Solution Set IX for 18.075-Fall 2004

SOLUTION SET IX FOR 18.075–FALL 2004 5. Boundary-Value Problems and Characteristic-Function Representations 5.6. Orthogonality of Characteristic Functions. 22. Reduce each of the following differential equations to the standard form λr)y = 0: dy d2 y (a) x dx 2 + 2 dx + (x + λ)y = 0, (b) (c) (d) d2 y dy + dx cot x + λy = 0, dx2 2 dy d y + a dx + (b + λ)y = 0 (a, b dx2 dy d2 y − ay + λy = x dx2 + (c − x) dx dy d dx (p dx ) + (q + constants), 0 (a, c constants). Solution. (a) x 2 d2 y dy dy 2d y + 2 + (x + λ)y = 0 ⇔ x + 2x + (x2 + λx)y = 0 dx2 dx dx2 dx d 2 dy ⇔ (x ) + (x2 + λx)y = 0 dx dx (b) dy dy d2 y d2 y + + cos x cot x + λy = 0 ⇔ sin x + λ sin x · y = 0 2 2 dx dx dx dx d dy ⇔ (sin x ) + λ sin x · y = 0 dx dx (c) 2 dy dy d2 y ax d y + a + a · eax + (b + λ)y = 0 ⇔ e + (b · eax + λ · eax )y = 0 dx2 dx dx2 dx d ax dy ⇔ (e ) + (b · aeax + λ · eax )y = 0 dx dx Date: December 1, 2002. 1 2 SOLUTION SET IX FOR 18.075–FALL 2004 (d) d2 y dy + (c − x) − ay + λy = 0 2 dx dx d2 y c dy −ae−x xc λe−x xc ⇔ e−x · xc 2 + e−x · xc ( − 1) +( + )y = 0 dx x dx x x d −x c dy ⇔ (e x ) + (−ae−x xc−1 + λe−x xc−1 )y = 0 dx dx x 23. By considering the characteristic functions of the problem d2 y + µ2 y = 0, y(0) = 0, αly 0 (l) + y(l) = 0, dx2 with α > 0, and using the results of Section 5.6, show that Z l sin(µ1 x) sin(µ2 x) dx = 0 0 when µ1 and µ2 are distinct positive solutions of the equation tan(µl) + αµl = 0. 2 d y 2 Solution. The general solution of dx 2 + µ y = 0, where µ 6= 0, is y = c 1 cos µx + c2 sin µx. µ2 (6= 0) is a characteristic value only if (1) y(0) = 0 ⇒ c1 cos 0 + c2 sin 0 = 0, and (2) αly 0 (l) + y(l) = 0 ⇒ αl[c1 cos(µx) + c2 sin(µx)]0 |x=l + [c1 cos(µl) + c2 sin(µl)] = 0 Equation (1) implies c1 = 0, c2 6= 0 and (2) gives αlµ cos(µl) + sin(µl) = 0 which is equivalent to (3) tan(µl) + αµl = 0 So the nonzero characteristic values of the given problem are the non-zero solutions of (3). We can easily see that if µ is a solution of (3) then y = sin(µx) is a solution of the given boundary-value problem. Thus, we conclude the set of nonzero characteristic values of the given problem is Λ = {µ2 real : µ > 0, tan(µl) + αµl = 0}, and for a µ 2 in Λ the corresponding characteristic function is sin(µx). Rl Now consider the integration 0 sin(µ1 x) sin(µ2 x) dx, where µ1 6= µ2 , µ1 , µ2 ≥ 0, tan(µ1 l)+ Rl aµ1 l = tan(µ2 l) + aµ2 l = 0. Clearly, if one of µ1 , µ2 = 0, then 0 sin(µ1 x) sin(µ2 x) dx = Rl 0 0 dx = 0. If both µ1 , µ2 6= 0 then sin(µ1 x) and sin(µ2 x) are characteristic functions of the given problem, corresponding to different characteristic values. Note that in our case, R1 Rl we have r(x) ≡ 1, so 0 sin(µ1 x) sin(µ2 x) dx = 0 r(x) sin(µ1 x) sin(µ2 x) dx = 0. 24. By considering the characteristic functions of the problem x d2 y dy + + µ2 xy = 0, y(0) finite, y(l) = 0, 2 dx dx SOLUTION SET IX FOR 18.075–FALL 2004 3 and using the results of Section 5.6, show that Z l xJ0 (µ1 x)J0 (µ2 x) dx = 0 0 when µ1 and µ2 are distinct positive roots of the equation J 0 (µl) = 0. 2 dy d y + X dx + X 2 y = 0. So the general solution of the given Solution. Let X = µx, then X 2 dX 2 equation is y = Z0 (X) = Z0 (µx) = c1 J0 (µx) + c2 Y0 (µx). Then µ2 (6= 0) is a characteristic value of the given boundary-value problem only if lim [c1 J0 (µx) + c2 Y0 (µx)] is finite and c1 J0 (µl) + c2 Y0 (µl) = 0. x→0 Since lim J0 (µx) = 1 and lim Y0 (µx) = −∞, the finiteness of the limit x→0 x→0 lim [c1 J0 (µx) + c2 Y0 (µx)] implies that c2 = 0 x→0 Hence c1 6= 0, and the second boundary condition implies J 0 (µl) = 0. On the other hand, it is clear that if µ > 0, and J0 (µl) = 0 then J0 (µx) is a solution of the given boundary problem. Now we conclude that the set of non-zero characteristic values of the given problems is Λ = {µ2 : J0 (µl) = 0}, and for µ2 in Λ the corresponding characteristic function is J0 (µx). Note that, in this problem, r(x) ≡ x, so for any two distinctive characteristic Rl values µ21 , µ22 we have 0 xJ0 (µ1 x)J0 (µ2 x) dx = 0. That is, if µ1 , µ2 > 0, µ1 6= µ2 and Rl J0 (µ1 l) = 0 = J0 (µ2 l), then 0 xJ0 (µ1 x)J0 (µ2 x) dx = 0. 5.7. Expansion of Arbitrary Functions in Series of Orthogonal Functions. P 31. Determine the coefficients in the representation f (x) = ∞ n=1 An sin(nx), (0 < x < π), in the following cases:   1 (x < π/2), 1 (a) f (x) = 1, (b) f (x) = x, (c) f (x) = (x = π/2),  2 0 (x > π/2) Rb Rb Solution. We know that An a r(x)[ϕn (x)]2 dx = a r(x)f (x)ϕn (x) dx since the functions sin(nx), n = 1, 2, 3, . . . are orthogonal for 0 < x < π. 4 SOLUTION SET IX FOR 18.075–FALL 2004 (a) Thus An π Z Z 2 [sin(nx)] dx = Z0 π sin(nx) dx, and 0 1 π [1 − cos(2nx)] dx 2 0 π  1 1 sin(2nx) x− 2 2n 0 π , and 2 [sin(nx)]2 dx = 0 = = Z π π sin(nx) dx = 0 = So An = Z π 1 − cos(nx) n 0 1 [1 − cos(nπ)] . n 1 − cos(nπ) = 2[1 − cos(nπ)]/nπ. nπ/2  (b) An Z π 2 [sin(nx)] dx = 0 Z π x sin(nx) dx, and 0 π π , from (a), and 2 0 π  Z π 1 1 x sin(nx) dx = − x cos(nx) + 2 sin(nx) n n 0 0 π = − cos(nπ). n Z [sin(nx)]2 dx = So An = − πn cos(nπ) π 2 2 2 = − cos(nπ) = (−1)n+1 · n n SOLUTION SET IX FOR 18.075–FALL 2004 5 (c) An π Z 2 [sin(nx)] dx = 0 Z π 2 sin(nx) dx, and 0 π π , from (a), and 2  π 2 1 sin(nx) dx = − cos(nx) n 0 nπ  1 1 − cos . = n 2 [sin(nx)]2 dx = 0 Z Z π 2 0 nπ  π 1 1 − cos / 2 2  n nπ nπ / . = 1 − cos 2 2 So An = 32. (a-c) Show that the series obtained in Problem 31 cannot be differentiated term by term. (Investigate the convergence of the result in each case.) Solution. P 2 (a) f (x) = 1 and f (x) = ∞ n=1 nπ [1 P−∞cos(nπ)] sin(nπx). Differentiating formally term 0 0 by term, we get f (x) = 0 and f (x) = n=1 2[1−cos(nπ)] cos(nπx). So we’re investigating ? 0= ∞ X n=1 2[1 − cos(nπ)] cos(nπx) in (0, π). Now for 0 < x < π, as n → ∞, cos(nπx) 6→ 0 and P for n odd, 2[1 − cos(nπ)] = 4 6= 0. Thus, as n → ∞, 2[1 − cos(nπ)] cos(nπx) 6→ 0, so ∞ n=1 2[1 − cos(nπ)] cos(nπx) diverges for all 0 < x < π and so indeed the series of Problem 31(a) cannot be differentiated term by term. P∞ (−1)n+1 n2 sin(nπx). Differentiating formally term by (b) f (x) = x and f (x) = n=1P 0 0 n+1 · 2π cos(nπx). So we’re investigating term, we get f (x) = 1 and f (x) = ∞ n=1 (−1) ? 1= ∞ X n=1 (−1)n+1 · 2π cos(nπx) in (0, π). Now for 0 < x < π, as n → ∞, cos(nπx) and certainly 2π(−1) n+1 6→ 0. Thus, as P∞ 6→ 0n+1 n+1 n → ∞, (−1) ·2π cos(nπx) 6→ 0, so n=1 (−1) ·2π cos(nπx) diverges for all 0 < x < π and so the series of Problem 31(b) cannot be differentiated term by term. 6 SOLUTION SET IX FOR 18.075–FALL 2004 0 0 (c) Differentiating
the series formally term by term, we get f (x) = 0 and f (x) = P∞ nπ n=1 2 1 − cos 2 cos(nπx). So we’re investigating ∞  X nπ  0= 2 1 − cos cos(nπx). 2 ? n=1 Now for 0 < x < π, as n → ∞, cos(nπx) 6→ 0
and for n = 4k + P 2, k = 0, 1, 2, . . .,
2(1 − ∞ nπ nπ ) = 4 = 6 0. Thus, as n → ∞, 2 1 − cos cos(nπx) → 6 0, so cos nπ n=1 2 1 − cos 2 cos(nπx) 2 2 diverges for all 0 < x < π and so, indeed, the series cannot be differentiated term by term. 33. (a) If the expansion f (x) = P∞ n=1 An sin(nx) Z π 0 is valid in (0, π), show formally that ∞ πX 2 An . [f (x)] dx = 2 2 n=1 (b) From this result, and from the results of Problems 31(a,b), deduce the following relations: 1 1 π2 1 1 1 π2 1 + + + · · · = , + + + · · · = . 12 32 52 8 12 22 32 6 Solution. (a) [f (x)] 2 ∞ X = An sin(nx) n=1 ∞ X = ! ∞ X Am sin(mx) m=1 ! An Am sin(nx) sin(mx). n=1,m=1 So since f (x) = Z 0 P∞ An sin(nx) is valid in (0, π), Z π X ∞ π An Am sin nx sin mx dx [f (x)]2 dx = n=1 0 = = n=1,m=1 ∞ X Z π n=1,m=1 0 ∞ X An Am n=1,m=1 But Rπ 0 An Am sin(nx) sin(mx) dx Z π sin(nx) sin(mx) dx 0 sin(nx) sin(mx) dx = 0 if m 6= n. So Z π Z π ∞ X [sin(nx)]2 dx [f (x)]2 dx = An An 0 = n=1 ∞ X π 2 n=1 0 A2n . SOLUTION SET IX FOR 18.075–FALL 2004 (b) From Problem 31(a) ⇒ 1 = Z π 2 1 dx = π = 0 = = π2 4 So = = π2 8 = 4 π n=1 ∞ X 1 [1 − cos(nπ)]. n2 n=1 ∞ X 1 [1 − cos(nπ)] n2 n=1 ∞ X 2 , (2m + 1)2 0 π (n = 2m + 1) 1 1 1 + 2 + 2 + ··· 2 1 3 5 Similarly, from problem 31(b) ⇒ x = Z [1 − cos(nπ)] sin(nπx), 0 < x < π. So 2 ∞  πX 2 [1 − cos(nπ)] 2 nπ n=1   ∞ 1 πX 4 1 · 1 − 2 cos(nπ) + [1 + cos(2nπ)] 2 π 2 n2 2 m=0 i.e. 2 n=1 nπ P∞ 7 π3 x dx = 3 2 = P∞ n=1 (−1) n=1 ∞ X n=1 So 6 · 2 n sin(nπx), 0 < x < π. So 2 ∞  πX n+1 2 (−1) · 2 n = 2π π2 n+1 1 . n2 ∞ X 1 1 1 1 = = 2 + 2 + 2 + ··· 2 n 1 2 3 n=1 34. Expand the function f (x) = 1 in a series of the characteristic functions of the boundaryd2 y 0 value problem dx 2 + λy = 0, y(0) = 0, ly (l) + ky(l) = 0 (k ≥ 0), over the interval (0, l). 8 SOLUTION SET IX FOR 18.075–FALL 2004 Solution. Solving d2 y dx2 + λy = 0, we get √ √ C1 sin( λ · x) + C2 cos( λ · x) and y(0) = 0 ⇒ C2 = 0 √ √ √ so y(x) = C1 sin( λ · x) and y 0 (x) = C1 λ · cos( λ · x) √ √ √ and ly 0 (l) + ky(l) = 0 ⇒ l λ · C1 cos( λ · l) + kC1 sin( λ · l) = 0 √ √ √ ⇒ l λ · cos( λ · l) + k sin( λ · l) = 0 √ √ l λ ⇒ tan( λ · l) = − k x i.e. tan(x) = − , x > 0. k y(x) =   So there are infinitely many solutions, one each on kπ − π2 , kπ + π2 and let λk be the kth solution. Thus, functions of this boundary-value problem are ϕ m (x) =
the characteristic √ √ sin λm · x , where λm > 0. This boundary-value problem is clearly Sturm-Liouville, and r(x) = 1 so the ϕ m ’s are orthogonal. So ∞ X 1 = f (x) = Am ϕm (x) and Am · Z Z So n=1 l Z l p p sin( λm · x) dx sin ( λm · x) dx = = 2 0 0 l p sin2 ( λm · x) dx = 0 = and Z l sin( 0 Thus, Am = λm · x) dx = p λm · x)  dx 2 0   p 1 1 sin(2l λm ) . l− √ 2 2 λm  p 1  √ 1 − cos(l λm ) λm 1 − cos(2
√
√ 1 − cos(l λm ) 2 1 − cos(l λm )
= √ √ √ √ . 2l λm − sin(2l λm ) l λm − 21 sin(2l λm ) √1 λm √1 4 λm p Z 1 l