Combination of Variable Solutions
Combination of variables solutions to partial differential equations are suggested whenever the
physical situation indicates that two independent variables can be combined to produce only one
independent variable. There are general methods for finding the appropriate combined variable.
However, in ChEN 430, you will be told what combined variable to use. The solution procedure
is to substitute this combined variable into the differential equation for the original independent
variables. If the method works, the resulting differential equation will involve only the
combined variable. In this way, you will have converted a partial differential equation into an
ordinary differential equation, which is then solved by solution methods for ordinary differential
equations.
Example
Consider the dependent variable θ that depends on independent variables t and x as specified in
the following partial differential equation and restricting conditions (initial and boundary
conditions):
∂θ
∂ 2θ
=γ
∂t
∂ x2
(1)
θ = θo
at t = 0
for 0 < x < ∞
(2)
θ = θ1
at x = 0
for t > 0
(3)
θ = θo
at x → ∞
for t > 0
(4)
In this problem, the variable θ is forced to depart from its initial value of θo. The distance x at
which departures from θo are felt grows as time increases. This suggests that it might be possible
to combine (ratio) x and t in some way to obtain a combined variable (η) that will transform the
partial differential equation in t and x into an ordinary differential equation in η. That is,
θ = f ( x , t ) = g (η)
where
η = h( x , t )
where f and g are functions that will be determined when the problem is solved. The functional
form of the combined variable, h(x,t) will be suggested in the problem statement.
For the example specified in Eqs. (1) through (4), the combined variable is expected to be:
η=
x
4γ t
(5)
The first step in the solution procedure is to obtain the terms specified by the partial differential
equation. The combination of variable solution will have worked if the resulting differential
equation involves only η.
First, find ∂θ/∂t in terms of η. Using chain rule,
∂θ dθ ∂η
=
∂t dη ∂t
(6)
1
where
∂η ∂
=
∂t ∂t
LM
MN
OP
PQ
x
=
4γ t
LM
N
x
1 1
− 3/ 2
2t
4γ
OP
Q
(7)
Consequently,
dθ
∂θ
x
=−
∂t
d η 2 t 4γ t
η dθ
=−
2t d η
(8)
Next find ∂2θ/∂x2 in terms of η. Begin by finding ∂θ/∂x. From chain rule,
∂θ dθ ∂η
=
∂ x dη ∂ x
where
∂η ∂
=
∂x ∂x
LM
MN
(9)
OP
PQ
x
=
4γ t
1
4γ t
(10)
Consequently,
∂θ dθ
=
∂ x dη
1
4γ t
(11)
η dθ
=
x dη
Differentiating Eq. (11) a second time yields:
LM
MN
∂2θ ∂ dθ
=
∂ x2 ∂ x d η
Since
∂
∂x
LM
MN
OP
PQ
LM
MN
d dθ
1
=
dη dη
4γ t
OP
PQ
∂η dθ ∂
1
+
4γ t ∂ x d η ∂ x
LM
MN
1
4γ t
OP
PQ
(12)
OP
PQ
1
=0
4γ t
Eq. (12) simplifies to give,
∂2θ d2θ 1
=
∂ x2 d η2 4γ t
(13)
Substituting Eqs. (8) and (13) into the differential equation (Eq. 1):
2
−
η dθ
1 d2θ
=γ
2t d η
4γ t d η2
(14)
which simplifies to yield the following ordinary differential equation:
d 2θ
dη 2
+ 2η
dθ
=0
dη
(15)
The recommended combination variable did transform the second-order partial differential
equation in x and t (Eq. 1) into a second-order ordinary differential equation in η.
The combination variable also must be introduced into the restricting conditions (Eqs. 2-4):
θ = θo
at t = 0
or
at η → ∞
(17)
θ = θ1
at x = 0
or
at η = 0
(18)
θ = θo
at x → ∞
or
at η → ∞
(19)
Notice that the conditions defined by Eq. (17) and (19) become the same. This seems reasonable
since the second-order differential equation requires only two conditions on η.
Eq. (15) is a linear, but non-constant coefficient equation. It can be solved by reduction of order.
Substitute for p = ∂θ/∂η into Eq. (15) to obtain:
dp
+ 2η p = 0
dη
(20)
z z
which can be separated and integrated:
dp
= − 2 η dη
p
(21)
to yield:
ln p = −η 2 + I1
(22)
After exponentiation Eq. (21) becomes:
p=
dθ
= I 1 exp −η 2
dη
e j
(23)
where I 1 = exp( I1 ) . Integrating Eq. (23) yields
z z
η
θ=
dθ = I 1
e j
exp −η 2 dη + I 2
(24)
0
Solving for I2 using Eq. (18) yields I2 = θ1. Solving for I1 using Eq. (17),
3
z
∞
θ o = I1
e j
exp −η 2 dη + θ 1
(25)
0
where
z
∞
π
e j
exp −η 2 dη =
(26)
2
0
After simplification, Eq. (25) yields:
b
I1 = θ o − θ 1
g
2
(27)
π
Substituting for I1 and I2 into Eq. (24):
b
θ = θ o −θ1
z
η
g
2
π
e j
exp −η 2 dη + θ 1
(28)
0
Recall that the error function [(i.e., erf(x)] is defined as:
z
x
2
erf ( x ) =
π
e j
exp − t 2 dt
(29)
0
As a consequence, Eq. (28) can be written simply as:
b
g
θ = θ 1 − θ 1 − θ o erf (η)
(30)
Substituting for the definition of η, we obtain the final result:
b
g d
θ = θ 1 − θ 1 − θ o erf x
4γ t
i
(31)
4