Today in Physics 217: separation of variables IV

Transcription

1 Today i Physics 27: separatio of variables IV Separatio i cylidrical coordiates Exaple of the split cylider (solutio sketched at right) More o orthogoality of trig fuctios ad Fourier s trick V = V V = E V 8 October 22 Physics 27, Fall 22

2 Separatio of variables i cylidrical coordiates I Griffiths proble 3.23, o hoework 6, you did the setup of the separatio solutio to all Laplace-equatio probles i cylidrical geoetry. Recall that the Laplace equatio i cylidrical coordiates is V V V V = s + + = s s s s φ z If you kow a priori that V does t deped o z ifiite cylider, boudary coditios idepedet of z the the last ter drops out V V s + s s s 2 2 = s φ 2 8 October 22 Physics 27, Fall 22 2

3 Separatio of variables i cylidrical coordiates (cotiued) V( s, φ) = S( s) Φ( φ) : So you tried a solutio of the for s d ds d Φ s + Sds ds 2 = = Φ dφ ad solved the resultig radial ad agular ordiary differetial equatios to obtai particular solutios: 2 d Φ 2 2 = Φ Φ ( φ) = Acos φ + Bsi φ dφ d ds 2 ad s s = S S( s) = Cr + Dr ds ds with a additioal radial solutio for = : d ds s s = S( s) = Cls + D ds ds 8 October 22 Physics 27, Fall

4 Separatio of variables i cylidrical coordiates (cotiued) Fro the periodicity of the agular solutio you also deduced that =,,2,3, The ost geeral solutio is a liear cobiatio of all of these solutios, for all values of, which you wrote as ( φ ) V s, = C ls+ D = ( Cs Ds )( Acos φ Bsi φ) You applied this solutio i a cocrete exaple, Griffiths proble 3.24, i which you were able to avoid the use of Fourier s trick. Now for oe i which you ca t. 8 October 22 Physics 27, Fall 22 4

5 The split cylider Exaple. A log, thi-wall coductig cylidrical tube with radius R a sall sectio of which is show at right is split i half legthwise. The two halves are isulated fro oe aother; oe is held at potetial V ad the other is grouded. Fid the potetial iside the tube. z y R V = V V = x 8 October 22 Physics 27, Fall 22 5

6 The solutio: V s, φ = C ls+ D ( ) The split cylider (cotiued) = ( Cs Ds )( Acos φ Bsi φ) Boudary coditios: iv. = V at s= R, φ = π ii. V = at s = R, φ = π iii. V fiite at s = Apply the last oe first: the ter Ds approaches ifiity at s = uless D =. Siilarly, C =. 8 October 22 Physics 27, Fall 22 6

7 More o trig fuctio orthogoality Now apply the first two boudary coditios. This will be coveietly doe i cocert with the applicatio of Fourier s trick to extract the rest of the coefficiets iside the su. Ad this will be a good place to fill i soe ore details about the orthogoality of the trig fuctios. Last Friday we showed that π π si y si ydy = δ Obviously we also have 2 si y si ydy = πδ. 8 October 22 Physics 27, Fall 22 7

8 More o trig fuctio orthogoality (cotiued) But what about cosies? It goes just like the derivatio for sies, startig with two itegratios by parts: = cosx cos xdx = cosx si x + si x si xdx Thus For =, Itegrate this by parts oce: 2 = π 2 π = si x cos x + cos x cos xdx 2 2 ( ) cosx cosxdx =, or cos x cosxdx =. 2 2 cos xdx = cos wdw. 8 October 22 Physics 27, Fall 22 8

9 More o trig fuctio orthogoality (cotiued) = 2 2 cos wdw = cos wsi w + si wdw Thus ad ( 2 2 ) = cos w + si w dw = dw = π cos xdx = π cosx cos xdx = πδ. 8 October 22 Physics 27, Fall 22 9

10 More o trig fuctio orthogoality (cotiued) The there are products of sies ad cosies to cosider: = cosx si xdx = si x si x si x cosxdx Recall the trig idetity ( ) si u+ v = si ucos v+ cosusi v: cosx si xdx si xdx cosx si xdx = ( + ) + ( ) cos + x cosx sixdx = =. + So, oce agai, this shows that ( ) cosx si xdx =. 8 October 22 Physics 27, Fall 22

11 More o trig fuctio orthogoality (cotiued) For the case =, recall the trig idetity 4π So it s zero whether = or ot. Suary: Now back to that cylider: 2cosusiu= si2 u: 4π cosx si xdx = si 2xdx = si udu = cos u = cosx cos xdx = si x si xdx =πδ, cos x si xdx =. 8 October 22 Physics 27, Fall 22

12 The split cylider (cotiued) What s left of the solutio: (, φ) ( cos φ si φ) V s = Cs A + B = Let s apply the first two boudary coditios, while extractig the A : ( ) = ( + ) V R, φ cosφdφ C R A cosφ B siφ cosφdφ = π V cos φdφ = CR ( Aπδ + ), as we just foud. = V π si φ = πa C R A C = 8 October 22 Physics 27, Fall 22 2

13 The split cylider (cotiued) Now work o the Bs by itegratig everythig with ( ) = ( + ) V R, φ siφdφ C R A cosφ B siφ siφdφ = π V si φdφ = CR ( + Bπδ), as we just foud. = V π cos φ = πb CR V ( ) = π BC R = if is eve, 2 if it's odd. si φ : 8 October 22 Physics 27, Fall 22 3

14 The split cylider (cotiued) So replace with 2+ i the su: = BC ( φ) 2V π R V s, = B C s siφ = 2V s = π R =,3,5,... = si φ 2+ 2V s = si ( 2 + ) φ. π 2+ R 8 October 22 Physics 27, Fall 22 4