PDE1-Assignment3-SolnsExtra

PDE1-Assignment3-SolnsExtra - Chapter 1 (a) The equilibrium...

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Unformatted text preview: Chapter 1 (a) The equilibrium will satisfy u“ = 0 on each piece of the rod. Integrating twice gives u] = 0,2: +d1 and u; = 6219+ 113. Since mm) = 0, d1: 0, and since u2(L1+ L2) = T we have i 1 1.4.6. d2 =T—OzlL1 +152)- Thus the equilibrium is defined piecewise by l u(£)_ 611' US$< L1 — T+Cz($—Ll—L2) L1<$<L1+L2. To solve for c. and c2 we use the fact that both it and new are continuous at :r 2 L1 to get c1L1 = T — (2ng and ma, = 52:32. Solving for CI and C2 we get c__§L and “L ‘ 1 _ itng +5215] 62 _ EiLz + 52151- 3 Thus T ‘ L ogr<L1 . um= “1% “a” L) ‘ m 33* 1’ 2 T+—__— L <a:<L +L. Klinierqu 1 i 2 (b) When f6] :2, flag: 1, L1 =3, L2=2andT: 10 we have “(m)_ 339:1: 0Sz<3 “ 10+37—(11775)3<:r<5. The sketch of u is shown in Figure 2. u l. 10 Chalk-0300 , 012345 “Figure 2: Solution of Exercise 1.4.6(b). Chapter 1 Section 1.5 1.5.1. Any function of the form Acosz + Bsina: solves the ODE. The condition u(0) = 0 implies A = 0, and the condition u(L) : 0 then implies Bsin(L) = 0. Thus, if L is any integer multiple of 7r, any function of the form Bsinsc is a solution, so solutions are not unique. Otherwise, if L is not an integer multiple of n, we must have B = 0, in which case the zero solution is the only solution, 1.5.2. (a) Suppose a and 1.! are both solutions. Let w = u — 1). Then to satisfies w" + 10’: 0 w’(0) = 111(0) 2 flora) +w(l)]. Solutions of the original problem are unique if and only if m = 0 is the only solution of this problem. This ODE has general solution to = 01+C'28’”. Since 10(0) = 01 +02 and w'(0) 2 mC'g, the first part of the boundary condition implies 01 = —ZC'2 and therefore w :Cg(*2+e‘”‘). Next, ' 1 gain) + mm] = gnaw—1 _ 202 + 028-1) = _021 which equals 10(0) and w’(0). Hence any multiple of —2 + e‘” is a solution of the above problem, so solutions of the original problem are not unique. (b) Integrating the equation from 0 to l and using the boundary conditions gives 1 I f u u : 13(5) +u(l) — (u'(0) +u(0)) = 0. Hence a solution can exist only if the integral off over [0,1] is zero. #2. {Io-w; ,‘S. -5 eascd- use dlS‘cn'm. (1.6.1.) +6 {oak ad. a m a (a) Since 0fo # anon = 22 — 1 - 1 = 3 > 0, the equation is hyperbolic. , ( 1.6.5. )Let u = veafifiy. Then o6 x — “mam; + um) (3,1 uI—e {mm [V1343 A" J ‘ fol fix) dw : / u’(sc) + u’(:r)da: = [1.513) Section 1 .6 my : enfifiWflv + fly) um = emwymflv + 2cmI + v”) uyy : eaz+fiy(fi2v + 2,811,, + Um.)- Ss K \ g (I g .9 A Chapter 1 Substituting into the equation gives (X7 eaztfiycum + 3vW + (2a — 2m + (613 + 24m + (a:2 + 3E2 — 20: + 24H + 5M = 0- To eliminate the '03 and 1)., terms, let or = 1 and ,3 = —4. Then the new equation is v” + Bow — 441} = 0. Letting y' = fig it follows that eye”: = 3n”, so “on + 'Uyry: — 441; = 0. Chapter 2 Section 2.1 2.1.1. By d’Alembert’s formula (2.1.8), 1 t 1 I-I-Ct t = — “+‘ $7“ ~f sinsd 11(59, ) 2 [e +e ] + 26 flat 3 _ l m at -ct _ i __ _ _ 23 (e + e ) 2C (cosh: + ct) cos($ ct)) r 1 = ex cosh(ct) + a sin($) sin(ct). C 2.1.3. The speed of the wave is c : x/T/p and the left hand edge of the disturbance begins at 1/2 — 0.. So the time it takes for the disturbance to travel to the position of the flea, 1/4, W = MUM _ a). C 2.1.6. If ct < a, then for any a the length of the interval (m w 015,512+ ct) is less than that of the interval (—0., a), and therefore the intersection of these intervals has length at most 2ct. Thus um, t) g 2% = t for all m. But at a.“ = 0 the interval (whet, $+et) = (uct, ct) is asubset of (—a,a), so the length of the intersection is exactly 2ct. Hence 110), t) = t = maxi. u(:r, t). On the other hand, if ct 2 a, then the length of the intersection is at most 2a, so u(z,t) S “E for all I. Since this maximum length is attained when x = 0, it follows that u(0, t) = a: : max: 11(22, t). So t< E infxuttfi) _ { t 2 nlfir‘i- 2.1.8. (a) Applying the product rule gives or : rur + u and "u,r = Tu” + 2'ur. Thus 2 2 v” = run = 6 (run + 211,) = c “U”. 2.1.9. Factoring the operator yields (a + came: — 43;)11 = 0. Set 1) = aw — 411;. Then vr+ot = 0, so it = h($— t) and U39 —4ui : My: #11). One solution of this equation is fins—fit), where f’(s) = %h(5). The general solution of the homogeneous equation um - 411‘ : 0 is 9(422 + it), so We t) : f(fl= e t) + 9(4E + t)- Chapter 4 Section 4.1 4.1.; e need to solve ut~kum=0 0<m<l,t>0 {u(a:,0)=1 0<x<l u(0,t) =0=u(l,t) t) 0. Looking for a solution of the form u(:r, t) = X (1:)T(t) implies T! X” TI X” E‘s—“vemv=* where ,\ is a constant. So we consider the eigenvalue problem {X”=—AX 0<m<l X(0) =0 =X(l). If A 2 [32 > 0, then X(a:) : Ccosfifiz) + Dsin(,8:s). The boundary conditions X(0) = 0 = X“) imply that C = 0 and [3,1 = (mr/l)2 for n : 1,2, . . .. All eigenvalues are positive (see text). Solving T1; _ (rm)? an 5 we see that z Tn(t) = Ana—WE) t Therefore, m I “(55: t) = EAR sin e Hui!) ‘ u(:c0):1=E(smfl+lsinfl+lsmw ) ’ 7r 1 3 l 5 4! implies A = { % 'rt odd “ 0 u even Therefore, 4.1.3. Using separation of variables, we look for a solution of the form u(:r, t) = X(n:)T(t). mg a function of this form into our PDE, we see that X and T must satisfy T! X” s-7=”' 32 Chapter 4 which implies that X must satisfy the eigenvalue problem X"=—/\X 0<az<l X{0)=O=X(£). The eigenfunctions are Xn(z) = sin with corresponding eigenvalues A“ = Now solving our equation for T, we have which implies Therefore, awkiymd¥aweh 71:1 Section 4.2 @Separation of variables leads to the eigenvalue problem X ” = —)\X 0 < m < l { X(0) = 0 = X’U). Looking for positive eigenvalues /\ = ,62 > 0 implies X(:c) = Ccosma) + Dsin(,6:c). The boundary condition X(O) = 0 implies C' = 0, while the boundary condition X’([) = 0 implies ,8 : (n + n/l. Therefore, _ {n+%)7r 2 _. (n+%)nz A" —— X,,(1:)i sm . If A = —72 < 0, then X($) : Acoshhcr) + Bsinhhx), and the boundary condition X(O) = 0 implies A = 0, so X(m) : Bsinhha‘). But the boundary condition X'U) = 0 implies Bcoshhl) = 0, so B = 0. Therefore there are no negative eigenvalues. If A = D, then X(:r) = A2: +13, and the boundary condition X(0) = 0 implies B : 0. But then the boundary condition X’U) = 0 implies A : 0. Therefore zero is not an eigenvalue, so all the eigenvalues are positive. Solving the equation T:1 = —A,,kT,, gives THU) = One—H“. Therefore, 00 l ‘ 2 q u($,t) = :0“ sin e—k(n+%) «21/1: n:D CE 33 Chapter 2 and therefore fl 2 °° (4)2231“ “segm- (c) Using the j = 0 and j = 1 terms above yields Q(,,,_i+; m _1( w ’ 2 fr? x/4kt 3 «41s ' (d) By Taylor‘s Theorem, the remainder takes the form 1 (ah/rm?)5 E 10 for some 2 between 0 and 3:. This is small for a fixed and t large. 2.4.17 Let u = e'bta/av. Then at = e‘maia(—bt2o +14) and u” = 6"”3/3'um, so substituting in o the equation a; — hum. + btau = 0 gives 11, —- kn” = 0. When t = O, u = 1) so the initial data is unchanged. So 1) is given by equation (2.4.8) and thus _3 eta/3 oo goo um) = asst/“we a. (/ZAJBJ Let o(y,t) = u(y + Vt, t). Then v¢(y, t) = Vnzw + Vt, t) + WU} + Vt,t), vy(y, t) = (gr-1— Vt, t) and oyyfy, t) = umfi; + Vt, t). Thus v.(y, 75) —~ kvw(y, t) = {My + Vt, t) — ku"(y + Vt, t) + Vux(y + Vt, t) = D, so 1) is a solution of the diffusion equation, with initial data 11(3), 0) = u(y, O) = My). Thus, by equation (2.4.6), vat) = w so — mum) dw. which means u(ac,t) = 11(m — Vt,t) =1“) S(:t — Vt —— w,t)¢»(w) dw. Section 2.5 2.5.1. Consider for instance the solution of the wave equation in one dimension with initial data Mm) = U and - _ 1 < 1 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. 14 Chapter 3 Section 3.1 3.1.2. Introduce the function 12(1‘, t) such that 12(33, t) : u(r, t) — 1. Then 1) satisfies vtvkofl=0 O<a§<oo U(m,0)=—-1 0<$<oo n(0,t)=0. We know by equation (3.1.6) that v is given by 1 00 Mat) = _ 41TH]; [e—(z—ywcmi _ {awn/4am” Making the change of variables 13 = (y -— :r)/\/4kt, we see that 1 fm e_(w_y)g/dmdy = 1 '30 e__p2 tip- 0 V 41rkt W ark/EH Similarly, making the change of variables q = (y + m)/V4kt, we see that 1 fm —(:c+y)’/4kt 1 °° — ’ e d = — e ‘7 d . m o y W? .st q Using these facts, we conclude that 1 2 o sc,t = —— a" tip ( ) fi nah/m 2 Wm _p. d : e— g p J? u = —6"rf(:r/\/4kt). We conclude that u(:r, t) E v(a‘,t) + 1 for :1: > 0 satisfies the desired problem. In particular, we conclude that 11(1):, t) = 1 — c5"rf(;r/\/4kt). 3.1.3. Let ¢5,,,m,,(:1:) be the even extension of Mac) to the whole real line. Let 'u(:r,t) be the solution of { 'Ug " kvm‘ : 0 'L'(.’.E, : Chuveucr)‘ Then w(a:,t) E v(s:, t) for 3: > 0 will be a solution of the heat equation on the halfuline with —oo<:c<oo,t>0 15 ...
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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-Assignment3-SolnsExtra - Chapter 1 (a) The equilibrium...

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