EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 15, No. 3, 2022, 1348-1362
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

On the existence, uniqueness and application of the
Finite difference method for solving Cauchy-Dirichlet

problem

Diogène Vianney Pongui Ngoma1,∗, Germain Nguimbi1, Vital Delmas Mabonzo2,
Bienaime Bervi Bamvi Madzou
1 Ecole Nationale Superieure Polytechnique, Marien Ngouabi University, Brazzaville, Congo
2 Ecole Normale Superieure, Marien Ngouabi University, Brazzaville, Congo

Abstract. In this paper we treat the existence, the uniqueness and the numerical resolution of
the problem at the elliptic limits case of the Cauchy-Dirichlet problem of the type the stationary
convection-diffusion equation. By applying the Lax-Milgram theorem, we proved the existence
and the uniqueness of the problem, then we solved the problem numerically by the finite difference
method. In addition, we solved the problem analytically using the method of variation of constants.
Finally, we performed a numerical simulation of said problem to approach the exact solution by
the numerical solution using the software Scilab

2020 Mathematics Subject Classifications: 65M06, 65M12, 65K05

Key Words and Phrases: Cauchy-Dirichlet Problem, convection-diffusion equation, Finite dif-
ference method, numerical simulation

1. Introduction

Partial differential equations (PDEs) are used in several fields, including engineering,
mechanics, physics, aeronautics, petroleum industry, but also in economics, chemistry, bi-
ology; in medicine ... they are defined in an open Ω of space Rn, n ≥ 1 (PFEs integration
domain) of border ∂Ω. They can be listed by type. Thus, the equations of the elliptical
type describe the phenomena of stationary diffusion, the equations of the parabolic type
describe the phenomena of diffusion and the equations of the hyperbolic type describe the
phenomena of transport at finite speed. As a general rule, it is difficult to find a unique
solution to an EDP without boundary conditions (additional information on the border
∂Ω). Thus, we speak of Dirichlet’s boundary condition, of Cauchy condition (that is to
say of Dirichlet type on a part of the edge ∂Ω and of Cauchy type on the other part).

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v15i3.4403

Email addresses: diogene.ponguingoma@umng.cg (D.V. Pongui Ngoma),
germain.nguimbi@umng.cg (G. Nguimbi), vital.mabonzo@umng.cg (V.D. Mabonzo)

https://www.ejpam.com 1348 © 2022 EJPAM All rights reserved.



D. V. Pongui Ngoma et al. , / Eur. J. Pure Appl. Math, 15 (3) (2022), 1348-1362 1349

Furthermore, a boundary problem is broken down into a PDEs (or ODE) inside the Ω do-
main and the boundary conditions of the problem. We speak in this case of the Dirichlet
problem, Cauchy problem, Cauchy-Dirichlet problem (problem whose boundary condi-
tions are a combination of a Cauchy condition and a Dirichlet condition). In addition,
it is necessary, to approach the theoretical (mathematical) study of these type of prob-
lems, to have a good knowledge of vector and functional analysis (normed vector spaces,
Lebesgue spaces), Hilbert analysis, matrix numerical analysis and also some knowledge of
distributions, Sobolev spaces , traces, Green’s formulas, the use of the Lax-Milgram theo-
rem and some fundamental inequalities (Cauchy-Schwartz, Hölder, Poincaré, etc.) which
emanate from the analysis and which are generally used in the framework of functional
analysis of PDE [2–6, 8]. The PDE numerical resolution represents a field of research
in mathematical sciences. There are several numerical methods of resolution available for
each type of boundary problem, classified by categories [1, 12–18]. Thus, we speak of finite
difference method, finite element method, finite volume method, etc. The manipulation of
these resolution methods and the implementation of numerical simulations depend on the
type of problem studied. However, incontestably few numerical methods have been used
to solve the Cauchy-Dirichlet problem. The object of our work is to propose a theoretical
(mathematical) and numerical study of the solution of the Cauchy-Dirichlet problem of
the type of stationary convection-diffusion equation by the method of finite differences
and to implement numerical simulations using the Scilab software, via the analytical and
numerical solutions obtained by solving the problem.

2. Cauchy- Dirichlet Problem

In mathematics, a Cauchy–Dirichlet problem is the problem of finding a function that
solves a specified partial differential equation (PDE) inside a given region that takes on
prescribed values at the boundary of the region. The problem finds its applications in
several fields of engineering sciences (for example in aeronautics) and is often presented in
the form: let Ω be a bounded non-empty domain of class C1 of Rn (n ≥ 1), f and c are
two functions defined respectively on Ω and on ∂Ω = Γ (the border of Ω).
The Cauchy-Dirichlet problem [2, 9, 10] is an elliptical boundary problem of solution
u = u(x, t) which presents itself in the following way:

−∇(µ∇u) + c(x)∇u = f(x), x ∈]0; 1[
u(0, t) = u(1, t) = 0

u(x, 0) = u0(x) = sin(19πx)

(1)

where −∇(µ∇u) and c(x)∇u represent respectively , the diffusion and convection
terms, f(x) = 5 cos(πx) + (x− 5), c(x) = sin(2x), µ > 0 fixed positive real.



D. V. Pongui Ngoma et al. , / Eur. J. Pure Appl. Math, 15 (3) (2022), 1348-1362 1350

Variational formulation of the boundary value problems

Let us transform the problem (1) into a variational problem As f ∈ L2(Ω) then

−µ∆u ∈ L2(Ω)

u ∈ H1(Ω)

u ∈ H1(Ω) and u = 0 on ∂Ω, then u ∈ H1
0 (Ω).

Now, let’s give the variational formulation of the problem (1)

Let v ∈ H1
0 (Ω)

Multiply the first equation of the system (1) by v and integrate member to member
on Ω ∫

Ω
−µ∆u.v dΩ+

∫
Ω
c∇u.v dΩ =

∫
Ω
f.v dΩ (2)

Let’s use Green’s formula [8, 11], then

µ

∫
Ω
∇u.∇v dΩ+

∫
Ω
c∇u.v dΩ =

∫
Ω
f.v dΩ (3)

We thus obtain the following variational formulation{
u ∈ H1

0 (Ω)

a(u, v) = L(v) ∀v ∈ H1
0 (Ω)

(4)

where a(u, v) = µ

∫
Ω
∇u.∇v dΩ+

∫
Ω
c∇u.v dΩ and L(v) =

∫
Ω
f.v dΩ

3. Existence and uniqueness of the solution

Let us prove the existence and the uniqueness of the solution u ∈ H1
0 (Ω) by Lax-

Milgram theorem [8, 9, 11].
let u ∈ V = H1

0 (Ω)

• Let us show that the bilinear form a is continuous
we have

| a(u, v) |=| µ
∫
Ω
∇u.∇v dΩ+

∫
Ω
c∇u.v dΩ | (5)

| a(u, v) |≤| µ |
∫
Ω
| ∇u.∇v | dΩ+

∫
Ω
| c∇u.v | dΩ (6)

Since
c ∈ L∞(Ω), then | c |≤ M



D. V. Pongui Ngoma et al. , / Eur. J. Pure Appl. Math, 15 (3) (2022), 1348-1362 1351

thus

| a(u, v) |≤| µ |
∫
Ω
| ∇u.∇v | dΩ+M

∫
Ω
| ∇u.v | dΩ (7)

According to the Cauchy-Schwarz inequality [8, 10, 11]

| a(u, v) | ≤| µ | (
∫
Ω
| ∇u |2 dΩ)

1
2 (

∫
Ω
| ∇v |2 dΩ)

1
2 +M(

∫
Ω
| ∇u |2 dΩ)

1
2 (

∫
Ω
| v |2 dΩ)

1
2

(8)

≤| µ |∥ ∇u ∥L2(Ω) . ∥ ∇v ∥L2(Ω) +M ∥ ∇u ∥L2(Ω) . ∥ v ∥L2(Ω) (9)

≤| µ |∥ u ∥H1
0 (Ω) . ∥ v ∥H1

0 (Ω) +Mc ∥ u ∥H1
0 (Ω) . ∥ v ∥H1

0Ω) (10)

≤| µ |∥ u ∥H1
0 (Ω) . ∥ v ∥H1

0 (Ω) +M ′ ∥ u ∥H1
0 (Ω) . ∥ v ∥H1

0Ω) (11)

≤ (| µ | +M ′) ∥ u ∥H1
0 (Ω) . ∥ v ∥H1

0 (Ω) (12)

| a(u, v) | ≤ C ∥ u ∥H1
0 (Ω) . ∥ v ∥H1

0 (Ω), where C =| µ | +M ′,whith M ′ = constant

(13)

therefore a is continue.

• Let us show that the bilinear form a is coercive
let v = u, then

a(u, u) = µ

∫
Ω
(∇u)2 dΩ+

∫
Ω
c∇u.u dΩ (14)

Assuming that, c ≥ c0 > 0, therefore

a(u, u) ≥ µ

∫
Ω
(∇u)2 dΩ+ c0

∫
Ω
∇u.u dΩ (15)

a(u, u) ≥ µ

∫
Ω
(∇u)2 dΩ (16)

a(u, u) ≥ µ ∥ u ∥2H1
0 (Ω) (17)

a(u, u) ≥ α ∥ u ∥2H1
0 (Ω),withµ = α (18)

therefore a is coercive. (19)

(20)

• Let us show that the linear form L is continuous on V = H1
0 (Ω)

we have, L(v) =

∫
Ω
f.v dΩ (21)

=|
∫
Ω
f.v dΩ | (22)



D. V. Pongui Ngoma et al. , / Eur. J. Pure Appl. Math, 15 (3) (2022), 1348-1362 1352

≤
∫
Ω
| f.v | dΩ (23)

According to the Cauchy-Schwartz inequality

| L(v) |≤ (

∫
Ω
| f |2 dΩ)

1
2 (

∫
Ω
| v |2 dΩ)

1
2 (24)

| L(v) |≤∥ f ∥L2(Ω) . ∥ v ∥L2(Ω) (25)

(26)

Since f ∈ L2(Ω) then ∥ f ∥L2(Ω)≤ M1 and ∥ v ∥L2(Ω)≤ C ∥ v ∥H1
0 (Ω)

therefore | L(v) |≤ M1C ∥ v ∥H1
0 (Ω) hence | L(v) |≤ β ∥ v ∥H1

0 (Ω) withβ = M1C

Thus L is continue.

All the hypotheses of the Lax-Milgram theorem being satisfied, we deduce that the
variational problem admits a unique solution u ∈ H1

0 (Ω).

4. Numerical resolution of the problem

In this section, we will try to solve the Cauchy-Dirichlet problem by a numerical
method, in particular the finite difference method. To do this, let’s use the Cauchy-
Dirichlet problem in dimension one, which is presented in the following way:{

−µu′′(x) + c(x)u′(x) = f(x), x ∈]0; 1[
u(0, t) = u(1, t)

(27)

One describes the method in three parts: choice of the mesh, choice of the numerical
scheme and discretization of the problem.

First step: Choice of the discretization.

We consider a subdivision, 0 = x0 < x1 < x2 < ... < xN < xN+1 = 1,
of the interval [0, 1], where N ∈ N.

For i = 0, ..., N , we put, ∆xi = xi+1 − xi, with ∆x = max
1⩽i⩽N

∆xi, the mesh step.

the first step of discretization consists in approximating the functions u, c and f at the
nodes xi, that is to say:

u(xi) ≃ ui, c(xi) ≃ ci, f(xi) ≃ fi

the problem (27) become {
−µ.u′′i + ci.u

′
i = fi,∀i = 1, ..., N

u(0) = uN+1

(28)



D. V. Pongui Ngoma et al. , / Eur. J. Pure Appl. Math, 15 (3) (2022), 1348-1362 1353

Second step: Construction of the numerical scheme
Assuming that, u ∈ C2([0; 1]),
then u admits a Taylor expansion in the neighborhood of xi in the form:

u(xi + 1)− u(xi +∆x) = u(xi)−
∆x

1!
u′(xi) +

∆x2

2!
u′′(xi) + 0(∆x3) (29)

u(xi − 1)− u(xi −∆x) = u(xi)−
∆x

1!
u′(xi) +

∆x2

2!
u′′(xi) + 0(∆x3) (30)

Going to the approximations, we have:

ui+1 = ui +
∆x

1!
u′i +

∆x2

2!
u′′i + 0(∆x3) (31)

ui−1 = ui −
∆x

1!
u′i +

∆x2

2!
u′′i + 0(∆x3) (32)

Adding the two equalities, to obtain the following expression

u′′i =
ui+1 − 2ui + ui−1

∆x2
(33)

The first derivative has been approximated using the forward finite difference method of
order 1, that is to say

(
∂u

∂x
)i ≃

ui+1 − ui
∆x

, (34)

with u0 = uN+1 = 0.
we have

−µu′′(x) + c(x)u′(x) = f(x), (35)

that is to say

−µ(
ui+1 − 2ui + ui−1

∆x2
) + ci(

ui+1 − ui
∆x

) = fi (36)

therefore,
−µ

∆x2
(ui+1 − 2ui − ui−1) +

ci
∆x

(ui+1 − ui) = fi (37)

Multiplying the equation (37) by ∆x2

µ , to obtain

(λci − 1)ui+1 + (2− λci)ui − ui−1 = λ∆xfi, with λ =
∆x

µ
(38)

for i = 1, (λc1 − 1)u2 + (2− λc1)u1 − u0 = λ∆xf1
for i = 2, (λc2 − 1)u3 + (2− λc2)u2 − u1 = λ∆xf2
for i = 3, (λc3 − 1)u4 + (2− λc3)u3 − u2 = λ∆xf3
...
for i = N , (λcN − 1)uN+1 + (2− λcN )uN − uN−1 = λ∆xfN



D. V. Pongui Ngoma et al. , / Eur. J. Pure Appl. Math, 15 (3) (2022), 1348-1362 1354

third step: writing Matrix
Taking into account the boundary conditions, we obtain the following linear system

(2− λc1)u1 + (λc1 − 1)u2 = λ∆xf1

−u1 + (2− λc2)u2 + (λc3 − 1)u3 = λ∆xf2

−u2 + (2− λc3)u3 + (λc3 − 1)u4 = λ∆xf3
...

−uN−1 + (2− λcN )uN = λ∆xfN

(39)

hence the matrix below:
2− λc1 λc1 − 1 0 0 ... 0
−1 2− λc2 λc2 − 1 0 ... 0

0 −1 2− λc3 λc3 − 1 ...
...

...
...

...
... ... λcN−1

0 0 ... −1 2− λcN




u1
u2
u3
...

uN

 = λ∆x


f1
f2
f3
...
fN

 (40)

Thus, the problem (40) boils down to the following linear system:

AλU = B (41)

It remains to verify that if Aλ is symmetric positive definite to prove the existence and
the uniqueness of the solution u of the system (41). That is to say:

At
λ = Aλ (1)

∀v ∈ RN,

Aλv.v > 0 (2)

Aλv.v = 0 =⇒ v = 0 (3)

let’s suppose that λci − 1 = −1, that is to say λci = 0 for all i ∈ {1, 2, 3, ..., N}, c = 0 and
λ ̸= 0.
The relation (1) is trivial because the matrix Aλ is tridiagonal with the values of the
overdiagonal which are equal to the values of the subdiagonal.
Admitting λci = 0 et v0 = vN+1 = 0, to get:

Aλv =


2 −1 0 ... 0
−1 2 −1 ... 0

0 −1 2
. . . 0

...
. . .

. . .
. . . −1

0 0 ... −1 2





v1
v2
v3
...

vN−1

vN





D. V. Pongui Ngoma et al. , / Eur. J. Pure Appl. Math, 15 (3) (2022), 1348-1362 1355

< Aλv, v >=


2v1 − v2 − v1 + 2v2 − v3

−v2 + 2v3 − v4
...

−vN−2 + 2vN−1 − vN
−vN−1 + 2vN





v1
v2
v3
...

vN−1

vN


;

< Aλv, v >= (2v1−v2)v1+(−v1+2v2−v3)v2+(−v2+2v3−v4)v3+(−vN−2+2vN−1−vN )vN−1+(−vN−1+2vN )vN

< Aλv, v >=
N∑
i=1

[−vi−1 + 2vi − vi+1]vi

< Aλv, v >=
N∑
i=1

(−vi−1vi) +
N∑
i=1

(2v2i ) +
N∑
i=1

(−vi+1vi)

or
N∑
i=1

(−vi+1vi) =

N+1∑
i=2

(−vi−1vi)

N+1∑
i=2

(−vivi−1) +
N∑
i=1

(−vi−1vi) =
N∑
i=1

(−vi−1vi)

and
N∑
i=1

(2v2i ) =

N∑
i=1

(−v2i + v2i−1) + v2N

thus

< Aλv, v >=
N∑
i=1

(−2vi−1vi) +
N∑
i=1

(v2i + v2i−1) + v2N

< Aλv, v >=
N∑
i=1

(v2i − 2vivi−1 + v2i−1) + v2N

< Aλv, v >= [

N∑
i=1

(vi − vi−1)
2 + v2N ] ≥ 0

the relation (3) gives:

< Aλv, v >= 0 =⇒


vi − vi−1 = 0

vi = 0
...

v1 = 0

=⇒ v = 0

(1),(2) et (3) being verified, then the problem (41) admits a unique solution.



D. V. Pongui Ngoma et al. , / Eur. J. Pure Appl. Math, 15 (3) (2022), 1348-1362 1356

5. Analytical resolution of the problem

To solve the problem analytically, we are going to use the method of
variation of constants which is done in three steps:

First step: Resolution of the associated homogeneous equation

Consider: f(x) = 5 cos(πx) + (x− 5), h(x) = sin(19πx) et c(x) = c0.

Thus, we obtain: {
−µu′′(x) + c0u

′(x) = 5 cos(πx) + (x− 5)

u(0) = 0, u(1) = 0

Solving the associated homogeneous equation

−µu′′(x) + c0u
′(x) = 0

to get the solution:
u(x) = c1 + c2e

βx

with β = co
µ , c1, c2 ∈ R

Second step: Variation of constants

Let us determine c1(x) and c2(x) such that:

u(x) = c1(x) + c2(x)e
βx, (42)

Let be the solution of the equation with second member. For that, it is a question of
solving the following system:{

c′1(x) + c′2(x)e
βx = 0

βc′2(x)e
βx = f(x)

a = −5 cos(πx)+(x−5)
µ

(43)

we obtain

c1(x) =
5

µβπ
sin(πx) +

1

µβ
(
x2

2
− 5x) + α1 with α1 ∈ R (44)

c2(x) =
1

co

[
−5

π2 + β2
(π sinπx− β cosπx) +

1

β2
+

1

β
(x− 5)

]
e−βx + α2 with α2 ∈ R

(45)
Thus, the relation (42) becomes:



D. V. Pongui Ngoma et al. , / Eur. J. Pure Appl. Math, 15 (3) (2022), 1348-1362 1357

u(x) =
5

coπ
sin(πx)+

1

co
(
x2

2
−5x)+

1

co

[
−5

π2 + β2
(π sinπx− β cosπx) +

1

β2
+

1

β
(x− 5)

]
+α1+α2e

−βx

(46)
Third step: General solution of the complete equation

Let’s find α1 and α2

Using the boundary conditions, we have:{
5β

co(π2+β2)
+ 1

coβ2 − 5
coβ

+ α1 + α2 = 0
−9
2c0

− 5β
c0(π2+β2)

+ 1
c0β2 − 4

c0β
+ α1 + α2e

β = 0
(47)

Solving the equation (47), to obtain

α1 =
1

co(eβ − 1)

[
−9

2
+

1− 4β + (5β − 1)eβ

β2
− 5β(1 + eβ)

β2 + π2

]
(48)

α2 =
1

co(eβ − 1)

[
9

2
+

10β

β2 + π2
− −1

β

]
(49)

Thus , replacing (49) and (48) in (46),

u(x) =
5

coπ
sin(πx) +

1

co
(
x2

2
− 5x) +

1

co

[
−5

π2 + β2
(π sinπx− β cosπx) +

1

β2
+

1

β
(x− 5)

]
+

+
1

co(eβ − 1)

[
9

2
(eβ − 1) +

1− 4β + (5β − 1)eβ

β2
+

5β(eβ − 1)

β2 + π2
− eβ

β

]
. (50)

6. Numerical simulation

The goal here is to represent on the same graph the solutions (50) and (41) exact and
numeric, respectively, taking into account the number of points N and the step h of the
finite difference method with the objective of making the two solutions converge (exact
and numerical). This simulation will be implemented in Scilab. The figure 1 illustrates
the exact solution (50) for µ = 1 and c0 = 3.



D. V. Pongui Ngoma et al. , / Eur. J. Pure Appl. Math, 15 (3) (2022), 1348-1362 1358

By fixing the number of points N = 5 in the finite difference method (41), we sought to

Figure 1: Representation of the exact solution.

vary the step h of the method to verify the numerical convergence of (41) on (50). (cf fig
2).

• Taking for step h = 0.001, on can notice that both exact and numerical solutions
converge almost everywhere numerically (figure 2 a).

• Taking for step h = 0.01, we note that at the beginning the two solutions (exact
and numerical) tend to move away and at a moment converge and move away very
quickly after ( figure ref fig2 b), which is explained by a weak numerical convergence.

• Taking for step h = 0.5, one can notice that both exact and numerical solutions
respectively tend not to converge except on almost negligible points, which is ex-
plained by a numerical divergence ( figure 2 c ).



D. V. Pongui Ngoma et al. , / Eur. J. Pure Appl. Math, 15 (3) (2022), 1348-1362 1359

(a) (b)

(c)

Figure 2: Representation of the exact and numerical solution for N = 5.

In the figure 3, let’s fix N = 10.

• By taking h = 0.001 and h = 0.01, we note that both exact and numerical solutions
respectively (Figure 3 d and Figure 3 e) converge very fast numerically.

• By taking h = 0.5, one can see that the exact and numerical solutions respectively
converge numerically almost everywhere.( Figure 3f)



D. V. Pongui Ngoma et al. , / Eur. J. Pure Appl. Math, 15 (3) (2022), 1348-1362 1360

(d) (e)

(f)

Figure 3: Representation of the exact and numerical solution for N = 10

The finite difference method being linked to a multiplicative term of the form 1
h2 ,

ensuring its convergence towards the exact solution requires a very large number of points
N and a very good choice of the step h of the method.

Conclusion

The resolution of the problem of Cauchy-Dirichlet remains a challenge to be taken up
of which several authors introduced the concept of very weak solution to a problem of
Cauchy for the elliptic equations. The Cauchy-Dirichlet problem is regularized by a non-
local boundary value problem whose solution is understood in this very weak sense [7].
In our this work we have numerically solved the Cauchy-Dirichlet problem by the finite
difference method. In addition to prove the existence and the uniqueness of the solution
to the linear system obtained we set λci − 1 = −1, i = 1, 2, ..., N so that the matrix is
symmetrical. We then solved the problem analytically using the method of the variation
of the constants. Finally, we have implemented numerical simulations in order to make
the numerical solution converge towards the exact solution, using the Scilab software.



REFERENCES 1361

Acknowledgements

The authors thank the anonym referees of European Journal of Pure and Applied
Mathematics, for their valuable comments and suggestions which have led to an improve-
ment of the presentation.

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