Hints and solutions to the Homework 1
1.2.3. Derive the heat equation for a rod assuming constant thermal properties
with variable cross-sectional area A(x) assuming no sources by considering
the total thermal energy between x=a and x=b.
Solution. We use the conservation of energy principle in an arbitrary
volume V of the rod:
rate of change of heat energy in V = heat energy flux though ∂V
+ heat energy generation in V.
As in the class, denote by e(x, t) the (unknown) heat energy density, φ(x, t)
the from-left-to-right energy flux through the rod cross-section. Since
no sources of energy is assumed, there is no generation of energy in V .
Applied to the volume V = ∪x∈(a,b) A(x) (the piece of rod between x = a
and x = b), the conservation of energy principle leads to the equality
Z
d b
e(x, t)A(x)dx = ψ(a, t)A(a) − ψ(b, t)A(b) ⇒
dt a
Z
Z b
d b
∂
(ψ(x, t)A(x)) dx ⇒
e(x, t)A(x)dx = −
dt a
∂x
a
Z b
Z b
∂
∂
e(x, t)A(x)dx = −
(ψ(x, t)A(x)) dx ⇒
a ∂t
a ∂x
Z b
∂
∂
e(x, t)A(x) +
(ψ(x, t)A(x)) dx = 0.
∂x
a ∂t
Since the interval (a, b) is an arbitrary subinterval of (0, L), the above
integral equality yields
∂
∂
e(x, t)A(x) +
(ψ(x, t)A(x)) = 0
∂t
∂x
for x ∈ (0, L), t ≥ 0.
The Fourier law for the heat flux and obvious manipulations lead to the
final equation for the temperature u(x, t):
∂u(x, t)
k ∂
∂u(x, t)
A(x)
=0
−
∂t
A(x) ∂x
∂x
for x ∈ (0, L), t ≥ 0.
1.2.9. Consider a thin one-dimensional rod without sources of thermal energy
whose lateral surface area is not insulated.
1
(a) Assume that the heat energy flowing out of the lateral sides per unit
surface area per unit time is w(x, t). Derive the partial differential
equation for the temperature u(x, t).
Solution. We start with the same arguments as in Ex.1.2.3 solution,
with the following differences: now A is a constant, and we have a heat
flux through lateral boundary with density w(x, t). To compute the total
flux at time t through the lateral boundary of a volume V = (a, b) ×
A, we should integrate w(x, t) over the lateral boundary of V , that is
Rb
Sw(x, t) dx, where S is the perimeter of the rods cross-section. One
a
√
computes S = 2πr and A = πr2 , so S = 2 πA. Thus the conservation
energy principle leads to the equality:
d
dt
Z b
d
dt
Z b
a
Z
b
b
Z
e(x, t)Adx = ψ(a, t)A − ψ(b, t)A +
a
⇒
√
2 πAw(x, t) dx
⇒
a
Z
e(x, t)Adx = −
a
√
2 πAw(x, t) dx
a
b
∂
(ψ(x, t)A) dx +
∂x
b
Z
a
Z b
Z b √
∂
∂
e(x, t)Adx = −
(ψ(x, t)A) dx +
2 πAw(x, t) dx ⇒
∂t
a ∂x
a
Z b
√
∂
∂
e(x, t)A +
(ψ(x, t)A) − 2 πAw(x, t)dx = 0.
∂x
a ∂t
The Fourier law for the heat flux and obvious manipulations lead to the
final equation for the temperature u(x, t):
r
∂u(x, t)
∂ 2 u(x, t)
π
−k
−
2
w(x, t) = 0
2
∂t
∂x
A
for x ∈ (0, L), t ≥ 0.
1.4.1. Determine the equilibrium temperature distribution for a one-dimensional
rod with constant thermal properties with the following sources and boundary conditions:
(c) Q = 0, ∂u
∂x (0) = 0, u(L) = T
Solution: We know that an equilibrium solution is time independent, u(x, t) = u(x), and so
obtain that
2
∂u
∂t
= 0. From the heat equation we
1 ∂u
∂2u
=
= 0.
∂x2
k ∂t
All possible solutions to the above are u(x) = C1 x + C2 . We find C1
and C2 by applying the boundary conditions,
∂u
∂x (0)
= 0, u(L) = T :
∂u
(0) = C1 implies C1 = 0
∂x
u(L) = C1 L + C2 implies C2 = T
Thus the answer is u(x) = T .
(e)
Q
K0
= 1, u(0) = T1 , u(L) = T2
Solution: As in item (c), we have
∂u
∂t
= 0. The heat equation
cρ ∂u
∂2u
Q
=
+
K0 ∂t
∂x2
K0
and equality
Q
K0
this are u(x) =
= 1 imply
2
− x2
∂2u
∂x2
+ 1 = 0. All possible solutions to
+ C1 x + C2 . We find C1 and C2 by applying the
boundary conditions, u(0) = T1 , u(L) = T2 :
u(0) = C2 implies C2 = T1
T2 − T1
L
L2
+ C1 L + C2 implies C1 =
+
u(L) = −
2
L
2
We find the final solution:
u(x) = −
x2
T2 − T1
L
+(
+ )x + T1 .
2
L
2
1.4.7. For the following problems, determine an equilibrium temperature distribution(if one exists). For what values of β are there solutions? Explain
physically.
(b)
∂u
∂t
=
∂2u
∂x2 , u(x, 0)
∂u
= f (x), ∂u
∂x (0, t) = 1, ∂x (L, t) = β
Solution: Similar to Ex1.4.1.(c), all possible solutions are u(x) = C1 x +
C2 Applying the boundary conditions,
∂u
∂x (0, t)
= 1, ∂u
∂x (L, t) = β gives us
the following
∂u
(0, t) = C1 implies C1 = 1
∂x
∂u
(L, t) = C1 implies C1 = β
∂x
3
Thus we obtain the restriction on β: β = 1. Integrating the heat equation
over x ∈ (0, L), we get
∂
∂t
so
L
Z
L
Z
udx =
0
0
∂u
dx =
∂t
L
Z
∂2u
∂u
∂u
dx =
(L, t) −
(0, t) = β − 1 = 0
2
∂x
∂x
∂x
0
RL
udx = C, where C is a constant. From the initial condition u(x, 0) =
RL
f (x), we determine C = 0 f (x)dx. Now we find the value of C2 from the
0
relations
Z
L
Z
L
(x + C2 )dx =
udx =
f (x)dx =
0
0
0
This gives C2 =
1
L
RL
0
f (x)dx −
L
2.
∂u
∂t
∂2u
∂x2
=
L2
+ LC2 .
2
The final solution is
1
u(x) = x +
L
1.4.11. Suppose
L
Z
Z
L
f (x)dx −
0
L
2
∂u
+ x, u(x, 0) = f (x), ∂u
∂x (0, t) = β, ∂x (L, t) = 7.
(a) Calculate the total thermal energy in the one-dimensional rod(as a
function of time).
Solution: The total energy is e(t) =
RL
0
e(x, t)dx = cρ
RL
0
u(x, t)dx.
We compute, employing the heat equation and boundary condition:
de
= cρ
dt
Z
0
L
∂u
dx = cρ
∂t
Z
0
L
L
L2
∂2u
∂u x2 + ) = cρ(7−β+ ).
( 2 +x)dx = cρ (
∂x
∂x
2 0
2
Integrating in time gives
Z
e(t) − e(0) =
0
where e(0) = cρ
RL
0
t
de
L2
= cρ(7 − β +
)t
dt
2
u(x, 0)dx = cρ
RL
0
f (x)dx. We find the solution:
L2
e(t) = cρ((7 − β +
)t +
2
4
Z
L
f (x)dx)
0