Example: Heat flow
Notes
Let us first solve the heat equation analytically before discussing
the numerical method to solve it.
Example: Heat equation in one spatial dimension.
∂T
∂2T
=κ 2,
∂t
∂x
where κ is the thermal diffusivity.
Open boundaries: T (x, t) defined on −∞ < x < +∞ and t ≥ 0.
Also, require that T (x, t) → 0 as x ± ∞
Initial value problem: T (x, t = 0) = Φ(x)
We will use the Fourier transform in order to solve this problem.
Solving Heat flow with an integral transform
We have two independent variables x, t. Apply Fourier transform
to spatial variables x at constant t:
Z +∞
1
F[T (x, t)] = √
T (x, t)e −ikx dx
2π −∞
Notes
What is the Fourier transform of ∂T /∂x ?
Z +∞
∂T −ikx
1
F[Tx ] = √
e
dx
2π −∞ ∂x
Integrating by parts, we obtain
−ikx
∞
Z +∞
e
ik
F[Tx ] = √ T (x, t)
+√
T (x, t)e −ikx dx
2π
2π −∞
−∞
Since T (x, t) vanishes as x → ±∞, the first term vanishes and we
have
F[Tx ] = ikF[T ]
Solving Heat flow with an integral transform
Notes
The Fourier transform of ∂T /∂t is simply
Z +∞
1
∂T −ikx
F[Tt ] = √
e
dx
2π −∞ ∂t
∂
=
F[T (x, t)]
∂t
using Leibniz rule.
Similarly,
∂2
F[T (x, t)]
∂t 2
2
F[Txx ] = (ik) F[T (x, t)] = −k 2 F[T (x, t)]
F[Ttt ] =
Solving Heat flow with an integral transform
Notes
Now apply Fourier transform to the heat equation (at constant t):
F[Tt ] = F[κTxx ]
∂
F[T ] = −k 2 κF[T ]
∂t
Note: Fourier transform method for solving this problem is
appropriate since T (x, t) vanishes at x → ±∞ according to the
BC.
Let us denote τ (k, t) = F[T (x, t)]
(1)
(2)
Solving Heat flow with an integral transform
Notes
∂τ (k, t)
= −k 2 κτ (k, t)
(3)
∂t
Equation (3) is now a simple ordinary differential equation. Heat
equation is much easier to solve in the Fourier domain!
The solution is
τ (k, t) = e −k
2 κt
τ (k, 0)
(4)
Still need to transform the initial condition T (x, 0) = Φ(x):
F[Φ(x)] = F[T (x, t = 0)] = τ (k, 0)
(5)
Solving Heat flow with an integral transform
Notes
Combining eqs. (4) and (5)
τ (k, t) = e −k
2 κt
(6)
F[Φ(x)]
In order to obtain solution in real space, apply the inverse Fourier
transform:
T (x, t) = F−1 [τ (k, t)] = F−1 [e −k
2 κt
F[Φ(x)]]
(7)
However, can use the convolution theorem on the right hand side.
Recall
F[f ⊗ g ] = F[f ]F[g ]
Therefore,
f ⊗ g = F−1 [F[f ]F[g ]]
(8)
Notes
Now apply this to our solution (eq. (7)):
2
Let F[f ] = e −k κt and F[g ] = F[Φ(x)].
It follows that g = F−1 [F[Φ(x)]] = Φ(x) by definition.
Also,
f
2
= F−1 [e −k κt ]
1
2
= √
e −x /(4κt)
2κt
(9)
(10)
In the last step we used the known inverse Fourier transform of a
Gaussian (see table of Fourier transforms - lecture 2):
1
2
2
F[e −αx ] = √ e −k /(4α)
2α
Solving Heat flow with an integral transform
Notes
Since
T (x, t) = f ⊗ g
(11)
according to eqs. (8) and (7) we have
T (x, t) =
=
=
1
2
e −x /(4κt) ⊗ Φ(x)
2κt
Z +∞
1
1
2
√
√
e −(x−ξ) /(4κt) Φ(ξ)dξ
2π −∞
2κt
Z +∞
1
2
√
e −(x−ξ) /(4κt) Φ(ξ)dξ
4πκt −∞
√
(12)
(13)
(14)
This is the fundamental solution of the heat equation with open
boundaries for an initial condition T (x, t = 0) = Φ(x).
Green’s function
Notes
The fundamental solution is the the convolution of the initial
conditions with the Green’s function.
Once the Green’s function of a linear PDE is known (for given
B.C.’s), it can be solved for arbitrary initial conditions via a
convolution integral (eq. (14)).
The Green’s function is the solution of the PDE for a delta
(impulse) function.
Let the initial condition be Φ(x) = δ(x). Then
Z +∞
1
2
T (x, t) = √
e −(x−ξ) /(4κt) δ(ξ)dξ
4πκt −∞
1
2
= √
e −x /(4κt)
4πκt
Note: T (x, t) → δ(x) as t → 0.
Heat diffusion from delta function impulse
Notes
The delta function at t = 0 becomes spread out as time goes on.
Finite difference methods
Notes
In the following, let’s see how to solve PDE’s using finite difference
methods.
For simplicity, assume u to be a function of two independent
variables x and t, only.
The continuous function u is then discretized on a grid in the x, t
plane where the points are separated by ∆x and ∆t, respectively.
Notation: u(x, t) → u(j∆x, n∆t), where j and n are integers.
Approximating the derivatives
In order to discretize the PDE we need to approximate the
derivatives appearing in the PDE by finite differences.
There are three possible approximations for the first order (partial)
derivative: ∂u(j∆x, n∆t)/∂x
The forward (Euler) difference:
n − un
uj+1
j
∆x
The backward difference:
n
ujn − uj−1
∆x
The centered difference:
n − un
uj+1
j−1
2∆x
Notes
Approximating the derivatives
The same applies for the temporal partial derivative, except that j
is kept constant. e.g. the forward difference is
Notes
ujn+1 − ujn
∂u(j∆x, n∆t)
≈
∂t
∆t
The finite differences follow from the Taylor expansion of u(x) (at
constant t):
u(x + ∆x) =
u(x − ∆x) =
1
u(x) + u 0 (x)∆x + u 00 (x)(∆x)2
2
1
+ u 000 (x)(∆x)3 + O(∆x)4
6
1
u(x) − u 0 (x)∆x + u 00 (x)(∆x)2
2
1
− u 000 (x)(∆x)3 + O(∆x)4
6
(15)
(16)
(17)
(18)
where in the last line ∆x → −∆x
Approximating the derivatives
Notes
From the Taylor series expansion we can find the local errors:
The forward (Euler) difference (eq.(17)):
u(x + ∆x) − u(x)
+ O(∆x)
∆x
The backward difference (eq.(18)):
u(x) − u(x − ∆x)
+ O(∆x)
∆x
The centered difference(subtract eq.(18) from eq. (17)):
u(x + ∆x) − u(x − ∆x)
+ O(∆x)2
2∆x
Note that the global (accumulated) error may be different from the
local one.
Approximating the derivatives
Notes
What about the second derivative? By adding the two Taylor
expansions (17) and (18) and solving for u 00 (x) we obtain
u 00 (x) =
u(x + ∆x) − 2u(x) + u(x − ∆x)
+ O(∆x)2
(∆x)2
(19)
In index notation, this reads for a partial second derivative at
constant t:
n − 2u n + u n
uj+1
∂u(j∆x, n∆t)
j
j−1
≈
2
∂x
(∆x)2
(20)
This is the second centered difference.
Similar expression for partial time derivative.
Discretizing the heat equation
Notes
Let us use a finite difference scheme to evaluate the heat equation
numerically in one spatial dimension.
Tt = κTxx
(21)
Again, assume open boundaries with initial condition:
T (x, t = 0) = Φ(x)
Apply forward difference to Tt and second centered difference to
Txx :
n − 2T n + T n
Tjn+1 − Tjn
Tj+1
j
j−1
=κ
(22)
∆t
(∆x)2
where Tjn = T (j∆x, n∆t).
Discretizing the heat equation - errors
Notes
There are two types of errors in this finite difference scheme:
Truncation errors: The local truncation error for the second
centered difference is ∆x 2 and for the time derivative ∆t.
These errors may accumulate and given an overall global
truncation error that may be bigger than the local one.
Round off error: This occurs in real computations, since the
computer only retains a certain number of digits for floating
point numbers. e.g. in C, float variables have 8 digit and
double have 16 digit precision. These round off errors also
accumulate. Always use double variables for finite difference
schemes!
Discretizing the heat equation
Notes
Solving for
Tjn+1
we obtain
n
n
Tjn+1 = (1 − 2s)Tjn + s(Tj+1
+ Tj−1
)
where s =
(23)
κ∆t
.
(∆x)2
This is an explicit difference scheme, since the values of the
(n + 1)th time step are explicitly given in terms of the value at
earlier times.
Example 1
Notes
Let’s look at a simple example. For simplicity, set s = 1. Then the
difference scheme simplifies to
n
n
Tjn+1 = Tj+1
− Tjn + Tj−1
(24)
Tjn=0
Suppose Φ(xj ) is a simple impulse:
= 1 for a particular point
j in space and zero everywhere else.
Can integrate this by “marching in time”:
Example 1
Notes
Something has gone wrong ! Expect the impulse to diffuse away.
According to the maximum principle of diffusion PDE’s, the
maximum of T (x, t) occurs either at the initial conditions t = 0 or
at the boundaries. In our example, this means that T (x, t) < 1 for
t > 0.
Instead the difference equations has given us an approximation
where the central point is 19 and growing.
It turns out that for this particular difference scheme ∆t and ∆x
cannot be chosen arbitrarily.
Example 2
Notes
Let’s consider another example of the heat equation with the
following boundary conditions:
Tt = Txx
for 0 < x < π, t > 0
T =0
at x = 0, π
The initial condition is
T (x, 0) = Φ(x) =
x
π−x
for 0 < x ≤ π/2
for π/2 < x < π
(25)
First, let’s solve this analytically using separation of variables.
Example 2
Notes
If the PDE is separable, then its solution can be expressed as
T (x, t) = A(x)B(t)
This allows us to turn the PDE into two ordinary differential
equations (ODE’s). Substitution yields
∂T
∂t
∂B(t)
A(x)
∂t
= κ
∂2T
∂x 2
= κB(t)
(26)
∂ 2 A(x)
∂x 2
(27)
Dividing by (A(t)B(x)κ) we obtain
1 ∂B(t)
1 ∂ 2 A(x)
= −k 2 =
κB(t) ∂t
A(x) ∂x 2
where −k 2 is the constant of separation. LHS and RHS are equal
for all x and t, which are independent of each other.
Example 2
Notes
The two ODE’s are
1 ∂B(t)
κB(t) ∂t
1 ∂ 2 A(x)
A(x) ∂x 2
= −k 2
(28)
= −k 2
(29)
The solutions are
A(x) = a cos(kx) + b sin(kx)
B(t) = ce −κk
2t
(30)
(31)
where a, b, c are constants of integration.
The boundary conditions T (x = 0, t) = T (x = π, t) = 0, require
a = 0 and k to be an integer.
Example 2
Notes
Therefore,
T (x, t) = bk sin(kx)e −κk
2t
(32)
where k is an integer.
Fourier analysis allows us to obtain the solution for our specified
initial condition:
T (x, t) =
∞
X
bk sin(kx)e −κk
2t
(33)
k=1
where
(
bk =
(−1)(k+1)/2
πk 2
0
for odd k
for even k
(34)
Example 2
Notes
This plot shows the initial temperature distribution Φ(x) and
T (x, t) for t = 3π 2 /80.
As t → ∞, T → 0 everywhere.
Numerically solving example 2
Notes
Now solve numerically using our difference scheme
n
n
Tjn+1 = (1 − 2s)Tjn + s(Tj+1
+ Tj−1
)
(35)
κ∆t
with s = (∆x)
2 = 5/11 = 0.45 and ∆x = π/20, where we have
discretized space with J + 1 = 21 points.
The discrete boundary and initial conditions are
u0n = uJn = 0 and uj0 = Φ(j∆x)
where j = 0, 1, 2, . . . J.
Numerically solving example 2
Notes
Agrees very well with analytical solution.
Numerically solving example 2
Notes
However, if we choose s = 5/9 = 0.55, the finite difference scheme
yields wild oscillations similar to what we obtained for the impulse
IC earlier.
Stability of finite difference scheme
Notes
In the following we will derive the stability criterion for this
difference scheme:
Let’s solve the difference scheme by separation of variables:
Tjn = Aj Bn
(36)
Substitution in the difference scheme (eq. (35)) yields
Aj Bn+1 = (1 − 2s)Aj Bn + s(Aj+1 Bn + Aj−1 Bn )
(37)
Dividing by Aj Bn we get
Aj+1 + Aj−1
Bn+1
= ξ = 1 − 2s + s
Bn
Aj
(38)
where ξ is the constant of separation.
Stability of finite difference scheme
Notes
Thus,
Bn+1
=ξ
Bn
(39)
Bn = ξ n B0
(40)
which has the solution
where B0 is some constant.
Now solve the spatial part:
1 − 2s + s
Aj+1 + Aj−1
=ξ
Aj
(41)
We wish to investigate the eigenmodes of the solution. Therefore,
assume a plane wave solution with wave number k (ignore
boundary conditions for this analysis):
Aj = e ikj∆x
(42)
Stability of finite difference scheme
Notes
Substituting yields
e ik(j+1)∆x + e ik(j−1)∆x
= ξ
e ikj∆x
ik∆x
−ik∆x
1 − 2s + s(e
+e
) = ξ
1 − 2s + s
1 − 2s + 2s cos(k∆x) = ξ
1 − 2s(1 − cos(k∆x)) = ξ
2
k∆x
1 − 4s sin
= ξ
2
For the difference scheme to be stable, require
|ξ(k)| ≤ 1 for all k
since the amplitude grows as ∝
ξn
(43)
(eq. (40))
Stability of finite difference scheme
Notes
Therefore,
2 k∆x
1 − 4s sin
≤1
2
for all k. This leads to the following stability criterion:
4s ≤ 2
2κ∆t
≤ 1
(∆x)2
Physically, this means that the maximum allowed time step ∆t is
the diffusion time across a cell of size ∆x (up to a numerical
2
factor). The diffusion time across some length λ goes as tλ ∼ λκ
von Neumann stability analysis
Notes
This is the von Neumann stability analysis:
The von Neumann analysis looks at the evolution of the
eigenmodes of the solution by substituting
ujn = ξ n e ikj∆x
into the difference equation, where ξ(k) is the amplification
factor
For the difference scheme to be stable, require
|ξ(k)| ≤ 1 + O(∆t) for all k
Otherwise, eigenmodes grow exponentially in time.
Notes
Notes
Notes