PDE1-Assignment2-ExtraSolns

PDE1-Assignment2-ExtraSolns - Chapter 1 Substituting into...

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Unformatted text preview: Chapter 1 Substituting into the equation gives Chapter 2 WWW“ + 3ny + (20: 2 2m + (5,6 + 24m, + (a2 + 3,32 — 20: + 245 + 5m 2 0, To eliminate the U3 and fly terms, let a 2 1 and x3 2 —4. Then the new equation is Section 2.1 “I: + 30w ‘ 44” = 0- d’Alembert’s formula (2.1.8), Letting y' 2 lg; it follows that 'u I . = 319 , so 'u + v : r 2 441) = D. 1 “a J5 y y vs; II I; y um, t) 2 [51“! + eT—C‘] + f sin .9 d5 .7: 26 —ct 1 E 1 a: at —c.! 1 as (e + e ) — 2—6: {cos{:c + ct) — cos(:1: — ct)} ‘ l 2 e” cosh(ct) + E sin(:1:) sin{ct). 2.1.3. The speed of the wave is c = T/p and the left hand edge of the disturbance begins at 11/2 — a. So the time it takes for the disturbance to travel to the position of the flea, 1/4, 15 M : Jp/TUM _ a). C 2.1.6. If ct < a, then for any a: the length of the interval (3 — ct, a: + ct) is less than that of the interval (2mm), and therefore the intersection of these intervals has length at most 2ct. Thus u(m, t) g % 2 t for all 3:. But at :E 2 0 the interval (as—ct, $+ct) 2 (—ct, ct) is a subset of (2:141), so the length of the intersection is exactly 2st. Hence u(0, t) 2 t 2 maxi. u{m,t). On the other hand, if ct 2 a, then the length of the intersection is at most 2:1, so u(m,t) : % for all (1:. Since this maximum length is attained when a: 2 0. it follows that u(0,t) 2 E = maxI u($, t). So C w 3 t < 9 i maxu{a:,t) 2 { a_ > 9": c - c' 5 2.1.8. (a) Applying the product rule gives “u, 2 Tu, + u and er, 2 run. + 2c... Thus . 2 I “u” 2 TU“ 2 c2 (mu.r + 2a,.) 2 c v". 2.1.9. Factoring the operator yields (61. + (9,)(6m — 46;)” 2 0. Set ’U 2 11¢ — 411,. Then or, +0: = 0, so 1) 2 Ma: 2 t) and ur — 4a: 2 Ma: — t). One solution of this equation is f{:c — t), where f’(s) 2 %h(s). The general solution of the homogeneous equation um — 4m 2 0 is g(4:c + t), so u(m,t) = f(:r — t} + 9(411: +15). Chapter 2 Chapter 2 The initial conditions imply flz) +g(4.1;) = 332 and —-f’(a:) +g'(4:r) = em. Differentiating the Thus first equation leads to the system f’lw) + 4y’(4w) —f’(w) + 9143) Solving this gives f’(m) = get — is” and g’(4:1:) = gm + gem. Thus f(1:) = $2 — is”, and J l l l l 11.71 T 0!er : C2 (alff _ 2arfirfr _ afinfr + a£flf12ffl+ {alf _ (half-1)). 23: ex. 1 3 5 9,05) = 11—03 + %le4u SO 9(3) = fi5 + 438‘”. The solution is therefore Section 2.3 l 4 1 4 ,t=_ _t2#_:4 ___4 2 _s+t/4 “(E ) 5(E ) 58 + 20( m +t) + Se 2.3. The maximum is u{0,0) =1 and the minimum is u(1,T) = —2kT. = :1. [er-Hid u exit] + $2 + its. 5 4 ‘ 2.3.4. (a) Since 1: equals zero on the lateral sides and the maximum value at time t = 0 is i u(1/2,0) = 1, the strong maximum principle implies that u(:r,t) < 1 for all t > 0 Section 2.2 l and 0 < a: < 1. Since the minimum value at time t = O is u(0,0) = 0, the strong ‘I minimum principle implies that u($,t) > 0 for all t > 0 and D < :5 < 1. 2'2'3' (b) Let o(z,t) = u{1 — Lt). Then 113(3, t) : #11::(1 — 1:, if), so (a) Let v(:s,t) = Me — y,t). Then UIALt) : “mu _ E! t) = at“ _ I)” 2 “Am, Q1 _ _ 2 __ 2 i Wait) — ant; — y’t) _ C umlx _ y! t) _ G 0mm, t). and thus 7; is also a solution of the wave equation. Since v(0,t) = u(l, t) = 0, v(l,t) = ' u(0,t) = 0 and v(:c, 0) : u(1 #1930) = 4(1 —:c)(1 — (1 —a:)) = 4:5(1—1‘), v is a solution of 2.2.5. Substituting u“ = czu'n-m 7 m1: : gum #— rut gives ' the diffusion equation with the same initial data and boundary data as u. Since solutions are unique, it follows that 11(113, t) = u($, t) for all t 2 0 and D S a: S 1. dE °° 5 E = f pulufl + Tuzuzt d3 5 (0) Since it vanishes as 2: = 0 and z = 1 for all t, we have “‘00 _ w 2 I 1 1 7 f Tutuxa: + T’Uizufl 7 p7"th ah: I u2($l dz = 2 him, It)”: (as) d1. _°° : dt 0 U 03 . _ _ 2 I 1 — [no T(utum)m pr‘ut dzc - : 2f u($,t]um(a:,t)d:c oo U 2 . 1-:1 1 = — 1" a dc < 0. p [00 t _ i = 2u(:i:,t}u$(m,t) “2] u§[m,t)dm w=u 1} Therefore the energy in non-increasing. (It is not necessarily the case that E is strictly ‘ 1 2 decreasing.) 1 = —2f u$(w,t)d$. u 2‘2'6‘ The integral cannot be zero since this would imply u1(:i:,t) = 0, which means that for (a) writing 1103:) = My)“; _ 300% we have each t, u(m,t) would be constant in .12. Since u(0, t) = 0 that constant would be zero. i‘ ‘ ‘ But by part {a}, u is positive for 0 < a: < 1. Thus u2(m, t) dc is strictly decreasing, “ti! = “(Tlfuft — filfl) since its derivative is negative. u. = a'(r)f(t - BM) - 0(T)fl’(r)f'(t — (RH) urr = ar"(7”)f(?3 - I300) ” 201'(r)fi’(1")f'(t ” 5W) — 0t(?“)i(3"(i")f'(t - 5M) + a(T)lfi’(T)]2f”(t - 150))- 10 11 Chapter 2 Chapter 2 he PDE is u. = mu”. 9.4.6. Let I = fume”! dm. Then 1 : ffe-v” dy. so et u = —25rt — 3:2. It is elementary to check that it is a solution. We use calculus to find 2 0° —1-1 00 -—y"’ its maximum and minimum in the rectangle R. To do this, we check the interior using . I : lo 6 LEA e dy u; = ex = 0 to get the point (0,0) where u = 0. Then we check each of the four sides l °° °° 1 a by taking one derivative in each case, and finally we consider the corners separately. l 2 j ( 8—3 dm) ew- dy Among all these points that are in the rectangle R, we look for the largest and the l “on o: smallest values of n. We find that the maximum value is u = 1 at the point (—1,1), l = f / edge—y] fix dy which is on the top of R, and the minimum value is u = —8 at (2,1), which is a corner . use “00 point. The maximum occurs on the top of R, which violates the (dilfusion) maximum = f / e'wgfl'gl rim sly. principle. 3 U ” (b) At any maximum point on the top, we must have at 2 0 and um S 0. At our maximum i Changing to polar coordinates, this becomes point (—1,1), u; = 2 > 0 and scum = (fil)(—2} I 2 > 0. There is no contradiction to E “l2 00 . fl/‘Z 1 the PDE ut = mu”, even if the inequalities were strict. (The purpose of introducing 6:172 I2 3 f / Err dr '19 : f _ in the text was to make the inequalities strict, but here this would not help.) 0 0 U 0° 1r/2 m=f 1&2? U D 2 4 la ThusI: 2. 2.4.8. For each fixed 3, 3(32, t) is a decreasing function of |${, so Sect' n 2.4 l. 2 max S :t,t = S(6,t} = e“5 "W. etting p = (y — $)/V4kt in equation (2.4.8) gives 55"”(‘3‘3 ( ) V 477’“ Now let t : 1/4ks. Then1 by L’Hopital‘s rule. I 1 a I u 32,: = __ 3"(”'ml “Mo! = ( ) 47TH 4 y = l' 1 “52/4” lim 1 fl lim 1 1 0 . nn 6 = — 2 = —fl = = —1—' (l—xflm e_p1 d 1 tall Jain-kt s—noo fi 85 a 5400 J1? Qfiazedis W? (—[umilflfi p 1 (i—zm/m _ a 1 bum/«5E 1 et u be the solution. Then (um). = {udmx = [kumhm 2 Hum)”, so um = e ‘” dp— 3—” (133 ution, and 0] = (fl—'Jfirz) = 0. Hence um is the zero function. Integrating 1 D l_ m 1 _£ _ 0 three times with respect to :1: gives u(3r,t) = A(t):r2 + B(t):r + C(t). Substituting uinto the = Egrf ~ §£IE( _ . diffusion equation u; = hum gives A'{t):::2 + B’(t}$ + C’(t) = Zia/Mt). 2.4.3. U ' t' 2.4. h 5mg equa 101‘ ( 8): we we Equating coeflicients gives A’ : B' : 0 and C’ = 2kA. Thus A and B are constant and C = QkAt + D, so u($,t) 2 Am? + Ba; + ZkAt + D. Now the initial condition u(:r, t) = LE2 1 im lies tA=1andB:D:0,sou(m,t}=m2+2ki. V41rkt Completing the square, the exponent becomes u(a:, t) = m f e—(wwdktesu dy. —oo £13.15) : %+ éé’rf (b) Since the Taylor series for e2 centered at z : 0 is 22:“ it follows that -'y2 + Zmya: l2kty _ 3:2 _ —(y — a: —— 6M)2 + 361::2t2 +12kxt 4kt — 4kt I So letting p = (y — 5r — 6kt)/\/4kt, we get 0° _ 3' '23" _1 Clo—"91:32 91:3 3—111: (up u{w,t)—fi1mepe t+ dp:e t+Il j; j! 12 13 Chapter 2 and therefore are) : —% i (c) Using the j = 0 and j = 1 terms above yields lect)=l+i w —1(I)3 ' 2 fi x/4kt 3 x/4kt ‘ (d) By Taylor’s Theorem, the remainder takes the form ; (Zn/ms ,fi. 10 ome 2 between 0 and 1:. This is small for a: fixed and t large. et u = FMS/av. Then a, = e'b‘3/3(—btzv +1») and um = KHz/31)“, so substituting e equation a, — kn“ + 515% = 0 gives u, — 1w” 2 0. When i = 0, u = 1) so the initial a is unchanged. So 1‘.) is given by equation (2.4.8) and thus _ 3 e brllfl oo thn'kt 790 t v(y,t) = My + Vt, t). Then tidy, t) = Vufiy + Vt, t} + u,(y + Vt,t), 1),,(y, t) = uz(y + Vt t) and vyy(y, t) = um(y + Vt, t). Thus U¢(y,t) — icvw(y,t) =Hu,(y + Vt,t) — hunky + Vt, t) + Vux(y + Vt, t) : 0] so 'u is a solution of the diffusion equation, with initial data v(y,0) = u(y,0) = Thus, by equation (2.4.6), u(:r, t) = e”(‘”"y)2/4k‘¢(y) (1y. v(y,t) = fig S(y — w,t)¢(w) do), which means 0,, u{:r,t) = v(.r — Vt, t) =/ 3(11: —— Vt i w,t)qfi(w)dw. ’W Section 2.5 2.5.1. Consider for instance the solution of the wave equation in one dimension with initial data (Mm) = 0 and i 1 < l which corresponds to the “hammer blow”. Since there is no boundary, the maximum prin- ciple would state that the maximum is attained initially. But the solution is zero initially, and takes on positive values for t > 0. Thus the maximum principle does not hold for the wave equation. Chapter Section 3.1 3.1.2. Introduce the function v(m, t) such that v(a:, t} = u(;r, t) — 1. Then 1; satisfies U,—kn._.,.,.=0 0<az<oo v($,0):71 0<m<oo v(0,t):0. We know by equation (3.1.6) that o is given by 1 0° 41—min!“ —(x+y)?/4ka ’U(.’JJ, t) = — [e — e ]dy. U m Making the change of variables p = (y — :1:)/\/4kt, we see that 1 m . 3 4k! 1 m 2 / e‘ia—V) 'f dy : 8—” dp. u x/47rkt W arm Similarly, making the change of variables q = (y + 3:)/\/4Ict, we see that \/ 1 fm ‘i‘i-ylirak‘ 1 ft:— — 2 e ” dy : A e ‘1 dq. thrift u \/"'—r 1/ 4k; Using these facts, we conclude that 1 Wm . v at = e— 8"" dp ( ) fi nth/m 2 :{Jm 9 = —— ’3" d w? .. e P = —é"1‘f(:c/V4kt). We conclude that u{a:,t) E n(:r,t) + 1 for :r > 0 satisfies the desired problem. In particular, we conclude that u(a:,t) : * €rf($/V4kt). 3.1.3. Let qb,,,,c,,(:r) be the even extension of flat) to the whole real line. Let v(:t, t) be the solution of { v, — kn” = 0 “($1 = ¢h=vnll($)‘ Then w($, t) E v(a:, t) for z > U will be a solution of the heat equation on the half-line with foo<$<oo,t>0 ...
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This note was uploaded on 01/17/2012 for the course AM 3815 taught by Professor Llll during the Spring '11 term at UWO.

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PDE1-Assignment2-ExtraSolns - Chapter 1 Substituting into...

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