Stability Analysis of Finite Dierence Schemes for the 1D Heat
Equation on an Unbounded Spatial Domain
Bachelor's Thesis
Written by: Réka Dávid
Analytical Mathematics
Advisor: Dr. Ágnes Havasi
Réka Dávid
Department of Applied Analysis and Computational Mathematics
Eötvös Loránd University, Faculty of Science, Institute of Mathematics
Eötvös Loránd University
Faculty of Science
2015
Acknowledgement
Firstly, I would like to express my sincere gratitude to my advisor Dr. Ágnes Havasi for the continuous
support, her patience, motivation, and immense knowledge. Her guidance helped me in all the time of
writing this thesis.
Besides my advisor, I would like to thank my family for supporting me throughout my education.
Contents
1 Introduction
1
2 The heat conduction problem
2
3 Semigroup of solution operators
5
4 Discretization of the heat conduction problem
8
4.1
Transfer operators r and p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2
Dierence Schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
5 Consistency, convergence, stability
5.1
14
Sucient and Necessary Conditions for Stability . . . . . . . . . . . . . . . . . . . . . . . 16
6 Fourier analysis
6.1
9
21
Transform operator F . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
7 Implicit schemes
26
8 Conclusion
29
A The source code of the programming task
II
B The main le
III
Chapter 1
Introduction
Ever since the early days of computing, the use of computers to solve scientic problems has been
a driving force. Most real physical processes are governed by partial dierential equations (PDEs). A
PDE is an equation stating a relationship between a function of two or more independent variables
and the partial derivatives of this function with respect to these independent variables. In many cases,
simplifying approximations are made to reduce the governing PDEs to ordinary dierential equations
(ODEs) or even to algebraic equations. However, because of the ever increasing requirement for more
accurate modeling, engineers and scientists are more and more involved to solve the actual PDEs that
govern the physical problem being investigated.
Most studies are devoted to numerically approximate the solution of partial dierential equations by
nite dierence methods. Our expectation for the numerical solution is that it converges to the exact
solution by renement of the grid. We will see that, if the dierence scheme is consistent with the PDE,
the stability is necessary and sucient for providing convergence. The aim of the present work is to
investigate conditions for stability of the one-dimensional heat
conduction problem when discretized with
nite dierence schemes on an unbounded spatial domain.
Wolfgang Hackbusch's The Concept of Stability in Numerical Mathematics [1] was followed for the main
structure of the thesis. First, we dene the one-dimensional heat conduction problem on two Banachspaces L2 (R) and C(R) of continuous functions in spatial variable x with the respective norms. Next, we
examine the exact solution of the problem and some of its qualitative properties. In the third chapter,
we introduce the operator semigroup background for the further investigations. Then, we move on to
the discretization of the heat conduction problem by nite dierence methods. We introduce the discrete
spaces l∞ and l2 and the transfer operators to provide mapping from the continuous spaces onto the
discrete ones. We also dene the dierence scheme in terms of that shift operator for the numerical
approximation and apply our considerations on a programming task. In Chapter 5, we present the classical
theory of Lax and Richtmyer stating that for a consistent scheme stability is sucient and necessary
for convergence. Then we derive some simple sucient and necessary conditions for stability. Next, we
introduce the theory of Fourier analysis, dene the Fourier synthesis and verify further conditions for
stability. The schemes examined so far in the thesis are, at best, conditionally stable. Hence, at last, to
obtain unconditionally stable schemes, we admit the implicit dierence schemes and some more conditions
for stability.
1
Chapter 2
The heat conduction problem
The
heat conduction problem describes the distribution of temperature in a given region over time.
We restrict our considerations to one spatial variable x. The changing in time is proportional to the
second derivative of the temperature in space, yields the parabolic equation
∂2u
∂u
=α 2
∂t
∂x
where α, the
for t > 0, x ∈ R
(2.1)
heat conduction coecient, is assumed to be constant. We restrict α to be 1 (otherwise
transform t → αt or x →
√
αx). Let B be a space of functions in spatial variable x. The desired solution
of the equation is denoted by u(t, x) of the independent variables x and t. By u(t) = u(t, ·) ∈ B we denote
the function partially evaluated at t. Hence, u(t)(x) and u(t, x) are equivalent notations. Let I = [0, T ]
be the interval in which t varies. Then, the domain of u(·, ·) is the set Σ := I × R with I = [0, T ] while
x ∈ R. The spatial domain R is chosen also for the periodic case as unbounded to avoid boundary
conditions. If the solutions are assumed to be 2π -periodic in x, we use the domain Σ := I × [0, 2π] with
u(t, 0) = u(t, 2π). It is possible to rewrite the problem to an abstract Cauchy problem with dierential
operator A. We assume that the domain of A is dened as DA := {v ∈ B : Av ∈ B is dened} ⊂ B ,
hence the partial dierential equation takes the following form: the task is to nd a continuous function
u : I → DA which solves the Cauchy-problem
(
du(t)
dt
= Au(t) for ∀t ∈ I
for some u0 ∈ DA ⊂ B
u(0) := u0
(2.2)
Focusing on the dierential operator A, we discuss the model
A :=
∂2
∂x2
As it is well known, the solution of problem (2.1) with a continuous initial function is given by
1
u(t, x) = √
4πt
Z
∞
u0 (ε) exp
−∞
− x−ε
4t
2
2 !
dε
for t > 0, x ∈ R.
(2.3)
The solution satises (2.1) and provides point-wise convergence limt→0 u(t, x) = u0 (x) for all x ∈ R.
The representation (2.3) shows that the solution is innitely often dierentiable at t > 0 although the
initial value u0 (x) is only continuous. However, the solution exists only for t > 0.
Our considerations are built on two Banach spaces, generally denoted by B :
• B := C(R) is the space of complex-valued and uniformly continuous functions with nite supremum
kukB = kuk∞ := sup{|u(x)| : x ∈ R}
• B := L2 (R) is the space of complex-valued,
measurable and square-integrable functions with a
qR
2
2
|v(x)| dx. This Banach space is also a Hilbert space with the
nite L -norm kukB = kuk2 :=
R
dot product
Z
hu, vi :=
u(x)v(x)dx.
(R)
For the 2π -periodic case, we can redene the spaces as
• Cper (R) := {v ∈ C(R) : v(x) = v(x + 2π) ∀x ∈ R}
• L2per (R) := {v ∈ L2 (R) : v(x) = v(x + 2π) for almost all x ∈ R}.
Further on, we will verify a qualitative property of the exact solution of the
heat equation, namely, that
its norm is not increasing in time. The next proposition is necessary to prove Lemma 2.3.
Proposition 2.1. The equality
√
Z
∞
4πt =
exp
−∞
!
−δ 2
dδ
4t
holds for t > 0.
Proof. Suppose that
(1) means applying polar-coordinate substitution
(2) means applying u = r2 substitution.
Denote by I the integral on the right-hand side of (1.3). Then
!
!
Z ∞Z ∞
−y 2
−(δ 2 + y 2 )
(1)
I :=
exp
dy =
exp
dδdy =
4t
4t
−∞
−∞ −∞
Z ∞ Z 2π
Z ∞
Z ∞
1
(2)
2 1
2 1
=
r · exp(−r
)dϕdr = 2π
r · exp(−r
)dr = π
exp(−u )du =
4t
4t
4t
0
0
0
0
∞
1 = π − 4t · exp(−u ) = 4tπ
4t 2
Z
∞
! Z
∞
−δ 2
dδ
exp
4t
−∞
0
√
From I = 4tπ the equation follows. Lemma 2.2
(∞-norm case).
Let u(t) ∈ B := C(R) be a solution of problem (2.1). The inequality
ku(t)k∞ 6 ku0 k∞ holds for ∀t > 0.
3
Proof. We suppose u : I × R → DA ⊂ C(R) and
magnitude of the solution (2.3). Then
1
|u(t)| = √
4πt
∞
Z
|u0 (ε)| exp
−∞
∈ C(R).
∂2u
∂x2
Let us dene c := ku0 k∞ and take the
!
def c Z ∞
−(x − ε)2
exp
dε 6 √
4t
4πt −∞
!
−(x − ε)2
2.1
dε = ku0 k∞ .
4t
We take the supremum of both sides, thus ku(t)k∞ 6 ku0 k∞ follows. Lemma 2.3
(2-norm case).
Let u(t) ∈ B := L2 (R) be a solution of problem (2.1). The inequality
ku(t)k2 6 ku(0)k2 holds for ∀t > 0.
Proof. We suppose that u : I × R → DA ⊂ L2 (R) and
Z
t00
Z
2
u(t)
t0
R
∂2
u(t)dxdt = 2
∂x2
Z
t00
h
u(t)
t0
∂2u
∂x2
∈ L2 (R),
i+∞
∂
u(t)
dt − 2
∂x
−∞
furthermore t00 > t0 > 0. Hence,
Z
t00
Z
t0
R
∂u(t) ∂u(t)
·
dxdt
∂x
∂x
where the rule of integration by parts was used for the inner integral. Since the solution decays towards
plus and minus innity, the rst term on the right-hand side is zero. On the other hand
00
ku(t
2
)k2
0
− ku(t
2
)k2
def
Z
Z
00 2
u(t ) dx −
=
R
Z
t00
=2
t0
Z
0 2
∂
u(t) u(t)dxdt = −2
∂t
R
t00
t0
t00
u(t ) dx =
R
Z
Z Z
Z
R
R
t0
∂
u(t)2 dtdx
∂t
∂u(t) ∂u(t)
·
dxdt 6 0
∂x
∂x
where we used the Newton-Leibnitz rule and the fact that the solution u is continuous, thus the integrals
can be interchanged. Our conclusion is that ku(t)k2 is weakly decreasing, i.e. ku(t)k2 6 ku0 k2 for ∀t >
0. 4
Chapter 3
Semigroup of solution operators
To analyze the consistency, convergence and stability of nite dierence methods applied for the heat
equation later, the notion of operator semigroups will be helpful. In this chapter, besides Hackbusch's
book [1], we rely on [11], [12] and chapter 9 of [5]. We know that solution u(t) ∈ B is strong or classical,
if it satises problem (2.1) in R with initial function u0 from B0 , where B0 := C ∞ (R) ∩ B , a dense subset
of B . For the abstract formulation of the problem, we can dene the
solution operator S(t) as
S(t) : u0 7→ u(t)
at xed t > 0. So, the solution operator S(t) maps the initial function u0 ∈ B0 to the solution at time
t. It is easy to show that S(t) is a linear mapping. Moreover, since ku(t)kB 6 ku0 kB , therefore S(t) is
bounded and continuous too. Hence, S(t) ∈ L(B0 , B) for all t > 0. The denition allows the derivation
of some properties of the solution from the properties of S(t).
The translation of the function u0 satises the property S(t + s) = S(t)S(s). So that, by using S(t), it
is possible to redene problem (2.1) to a general starting point s > 0.
Proposition 3.1. The property S(t)S(s) = S(t + s) holds for all t, s > 0:
Proof. Suppose the case of strong solution, thus S(t) ∈ L(B0 , B) and S(τ )u0 = u(τ ). Set u(s) equal to
S(s)u0 := û0 for a xed s > 0. Stepping further, û(t) := u(t + s) = S(t)û0 . Then,
∂ û(t)
∂u(t + s) def
=
= Au(t + s) = Aû(t)
∂t
∂t
is problem (2.1) with a possible starting point û0 . It yields to the property
S(t + s)u0 = u(t + s) = û(t) = S(t)û0 = S(t)S(s)u0 =⇒ S(t + s) = S(t)S(s). Next, we show the property that S(t) and A commute, which results in the fact that on DA S(t)
maps into DA . We can call {S(t), 0 > t} a one-parameter semigroup with neutral element S(0) = I and
A as the innitesimal generator of the semigroup.
5
Proposition 3.2.
A
and S(t) commute, i.e. AS(t) = S(t)A on B0 .
Proof. Consider a strong solution u(t) = S(t)u0 for u0 ∈ B0 . Then,
1
du(t)
u(t + h) − u(h) def
3.1
= lim [S(t + h) − S(t)]u0 =
= lim
h→0 h
h→0
dt
h
1
u(h) − u0
∂u0
= S(t) lim [S(h) − S(0)]u0 = S(t) lim
= S(t)
= S(t)[Au0 ]
h→0 h
h→0
h
∂t
(2.2)
A[S(t)u0 ] = Au(t) =
where we used that S(0) = I and the continuity of S(t) on B0 . AS(t) = S(t)A follows. Since S(t)[Au0 ] ∈ B is dened for all u0 ∈ B0 , also A[S(t)u0 ] has this property. This proves that
S(t) : DA → DA for t > 0. So far, S(t) was dened on B0 and it was satisfying:
• S(t + s) = S(t)S(s)
t, s > 0,
• S(0) = I,
• Av = limt→0
S(t)v−v
t
for ∀v ∈ DA .
In the case of B = C(R), the elements of which are uniformly continuous, limt→0 kS(t)v − vk = 0 also
follows for all v ∈ B0 . Another notation of S(t) is eAt . Now, we recall the density of B0 and that
ku(t)kB 6 ku0 kB has been proved for the heat conduction problem. We infer that operator S(t) can be
extended uniquely and continuously onto B .
Theorem 3.3. Suppose S(t) ∈ L(B0 , B) and B0 is dense in B . Then, S(t) can be extended uniquely
and continuously onto B. Furthermore, the extended Se (t) ∈ L(B, B) and the original S(t) ∈ L(B0 , B)
have equal norms, i.e. sup{kS(t)vkB / kvkB : v ∈ B0 \{0}} = sup{kSe (t)vkB / kvkB : v ∈ B\{0}}.
The point of the extension to B is that, due to the density of B0 ⊂ B , for all v ∈ B there exists a
sequence (v0,n ) ⊂ B0 for which lim v0,n = v . Since v0,n ∈ B0 , ∀n ∈ N, S(t) can be applied to each element
v0,n of this sequence, therefore the extension of S(t) onto B can be dened as: S(t)v := lim S(t)v0,n ,
where v ∈ B . To continue the thread, the existence of a strong solution (2.3) cannot be always guaranteed
for u0 ∈ B\B0 , but there is a resulting function for any u0 ∈ B such that u(t) := S(t)u0 ∈ B is a
weak
generalized solution. The weak solution is a function for which the derivatives may not all exist but
which nonetheless satises the dierential equation in some precisely dened sense, namely the variational
equation ; we multiply the equation (2.1) by a smooth test function v ∈ C ∞ (R) and integrate both sides
or
over R.
In the generalized case, the solutions are u(t) := S(t)u0 ∈ B with u0 ∈ B . We would like to provide
that u(t) = S(t)u0 tends to u0 ∈ B while t → 0. It can be shown that if u0 is uniformly continuous,
then for the solution u(t) of the heat conduction problem u → u0 uniformly on R. In other words,
ku(t) − u0 kB → 0, so limt→0 k[S(t) − S(0)]u0 kB = 0. Therefore, in this case the semigroup has the
property limt→0 k[S(t) − S(0)]u0 kB = 0 too, i.e. the semigroup is continuous. Proposition 3.1 of the case
where u0 ∈ B0 is also true for the presented one, since B0 is dense in B . The range t > 0 may be reduced
to an interval [0, τ ] with τ > 0, which leads to the following
Proposition 3.4. Suppose that Kτ
:= sup06t6τ kS(t)kB→B < ∞
dte
kS(t)kB←B 6 Kτ τ
6
for τ > 0. Then,
for any t > 0
where dxe := min{n ∈ Z : x 6 n}.
Proof. We will divide the proof into three cases:
(1). Suppose that t = 0, then
ltm
τ
t
=0
thus Kτd τ e = 1 and S(t) = I yield to kS(t)kB←B = 1.
(2). Suppose that t ∈ (0, τ ]. Then
ltm
τ
=1
def
thus kS(t)kB→B 6 sup kS(t)kB→B = Kτ1 .
06t6τ
(3). Suppose that t > 0 is out of the interval [0, τ ]. There exists n ∈ Z and δ with 0 < δ < τ so that
t = nτ + δ . Then, by the semigroup property (3.1) we have
kS(t)kB→B = kS(δ)S(τ )n kB→B .
Now, t = nτ + δ implies
n=
t
t−δ
6
τ
τ
and since kS(0)kB→B = 1 yields Kτ > 1, we have
dδe
dτ e
n+d τδ e
kS(t)kB→B 6 Kτ τ (Kτ τ )n = Kτ
d nττ+δ e
= Kτ
dte
= Kτ τ .
For each case, the inequality with τ > 0 follows. Further on in the next chapters, we shall refer to the inequality with τ = T and t ∈ I = [0, T ], hence
d Tt
e = 1. Moreover, S(t) is supposed to be uniformly bounded on I = [0, T ]:
for ∀t ∈ I = [0, T ].
kS(t)kB←B 6 KT
(3.1)
For the heat equation, the condition (3.1) is satised with KT = 1, since we showed that the norm of the
solution was not increasing in time and kS(t)kB←B = supu0 6=0 {ku(t)kB / ku0 kB } 6 ku0 kB / ku0 kB = 1.
7
Chapter 4
Discretization of the heat conduction
problem
We move on to the discretization of the
heat conduction problem for numerical approaches. In this
chapter, our considerations build on chapter 3 of [9]. The properties of the generalized case resulted in a
uniformly continuous mapping from u0 ∈ B into u(t) ∈ B on the bounded interval t ∈ [0, T ]. Since the
spatial variable x is dened on R, it is possible to create an innite grid of step size ∆x > 0 as
G∆x := {x = v∆x : v ∈ Z}.
The time variable ranges in the interval [0, T ], so we dene the nite grid of step size ∆t > 0 as
I∆t := {t = µ∆t 6 T : µ ∈ N0 }.
Correspondingly, from the Cartesian product of both grids, it is also possible to dene the rectangular
grid
Σ∆t
∆x := {(t, x) ∈ Σ :
x
t
∈ Z and
∈ N0 }.
∆x
∆t
(4.1)
where we used the notation Σ := I × R. The step sizes ∆t and ∆x are chosen independently, but
are connected by λ =
U:
Σ∆t
∆x
∆t
∆x2 .
Using the rectangular grid as the domain, we can dene the grid function
→ C with the notation
Uvµ := u(µ∆t, v∆x)
for (µ∆t, v∆x) ∈ Σ∆t
∆x
(4.2)
For a xed parameter t = µ∆t, the innite sequences on Σ∆t
∆x are
U µ := {u(µ∆t, v∆x) : v ∈ Z} = {Uvµ }v∈Z
Uv := {u(µ∆t, v∆x) : µ ∈ N0 } = {Uvµ }µ∈N0
8
(4.3)
The set of two-sided innite grid functions together with the operations of addition and scalar multiplication form a linear space, on which we will use two dierent norms, leading to the normed spaces
l2 and l∞ :
s
kU kl2 :=
∆x
X
|Uv |2
for the space l2 , a Hilbert-space
v∈Z
(4.4)
for the space l∞ , a Banach-space
kU kl∞ := sup{|Uv | : v ∈ Z}
4.1 Transfer operators r and p
The continuous Banach space can be mapped onto the linear space of grid functions via a transfer
operator. The transfer operator r is dened as a
restriction
r := r∆x : B → lp ,
where ∆x is the step size of the innite grid G∆x . We need to distinguish the mapping in the sense of
B = C(R) ∨ L2 (R). An obvious choice of ru ∈ l∞ is
u ∈ C(R) =⇒ ru ∈ l∞ with (ru)j := u(j∆x)
for j ∈ Z
By substituting it into (4.4), the ∞-norm can be calculated as the supremum of the absolute values of the
discrete (ru)j points. A choice ru ∈ l2 is not that self-evident, since L2 (R) functions have no well-dened
point evaluations. In this case, the point evaluation is replaced by the integral mean value of the function
on the sub-interval [xj − ∆x/2, xj + ∆x/2] around the grid point xj :
1
u ∈ L (R) =⇒ ru ∈ l with (ru)j :=
∆x
2
2
Z
(j+ 21 )∆x
u(x)dx
(j− 12 )∆x
for j ∈ Z
We get the 2-norm as the square root of the sum of discrete squared integrals multiplied by ∆x. We can
suppose that the norm of r is bounded with respect to the norms k·kl∞ ←C(R) and k·kl2 ←L2 (R)
kr∆x klp ←B 6 Cr
for ∀∆x > 0
(4.5)
We show that the restrictions considered above satisfy the condition (4.5) with Cr := 1. In the case of
the 2-norm we use the
Theorem 4.1
true that
Cauchy-Schwarz inequality :
(The Cauchy-Schwarz inequality).
For all vectors x and y of a dot product space it is
|hx, yi|2 6 hx, xihy, yi,
Equivalently, by taking the square root of both sides, and referring to the norms of vectors generated by
the dot product, the inequality is written as
|hx, yi| 6 kxk kyk .
Moreover, the two sides are equal i x and y are linearly dependent.
9
Lemma 4.2. The restrictions r satisfy the condition (4.5) with Cr
norms.
=1
with respect to both considered
Proof. The basic idea is that from B → lp we need to prove the inequality krukl
kr∆x kl ←B 6 1 is also satised.
(1). For p = ∞, it is possible to use the denition of supremum of function u, then:
p
6 kukB ,
thus
p
def
krukl∞ = sup{|u(j∆x)| : j ∈ Z} 6 sup{|u(x)|}, x ∈ R} = kukB=C(R)
where u(x) is the continuous function and u(j∆x), j ∈ Z represents a subset of function values.
(2). For p = 2,
2
krukl2
= ∆x
1 X |(ru)j | =
j
j
∆x
X
2
Z
(j+ 12 )∆x
(j− 21 )∆x
2
u(x)dx ,
we apply the Cauchy-Schwarz inequality for the interior of the sum, then:
2
krukl2
6
X Z
j
(j+ 21 )∆x
|u(x)|2 dx =
(j− 12 )∆x
So far, we have been dealing with
Z
2
|u(x)|2 dx = ku(x)kB=L2 . R
restriction operator r. The prolongation p acts in the reverse
direction, such that the operator maps the discrete space onto the continuous space, so
p := p∆x : lp → B.
We can also suppose that p is bounded with respect to the norms k·kB:=C(R)←l∞ and k·kB:=L2 (R)←l2 as
kp∆x kB←lp 6 Cp
for ∀∆x > 0.
(4.6)
Furthermore, suppose rp = I , which indicates that p is a right-inverse of r. We also need to distinguish
the mapping in the sense of l∞ and l2 .
Lemma 4.3. The prolongations with the following choices of p satisfy condition (4.6) with Cp = 1:
(1) p is a piecewise linear interpolation:
v ∈ l∞ 7→ pv ∈ C(R)
with (pv)(x) = ϑvj + (1 − ϑ)vj+1
where x = (j + ϑ)∆x, j ∈ Z, ϑ ∈ [0, 1)
(2) p is a piecewise constant interpolation:
v ∈ l2 7→ pv ∈ L2 (R) with (pv)(x) = vj
h
1
1 where x ∈ j − ∆x, j + ∆x , j ∈ Z.
2
2
Proof. The basic idea is that from lp → B we need to prove inequality kpvkB 6 kvkl , thus kr∆x kB←l
1 is also satised. During the poof, we use the triangle inequality and that function v is continuous.
p
10
p
6
In case (1):
k(pv)(x)kB=C(R) = sup sup{|ϑvj + (1 − ϑ)vj+1 | : ϑ ∈ [0, 1)} 6
j∈Z
6 sup(max{|vj |, |vj+1 |}) = sup{|vj |} = kvkl∞
j
j
where, we use the fact that pv is linear and inhomogeneous on interval [xj , xj+1 ], so its supremum on
the subinterval equals sup{|vj |, |vj+1 |}.
In case (2):
Z
2
|(pv)(x)|2 dx =
k(pv)(x)kB=L2 (R) =
XZ
R
6
X
j∈Z
Z
(j+ 12 )∆x
j∈Z
! Z
1dx
(j− 12 )∆x
(j+ 12 )∆x
!
(j+ 21 )∆x
1|(pv)(x)|2 dx
(j− 21 )∆x
2
|vj | dx
= ∆x
(j− 12 )∆x
XZ
j∈Z
(j+ 12 )∆x
(j− 12 )∆x
2
|vj |2 dx = kvkl2
then taking the square root of both sides gives the desired form. We say lp with p ∈ {2, ∞} is suited to B if condition (4.5) and (4.6) hold for ∀∆x > 0.
4.2 Dierence Schemes
The next task is to dene the time stepping operator that maps U µ+1 to U µ . In this section, we rely
on lecture 3 of [10], chapter 6 of [8] and chapter 2 of [7]. We prescribe U 0 via Uv0 = ru0 . The
dierence scheme is dened as
Uvµ+1 =
X
µ
aj Uv+j
,
explicit
(4.7)
j∈Z
where aj ∈ R are xed constants. In practice
describes a linear mapping U → U
µ
µ+1
P
j∈Z
is a nite sum, all aj vanish. The dierence scheme
and motivates to dene the linear
C : lp → lp where (CU )v :=
X
aj Uv+j
dierence operator
for U ∈ lp
(4.8)
j∈Z
where p ∈ {2, ∞} and C j : U µ 7→ U µ+j . As a result, a shorter form is U µ+1 = CU µ , and a µ-fold
application leads to U µ = C µ U 0 . The coecients aj are assumed to be scalars. If the scalar equation
d
dt u
= Au for u : I × R 7→ R is replaced by a vector-valued equation, the coecients aj become N × N
matrices. The dierence operator C (and in this case aj ) may depend on the parameter λ and step size
∆t, thereby also on ∆x via λ =
∆t
∆x2 .
Thus, C will be denoted as C = C(λ, ∆t) for the rest of the chapter.
On the other hand, we can dene the
shift operator
Ej : lp 7→ lp , where (Ej U )v := Uj+v .
11
Since the shift operator does not change the lp -norm of U , therefore kEj klp ←lp is equal to 1. The operator
C can be replaced by
C :=
X
aj Ej =
j∈Z
X
(4.9)
Cj .
j∈Z
A simple representative example of a ∆t-dependent dierence operator can be given. Let operator C(λ) be
a suitable dierence
operator for A =
∂2
∂x2
of the equation 2.1. Then the dierential
operator A =
∂2
∂x2
+b
can be discretized by C (λ, ∆t) := C(λ) + ∆t · b.
0
In the case of A =
∂2
∂x2 ,
the dierence quotient
u(t + ∆t, x) − u(t, x)
∆t
for
∂u
∂t
and the second dierence quotient
u(t, x − ∆x) − 2u(t, x) + u(t, x + ∆x)
∆x2
are obvious choices and lead together with λ =
∆t
∆x2
for
∂2u
∂x2
to
2u(t, x) − u(t, x − ∆x) − u(t, x + ∆x)
u(t + ∆t, x) − u(t, x)
=
∆t
∆x2
hence
µ
µ
Uvµ+1 = λUv−1
+ (1 − 2λ)Uvµ + λUv+1
,
and the presented
dierence scheme is as follows:
C : lp 7→ lp ,
(CU )v = λUv−1 + (1 − 2λ)Uv + λUv+1
for U ∈ lp .
(4.10)
Next, we will apply this explicit scheme in a programming task with dierent values of the parameter λ,
and investigate its convergence.
The programming task. Consider the heat conduction problem with initial function
u(x, 0) := 1.
Since the second derivative with respect to x is equal to zero and so the time derivative of the temperature
is zero by the heat equation, the exact solution of the problem is u(x, t) ≡ 1. We perturb the initial
function at x = 0 by a small value, namely, 1e − 5, and solve the problem by using the above explicit
p
∆t/λ). We expect that the numerical
method with ∆t = 0.01 for λ = 0.45 and λ = 0.7 (and ∆x =
solution will not dier very much from the constant 1 solution of the unperturbed problem. It is important
to mention that we set the task of the problem on an innite grid for spatial variable x, however to
calculate the numerical solution, we have to restrict the computation to the grid points of a nite
interval. Here we use the interval [−5, 5] and the time interval was chosen as [0, 1].
12
The source code of the task can be found in Appendix A and Appendix B. Since the applied explicit
formula relies on one more neighboring point at both sides of the chosen spatial interval [xmin, xmax],
we have to start the computation from the wider interval [xmin − ∆x · tmax/∆t, xmax + ∆x · tmax/∆t].
Figure 4.1 shows that the solution is spoiled for λ = 0.7, even if we rene the grid, however for λ = 0.45 the
numerical solution is very close to the constant 1 solution that we would obtain by using the unperturbed
constant 1 initial condition. The dierent colors belong to dierent time layers, as it can be seen below.
Test with dt=.01, lambda=0.45
2
ts = 1
ts = 20
ts = 50
ts = 70
ts = 100
u(x,t)
1.5
1
0.5
0
-5
2
-4
-3
-2
-1
0
1
2
3
4
×10 19
ts = 1
ts = 90
ts = 95
ts = 97
ts = 100
u(x,t)
1
0
-1
-2
-5
2
-4
-3
-2
-1
0
1
2
3
4
×10 248
ts = 1
ts = 990
ts = 995
ts = 997
ts = 1000
0
-1
-2
-5
5
x
Test with dt=.001, lambda=0.7
1
u(x,t)
5
x
Test with dt=.01, lambda=0.7
-4
-3
-2
-1
0
1
2
3
4
5
x
Figure 4.1: The results of the programming task for dierent values of parameter λ.
In the legends, ts should be understood as the number of time-steps of length ∆t that were made during
the calculation. In the following we would like to get an explanation to the bad behavior of the numerical
solution in this example. We will see that for certain choices of the parameter λ instability arises, which
prevents the numerical solution from being convergent.
13
Chapter 5
Consistency, convergence, stability
Let B0 ⊂ DA be a dense subset of B and lp with p ∈ {2, ∞} be suited to B . Furthermore, we denote
the solution by u(t) = S(t)u0 where u0 ∈ B0 . The local discretization error E is dened by
E(t) :=
1
[ru(t + ∆t) − C(λ, ∆t)ru(t)].
∆t
The dierence scheme C(λ, ∆t) is called consistent with respect to the linear spaces lp with p ∈ {2, ∞}
if for all u0 ∈ B0
sup{kE(t)klp : 0 6 t 6 T − ∆t} → 0
as ∆t → 0.
Using error E, the latter condition is equivalent to
sup
06t6T −∆t
k[rS(∆t) − C(λ, ∆t)r]S(t)u0 klp = O(∆t)
for all u0 ∈ B0 .
Next, the denition of convergence and stability refer to the whole Banach space B , not only to a dense
subset B0 . Now, we can suppose the generalized solution denoted by u(t) = S(t)u0 for all u0 ∈ B . The
dierence scheme C(λ, ∆t) is called convergent with respect to lp if
kru(t) − C(λ, ∆t)µ ru0 klp → 0
for ∆t → 0 and µ∆t → t ∈ I = [0, T ].
The dierence scheme is called stable with respect to lp if
sup{kC(λ, ∆t)µ klp ←lp : ∆t > 0, µ ∈ N0 , 0 6 ∆t 6 T } < ∞.
In this case, the stability constant is dened by
K = K(λ) := sup{kC(λ, ∆t)µ klp ←lp : ∆t > 0, µ ∈ N0 , 0 6 ∆t 6 T }.
If the stability condition holds only for certain values of λ, the scheme is called
otherwise unconditionally
(5.1)
conditionally stable,
stable. We will see that the schemes examined so far are, at best, conditionally
stable. Before, we show that consistency and stability with some technical assumptions imply convergence,
so stability is sucient for convergence.
14
Theorem 5.1 (convergence theorem). Suppose
•
r is bounded with respect to lp ,
• S(t)
satises assumption (3.1),
• lp
stability of the dierence scheme C(λ, ∆t),
• lp
consistency,
then the dierence scheme is convergent with respect to lp .
Proof. We divide the proof into two cases, based on u0 ∈ B0 or u0 ∈ B .
Case (1): we suppose u0 ∈ B0 and dene u(t) = S(t)u0 . We split the discretization error E as
ru(t) − C(λ, ∆t)µ ru0 = r[u(t) − u(µ∆t)] + [rS(∆t)µ − C(λ, ∆t)µ r]u0 ,
and use the telescopic sum
rAµ − B µ r =
µ−1
X
B v [rA − Br]Aµ−v−1
v=0
with A := S(∆t), B := C(λ, ∆t) and (4.5) and (5.1). Then
kru(t) − C(λ, ∆t)µ ru0 klp 6 krklp ←B ku(t) − u(µ∆t)kB
+
µ−1
X
k{rS(∆t) − C(λ, ∆t)r}u((µ − v − 1)∆t)klp 6
v=0
6 Kr ku(t) − u(µ∆t)kB +
µ−1
X
K(λ)∆t kE((µ − v − 1)∆t)klp .
v=0
Since u0 ∈ B0 ⊂ DA , the solution is strong, therefore continuous. The rst term is a zero sequence, which
tends to zero while µ∆t → t. Because of the consistency assumption, the local discretization error tends
to zero uniformly, i.e. ∀ε > 0 ∃∆t > 0 for which kE((µ − v − 1)∆t)klp 6 ε. We infer that the telescopic
sum is bounded with K(λ)T ε, which yields that the whole sum tends to zero and proves convergence in
the case of an initial function u0 ∈ B0 .
Case (2): we suppose a general initial function u0 ∈ B and a given ε > 0. One nds u∗0 ∈ B0 that
is suciently close to u0 ∈ B , so that ku0 − u∗0 kB 6 ε/[3Kr max kS(t)kB←B , K(λ)]. The associated
solution u∗ (t) = S(t)u∗0 satises
kr[u(t) − u∗ (t)]klp 6 Kr kS(t)[u0 − u∗0 ]kB 6 Kr kS(t)kB←B ku0 − u∗0 kB 6
ε
3
and
kC(λ, ∆t)µ r[u0 − u∗0 ]klp 6 K(λ)Kr ku0 − u∗0 kB 6
ε
.
3
Together with kru∗ (t) − C(λ, ∆t)µ ru∗0 klp 6 3ε from case (1) for suciently small ∆t and t−µ∆t, it follows
that kru(t) − C(λ, ∆t)µ ru0 klp 6 ε. Therefore, convergence is shown for a general initial function. 15
Theorem 5.2 (stability theorem). Choose lp to be suited to B , so that conditions (3.1), (4.5) and (4.6)
hold. Then, lp convergence implies lp stability.
The latter theorem implies that stability is also necessary for convergence. The point of necessity of
stability for providing convergence is that there are not any sequences ∆tv > 0, µv ∈ N0 with 0 6 µv ∆tv 6
T for which the k·klp ←lp norm of the dierence scheme is not bounded. Due to the compactness of I =
[0, T ], there is a subsequence with (µv ∆tv ) → t ∈ I , which yields that kru(t) − C(λ, ∆tv )µv ru0 klp → 0
for all u0 ∈ B . Therefore the lp -norm of C(λ, ∆tv )µv ru0 is bounded for suciently large values of ν . One
concludes that Cv := C(λ, ∆tv )µv r is a point-wise bounded sequence of operators, therefore uniformly
bounded. Since p is a bounded inverse of r, the boundedness of kC(λ, ∆tv )µv klp ←lp also follows.
The conclusion of theorems 5.1 and 5.2 is that if we suppose (3.1), (4.5), (4.6) and lp consistency, then
lp convergence and lp stability are equivalent. In other words, if the dierence scheme is consistent with
the PDE, then stability is necessary and sucient for providing convergence.
So far, we restricted the analysis to the lp space with p ∈ {2, ∞}. A more involved analysis is necessary
to describe the properties for lp with 2 < p < ∞. On the one hand, for p > 2 the Hilbert structure is lost
and certain properties also change between the case p < ∞ and p = ∞. On the other hand, if stability
estimates hold for both p = 2 and p = ∞, these bounds imply corresponding estimates for the lp and Lp
setting for 2 < p < ∞. In the following lemma, let k·kp←p be the operator norm of L(lp , lp ) or L(Lp , Lp ).
Lemma 5.3
(Riesz-Thorin theorem).
Assume 1 6 p1 6 q 6 p2 6 ∞. Then
β
α
k·kq←q 6 k·kp1 ←p1 k·kp2 ←p2
with α =
q−p1
p2 −p1
and β =
p2 −q
p2 −p1
.
It is clear to see that α + β = 1. As a conclusion, p1 stability k·kp1 ←p1 6 M1 and p2 stability
k·kp2 ←p2 6 M2 imply q stability in the form
k·kq←q 6 M := M1α M2β .
Hence, if a criterion yields both l2 and l∞ stability, then lp stability holds for all 2 6 p 6 ∞.
5.1 Sucient and Necessary Conditions for Stability
The following results belong to the classical stability theory of Lax-Richtmeyer [13].
We assume that kC(λ, ∆t)klp ←lp 6 1 + Kλ ∆t. Then
µ
kC(λ, ∆t)µ klp ←lp 6 kC(λ, ∆t)klp ←lp 6 (1 + Kλ ∆t)µ 6 (eKλ ∆t )µ 6 eKλ T ,
therefore the dierence scheme is lp stable with stability constant K(λ) := eT Kλ . The coecients aj of
C(λ, ∆t) can be used to estimate kC(λ, ∆t)klp ←lp as
X
kC(λ, ∆t)klp ←lp = aj Ej j
lp ←lp
16
6
X
j
|aj | kEj klp ←lp =
X
j
|aj |.
Combining the previous statements, we can state the following criterion for stability.
Proposition 5.4. We suppose that j |aj | 6 1 + Kλ ∆t for all
C(λ, ∆t) is stable with stability constant K(λ) := eK T .
P
∆t > 0.
Then, the dierence scheme
λ
If aj > 0 holds for the coecients, a dierence scheme is called positive and it maps non-negative
P
initial functions into non-negative solutions. The positive dierence schemes with j aj = 1 + O(∆t)
are stable with respect to the l2 and l∞ norms. To mention an example, for λ ∈ (0, 1\2] the scheme
(4.10) is positive, and the latter criterion ensures l2 and l∞ stability. We can also formulate a criterion
for instability.
Proposition 5.5. We suppose that aj > 1 + ∆tc(∆t) with lim∆t→0 c(∆t) = ∞. Then the dierence
scheme is unstable with respect to the l2 and l∞ norms.
P
In the present chapter, we only assume a necessary statement for proving 5.5 in chapter 6. We suppose
that lim∆t→0 c(∆t) = ∞. Then
sup{[1 + ∆tc(∆t)]µ : µ ∈ N0 , ∆t > 0, µ∆t 6 T } = ∞.
(5.2)
Later on, we will use that (5.2) also implies that Cp := sup∆t>0 c(∆t) < ∞ is possible with setting
c(∆t) :=
√
µ
K−1
∆t
for any K > 0 and all positive integer µ 6 T /∆t. Therefore, the µth-root of K satises
the inequality
√
µ
K 6 1 + Cp ∆t.
(5.3)
An interesting question is whether a stable scheme remains stable after a perturbation. We show that
the stability of C(λ, ∆) implies the stability of the scheme after perturbation in both cases that C(λ, ∆t)
and perturbation D(λ, ∆t) are commutative operators or not.
Lemma 5.6 (perturbation lemma). Let C(λ, ∆t) be lp stable with stability constant K(λ). Suppose that
a perturbation operator D(λ, ∆t) is bounded by
kD(λ, ∆t)klp →lp 6 CD ∆t.
Then
C 0 (λ, ∆t) := C(λ, ∆t) + D(λ, ∆t)
is again lp stable with stability constant
K 0 (λ) 6 K(λ)eK(λ)CD T ,
where I = [0, T ] is the given interval. This result holds also in the case where C(λ, ∆t) and D(λ, ∆t) are
non-commutative operators, which is interesting in the case of matrix-valued coecients aj .
17
A simple application of the
perturbation lemma is the following one. Let us consider the dierential
operator A in du/dt = Au be A = A1 + A0 , where A1 contains derivatives of at least rst order, while
A0 u = a0 u is the term of order zero. The discretization also yields C(λ, ∆t) = C1 (λ, ∆t) + C0 (λ, ∆t),
where a consistent discretization of C0 can be estimated as kC0 (λ, ∆t)klp ←lp = O(∆t). So that, by the
perturbation lemma the stability of C1 implies the stability of C . Therefore, it suces to investigate A
without terms of order zero. In this case, A1 = 0, where 1 ∈ l∞ the constant function with value 1, shows
that u := 1 is a solution of the heat conduction problem. This implies the special consistency condition
P
j aj = 1 for the coecients of C(λ, ∆t) in (4.10).
Next, we introduce the spectral radius
ρ(A) := sup{|λ| : λ singular value of A}.
Using section 6.5 of [5], we dene λ as regular value of A, if λI −A is bijective, and the inverse (λI −A)−1 ∈
L(B, B) exists. Otherwise, λ is a singular value of A, i.e. the inverse (λI −A)−1 ∈ L(B, B) does not exist.
Furthermore, p(Aµ ) equals p(A)µ and p(A) 6 kAk is valid for any associated norm. These properties
with (5.3) yield a necessary condition for stability with respect to the l2 and l∞ norms as
(5.4)
ρ(C(λ, ∆t)) 6 1 + O(∆t).
In the case that the operator C of the dierence scheme is normal, it is possible to show that the spectral
radius is equal to the l2 norm of C(λ, ∆t) and the l2 stability is equivalent to the latter necessary
condition (5.4). Let C(λ, ∆t) be denoted by C .
Proposition 5.7. We suppose that C ∈ L(l2 , l2 ) is normal, i.e. C commutes with the adjoint operator
C ∗ . Then p(C) = kCkl ←l holds and l2 stability is equivalent to ρ(C) 6 1 + O(∆t).
2
2
Proof. First, we prove that C 2 l
2 ←l2
= kC ∗ Ckl2 ←l2 .
From
hCCu, CCuil2 = hC ∗ CCu, Cuil2 = hCC ∗ Cu, Cuil2 = hC ∗ Cu, C ∗ Cuil2
it follows that
2 2
C 2 2 =
l ←l
2
sup hC ∗ Cu, C ∗ Cuil2 = kC ∗ Ckl2 ←l2 .
sup hCCu, CCuil2 =
kukl2 =1
kukl2 =1
Since also
kC ∗ Ckl2 ←l2 =
sup
hu, C ∗ Cvi =
kukl2 =kvkl2 =1
2
kCkl2 ←l2 > C 2 l2 ←l2 is shown.
equality C 2 2 2 = kCk22 2 is
l ←l
l ←l
sup
2
hCu, Cvil2 >
kukl2 =kvkl2 =1
sup hCu, Cuil2 = kCkl2 ←l2 ,
kukl2 =1
Using the property of the operator norm C 2 l2 ←l2 6 kCk2l2 ←l2 , the
proved. Therefore, kC n kl2 ←l2 = kCknl2 ←l2 also follows for all n = 2k ,
where k ∈ N.
Since C is a bounded linear operator, we can use Gelfand's formula:
ρ(C(λ, ∆t)) = limn→∞
18
p
n
kC n kl2 ←l2 .
Applying the results of (1), the equality ρ(C) = kCkl2 ←l2 follows. We can use that (5.4) is necessary,
while kCkl2 ←l2 6 1 + O(∆t) is sucient. Since we showed ρ(C) = kCkl2 ←l2 , both inequalities are
identical. One may ask whether a similar condition holds for more general operators. For this purpose, we
introduce the
almost normal operators : C(λ, ∆t) is almost normal if
2
kC(λ, ∆t)C ∗ (λ, ∆t) − C ∗ (λ, ∆t)C(λ, ∆t)kl2 ←l2 6 M (∆t)2 kC(λ, ∆t)kl2 ←l2
for some constant M < ∞.
Proposition 5.8. If C(λ, ∆t) is almost normal, then l2 stability is equivalent to the estimate
kC(λ, ∆t)kl2 ←l2 6 1 + O(∆t).
Proof. Since we stated that kC(λ, ∆t)kl ←l 6 1+O(∆t) is sucient, it is only needed to prove necessity.
Again, C(λ, ∆t) is denoted by C . First we prove that the (C ∗ )µ C µ can be reordered into (C ∗ C)µ in at
most µ2 /2, where one step consists of C ∗ C 7→ CC ∗ . In the case of µ = 1, no permutation is required and
0 6 12 /2 proves the start of the induction approach. Suppose the induction hypothesis for µ. Then
2
2
(C ∗ C)µ+1 ,→µ2 \2 (C ∗ )µ C µ C ∗ C ,→µ (C ∗ )µ+1 C µ+1 ,
prove the hypothesis. Next, we show that the interchange perturbs the operator norm
and µ2 + 1 6 (µ+1)
2
∗
at most by M (∆) kCk2µ
l2 ←l2 . In the following formula Cj , 0 6 j 6 2µ are either C or C and we focus
on one interchange. Furthermore, we use that kC ∗ kl2 ←l2 = kCkl2 ←l2 . Then
2
2
kC1 ..Cv CC ∗ Cv+3 ..C2µ − C1 ..Cv C ∗ CCv+3 ..C2µ kl2 ←l2 6
6 kC1 ..Cv kl2 ←l2 kCC ∗ − C ∗ Ckl2 ←l2 kCv+3 ..C2µ kl2 ←l2 6
"
#
def
v
2
2µ−v−2
6 kCkl2 ←l2 M (∆t)2 kCkl2 ←l2
kCkl2 ←l2
= M (∆t)2 Cl2µ
2 ←l2 .
for all 0 6 v 6 2µ − 2. Using the aforementioned,
2
kC µ kl2 ←l2 =
sup hu, (C ∗ )µ C µ uil2 >
kukl2 =1
sup hu, (C ∗ C)µ uil2 −
>
kukl2 =1
M
2µ
2
(µ∆t)2 kCkl2 ←l2 kukl2 =
2
"
#
M
2µ
2
= 1−
(µ∆t) kCkl2 ←l2 .
2
√
Restricting µ and ∆t by µ∆t 6 min{1\ M , T }, we obtain 1 − M2 (µ∆t)2 > 21 . Then
def
1
2µ
2
kCkl2 ←l2 6 kC µ kl2 ←l2 6 K(λ)2 ,
2
so that kCk 6
p
2µ
2K(λ)2
and from (5.3), inequality kCkl2 ←l2 6 1 + O(∆t) follows. These inequalities are the most convenient conditions proving stability.
19
However, even if kC(λ, ∆t)kl2 ←l2 > c > 1, it may happen that the powers still stay bounded for a
constant K , kC(λ, ∆t)µ kl2 ←l2 6 K . In this case one may try to nd an equivalent norm which behaves
P
2
easier. Since the square kU kl2 = i∈Z |Ui |2 is a quadratic form, one way is to introduce another quadratic
form Q(U ) such that
1
2
2
kU kl2 6 Q(U ) 6 K1 kU kl2
K1
which describes the equivalence of the norms
p
p
k·kl2 and Q(·). We may put the question of whether
the stability property holds for Q(U ). Let us suppose that the growth of Q(·) for one time step U µ+1 =
C(λ, ∆t)U µ is limited by
!
0 6 Q(U
µ+1
2
kU µ kl2
µ
) − Q(U ) 6 K2 ∆t
2
+ U µ+1 l2
+ K3 ∆t,
then for ∆t 6 1/(2K1 K2 ), the norms Q(U µ ) and kU µ kl2 stay bounded. More precisely
(
2
kU µ kl2
6
(2K12
2
K3
− 1) U 0 l2 +
2K2
)
20
1 + K∆t
1 − K∆t
!µ
−
K3
2K2
for µ ∈ N.
Chapter 6
Fourier analysis
To derive further conditions for stability, we consider f as a 2π -periodical function in L(R)22π , meaning
the 2π -periodic extension of L2 (0, 2π) to R. In general, we call a function α-periodical if
f (x + α) = f (x)
for any x ∈ R,
where in our case α = 2π . We call a series trigonometric if it has the form
∞
X
[ak cos(kx) + bk sin(kx)],
k=1
where the partial sum of the 2n + 1 terms is called trigonometric polynomial of order n
n
Tn (x) :=
a0 X
[ak cos(kx) + bk sin(kx)].
+
2
k=1
Using chapter 5 of [2] and the notes of [4] to understand the periodical sign, we start by assuming that
there exists a series of sine waves with wave numbers k ∈ N, which converges to function f as its sum
on R. So that, the task is to nd coecients b1 , . . . bn , . . . for the following problem
f (x) = b1 sin(x) + . . . + bn sin(nx) + . . .
for any x ∈ R.
Such a series, consisting only of sine waves and summing up to f , does not alway exist. So one may apply
the linear combination of {sin(kx), k = 1, 2, 3, . . .} and {cos(kx), k = 0, 1, 2, . . .} to provide solvability for
all α-periodic functions, at 'almost each' x. In the theory of the Fourier series, our aim is to approximate
function f with
trigonometric polynomials. It is possible to show that the trigonometric series dened
above may converge uniformly to f only if the coecients are dened as
1
ak :=
π
bk :=
1
π
Z
2π
f (x) cos(kx)dx
for k = 0, 1, 2, ..
f (x) sin(kx)dx
for k = 1, 2, 3, ..
0
Z 2π
0
21
and this is called the Fourier series of f . If ak , bk are chosen to be the above Fourier coecients, then
2π
Z
(f (x) − Tn (x))2 dx
0
reaches its minimum. The next question is that the setting of such a series to f under which conditions
provides the uniform convergence and generates f . In the case of a suciently smooth 2π -periodic
function, the Fourier series converge uniformly with respect to k·k∞ to f . For general L2 functions, the
convergence holds only in the sense of L2 -norm. There are various conditions for the uniform convergence.
More details can be found in chapter 1 of [3] and chapter 6 of [6]. Furthermore, using the Euler formula
√
and ordering Tn (x) via exp(ix) with i = −1, it can be rewritten as
n
X
Tn (x) =
ck exp(−ikx),
k=−n
with
ck :=
1
2π
2π
Z
f (x) exp(ikx)dx,
0
where c−k = ck for k ∈ {1, 2 . . .}. We conclude that it is possible to nd connections with the ndimensional Euclidean space. For example, the Fourier coecients can be specied as the components of
a vector, based on an orthonormal basis. We suppose that
Tn (x) :=
Tn0 (x) :=
n
X
k=−n
n
X
ck exp(−ikx)
c0k exp(−ikx),
k=−n
then
1
2π
2π
Z
Tn (x)Tn0 (x)dx =
0
n
X
ck c0k .
k=−n
Therefore, the multiple integration of two trigonometric polynomials is equal to the dot product of the
representative coecients. More generally, it is well-known that for a function f ∈ L2 (0, π) Parseval's
equality
1
2π
Z
2π
|f (x)|2 dx =
0
∞
X
|ck |2
k=−∞
holds. In the following, the associated Fourier series is
Z 2π
1
1 X
√
ϕα exp(iαx) with ϕα := √
f (x) exp(−iαx)dx.
2π α∈Z
2π 0
22
6.1 Transform operator F
According to the Parseval equality, with the notation introduced before:
(6.1)
kf kL2 (0,2π) = kϕkl2 .
On the one hand, the transfer from f ∈ L2 (0, 2π) to its Fourier coecients ϕ ∈ l2 is the Fourier
analysis,
and the operator of which will be denoted by F:
F : f 7→ ϕ.
On the other hand, it is also possible to dene the inverse mapping, which is called the Fourier
denoted by F
−1
synthesis,
. In the following, we rely on chapter 5 of [6].
For p = 2, we showed that the solutions U µ of the dierence scheme were l2 sequences, so that we can
apply the Fourier synthesis and denote the associated 2π -periodic function by Û µ ∈ L2 (0, 2π) as
Û µ := F −1 U µ where
1 X µ
Uα exp(iαξ)
Û µ (ξ) = √
2π α∈Z
for ξ ∈ R
with the coecients (Uαµ )α∈Z . Therefore, it is possible to derive further stability conditions in terms of
the obtained for L2 (0, 2π) → L2 (0, 2π) operators. First, one may put the question of how the dierence
operator C(λ, ∆t) = C should be transformed for Û µ . We use the dierence scheme U µ+1 = CU µ , then
Û µ+1 = F −1 U µ+1 = F −1 CU µ = F −1 CFF −1 U µ = Ĉ Û µ
with Ĉ := F −1 CF .
Using isometry (6.1) and the denitions of operators F and F −1 one can prove three important assertions:
1. kFkl2 ←L2 (0,2π) = F −1 L2 (0,2π)←l2 = 1, i.e. F and F −1 are unitary,
2. for all functions M ∈ L(l2 , L2 (0, 2π)), the kM kL2 (0,2π)←l2 = kM FkL2 (0,2π)←L2 (0,2π) holds,
3. for all functions D ∈ L(l2 , l2 ), the kDkl2 ←l2 = F −1 DL2 (0,2π)←l2 holds.
To verify these three points, we need to restrict f as a xed function on L2 (0, 2π). Furthermore, let the
Fourier coecients of f be denoted by ϕ. Then,
−1 F ϕ 2
= kf kL2 (0,2π) = kϕkl2 = kFf kl2
L (0,2π)
yields the rst point, as F and F −1 do not change the norm values. Since Ff = ϕ, the norms of M Ff
and M ϕ are equal and using (6.1), the equality of the norms of M F and M also follows. For the third
point, the following equalities hold:
−1
F Dϕ 2
= F −1 γ L2 (0,2π) = kgkL2 (0,2π) = kγkl2 = kDϕkl2
L (0,2π)
where we use the fact that operator D maps ϕ to another sequence, denoted by γ . In this manner, the
norms of F −1 Dϕ and F −1 γ are equal. Furthermore, there exists a function g , for which γ is the set of
23
its Fourier coecients. The two extremes of the equalities imply the third point.
As an application of the aforesaid three points, we can state that the norms of C µ and Ĉ µ are equivalent
since
1.
2. 3. kC µ kl2 ←l2 = F −1 C µ L2 (0,2π)←l2 = F −1 C µ F L2 (0,2π)←L2 (0,2π) =
.
= [F −1 CF]µ L2 (0,2π)←L2 (0,2π) = Ĉ µ (6.2)
L2 (0,2π)←L2 (0,2π)
Therefore, an equivalent denition of l2 stability with the stability constant K(λ) also holds for Ĉ(λ, ∆t)
as
µ
Ĉ L2 (0,2π)←L2 (0,2π)
for ∀µ ∈ N0 , ∆t > 0 with µ∆t 6 T.
6 K(λ)
For a concrete determination of Ĉ , using (4.8) and (4.9), we consider a term Cj = aj Ej from C =
as (Cj U )v := aj Uv+j for U ∈ l . The
2
Fourier synthesis of Cj U = (aj Uα+j )α∈Z yields
(6.3)
P
j
Cj
aj X
aj X
def. 1 X
F −1 Cj U = √
(aj Uα+j ) exp(iαξ) = √
Uα+j exp(iαξ) = √
Uβ exp(i(β − j)ξ) =
β=α+j
2π α∈Z
2π α∈Z
2π β∈Z
=
aj exp(−ijξ) X
√
Uβ exp(iβξ) = aj exp(−ijξ)Û .
2π
β∈Z
Since Ĉj Û = F −1 Cj U holds with Û = F −1 U , Ĉj can be dened as aj exp(−ijξ), i.e. the linear mapping
P
Ĉj is the multiplication by the function aj exp(−ijξ). Since the equality Ĉ = j∈Z Ĉj also holds, we can
P
obtain that the Fourier transformed operator is Ĉ = j∈Z aj exp(−ijξ). Other words, the application of
Ĉ corresponds to the multiplication by the trigonometric polynomial
G(ξ) :=
X
(6.4)
aj exp(−ijξ)
j∈Z
which is called the
characteristic function of C . We need to note that parameters λ and ∆t are omitted
in G(ξ), however aj depends on λ and ∆t. We state that the identity (6.2) becomes
kC µ kl2 ←l2 = sup{|G(ξ)|µ : ξ ∈ R},
and it yields
!µ X
=
aj exp(−ijξ) = sup{|G(ξ)|µ : ξ ∈ R} =
j∈Z
L2 (0,2π)←L2 (0,2π)
X
µ ∞ µ
= sup{|G(ξ)| : ξ ∈ R}µ = aj exp(−ijξ) = Ĉ 2
.
2
µ
Ĉ j∈Z
∞
L (0,2π)←L (0,2π)
The latter equality is valid only for scalar aj . Together with (6.3) it leads to the fact that the boundedness
of the absolute value of the characteristic function is a necessary and sucient condition for the stability
of the dierence scheme (4.8).
24
Theorem 6.1. The dierence scheme (4.8) formed by C(λ, ∆t) is l2 stable if and only if the characteristic
P
function G(ξ) = j∈Z aj exp(−ijξ) satises the following estimate with a suitable Kλ :
|G(ξ)| 6 1 + Kλ ∆t
for all ξ ∈ R.
Proof. It is clear that the inequality is sucient for the stability of C , since
kCkl2 ←l2 = Ĉ L2 (0,2π)←L2 (0,2π)
= kGk∞ 6 1 + Kλ ∆t.
So that, we only need to prove the necessity. Similarly to (5.3), set c(∆t) := (G(ξ) − 1)/∆t. If there is
no constant Kλ satisfying |G(ξ)| 6 1 + Kλ ∆t, then c(∆t) → ∞ follows for ∆t → 0. Using Theorem 6.1, we are also able to prove Proposition 5.5 for l2 instability of C (cf. Chapter 5).
P
Since inequality j∈Z aj > 1 + ∆tc(∆t) with lim∆t→∞ c(∆t) = ∞ is supposed and kGk∞ > |G(0)| =
P
| aj | holds, therefore |G(ξ)| is unbounded and yields the l2 instability. This analysis refers to l2 , however
the characteristic function has also consequences for l∞ stability. The negation of this statement is: if
the dierence scheme is l2 unstable, it is also l∞ unstable.
Lemma 6.2. Let
stability:
be of the form (4.8) with constant coecients aj . Then l∞ stability implies l2
C
kC µ kl∞ ←l∞ > kGµ k∞ = kCµkl2 ←l2
Proof. Choose the special initial value
equality |G(ξ)| = kGk∞ . Note that U 0
U 1 = CU 0 with
Uv1 =
X
0
aj Uv+j
=
j∈Z
X
for all µ ∈ N0 .
with Uv0 = exp(ivξ), where ξ ∈ R is characterized by the
0
∈ l∞ so U 0 l∞ = kU kl∞ = 1. The application of C yields
U0
aj exp(i(v + j)ξ) = exp(ivξ)
j∈Z
X
aj exp(ijξ) = U 0 G(ξ),
j∈Z
hence C µ U 0 = Gµ (ξ)U 0 . Then,
0
U ∞ kC µ k ∞ ∞ > C µ U 0 ∞ = Gµ (ξ)U 0 = |Gµ (ξ)| U 0 ∞ = kGµ k U 0 ∞
∞
l ←l
l
l
l
l
∞
and the assertion follows. A possible application of Theorem 6.1 and Lemma 6.2 is the following. Because of
G(ξ) := λ exp(iξ) + 1 − 2λ + λ exp(−iξ) = 1 − 2λ(1 − cos(ξ))
and |G(ξ)| reaches its maximum at π , kG(ξ)k∞ = |G(π)| = |1 − 4λ|. So the dierence scheme (4.10) is
conditionally stable for λ ∈ (0, 1/2] but unstable for λ > 1/2 (with respect to l2 and l∞ ).
25
Chapter 7
Implicit schemes
All the schemes using the explicit form U µ+1 = C(λ, ∆t)U µ examined so far are conditionally stable.
In order to obtain
the case of A =
unconditionally stable schemes, one must admit implicit dierence schemes. Now, in
∂2
∂x2 ,
let the dierence quotient at time level t + ∆t be approximated as
u(t + ∆t, x) − u(t, x)
∆t
for
∂u
∂t
while the second dierence quotient is
u(t + ∆t, x − ∆x) − 2u(t + ∆t, x) + u(t + ∆t, x + ∆x)
∆x2
which lead together with λ =
∆t
∆x2
for
∂2u
∂x2
to
µ+1
µ+1
−λUv−1
+ (1 + 2λ)Uvµ+1 − λUv+1
= Uvµ .
(7.1)
So that, one may obtain an implicit scheme of the form
C1 (λ, ∆t)U µ+1 = C2 (λ, ∆t)U µ
(7.2)
where in the present case the operators C1 (λ, ∆t) = C1 and C2 (λ, ∆t) = C2 are given by
(C1 U )v =
X
a1,j Uv+j
with a1,−1 = a1,1 = −λ, a1,0 = 1 + 2λ
−16j61
(C2 U )v =
X
a2,j Uv+j
with a2,0 = 1,
j∈Z
and all other coecients are equal to 0, i.e C2 is the identity operator. Since one would like to solve (7.2)
with respect to U µ+1 , therefore the existence of the inverse C1−1 is to be investigated.
26
Lemma 7.1. (a) If the coecients a1,j of C1 satisfy the inequality
!
X
|a1,j | /|a1,0 | 6 1 − ε < 1
for some ε > 0
j∈Z\{0}
then the inverse exists and it satises C1−1 lp ←lp 6 ε|a11,0 | .
(b) The inverse C1−1 exists in L(l2 , l2 ) i the characteristic function G1 (ξ) of C1 satises an inequality
|G1 | > η > 0. The norm equals
−1
−1 C 2 2 = inf |G1 (ξ)|
.
1
l ←l
ξ∈R
Proof. Case (a). The equation C1 V
Q(V ) :=
is equivalent to the xed-point equation Q(V ) = V with
=U
1
(U − C10 V )
a1,0
with C10 = C1 − a1,0 I.
The contraction constant of Q(V ) = V for p ∈ {2, ∞} is the following:
kQ(V1 ) − Q(V2 )klp > k1/a1,0 (U − C10 V1 ) − 1/a1,0 (U − C10 V2 )klp =
Using kCklp ←lp 6
P
j
|aj |
kC10 klp
kV1 − V2 klp .
|a1,0 |
(cf. Chapter 5):
kC1 klp ←lp
6
|a1,0 |
P
j∈Z\{0}
|a1,j |
|a1,0 |
61−ε
is valid, so that a unique inverse exists. Furthermore, the estimate
kV klp 6
1
|a1,0 |
kU klp + (1 − ε) kV klp
1
−1 p p has the bound 1/(ε|a1,0 |).
shows that ε kV klp 6 |a1,0
| kU klp . so that C
l ←l
Case (b). We suppose that Ĉ1 = F −1 C1 F is the Fourier transformation of C1 , and Ĉ1 is the operator
Û 7→ G1 (ξ)Û . Obviously, the multiplication operator M ∈ L(l2 , l2 ) with M Û := (1/G1 (ξ))Û is the inverse
of Ĉ1 . The norm of M is:
kM kl2 ←l2 = Ĉ1−1 2
l ←l2
= k1/G1 k∞ .
One may verify that the norm of Ĉ1−1 can not be nite if inf{|G1 (ξ)|} equals zero. So that, there exists
η > 0 for which |G1 (ξ)| > η for anyξ ∈ R. Moreover, since the Fourier transform does not change the
l2 norm, the equality C1−1 l2 ←l2 = Ĉ1−1 2 2 = 1/ inf ξinR {|G1 (ξ)|} follows. l ←l
27
Using the latter lemma, it is easy to show that the scheme (7.1) is l2 -stable for all λ = ∆t/∆x2 ,
i.e. unconditionally stable. Since the characteristic function of C1 is G1 (ξ) := −λ exp(iξ) + 1 + 2λ −
λ exp(−iξ) = 1 + 2λ(1 − cos(x)) > inf |G1 (ξ)| = 1 and C2 = I . we obtain C = C1−1 C2 = C1−1 . Hence,
stability follows by C −1 2 2 = 1 6 1 + Kλ ∆t for a suitable Kλ . For a concrete determination, we
1
dene
C1−1
l ←l
as an innite operator
C1−1 :=
X
aj (λ)Ej
j∈Z
1
with aj := √
1 + 4λ
i.e the implicit scheme is identical to Uvµ+1 =
P
j∈Z
2λ
√
2λ + 1 + 1 + 4λ
!|j|
µ
. In general, it is also possible to assume that
aj Uv+j
the scheme (7.2) is l2 stable if and only if the characteristic function
G(ξ) := G2 (ξ)/G1 (ξ)
satises Theorem 6.1. A possible application of the latter statement is represented in the following. First,
we introduce the
theta scheme, a modication of the schemes used so far as
µ+1
µ+1
µ
µ
−λθUv−1
+ (1 + 2λθ)Uvµ+1 − λθUv+1
= λ(1 − θ)Uv−1
+ (1 − 2λ(1 − θ))Uvµ + λ(1 − θ)Uv+1
.
It is easy to see that (7.3) is the presented
(7.3)
explicit one for θ = 0 and the implicit one for θ = 1. The
above dened scheme satises the following lemma:
Lemma 7.2. The scheme (7.3) is unconditionally
θ ∈ [0, 1/2) it is conditionally stable for
l2
stable for θ ∈ [1/2, 1], whereas in the case of
λ 6 1/(2(1 − 2θ)).
In all stable cases, kCkl2 ←l2 = 1 holds.
Proof. The idea is to dene a second dierence operator D with the scheme (DU )v = −Uv−1 + 2Uv −
Uv+1 . The characteristic function of D using Euler's formula is GD (ξ) = 2 − 2 cos(ξ) = 4 sin2 (ξ/2). In
the case of scheme (7.3), the operators C1 and C2 change to
C1 = I + λθD
with G1 (ξ) = 1 + λθGD (ξ)
C2 = I − λ(1 − θ)D
with G2 (ξ) = 1 − λ(1 − θ)GD (ξ),
The characteristic function G(ξ) := G2 (ξ)/G1 (ξ) is monotonically decreasing with respect to GD (ξ), so
that the maximum of |G(ξ)| is taken at GD (0) = 0 or GD (π) = 4, thus kGk∞ = max{G(0), −G(π)}. On
the one hand, if θ ∈ [1/2, 1], then −G(π) 6 1 for any λ and kG(ξ)k∞ = 1 provides stability. On the other
hand, if θ ∈ [0, 1/2), then the choice of λ 6 1/(2(1 − 2θ)) ensures the estimate −G(π) 6 1 and provides
conditional stability. 28
Chapter 8
Conclusion
The starting point of our considerations was to rewrite the heat conduction problem to an abstract
Cauchy problem with dierential operator A on two Banach spaces of functions in spatial variable x
with respect to the associated norms. We mapped the initial function to the solution u(t, x) through a
uniformly bounded and continuous solution operator S(t). The solution operators for all t > 0 formed
an operator semigroup with generator operator A. For the numerical approximations, we dened a
rectangular grid on the space-time domain of the problem. To map the continuous Banach space onto
the linear space of grid functions, we used transfer operators r and p. After that, it was possible to
introduce an explicit dierence scheme with linear dierence operator C . On a numerical example we
observed that the scheme provided a good solution only for certain values of λ = ∆t/∆x2 . We sought the
answer to the question why this instability occurred and what conditions of stability could be applied.
Our investigation was based on Lax's equivalence theorem, which states that if the dierence scheme
is consistent with the PDE, then stability is necessary and sucient for providing convergence. Then,
we saw via the perturbation lemma that a stable scheme remained stable after a perturbation when a
norm of the perturbation operator was bounded. Using the lemma, we derived that it was sucient to
investigate operator A without terms of order zero. Introducing the spectral radius made it possible to
derive a necessary condition for normal and a sucient one for almost normal dierence schemes. We
introduced the Fourier analysis, synthesis and Fourier transform operator F to prove further conditions
for L2 (0, 2π) → L2 (0, 2π) operators. One could show that the application of the transformed dierence
operator Ĉ corresponded to multiplication by the characteristic function of C . Using this function, it was
possible to prove further sucient and necessary criteria for stability. To obtain unconditionally stable
schemes, we admitted the implicit dierence schemes with their stability conditions. At the end, in order
to close our arguments, we considered the general theta scheme, a modication of the used schemes, and
showed that it was conditionally or unconditionally stable under some assumptions.
29
List of Figures
4.1
The results of the programming task for dierent values of parameter λ. . . . . . . . . . . 13
I
Appendix A
The source code of the programming
task
function [ x , u ] = prog ( dt , l )
2 %GENERAL INFORMATIONS
3 xmax = 5; % Maximum length
4 xo = − 5;
5 tmax = 1; % Maximum time
6 to = 0;
1
7
8
9
10
dx = sqrt ( dt/ l ) ;
x = (xo −(dx ∗ tmax)/dt ) : dx : ( xmax+(dx ∗ tmax)/dt ) ;
t = to : dt : tmax ;
11
%INITIAL VALUE:
u (: ,1) = ones (1 , s i z e (x ,2) ) ;
14 u( round (( end − 1)/2) ,1)=1+(1e − 5) ;
12
13
15
16
%IMPLEMENTATION:
for j = 1: s i z e ( t ,2) −1
19
for i = ( j +1) : s i z e (u ( : , j ) ,1)− j
20
u( i , j +1) = l ∗ u( i −1, j )+(1−2∗ l ) ∗ u( i , j )+l ∗ u( i +1, j ) ;
21
end
22 end
17
18
23
24
end
II
Appendix B
The main le
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
clear a l l
close a l l
tic
[ x1 , u1 ] = prog (.01 ,0.45) ;
[ x2 , u2 ] = prog (.01 ,0.7) ;
[ x3 , u3 ] = prog (.001 ,0.7) ;
toc
subplot (3 ,1 ,1) ;
plot (x1 , u1 (: ,1) ,x1 , u1 (: ,20) ,x1 , u1 (: ,50) ,x1 , u1 (: ,70) ,x1 , u1 (: ,100) )
xlabel ( 'x ' , ' FontSize ' ,12) ;
ylabel ( 'u(x , t ) ' , ' FontSize ' ,12) ;
t i t l e ( ' Test with dt =.01 , lambda=0.45 ' , ' FontSize ' ,12) ;
legend ( ' ts = 1 ' , ' ts = 20 ' , ' ts = 50 ' , ' ts = 70 ' , ' ts = 100 ' )
axis ([ − 5 5 0 2])
subplot (3 ,1 ,2) ;
plot (x2 , u2 (: ,1) ,x2 , u2 (: ,90) ,x2 , u2 (: ,95) ,x2 , u2 (: ,97) ,x2 , u2 (: ,100) )
xlabel ( 'x ' , ' FontSize ' ,12) ;
ylabel ( 'u(x , t ) ' , ' FontSize ' ,12) ;
t i t l e ( ' Test with dt =.01 , lambda=0.7 ' , ' FontSize ' ,12) ;
legend ( ' ts = 1 ' , ' ts = 90 ' , ' ts = 95 ' , ' ts = 97 ' , ' ts = 100 ' )
axis ([ − 5 5 −2e19 2e19 ] )
25
26
27
28
29
30
31
32
33
subplot (3 ,1 ,3) ;
plot (x3 , u3 (: ,1) ,x3 , u3 (: ,990) ,x3 , u3 (: ,995) ,x3 , u3 (: ,997) ,x3 , u3 (: ,1000) )
xlabel ( 'x ' , ' FontSize ' ,12) ;
ylabel ( 'u(x , t ) ' , ' FontSize ' ,12) ;
t i t l e ( ' Test with dt =.001 , lambda=0.7 ' , ' FontSize ' ,12) ;
legend ( ' ts = 1 ' , ' ts = 990 ' , ' ts = 995 ' , ' ts = 997 ' , ' ts = 1000 ' )
axis ([ − 5 5 −2e248 2e248 ] )
III
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A GENERAL THEOREM ON THE CONVERGENCE OF OPERATOR SEMI-
TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 148, March
1970
[12] Dave Gilliam,
Semigroups of Operators in Banach Space.
Department of Mathematics Texas Tech
University, Lubbock
[13] P.D. Lax, R.D Richtmyer,
Survey of the stability of linear nite dierence equations.
Appl. Math. 9, 267-293 (1965)
IV
Comm. Pure