Solution Manual for Applied Partial Differential Equations with Fourier Series and Boundary Value Problems, 5th Edition (Haberman, 2012), Chapter 1-14 | All Chapters

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SOLUTION MANUAL

Applied Partial Differential Equations with Fourier Series and Boundary Value
Problems (Classic Version)

by Richard Haberman
5th Edition
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All Chapters Included


All Answers Included




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Chapter 1. Heat Equation
All Answers
Section 1.2
1.2.9 (d) Circular cross section means that P = 2πr, A = πr2, and thus P/A = 2/r, where r is the radius.
Also γ = 0.
1.2.9 (e) u(x, t) = u(t) implies that
du 2h
cρ =− u.
dt r
The solution of this first-order linear differential equation with constant coefficients, which satisfies the
initial condition u(0) = u0, is
2h
u(t) = u0 exp t−
.
cρr

Section 1.3
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1.3.2 ∂u/∂x is continuous if K0(x0−) = K0(x0+), that is, if the conductivity is continuous.

Section 1.4
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1.4.1 (a) Equilibrium satisfies (1.4.14), d2u/dx2 = 0, whose general solution is (1.4.17), u = c1 + c2x. The
boundary condition u(0) = 0 implies c1 = 0 and u(L) = T implies c2 = T/L so that u = Tx/L.
1.4.1 (d) Equilibrium satisfies (1.4.14), d2u/dx2 = 0, whose general solution (1.4.17), u = c1 + c2x. From
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the boundary conditions, u(0) = T yields T = c1 and du/dx(L) = α yields α = c2. Thus u = T + αx.
1.4.1 (f) In equilibrium, (1.2.9) becomes d2u/dx2 = −Q/K0 = −x2 , whose general solution (by integrating
twice) is u = −x4/12 + c1 + c2x. The boundary condition u(0) = T yields c1 = T , while du/dx(L) = 0
yields c2 = L3/3. Thus u = −x4/12 + L3x/3 + T .
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1.4.1 (h) Equilibrium satisfies d2u/dx2 = 0. One integration yields du/dx = c2, the second integration
yields the general solution u = c1 + c2x.
x = 0 : c2 — (c1 − T ) = 0
x = L : c2 = α and thus c1 = T + α.
Therefore, u = (T + α) + αx = T + α(x + 1).
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1.4.7 (a) For equilibrium:
d2 u x2 du
=
dx2 −1 implies u = − + c1x + c2 and = −x + c1.
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2 dx
From the boundary conditions (0) = 1 and (L) = β, c1 = 1 and −L + c1 = β which is consistent
du du
dx dx 2
only if β + L = 1. If β = 1 —L, there is an equilibrium solution (u =− 2 x + x + c2). If β /= 1 − L,
there isn’t an equilibrium solution. The difficulty is caused by the heat flow being specified at both
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ends and a source specified inside. An equilibrium will exist only if these three are in balance. This
balance can be mathematically verified from conservation of energy:
∫ ∫L
d L du du
cρu dx = − (0) + (L) + Q0 dx = −1 + β + L.
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dt 0 dx dx 0

If β + L = 1, then the total thermal energy is constant and the initial energy = the final energy:
∫L ∫L
x2
f (x) dx = — + x + c2 dx, which determines c2.
0 0 2

If β + L = 1, then the total thermal energy is always changing in time and an equilibrium is never
reached.




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Section 1.5
1.5.9 (a) In equilibrium, (1.5.14) using (1.5.19) becomes drd rdu dr
= 0. Integrating once yields rdu/dr = c1
and integrating a second time (after dividing by r) yields u = c1 ln r + c2. An alternate general solution
is u = c1 ln(r/r1) + c3. The boundary condition u(r1) = T1 yields c3 = T1, while u(r2) = T2 yields c1
= (T2 − T1)/ ln(r2/r1). Thus, u = 1 [(T2 − T1) ln r/r1 + T1 ln(r2/r1)].
ln(r2/r ) 1

1.5.11 For equilibrium, the radial flow at r = a, 2πaβ, must equal the radial flow at r = b, 2πb. Thus β = b/a.
1.5.13 From exercise 1.5.12, in equilibriumdrd r2drdu = 0. Integrating once yields r2du/dr = c1 and integrat-
and u(1) = 0 yields 80 = −c1/4 + c2 a2nd 0 = −c1 + c2. Thus c1 = c2 = 320/3 or u = 320 1− 1 .
ing a second time (after dividing by r ) yields u = −c1/r + c2. The boundary conditio3ns u(4)r = 80
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Chapter 2. Method of Separation of Variables
Section 2.3
2.3.1 (a) u(r, t) = φ(r)h(t) yields φdh = kh d
rdφ . Dividing by kφh yields 1 dh
= 1 d
rdφ = −λ or
dt r dr dr kh dt rφ dr dr
dh
dt
= −λkh and 1 d
r dr
r dφdr = −λφ.
2 2 2
= −λ or
φ 1d φ
2.3.1 (c) u(x, y) = φ(x)h(y) yields hd + φd h
= 0. Dividing by φh yields
= − 1 d2h2
dx2 dy2 φ dx2 h dy
d2 φ 2

dx2 = −λφ and d h
dy2 = λh.
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2.3.1 (e) u(x, t) = φ(x)h(t) yields φ(x)dh = kh(t) d φ . Dividing by kφh, yields 1 dh
= 1 d φ
= λ.
dt dx4 kh dt φ dx4
2 2 2 2
2.3.1 (f) u(x, t) = φ(x)h(t) yields φ(x)dt
d h 2 d φ
2 = c h(t) dx2 . Dividing by c φh, yields c2h dt2
2 1 d h
= 1d φ
φ dx2 = −λ.

2.3.2 (b) λ = (nπ/L)2 with L = 1 so that λ = n2π2, n = 1, 2, . . .
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2.3.2 (d)
(i) If √λ > 0 ,√φ = c cos √λx + c sin √λx. φ(0) = 0 implies c = 0, while dφ (L) = 0 implies
c λ cos λL = 01. Thus = 2+ ( =1 2 ).
√ 2 1 dx
−π/
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2 λL nπ n , ,...
(ii) If λ = 0, φ = c1 + c2x. φ(0) = 0 implies c1 = 0 and dφ/dx(L) = 0 implies c2 = 0. Therefore λ = 0
is not an eigenvalue. √ √
(iii) If λ < 0, let λ = −s and φ = c cosh sx + csxs.inφh c (0) = 0 implies d φ=/ d0x aLnd ()=0
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1
√ √
implies c2 s cosh sL = 0. Thus c2 = 0 and hence there are no eigenvalues with λ < 0.
2.3.2 (f) The simpliest method is to let x′ = x − a. Then d2φ/dx′2 + λφ = 0 with φ(0) = 0 and φ(b − a) = 0.
Thus (from p. 46) L = bΣ−a and λ = [nπ/(b − a)]2 , n = 1, 2, . . ..
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∞
2.3.3 From (2.3.30), u(x, t) = n=1 B sin
nπx −k(nπ/L) t
e . 2The initial condition yields
n
Σ∞
n=1 n
L
n ∫L
2 cos 3Lπx
= B sin
. ∫From (2.3.35), B =
nπLx
Σ 2L 02 cos 3 π Lx sin nπxL dx.
2.3.4 (a) Total heat energy = L
cρuA dx = cρA ∞ B e−k( nπ )2 t 1−cos nπ , using (2.3.30) where B
0 L
satisfies (2.3.35). n=1 n L nπ n

2.3.4 (b)
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heat flux to right = −K0∂u/∂x
total heat flow to right = −K 0 A∂ u. /∂x
heat flow out at x = 0 = K0A∂u .
∂x x.=0
∂u .
heat flow out (x = L) = −
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K ∂x0A
x=L ∫L .L
2.3.4 (c) From conservation of thermal energy, d
u dx = k ∂u = k ∂u (L) − k ∂u (0). Integrating from
dt 0 ∂x .0 ∂x ∂x
t = 0 yields ∫ ∫ L ∫
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L
u(x, t) dx −
t
∂u (L) − ∂u (0)
u(x, 0) dx = k dx .
0 ∂x
0 ` ˛¸ x ` 0 ˛∂ ¸
x x ` ˛¸
` ˛¸ x initial heat integral of integral of
heat aet tnergy
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flow in at flow out at
energy
x=L x=L
2u √ √
2.3.8 (a) The general solution of kd = αu (α > 0) is u(x) = a cosh αx + b sinh αx. The boundary
dx2 k k
condition u(0) = 0 yields a = 0, while u(L) = 0 yields b = 0. Thus u = 0.




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