/compile/output.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 8, No. 2, 2015, 239-254

ISSN 1307-5543 – www.ejpam.com

Gaussian Radial Basis Functions for the Solution of an Inverse

Problem of Mixed Parabolic-Hyperbolic Type

Farzaneh Parzlivand, Alimardan Shahrezaee ∗

Department of Mathematics, Alzahra University, Vanak, Tehran, Iran

Abstract. In this paper, we consider an inverse problem of mixed parabolic-hyperbolic type. This in-

verse problem related to finding the unknown right-hand side of the equation of mixed parabolic-

hyperbolic type in a ectangular domain. We proposed a n umerical approach to solve this problem.

This method is a combination of collocation method and Gaussian radial basis functions (GA-RBFs).

The operational matrix of derivative for GA-RBFs is introduced. The operational matrix of derivative is

utilized to reduce the problem to a set of algebraic equations. Using this method, a rapid convergent

solution is produced which tends to the exact solution of the problem. The accuracy of the method is

tested in term of RMS error. Some examples is included to demonstrate the validity and applicability

of the technique.

2010 Mathematics Subject Classifications: 35M10, 65N21, 65N35

Key Words and Phrases: Mixed parabolic-hyperbolic problem; inverse problem; RBFs; collocation

method

1. Introduction

Let us consider the equation

Lu=

¨

ut − ux x + B2u= f1(x); > 0,

ut t − ux x + B2u= f2(x); < 0,
(1)

of mixed parabolic-hyperbolic type with the unknown right-hand side in the rectangular do-

main D = {(x , t)|0 < x < 1, −α < t < β}, where B ≥ 0, α > 0 and β > 0 are given real

numbers and let us pose the following inverse problem.

Inverse problem. For any nonnegative integer m let Cm(D) denote the vector space consisting

of all functions µ which, together with all their partial derivatives Dαµ of orders |α| ≤ m, are

∗Corresponding author.

Email addresses: fparzlivand@gmail.com (F. Parzlivand), ashahrezaee@alzahra.ac.ir (A. Shahrezaee)

http://www.ejpam.com 239 c© 2015 EJPAM All rights reserved.



F. Parzlivand, A. Shahrezaee / Eur. J. Pure Appl. Math, 8 (2015), 239-254 240

continuous on D. In the domain D, it is required to find the functions u(x , t), f1(x) and f2(x)

satisfying the following conditions:

u(x , t) ∈ C1(D̄), ut(x , t) ∈ C2(D−)∩ C
2,1
x ,t (D+), (2)

f1(x) ∈ C(D+), f2(x) ∈ C(D−), (3)

Lu(x , t) =

¨

ut − ux x + B2u= f1(x); > 0,

ut t − ux x + B2u= f2(x); < 0,
(x , t) ∈ D+ ∪ D−, (4)

u(0, t) =g(t); −α≤ t leqβ , (5)

u(1, t) =h(t); −α≤ t leqβ , (6)

u(x ,−α) =ψ(x); 0≤ x ≤ 1, (7)

ut(x ,−α) =q(x); 0≤ x ≤ 1, (8)

u(x ,β) =ϕ(x); 0≤ x ≤ 1, (9)

where g(t), h(t), q(t),ψ(x) andϕ(x) are given sufficiently smooth functions, D− = D∩{t < 0}
and D+ = D ∩ {t > 0}. The existence and uniqueness of the solution of this problem are

discussed in [30].

The first fundamental research on the theory of mixed type equations are works of F. Tri-

comi, and S. Gellerstedt, which were published in the 1920’s. Due to the research of F.I. Frankl,

I.N. Vekua, M.A. LavrentâĂŹev and A.V.Bitsadze, K.I. Babenko, P. Germain and R. Bader, M.

Protter, K. Morawetz, M.S. Salakhitdinov, T.D. Djuraev, A.M. Nakhushev, V.N. Vragov and many

other authors, this theory became one of the main directions of the modern theory of partial

differential equations [1].

The necessity of the consideration of the parabolic-hyperbolic type equation was specified

in 1959 by I. M. Gelfand [7]. He considered the problem on the motion of a gas in a channel

surrounded by a porous medium; the motion is described by the wave equation in the channel

and by the diffusion equation outside the channel.

At present, the most complete results have been obtained in the study of direct problems

for equations of mixed type. For example, boundary-value problems for equations of mixed

parabolic-hyperbolic type were studied in [2, 25]. In recent years, in [27, 28], a new approach,

the spectral expansion method, was proposed for justifying the existence and uniqueness of

solutions of direct problems for mixed-type equations. Via such a method, inverse problems

for equations of mixed parabolic-hyperbolic type were solved in [25, 27]. The existence and

uniqueness of the solution of these problems and more applications are discussed by several

authors [12, 20, 26–29, 31]. However, the theory of the numerical solution of this problem is

far from satisfactory.

In this paper, we proposed a numerical technique to solve this problem. This method is a

combination of collocation method and GA-RBFs as a truly meshless method. The use of RBFs

as a meshless method for numerical solution of partial differential equations is based on the

collocation scheme. Due to the collocation technique, this method does not need to evaluate

any integral. The main advantage of numerical procedures which use radial basis functions

over traditional techniques is the meshless property of these methods.



F. Parzlivand, A. Shahrezaee / Eur. J. Pure Appl. Math, 8 (2015), 239-254 241

RBFs are used actively for solving partial differential equations (PDEs) and ordinary dif-

ferential equations (ODEs). For example see [10, 13, 21]. Also some applications of these

functions in solving inverse problems can be found in [9, 17, 18, 23]. Our approach in the

current paper is different. We introduce a direct computational method to solve the problem.

This method consists of reducing the problem to a set of algebraic equations by expanding the

candidate function as GA-RBFs with unknown coefficients.

2. Radial Basis Functions

For the last years, the RBFs method was known as a powerful tool for the scattered data

interpolation problem. The main advantage of numerical methods which use radial basis func-

tions is the meshless characteristic of these methods. The use of radial basis functions as a

meshless method for the numerical solution of ODEs and PDEs is based on the collocation

method.

Recently, RBFs was extended to solve various ODEs and PDEs including the nonlinear Klein-

Gordon equation [4], high order ODEs [19], regularized long wave (RLW) equation [11], the

case of heat transfer equations [22], Hirota-Satsuma coupled KdV equations [14] and second-

order parabolic equation with nonlocal boundary conditions [5].

A radial basis function is a real-valued function whose value depends only on the distance

from the origin, so that φ(x) = φ(‖x‖); or alternatively on the distance from some other point

xi , called a center, so that φi(x) = φ(x,xi) = φ(‖x− xi‖). Any function φ that satisfies the

property φ(x) = φ(‖x‖) is a radial function. The norm is usually Euclidean distance, although

other distance functions are also possible [8]. Commonly used types of radial basis functions

include (writing r = ‖x− xi‖):
• Gaussian (GA): φ(r) = e−(ǫr)2

• Multiquadric (MQ): φ(r) =
p
ǫ2 + r2

• Inverse quadratic (IQ): φ(r) = 1
ǫ2+r2

• Inverse multiquadric (IMQ): φ(r) = 1p
ǫ2+r2

• Thin plate spline (TPS): φ(r) = r2 ln(r)

where ǫ is a free positive parameter, often referred to as the shape parameter, to be specified

by the user. Despite many research works which are done to finding algorithms for selecting

the optimum values of ǫ [3, 6, 24], the optimal choice of shape parameter is an open problem

which is still under intensive investigation.

2.1. Function Interpolation

Let x1, x2, · · · , xN ∈ Ω ⊂ R be a given set of scattered data and φi(x) = φ(‖x − x i‖),
i = 1, . . . , N be a set of RBFs. The function s(x), s : R→ R, to be interpolated can be repre-



F. Parzlivand, A. Shahrezaee / Eur. J. Pure Appl. Math, 8 (2015), 239-254 242

sented by RBFs as [16]:

s(x)≃
N
∑

j=1

λ jφ j(x) = Λ
TΦN (x), (10)

where the coefficient vector Λ and RBF vector ΦN (x) are given by:

Λ =[λ1,λ2, . . . ,λN ]
T , (11)

ΦN (x) =[φ1(x),φ2(x), . . . ,φN (x)]
T , (12)

respectively.

2.2. The Operational Matrix of Derivative

Suppose φi(x) is the Gaussian radial basis function, e.i. φi(x) = e−ǫ
2(x−x i)

2

and x ∈ R.

The differentiation of vectors ΦN (x) in (12) can be expressed as [16]:

Φ′N (x) = DN (x)ΦN (x), (13)

where DN (x) is N × N operational matrix of derivative for radial basis function. The matrix

DN (x) can be obtained as:

Φ′N (x) = [φ
′
1(x),φ

′
2(x), . . . ,φ′N (x)]

T =











−2ǫ2(x − x1)φ1(x)

−2ǫ2(x − x2)φ2(x)
...

−2ǫ2(x − xN )φN (x)











. (14)

Comparing (13) and (14), we can write:

DN (x) =











−2ǫ2(x − x1) 0 · · · 0

0 −2ǫ2(x − x2) · · · 0
...

...
. . .

...

0 0 · · · −2ǫ2(x − xN )











(15)

Also:

Φ′′N (x) =











−2ǫ2 + 4ǫ4(x − x1)
2 0 · · · 0

0 −2ǫ2 + 4ǫ4(x − x2)
2 · · · 0

...
...

. . .
...

0 0 · · · −2ǫ2 + 4ǫ4(x − xN )
2











ΦN (x).

So we have:

Φ′′N (x) = (PN + D2
N (x)ΦN (x), (16)

where:

PN =











−2ǫ2 0 · · · 0

0 −2ǫ2 · · · 0
...

...
. . .

...

0 0 · · · −2ǫ2











. (17)



F. Parzlivand, A. Shahrezaee / Eur. J. Pure Appl. Math, 8 (2015), 239-254 243

2.3. Error Bound

Suppose that H = L2([0,1] × [−α,β]) and {φ1(x , t),φ2(x , t), . . . ,φN (x , t)} ⊂ H be the

set of Gaussian radial basis functions and Y = span{φ1(x , t),φ2(x , t), . . . ,φN (x , t)}; and y

be an arbitrary element in H. Since Y is a finite dimensional vector space, y has the unique

best approximation out of Y such as y0 ∈ Y , that is:

∀g ∈ Y,‖y − y0‖ ≤ ‖y − g‖.
Since y0 ∈ Y , there exist unique coefficients γ1,γ2, . . . ,γN such that:

y ≃ y0 =

N
∑

i=1

γiφi(x , t).

Theorem 1. Let H be a Hilbert space and Y be a closed subspace of H such that dimY <∞ and

{y1, y2, . . . , yN} is any basis for Y . Let y be an arbitrary element in H and y0 be the unique best

approximation to y out of Y . Then [15]

‖y − y0‖2 =
G(y, y1, . . . , yN )

G(y1, y2, . . . , yN )
≤ G(y) = ‖y‖2,

where

G(y1, y2, · · · , yN ) =

�

�

�

�

�

�

�

�

�

< y1, y1 > < y1, y2 > · · · < y1, yN >

< y2, y1 > < y2, y2 > · · · < y2, yN >
...

...
. . .

...

< yN , y1 > < yN , y2 > · · · < yN , yN >

�

�

�

�

�

�

�

�

�

.

The determinant G(y1, y2, · · · , yN ) is called the Gram determinant of y1, y2, . . . , yN .

3. The Employed Transformation

For solving the inverse problem (2)-(9), at first we transform it into a direct problem then

use the proposed method. Let us once differentiate equations (4)-(9) with respect to t. Then

for the function v = ut we obtain the equation:

Lv(x , t) =

¨

vt − vx x + B2v = 0; t > 0,

vt t − vx x + B2v = 0; t < 0,
(18)

and from conditions (5) and (6) we get:

v(0, t) = ut(0, t) =g ′(t); −α≤ t ≤ β , (19)

v(1, t) =ut(1, t) = h′(t); −α≤ t ≤ β . (20)

Using condition (8), we see that:

v(x ,−α) = ut(x ,−α) = q(x); 0≤ x ≤ 1. (21)



F. Parzlivand, A. Shahrezaee / Eur. J. Pure Appl. Math, 8 (2015), 239-254 244

The problem (2)-(9) may be divided into two separate problems. The first problem is the

following parabolic problem:

vt − vx x + B2v =0; 0< x < 1, 0< t < β , (22)

v(0, t) =g ′(t); 0≤ t ≤ β , (23)

v(1, t) =h′(t); 0≤ t ≤ β . (24)

The second problem is a hyperbolic problem as follows:

vt t − vx x + B2v =0; 0< x < 1, −α < t < 0, (25)

v(x ,−α) =q(x); 0≤ x ≤ 1, (26)

v(0, t) =g ′(t); −α≤ t ≤ 0, (27)

v(1, t) =h′(t); −α≤ t ≤ 0. (28)

Therefore, for solving the inverse mixed parabolic-hyperbolic problem (2)-(9), we shall

investigate the direct parabolic problem (22)-(24) and the direct hyperbolic problem (25)-

(28).

4. Numerical Procedures

In this section, the GA-RBFs method are used for solving the mixed parabolic-hyperbolic

problem (2)-(9). In order to use the GA-RBFs for solving this problem, we shall investigate

the direct parabolic problem (22)-(24) and direct hyperbolic problem (25)-(28).

4.1. Application of CA-RBFs in the Parabolic Problem (22)-(24)

For this problem, let x i =
i−1
N−1 , i = 1,2, . . . , N and t j = β

j
M , j = 1,2, . . . , M . And supposed

that {(x i , t j), i = 1,2, . . . , N , j = 1, . . . , M} be a set of scattered nodes. Note that u(x , 0) is

unknown so that boundary t = 0 does not discretization. Then the solution of the problem

(22)-(24) by using GA-RBFs is considered as follows:

v(x , t) ≃
N
∑

i=1

M
∑

j=1

vi jφi j(x , t) =

N
∑

i=1

M
∑

j=1

vi je
−ǫ2((x−x i)

2)+(t−t j)
2)

=

N
∑

i=1

M
∑

j=1

vi jφi(x)φ j(t)

=ΦT
N (x)VΦM (t), (29)

where φi(x) is the GA-RBF on [0,1], e.i. φi(x) = e−ǫ
2(x−x i)

2

, φ j(t) is the GA-RBF on [0,β]

and the unknown matrix V is N ×M and can be shown as:










v11 v12 · · · v1M

v21 v22 · · · v2M
...

...
. . .

...

vN1 vN2 · · · vN M













F. Parzlivand, A. Shahrezaee / Eur. J. Pure Appl. Math, 8 (2015), 239-254 245

By using (13) we have:

vt(x , t) =
∂

∂ t
ΦT

N (x)VΦM (t) = Φ
T
N (x)VΦ

′
M (t) = Φ

T
N (x)V DM (t)ΦM (t). (30)

Also, from (16) we obtain:

vx x(x , t) =
∂ 2

∂ x2
ΦT

N (x)VΦM (t) = Φ
′′T
N (x)VΦM (t) = Φ

T
N (x)(PN + D2

N (x))
T VΦM (t). (31)

Using (29)-(31) in (22), we obtain:

ΦT
N (x)V DM (t)ΦM (t)−ΦT

N (x)(PN + D2
N (x))

T VΦM (t) + B2ΦT
N (x)VΦM (t) = 0. (32)

And using (29) in (23)-(24) yields:

ΦT
N (0)VΦM (t)− g ′(t) = 0, (33)

ΦT
N (1)VΦM (t)− h′(t) = 0. (34)

The collocation technique is used for finding the unknown matrix V . We collocate (32) in

(N − 2)×M points Ω1 = {(xk, ts)|k = 2,3, . . . , N − 1, s = 1,2, . . . , M}, we get:

ΦT
N (xk)V DM (ts)ΦM (ts)−ΦT

N (xk)(PN+D2
N (xk))

T VΦM (ts)+B2ΦT
N (xk)VΦM (ts) = 0; (xk, ts) ∈ Ω1.

(35)

Now, collocation (33) and(34) in M points

Ω2 = {(xk, ts)|k = 1, s = 1,2, . . . , M} and Ω3 = {(xk, ts)|k = N , s = 1,2, . . . , M}, respectively,

yields:

ΦT
N (0)VΦM (ts)− g ′(ts) = 0; (xk, ts) ∈ Ω2 (36)

ΦT
N (1)VΦM (ts)− h′(ts) = 0; (xk, ts) ∈ Ω3 (37)

Equations (35)-(37) give a N × M system of linear algebraic equations with the N × M

unknown coefficients vi j . Solving this system, the unknown function of v(x , t) on t ∈ [0,β]

can be found.

4.2. Application of GA-RBFs in the Hyperbolic Problem (25)-(28)

Now, let x i =
i−1
N−1 ; i = 1,2, . . . , N , and t j = −α j−1

M−1 , j = 1,2, . . . , M . The unknown function

v(x , t) in (25)-(28) can be approximated as:

v(x , t)≃
N
∑

i=1

M
∑

j=1

wi jφi j(x , t) =

N
∑

i=1

M
∑

j=1

wi je
−ǫ2((x−x i)

2)+(t−t j)
2)

=

N
∑

i=1

M
∑

j=1

wi jφi(x)φ j(t)

=ΦT
N (x)WΦM (t), (38)



F. Parzlivand, A. Shahrezaee / Eur. J. Pure Appl. Math, 8 (2015), 239-254 246

where φi(x) is the GA-RBF on [0,1], φ j(t) is the GA-RBF on [−α, 0] and the unknown matrix

W is N ×M and can be shown as:











w11 w12 · · · w1M

w21 w22 · · · w2M
...

...
. . .

...

wN1 wN2 · · · wN M











By using (16) and (30) we have:

vt t(x , t) =
∂ 2

∂ t2
ΦT

N (x)WΦM (t) = Φ
T
N (x)WΦ

′′
M (t) = Φ

T
N (x)W (PM + D2

M (t))ΦM (t). (39)

Using (39), (30) and (31) in (25), we can write:

ΦT
N (x)W (PM + D2

M (t))ΦM (t)−ΦT
N (x)(PN + D2

N (x))
T WΦM (t) + B2ΦT

N (x)WΦM (t) = 0. (40)

And using (39) in (26)-(28) yields:

ΦT
N (x)VΦM (−α)− q(x) = 0, (41)

ΦT
N (0)VΦM (t)− g ′(t) = 0, (42)

ΦT
N (1)VΦM (t)− h′(t) = 0. (43)

The collocation technique is used for finding unknown matrix W . We collocate Equation

(40) in (N − 2)× (M − 1) points γ1 = {(xk, ts)|k = 2,3, . . . , N − 1, s = 2,3, . . . , M}, we get:

ΦT
N (xk)W (PM + D2

M (ts))ΦM (ts)−ΦT
N (xk)(PN + D2

N (xk))
T WΦM (ts) + B2ΦT

N (xk)WΦM (ts) = 0,

(44)

for (xk, ts) ∈ γ1. By collocation (41) in N points γ2 = {(xk, ts)|k = 1,2, . . . , N , s = 1}, we have:

ΦT
N (xk)VΦM (−α)− q(xk) = 0; (xk, ts) ∈ γ2. (45)

And collocation (42) and(43) in (M − 1) points γ3 = {(xk, ts)|k = 1, s = 2,3, . . . , M} and

γ4 = {(xk, ts)|k = N , s = 2,3, . . . , M}, respectively, yields:

ΦT
N (0)VΦM (ts)− g ′(ts) = 0; (xk, ts) ∈ γ3, (46)

ΦT
N (1)VΦM (ts)− h′(ts) = 0; (xk, ts) ∈ γ3. (47)

Equations (44)-(47) give a N × M system of linear algebraic equations with the N × M

unknown coefficients wi j . Solving this system, the unknown function of v(x , t) on t ∈ [−α, 0]

can be found.



F. Parzlivand, A. Shahrezaee / Eur. J. Pure Appl. Math, 8 (2015), 239-254 247

4.3. Solve u(x , t), f1(x) and f2(x) from v(x , t)

In order to recover u from v in [0,β], by integration both sides of v(x , t) = ut(x , t) with

respect to the variable t from t to β , we obtain:

u(x , t) = u(x ,β)−
∫ β

t

v(x , s)ds = ϕ(x)−
∫ β

t

v(x , s)ds. (48)

And for recover u in [−α, 0], by integration both sides of v(x , t) = ut(x , t) with respect to the

variable t from −α to t, we can write:

u(x , t) = u(x ,−α) +
∫ t

−α
v(x , s)ds =ψ(x) +

∫ t

−α
v(x , s)ds. (49)

By using (4), we get:

v(x ,β) = ut(x ,β) = ux x(x ,β)− B2u(x ,β) + f1(x), (50)

therefore, from (9), we have:

f1(x) = v(x ,β)−ϕ′′(x) + B2ϕ(x). (51)

Also, using (4) and (7), we get:

f2(x) = vt(x ,−α)−ψ′′(x) + B2ψ(x). (52)

Therefore, having approximation solution of v (ṽ) is determined, then approximation so-

lutions of u, f1 and f2 can be obtain as follow:

ũ(x , t) =ϕ(x)−
∫ β

t

ṽ(x , s)ds; t > 0, (53)

ũ(x , t) =ψ(x) +

∫ t

−α
ṽ(x , s)ds; t < 0, (54)

f̃1(x) =ṽ(x ,β)−ϕ′′(x). (55)

f̃2(x) =ṽt(x ,−α)−ψ′′(x). (56)

5. Test Examples

In this section two examples are presented to demonstrate the applicability and accuracy

of the method. These tests are chosen such that their analytical solutions are known. But

the method developed in this research can be applied to more complicated problems. The

numerical implementation is carried out in Maple 13.

We tested the accuracy and stability of the method presented in this paper by performing

the mentioned method for different values of N and M . To study the convergence behavior of



F. Parzlivand, A. Shahrezaee / Eur. J. Pure Appl. Math, 8 (2015), 239-254 248

the RBFs method, we applied the following law:

The root mean square (RMS) is described using:

RMS(u) =

√

√

√

√

1

N M

N
∑

i=1

M
∑

j=1

|u(x i , t j)− ũ(x i , t j)|2,

RMS( f1) =

√

√

√

√

1

N

N
∑

i=1

| f1(x i)− f̃1(x i)|2,

RMS( f2) =

√

√

√

√

1

N

N
∑

i=1

| f2(x i)− f̃2(x i)|2.

For simplicity, we set α= β = 1 in all following examples

5.1. Example 1

We solve the problem (2)-(9) with B = 1 and:

g(t) =

¨

1+ exp(−2t); 0< t < 1,

cos(
p

2t); −1< t < 0,

h(t) =

¨

(1+ exp(−2t)) cos(1); 0< t < 1,

cos(
p

2t) cos(1); −1< t < 0,

q(x) =−p2sin(
p

2) cos(x),

ϕ(x) =(1+ exp(−2)) cos(x),

ψ(x) = cos(
p

2) cos(x).

The exact solution of this problem is:

u(x , t) =

¨

(1+ exp(−2t)) cos(x); 0< t < 1,

cos(
p

2t) cos(x); −1< t < 0,
,

¨

f1(x) = 2cos(x),

f2(x) = 0.

Tables 1 and 2 show the absolute values of error for u at t = 0.5 and t = −0.5 for different

values of M and N , using the method presented in Section 3, respectively. The corresponding

results obtained for f1(x) and f2(x) are presented in Tables 3 and 4, respectively. Also, Table

5 shows the RMS error values for f1(x), u(x , t) on the interval t ∈ [0,1] and x ∈ [0,1], f2(x)

and u(x , t) on the interval t ∈ [−1,0] and x ∈ [0,1] for various values of N and M . It can

be obtained from results obtained that the accuracy increases with the increase of the number

of collocation points. In addition, the graphs of the error functions |u(x , t) − ũ(x , t)| on the

interval t ∈ [0,1] and t ∈ [−1,0], | f1(x)− f̃1(x)| and | f2(x)− f̃2(x)| are plotted in Figure 1.



F. Parzlivand, A. Shahrezaee / Eur. J. Pure Appl. Math, 8 (2015), 239-254 249

Table 1: Absolute Values of Error for u from Example 1 with t = 0.5 and ǫ = 0.1.

x N = 6, M = 5 N = 7, M = 7 N = 8, M = 10 N = 11, M = 11

0.0 2.8×10−5 2.1×10−7 7.0×10−9 7.9×10−10

0.2 5.0×10−5 2.8×10−7 1.2×10−8 1.3×10−10

0.4 6.6×10−5 4.9×10−7 1.7×10−8 1.7×10−10

0.6 7.5×10−5 5.5×10−7 2.0×10−7 1.9×10−10

0.8 7.8×10−5 5.6×10−7 2.1×10−9 1.9×10−10

1.0 7.4×10−5 5.3×10−7 2.0×10−9 1.8×10−10

Table 2: Absolute Values of Error for u from Example 1 with t = −0.5 and ǫ = 0.1.

x N = 6, M = 5 N = 7, M = 7 N = 8, M = 10 N = 11, M = 11

0.0 1.2×10−5 5.9×10−8 6.8×10−10 2.5×10−10

0.2 7.8×10−3 1.6×10−3 3.5×10−5 2.1×10−5

0.4 1.2×10−2 2.9×10−3 5.7×10−5 3.4×10−5

0.6 1.2×10−2 2.6×10−3 5.7×10−5 3.4×10−5

0.8 7.8×10−3 1.6×10−3 3.5×10−5 2.1×10−5

1.0 6.9×10−6 3.4×10−8 7.3×10−10 1.5×10−10

Table 3: Absolute Values of Error for f1(x) from Example 1 with ǫ = 0.1.

x N = 6, M = 5 N = 7, M = 7 N = 8, M = 10 N = 11, M = 11

0.2 2.8×10−4 2.9×10−6 8.8×10−9 1.5×10−9

0.4 3.1×10−4 2.6×10−6 1.4×10−8 8.3×10−10

0.6 2.8×10−4 2.2×10−6 1.1×10−8 7.5×10−10

0.8 2.4×10−4 1.8×10−6 3.9×10−9 4.4×10−10

Table 4: Absolute Values of Error for f2(x) from Example 1 with ǫ = 0.1.

x N = 6, M = 5 N = 7, M = 7 N = 8, M = 10 N = 11, M = 11

0.2 6.9×10−2 1.6×10−2 2.1×10−3 3.5×10−4

0.4 1.1×10−1 2.7×10−2 3.5×10−3 5.7×10−4

0.6 1.1×10−1 2.7×10−2 3.5×10−3 5.7×10−4

0.8 6.9×10−2 1.6×10−2 2.1×10−3 3.5×10−4



F. Parzlivand, A. Shahrezaee / Eur. J. Pure Appl. Math, 8 (2015), 239-254 250

Table 5: RMS errors for u, f1 and f2 for Example 1 with ǫ = 0.1.

Error N = 6, M = 5 N = 7, M = 7 N = 8, M = 10 N = 11, M = 11

RMS(u(x , t)), t ∈ [0,1] 2.402E–04 1.256E–05 3.511E–07 1.501E–08

RMS(u(x , t)), t ∈ [−1,0] 1.104E–02 2.220E–03 4.749E–04 2.951E–05

RMS( f1(x)) 2.248E–04 2.063E–06 1.600E–08 7.129E–10

RMS( f2(x)) 8.004E–02 1.877E–02 4.017E–03 2.499E–04

0

0.2

0.4

0.6

0.8

1

x

0

0.2

0.4

0.6

0.8

1

t

0

2e–07

4e–07

6e–07

8e–07

1e–06

1.2e–06

1.4e–06

1.6e–06

(a) Graph of |u(x , t)− ũ(x , t)| on t ∈ [0, 1].

0

0.2

0.4

0.6

0.8

1

x

–1

–0.8

–0.6

–0.4

–0.2

0

t

0

2e–05

4e–05

6e–05

8e–05

0.0001

(b) Graph of |u(x , t)− ũ(x , t)| on t ∈ [−1, 0].

0

0.0005

0.001

0.0015

0.002

0.0025

0.003

0.0035

0 0.2 0.4 0.6 0.8 1

x

(c) Graph of | f1(x)− f̃1(x)|.

0

1e–08

2e–08

3e–08

4e–08

5e–08

6e–08

7e–08

0.2 0.4 0.6 0.8 1

x

(d) Graph of | f2(x)− f̃2(x)|.
Figure 1: Graph of absolute error for u, f1 and f2 by using GA-RBF for Example 1 with N =

8, M = 10 and ǫ = 0.1.

5.2. Example 2

We solve the problem (2)-(9) with B = 0 and:

g(t) =

¨

0; 0< t < 1,

0; −1< t < 0,

h(t) =

¨

−3; 0< t < 1,

2; −1< t < 0,

q(x) =2x ,



F. Parzlivand, A. Shahrezaee / Eur. J. Pure Appl. Math, 8 (2015), 239-254 251

ϕ(x) =x3 − 3x ,

ψ(x) = sin(x) + 2x ,

The exact solution of this problem is:

u(x , t) =

¨

x3 − 3x t; 0< t < 1,

sin(x) + 2x t; −1< t < 0,
,

¨

f1(x) = −9x ,

f2(x) = sin(x).

Table 6 shows the RMS error values for u(x , t) f1(x) and f2(x) for various values of N

and M . Also, the graphs of the error functions |u(x , t)− ũ(x , t)| on the interval t ∈ [0,1] and

t ∈ [−1,0] are plotted in Figure 2.

Table 6: RMS Errors for u, f1 and f2 for Example 2 with ǫ = 0.1.

Error N = 6, M = 5 N = 7, M = 7 N = 8, M = 10 N = 11, M = 11

RMS(u(x , t)), t ∈ [0,1] 4.698E–10 1.541E–10 1.835E–13 4.716E–15

RMS(u(x , t)), t ∈ [−1,0] 4.704E–08 1.112E–08 2.650E–10 1.111E–12

RMS( f1(x)) 1.222E–09 2.565E–10 4.910E–13 2.001E–14

RMS( f2(x)) 3.697E–007 1.091E–07 7.141E–10 8.304E–11

0

0.2

0.4

0.6

0.8

1

x

0

0.2

0.4

0.6

0.8

1

t

0
2e–10
4e–10
6e–10
8e–10
1e–09

1.2e–09
1.4e–09
1.6e–09

(a) Graph of |u(x , t)− ũ(x , t)| on t ∈ [0, 1].

0

0.2

0.4

0.6

0.8

1

x

–1

–0.8

–0.6

–0.4

–0.2

0

t

0

2e–08

4e–08

6e–08

8e–08

1e–07

(b) Graph of |u(x , t)− ũ(x , t)| on t ∈ [−1, 0].

Figure 2: Graph of absolute error for u by using GA-RBF for Example 2 with N = 6, M = 5 and

ǫ = 0.1.

6. Conclusion

In this paper, we presented a numerical scheme for solving an inverse problem to a class

of mixed parabolic-hyperbolic equation. This inverse problem related to finding the unknown

right-hand side of the equation of mixed parabolic-hyperbolic type in a rectangular domain.

The GA-RBFs were employed. This method is meshless, so that unknown boundary does not

require to discretization. Using this method, a rapid convergent solution is produced which

tends to the exact solution of the problem. Illustrative examples are included to demonstrate

the validity and applicability of the technique.



REFERENCES 252

ACKNOWLEDGEMENTS We would like to thank the anonymous referees for their valuable

comments and helpful suggestions to improve the final manuscript.

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