Brief Notes For Math 3710

Transcription

1 Brief Notes For Mth 371 Afshin Ghoreishi Fll 13

2 Contents Prefce iii -1 Hndouts Hndouts Introduction 15 1 Fourier Series Periodic Functions nd Fourier Series Aritrry Period nd Hlf-Rnge Expnsions Convergence of Fourier Series Uniform Convergence Opertions on Fourier Series The Het Eqution 35.1 Derivtion nd Boundry Conditions Stedy-Stte Tempertures Exmple: Fixed End Tempertures Exmple: Insulted Br Exmple: Different Boundry Conditions Exmple: Convection Sturm-Liouville Prolem Generlities on the Het Conduction Prolem The Wve Eqution The Virting String Solution of the Virting String Prolem D Alemert s Solution Generlities on the One-Dimensionl Wve Eqution The Potentil Eqution Potentil (Lplce) Eqution Potentil in Rectngle Further Exmples for Rectngle Potentil in Disk Clssifiction of Prtil Differentil Equtions Prolems in Severl Dimensions Two-Dimensionl Het nd Wve Equtions Solution of the Two-Dimensionl Het Eqution Prolems in Polr Coordintes Bessel s Eqution Temperture in Cylinder Virtion of Circulr Memrne

3 CONTENTS ii 5.9 Sphericl Coordintes; Legendre Polynomils Legendre s Eqution nd Polynomils Some Applictions of Legendre Polynomils

4 Prefce This is preliminry collection of rief notes nd hndouts for Mth 371. It is written to mtch the 6th edition of Powers textook. This is not replcement for your own course notes! However, if you print this version nd ring it to clss, you cn dd clss notes to it (spce hs een lloted for this) to mke complete set of course notes. You cn otin this document from the we site for this course in

5 Chpter -1 Hndouts (Chpter nd Appendix in the Course Textook) The chpter zero of your textook contins review of ordinry differentil equtions nd in the ppendix you find mthemticl references. Red them, s needed. This chpter contins the hndouts for the course which includes textook corrections nd review of oth ODE s nd mthemticl references we need in this course. -.9 Hndouts A Prtil List of Corrections to the Boundry Vlue Prolems y Dvid L. Powers, Sixth Edition Loction Originl Correction 1. Chpter 1, Misc Exer 7(f) f(x) = x f(x) =. Chpter 1, Misc Exer 8 f(x) = f(x) = x 3. Sec.1, Exer 3 Wht is g Wht is the replcement of A xg 4. Sec 3., Exer 1 f(x) is s in Eq. (11) f(x) is s in the exmple in pge Sec 3.4, Exer 8 u(x, t) = f(x) u(x, ) = f(x) 6. Pge 33, Eqution (3) φ(r, π) φ(r, θ) 7. Pge 381, Eqution for L(f (t)) 8. Pge 457, Solution to Exer 3, Sec Pge 466, Solution to Exer 3, Sec.1 1. Pge 471, Solution to Exer 5, Sec Pge 49, Solution to Exer 1, Chp 5, Misc Exer 1. Pge 491, Solution to Exer 3, Chp 5, Misc Exer = f() + = f() = f () + = f () 1 for < x < 1 for 1 < x Eq. (6) Eq. (9) u(.5, 1./c) =.α 1 for < λ < 1 for 1 < λ u(.5, 1./c) = α/ µ m = mπ µ m = mπ/ m=1

6 CHAPTER -1. HANDOUTS Review, Identities, Formuls nd Theorems Let n, n, m, m, k, k, l, p, q nd q e nonnegtive integers, unless stted otherwise. Let z e nonnegtive rel numer, unless stted otherwise. Trigonometric Identities 13. sin cos = 1 [sin( + ) + sin( )] 14. sin sin = 1 [cos( ) cos( + )] 15. cos cos = 1 [cos( + ) + cos( )] Hyperolic Functions 16. sinh x = ex e x 17. cosh x = ex +e x 18. tnh x = sinh x cosh x 1. csch x = 1 sinh x 19. coth x = cosh x sinh x. sech x = 1 cosh x. sinh( x) = sinh x 3. cosh( x) = cosh x 4. cosh x sinh x = tnh x = sech x 6. sinh(x ± y) = sinh x cosh y ± cosh x sinh y 7. cosh(x ± y) = cosh x cosh y ± sinh x sinh y d 8. dx (sinh x) = cosh x 9. d dx (cosh x) = sinh x 3. d dx (tnh x) = sech d x 31. dx (coth x) = csch x d 3. dx (sech x) = sech x tnh x 33. d dx (csch x) = csch x coth x ˆ ˆ ˆ ˆ x sin x dx = 1 sin x x cos x + C 35. x sin x dx = 3 cos x + x sin x 1 x cos x + C Integrls ˆ x cos x dx = 1 cos x + x sin x + C x cos x dx = sin x + x cos x x sin x + C ˆ e x sin x dx = ex ( sin x cos x) + C 39. e x cos x dx = + ex ( cos x + sin x) + C + 4. ˆ sin nπx { mπx sin dx =, if n = m, otherwise 4. For positive integers n nd m, 43. For positive integers n nd m, ˆ ˆ sin nπx sin nπx ˆ ˆ Definite Integrls cos (n 1)πx sin (n 1)πx { mπx, if n = m sin dx =, otherwise mπx cos dx = 41. ˆ cos nπx cos (m 1)πx dx = sin (m 1)πx dx = 45. ˆ cos nπx mπx cos dx = {, if n = m, otherwise {, if n = m, otherwise mπx cos dx =, if n = m =, if n = m, otherwise, if n = m =, if n = m, otherwise 47. For < α 1 < α < zeros of J z (x) nd z, ˆ J z ( αmr )J z( α mr ) r dr = { J z+1 (α m), if m = m, otherwise

7 CHAPTER -1. HANDOUTS For < β 1 < β < zeros of J (x), ˆ ˆ J ( βmr )J ( β mr J ( βmr ) r dr = nd ) r dr = { J (β m), if m = m, otherwise 49. For < β 1 < β < zeros of J z(x) nd z >, ˆ J z ( βmr )J z( β mr ) r dr = { [J z (β m ) J z 1 (β m )J z+1 (β m )], if m = m, otherwise 5*. 51*. ˆ π ˆ 1 (n+m)! n+1 (n m)!, { Pn m (cos φ)pn m(cos φ) sin φ dφ = if n = n m, otherwise 1 (n+m)! n+1 (n m)!, { Pn m (s)pn m(s) ds = if n = n m, otherwise * For m =, P m k (s) = P k(s) nd (n+m)! (n m)! = 1. Definite Doule Integrls ˆ ˆ ˆ ˆ ˆ ˆ sin nπx sin nπx cos nπx mπy sin sin pπx mπy cos sin pπx mπy cos cos pπx qπy sin dy dx = qπy cos dy dx = qπy cos dy dx = { 4, if n = p nd m = q, otherwise 4, if n = p nd m = q =, if n = p nd m = q, otherwise, if n = m = p = q = 4, if n = p nd m = q =, if n = p = nd m = q, if n = p nd m = q, otherwise Suppose < α 1 < α < re zeros of J z (x). 55. ˆ ˆ J z ( αmr )J z( α mr ) r cos kπθ cos kπθ dθ dr = J z+1 (α m), if m = m nd k = k = 4 J z+1 (α m), if m = m nd k = k, otherwise ˆ ˆ ˆ ˆ ˆ ˆ J z ( αmr J z ( αmr J z ( αmr )J z( α mr )J z( α mr )J z( α mr L ) r sin kπθ ) r cos kπθ ) r sin kπθ sin kπθ dθ dr = cos kπθ dθ dr = sin kπθ dθ dr = { 4 J z+1 (α m), if m = m nd k = k, otherwise J z+1 (α m), if m = m nd k = k = J z+1 (α m), if m = m nd k = k, otherwise { J z+1 (α m), if m = m nd k = k, otherwise

8 CHAPTER -1. HANDOUTS 4 59*. 6*. 61*. 6*. ˆ π ˆ ˆ π ˆ ˆ π ˆ ˆ π ˆ P m n (cos φ)p m n P m n (cos φ)p m n P m n (cos φ)p m n P m n (cos φ)p m n (cos φ) sin φ cos kπθ (cos φ) sin φ sin kπθ (cos φ) sin φ cos kπθ (cos φ) sin φ sin kπθ cos kπθ dθ dφ = sin kπθ dθ dφ = cos kπθ dθ dφ = sin kπθ dθ dφ = (n+m)! n+1 (n m)! (n+m)! n+1 (n m)!, if n = n m nd k = k =, if n = n m nd k = k, otherwise { (n+m)! n+1 (n m)!, if n = n m nd k = k, otherwise 4 (n+m)! n+1 (n m)! (n+m)! n+1 (n m)!, if n = n m nd k = k =, if n = n m nd k = k, otherwise { (n+m)! n+1 (n m)! * It lso holds if k = m. For m =, P m k (s) = P k(s) nd (n+m)! (n m)! = 1. Suppose < α 1 < α < re zeros of J z (x). 63**. 64**. 65**. 66**. ˆ ˆ π ˆ ˆ ˆ π ˆ ˆ ˆ π ˆ ˆ ˆ π ˆ J z ( αmρ J z ( αmρ J z ( αmρ J z ( αmρ )J z( α mρ (k+l)! k+1 (k+l)! (k+1) Definite Triple Integrls, if n = n m nd k = k, otherwise ) ρ P k l(cos φ)p l qπθ (cos φ) sin φ cos k cos qπθ dθ dφ dρ = (k l)! J z+1 (α m), if m = m, k = k l nd q = q = (k l)! J z+1 (α m), if m = m, k = k l nd q = q, otherwise )J z( α mρ ) ρ P k l(cos φ)p l qπθ (cos φ) sin φ sin k sin qπθ dθ dφ dρ = { (k+l)! (k+1) (k l)! J z+1 (α m), if m = m, k = k l nd q = q, otherwise )J z( α mρ ) ρ P k l(cos φ)p l qπθ (cos φ) sin φ cos k cos qπθ dθ dφ dρ = (k+l)! k+1 (k l)! J z+1 (α m), if m = m, k = k l nd q = q = (k+l)! k+1 (k l)! J z+1 (α m), if m = m, k = k l nd q = q, otherwise )J z( α mρ ) ρ P k l(cos φ)p l qπθ (cos φ) sin φ sin k sin qπθ dθ dφ dρ = { (k+l)! k+1 (k l)! J z+1 (α m), if m = m, k = k l nd q = q, otherwise ** For l =, P l n(s) = P n (s) nd (k+l)! (k 1)! = First Order Liner ODE: y + f(x)y = g(x) Ordinry Differentil Equtions f(x)dx Integrting Fctor: µ(x) = e with C =, µ(x)y (x) + µ(x)f(x)y(x) = µ(x)g(x) = ˆ ˆ d dx [µ(x)y(x)] = µ(x)g(x) = µ(x)y(x) = µ(x)g(x) dx + C = y(x) = 1 µ(x)g(x) dx + Or, integrting fctor: µ(x) = e xx f(t)dt nd y(x) = 1 µ(x) ˆ x µ(x) x µ(t)g(t) dt + y(x ) µ(x) C µ(x)

9 CHAPTER -1. HANDOUTS First Order Seprle ODE: dy dx = g(x) h(y) ˆ ˆ Implicit Solution: h(y) dy = g(x) dx = H(y) = G(x) + C with H = h nd G = g Or, ˆ y y(x ) h(t) dt = ˆ x x g(t) dt 69. Exct ODE: M(x, y) + N(x, y) dy M = is clled exct if dx y = N x Implicit Solution: F (x, y) = C where F x = M nd F y = N Strt with F F x = M or y = N integrte with respect to x or y, respectively, then differentite with respect to the other vrile, nd use the other eqution to find the remining function of y or x. 7. Second Order Liner ODE with Constnt Coefficients: y + y + cy = Chrcteristic Eqution: r + r + c = with solutions r 1 nd r c 1 e r1x + c e rx, if r 1 nd r re rel-vlued nd unequl y(x) = c 1 e r1x + c x e r1x, if r 1 = r c 1 e λx cos µx + c e λx sin µx, if r 1, r = λ ± µi If r 1, r = ±r, then y(x) = c 1 e rx + c e rx or y(x) = c 1 cosh rx + c sinh rx or y(x) = c 1 cosh r(x x ) + c sinh r(x x ) or y(x) = c 1 sinh r(x x ) + c sinh rx or y(x) = c 1 cosh r(x x ) + c cosh rx 71. Second Order Liner Nonhomogeneous ODE: y + p(x)y + q(x)y = g(x) Generl Solution: y(x) = y h (x) + y p (x) where the homogeneous solution y h (x) = c 1 y 1 (x) + c y (x) is the generl solution to the homogeneous eqution y + p(x)y + q(x)y =, while y 1 nd y re two linerly independent solutions of the sme homogeneous eqution, nd the prticulr solution y p (x) is solution to the nonhomogeneous eqution y + p(x)y + q(x)y = g(x). Method of Vrition of Prmeters: y p (x) = u 1 (x)y 1 (x) + u (x)y (x) where u 1 (x) = y (x)g(x) W (x), u (x) = y 1(x)g(x) W (x) nd the Wronskin W (x) = y 1 (x)y (x) y (x)y 1 ˆ ˆ (x). y p (x) = y 1 (x) dx + y (x) dx or y p (x) = y 1 (x) ˆ x y (x)g(x) W (x) y (t)g(t) W (t) x dt + y (x) ˆ x 7. Cuchy-Euler Eqution: x y + αxy + βy = y 1 (x)g(x) W (x) y 1 (t)g(t) W (t) x Indicil Eqution: p(p 1) + αp + β = with solutions p 1 nd p c 1 x p 1 + c x p, if p 1 nd p re rel-vlued nd unequl y(x) = (c 1 + c ln x ) x p 1, if p 1 = p x λ [c 1 cos(µ ln x ) + c sin(µ ln x )], if p 1, p = λ ± µi 73. x d φ dx + xdφ dx n φ = nd φ() ounded = φ(x) = x n for n =, 1, dt

10 CHAPTER -1. HANDOUTS d [ dx s(x) dφ dx 75. φ + λφ = = λ = ] q(x)φ + λp(x)φ = = λ = C φ φ ˆn ds + R Ryleigh Quotients ˆ s(x)φ(x) dφ r r + l R φ da φ da dx ˆ l r l [ ] s(x)( dφ dx ) + q(x)φ (x) dx φ (x)p(x) dx Lgrnge s Identity nd Green s Formul For L(φ) = d [ s(x) dφ ] q(x)φ, dx dx 76. ul(v) vl(u) = d [ ( s(x) u(x) dv )] dx dx v(x)du dx ˆ r [ 77. [ul(v) vl(u)] dx = s(x) u(x) dv ] r dx v(x)du dx R R Ω l u v da = C u v ˆn ds (u v v u) da = C (u v v u) dv = Ω R Green s Identities u v da (u v v u) ˆn ds (u v v u) ˆn ds Eigenvlue Prolems 81. d φ dx = λφ, φ() = nd φ() = = λ = ( nπ ), φ(x) = sin nπx for n = 1,, 8. d φ dx = λφ, dφ dx 83. d φ dx = λφ φ( ) = φ() dφ dφ ( ) = dx dx () () = nd dφ dx () = = λ = ( nπ ), φ(x) = cos nπx for n =, 1, = λ = ( nπ ) φ(x) = cos nπx nd sin nπx l for n =, 1, d φ dx = λφ φ() = dφ dx () = d φ dx = λφ dφ dx () = φ() = = λ = [ (n 1)π ] φ(x) = sin (n 1)πx = λ = [ (n 1)π ] φ(x) = cos (n 1)πx for n = 1,, for n = 1,,

11 CHAPTER -1. HANDOUTS 7 x d φ dx xdφ dx + (λx n )φ = φ() ounded φ() = [ d ρ df ] + [λρ n(n + 1)]f(ρ) = dρ dρ 87. f() ounded f() = x d φ dx + xdφ dx + (λx n )φ = 88. φ() ounded = dφ dx () = = λ = ( αm ) φ(x) = J n ( αmx ) for < α 1 < α < zeros of J n (x) = λ = ( α k ) f(ρ) = ρ 1 J n+ 1 ( α kρ ) for < α 1 < α < zeros of J n+ 1 n =, λ =, φ(x) = 1 n >, λ = ( βm ), φ(x) = J n ( βmx ) for < β 1 < β < zeros of J n(x) [ d sin φ dg ] ) + ( µ sin φ m g(φ) = µ = n(n + 1) 89*. dφ dφ sin φ = for n = m, m + 1, g() nd g(π) ounded g(φ) = Pn m (cos φ) (1 s ) d φ ( ) 9*. ds sdφ ds + µ m φ = µ = n(n + 1) 1 s = for n = m, m + 1, φ( 1) nd φ(1) ounded φ(s) = Pn m (s) * For m =, Pn m (s) = P n (s). (ρ) φ x + φ = λφ(x, y) y φ(, y) = φ(, y) = φ(x, ) = φ(x, ) = φ x + φ = λφ(x, y) y φ φ (, y) = (, y) = x x φ y φ (x, ) = (x, ) = y φ x + φ = λφ(x, y) y φ(, y) = φ(, y) = φ φ (x, ) = (x, ) = y y φ x + φ = λφ(x, y) y φ φ (, y) = (, ) = x x φ(x, ) = φ(x, ) = Two-Dimensionl Eigenvlue Prolems = λ = ( nπ ) + ( mπ ) φ(x) = sin nπx mπy sin = λ = ( nπ ) + ( mπ ) φ(x) = cos nπx mπy cos = λ = ( nπ ) + ( mπ ) φ(x) = sin nπx mπy cos = λ = ( nπ ) + ( mπ ) φ(x) = cos nπx mπy sin for n = 1,, nd m = 1,, for n =, 1, nd m =, 1, for n = 1,, nd m =, 1, for n =, 1, nd m = 1,, 95. φ x + φ = λφ(x, y) y φ(, y) = φ(, y) = φ (x, ) = φ(x, ) = y = λ = ( nπ ) + [ (m 1)π ] φ(x) = sin nπx (m 1)πy cos for n = 1,, nd m = 1,,

12 CHAPTER -1. HANDOUTS φ x + φ = λφ(x, y) y φ(, y) = φ(, y) = φ(x, ) = φ y (x, ) = φ x + φ = λφ(x, y) y φ φ (, y) = (, y) = x x φ (x, ) = φ(x, ) = y φ x + φ = λφ(x, y) y φ φ (, y) = (, y) = x x φ(x, ) = φ (x, ) = y = λ = ( nπ ) + [ (m 1)π ] φ(x) = sin nπx sin (m 1)πy = λ = ( nπ ) + [ (m 1)π ] φ(x) = cos nπx (m 1)πy cos = λ = ( nπ ) + [ (m 1)π ] φ(x) = cos nπx (m 1)πy sin for n = 1,, nd m = 1,, for n =, 1, nd m = 1,, for n =, 1, nd m = 1,, φ x + φ = λφ(x, y) y φ 99. (, y) = φ(, y) = x φ(x, ) = φ (x, ) = y φ x + φ = λφ(x, y) y φ 1. (, y) = φ(, y) = x φ (x, ) = φ(x, ) = y φ x + φ = λφ(x, y) y 11. φ(, y) = φ (, y) = x φ(x, ) = φ (x, ) = y 1. Green s Theorem (vector version) = λ = [ (n 1)π ] + [ (m 1)π ] φ(x) = cos (n 1)πx sin (m 1)πy = λ = [ (n 1)π ] + [ (m 1)π ] φ(x) = cos (n 1)πx cos (m 1)πy = λ = [ (n 1)π ] + [ (m 1)π ] φ(x) = sin (n 1)πx sin (m 1)πy Supporting Theorems for n = 1,, nd m = 1,, for n = 1,, nd m = 1,, for n = 1,, nd m = 1,, Let R e region in R ounded y piecewise-smooth, simple closed curve C with counterclockwise orienttion. Let F e vector field whose components hve continuous prtil derivtives on n open region contining R, then F da = F ˆn ds. 13. Divergence Theorem Let Ω e simple solid region in R 3 nd let Ω e its oundry with the outwrd orienttion. Let F e vector field whose components hve continuous prtil derivtives on n open region contining Ω, then F dv = F ˆn ds. Ω 14. If function f is continuous, f(x) nd f(x) for x, then Ω R ˆ C f(x) dx >.

13 CHAPTER -1. HANDOUTS For continuous nonnegtive function f if 16. Uniform Convergence Definition ˆ f(x) dx =, then f(x) = for x. The sequence of functions f n : D R, n = 1,,, is sid to converge uniformly to the function f : D R if for every ɛ >, there is nturl numer N such tht for ll x D we hve f n (x) f(x) < ɛ for ll n N. 17. Weierstrss M Test (A test for uniform convergence.) Suppose for ech function f n : D R, n = 1,,, there exists constnt M n with f n (x) M n for ll x D, nd M n converges. Then f n converges uniformly. 18. Interchnging Limit nd Integrl Suppose functions f n : [, ] R, n = 1,,, re continuous nd converge uniformly to [ˆ ] ˆ [ ] ˆ function f : [, ] R. Then lim f n (x) dx = lim f n(x) dx = f(x) dx. n n 19. Interchnging Integrl nd Summtion Suppose functions f n : [, ] R, n = 1,,, re continuous, nd Then ˆ [ ] f n (x) dx = [ˆ ] f(x) dx. 11. Interchnging Differentition nd Summtion Suppose functions f n, n = 1,,, re continuously differentile, nd [ f n converges uniformly. Then d ] f n (x) = dx f n converges uniformly. f n converges pointwise, [ ] d dx f n(x) Leiniz Integrl Rule (Interchnging differentition nd integrtion with respect to different vriles.) Suppose functions f(x, y) nd f (x, y) re continuous on [, ] [c, d]. Then y [ˆ d ] ˆ [ ] f(x, y) dx = f(x, y) dx. dy y 11. If f(x) = A If f(x) = 114. If f(x) = + n = 1 ˆ B n sin nπx A n cos nπx n cos nπx f(x) cos nπx Fourier Series for < x <, then A = 1 for < x <, then B n = + n sin nπx dx nd n = 1 ˆ ˆ ˆ f(x) dx nd A n = f(x) sin nπx dx for < x <, then = 1 f(x) sin nπx dx ˆ ˆ f(x) dx, f(x) cos nπx dx 115. If f(x) = A n cos (n 1)πx for < x <, then A n = ˆ f(x) cos (n 1)πx dx

14 CHAPTER -1. HANDOUTS If f(x) = B n sin (n 1)πx for < x <, then B n = ˆ Generlized Fourier Series f(x) sin (n 1)πx dx 117. Suppose < α 1 < α < re zeros of J z (x). If f(r) = m=1 m J z ( α mr ) for < r <, then m = ˆ f(r)j z ( αmr )r dr ˆ 118. Suppose < β 1 < β < re zeros of J (x). If f(r) = + m J ( β mr ) for < r <, then = m = ˆ ˆ m=1 f(r)j ( βmr )r dr J ( βmr )r dr = ˆ f(r)j ( βmr )r dr J (β m) 119. Suppose < β 1 < β < re zeros of J z(x) nd z >. If f(r) = m J z ( β mr ) for < r <, then m = 1. If f(φ) = n = 11. If f(φ) = n = 1. If f(φ) = k = m=1 ˆ f(r)j z ( βmr )r dr ˆ Jz ( βmr )r dr = ˆ f(r)j z ( βmr )r dr Jz ( αmr )r dr = ˆ [J z (β m ) J z 1 (β m )J z+1 (β m )] n Pn m (cos φ) for < φ < π nd m >, then n=m ˆ π ˆ π f(φ)p m n (cos φ) sin φ dφ = n+1 [Pn m (cos φ)] sin φ dφ n P n (cos φ) for < φ < π, then n= ˆ π ˆ π ˆ π ˆ π (n m)! (n+m)! ˆ π f(r) r dr nd f(φ)p m n (cos φ) sin φ dφ f(φ)p n (cos φ) sin φ dφ = n+1 [P n (cos φ)] f(φ)p n (cos φ) sin φ dφ sin φ dφ k P k (cos φ) for < φ < π, then k= ˆ π f(φ)p k (cos φ) sin φ dφ = (4k + 1) [P k (cos φ)] sin φ dφ ˆ π f(φ)p k (cos φ) sin φ dφ ˆ f(r)j z ( αmr )r dr J z+1 (α m)

15 CHAPTER -1. HANDOUTS If f(φ) = k = k P k 1 (cos φ) for < φ < π, then k=1 ˆ π ˆ π f(φ)p k 1 (cos φ) sin φ dφ = (4k 1) [P k 1 (cos φ)] sin φ dφ ˆ π f(φ)p k 1 (cos φ) sin φ dφ 14. If f(x, y) = B nm = If f(x, y) = A = 1 A m = 16. If f(x, y) = C n = 17. If f(x, y) = m=1 ˆ ˆ n= m= ˆ ˆ ˆ ˆ m= ˆ ˆ n= m=1 B nm sin nπx f(x, y) sin nπx A nm cos nπx Doule Fourier Series mπy sin mπy sin dy dx mπy cos f(x, y) dy dx, A n = f(x, y) cos mπy C nm sin nπx f(x, y) sin nπx C nm sin mπx for (x, y) (, ) (, ), then for (x, y) (, ) (, ), then ˆ ˆ dy dx nd A nm = 4 mπy cos f(x, y) cos nπx ˆ ˆ dy dx, f(x, y) cos nπx for (x, y) (, ) (, ), then dy dx nd C nm = 4 nπy cos (x, y) (, ) (, ), then C m = 1 C nm = 18. If f(x, y) = C nm = If f(x, y) = C nm = If f(x, y) = C m = ˆ ˆ m=1 ˆ ˆ f(x, y) sin mπx C nm cos m=1 ˆ ˆ n= m=1 ˆ ˆ cos nπy (n 1)πx f(x, y) cos (n 1)πx C nm sin nπx f(x, y) sin nπx C nm cos nπx f(x, y) cos (m 1)πy + cos m=1 ˆ ˆ ˆ ˆ D nm sin mπx f(x, y) sin nπx nπy cos f(x, y) sin mπx dy dx, dy dx nd D nm = (m 1)πy cos (m 1)πy (m 1)πy cos (m 1)πy cos dy dx (m 1)πy cos dy dx dy dx nd C nm = 4 ˆ ˆ mπy cos dy dx mπy cos dy dx for f(x, y) sin mπx for (x, y) (, ) (, ), then for (x, y) (, ) (, ), then for (x, y) (, ) (, ), then ˆ ˆ f(x, y) cos nπx sin nπy dy dx (m 1)πy cos dy dx

16 CHAPTER -1. HANDOUTS If f(x, y) = C nm = If f(x, y) = C m = 133. If f(x, y) = C nm = 4 m=1 ˆ ˆ n= m=1 ˆ ˆ m=1 ˆ ˆ C nm sin nπx f(x, y) sin nπx C nm cos nπx f(x, y) sin (m 1)πy C nm sin k= m=1 (m 1)πy sin (m 1)πy sin dy dx (m 1)πy sin (n 1)πx f(x, y) sin (n 1)πx dy dx nd C nm = 4 sin for (x, y) (, ) (, ), then (m 1)πy sin (m 1)πy for (x, y) (, ) (, ), then dy dx ˆ ˆ Generlized Doule Fourier Series f(x, y) cos nπx for (x, y) (, ) (, ), then Suppose < α 1 < α < re zeros of J z (x) If f(r, θ) = A mk J z ( α mr kπθ ) cos for (r, θ) (, ) (, ), then 135. If f(r, θ) = 136. If f(r, θ) = 137*. If f(θ, φ) = A m = J z+1 (αm)ˆ A mk = 4 J z+1 (αm)ˆ k=1 m=1 B mk = 4 J z+1 (αm)ˆ k= m=1 ˆ ˆ f(r, θ)j z ( αmr f(r, θ) cos kπθ )r dθ dr nd J z( αmr L )r dθ dr B mk J z ( α mr kπθ ) sin for (r, θ) (, ) (, ), then ˆ A mk J z ( α mr f(r, θ) sin kπθ ) cos kπθ + J z( αmr )r dθ dr k=1 m=1 (r, θ) (, ) (, ), then A m = 1 J z+1 (αm)ˆ A mk = J z+1 (αm)ˆ B mk = J z+1 (αm)ˆ A n = n+1 A nk = n+1 k= n=m ˆ π ˆ (n m)! (n+m)! ˆ π ˆ (n m)! (n+m)! ˆ ˆ f(r, θ) cos kπθ f(r, θ) sin kπθ B mk J z ( α mr kπθ ) sin for ˆ J z( αmr )r dθ dr nd J z( αmr )r dθ dr f(r, θ)j z ( αmr )r dθ dr, A nk Pn m (cos φ) cos kπθ for (θ, φ) (, ) (, π), then f(θ, φ)p m n (cos φ) sin φ dθ dφ nd f(θ, φ) cos kπθ P m n (cos φ) sin φ dθ dφ * It lso holds if k = m. For m =, P m n (s) = P n (s) nd (n m)! (n+m)! = 1. (m 1)πy sin dy dx

17 CHAPTER -1. HANDOUTS *. If f(θ, φ) = B nk = n+1 139*. If f(θ, φ) = A nk Pn m (cos φ) sin kπθ for (θ, φ) (, ) (, π), then k= n=m ˆ π ˆ (n m)! (n+m)! k= n=m f(θ, φ) sin kπθ A nk P m n (cos φ) cos kπθ (θ, φ) (, ) (, π), then A n = n+1 4 A nk = n+1 B nk = n+1 ˆ π ˆ (n m)! (n+m)! ˆ π ˆ (n m)! (n+m)! P m n (cos φ) sin φ dθ dφ + B nk Pn m (cos φ) sin kπθ for k=1 n=m ˆ π ˆ (n m)! (n+m)! f(θ, φ) cos kπθ P n m (cos φ) sin φ dθ dφ nd f(θ, φ) sin kπθ P n m (cos φ) sin φ dθ dφ * It lso holds if k = m. For m =, P m n (s) = P n (s) nd (n m)! (n+m)! = 1. Suppose < α 1 < α < re zeros of J z (x). 14**. If f(ρ, θ, φ) = A mkq J z ( α mρ q= k=l m=1 k+1 (k l)! then A mk = Jz+1 (αm) (k+l)! A mkq = 141**. If f(ρ, θ, φ) = then B mkq = 14**. If f(ρ, θ, φ) = (k+1) (k l)! Jz+1 (αm) (k+l)! q=1 k=l m=1 Generlized Triple Fourier Series ˆ ˆ π ˆ ˆ ˆ π ˆ ˆ (k+1) (k l)! Jz+1 (αm) (k+l)! q= k=l m=1 q=1 k=l m=1 )P k l qπθ (cos φ) cos f(ρ, θ, φ)j z ( αmρ f(ρ, θ, φ)j z ( αmρ f(θ, φ)p m n (cos φ) sin φ dθ dφ, for (ρ, θ, φ) (, ) (, ) (, π), )ρp k l (cos φ) sin φ dθ dφ dρ nd )ρp k l qπθ (cos φ) sin φ cos dθ dφ dρ B mkq J z ( α mρ )P k l qπθ (cos φ) sin for (ρ, θ, φ) (, ) (, ) (, π), ˆ π ˆ f(ρ, θ, φ)j z ( αmρ A mkq J z ( α mρ )P k l qπθ (cos φ) cos + for (ρ, θ, φ) (, ) (, ) (, π), then k+1 (k l)! A mk = Jz+1 (αm) (k+l)! k+1 (k l)! A mkq = Jz+1 (αm) (k+l)! k+1 (k l)! B mkq = Jz+1 (αm) (k+l)! B mkq J z ( α mρ )P k l qπθ (cos φ) sin ˆ ˆ π ˆ ˆ ˆ π ˆ ˆ ˆ π ˆ ** For l =, P l k (s) = P k(s) nd (k l)! (k+l)! = 1 f(ρ, θ, φ)j z ( αmρ f(ρ, θ, φ)j z ( αmρ f(ρ, θ, φ)j z ( αmρ )ρp l k )ρp k l qπθ (cos φ) sin φ sin dθ dφ dρ (cos φ) sin φ dθ dφ dρ, )ρp k l qπθ (cos φ) sin φ cos dθ dφ dρ nd )ρp k l qπθ (cos φ) sin φ sin dθ dφ dρ

18 CHAPTER -1. HANDOUTS 14 Suppose < α 1 < α < re zeros of J z (x). 143*. If f(ρ, θ, φ) = q= k=l m=1 q=1 k=l m=1 A mkq ρ 1 Jz+ 1 B mkq ρ 1 Jz+ 1 for (ρ, θ, φ) (, ) (, ) (, π), then A mk = k+1 J z+ 3 A mkq = k+1 J z+ 3 B mkq = k+1 J z+ 3 ˆ ˆ π ˆ (k l)! (α m) (k+l)! ˆ ˆ π ˆ (k l)! (α m) (k+l)! ˆ ˆ π ˆ (k l)! (α m) (k+l)! * For l =, P l k (s) = P k(s) nd (k l)! (k+l)! = 1 ( α mρ )P k l qπθ (cos φ) cos + f(ρ, θ, φ)j z+ 1 f(ρ, θ, φ)j z+ 1 ( α mρ )P k l qπθ (cos φ) sin f(ρ, θ, φ)j z+ 1 ( αmρ )ρ 3 Pk l (cos φ) sin φ dθ dφ dρ, ( αmρ )ρ 3 Pk l qπθ (cos φ) sin φ cos dθ dφ dρ nd ( αmρ )ρ 3 Pk l qπθ (cos φ) sin φ sin dθ dφ dρ

19 Chpter Introduction (Not corresponding to the course textook) Definitions: 1. A prtil differentil eqution (PDE) is n eqution F (u, u x, u y,, u xx, u xy, ) = G(x, y, ) involving independent vriles x, y,, function u of these vriles nd the prtil derivtives u x, u y,, u xx, u xy,, of the function. Also, functions of independent vriles my e used s coefficients for function u nd its prtil derivtives.. The order of PDE is the order of the prtil derivtive of highest order ppering in the eqution. 3. A function u(x, y, ) is clled solution of the PDE if the PDE ecomes n identity in the independent vriles when u nd its prtil derivtives re sustituted in the PDE. 4. A PDE is clled homogeneous if G, (no independent vrile ppers y itself). 5. A PDE is clled liner if for ll constnts α nd β nd functions u nd v we hve F (w, w x, w y,, w xx, w xy, ) = αf (u, u x, u y,, u xx, u xy, ) + βf (v, v x, v y,, v xx, v xy, ) where w = αu + βv. Exmples: 1. t u x = is nonliner homogeneous first-order PDE. For nonlinerity, show F (αu + αv) αf (u) + βf (v) for prticulr vlues of α, β nd functions u nd v, where F (u) = t u x. Clssroom discussion!. u xx + u yy = 6x is liner nonhomogeneous second-order PDE nd u(x, y) = x 3 + x y nd u(x, y) = x 3 + e x cos y re two solutions of it.

20 CHAPTER. INTRODUCTION 16 Clssroom discussion! 3. u xx = 1 k u t is homogeneous liner PDE of order. Here F (u) = u xx 1 k u t. Clssroom discussion! Exmples: 1. Find the solution u(x, y) of x y sin x =. For how to solve seprle ODE see the Review, Identities, Formuls nd TheoremsHndout. Clssroom discussion!. Find the solution u(x, y) of u xx u = which stisfies the uxiliry conditions; u(, y) = y + 6 nd u x (, y) = y. For how to solve nd order liner ODE with constnt coefficients see the Review, Identities, Formuls nd TheoremsHndout. Clssroom discussion!

21 CHAPTER. INTRODUCTION 17 Exercises: 1. In exmple (1) ove, hve we found ll of the solutions?. Find the solution u(x, y, z) of u x y sin x =. 3. Find the solution u(x, y) of u u xy + u x u y =. Hint: Notice tht it is n exct ODE with respect to y derivtive whose solution leds to seprle eqution. For how to solve n exct ODE see the Review, Identities, Formuls nd TheoremsHndout. Sometimes we cn find infinitely mny solutions. For exmple consider u x + u y =. Functions u n (x, y) = (x y) n, n =, 1,, stisfy the PDE, so perhps their infinite liner comintion u(x, y) = n= c nu n (x, y) will lso e solution of this PDE. Notice tht if we tke c n = 1 n!, then u(x, y) = (x y) n n= n! = e x y which is solution of our PDE. The typicl prolem is to find solution of PDE which stisfies certin uxiliry conditions, for exmple; Auxiliry Conditions u xx = 1 k u t, < x <, } t > (Het Eqution) u(, t) = T, t > Boundry Conditions u(, t) = T 1, t > u(x, ) = f(x), < x < Initil Condition Our min solution technique will e the method of seprtion of vriles, lso clled product method nd Fourier s method. Exmple. Solve u t u u x = y seprtion of vriles. Assume u(x, t) = X(x)T (t), plug into the PDE nd simplify to get X (x) = T (t) T (t). Since the left hnd side (LHS) is function of x nd RHS is function of t, this equlity will hold only if they re equl to constnt, sy λ. Solve X (x) = λ nd T (t) = λ nd plug ck in the function u. Clssroom discussion! T (t) Exercise. Solve u t = u xx y the method of seprtion of vriles. Hint: Consider the cses λ >, λ = nd λ <.

22 Chpter 1 Fourier Series (Fourier Series nd Integrls in the course textook) 1.1 Periodic Functions nd Fourier Series Definition. A function f is sid to e periodic with positive period p if 1. f(x) hs een defined for ll x, nd. f(x + p) = f(x) for ll x. For periodic function f with period p, it is esy to show tht f(x np) = f(x) = f(x+np) for n =, 1, nd thus periodic function, defined s ove, hs mny periods! Clssroom discussion! Exmples: 1. PUT GRAPH HERE!. PUT GRAPH HERE! 3. Functions sin x nd cos x re π-periodic. Functions sin πx p of the function sin 5πx 3 is π π = nd cos πx p hve period p nd the period Let f e -periodic function (of period ). In this chpter we wnt to find constnts, n, n, n = 1,, such tht ( f(x) = + n cos nπx + n sin nπx ). Orthogontlity Reltions Assume m nd n re nonnegtive integers, unless stted otherwise. ˆ π ˆ π π ˆ π π ˆ π π π ˆ π π sin nx dx = for ll n cos nx dx = {, n π, n = sin nx cos mx dx = for ll n nd m {, n m sin nx sin mx dx = π, n = m cos nx cos mx dx =, n m π, n = m π, n = m =

23 CHAPTER 1. FOURIER SERIES 19 See Review, Identities, Formuls nd Theorems. These re esy to check, (do it). We will need the following identities. See Review, Identities, Formuls nd Theorems. sin cos = 1 (sin( + ) + sin( )) cos cos = 1 (cos( + ) + cos( )) sin sin = 1 (cos( ) cos( + )) We will use the first identity to prove the third orthogonlity reltion ove. ˆ π π sin nx cos mx dx = ˆ π π This cn e treted in more generl form. ( 1 sin(n + m)x + 1 sin(n m)x) dx =, using the first orthogonlity reltion. Definitions: 1. Function f(x) is n even function if f( x) = f(x) or equivlently its grph is symmetric out y-xis.. Function f(x) is n odd function if f( x) = f(x) or equivlently its grph is symmetric out the origin. Then we hve ˆ ˆ (even function) dx = (odd function) dx = ˆ (even function) dx nd cos x is n even function, while sin x is n odd function. We cn lso think of even s + nd odd s - in the following sense: even odd = odd, odd odd = even, even even = even, even + even = even nd odd + odd = odd. Ech function cn e written s the sum of n even function nd n odd function. f(x) = 1 (f(x) + f( x)) + 1 (f(x) f( x)) } {{}} {{} even odd Therefore since sin nx is odd nd cosmx is even, their product is odd nd so π π sin nx cos mx dx =. Now suppose f is π-periodic function nd tht f(x) = + ( n cos nx + n sin nx). By interchnging the order of integrtion nd infinite sum, we cn show the following. = 1 ˆ π f(x) dx = π π {, f odd f(x) dx, f even 1 π π For ny fixed integer vlue m = 1,,, m = 1 ˆ π {, f(x) cos mx dx = f odd π, nd π f(x) cos mx dx, f even m = 1 π π ˆ π π Clssroom discussion! π π { f(x) sin mx dx = π f(x) sin mx dx, f odd, f even

24 CHAPTER 1. FOURIER SERIES

25 CHAPTER 1. FOURIER SERIES 1 Definition. Fourier series (F.S.) of π-periodic function f(x) is defined s f(x) + ( n cos nx + n sin nx) where = 1 π π π f(x) dx, n = 1 π π π f(x) cos nx dx nd n = 1 π π π f(x) sin nx dx for n = 1,,. Question: Does this series ctully represent function f(x)? Exmple. Find F.S. of f(x) = x for π < x < π nd f(x + π) = f(x). f(x) is n even function with period π nd the grph s shown. PUT THE GRAPH HERE. So, n = nd = 1 π π π π π f(x) dx == 1 π π f(x) dx = = π, n = 1 π x cos nx dx = = πn (cos nπ) 1 = nd n 1 = 4 (n 1) f(x) π 4 π Clssroom discussion! π π f(x) cos nx dx = π π f(x) cos nx dx = { 4, n odd πn, n even, using integrtion-y-prts. Or, n = for n = 1,,. Thus 1 cos(n 1)x. (n 1)

26 CHAPTER 1. FOURIER SERIES 1. Aritrry Period nd Hlf-Rnge Expnsions Definition. Let f e periodic function of period, then its F.S. is f(x) + ( n cos nπx + n sin nπx ) where = 1 f(x) dx, n = 1 nπx f(x) cos dx nd n = 1 nπx f(x) sin dx for n = 1,,. Exercise. As we did for the π-periodic functions, derive ove equtions for, n, n, n = 1,,. Wht { if f is defined only on finite intervl, sy (, )? For exmple, let s find the F.S. of f(x) = 1, < x <. f(x) is not defined on ll of the R (rel numers), so to use the ove formuls, 1, < x < we extend f to ll of R in such wy tht it will hve period. Cll this new function f. F.S. of f on (, ) is the F.S. of f on (, ). PUT GRAPH HERE Note. More often thn not we simply cll this new function f gin. n π x f(x) sin Since f is n odd function, = n = nd n = {, n even 4 4 n π, n odd or n = nd n 1 = (n 1)π, n = 1,,. Thus f(x) Clssroom discussion! dx = = n π (cos nπ 1) = 4 (n 1)π (n 1)πx sin. If function f(x) is not defined on the intervl (, ), then we mke either n odd or n even extension of f to ll of R. Exmples: Mke even nd odd extensions of the following functions to the entire rel-numer line. 1. f(x) = x, < x < 1. f(x) = sin x, < x < π 3. f(x) = (x ), < x < 3 PUT ALL GRAPHS HERE These extensions re not unique, for exmple, for exmple 3 we could use the following extensions. PUT GRAPHS HERE Since in either cse evlution of s nd s will only involve integrtion on (, ), we cll the F.S. for these cses Hlf-Rnge Expnsion. Definitions: Let f e function defined on (, ). { f(x), < x < 1. The odd extension of f to (, ) is f o (x) = f( x), < x <. The F. sine series of f is the F. series of f o. { f(x), < x <. The even extension of f to (, ) is f e (x) = f( x), < x <. The F. cosine series of f is the F. series of f e. Exmples: Consider the function f, in exmple ove. 1. Find the F. cosine series of f. Using f e, the even extension of f to ll of R, we hve

27 CHAPTER 1. FOURIER SERIES 3 n = = π n = π n = ˆ π ˆ π sin x dx = = π sin x cos nx dx = = 4 π(4n 1), n 1 =, n = 1,, f(x) π Clssroom discussion! 4 (4n cos nx 1)π, n = 1, n > 1 nd n odd = 4 π(n 1), n > 1 nd n even. Find the F. sine series of f. The odd extension of f to ll of R is sin x itself. So, F. sine series of f should e f(x) sin x. Question. At x = π, sin x = 1 nd 4 (4n 1)π cos nx = { x Exercise. Find the F. sine series of f(x) =, < x x, x <. Answer: f(x) 8 π sin n π n Definitions: Let function f e defined on (, ). sin n π x = 8 ( 1) n+1 π sin (n 1) π x (n 1) 4( 1) n (4n 1)π. Is π 4 π 1. The Fourier sine series of f is n sin n π x where n = f(x) sin n π x dx.. The Fourier cosine series of f is + n cos n π x where = 1 ˆ f(x) dx nd n = ˆ f(x) cos n π x dx. ˆ ( 1) n (4n 1) = 1? 1.3 Convergence of Fourier Series Definition. Let f(x) e function nd x R. We sy tht lim f(x) exists if x x 1. left limit exists f(x ) = lim f(x) = lim f(x + h) = lim f(x + h) exits, x x h h h<

28 CHAPTER 1. FOURIER SERIES 4. right limit exists f(x + ) = lim f(x) = lim f(x + h) = lim f(x + h) exits, nd x x + h + h 3. the ove two limits re equl f(x ) = f(x+ ). Then lim x x f(x) = f(x ) = f(x+ ). Definition. Function f(x) is continuous t x if 1. f(x ) exits,. lim f(x) exits, nd x x 3. f(x ) = lim f(x). f(x ) = f(x+ ) = f(x ) x x Definition. A function is continuous (everywhere) if it is continuous t ech point. Types of Discontinuity t x - 1. Removle discontinuity; f(x ) = f(x+ ) f(x ), (f(x my not e defined.). Jump discontinuity; f(x ) f(x+ ), ut oth exist. 3. Bd discontinuity; f(x ), f(x+ ) or oth fil to exist. Exmples: 1.. PUT GRAPHS HERE h> PUT GRAPHS HERE In the cse of removle discontinuity t x, if we redefine our function t x to e f(x ) = lim x x f(x), then this new function is continuous t x, hence the terminology removle discontinuity. Definition. A function is sectionlly continuous on finite intervl (, ), if it is ounded nd continuous on (, ), except possily for finite numer of jumps nd removle discontinuities. Definition. A function is sectionlly continuous if it is sectionlly continuous on every finite intervl. Remrk. Another nme for sectionlly continuous is piecewise continuous. Exmples: Addition to the prts of the lst exmple. 1. Function f is not sectionlly continuous on (, x ), so it is not sectionlly continuous. 4. Function f is sectionlly continuous on ech finite intervl, so it is not sectionlly continuous. If function defined on finite intervl is sectionlly continuous, then its periodic (odd or even) extension is lso sectionlly continuous, for exmple consider PUT GRAPHS HERE Definition. A function f is sectionlly smooth if 1. f is sectionlly continuous,. f (x) exists for every x, except perhps t finite numer of points, nd 3. f (x) is sectionlly continuous. Exmples: 1. f(x) = x 1 3, 1 < x < 1. f is continuous on, f is not continuous on nd therefore f is not sectionlly smooth. Clssroom discussion!

29 CHAPTER 1. FOURIER SERIES 5. f(x) = x, 1 < x < 1. f is continuous, lthough f () does not exist f is sectionlly continuous nd therefore f is sectionlly smooth. Clssroom discussion! Theorem (Convergence Theorem, Function Hypotheses). If f(x) is sectionlly smooth nd periodic with period, then t ech point x the F.S. corresponding to f converges nd + ( n cos nπx + n sin nπx ) = 1 ( f(x ) + f(x + ) ). Remrk. If f is continuous t x, then 1 (f(x ) + f(x + )) = f(x). 4( 1) n Exmple. Show tht 4n 1 = π. Apply the ove theorem to the F. cosine series of f(x) = sin x, < x < π nd then plug in x = π. Clssroom discussion!

30 CHAPTER 1. FOURIER SERIES 6 Exmple. The grph of periodic function f is shown elow. Drw the grph of its F. series. PUT GRAPH HERE Clssroom discussion! Remrk. The ove exmples show the power of the Convergence Theorem. ( 1) n Exercises: 1. Use the F.S. of f(x) = sin x to show tht 4n 1 = 1 π 4.. Use the F. cosine series of f(x) = x ( 1) n+1, < x < π, to show tht 3. Use the F. cosine series of f(x) = x 4, < x < π, nd 4. Consider the -periodic function f with f(x) = (Do not compute its F.S. coefficients.) { 1, x < n = π 1 nd 1 n = π 6 to show tht 3, x < 1 n 4 = π n = π 6.. Find its F. series grphiclly. It is lso useful to stte convergence theorems for the F. series when f is defined on (, ) nd oth F. sine nd cosine series when f is defined on (, ). We need to find the conditions for which the desired extension of f meets the hypotheses of the Convergence Theorem. We must lso py specil ttention to the endpoints. Exercises: Fill in the lnk. 1. Let f(x) e function defined on (, ). If f is, then F.S. corresponding to f converges nd + ( n cos nπx + n sin nπx ) = 1 ( f(x ) + f(x + ) )

31 CHAPTER 1. FOURIER SERIES 7 for < x <. At x = ±, the F.S. converges to.. Let f e function defined on (, ). If f is, then F. sine series corresponding to f converges nd n sin nπx = 1 ( f(x ) + f(x + ) ) for < x <. At x = nd x =, the F. sine series converges to. 3. Let f(x) e function defined on (, ). If f is, then F. cosine series corresponding to f converges nd + n cos nπx = 1 ( f(x ) + f(x + ) ) for < x <. At x =, the F. cosine series converges to converges to. 1.4 Uniform Convergence. At x =, the F. cosine series Definitions: Consider functions f n (x), n = 1,, defined on the intervl I. N 1. We sy tht f n (x) converges to f(x) pointwise in the intervl I if t ech point x in I, lim f n (x) f(x) N =. N f n (x) is clled the Nth prtil sum of. We sy tht f n (x) nd is denoted y S N (x); sum of the 1st N term. f n (x) converges to f(x) uniformly in the intervl I if lim Mx S N(x) f(x) =. N x I Note. This mximum might not exist, in tht cse we use supremum; the lest numer greter thn S N (x) f(x) for every x, in plce of it. Lemm. If f n (x) converges uniformly to, sy, f(x) nd if f n (x) re continuous functions, then f(x) is lso continuous. Exmples: Exmine convergence of F.S. of following functions grphiclly. { 1, < x < π 1. f(x) = 1, π < x < 4 f(x) sin(n 1)x (n 1)π PUT ALL GRAPHS HERE F.S. does not converge uniformly to f(x).. g(x) = x, π < x < π g(x) π 4 1 cos(n 1)x π (n 1) F.S. converges uniformly to g(x). Theorem 1. (Convergence Theorem, F. Coefficients Hypotheses) ( Consider the series + n cos nπx + n sin nπx ). If ( n + n ) converges then this series converges uniformly (nd hence to continuous function) nd if it is the F.S. of the function f(x), it converges uniformly to f(x).

32 CHAPTER 1. FOURIER SERIES 8 Exmple. Use the ove theorem to show tht the F.S. of g(x) in the lst exmple converges uniformly to g(x). Clssroom discussion! Remrk. Notice tht for the function f in the lst exmple, ( n + n ) = converge nd hence the ove uniform convergence theorem does not hold. 4 (n 1)π does not The following theorems stte conditions under which F.S. of function converges uniformly. These conditions do not involve the F.S. itself. Theorem. If function f(x) is periodic, continuous, nd hs sectionlly continuous derivtive, then F.S. corresponding to f converges uniformly to f on the entire rel xis. Theorem 3. Let f(x) e function defined on (, ) such tht 1. It is continuous, ounded (dd f(x) < M for ll x nd some M > ) nd hs sectionlly continuous derivtive, nd. f(( ) + ) = f( ). Then the F.S. of f converges uniformly to f on the intervl (, ). (F.S. converges to f(( ) + ) = f( ) t x = ±.) Theorem 4. Let f(x) e function defined on (, ) such tht 1. It is continuous, ounded, nd hs sectionlly continuous derivtive, nd. f( + ) = f( ) = Then the F. sine series of f converges uniformly to f in the intervl (, ). (F. sine series converges to zero t x = nd x =.) Theorem 5. Let f(x) e function defined on (, ) such tht it is continuous, ounded, nd hs sectionlly continuous derivtive. Then the F. cosine series of f converges uniformly to f in the intervl (, ). (F. cosine series converges to f( + ) t x = nd to f( ) t x =.) Exercises: 1. Show tht f(x) = x, π < x < π, stisfies the hypothesis of theorem 3.. Show tht f(x) = sin x, < x < π, stisfies the hypothesis of oth theorems 4 nd 5. Wht re its F. sine nd cosine series? Are they equl? 1.5 Opertions on Fourier Series Let f(x) e -periodic function nd + Theorem 1. ( n cos nπx + n sin nπx ) its Fourier series. The F.S. of function cf(x) hs coefficients c, c n nd c n, where c is ny constnt.

33 CHAPTER 1. FOURIER SERIES 9 Theorem. The Fourier coefficients of the sum f(x) +g(x) re the sums of the corresponding coefficients of F.S. of f(x) nd g(x). Exercise. Prove theorems 1 nd, y direct computtion of F. coefficients of cf(x) nd f(x) + g(x). Theorem 3. If f(x) is -periodic sectionlly continuous function, then F.S. of f my e integrted term y term. Tht is,ˆ d f(x) dx = c c c ˆ d c dx + ˆ d c ( n cos nπx c + n sin nπx ) dx. Theorem 4. If f(x) is -periodic sectionlly continuous function nd function g(x) is lso sectionlly continuous on (c, d), then ˆ d ˆ d ˆ d ( f(x)g(x) dx = g(x) dx + n cos nπx + n sin nπx ) g(x) dx. Remrks: 1. The hypotheses in theorems 3 nd 4 re weker thn tht of the Convergence Theorem. We re not requiring convergence of the F.S. to the function!. The term-y-term integrtion of F.S. my not result in nother F. series! Theorem 5. (Uniqueness Theorem) If f(x) is periodic nd sectionlly continuous, then its F.S. is unique. Remrk. If f(x), < x < is sectionlly continuous, then its F. sine nd cosine series re unique. Exmple. Consider f(x) = x, π < x < π. Its F.S. is f(x) = π 4 π due to convergence theorem). 1. Find F.S. of h(x) = π. Evlute 3. Evlute ˆ x ˆ x 8 π 4 f(x), π < x < π. h(t) dt, y use of theorem 3. h(t) dt, < x < π directly. 4. Find F. sine series of g(x) = π 8 x(π x), < x < π. 5. Show tht π π g(x) sin(n 1)x dx = 1, n = 1,,. (n 1) 3 6. Show tht ( 1) n+1 (n 1) 3 = π (n 1) cos(n 1)x (equlity is

34 CHAPTER 1. FOURIER SERIES 3 Clssroom discussion! In the following exercise, we will see some issues tht rise from term-y-term integrtion of F.S., including not eing F. series. Exercise. The F. sines series of f(x) = x, < x < π, is x = due to the convergence theorem.) 1. Show x = 4( 1) n n + 4( 1) n n cos nx, < x < π. ( 1) n+1 n sin nx, x < π. (Equlity is

35 CHAPTER 1. FOURIER SERIES 31 4( 1). Find the vlue of n y noticing tht it is the constnt in the F. cosine series of g(x) = x, n < x < π. 3. Using erlier prts, show tht the F. cosine series of g(x) = x, < x < π, is g(x) = π 3 + 4( 1) n cos nx, x π, nd use it to show x 3 = π 1( 1) x + n sin nx, < x < π. (Note: The n n 3 series π x + 1( 1) n n 3 sin nx is not F. series!) 4. Use erlier prts to find the F. sine series of h(x) = x 3, < x < π. The following exercise demonstrtes very interesting property of F. coefficients. Exercises: 1. Let f e n odd periodic sectionlly continuous function with period. Show tht ˆ f (x) dx = n, where n s re the coefficients of F. sine series of f. 1 Remrk. This is form of Prsevl s equlity.. Let f e n odd periodic sectionlly continuous function with period. Show tht lim n =, where n n s re the coefficients of F. sine series of f. Hint: Divergence Test - If c n converges, then lim c n =. n (If lim c n, then c n diverges.) n We will use the following result when we pply comprison theorem to series involving F. coefficients. Lemm. If sequence { n } converges then it is ounded. Tht is, there exists numer M > such tht n < M for every n. The following exmple demonstrtes tht term-y-term differentition of F. series is not lwys possile. Exmple. The F. sines series of f(x) = x, < x <, is x = nπ ( 1)n+1 sin nπx, x <. (Equlity is due to the convergence theorem.) Show tht the F. series of f (x) is not the term-y-term differentited F. series of f(x). Clssroom discussion! Theorem 6. (Term-y-Term Differentition Theorem, Function Hypotheses) If f(x) is -periodic, continuous, nd sectionlly smooth, then the term-y-term differentited F.S. of f(x) converges to f (x) t every point x where f (x) exists. f (x) = ( nπ n sin nπx + nπ n cos nπx ), where f (x) exists.

36 CHAPTER 1. FOURIER SERIES 3 Exmple. Find the derivtive of π 4 π π < x < π. Clssroom discussion! 1 (n 1) cos(n 1)x using the fct tht it is the F.S. of f(x) = x, Theorem 7. (Term-y-Term Differentition Theorem, F. Coefficients Hypotheses) If f(x) is periodic ( with F. coefficients, n nd n, nd if the series n k n + n k n ) converges for n integer k 1, then f hs continuous derivtives f,, f (k) whose F.S. re the corresponding term-y-term differentited series of f. Remrk. Suppose the ove theorem holds. Then, the F. coefficient of f (k) (x) re ±( nπ )k n nd ±( nπ )k n. ( Also, ± ( nπ )k n + ± ( nπ )k n ) = ( π ( )k n k n + n k n ) converges. This mens tht not only the F.S. of f (k) (x) is otined y k term-y-term differentition of the F.S. of f(x), ut lso tht the F.S. of f (k) (x) is equl to the function f (k) (x) itself, due to theorem 1 in section 1.4. Exmple. Given u(x) Me nt sin nx, where M nd t re fixed positive constnts. Find the F. series of d u dx. Discuss the convergence of the F. series of d u dx. Clssroom discussion! The theorems in this section for interchnging the order of summtion nd integrtion or differentition re for F. series only nd re sed on the following generl results. Theorem 8. (Interchnging Differentition nd Summtion) See Review, Identities, Formuls nd The- Theorem 9. orems. (Interchnging Integrl nd Summtion) See Review, Identities, Formuls nd Theorems.

37 CHAPTER 1. FOURIER SERIES 33 A convenient wy to show uniform convergence is through Weierstrss M Test. Theorem 1. (Weierstrss M Test) See Review, Identities, Formuls nd Theorems. Note. Theorems 8-1 re not in the textook, ut re stted in Review, Identities, Formuls nd Theorems. In the prolem elow, we need to find the derivtive of series which is not F.S. with the respect to the vrile we must differentite. In this cse, we will use the ove theorem for interchnging differentition nd summtion. Exmple. (Mthemticl Justifiction) Let f e n odd, periodic, sectionlly smooth function with F. sine series coefficients n, n = 1,,. Show tht the function defined y u(x, t) = n e nt sin nx stisfies the following.. u x = t, < x < π, t >. u(, t) = u(π, t) =, t > c. u(x, ) = 1 (f(x ) + f(x + )), < x < π Clssroom discussion!

38 CHAPTER 1. FOURIER SERIES 34

39 Chpter The Het Eqution.1 Derivtion nd Boundry Conditions We wnt to otin the eqution governing the flow of the het in thermlly conducting rod whose solution gives the temperture t ny given position on the rod t ny given time. Assume the rod hs uniform cross section nd tht the temperture does not vry form point to point in cross section. Therefore the temperture in the rod will only depend on position x nd time t. PUT GRAPH HERE We will mke use of the following. 1. The lw of Conservtion of Energy - The mount of het which enters region plus wht is generted inside is equl to the mount of the het which leves plus the mount stored; Het in + Het generted = Het out + Het stored. This is eqully vlid in terms of rtes of het per unit of time insted of mounts of het.. The rte of het stored in ody is proportionl to the mss of tht ody nd to the rte of chnge of temperture. 3. Fourier s Lw - Het flows in the direction of decresing temperture t rte proportionl to the grdient of the temperture, (so het flow is positive when temperture grdient is negtive.) Nottion - H = Het: clorie, Joule,... ; t = time: second,...; T = Temperture: C, F,... ; m = Mss: grm, slug,... ; L = length: cm, ft,... ; ρ = Density = volume mss : grm,... ; cm 3 c = Het Cpcity per unit of Mss: cl grm C,... ; κ ( kpp ) = Therml Conductivity: cl sec cm C,... ; k (smll letter k) = Diffusivity = κ c ρ : cm sec. Consider slice of the rod which lies etween x nd x + x. Let q(x, t) e the rte of het flow t point x nd time t: cl,..., nd ssume q is positive when het flow to the right. Let u(x, t) e the temperture sec cm t point x nd time t: C,.... PUT GRAPH HERE The rte t which het is entering the slice through the surfce t x with re A is Aq(x, t): cl sec,..., nd the rte t which het is leving the slice through the surfce t x + x is Aq(x + x, t): cl sec,....

40 CHAPTER. THE HEAT EQUATION 36 The het stored in the slice is c ρ A x }{{} mss (x, t) } t {{} rte of chnge of temp : cl sec,.... If the rte of the het generted per unit of volume is g: cl in the slice is A x g: sec,.... Thus, y the lw of Conservtion of Energy we hve A q(x, t) + A x g = A q(x + x, t) + c ρ A x q(x + x, t) q(x, t) g = c ρ x t Tking the limit of oth sides s x we get q g = c ρ x t. cl sec cm 3,..., then rte t which het is generted t, cl sec By the Fourier s lw q = κ q x, nd so x = κ u. Thus x u x + g κ = c ρ κ t, t >. If no het is generted, g =, nd letting k = κ c ρ u x = 1 k, < x <, t >. t we get, nd so Now, u = 1 x k t, < x <, t >, descries the temperture u in rod of length with uniform properties nd cross section, in which no het is generted, nd whose cylindricl surfce is insulted. This eqution hs mny solutions: u(x, t) = x + kt, u(x, t) = e kt sin x. We wnt to hve unique solution, therefore we plce uxiliry conditions on our PDE: 1. The initil temperture distriution in the rod, (clled initil condition, I.C., or initil vlue, I.V.).. Wht is hppening t the ends of the rod, (oundry condition, B.C., or oundry vlue, B.V.)? I.C.: u(x, ) = f(x), < x <. B.C.: 1. Dirichlet B.C. (lso clled B.C. of the 1st kind) u(, t) = T, u(, t) = T 1, t > (fixed, end tempertures). Neumnn B.C. (lso clled B.C. of the nd kind) x (, t) = φ 1(t), x (, t) = φ (t), t > φ 1 or φ = corresponds to n insulted surfce t the end x = or x =, respectively. 3. Roin B.C. (lso clled B.C. of the 3rd kind) c 1 u(, t) + d 1 x (, t) = ψ 1(t), t > c u(, t) + d x (, t) = ψ (t), t > For convection t the end x =, h u(, t)+κ x (, t) = h T cl 1(t), where h is convection coefficient: cm sec C,..., nd T 1 (t) is the temperture of the medium surrounding the end t x =. Similrly for convection t

41 CHAPTER. THE HEAT EQUATION 37 the end x =, h u(, t) κ x (, t) = h T (t). Clssroom discussion! If B.C. involves more thn one oundry point, it is clled mixed oundry condition. For exmple: u(, t) = u(, t) nd x (, t) = x (, t). Of course, there re mny more possile oundry conditions. An Initil Vlue - Boundry Vlue Prolem: u x = 1 k t, < x <, t > u(, t) = T, t > hu(, t) + κ x (, t) = h T 1, t > u(x, ) = f(x), < x < There is exctly one nd only one solution to complete I.V. - B.V. prolem.. Stedy-Stte Tempertures The stedy-stte temperture distriution is time independent function v(x) which is solution of the time independent het eqution tht stisfies the B.C. s. Physiclly, when het conduction through ody is left undistured for long time, the vrition of the temperture with respect to time dies out nd we chieve stedy-stte temperture distriution. In this cse we expect lim u(x, t) = v(x) nd lim t t t =. Exmples: Stte nd solve (find) the stedy-stte prolem corresponding to ech of the following. 1. u x = 1 k t, < x <, t > u(, t) = T, u(, t) = T 1, t > u(x, ) = f(x), The S-S prolem is d v dx =, v() = T, v() = T 1 < x < < x < nd its solution is v(x) = (T 1 T ) x + T. Clssroom discussion!