homework 13 solutions - Section 9.5 15 We need only...

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Unformatted text preview: Section 9.5 15. We need only calculate the coefficients in the usual zero—endpoint series n27r2kt . n7z'x sm u(x,t) = 2b,, exp(— L2 n=l Forthe function f(x)EA for O<x<L/2, f(x)EO for L/2<x<L we calculate the Fourier sine coefficient 1/ 2 for n odd, 4A . 2 "71' L/2 b" = Asinn—fliidx = —sm —4— = —-x 1 for n=2,6,10,---, ° "7’ n” 0 for n=4,8,12,---. 1'7. (a) 523%? = 0 implies u”($) = Cm + D. Then, u”(0) = A and u”(L) = B forces D = A, C, B_A, so 1133(1)) = 3—242: + A. ll (b) Since % = [egg—1%, 11(0,t) = A, u(L,t) : B, and u(:L',0) = f(;c), it is clear that, if u” = u use, then 2 073;: = k—WB; , u“(0,t) = utt(L,t) = 0, and name) = f(m) F anus). (c) By (31) and (30), u (:c t) = i1) exp(—n27r2kt/L2)sin(llfl) ft 5 "=1 Tl L a where L b" = [f(:c) _ u”(x)]sin(9—:l)dx. Therefore, ‘ ‘_ _ ~00 221 2 , n7rrc B-A' u(.v,t) _ Imam) + u”(a:,t) M “2:311”, exp(—n 7r kt/L )sm(T) + L a: +A 19. Since heat content : density >< specific heat >< change in temperature, the rate of temperature change at (at) due to this extra generated heat is equal to the rate of generated heat at (:c,t) divided by the product of the density and the specific heat. 24° We search for un(:v,t) = Xn(:c)Tn(t) solving Lian-(Lt) = kggfgl(x,t), un(0,t) = Egg-(Lt) = 0. Plugging the guess into the PDE, Xn(%)T/1(t) = kXif(1=)Tn(t)7 so 1 T12“) * X510”) ana) _ Xnm' Since neither side depends on 1' or t, both sides are equal to a constant, say, ——An. Then, X,’,’(x)+Aan = 0, Xn(0) 2 XML) 2 0 and T,’,(t)+kAnTn(t) = 0. The only way to solve the equations for X1, is if An > 0 and JAIL = "—2"— for n odd, and Xn(:v) = sin("T"L£). The equation for Tn(t) is solved by Tn(t) = e*k"2"2‘/(2L)2. Notice that if u is any linear combination of the u”, then we will have —‘Z,—'t‘(12,t) : kg:'2‘(19,t), u(0,t) : %(L,t) = 0. So if we take = Z Cnun(w,t) = Z Cn sin(n7r$ )e_kn2"2t/(2L)2, n odd 11 odd 2L all we need to do is enforce that = u(:c,0) = Z enun(x,t) = Z cn sin(%7~rg— . n odd n odd That is, the Cu must be precisely the coefficients of the odd half—multiple sine series for f, as given in Section 9.3 Problem 21. Section 9.6 13. If y(x, t) = F (x + at) = F (u) with u = x + at, then the chain rule gives 91 = 545% = F'(u).1 = F(x+at); 6x du 6x 93’. = 5151“ = F'(u)-a = aF’(x+at) = c191; at du 61 6x 62y dF’au n n g = = F(u)-1 = F(x+at); 2 I 2 6—2: = aiF—g = a-F”(u)-a = azF"(x+at) = a22%. 6t du at 6x 14. y(0, t) = fiF (at) + F (—at)] = %[F (at) — F (at)] = 0 y(L, t) = —2L[F(L + at) + F(L — at)] = %[F(L + at) — F(—L + at)] = %[F(2L + (—L +at)) — F(—L + at)] = 0 y(x,0) = H1706) + F (9‘)] = F (x) y,(x,t) = %[aF’(x + at) — aF'(x - at)] y.(x, 0) = %[aF’(x) — aF'(x)] = o 15. If y(x,0) = 0 then the fundamental theorem of calculus gives y(x,t) = y(x,t) — y(x,0) = £y,(x,r) d1 = Lg—[Gu + at) + G(x — a1)] dz: Letu=z+ar,v=m—a7‘. Then 1 I+at z = ~2—a I G(u)du + A-“ G(v) d1) 1 z+at = E; G(u) du 3—1” = 0 G(u)du+/Ox+atG(u)du] x—at 1 = E; [—H(x — at) + H(x + at)]- Evidently, Equation (37) has a typo. 16. If u = x + at, v = x — at then we solve readily for x = 7(u + v), t = fi(u — v). Hence 3y a 1 1 53’ 59‘ 5y at 10y 1 5)’ —-— = —— —- + ,—a — :: —— —— = —— au 8u[y(2(u ‘02 (u v» 6x6u+6t6u 26x+2a at 5y a 1 1 ‘9an 5y at 15y 1 5y — = — —— + 3—— — = ———-—+~———-—- = ———-————; av 6v[y(2(u v) “(u M] ax av 6t av 26x 2a 6t 62y _ EGQ+L§VJ avau 6v 2 6x 2a at = li£lQ+LQ]_LE[lQ+L§X] 26x 26x 2aat 2a6t26x 21161 = __—2‘+—"—_"——_—2_-T _ 2 4 6x 4a 6x6t 4a atax 4a at 4a lazy 1 62y 1 62y 1 62y 1 {62y (120212]: 0 avau ' av E 6y / au = G(v), an arbitrary function of v. Finally, antidifferentiation with respect to u gives y = F(u) + G(v) = F(x + at) + G(x — at). 2 Now if a y - 6y) = 0 then antidifferentiation with respect to v gives 19. The general solution of the second-order ordinary differential equation a2 y" = g is a second-order polynomial in x with leading coefficient g/ 2a2. But the polynomial ¢(x) = gx(x — L)/ 2a2 has this leading coefficient and satisfies the endpoint conditions y(0) = y(L) = 0- 20. If y(x,t) = v(x,t)+¢(x), then y,,=v,, and 62 y 62v ,, 82v g 62 y 62v 6x2 = g” (x) = 5x77, 5" “zy' = The transformation of the boundary conditions is straightforward. By Equation (23) and (22), °° t r. v(w, t) z n; An cos( Sin(1?%£), where 2 n7rx L An = —¢($)sin(-—L.«)dx. Notice that sin(%?) is odd with respect to the line as = L/2 when n is even. Also notice that, from Problem 197 = —— L) is even with respect to the line 1- = L/2. Therefore the integrand in An is odd with respect to the line at = L / 2 when n is even7 and hence An = 0 for even 11. Then , t / . v(1€,t): Z An cos(mla )sinU—jgi), 11 odd so t y(x,t) : ¢u(1-) + v(.1:,t):: ¢(1-) + Z An cos( Tma )sin(fl). 72 odd L L Fix x and notice that cos( "’2‘”) 2 1 for all n if t : 07 and cos( "7131“) = ~1 for all odd n ift : L/a. Therefore, y(x,t)) = Mr) 7+ EMA" and?) = ¢(x) e war) = 0 and n y(x,L/a)=q§(1')— Z Ansin( 71 odd 2“) : W) + 9W”) 2 2W). This establishes that y(at, t) covers all the values between 0 and but it does not establish that y(x,t) can never be outside this range. ...
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