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Numerical Analysis of Differential Equations
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Numerical Solution of Parabolic Equations
6 Numerical Solution of Parabolic Equations
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Numerical Analysis of Differential Equations
6.1
The One-Dimensional Model Problem
We consider the following initial boundary value problem (IBVP) modelling
heat flow in a thin rod, i.e., in one space dimension:
x ∈ (0, 1), t > 0,
(6.1a)
u(0, t) = g0 (t),
t > 0,
(6.1b)
u(1, t) = g1 (t),
t > 0,
(6.1c)
u(x, 0) = u0 (x),
x ∈ [0, 1]
(6.1d)
ut = κuxx ,
with given (constant) heat conductivity κ (which we set to one in the following) as well as (possibly time-dependent) Dirichlet boundary values g0 , g1
and initial data u0 .
6.1 The One-Dimensional Model Problem
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• Steady-state version:
uxx = 0,
u(0) = g0 ,
u(1) = g1 .
• Analogous IBVP in 2D and 3D:
ut = ∆u
+
initial and boundary data.
• Related: linear ordinary differential equation
ut = Au,
u : t 7→ u(t) ∈ Rn .
Here linear differential operator ∂xx in place of matrix A.
6.1 The One-Dimensional Model Problem
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Series solution by separation of variables: in special cases an analytic representation of the (exact) solution may be constructed using the
technique of separation of variables. This is helpful for checking numerical approximations and provides important insight into the structure of the
solution.
Inserting the special trial solution u(x, t) = f (x)g(t) into the PDE ut = uxx
results in
0
00
f g = f g,
i.e.
f 00
g0
=
= const. =: −k 2 .
g
f
For each value of k we obtain a solution
2
uk (x, t) = e−k t sin(kx)
of ut = uxx . The boundary condition u(0, t) = u(1, t) = 0 constrains k to the
discrete values
k = km := mπ,
m ∈ N.
6.1 The One-Dimensional Model Problem
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Due to the linearity and homogeneity of ut = uxx any linear combination of
these solutions is also a solution. If we succeed in finding coefficients am
in such a way that
∞
X
u0 (x) =
am sin(mπx),
m=1
then the series
u(x, t) :=
∞
X
am e−m
2
π2 t
sin(mπx)
m=1
solves the complete IBVP (6.1).
Since the functions {sin(mπx)}∞
m=1 form a complete orthogonal system of
the function space L2 (0, 1), this is possible for all u0 ∈ L2 (0, 1).
The coefficients are given by
Z
am = 2
1
u0 (x) sin(mπx) dx.
0
6.1 The One-Dimensional Model Problem
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Example:
1
u0 (x) = 1 − 2|x −
1
2 |.
Here the coefficients are
am
8
mπ
= 2 2 sin
,
m π
2
m ∈ N.
N=1
N=3
N=5
N=19
u0(x)
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.2
0.4
0.6
0.8
1
x
P
Partial sums N
m=1 am sin(mπx)
of the Fourier series of u0 .
6.1 The One-Dimensional Model Problem
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0
10
t=0
t=1/2
−20
10
−40
1
−60
0.8
−80
0.6
−100
0.4
−120
0.2
10
10
10
10
10
0
1
2
3
4
5
6
7
m
Decay of first 4 nonzero Fourier
coefficients of u0 .
0
1
0
0.2
0.5
0.4
0.6
0
x
0.8
1
t
Fourier series solution of IBVP (6.1) with
κ = 1 and u0 (x) = 1 − 2|x − 21 |
in domain (x, t) ∈ [0, 1] × [0, 1].
6.1 The One-Dimensional Model Problem
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We first discretize the IBVP (6.1) in the spatial variable x only, leaving time
t continuous. To this end we proceed as in the elliptic case and introduce
the grid points
0 = x0 < x1 < · · · < xJ < xJ+1 = 1
using a fixed grid spacing ∆x = 1/(J + 1) and approximate
uxx |x=xj ≈ [A∆x u]j ,
with
A∆x
j = 1, 2, . . . , J,
1
tridiag(1, −2, 1).
=
∆x2
(6.2)
If u = u(t) denotes the vector with components
uj (t) ≈ u(xj , t),
6.1 The One-Dimensional Model Problem
t > 0,
j = 1, 2, . . . , J,
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then (6.1) is transformed into the semi-discrete system of ODEs
u 0 (t) = A∆x u(t) + g (t),
(6.3a)
u(0) = u0 ,
(6.3b)
with [u0 ]j = u0 (xj ), j = 1, 2, . . . , J as well as
g (t) = 1/(∆x2 )[g0 (t), 0, . . . , 0, g1 (t)]> ∈ RJ .
We can now solve (6.3) with known numerical methods for solving ODEs.
Introducing the fixed time step ∆t > 0, we set
Ujn ≈ [u(tn )]j ≈ u(xj , tn ),
tn = n∆t.
The approximation of the solution of a time-dependent PDE as a system of
ODEs along the “lines” {(xj , t) : t > 0} is known as the method of lines.
6.1 The One-Dimensional Model Problem
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Applying the explicit Euler method to (6.3) (setting g0 (t) = g1 (t) = 0 for
now) leads to
Ujn+1
=
Ujn
∆t
n
n
n
Uj−1 − 2Uj + Uj+1 ,
+
2
∆x
1 ≤ j ≤ J,
n = 0, 1, 2 . . . .
This corresponds to the finite difference approximation
n
n
Ujn+1 − Ujn
Uj−1
− 2Ujn + Uj+1
=
2
∆t
∆x
|
{z
} |
{z
}
≈ut
(6.4)
≈uxx
of the differential equation (6.1a).
In matrix notation:
U n+1 = (I + ∆t A∆x )U n ,
n = 0, 1, 2, . . . ,
with U n = [U1n , U2n , . . . , UJn ]> .
6.1 The One-Dimensional Model Problem
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We define the local discretisation error of the difference scheme (6.4) to
be the residual obtained on inserting the exact solution into the difference
scheme:
d(x, t) :=
u(x, t + ∆t) − u(x, t) u(x − ∆x, t) − 2u(x, t) + u(x + ∆x, t)
−
.
2
∆t
∆x
Using Taylor expansions in (x, t) one easily obtains:
2
∆t ∆x
deE (x, t) =
−
uxxxx + O(∆t2 ) + O(∆x4 ).
2
12
(6.5)
Terminology: the explicit Euler method for solving the heat equation is
consistent of first order in time and of second order in space.
6.1 The One-Dimensional Model Problem
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Besides the asymptotic statement (6.5) we also require upper bounds for d.
Truncating the Taylor expansions with a remainder term, we obtain in time
∆t2
u(x, t + ∆t) = (u + ∆t ut ) |(x,t) +
utt (x, τ ),
2
τ ∈ (t, t + ∆t).
Proceeding analogously in x yields
∆x2
∆t
utt (x, τ ) −
uxxxx (ξ, t),
d (x, t) =
2
12
eE
and we obtain, setting µ :=
ξ ∈ (x − ∆x, x + ∆x),
∆t
∆x2 ,
∆t
∆x2
∆t
eE
|d (x, t)| ≤
Mtt −
Mxxxx =
2
12
2
1
Mtt +
Mxxxx ,
6µ
(6.6)
assuming |utt | ≤ Mtt und |uxxxx | ≤ Mxxxx on [0, 1] × [0, T ].
6.1 The One-Dimensional Model Problem
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Applying the implicit Euler method to (6.3) one obtains, in place of (6.4),
the implicit difference scheme
n+1
n+1
Ujn+1 − Ujn
Uj−1
− 2Ujn+1 + Uj+1
=
.
∆t
∆x2
(6.7)
The calculation of u n+1 from u n is seen to require the solution of a linear
system of equations with coefficient matrix I − ∆tA∆x .
Here Taylor expansion results in a local discretization error of
2
∆t ∆x
+
uxxxx + O(∆t2 ) + O(∆x4 ).
diE (x, t) = −
2
12
6.1 The One-Dimensional Model Problem
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Applying instead the trapezoidal rule, which for an ODE y 0 (t) = f (t, y (t))
is given by
y
n+1
∆t n
n+1
=y +
f (tn , y ) + f (tn+1 , y
)
2
n
yields another implicit scheme
n
n+1
n+1
n+1 n
n
Ujn+1 − Ujn
U
−
2U
+
U
U
−
2U
+
U
1
j−1
j
j+1
j−1
j
j+1
=
+
,
2
2
∆t
2
∆x
∆x
which in this context is known as the Crank-Nicolson schemea . This method
is also implicit, requiring in each time step the solution of a linear system of
equations with the coefficient matrix I − ∆t
2 A∆x .
For Crank-Nicolson (CN) there holds
dCN (x, t) = O(∆t2 ) + O(∆x2 ).
a J.
C RANK AND P. N ICOLSON (1947)
6.1 The One-Dimensional Model Problem
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6.2
Convergence
All three methods considered so far are consistent, i.e., at all points (x, t)
of the domain we have d(x, t) → 0 as ∆x → 0 and ∆t → 0.
To analyze their convergence, we proceed as in the case of numerical
methods for ODEs and consider a finite time interval t ∈ [0, T ], T > 0
as well as a sequence of grids with grid spacings ∆x → 0, ∆t → 0 and
determine whether at every fixed grid point (xj , tn ) also the global error
u(xj , tn ) − Ujn tends to zero uniformly.
A sequence of grid spacings {(∆x)k , (∆t)k } can approach the point (0, 0)
in different ways. The following figure shows different “refinement curves”
in the (∆x, ∆t)-plane.
We will see that the explicit Euler method converges only if the refinement
satisfies
∆t
1
µ :=
≤ .
∆x2
2
6.2 Convergence
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0.5
∆ t = ∆ x2/4
2
0.45
∆ t = ∆ x /2
0.4
∆ t = ∆ x2
∆t=∆x
∆ t = ∆ x/2
0.35
∆t
0.3
0.25
0.2
0.15
0.1
0.05
0
0
0.1
0.2
0.3
∆x
0.4
0.5
0.6
0.7
Different refinement curves ∆t, ∆x → 0:
red: ∆t/∆x2 = const, blue: ∆t/∆x = const.
6.2 Convergence
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Theorem 6.1 If a sequence of grids satisfies
(∆t)k
1
µk =
≤
(∆x)2k
2
for all k sufficiently large,
and if for the corresponding sequences {jk } and {nk } there holds
nk (∆t)k → t ∈ [0, T ],
jk (∆x)k → x ∈ [0, 1],
then, under the assumption that |uxxxx | ≤ Mxxxx uniformly in [0, 1] × [0, T ],
the approximations generated by the explicit Euler method (6.4) converge
to the exact solution u(x, t) as k → ∞ uniformly on [0, 1] × [0, T ].
6.2 Convergence
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−2
10
expl. Euler dx=1/10
expl. Euler dx=1/20
Crank Nicolson, dx = 1/10
Crank Nicolson, dx = 1/20
−3
10
−4
En
10
−5
10
−6
10
−7
10
0
0.2
0.4
0.6
0.8
1
t
n
Error E(tn ) = maxj |un
j − Uj | of numerical solution of model problem(6.1) with
g0 = g1 = 0 and u0 (x) = x(1 − x) obtained with explicit Euler method with
∆t = ∆x2 /2 and the Crank-Nicolson method with ∆t = ∆x/2.
6.2 Convergence
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6.3
Stability
In the method of lines, an ODE solver is applied to the semidiscrete system
(6.3). Recalling the regions of absolute stability of the explicit and implicit
Euler and trapezoidal methods, we are led to investigate the behavior of
the eigenvalues of the matrix A∆x in (6.2) as ∆x → 0.
2
2
2
1.5
1.5
1.5
1
1
1
0.5
0.5
0.5
0
0
0
−0.5
−0.5
−0.5
−1
−1
−1
−1.5
−1.5
−1.5
−2
−3
−2
−1
explicit Euler
b =1+λ
b
R(λ)
6.3 Stability
0
1
−2
−1
0
1
implicit Euler
b = 1
R(λ)
b
1−λ
2
3
−2
−2
−1
0
1
2
trapezoidal
b
b = 1+λ/2
R(λ)
b
1−λ/2
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The eigenvalues of A∆x (cf. (5.7)) are given by
λj = −
4
2 jπ∆x
sin
,
∆x2
2
j = 1, 2 . . . , J.
The eigenvalue of largest magnitude is
4
−4
2 π
2
2
λJ = −
sin
+ π + O ∆x .
(1 − ∆x) =
∆x2
2
∆x2
Therefore, in order to satisfy
|R ∆t λj | ≤ 1
for all j = 1, 2, . . . , J
in the explicit Euler method, it is necessary that
∆t
1
≤
.
∆x2
2
(6.8)
Both implicit Euler and the trapezoidal rule are absolutely stable methods. Therefore, as all eigenvalues of A∆x lie in the left half plane, the requirement of absolute
stability places no constraints on the time step ∆t.
Since the CN scheme (trapezoidal rule) is consisten of order 2 in both x and t, this
suggests a time step ∆t ≈ ∆x.
6.3 Stability
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Absolute stability describes the behavior of a numerical approximation u n
at time tn of the solution of an ODE as n → ∞. The appropriate stability
concept for the convergence analysis of numerical methods for solving
IVPs for PDEs was developed by Laxa and Richtmyerb .
All three methods considered so far can be written in the formc
U n+1 = B(∆t)U n + g n (∆t),
n = 0, 1, . . .
(6.9)
with different matrices B(∆t) given in each case by
B eE (∆t) = I + ∆t A∆x ,
(6.10a)
B iE (∆t) = (I − ∆t A∆x )−1 ,
−1 ∆t
∆t
B CN (∆t) = I −
A∆x
I+
A∆x .
2
2
(6.10b)
(6.10c)
a P ETER
D. L AX (∗ 1906)
b R OBERT DAVIS R ICHTMYER (1910–2003)
c We assume that ∆t is chosen as a fixed given function of ∆x.
6.3 Stability
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A linear finite difference scheme of the form (6.9) is called Lax-Richtmyer
stable, if for any fixed stopping time T there exists a constant KT > 0 such
that
kB(∆t)n k ≤ KT ,
∀∆t > 0 and n ∈ N, n∆t ≤ T.
Theorem 6.2 (Lax-Richtmyer equivalence theorem) A consistent linear
finite difference scheme (6.9) is convergent if, and only if, it is Lax-Richtmyer
stable.
Examples: For the explicit Euler method applied to (6.1) the resulting
matrix (6.10a) is normal, implying kB(∆t)k2 = ρ(B(∆t)). If condition (6.8)
is satisfied, we have kB(∆t)k2 ≤ 1, implying that the explicit Euler method
is Lax-Richtmyer stable and therefore convergent.
The implicit Euler and Crank-Nicolson schemes are also Lax-Richtmyer
stable but without constraints on the mesh ratio or time step.
6.3 Stability
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Remark 6.3 In all three schemes considered so far we were able to show
the stronger statement kB(∆t)k ≤ 1, which is sometimes called strong stability. However, strong stability is not necessary for Lax-Richtmyer stability.
More precisely, for Lax-Richtmyer stability it is necessary that there exist a
constant α such that
kB(∆t)k ≤ 1 + α∆t
(6.11)
for all sufficiently small time steps ∆t, which is evident from the inequality
kB(∆t)n k ≤ (1 + α∆t)n ≤ eαT ,
6.3 Stability
∀ n∆t ≤ T.
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6.4
Von-Neumann Analysis
A formal procedure based on Fourier analysis for determining stability
bounds was introduced by von Neumann a .
In general, von Neumann analysis only yields necessary conditions, which
are also sufficient only in special cases, among these linear PDEs with constant coefficients, unbounded spatial domain (Cauchy problem) or periodic
boundary conditions, uniform grid.
Fundamental fact: in the continuous case (x ∈ R) the functions eiξx , ξ ∈ R,
are eigenfunctions of the differential operator ∂x , i.e.,
∂x eiξx = iξ eiξx ,
and thus are also eigenfunctions of any linear differential operator with
constant coefficients.
a J OHN VON
6.4 Von-Neumann Analysis
N EUMANN (1903–1957)
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On the unbounded spatial grid {xj = j∆x}j∈Z the grid function
vj := eiξxj = eiξj∆x ,
j ∈ Z,
is an eigenfunction of any difference operator with constant coefficients.
Example: For the central difference operator δ0 defined as
(δ0 v)j := (vj+1 − vj−1 )/(2∆x) there holds
eiξ(j+1)∆x − eiξ(j−1)∆x
eiξ∆x − e−iξ∆x iξj∆x
δ0 vj := (δ0 v)j =
=
e
2∆x
2∆x
i
=
sin(ξ∆x) vj
∆x
The power series expansion
(ξ∆x)3
i
i
sin(ξ∆x) =
ξ∆x −
+ O (ξ∆x)5
= iξ + O(∆x2 ξ 3 )
∆x
∆x
6
shows that the eigenvalue of δ0 approximates the corresponding eigenvalue of ∂x .
6.4 Von-Neumann Analysis
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Any grid function on {xj = j∆x : j ∈ Z} can be represented as the superposition
π/∆x
Z
1
vj =
2π
−π/∆x
of “grid waves” eiξj∆x , ξ ∈ [ −π
,
∆x
vb(ξ) = ∆x
vb(ξ)eiξj∆x dξ,
π
]
∆x
∞
X
j∈Z
with Fourier coefficients vb(ξ) given by
−iξj∆x
vj e
,
j=−∞
h −π
π i
.
ξ∈
,
∆x ∆x
If the magnitude of grid functions and their Fourier coefficients are measured by
kvk =
∆x
∞
X
!1/2
|vj |2
Z
and
!1/2
π/∆x
|b
v (ξ)|2 dξ
kb
vk =
,
−π/∆x
j=−∞
then Parseval’s identity states that
kb
vk =
6.4 Von-Neumann Analysis
√
2πkvk.
(6.12)
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The bound kBk ≤ 1 + α∆t requires verifying
kBuk ≤ (1 + α∆t)kuk
∀u.
According to (6.12), an equivalent condition is
c ≤ (1 + α∆t)kb
kBuk
uk
∀b
u,
(6.13)
which is easier to show because the Fourier coefficients are decoupled.
For a one-step method one obtains
u
bn+1 (ξ) = g(ξ) u
bn (ξ)
with a so-called amplification factor g(ξ). If we are able to show that
|g(ξ)| ≤ 1 + α∆t.
with a constant α which is independent of ξ, then this implies (6.13).
6.4 Von-Neumann Analysis
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Examples:
(a) For the explicit Euler scheme applied to (6.3) we obtain the amplification
factor
4∆t
2 ξ∆x
g(ξ) = 1 −
.
sin
∆x2
2
(b) For Crank-Nicolson,
1 + z(ξ)
g(ξ) =
,
1 − z(ξ)
6.4 Von-Neumann Analysis
∆t z(ξ) =
−1 + cos(ξ∆x) ≤ 0
2
∆x
∀ξ.
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