# Abstract

We analyze a second-order accurate finite difference method for a spatially periodic convection-diffusion problem. The method is a time stepping method based on the Strang splitting of the spatially semidiscrete solution, in which the diffusion part uses the Crank–Nicolson method and the convection part the explicit forward Euler approximation on a shorter time interval. When the diffusion coefficient is small, the forward Euler method may be used also for the diffusion term.

## 1 Introduction

In this paper we shall
consider the numerical solution of the following convection-diffusion problem
in the cube

under periodic boundary conditions,
where the positive definite

Equation (1.1) is a special case of the initial-value problem for the operator equation

where

The solution of (1.2) may be formally expressed as

To discretize such an equation in time,
a common approach is to split
*k* a time step one introduces

which thus locally involves solutions of

The three exponentials on the right in (1.3) are then approximated
by rational functions of

We note that if *a* and *b*
are independent of *x*, then

In the method that we study in this paper,
we begin by discretizing (1.1) in the spatial variables.
We let *M* is a positive integer, and define
a corresponding uniform mesh

For *M*-periodic vectors *u* with elements

Here *v* the restriction
of *V* to

where the *A* and *B*
correspond to the differential
operators

In the one-dimensional case we may take, with

where

The solution of (1.6), the spatially semidiscrete solution, is

and we shall see that the error in this approximation is

and finally approximate the three exponential factors.

We would like the time discretization
error to match that of the discretization in space. For a
second-order time discretization method, this will require
*N* to be relatively small, we will consider methods with *h* and *k* of the same order.

For the approximation of
*b* is constant and thus
*B* skew-symmetric, we note that

Here and below, *k* should be of the same order as *h*, this makes it natural to choose *k* into *N*
subintervals of length

Thus the diffusion part of the equation is approximated
on intervals of length *k* and the convection part on intervals
of length

In fact, in the successive time stepping,
only the matrix

The analysis sketched above is carried out in Section 2.
The analysis
will use discrete Sobolev norms.
After this, in Section 2,
we discuss the case when equation (1.1)
contains a small diffusion coefficient ε.
In this case we are able to show
that if

## 2 Basic Error Analysis

For the periodic problem in

Further, we set

We shall also use

We shall write *U*, i.e. by

Consider now the matrices *A* and *B* in our convection-diffusion problem
(1.2).
They satisfy, with *C* independent of *h*,

Setting *U* at the mesh-points *U* at

and

expressing, in particular, that (1.5) is a second-order approximation of (1.1).

We note that, for

and we may conclude from (2.1) that

The matrix *A* is positive semidefinite, with

Further, it follows easily from the definition of *A* that

For

and that then

Here

Note also that *v*.

We begin with the straightforward standard analysis of the spatially semidiscrete problem (1.6), which we include for completeness. We first show the stability of the solution operator of (1.6) in discrete Sobolev norms.

## Lemma 2.1.

*Let *

*h*,

## Proof.

Let

Here, by (2.4),

Further, since

Therefore, by (2.9),

Hence, using (2.6) and summing over

or, with

from which the lemma follows. ∎

Note that the special case of

As a consequence, we have the following second-order error estimate.

## Theorem 2.1.

*We have, for the solutions of (1.6) and (1.1),
with *

## Proof.

Setting

where

and hence

where

Turning to the analysis of the time discretization, we first
show the stability of

## Lemma 2.2.

*For any *

*h*,

*Here we may choose *

## Proof.

Let

where

We now show the following error estimate for the Strang splitting.

## Lemma 2.3.

*We have, with C independent of h and k,*

## Proof.

Setting

Here,
for

which shows the lemma. ∎

We now turn to the time stepping operator

## Lemma 2.4.

*Let *

*Further, *

## Proof.

Since

Hence

which shows (2.13)
since

We start the analysis of the time discretization error with the following.

## Lemma 2.5.

*Let M be a square matrix, and
assume *

*and*

*If also *

## Proof.

By Taylor expansion we have

and hence

Estimate (2.16) follows analogously. For (2.17), we use (2.16) together with

where

We now show the following error estimate for

## Lemma 2.6.

*Let *

## Proof.

Since

By Lemma 2.5, we have

By Lemma 2.2,

which completes the proof. ∎

We now show the following error estimate for

## Lemma 2.7.

*Let *

## Proof.

In view of Lemma 2.3, it remains to show

We have

Here, by the above lemmas,

which completes the proof. ∎

We can now prove the following error estimate:

## Theorem 2.2.

*Let *

## Proof.

We write

Using Lemmas 2.4,
2.7 and 2.1, we obtain, for

Since

## 3 The Case of a Small Diffusion Coefficient

In this section,
we consider the variant of problem (1.1) with
a small diffusion coefficient

The corresponding semidiscrete system (1.6) may then be written

where
*A* and *B* are as before.
We shall see that (3.1) is
stable, and satisfies an *k* small, or more precisely, if
*A* part of the
time stepping operator, and with weaker
regularity requirements than earlier. Also, we do not need to use the
symmetric Strang splitting, and consider now, with

We note the inverse inequality

As in Section 2, we first attend to the spatially semidiscrete problem.

## Lemma 3.1.

*Let *

## Proof.

Following the steps in the proof of Lemma 2.1, we have,
for

Here, as in the proof of Lemma 2.1,

and thus, by (2.6), with

This implies (2.11), with *C* independent of ε
and *h*, and thus
completes the proof.
∎

In the same way as in Section 2, the stability shows the following error estimate.

## Theorem 3.1.

*We have, for the solutions of (3.1) and (1.1),
with *

In the analysis of the time discretization we begin with the analogue of Lemma 2.3.

## Lemma 3.2.

*We have, with *

## Proof.

With

Here

Using (3.2), (2.5) and the boundedness of the exponentials, we find

Further,

Together these estimates complete the proof of the lemma. ∎

We now turn to the time stepping operator

## Lemma 3.3.

*If *

*If also *

## Proof.

We note that, since *A* is positive semidefinite,

and thus (3.4) holds by (2.7). Hence, by Lemma 2.4,

We now turn to the error analysis and show the following.

## Lemma 3.4.

*If *

## Proof.

In view of Lemma 3.2 it remains to show

We first note that by Lemma 2.5,

We write

Here by (3.5) and Lemma 2.5, and by (3.4) and Lemma 2.4,

which completes the proof ∎

The following is the resulting error estimate.

## Theorem 3.2.

*If *

## Proof.

Using the analogue of (2.18), we find

As in Section 2, our error estimates in Theorems 3.1 and 3.2 together show a total error estimate.

## Theorem 3.3.

*With *

## 4 Numerical Illustrations

In this section we present some numerical computations to illustrate our error estimates. We restrict ourselves to the one-dimensional version of (1.1),

As before we shall
choose *M* and *N* are positive integers, and study the
effect of doubling these integers.

We begin with the simple case

M | N | Ratio | |||

20 | 4 | 0.01670 | 0.01199 | 0.02494 | |

40 | 8 | 0.00423 | 0.00300 | 0.00621 | 4.01 |

80 | 16 | 0.00106 | 0.00075 | 0.00155 | 4.01 |

160 | 32 | 0.00027 | 0.00019 | 0.00039 | 3.97 |

320 | 64 | 0.00007 | 0.00005 | 0.00010 | 3.90 |

We recall that in the case that *a* and *b* are constant,
the matrices *A* and *B* involved in our method
commute, and consequently the splitting error
given in Lemma 2.3 vanishes. In order to also consider
a situation when this does not happen, we let
*A* and *B* do not commute in this case,
we consider the corresponding continuous operators

and find, after some effort,

Thus *A* and *B*.
The exact solution *U* is taken to be the
semidiscrete solution with

M | N | Ratio | |||

20 | 4 | 0.02641 | 0.01419 | 0.03323 | |

40 | 8 | 0.00651 | 0.00356 | 0.00817 | 4.07 |

80 | 16 | 0.00163 | 0.00089 | 0.00203 | 4.02 |

160 | 32 | 0.00041 | 0.00022 | 0.00051 | 3.98 |

320 | 64 | 0.00010 | 0.00005 | 0.00013 | 3.92 |

We finally consider a numerical example for Section 3,
for which we use (4.1) with

M | N | Ratio | |||

20 | 4 | 0.02872 | 0.05381 | 0.06237 | |

40 | 8 | 0.00721 | 0.01365 | 0.01555 | 4.01 |

80 | 16 | 0.00180 | 0.00342 | 0.00388 | 4.00 |

160 | 32 | 0.00045 | 0.00085 | 0.00097 | 4.00 |

320 | 64 | 0.00011 | 0.00021 | 0.00024 | 4.04 |

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**Received:**2017-11-17

**Revised:**2018-04-16

**Accepted:**2018-06-07

**Published Online:**2018-07-12

**Published in Print:**2019-04-01

© 2019 Walter de Gruyter GmbH, Berlin/Boston

This work is licensed under the Creative Commons Attribution 4.0 Public License.