MODULE 8: THE FOURIER TRANSFORM METHDOS FOR PDES
Lecture 3
10
Heat Flow Problems
In this lecture we shall study some applications of the Fourier transform in solving the
heat flow problems where the spatial domain is infinite or semi-infinite.
1
Heat flow problem in an infinite rod
Consider the heat flow in an infinite rod where the initial temperature is u(x, 0) = f (x).
We shall prove that if the function f (x) is continuous and either absolutely integrable i.e.,
∫ ∞
|f (x)|dx < ∞
−∞
or bounded (i.e., |f (x)| ≤ M ∀x), then the following IVP problem has a solution u(x, t)
which is continuous throughout the half-plane t ≥ 0, −∞ < x < ∞.
PDE:
IC:
ut (x, t) = α2 uxx (x, t),
u(x, 0) = f (x),
−∞ < x < ∞, t > 0,
−∞ < x < ∞,
(1)
(2)
with u(x, t), ux (x, t) → 0 as x → ±∞, t > 0.
The stepwise solution procedure is given below.
Step 1. (Transforming the problem to an IVP in ODE )
We apply FT F to the PDE (1) and IC (2) and use the properties of FT to reduce the
given Cauchy problem to an IVP for an ODE. Let
F[u] = û(ω, t)
F[f (x)] = fˆ(ω).
Taking the FT of both sides of the PDE (1) and IC (2) with respect to the x variable, we
obtain
F[ut ] = α2 F[uxx ]
F[u(x, 0)] = F[f (x)].
Using the properties of the FT
F[ut ] =
d
û(ω, t),
dt
F[uxx ] = −ω 2 û(ω, t)
we have
d
û(ω, t) = −α2 ω 2 û(ω, t),
dt
û(ω, 0) = fˆ(ω).
(3)
(4)
MODULE 8: THE FOURIER TRANSFORM METHDOS FOR PDES
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Step 2. (Solving the transformed problem)
Note that (3) is a first-order IVP for an ODE in t for each fixed ω. The solution to this
problem is given by
2 2
û(ω, t) = fˆ(ω)e−α ω t .
(5)
Step 3. (Finding the inverse transform)
To find the solution u(x, t), we take inverse transform, with t fixed, to obtain
u(x, t) = F −1 [û(ω, t)]
= F −1 [fˆ(ω)e−α
2 ω2 t
].
Step 4. (Using convolution property of the inverse FT )
Using the convolution property of F −1 , we write
u(x, t) = F −1 [fˆ(ω)e−α
2 ω2 t
]
= F −1 [fˆ(ω)] ∗ F −1 [e−α ω t ]
]
[
2
1
−( x 2 )
= f (x) ∗ √
e 4α t
2α2 t
∫ ∞
(x−ω)2
1
√
f (ω)e− 4α2 t dω.
=
2 α2 πt −∞
2
2
REMARK 1.
• Note that integrand is made up of two terms i.e., the initial temperature f (x) and
the function
(x−ω)2
1
G(x, t) = √
e− 4α2 t .
2 α2 πt
The function G(x, t) is called Green’s function or impulse-response function which
is the temperature response to an initial temperature impulse at x = ω.
• The major drawback of the FT method is that all functions can not be transformed.
Only functions that damp to zero sufficiently fast as |x| → ∞ have FTs.
2
Heat flow problem in a semi-infinite rod
Consider the heat flow in a semi-infinite region with the temperature prescribed as a
function of time at x = 0.
MODULE 8: THE FOURIER TRANSFORM METHDOS FOR PDES
12
EXAMPLE 2. Solve the problem
PDE:
BC:
IC:
ut (x, t) = α2 uxx (x, t),
0 < x < ∞, t > 0
u(0, t) = b0 t > 0,
(7)
−∞ < x < ∞,
u(x, 0) = 0,
(6)
(8)
with u(x, t), ux (x, t) → 0 as x → ∞.
Since 0 < x < ∞, we may wish to use a transform. Since u is specified at x = 0, we
should try to use Fourier sine transform (and not the Fourier cosine transform). We solve
this problem with the following steps.
Step 1. (Transforming the problem)
Notice that u is specified at x = 0. Let Fs [u] = ûs (ω, t). Now taking FST of both sides of
(6) and noting the following properties of FST
√ ∫ ∞
2
Fs [ut ] =
ut (x, t) sin(ωx)dx
π 0
]
[√ ∫
d
2 ∞
=
ut (x, t) sin(ωx)dx
dt
π 0
=
=
d
Fs [u]
dt
d
ûs (ω, t).
dt
and
√
2
ωu(0, t)
π
√
2
= −ω 2 ûs (ω, t) +
ωu(0, t)
π
√
2
= −ω 2 ûs (ω, t) +
b0 ω,
π
where in the last step we have used BC u(0, t) = b0 , we arrive at the ODE
(
)
√
d
2
2
2
ûs (ω, t) = α −ω ûs (ω, t) +
b0 ω .
dt
π
Fs [uxx ] = −ω Fs [u] +
2
Next, taking FST of the IC (8), we obatin
Fs [u(x, 0)] = Fs [0] −→ ûs (ω, 0) = 0.
Thus, we transform the original problem (6)-(8) to an IVP in ODE:
√
d
2 2
2 2
ûs (ω, t) + α ω ûs (ω, t) =
α b0 ω,
dt
π
ûs (ω, 0) = 0.
MODULE 8: THE FOURIER TRANSFORM METHDOS FOR PDES
13
Step 2.(Solving the transformed problem)
Using the standard method of solving ODE, the solution is given by
√
2 b0
2 2
ûs (ω, t) =
(1 − e−ω α t ).
πω
(9)
Step 3. (Finding the Inverse Transform)
Applying the inverse FST to both sides of (9), we find that
[√
]
2
b
2
2
0
u(x, t) = Fs−1 [ûs (ω, t)] = Fs−1
(1 − e−ω α t )
πω
∫ ∞
2
sin(ωx)
2 2
=
b0
(1 − e−α ω t )dω
π
ω
[ 0
]
x
= b0 erf c( √
) ,
2α2 t
where erf c(y) is the complementary error function given by
√ ∫ ∞
2
2
erf c(y) =
e−τ dτ.
π y
Hence, the solution of the heat conduction problem is
(
)
x
u(x, t) = b0 erf c √
.
2α2 t
Practice Problems
1. Solve the following IVP:
ut = uxx , −∞ < x < ∞, t > 0,
u(x, 0) = 2x, −∞ < x < ∞,
u(x, t), ux (x, t) → 0 as x → ±∞, t > 0.
2. Let f (x) ∈ C(R) be an odd function. If f (x) is absolutely integrable on R and
f (x) → 0 as x → ±∞ then show that the unique continuous solution of the problem
ut = α2 uxx , −∞ < x < ∞, t > 0,
u(x, 0) = f (x), −∞ < x < ∞,
u(x, t), ux (x, t) → 0 as x → ±∞, t > 0.
is also odd in the variable x. Show that the conclusion is false if the BC is dropped.
MODULE 8: THE FOURIER TRANSFORM METHDOS FOR PDES
3. Apply appropriate FT to solve the IBVP:
ut = uxx , x > 0, t > 0,
u(x, 0) = x, x > 0,
ux (0, t) = e−t , t > 0,
u(x, t), ux (x, t) → 0 as x → ∞, t > 0.
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