MATH 38:Partial Differetial Equatios Solutios to The Secod Midterm DEGENERATE ELLIPSOID COORDINATES. Problem PROOF: Give csih si cos, y csihsi si, z ccoshcos, where c is a scalig costat,,,. r We compute U, U, ad U from the defiitio: Ui, ui where r csihsi cos, csihsi si, ccoshcos U ccoshsi cos, ccoshsi si, csihcos U csihcos cos, csihcos si, ccoshsi U csihsi si, csihsi cos, We ow verify that UU UU UU orthogoality, by Hildebrad Eq. 49 We first remark the basic trig ad hyperbolic trig idetities cos si, cosh sih, U U c cosh sihsi cos cos c coshsihsi cos si c coshsihsi cos coshsihsi cos cos si c factorig out cosh sih si cos c coshsihsi cos UU c sih cos si cossi c si cos si cossi c sih cos si cossi UU c coshsihsi cossi c coshsihsi cossi c coshsihsi cossi Hece Eq. 49 is verified ad therefore the system of coordiates,, is orthogoal. By defiitio, ds hd hd hd where by defiitio, hi UiU i. h c cosh si cos c cosh si si c sih cos c c cosh si cos si c sih cos cosh si sih cos c sih si sih si si sih si sih sih si c sih si c h c sih cos cos c sih cos si c cosh si sih cos cos si c cosh si c c c sih cos cosh si sih si sih si c c sih sih si si sih si c sih si. Hece h h h c sih si si c sih si cos c sih si si cos c sih asi ds c sih si d d c sih a si d QED
MATH 38:Partial Differetial Equatios Solutios to The Secod Midterm Problem hh hh hh Give that u u u u, ad otig that h h, hhh h h h h u h u h u u hh hh hh h Ad usig our results for the hi 'si problem, h h c sih si, ad h c sih si, Notig that i geeral, qjri qj ri for i j, we may rewrite the epressios i i i i h u csih si u c sih si csihsi hh hh csi sih u sih u c sih si csih si c sih si sih h u csihsi u c sih si csih si csih si u si u c sih si csih si c sih si si h c sih si u u h c sih si csihsi csihsi c sih si u c sih si csih si c c hh u sihsi sih si Hece, eplicitly, our Laplacia is: u sih u si u u c sih si sih sih si si c sih si a c which is ideed of the form: ua A ua A ua A u a where a, a, ad a, c sih si sih c sih si si c sih si where A sih, A si,ad A
MATH 38:Partial Differetial Equatios Solutios to The Secod Midterm TCHEBYCHEFF POLYNOMIALS. For the followig problems o Tchebycheff polyomials, I fid it more coveiet to thik i terms of cos, ad hece si U. However, I will be aswerig the problems as required by si makig the correspodece cos i all the problems. Problem 3 si PROOF: arccos Give that U, we show that Uis ideed a polyomial i. Note that if we make the chage of variable cos, we may rewrite the above as: si arccos cos si U, for cos cos si which is defied for, correspodig to the variable,. To show that this is ideed a polyomial i, we write the first few terms eplicitly otig that cos, si si U si si si si sicos U cos si si si U si si 3 si cos cos si si cos cos si cos si si si si cos cos 4cos 4 We ow claim that si may always be writte i the form si si P cos where P cos is a polyomial i cos m m m m m. The secod term is of the form si Pcos m P PROOF: By way of iductio, we see that the base case si si is trivially true. Now, suppose that si is true, we show that si holds by iductio. Rewritig, si si cos cos si by the iductive hypothesis. It remais to show that si cos is also of the form si cos. 3
MATH 38:Partial Differetial Equatios Solutios to The Secod Midterm Now, we remark that cos m cos m cos sisi m. The secod term is of the form P m m m m si cos after applyig the iductive hypothesis to si. This however, is precisely a polyomial i P cos sice we may use the idetity cos si to write si P cos cos P cos, which is a polyomial i cos, as claimed. The first term cos cos may be broke dow the same way ito: cos cos si si cos, ad subsequetly, without aother formal iductive proof, we ca decompose the first term util we obtai terms with order cos. m P Recall that the terms sisi cos,as proved previously m P Therefore the first term will be a polyomial i cos ad hece cos cos m P P Cosequetly, the first term si cos is also of the form si cos, whece, by way of mathematical iductio, si si cos. P Sice si si P cos as proved i the above claim, we coclude that si si cos U si si Pcos P sice cos, provig the claim. QED We ow seek to prove the recurrece relatio U U U PROOF: si si cos cos si Eq si si cos cos si Eq Eq Eq si si Remark that by the trigoometric idetity: si m si cos m cos si m Addig ad, we obtai: si si si cos Dividig the equality by si, we obtai: si si si cos si si Sice we have show earlier that U, for cos si which is defied for, correspodig to the variable,, we see that: ad reamig the idices U U U U U U, as claimed. U U U by each by a icremet of, QED 4
MATH 38:Partial Differetial Equatios Solutios to The Secod Midterm Problem 4 si arccos si Give that U, for cos, we first establish some useful trig idetities: si si si cos cos cos si 3 si cos cos si sicos cossi cos sicos cos si4cos 3 cos3 cos cos si si coscos sisi cos cos cos si cos 3 3 3 3 cos cos cos cos cos cos cos cos 4cos 3cos We ow evalua te U ad U from the idetities, otig that cos : 5 6 3 si4cos 4cos 3cos si 5 si 6 si 3 cos3 U5 si si si si 5 3 3 5 3 6cos cos 4cos 3cos3cos 3cos 6cos 5 3 3 3 6 si 6 si 7 sicos6 cossi 6 U6, ad usig our result of si 6 from evaluatio of U5 si si si 5 3 sicos 3 cossi3cos 3cos 6cos si 3 4cos 3cos cos 3cos 5 3cos 3 6cos 6 4 6 4 6cos 4cos 9cos 3cos 3cos 6cos 6 4 6 4 3cos 48cos 8cos 3cos 3cos 6cos 6 4 64cos 8cos 4cos 6 4 64 8 4 5 3 U5 3 3 6 6 4 U6 64 8 4 5
Problem 5 PROOF: We wat to prove: U t t t si MATH 38:Partial Differetial Equatios Solutios to The Secod Midterm si arccos si Give that U, for cos, we write: si U t t si t si si i i Im U t e t sice si Im e si *** i i i 3i i i i Epadig the series eplicitly, e t e e t e t... e e t e t... # By the biomial epasio, i i i... Hece, idetifyig e t, e te t... i e t i i e i Therefore, # simplifies ito: e t, ad by Euler's formula, e cos isi i e t i i i e e equatig real ad imagiary e t tcos isi tcositsi parts i the deomiator Multiplyig the umerator ad deomiator by the comple cojugate of the deomiator, i e tcos itsi i e t cos isi tcositsi t cos t si tcos t si i cos tcositcos si isi tcostsi e t tcos t si cos tcostsi itcossi si tcos i t cos tcostsi itcossi si tcos U t Im si tcos t si tcossi si tcos tcos tcos U t si tcos t si tcos t si e tcos t si ad substitutig back ito ***, we have: U t tcos t cos ad otig that cos U t t t t t t t U t t t, as claimed. equatig real ad imagiary parts i the deomiator epressig i terms of cos usig cos si factorig out si 6
MATH 38:Partial Differetial Equatios Solutios to The Secod Midterm Problem 6 si arccos Give that U, We first evaluate U. si arccos U si sice arccos. We rewrite this i the followig more suggestive otatios by cosiderig: m eve ad m odd for mz. m Um sim m For odd, For eve, Um m m m m m sice si m, cos m mz ad si si si cos cos si cos U We ow evaluate si arccos si U sice arccos. We see that this epressio is meaigless uless we see the behavior of the limits via L'Hopital's rule. f f ' L'Hoptial's rule states that for fuctios f ad g, if lim f lim g, the lim lim. c c c g c g' By the Chai rule, we evaluate the derivatives of si arccos ad d d si arccos cosarccos sice arccos d d d d d si arccos m cos arccos si arccos Cosequetly, lim lim d lim d d si arccos cosarccos cos lim lim sice cos U 7
MATH 38:Partial Differetial Equatios Solutios to The Secod Midterm Problem 7 / PROOF: We wat to prove for k that U k U d si arccos si Usig the defiitio U, for cos, si cos cos lower lower We let cos be the chage of variables,, d sid cos upper upper cos / / si k si Uk U cos si d d si si Usig the idetity cos si ad flippig the itegratio limits, k si si / U U d k d Ad usig the fact that a product of sies ca be epressed as a combiatio of cosies, i.e. cos cos cos cos si si cos cos si si ab ab a b a b a b a b si asi b si asi b Therefore idetifyig a k ad b, we have: / U cos cos k U d k k d k k / si si U k U d k k Sice k, Zad si a az, we coclude that k / U U d, as claimed for k QED / / si For k, we evaluate U cos si d d si Usig the idetity cos si ad flippig the itegratio limits, / U k U d d / k cos / si k / Uk U d si, ad usig the idetity cos si, U U d d U U d, agai usig the fact that si a az, 8
MATH 38:Partial Differetial Equatios Solutios to The Secod Midterm 3 FOURIER TRANSFORMS Recall that the Fourier trasform of a fuctio f is give by: ˆ it f e f tdt Problem 8 t, t Give f t, the Fourier trasform of f t is give by:, otherwise fˆ e f t dt cos t isi t f t dt. it f t is a eve fuctio i t, ad hece cost Remark, however that fˆ cos tf t dt cos tf t dt Itegratig by parts, we obtai: ˆ f tsit sitdt cost ˆ f cos cos ˆf t dt Problem 9 t e, t Give f t, the Fourier trasform of f t is give by: t e, t ˆ it it t it t i t i t f e f t dt e e dt e e dt e dt e dt it it ˆ e e f i i Now, refer back to the fuctio f t, ad recall that i the first itegral, t, hece the term i t i t, therefore e e. Similarly, i the secod itegral, t, hece the term e it it i, therefore e t t e. i i ˆ f i i i i i i fˆ 9