi( r') d r = e U( r') Ψ( ') d r, 4 r π r r' f θφ,!. This yields a itegral epressio for the scatterig amplitude ( ) Eercise: Simplify the first Bor approimatio i the case that Aswer: I the followig figure, oe fids: r >> r'. / r r ' r ' r r = r + r r r = r + r r (a) r r ' r r r r ' r rˆ r ', r r r ' ' ' 3

24 / r r ' r ' r r ' = + (b) + r r ' r r r r r r r From the above two equatios we have: ik r r ' ik ( r rˆ r ') ikr ik ( rˆ r ') ikr e e e e e ikf r' = = e r r ' r r r The ikr i kf r' 3 e Ψ( r) ϕi( r) + e U( r') i( r') d r' 4 ϕ π r The term i curly bracket is called the Scatterig amplitude f B ( θφ, ), ' 3 (, ) ik f r f B θφ = e U( r') i( r') d r' = < f U i > 4 ϕ ϕ ϕ π 4π ik i r ' Usig ϕ i ( r') = e, the iqr ' 3 f B( θφ, ) = e U( r') d r', = i f 4π q k k q mometum trasfer. I case of elastic collisio: ki = kf ( ) ( ) q = ki kf = ki + k f kk i fcosθ = k cosθ = 4k si ( θ / ) H.W. For cetral potetial, use the relatio q r ' =qr'cos θ ' to derive the relatio: µ f B ( θ ) = V ( r ')si qr ( ') r ' dr ' q 0 Aswer: µ iqr ' 3 f B ( θ ) = e V ( r') d r' π π µ iqr 'cos θ ' µ = = ϕ θ π V ( r ') r ' dr ' d ' e d co s' V ( r ')si qr ( ') r ' dr ' q π si( qr ') qr ' br ± i q r e 4π H.W. Do eercise for Yukawa potetial. You may eed I = dr =. r b + q 4

25 I br ± i q r e 4π = dr = r b + q I 8πb I = e dr = = b ( b + q ) br ± i q r ± i q r e 4π I4 = dr = e r - r' q b( r + r ) b( r + r ) 5 5 r r b b( r+ r) e 5π I = drdr =, r = r -r 6 5 r 8b 7 t ; ± iq r' e e π I = dr dr = dr dr = I = i ωt e dt + τ π = e τ + ω br + i ωr 8 = = I e dr b ωτ ; 5