Applied partial differential equations with fourier series and boundary value problems 5th edition r
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Solutions Manual for Applied Partial Differential Equations with Fourier
Series and Boundary Value Problems 5th Edition by Richard
Haberman
Full clear download (no formatting errors) at:
//downloadlink.org/p/solutions-manual-for-applied-partial-differential-
equations-with-fourier-series-and-boundary-value-problems-5th-edition-by-
richard-haberman/
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
2
h
cρ dt
=
− r u
.
The solution of this first-order linear differential equation with
constan
t
coefficients, which satisfies
the
initial condition u(0)
=
u0 , is
u(t) =
u0
exp
·
2h
¸
− cρr
t .
Section 1.3
1.3.2
∂u/∂x
is continuous if
K
0
(
x
0
−)
=
K
0
(
x
0
+),
that
is, if the
conductivity
is con
tin
uous.
Section 1.4
1.4.1 (a)
Equilibrium
satisfies (1.4.14), d2
u/dx
2
=
0, whose general solution is (1.4.17), u
=
c1
+
c
2
x. The
boundary
condition u(0)
=
0 implies c1
=
0 and u(L)
=
T implies c2
=
T
/L
so
that
u
=
T
x/L
.
1.4.1 (d)
Equilibrium
satisfies (1.4.14), d2
u/dx
2
=
0, whose general solution (1.4.17), u
=
c1
+
c
2
x.
F
rom
the
boundary
conditions, u(0)
=
T yields T
=
c1 and
du/dx(L) =
α yields α
=
c
2
. Thus u
=
T
+
α
x
.
1.4.1 (f ) In equilibrium, (1.2.9) becomes d2
u/dx
2
= −Q/K
0
= −x
2
, whose general solution (by
in
tegrating
t
wice) isu
=
−
x
4
/
12
+
c1
+
c
2
x.
The
boundary
condition u(0)
=
T yields c1
=
T , while
du/dx(L) =
0
yields c2
=
L3
/
3. Thus u
=
−
x
4
/
12
+
L3
x/
3
+
T
.
1.4.1 (h)
Equilibrium
satisfies d2
u/dx
2
=
0. One
integration
yields
du/dx =
c
2
, the second
in
tegration
yields the general solution u
=
c1
+
c
2
x
.
x =
0 : c2 −(c1 − T )
=
0
x =
L : c2
=
α and thus c1
=
T
+
α
.
Therefore, u
=
(T
+
α)
+ αx =
T
+ α(x + 1).
1.4.7 (a) For
equilibrium:
d2
u
dx
2
x
2
=
−1 implies u
=
−
2
+
c
1
x +
c2
and
du
dx
= −x +
c
1
.
From the
boundary
conditions
du
(0)
=
1 and
du
(L)
=
β, c1
=
1 and −L
+
c1
=
β which is
consisten
t
dx dx
2
only if β
+
L
=
1. Ifβ
=
1 − L, there is an equilibrium solution (u
=
−
x
+ x +
c
2
). Ifβ
=
1 −
L
,