-1-
Solutions of Assignment Two
Question 1
Consider the steady state conduction problem in the rectangular region as shown. One side of
the region is insulated; two sides are maintained at a constant temperature T1; while the forth
side is at a constant temperature T2 ≠ T1 .
(a) Set up the equations defining the temperature distribution T(x,y).
y
(b) Solve the equations in (a).
2 T1 T1
x
T2
1
Solution.
(a). The equations governing the temperature distribution for the given problem
include the two-dimensional steady state heat conduction equation and boundary
conditions as follows.
Heat conduction equatios
∂ 2T ∂ 2T
+ =
0
∂x 2 ∂y 2
Boundary conditions
=
T (0, y ) T=
1, T (1, y ) T1, ( 1)
∂T
= ( x, 2) 0,=
T ( x, 0) T2,
∂y
(b) To solve the above boundary value problem by the method of separation of variables,
we first need to reduce the number of non-homogeneous boundary conditions from
three to one. For this purpose, we introduce the transformation
θ= T − T1 (2)
Substituting Equation (2) into Equation (1), we obtain the transformed differential
equation and BCs as follows
∂ 2θ ∂ 2θ
+ =
0
∂x 2 ∂y 2
θ (0, y ) 0,=
= θ (1, y ) 0, (3)
∂θ
( x, 2) = 0, θ ( x, 0) = T2 − T1 := θ 0
∂y
-2-
We now apply the separation of variables technique to solve the problem. Let
θ ( x, y ) = X ( x)Y ( y ) (4)
Substituting the above into Equation (3)1 and dividing by XY, we obtain
X ′′ Y ′′
= − (5)
X Y
For the above equality to hold for any x or y, both sides have to be equal to the same
constant, say β. Thus, we have
X ′′ − β X =
0
(6)
Y ′′ + β Y =
0
The partial differential equation has thus been reduced to two ordinary differential
equations. From the homogeneous boundary conditions in (3), we can derive some
boundary conditions for X and Y , as detailed below
θ (0, y ) =X (0)Y ( y ) =0 ⇒ X (0) =0 or otherwise trivial sol (7)
θ (1, y ) = X (1)Y ( y ) = 0 ⇒ X (1) = 0 (8)
∂θ
( x, 2) = X ( x)Y ′(2) =0 ⇒ Y ′(2) = 0 (9)
∂y
Now we proceed to find non-trivial solutions of (6) satisfying conditions (7)-(9).
For β = λ 2
X ′′ − λ 2 X =
0
⇒ X = A1eλ x + A2 e − λ x
Y ′′ + λ 2
Y =0
From the boundary conditions (7) and (8):
X (0) = A1 + A2 = 0
λ −λ
⇒ A1 =A2 =0 - trivial solution
X (1) =A1e + A2 e =0
For β = 0
X ′′ =0 ⇒ X =B1 + B2 x
From the Boundary conditions (7) and (8):
= B=
X (0) 1 0
- trivial solution
= B=
X (1) 2 0
For β = −λ 2
X ′′ + λ 2 X =
0
Y ′′ − λ Y =
2
0
⇒ X = C1 cos λ x + C2 sin λ x, Y = C3eλ y + C4 e − λ y (10)
-3-
We shall now apply the boundary conditions, (7)-(9), to determine λ and some
constants.
= C=
X (0) 1 0
λ 0
= C2 sin=
X (1) λ nπ as C2 ≠ 0 or otherwise trivial sol
⇒ =
Y ′(2)= ( λC e 3
λy
− λ C4 e − λ y ) y =2
(
= λ C3e 2 λ − C4 e −2 λ ) ⇒ C=
4 C3e 4 λ
C2 sin(nπ x)
∴ X ( x) =
( )
Y ( y ) = C3 eλ y + eλ (4− y ) = C3 e nπ y + e nπ (4− y ) ( )
Hence a desired solution may be expressed as
θ n ( x,= = Cn ( e nπ y + e nπ (4− y ) ) sin(nπ x) , n=1, 2,...
y ) XY
As the problem is linear, a more general solution may be obtained from a
superposition of the form
∞
θ ( x, y )
= ∑ C (e π
n =1
n
n y
)
+ e nπ (4− y ) sin ( nπ x ) (11)
To determine Cn , we now apply the remaining boundary condition (the non-
homogeneous boundary condition) (3)5,
∞
θ ( x, 0)
= θ=
0 ∑ C (1 + e π ) sin ( nπ x ) .
n =1
n
4n
(12)
Now, the right hand side of (12) can be considered as the Fourier sine series
expansion of the function f(y)=θ0 . Thus
2θ 0 2θ 0
( )
2 ∫ θ 0 sin ( nπ x ) dx [ − cos nπ x=
]0
1
Cn 1 + e 4 nπ = 1 − (−1) n
1
=
0 nπ nπ
2 1 − (−1) n θ 0
∴ Cn = 4 nπ
nπ 1 + e( )
Substituting the above into Equation (11), we then obtain the final solution
2θ 0 1 − (−1) n nπ y nπ (4− y )
∞
=θ ( x, y ) ∑ e +e (
sin ( nπ x ) ) (13)
π n =1 n 1 + e 4 nπ ( )
Thus
2 (T2 − T1 ) ∞ 1 − (−1) n nπ y nπ (4− y )
T =θ + T1 =T1 + ∑ e +e( )
sin ( nπ x ) .
π n =1 n 1 + e(4 nπ
)
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Question 2 Solution
(a) The initial traffic density is as shown in the figure below
ρ
ρm / 2
x
ξl 0 ξ L ξr
V Vm (1 − ρ 2 / ρ m
(b) From the given velocity - traffic density model= 2 , we have
)
traffic flux ρ ) ρ=
F (= V Vm ( ρ − ρ 3 / ρ m2 )
3ρ 2
∴ ′
F (ρ ) =
Vm 1 − 2 .
ρm
At t=0 and x= ξl < 0, ρ = ρm / 2 and thus the characteristic that intersects the x-axis at ξl
satisfies the equation
dx 3ρ m2 / 2 1
= F ′( ρ / 2) =
Vm 1 − =− Vm
dt
m
ρm
2
2
x = ξ
t =0 l
1
− Vmt + ξl
⇒ x=
2
x ξ r > L , ρ=0 and thus the characteristic that intersects the x-axis at ξ r is defined by
At t=0 and =
dx
′(0) Vm ,
= F= x t =0 = ξ r
dt
∴ x =Vmt + ξ r
On the characteristic that intersects the x-axis at ξ where 0 < ξ < L , the traffic density is
ρm ( L − ξ )
ρξ = (A3.1)
2L
and the equations defining the characteristic are
dx 3ρξ2 3ρ m2 ( L − ξ ) 2 / (2 L2 ) 3( L − ξ ) 2
= F ′( ρ ξ ) =V
m 1 − 2
=
m
V 1 − m 1 −
=V
dt ρ m ρ 2
m 2 L2
x t =0 = ξ
3( L − ξ ) 2
⇒ x = Vm 1 − 2 t +ξ
2 L
-7-
The above equation can be solved to yield ( L − ξ ) in terms of x and t and consequently from (A3.1)
determine ρ in terms of x and t. For this purpose, let η= L − ξ to get
3η 2
x Vm 1 − 2 t + L − η
=
2L
3Vm t
⇒ 2
η 2 + η + x − Vm t − L =0
2L
3V t 6V t
−1 ± 1 − 4 m2 ( x − Vm t − L) −1 ± 1 + m2 ( L + Vm t − x)
=⇒ η = 2L L
2
3Vm t / L 3Vm t / L2
Since η = L − ξ > 0 , we must choose the positive sign in the solution for η . Hence, from (A3.1),
the density, as a function of x and t, is given by
6Vm t
−1 + 1 + 2 ( L + Vm t − x)
ρ mη
ρ (=
x, t ) ρ (ξ=
, 0) = ρm L
2L 3 2Vm t / L
Based on the above results, the characteristics are sketched in the figures below.
t
x=-Vmt/2 x=Vmt+L
ρm / 2 ρ=0
x
O L
(b) In summary, the density function at later times (t>0) is as follows
1
ρm / 2 x < − Vmt
2
6Vm t
−1 + 1 + L2 ( L + Vm t − x) 1
=ρ ρm − Vmt < x <Vmt + L
3 2V t / L 2
m
0 x >Vmt + L
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