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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 _ (half1)). 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. [erHid 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'nm 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 nonincreasing. (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 : ffev” dy. so et u = —25rt — 3:2. It is elementary to check that it is a solution. We use calculus to ﬁnd 2 0° —11 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 — $)/V4kt in equation (2.4.8) gives 55"”(‘3‘3 ( ) V 477’“ Now let t : 1/4ks. Then1 by L’Hopital‘s rule. I 1 a I
u 32,: = __ 3"(”'ml “Mo! =
( ) 47TH 4 y = l' 1 “52/4” lim 1 ﬂ lim 1 1 0
. nn 6 = — 2 = —ﬂ =
= —1—' (l—xﬂm e_p1 d 1 tall Jainkt s—noo ﬁ 85 a 5400 J1? Qﬁazedis
W? (—[umilﬂﬁ p
1 (i—zm/m _ a 1 bum/«5E 1 et u be the solution. Then (um). = {udmx = [kumhm 2 Hum)”, so um
= e ‘” dp— 3—” (133 ution, and 0] = (fl—'Jﬁrz) = 0. Hence um is the zero function. Integrating
1 D l_ m 1 _£ _ 0 three times with respect to :1: gives u(3r,t) = A(t):r2 + B(t):r + C(t). Substituting uinto the
= Egrf ~ §£IE( _ . diffusion equation u; = hum gives
A'{t):::2 + B’(t}$ + 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‘iylirak‘ 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($, t) E v(a:, t) for z > U will be a solution of the heat equation on the halfline with foo<$<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.
 Spring '11
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