This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 Chalk0300 , 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 + 0281) = _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. {Iow; ,‘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 IICt
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”
s7=”' 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) = (gr1— 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 ...
View
Full Document
 Spring '11
 llll

Click to edit the document details