The discrete analogue of the method of quickest descent for an inverse acoustic problem in
case of a smooth source
G. Tyulepberdinova, G.Gaziz, N. Kerimbayev, S. Abdykarimova
al-Farabi Kazakh National university, Almaty, Kazakhstan
Abstract
The article considers the discrete analogue of the method of quickest descent for an inverse Acoustics
problem in case of a smooth source. The authors derived the gradient of functional in differential and
discrete cases, described the algorithm of solving a problem, and compared gradients of functional in
continuous and discrete cases. In the article the improved estimates of the rates of convergence of gradientbased methods are obtained, which are very important for practice because they provide with the
possibility to make input data errors consistent with the iteration number. There is a practical application
of the proposed new method of deriving the gradient of functional for an Acoustics discrete problem, for it
provides with calculations that are more accurate. The theoretical importance of the method is the
developed technique of deriving estimates and gradients of functional at a discrete level.
Key words: inverse problem, acoustics problem, D'Alembert formula, Landweber iteration,
discrete analogue, gradient, dual problem.
1 Introduction
In Acoustics, inverse problems are understood as restoration of sound sources or heterogeneousness
(inhomogeneous) characteristics, which scatter a primary field on the basis of measuring the primary or
scattered acoustic field. Historically, first solving methods used to be based on approximation of a single
scattering and smooth variation of scattering characteristics. However, the assumption used in these
approximations seriously constrain the field of their application. Further investigations connected with
taking into account effects of multiple scatterings, have shown that the inverse scattering problem is
incorrect and nonlinear in unknown functions. One of methods of solving inverse problems, which
consider multiple scatterings, is the iteration method.
The objective of this work is a proper study and derivation of the gradient of functional in
the differential and discrete analogues using the method of quickest descent for an inverse
acoustics problem in case of a smooth source.
A.L. Milman dedicated his investigations in the sphere of scattering theory for a wave
equation in two-dimensional space. The paper “The Inverse Scattering Problem for Acoustic
Centro Symmetrical Finite Objects in Two-dimensional Space”, presents Scattering theory for a
wave equation in two-dimensional space perturbed by a finite function of a radial variable. There
has been shown that this compression has one-dimensional defect subspaces, and its
characteristic operator-function is a meromorphic function (1990).
J.R. McLaughlin, P.E. Sacks, M. Somasundaram in their publication “Inverse Scattering
in Acoustic Media Using Interior Transmission Eigenvalues” consider well-known inverse
problems for acoustic media consisting in the determination of the sound speed of the medium
using far field data. A precise formulation is as follows. The paper considers the Helmholtz
equation, where the sound speed, с (z), satisfies some condition for some constant c0  0 (1997).
Works by A.S. Blagoveschenskiy and D.A. Fedorenko are dedicated to the inverse
problem for acoustics equations in the weekly horizontally inhomogeneous medium (2008).
The paper by V.A. Barkhatov, “Solution of the One-dimensional Inverse Acoustic
Problem with Allowance for Velocity Dispersion and Frequency-dependent Wave Attenuation”,
considers a one-dimensional inverse acoustics problem. The solution of this inverse problem
determines the dependence of the reflecting coefficient of echo-signals, which a defectoscope
receives and processes at a distance. The solution of a one-dimensional inverse acoustics
problem is considered taking into account velocity dispersion and attenuation dependent on wave
frequency. Besides, the paper presents the results of numerical experiments(2009).

2. Formulation of the problem
Let us consider the one-dimensional acoustic equation:

1

1
w
(
z
,
t
)

w
(
z
,
t
)

,
t
с 2 tt


z

z  0, t  0

(1)

Here c(z) > 0 is the sound speed, p(z) > 0 is the density, ω(z,t) is the pressure. Equation (1)
describes the propagation of acoustic waves of a small amplitude in a subspace satisfying the
following conditions

w( z, t ) |t 0  0,

(2)

wz (0, t )   (t ), t  0,

(3)

w(0, t )  f (t ), t  0,

(4)

where θ(t) is a smooth function. It is well known that using an additional information in the form
(4), it is difficult to determine the two functions, c(z) and p(z), simultaneously. Therefore, we
formulate the inverse problem of determining one function. For this purpose, we introduce a new
variable and function.
z

x   ( z) 



d
, z   1 ( x)
c( )

0

v( x, t )  w( z, t )  w( 1 ( x), t )
h( x)   ( z)   ( 1 ( x)), g ( x)  c( z)  c( 1 ( x)).
Then (1) – (4) will have the forms

vtt ( x, t )  v xx ( x, t ) 

 ( x)
v ( x, t ), x  0, t  0,
 ( x) x

(5)

v( x, t ) |t 0  0,

(6)

vx (0, t )   , t  0,

(7)

v(0, t )  f (t ), t  0.

(8)

where σ = g(x)h(x) is the acoustic impendence,   с(0).
Further (V.G. Romanov), we reduce equation (5) to the following form:

utt ( x, t )  u xx ( x, t )  q( x)u( x, t ),

(9)

 1

u ( x, t )  v( x, t ) exp ln  ( x),
 2


(10)

where

1
 1  ( x ) 
q( x)   ln  ( x)  
2
4  ( x) 

2

(11)

Further, in order to make investigation of the inverse problem at a discrete level more
convenient, taking into account the above reduction, we consider the following problem. We will
determine q(x) using the following expressions

u~tt ( x, t )  u~xx ( x, t )  q( x)u~( x, t ), x  R, t  0

(12)

u~( x,0)  0 ( x), u~t ( x,0)  1 ( x), x  R,

(13)

u~(0, t; q)  2 (t ), u~t (0, t )  0, t  0.

(14)

Let us consider the equivalent inverse problem (S.I. Kabanikhin, K.T. Iskakov):

utt ( x, t )  u xx ( x, t )  q( x)u( x, t ), x  R, t  0
u( x,0)  0( x)  q( x)0 ( x), ut ( x,0)  0, x  R,
u(0, t; q)  f (t ), u x (0, t )  0, t  0.

(15)
(16)
(17)

The expressions (15) – (17) follow from (12) – (14), if we substitute

u  u~tt ,

f (t )  2(t ), 1 (t )  0.

When using the optimization method for solving an inverse problem, a criterion functional
is minimized, for example a criterion functional of the following form:
T

J ( p) 



u (0, t; p)  f (t )2 dt.

(18)

0

The notation u(0,t;p) indicates that u(x,t) is a solution of the direct problem (15) – (17) at fixed
р(х). The minimization of the functional J(p) is realized by the method of quickest descent:

 

p ( n1) ( x)  p ( n ) ( x)   nJ p ( n ) ,
in which the descent factor, n , is determined using the condition

J  p ( n)   n J ( p ( n) )  inf J  p ( n )   nJ ( p ( n ) ),  n  0.
 0

Here J ( p ) is the gradient of functional, n is the number of iteration.
(n )

3. Optimization method at differential level
Let us formulate the inverse problem under study concretely. In the domain
D  [0, L]  [0, T ] it is required to determine the function q(x) using the expressions
utt  u xx  q( x)u, ( x, t )  D

(20)

u ( x,0)  q( x),

ut ( x,0)  0

(21)

u x (0, t )  0,

u ( L, t )  0

(22)

and the given additional information of the form
u0, t; q   f t 

(23)

For illustrative purposes, we choose a rectangular domain D.
It is easy to see that the expressions (20) – (22) can be easily obtained from (15) assuming
that  0 ( x)  1 .
Let р(х) be the approximate solution of the inverse problem. As before, we consider the
discrepancy functional
T

J ( p) 



u (0, t; p)  f (t )2 dt

(24)

0

The essence of the optimization method is the following: we define the initial approximation
p ( 0) ( x), we determine the successive approximations using (19).
Let us get down to deriving the formula of the gradient for the functional (24) using the
method analogous to those in the works (S.I.Kabanikhin and others, G.A.Tyulepberdinova,
A.E.Yerdeneyeva).
Let us consider the increment p( x)  p( x) и u  u( x, t; p  p)  u( x, t; p) , then for
the increment u ( x, t ), neglecting second-order small quantities, we get the following problem
(A.T.Nurseitova, D.B.Nurseitov, G.A.Tyulepberdinova):

 utt  u xx  up  pu, ( x)  D

(25)

 u( x,0)  p, ut  0,

(26)

 u x (0, t )  0, u( L, t )  0,

(27)

Now we multiply both parts of the equation (25) by the function ψ(x,t), and integrating over the
domain D, we obtain

T L

T L

 ( x, t)u dxdt   ( x, t)u dxdt
tt

xx

0 0

0 0

T L



T L

 ( x, t)u( x, t)p( x)dxdt   p(x)u( x, t) ( x, t)dxdt
0 0

(28)

0 0

Denoting by S1 , S 2 , S3 , S 4 each item of the latter expression and applying integration by parts to
S1 , S 2 , we obtain:
T L

S1 

L



 ( x, t )utt dxdt 

0 0


0

L




0


ut




ut |T0




  t ( x, t )ut dt  dx

0

T

T
0



T


T

  t u |0   tt  x, t u ( x, t )dt  dt


0





L



  x,T u x,T   x,0u x,0  x,T ux,T 
t

t

t

0

T



  t x,0u x.0   tt  x, t u x, t dt dx
0

Using the condition (26) and assuming that  x, T   0,
T L

S1 

 t x, T   0, we have

T

 x,t ux,t dxdt   x,0pdx
tt

t

0 0

0

Let us reduce
T L

S2 

T

 x, t u dxdt  
xx

0 0

0

L


u x |0L   x  x, t u x dx  dt


0





(29)

T




0


u x |0L



L


L

  xu 0   xx x, t u  x, t dx  dt


0





T



  L, t u L, t   0, t u 0, t   L, t uL, t 
x

x

x

0

L



  x 0, t u 0, t    xx  x, t u  x, t dxdt
0

 x 0, t   2u0, t; p   f t ,

Using the condition (27) and assuming that  L, t   0,
we obtain
T L

S2 



T

xx

x, t u( x, t )dtdx 

0 0

 2u0, t; p  f t u(0, t)dt
0

Further, we leave the formulae for S3 , S 4 alone, that is
T L

S3  S 4  

 x,t ux, t pxdxdt 
0 0

T L



 pxux,t  x, t dxdt
0 0

Using (29) and (30) the formula (28) will have the form
T L

T L

 x,t ux,t dtdx  
tt

0 0

 2u 0, t ; p   f t u 0, t dt 
0

x, t ux, t dxdt 

0 0

T



xx

T L

 x,t ux,t  pxdxdt 
0 0

T L



 x,t ux,t pxdxdt
0 0

Assume that  tt   xx  px  , then in the latter expression we have

(30)

T

T L

0

0 0

 2u0, t; p   f t u0, t dt    x, t ux, t px dtdx .

The left part is equal to the increment of functional J  p   J  p  p   J  p , and the right
part is its gradient. Therefore, the gradient of functional is equal to
T



J  p    x, t u x, t dt

(31)

0

Thus, in the process of deriving the formula for finding the gradient of a functional, we get
an auxiliary dual problem for  x, t  , which has the form

x, t   D

(32)

 x, T   0,  t x, T   0,

(33)

 x 0, t   2u0, t; p   f t ,

(34)

 L, t   0

(35)

 tt   xx  px  ,

Let us introduce the general scheme of the optimization method at a differential level
(A.T.Nurseitova, D.B.Nurseitov, G.A.Tyulepberdinova):
1 Define the initial approximation p ( 0 ) ( x), solve the direct problem (20) – (22)
assuming that q( x)  p ( 0) ( x) , and find u ( 0) ( x, t; p ( 0) ( x)).
2 Calculate the value of the functional (24), if it attained the minimum then take
(0)
p ( x) as an approximate solution of the inverse problem, if it did not then go to the next step.
3

Calculate the boundary value condition (34) when p( x)  p ( 0) ( x) , and solving the

problem (32) – (35), we obtain its solution  ( x, t; p ( x)).
4 Find the gradient of the functional using the formula (31).
(1)
5 Using the formula (19) we find the next approximation p ( x).
6 Again calculate the value of the functional (24), if it attained the minimum then we
 
 
take p( x)  p (  ) ( x) as an approximate solution, if it did not then assuming p 0 x   p  x 
we return to step 3.
( 0)

( 0)

4. Discrete analogue of the optimization method
Let p xi  be an approximate solution of the inverse problem. We approximate the
problem (20) – (22) using the following difference scheme:

ytt  yˆ xx  pi yˆ , ( xi , t j )  wh ,

(36)

yi0  pi ,

yt0.i  0, i  0, 1, 2, ..., N ,

(37)

y xj,0  0,

y Nj  0,

(38)

j  2, 3, , M .

Here wht  wh  wr is a net analogue of the domain D = [0, L] х [0, Т],

wh  xi  ih,

i  0,1, 2,...,N , h  L / N ,

w  t j  j ,

yij  yxi , t j ,

j  0,1, 2...,M ,   T / M .

yˆ  yij 1  yxi , t j 1  are net approximations of the function u(x,t).

Let us assume that for the difference direct problem (36) – (38) there is an additional
information. Let

y0j  f (t j ),

j  2, 3, , M .

(39)

Let us consider the discrete analogue of the functional (24) of the form

[ y { p }  f
M

J [ p]  

j
0

j 2

].

i

(40)

j 1

pi  pi , and yi  yij { pi  pi }  yij { pi }.
further we will use the designation: yij  { pi } , which shows a particular dependence of the net
function yij on the desired coefficient pi . It is not difficult to obtain the following difference
problem subject to the increment yij :
(41)
ytt  yˆ xx  yˆp  pyˆ , ( xi , t j )  wh ,
Let

us

define

the

increment

yi0  p, yt0.i  0,
y xj,i  0, y Nj  0.
Let us multiply both parts of the difference expression (41) by the net function
summing up over j from 1 to М – 1 and over the index i from 1 to N, we obtain:
M 1 N 1


j 1

y  i
j
tt i

M 1 N 1

j



i 1


j 1

yˆ xx i

i 1

M 1 N 1

j




j 1

 yˆpi  pyˆi i j

(42)
(43)

 i j and

(44)

i 1

We will denote the items by S1 , S 2 , S3 consecutively in that expression and modify them
separately:

S1 

1



N 1 M 1


i 1

j 1

 y j 1  y j y j  y j 1  j

i 








1

2

N 1

, y  ,y
i 1

Here

y i  y j 1  y i , y j  y j 1

 y, v  

M 1


j 1

yiv j ,

 y, v  


M

y jv j

j 1

We apply to S1 difference analogues of integration by parts:

 y, v   v, y  y M v M  y 0 v1
 y, v   v, y   y M v M 1  y 0 v 0

Then the expression for S1 will have the form:

S1 



1



2

1



2

 y,   y 
M



  0y1  y,     M y M 1   0y 0 

M

i 1

N 1

y,    y,    y 
M

  0y1   M y M 1   0y0 

M

i 1

N 1



N 1


1





M 1

 
i 1

j 0

 i j 1  i j
yi 



 1
 
 

j


M

j 1

  j   j 1 


yi 




j

 y M  M   0y1   M y M 1   0y
Further (G.A.Tyulepberdinova), we bring to summing up over j from 1 to М – 1, we
obtain:
N 1

S1 

1



M 1

 
i 1

j 1

 i j 1   i j j 1
yi 



M 1


j 1

 i j   i j 1 j
yi  yi0  i1   i0  


 yiM  iM   iM 1   yiM  iM   i0yi1   iM yiM 1  i  i0
0



After combining like terms and using the difference derivative for i , we have:
j

N 1 M 1

N 1



y t t 

S1 

i 1

j

j 1

y

M

 M 1   M y M  y 0 1   0y1 

i 1

Adding to the second term and then subtracting yi0i0 , yiM iM , we obtain
N 1 M 1

S1 

N 1

y   y    
j

0
i

tt

i 1

j 1

1
i

i 1

0
i



 i0 yi1  yi0   yiM iM 1  iM   iM yiM  yi0 

Using the condition (42) and assuming that



M
i

S1 

 0, 

 0, we obtain



y j t t

M
tt
N 1 M 1

i 1

j 1

We translate the expression for S 2 . Expanding difference derivatives we have:
M 1

S2 

j 1

1
 2
h

N 1


M 1

1
h

i 1

 yˆ i 1  yˆ i yˆ i  yˆ i 1  j


i
h
h



N 1

 yˆ 
i

j 1

i 1

j
i

 yˆ ii j 

(45)

Using difference analogues of integration by parts, we have

1
S2  2
h

M 1

 (yˆ ,  ]   y
j

i

i

i

N

N

 0y1 

j 1

 [yˆ i , i j ) y N 1 N  0y0 
Expanding scalar products, we obtain:
M 1





N 1

 yˆ ,   yˆ 

1
S2  2
h

j 1

N

j

i

i

i

i 1

j
i

i 0

  N y N  0y1  y N 1 N  0y0 
Expanding the sums and combining them again, we obtain:

1
S2  2
h

M 1





N 1

 yˆ 
j 1

i

j
i

 i j    Nj yˆ N

i 1

  0jyˆ1  yˆ N 1 Nj   0jyˆ 0  yˆ N  Nj   Nj 1   yˆ 0  1j   0j
1
 2
h

M 1





N 1

 
j 1

i 1




j
j
j
j
j
j
ˆ
ˆ
ˆ
ˆ
ˆ
yi i   i   0 y1  y N 1 N   N 1y N  y01 


Substituting j   j  1, we have

1
S2  2
h





N 1

 
M

j  2





y ij i  i  

(46)

i 1





  N 1y Nj   y0j 1  y Nj 1 N  0y1j 



(47)

Let us consider the last expression
N 1 M 1

 yˆp  p yˆ 

S3  

i

i 1

i

i

j
i

j 1

If we substitute j' = j + 1 then
N 1

S3  

 y 

N 1 M 1

M

j

i 1

j  2

pi 

j  1
i

 p 
i

i 1

y

j  1
i

j  2

Let us reduce it to the common summation over the index j. For this aim, let us consider the
formula (45) for the expression for S1 . We have
N 1 M 1

S1 

N 1

2
1
 M 1
i1  i0 
j
1 1  i  i


y tt  y 




 j 2



y    
j

tt

i 1

j 1

i 1

(48)

Let us consider the following auxiliary dual problem:





tt   xx  pii , i  1, 2 ,...,N  1,

j  M  1, M  2,...,1,

iM  0, tM,i  0, i  0, 1, 2, ..., N ,

 xj,0  2[ y0j  f 0 j ],

(49)
(50)

 Nj  0,
j  M  2, M  1,...,0,
j  M  2, M  1,...,0.

(51)

(52)
If we use the conditions (42) and (43), and the problem (49) – (52), then the expression (44) will
have the form
N 1


i 1

 y

1  i2  i1 i1   i0 
2
y 

 
 
1

M



N 1 M 1

 f 0 y 

j
0

j

j 2

j
0


i 1

y ji j 1pi

j 2

Which implies that

J  p   h

N 1

M 1

N 1

p  y   
j 1

j

i

i 1

i

j 2

i 1

 i2  i1 i1  i0 

pi 






Then the gradient of the functional has the form:
M 1
1   2   i1  i1   i0 
.
J  p     yi i j 1   i


 

j 2


5 Conclusion
In this paper, the method of quickest descent applied to a one-dimensional inverse acoustic
problem at a discrete level is considered. The gradient of functional at a discrete level is
obtained. The algorithm of solving a problem using the optimization method at a differential
level for a one-dimensional inverse acoustic problem is described. For numerical results (Напке
М., Neuhauer A., Scherzer О.) of the simplified formulation of the inverse problem, the
gradients of functional in continuous and discrete cases were compared in this article. The
improved estimates of the rates of convergence of gradient-based methods considered were obtained. They
are very important for practice because they provide with the possibility to make input data errors
consistent with the iteration number. The practical application of this proposed new method of deriving
the gradient of functional for an Acoustics discrete problem is that it makes calculations more accurate.
The theoretical importance of the proposed method is the development of the technique of deriving
estimates and gradients of functional at a discrete level.
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