Lecture 4 Coformal Mappig ad Gree s Theorem Today s topics. Solvig electrostatic problems cotiued. Why separatio of variables does t always work 3. Coformal mappig 4. Gree s theorem The failure of separatio of variables. Let s try to solve the followig problem by separatio of variables φ φ( S ) φ( S ). Use separatio of variables i cylidrical coordiates φ φ r + r r r r r 3. We assume that φ( r, θ) ( a r + b r ) cos θ
4. To satisfy the boudary coditio o S we require φ(, θ ). Therefore a b ad φ reduces to φ( r, θ) b ( r r ) cos θ [ ] 5. To satisfy the boudary coditio o S we require φ r ( θ), θ 6. We eed to calculate r ( θ ). This is doe by otig that the equatio describig S is give by ( ) x δ + y a 7. Trasform to cylidrical coordiates: x r cos θ y r si θ δr cosθ+ δ a 8. The equatio for the surface becomes. 9. Solve for r r r r ( θ) δ θ+ ( a δ θ) / r cos si. Retur to the boudary coditio. We determie the coefficiets by Fourier [ ] aalysis. Multiply φ r ( θ), θ by ( /π) cosmθdθ. The boudary coditio becomes b b π π ( r r ) cosθcosmθdθ cosmθd π π θ. The right had side is easily evaluated RHS δ m. The left had side becomes LHS M b m π Mm [ r ( θ) r ( θ) ] cosθcos θ π m dθ
3. The problem is thus reduced to a simple liear algebra problem for the coefficiets b M b v [,,,... ] b M v 4. We simply have to ivert the matrix M to fid the coefficiets b. This is i geeral a very simple umerical procedure. 5. Oe might therefore thik that separatio of variables should work i a wide variety of cases, the oly added difficulty beig the eed to umerically ivert a matrix. Not a very big deal. 6. This is a icorrect coclusio. What happes if you try this procedure? For small values of δ the procedure works well ad a correct set of b are calculated. 7. However, above a certai critical value of δ your PC will tell you that the iverse matrix does ot exist. You caot determie the values of b. No amout of facy umerical tricks will help. The iverse truly does ot exist. 8. What has goe wrog? We have tried to write the solutio usig the wrog class of expasio fuctios. Eve though the expasio fuctios satisfy φ their radius of covergece does ot iclude the etire domai of iterest. 9. Below is a simple ituitive example that shows the problem. For the problem uder cosideratio assume a large value of δ so that origi. S does ot eclose the. To satisfy the boudary coditio o we require both the r ad solutios. However, the r S r solutios diverge at the origi which is cotaied i the regio where we eed a solutio.. Although the potetial at r is fiite ad smooth we are attemptig to describe this behavior by summig over a ifiite set of diverget fuctios. This will ot work ad is the source of the problem. 3
. OK ca we fid a better set of expasio fuctios? I geeral the aswer is NO!!! Oly for special simple geometries with a high degree of symmetry do expasio techiques based o separatio of variables work. 3. What do we do? A elegat procedure that works i certai cases makes use of coformal mappig techiques. A very geeral procedure makes use of Gree s theorem. The key poit of the Gree s fuctio procedure is that the method guaratees the existece of good surface expasio fuctios, thus always leadig to the desired solutio. Coformal mappig. The coformal mappig procedure ca be used to covert a complicated geometry ito a much simpler geometry.. The mappig fuctio is defied as w f ( z) where w u + iv ad z x + iy are complex variables. 3. The procedure works extremely well i -D geometries. The reaso, as show shortly, is that the -D Laplace s equatio coordiates to φ + φ. uu vv φ xx + φ yy trasforms i the ew 4. I the ew coordiates φ also satisfies Laplace s equatio. The beauty is that i the coordiate system the geometry is much simpler ad it is thus much easier to fid a solutio. 5. What are the mai limitatios o the coformal mappig techique? 6. First, it oly works well i a -D geometry. Furthermore the geometry must be such that Laplace s equatio ca be writte as r, θ cylidrical system φrr + φr / r + φ θθ / r ). φ xx + φ yy (or the equivalet 7. It does ot, for istace, work for the -D rz, system where Laplace s equatio has the form φ + φ / r + φ. This equatio caot be trasformed ito the form φ xx + φ yy rr r zz ad the procedure is ot very useful. 8. The secod problem is that eve if the geometry of iterest satisfies we must still fid the mappig fuctio. For certai relatively simple cases the mappig fuctio ca be foud aalytically. φ xx + φ 9. I more complicated cases the mappig fuctio ca be determied umerically. I fact there are stadard umerical packages which carry out this task. yy 4
. Cosequetly fidig the mappig fuctio ca be slightly icoveiet but is possible i a wide variety of cases as log as Laplace s equatio i the relevat geometry ca be writte as φ + φ. xx yy Review of complex variables. The key poit that we eed to prove is that the Laplacia i xy, coordiates trasforms ito the Laplacia i uv, coordiates for a coformal trasformatio.. To do this we eed the Cauchy-Riema equatios which are obtaied as follows. 3. Cosider the coformal mappig w f ( z) where w u + iv ad z x + iy. 4. The derivatio is as follow w u + iv f z ( ) u v df z df + i x x dz x dz u v df z df + i i y y dz y dz 5. Now equate the real ad imagiary parts of the two expressios for yieldig the Cauchy-Riema equatios df / dz u x u y v y v x 6. Straightforward differetiatio of these relatios shows that both u ad v satisfy Laplace s equatio. u u v v x y x y + + 7. The ext step is to trasform Laplace s equatio i xy, coordiates ito uv, coordiates. 5
8. For a geeral trasformatio (ot ecessarily a coformal trasformatio) it follows that if u u( x, y) ad v v( x, y), applicatio of the chai rule for derivatives leads to φ φ u + φ u v + φ v + φu + φv xx uu x uv x x vv x u xx v φ φ u + φ u v + φ v + φu + φv yy uu y uv y y vv y u yy v yy xx 9. We ow add the equatios together, assume the trasformatio correspods to a coformal mappig, ad make use of the Cauchy-Riema equatios. We fid df φxx + φyy ( φuu + φvv ) dz df dz x x x u + iv u + v x. Therefore if φ satisfies Laplace s equatio i xy, coordiates it also does so i uv, coordiates. φ u φ v +. This is the beauty of the coformal mappig. The mappig fuctio. Cosider ext the shifted circle problem previously discussed.. What we wat to fid is a mappig fuctio that has two properties: () it maps the outer uit circle i the xy, plae ito the uit circle i the uv, plae, ad () it maps the ier shifted circle i the xy, plae ito a cetered circle i the uv, plae. 6
3. If we ca fid such a mappig the our task will be to solve Laplace s equatio i a coordiate system where the boudaries are two cocetric circles, usually a very simple task. 4. The correct mappig fuctio is give by w z α αz where α is a free parameter to be determied. 5. Let s first check the mappig of the uit circle. I the xy, plae the uit circle i is give by x cos θ, y si θ which is equivalet to z e θ. I the uv, plae we let u Rcos χ, v Rsi χ which is equivalet to w Re iχ. 6. We eed to verify that the mappig leads to R, a uit circle i the uv, plae. Straightforward substitutio shows that iχ cos θ α+ i si α Re αcosα iαsiα 7. Multiply each side of the equatio by its complex cojugate. R αcosθ+ α αcosθ+ α 8. The uit circle maps ito the uit circle for ay value of α. 9. Cosider ow the ier shifted circle x δ + acos θ, y asi θ. This maps ito 7
R a + ( δ α) + a( δ α) cosθ αa + ( αδ ) αa( αδ ) cosθ a + ( δ α) + A cosθ αa + ( αδ ) + C cosθ where a ( δ α) A a + ( δ α) αa ( αδ ) C αa + ( αδ ). I order for the surface to be a cocetric circle i the uv, plae we require that the mappig lead to R R cost There caot be ay θ depedece i R.. To satisfy this costrait we must choose α such that A C. The the θ depedece will cacel out.. After a slightly tedious calculatio we obtai a quadratic equatio for α. α + α+ δ δ a. The root that correspods to R < is give by δ + a δ + a α 4 δ δ / 3. Also, keep i mid that for topologically cosistet solutios to exist δ, a must lie i the rages < δ < ad < a < δ. 8
4. For a give δ, a we have ow determied α ad the mappig is completely defied. The solutio. The solutio to the problem i the uv, plae is easily foud. We let u Rcos χ, v Rsi χ. Laplace s equatio ad the boudary coditios become φ φ R φ() φ ( R ) R R + R R χ. Because of the χ symmetry i the boudary coditios it follows that the solutio is of the form φ( R, χ) φ( R). 3. The solutio is give by l R φ ( R) l R 4. We ca trasform back ito the xy, plae by otig that R R ( ) x α + y ( αx ) + α y ( δ α) + a ( αδ ) + α a 5. To visualize the potetial surfaces ote that a φ cost. surface correspods to a R cost. surface. The relatio givig R R( x, y) ca be easily iverted yieldig 4 ( R ) ( R )( α R ) y α R ( α R ) α x + 6. The potetial surfaces i the xy, plae are a sequece of shifted circles 7. This completes the solutio. 9
Gree s theorem. What is the purpose of Gree s theorem?. I a ifiite space problem Gree s theorem coverts a PDE ( φ ρ/ ε ) ito a simple itegral evaluatio. This is good but we already kow how to do this. 3. For boudary value problems Gree s theorem coverts a PDE ( φ ρ/ ε ) ito a simple itegral evaluatio. This is good but i geeral it is difficult to fid the Gree s fuctio with fiite boudaries. 4. However, if oe kows the Gree s fuctio for a fixed set of boudaries, it is easy to calculate the solutio for a wide rage of boudary coditios. 5. For boudary value problems Gree s theorem usig the simple free space Gree s fuctio coverts the PDE ( φ ρ/ ε ) ito a itegral equatio. This is good because we kow the Gree s fuctio, but upleasat because we have to solve a itegral equatio. 6. For simple geometries the itegral equatio is easy to solve. This is good. 7. I geeral geometries, solvig the itegral equatio is ofte the best approach. This may be hard to believe but it is true. Derivatio of Gree s theorem. Cosider two fuctios φ ad G.. Note the followig idetity ( φ G G φ) φ G + φ G G φ G φ φ G G φ 3. Now itegrate over a closed volume usig the divergece theorem (We could easily iclude multiple boudaries but restrict the aalysis to oe boudary for simplicity) A dr A ds S
4. Apply this to our idetity ( φ φ ) r ( φ φ ) G G d G G ds S 5. Now assume that G is the Gree s fuctio satisfyig ( r r) ( ) ( ) ( z z) G δ δ x x δ y y δ 6. For the momet let s ot worry about the boudary coditios o G. Also ote that r is the observatio poit ad r is the itegratio poit. 7. Note that the delta fuctio is defied such that For a iterior poit δ ( ) d r r r For a exterior poit / For a surface poit 8. Now assume that φ satisfies Poisso s equatio: 9. Evaluate the terms i Grees theorem φ ρ/ ε
S ( r ) G( r r) dr ( r ) ( r r) dr () r φ φ δ αφ Iterior poit α Exterior poit / Surface poit G( r r) φ( r ) dr G( ) ρ( ) d ε r r r r ( φ φ ) φ( ) G φ G G ds S G ds S. Combie terms G φ αφ() r G( ) ρ ( ) d φ( S ) G ds ε r r r r + S. This is Gree s theorem. Admittedly it does t look very helpful i its preset form.. To show how it ca be helpful let s calculate the ifiite space Gree s fuctio for a sphere ad a cylider. Free space spherical Gree s fuctio. G satisfies: G δ( r r) δ( x x ) δ( y y ) δ( z z ) ( ) + ( ) + ( ). Try a solutio of the form G G( u) where u x x y y z z 3. A short calculatio shows that G satisfies δ G u u u u u ( ) 4. The solutio for G for u is give by c ( ) c G u c ( ) for G u + u
5. The value of c is foud by itegratig aroud a small sphere located at u. The terms are G Gd G ds u si d d 4πc u r θ θ φ δ ( ud ) r 6. Therefore c /4π ad G 4π r r ( r r) 7. This is the spherical Gree s fuctio. Free space cylidrical Gree s fuctio. The cylidrical Gree s fuctio is foud i a similar way. satisfies G x G + x x y y y G δ( ) δ( u ( x x) + ( y y) 3. Let G G( u) where G 4. satisfies u δ u u u u G ( ) ) 5. The solutio is G( u) c l u + c c l u for G( ) 6. Fid c by itegratig over a small circle located at u. The separate terms are G Gdr GdS u dθ dz πlc z u ( ) r ( ) ( ) δ u d δ x x δ y y dx dy dz L z 7. Therefore c /π ad the Gree s fuctio is give by G l l ( ) ( u π 4π x x + y y) 3
Usig Gree s theorem for ifiite space problems (i.e. o coductors). Cosider a 3-D domai where the volume exteds to ifiity. This is the case of the spherical Gree s fuctio.. Recall that for localized charges (i.e. o sources at ) both φ ad G vaish at ifiity. 3. Cosequetly both terms i the surface itegral vaish S G φ φ( S ) G ds as r 4. For ay poit withi the domai we must take α ad Gree s theorem reduces to φ r () πε 4 ( r ) ρ dr r r 5. This is correct but we already kew the result from previous lectures. 6. A similar result follows from the cylider φ r r l r r ds () ρ( ) πε 7. Here r x e + y e, r x e + y e, ds dx dy are two dimesioal fuctios. x y x y The total lie charge desity λ is give by ( ) ds Coul / λ ρ r m. 8. So far we have doe a lot of work verifyig what we already kow. I the ext lecture we will lear how to use Gree s theorem to solve far more difficult problems. 4