Assign4-Solns-NotHandin

# Assign4-Solns-NotHandin - Chapter 5 Thus 2_4 lsmxl = E 2...

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Unformatted text preview: Chapter 5 Thus . 2 __4 lsmxl = E + 2 “(m2 _ 1) cosma: m 0V9“. Evaluating this at a: = 0 and letting m = 271 gives 4n2q1‘2' n:1 Evaluatin at m = W/g ives —F '2 2, g g at the pattern ﬁ/Z, ﬁ/Z, 1 1 so 4 x/i 1 n - - — m h n H __. _._.— 1 + - ( 1) 2 W = isn‘mﬁM) w2( 3 5 7 Z 2_ —_—. dd Mr H _ A “:1 4n 1 4 n 0 .. . n r Therefore ( I I 3 5 r h 'e'sine series for ¢(:c) = a; is 00 1251'- _ (—1)m+12£ . mire: coefﬁcients of the sine series for I!) are i i a: _ Z T 8m 3 e m:l 1 f I ( ) E: equation Integra g Illlwb — erm gives .. i} ll 6 t 0 Y 1 4 1 diﬁ oo _._ mcoSW’l‘m‘) ; 1 —1 “212 = ———?—cos(“mml932 + mrfo —3:2 "—t C + Z ( 3 2 cos 7mm mr 0 1 4 1 mmﬁm) dm . 2 "121 m 7r l 4 - a; "'—" S i ___g_ cos(n’ffl + W smmﬂm) o ngﬁz 0 The constant 0 must be TM 4 -— l. 1 1 l2 __ ____2_ costnfr) + "n37r'3 [008(n1l') l C = EA” _ % \$352 dz]: = _’ 7171' 0 4 'n, a _3_(—1)“ 175W” 1]” so ﬂ mr n W 1 £2 00 (—1)m2ll2 mm: _ 2 —- . a: H — + 2 cos m 4 n , 1] 5111mm)- 2 6 m27r2 5" 2 ( 2 (—1)” “HR—1) m=1 ct = ——-— 7‘ 7T “:1 mr (b) Evaluating at :1: = 0 yields i (_1)m+1 _ 7T2 = 0, 1 if d j 3 mzl m2 _ 12 A0 :: — ism ml at 71' j 11' ~1r .' 2 1r I 3: d3; 5.1.9. ‘e ¢(:v) = 0 and Mrs) r: (1032 a: = % + \$003 233, the coefﬁcients in the cosine series For m > 01 7' . m‘cosmw d3 = —- f Slnmcosm ,ar An = 0 for all n, and the coefﬁcients in the cosine series for 1/) are Bu = 1, 82 3 ﬁ ; ' Am i ‘ ] ism 7T 0 and Bn = 0 for all other H. Hence equation (4.2.?) implies ii 7? —1r . [.81 becomes -l 1 ' (b + a) " 81110” a» SO the mteg 1 1) 1“ 1 1 ' = " SlIl _. {E _ ‘ Now SIDGCOSb 2( 1) d 1 {fir—i COS(m + D3: + E 008(7’n 0 u(:c, t) — it + 4—; sm 2ct cos 2a". .5 1T . _. - __ a: '11 = ‘- m + 5 l 171%?) m even' AA Chapter 5 5.3.12. Let u = 9(a), dv = f”(a:) dzc. Then du = g’(3:) dz: and v = f’(m), so integrating by parts gives Abf'Tmlgbzl dw = [Mﬂf’hﬂ]: - Abf'(\$)g’(:r) dag. Section 5.4 I ﬁﬂ'wganmwﬂ 5.4.1. - eometrlc ser1es has ratio —3:2. The N “l partial sum is therefore 1 _ (_\$2)N+1 SN = W‘— For each a: E (—1, 1) this converges to 1 / (1 + 3:2), so the series does converge pointwise. (b) The series does not converge uniformly on (—1,1). For suppose N is odd. Then it is possible to choose 3:... 6 (“1,1) so that 1 — (—zri)N+1 < 1/4, and thus SN(m*) < 1/4. Since 1/(1 + 1:2) > 1/2 for all a: 6 (—1,1), it follows that |SN(:::*) — 1/(1+a:2)| > 1/4 mmMWMﬂ—UﬂﬁﬂbﬂMmﬂmmmMWﬁﬂ—Uﬂ+ﬁﬂ#0 * 3 (c) Since /“\1—(~th+1_ 1 1+\$2 1+:1r:2 as N —> 00, the series does converge in the L2 sense. 5.4.6. For 77. >1, 7: f cos a: sin(m:) d2: 0 1T 3 7f . ; sm((n +1)a:)+ sin((n — 1)::3) d3: (— cosun + nan) + — caste — me) O 11' ’ﬂ'(n +1) 17(n —- 1) 1+ (—1)“ 1+ (—1)” 7T(n+1) + 7r(n-1)) 0 52 Chapter 6 J ’1 f r 6.Quess that u = Aw? + Bally + Cy2 + D3: + Ey + F. Then an + aw = 2A + 20. If u ‘ r Inonic, then C = —A. The boundary conditions lead to the equations By+D=—a Bm+E=b 2Aa+By+D20 By+20b+E=0. The ﬁrst two equations imply B = 0, D = —-a and E = b. Plugging these results into the last two equations gives A = % and C : —~%. Thus u(:r, y) = arr? — yz) — as + by + F, where F is arbitrary. 6.2.3. Separating variables gives 3% + Y7: = 0, so X” = AX and Y” = —)\Y for some constant A. Since Y satisﬁes the boundary conditions Y’ (O) = Y’ (71') = 0, we have Yn = _ cos my and An = 112. The solutions of X” = an with boundary condition X (0) = 0 are “a: I;J‘ Xn = sinh m: for n > 0 and X0 = 3:, so we have the expansion 00 u(:l:, y) = Aom + Z An sinh m: cos ny. 11:1 The boundary condition u(1r, y) = %(1 + cos 2y) then implies 1 00 5(1 + cos 23;) = A017 + Z An sinh mr cos ny. n=1 I‘lence A01 = %, A2 : and A” = 0 for all other n, so the solution is u(m,y) ﬁx + m s1nh 29: cos 2y. 6.2.6. Separating variables gives H' + Y” + Z” — X Y z _ Each of the three terms is constant, so X" = —/\X yr! ___ Z" — (A + ,u)Z The boundary conditions imply X’(0) = X’ (1) = Y’(0) = Y’(1) = Z’(0) = 0. Th: Xn(:r) 2 005mm: and A” = ngwg for n = 0,1,2,...; Ym(y) = cosmqry and um = 271’” m = 0. 1, 2. - - -‘, and Zmn(z) = coshwm2 + nng). Writing the solution as a series 00 00 u(:L', y, z) = Z 2 Am” cos mm: cos may cosh(v m2 + ngvrz), m=U 11:0 68 Chapter 6 Section 6.3 6.3.1. (a) By the Maximum Principle, u attains its m_a_x'1mum o of 3 sin 26 + 1 is 4, the maximum of u on D is 4. n the boundary. Since the maximum (b) By the Mean Value Property, the value of u at the origin is the average of 3sin 29 + 1 on the circumference. So 2 u(0,0)=—1—- 3sin28+1d9=1. 1+ 3sin9, A0 = 2, 31 : 3/0. and all other n the full Fourier series for ME?) = u(“r,9) = 1 + %rsin6. In rectangular I 6.3.2. ts are zero. Thus by equation (6.3.10), - t . dinates u(w,'y) = 1 + 3W“- Section 6.4 6.4.7). The condition at inﬁnity implies 6.4.1. The solution takes the form of equation ( ,, into An and 8,, gives 0,, = 0 for n 2 1 and D0 : 0. Absorbing the coefﬁcients D 00 uh", 6) = \$00 + 27"" (A,1 cos 77.6 + 3,, sin n6). 11:1 The condition at "r = 0. implies 00 1+ Ssinﬂ = \$00 + gay—“(An cosnﬂ + Bn sinnB). Thus 00 = 2, An = 0 for all n, 31 = 3a. and B,, = 0 for all other n. Hence u(r, 6) = 1+1} sin 6‘.: In rectangular coordinates, u(a:, y) = 1 + \$2+y2- - 6.4.3. The solution takes the form 1 m u(’r, 6) = E(Co + DD log T) + 2(Cnr” + Dnr‘") cos n9 + (Anrn + Bur“) sin n9. n=1 Atr=aandr=bwehave _ 1 0° 9(9) = 5(a) + Do log a) + 2(Cna” + Duct—n) cos M? + (Anon + Bna'”) sin n9 n=1 1 oo h(t9) = 5(00 + D9 log b) + 2(Cnb” + Dnb‘“) cos of) + (Anbn + Bub—n) sin n6. 71:1 70 ...
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