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## Partial Differential Equations

by: Vernie McCullough

40

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5

# Partial Differential Equations MATH 467

Vernie McCullough

GPA 3.58

Staff

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COURSE
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KARMA
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## Popular in Mathematics (M)

This 5 page Class Notes was uploaded by Vernie McCullough on Thursday October 15, 2015. The Class Notes belongs to MATH 467 at Millersville University of Pennsylvania taught by Staff in Fall. Since its upload, it has received 40 views. For similar materials see /class/223531/math-467-millersville-university-of-pennsylvania in Mathematics (M) at Millersville University of Pennsylvania.

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Date Created: 10/15/15
Orthogonality In many of the exercises we will solve this semester we will write a function f as a trigonometric series of the form fw ZAnsin n1 A natural question to ask is How do we calculate the coef cients An The method is based on the orthogonality of the trigonometric functions Throughout this document we will make the following assumptions 0 Function fw is de ned on the interval 07 L o The trigonometric series converges uniformly to f To start off7 consider the following integrals L 717 m7rw 7 d 1 0 sin L s1n L w Taking the integral in Eq 1 and applying the productto sum formula 717 Sin m7rw 1 COS n 7 m7rw 7 COS n m7rw L L 7 L L sin 7 we have when 71 7E m L L 7mm m7rw 1 n7m7rw nm7rw O s1nTs1anw iO cos77cos7dw 1 L n7m7rw L nm7rw L 7 sin 7 sin 2 n7m7r L nm L 0 7 L ltlt gt gt 1 ltlt gt gt 7 27F ms1n n m ms1n n m7r 7 0 In the case that n m we have L 7mm m7rw L i 2 7mm sin 7 sin dag sin 7 dw 0 L L 0 L L 2 7 cosdw Thus to summarize ifnm L i OsinnL sindw 1fn7 mj 2 To draw in some of the language of linear algebra7 suppose we think of each sin m for n E N as a vector in the linear space of functions de ned on the interval 0L satisfying the condition that un0 unL 0 Integration over the interval 0L will serve as our inner product Typically the inner product of two vectors an and um will be denoted Thus the inner product can be de ned as L mum 0 unwumwdw By the results derived above7 when n 7E m the inner product of vectors an and um is zero or in the language of linear algebra orthogonal The condition expressed in Eq 2 is sometimes referred to as an orthogonality condition We will use the orthogonality condition to calculate the coef cients An Multiply both sides of the trigonometric series by sin m2 and integrate from O to L L L L fw sin mm dw L sin mm film sin y div 0 L O L n1 L L 00 7mm n7rw O An s1n L s1n T dw Since we have assumed the trigonometric series converges uniformly on 07 L then we may exchange the order of integration and summation Thus we have L An sin mm sin dw i147 L sin mm sin m dw 0 n1 L L 1 0 L L L 5 0 m7rw m7rw mm fws1n7 sin 2 AnSIDT n1 Am by the orthogonality condition Therefore for all n E N L An 0 fwsin 6170 1 Bessel s Equation of Order One Consider Bessel s equation of order one x221 xy ac2 71y 0 If we let Px x2 x and Rx x2 7 1 then we can easily see that 130 0 and thus are 0 is a singular point and xQx7 x 7 7 7 mm wrmm p0 x2Rx 2x271 2 513 pm ng x2 ng lxl am and hence are 0 is a regular singular point The indicial equation will have the form FTTT71p0Tq0TT71T71T271 Thus the indicial roots are T1 1 and 7 2 71 Our goal is to nd two linearly independent solutions to the ODE We can always nd a series solution corresponding to the larger of the two indicial roots so we could look for a solution of the form y1x 20 a xn1i However since we also want a second linearly independent solution we will follow the advice of the textbook authors who suggest that we obtain a recurrence relation involving the general formula anT see page 277 Thus we will look for a solution of the form End 17957 T Begin by differentiating this solution and substituting it into the di erential equation and combine the series into a single series Then re index oo oo 00 x2 Tn T 7 1anacnT 2 x Tanac 39quot 1 x2 7 1 Z ananrr 0 n0 n0 n0 oo oo oo oo 01 Tn T 71a xnr Ta xnr Z anxnr2 7 Zanxwrr 0 n0 n7 n0 n0 oo oo Tn T 71n T 7 llanxwrr Za xwrp 0 n0 n0 oo oo T2 7 llanxwrr Za xn u 0 n0 n0 oo 00 Z FT nanxquotr Z an72xquot 2r2 0 n0 n720 oo 00 Z FT na xnr Z an72xnr 0 n0 n2 oo oo FTaoacT FT 1alacTJrl ZFU na xnr Z an72xnr 0 n2 n2 oo FTaoacT FT 1alacTJrl ZF7 nan 17QlacwrT 0 n2 Since T 71 or T 1 the rst term above is zero The second term has coefficient a1FT 1 11T2 2T which must match the coefficient of xr1 on the righthand side of the equation The quadratic part r2 2r y 0 since r 71 or r 1 thus it must be the case that 11 0 For n 2 2 we have derived the following recurrence relation 7 17720 170 iFltn T Since 110 0 then 0 130 150 127440 We then can let T 1 the larger of the exponents of singularity To keep the discussion simple let a0 I recall that 10 can be any arbitrary nonzero value Then 1 1 1 121 7m m 1 7 121 7 1 7 a4 7 W W39Q W 71quot a2n1 22Wn 1lnl Thus the rst solution has the form 7 00 1quot n1 2106 7 727 22quotn 1lnlx According to Theorem 571 the second solution Will have the form 1 2 c r x 31295 ay1xln x x Where the coefficients c r2 satisfy the formula d WW g T Wanmllr z for n 12H and a lim r 7 r2aNr r7gtr2 Where N r1 7 r2 Let s apply these formulas in this example First N 2 1 7 71 r1 7 r2 so 7 7 7110 7 7r1 7 7r1 7 717 l a BEWHWW Billv lpv 2 T131110 22 71 r1391 T2 4T 3 4413111 T 3 2 When n is an odd integer 170 0 Which implied c r2 0 When n is odd When n is an even integer say 71 2m then CQmT 7 hen 213725 v a2m767 l llml Slsilsils S9 0 1 71m1a0 Hicn1 F0 2k 2 inm ghwUWl vWiilWMl lt1gtm1 lWPWl 1gt lWWl Milwml UwngMl This could be very cumbersome to differentiate in generali We can calculate a few examples and look for a pattern in the derivatives 1 m CQTi dr r3 7r32 This implies C271 141 ii 1 7 3r13 W i an ltltT3gtltltT4gt271gtgt ltr3gt2ltr5gt2 This implies 6471 75641 While the following is far from obvious it is nevertheless true 75 1gt17Ei1211 6471 a 7 242111 24211 C iii 1 7 5r252r127 6 7 dr r3r4271r6271 7 r32r52r72 This implies 3 2 6671 5 Ek12k1 1152 26312 1 Using proof by induction then we can see that in general 221 2211 CMH Hlm WWW1107171 for m 121111 Hence 1 7 n E 2211 n W i m 11 a

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