NUMERICAL METHODS
FOR MATHEMATICAL PHYSICS INVERSE PROBLEMS
Lecture 6. Inverse problems for ordinary differential equations
We know that the inverse problems can be transformed to the problems of finding of extremum. So the practical
methods of inverse problems theory are based on the optimization methods. The problems of the minimization of the
functionals can be solved by means of the gradient methods. We know how we can use the gradient methods for
linear abstract inverse problems. Now we consider this technique for the analysis of the inverse problems described
by ordinary differential equations.
6.1. Differential equation with unknown absolute term
We consider the system be described by the equation
y  ay  v
(6.1)
with initial condition
(6.2)
y(0)  y 0 ,
0
where y  y (t ) is the state function, a  0 and y are known parameters, and v  v(t ) is the
unknown function. For all smooth enough function v the problem (6.1), (6.2) has a unique
solution y  y[ v ]. We have also the equality
(6.3)
y (t )  z (t ), t0  t  T ,
where z  z (t ) is a given function (result of measuring). Determine the following inverse
problem.
Problem 6.1. Find the function v such that the respective state function y[ v ] satisfies the
equality (6.3).
This problem can be transformed to the optimization control problem. Determine the
functional
T
I (v)    y(t )  z (t ) dt.
2
t0
Problem 6.1'. Minimize the functional I.
We know the relation between these problems. If u is a solution of Problem 6.1, then it
satisfies the equality (6.3). So the respective value of the functional I is zero. Therefore this is the
minimum of this functional because it does have any negative value. Thus the solution of the
inverse problem is the solution of the optimization control problem. Then let u be a solution of
Problem 6.1'. We can have two different cases. If the value I (u ) is zero, then the equality (6.3)
is true; so the solution of the optimization control problem is the solution of the inverse problem.
If the minimum of the functional is positive, then the inverse problem is insolvable, else there
exist a parameter v with more small value than minimum. However the solution of Problem 6.1'
can be chosen an approximate solution of Problem 6.1 because the equality (6.1) is realized in
the best form. Hence we can transform the given inverse problem to the optimization control
problem
We would like to use gradient method for solving Problem 6.2. The general step is the
determination of the functional derivative. Determine the value
T
I (u   h)    y[u   h]  z  dt ,
2
t0
where u and h are functions,  is a number. So we can find the difference
T
I (u   h)  I (u )  
 y[u   h]  z 
2

  y[u ]  z  dt 
2
t0
T
   y[u   h]  z    y[u ]  z  y[u   h]  z    y[u ]  z  dt 
t0
T
(6.4)
T
 2   y[u ]  z  y[u   h]  y[u ] dt    y[u   h]  y[u ] dt.
2
t0
t0
Then we would like to determine of the difference y[u   h ]  y[u ] by means of the problem
(6.1), (6.2).
Consider this system for the functions v  u   h and v  u . We have
d
(6.5)
 y[u h] y[u]  a  y[u h] y[u]   h,
dt
(6.6)
 y[u h] y[u] t 0  0.
Multiply the equality (6.5) by an arbitrary function  and integrate by t. We obtain
T
T
d
y
[
u


h
]

y
[
u
]

a
y
[
u


h
]

y
[
u
]

dt






 dt
 hdt.
0


(6.7)
0
After the integration by parts we get
T
T
T
d
 dt  y[u  h] y[u ]  dt   y[u  h] y[u ]  0      y[u  h] y[u ] dt 
0
0
T
  y[u  h] y[u ]  t T      y[u  h] y[u ] dt
0
because of the equality (6.6). So we obtain the equality
T
     a   y[u  h] y[u ] dt   y[u  h] y[u ]
T
t T
   h dt .
0
(6.8)
0
Chose the function  equal to the solution p of the differential equation
 p(t )  ap(t )  g (t ), 0  t  T
with final condition
p (T )  0,
where
2  y[u ]  z  , if t  t0 ,
g (t )  
0,
if t  t0 .

Therefore the relation (6.8) can be transformed to the equality
T
T
t0
0
(6.9)
(6.10)
(6.11)
2  y[u ]  z  y[u   h]  y[u ] dt    hpdt.
Put it to the equality (6.4). We get
T
T
I (u   h)  I (u )    hpdt    y[u   h]  y[u ] dt.
2
0
(6.12)
t0
Prove that the second term at the right side of this equality has second order with respect to
the parameter . Return to the equality (6.7). Chose     y[u   h]  y[u ] here. We get
d
2
  dt dt  a   dt    h dt.
Using the equality
T
T
T
0
0
0
d 1 d 2

 ,
dt 2 dt  
we transform the previous equality. We get

T
T
 (T )  a   2 dt    h dt
2
0
(6.13)
0
because of the equality  (0)  0 by (6.6).
We know the inequality
h 
 2 h2

2
 a 
  a  2 h  a  0.
a

2
So we
1
 2 h2
.
2
2a
Using this inequality, we transform the formula (6.13).
T
T
1
2 T 2
2
2
2
 (T )  a   dt  a   dt   h dt.
2 0
2a 0
0
Then we have
T
2
2 T 2
2
2
 (T )    dt  2  h dt.
a
a 0
0
We obtain
T
T
T
T
2
2 T 2
2
2
2
2
2
  y[u   h]  y[u] dt  t  dt  0  dt  a  (T )  0  dt  a2 0 h dt.
t0
0
 h  a 2 
Devise this equality by . After the passing to the limit as   0 we get
T
1
 T 2
2
0  lim   y[u   h]  y[u] dt  lim 2  h dt  0.
 0 
 0 a
t0
0
Hence
lim
 0
1
T
 y[u   h]  y[u]

2
dt  0.
(6.14)
t0
Devise the equality (6.12) by . We obtain
T
I (u   h)  I (u )
1
  phdt 

0
T
 y[u   h]  y[u]

2
dt.
t0
After the passing to the limit with using the equality (6.14) we have
T
I (u   h)  I (u )
lim
  I (u ), h    phdt.
 0

0
Hence we find the derivative of the functional I at the point u
I (u )  p.
(6.15)
We know the gradient methods for solving the problem of the minimization of a functional I
on a unitary space
vk 1  vk  k I (vk ), k  0,1,...,
(6.16)
where  k is a positive iterative parameter. Using formula (6.15), we can determine the gradient
method for our problem. Let k-iteration value vk of the control is known. Then we can find the
appropriate value of the state function yk from Cauchy problem
yk  ayk  vk ,
(6.17)
yk (0)  y 0 .
(6.18)
Next step is the determination of the adjoint state pk from the adjoint equation (6.9), (6.10)
 pk (t )  apk (t )  gk (t ), 0  t  T ,
(6.19)
pk (T )  0,
(6.20)
where
2  yk  zk  , if t  t0 ,
g k (t )  
(6.21)
0,
if t  t0 .

The next iteration of the control can be determined by the formula (6.16) with using the formula
(6.15) of the functional derivative. We get
vk 1  vk  k pk , k  0,1,... .
(6.22)
Hence we have the iterative algorithm (6.17) – (6.22) for solving our problem.
6.2. System of differential equations with unknown absolute term
We consider now the following system of differential equations
 y1  a11 y1  a12 y2  v
,

 y2  a21 y1  a22 y2  f
(6.23)
with initial conditions
yi (0)  yi0 , i  1, 2,
(6.24)
where y  y(t )   y1 (t ), y2 (t )  is the state function, aij and y are known parameters, i, j  1, 2,
0
i
f  f (t ) is the unknown function, and v  v(t ) is the unknown function. For all smooth enough
function v the problem (6.23), (6.24) has a unique solution y  y[ v ]. We have also the equality
(6.25)
y2 (t )  z (t ), 0  t  T ,
where z  z (t ) is a given function (result of measuring). Determine the following inverse
problem.
Problem 6.2. Find the function v such that the respective state function y[ v ] satisfies the
equality (6.25).
This problem can be transformed to the optimization control problem. Determine the
functional
T
I (v)    y2 (t )  z (t )  dt.
2
0
Problem 6.2'. Minimize the functional I.
Find the derivative of this functional. Determine the value
T
I (u   h)    y2 [u   h]  z  dt ,
2
0
where u and h are functions,  is a number. Find the difference
T
I (u   h)  I (u )  
 y [u   h]  z    y [u]  z   dt 
2
2
2
2
0
T
   y2 [u   h]  z    y2 [u ]  z  y2 [u   h]  z    y2 [u ]  z  dt 
0
T
T
 2  y2 [u ]  z  y2 [u   h]  y2 [u ] dt    y2 [u   h]  y2 [u ] dt.
2
0
0
(6.26)
Determine of the difference y2 [u   h]  y2 [u] by means of the problem (6.23), (6.24).
Consider this system for the functions v  u   h and v  u . We have
1  a111  a122   h
(6.27)
,




a


a


f
 2 21 1 22 2
(6.28)
i (0)  0, i  1, 2,
where i  yi [u   h]  yi [u], i  1, 2. Multiply the i-th equality (6.27) by an arbitrary function
i, add by i, and integrate by t. We obtain
T
2 T
2 T
di

dt

a


dt


(6.29)

 dt i i
 ij j i
 h1dt ,
i 1
, j 1
0
0
0
After the integration by parts we get
T
T
T
T
di


dt






dt




i i 0
i i t T
 dt i
 ii
 ii dt
0
0
0
because of the equality (6.28). So we obtain the equality
2
2 T
i 1
i 1 0
T
T
i , j 1 0
0
2
 i i t T    ii dt    aij j i dt    h1dt.
Change the order of the addition; we get
2 T
2


2
T
i i t T     i   a ji j i dt    h1dt.
(6.30)


0
Chose the functions 1, 2 equal to the solution p1, p2 of the system of the differential
equations
 p1  a11 p1  a21 p2  0
,
(6.31)

 p2  a12 p1  a22 p2  2  y2 [u ]  z 
with final condition
pi (T )  0, i  1, 2.
(6.32)
Therefore the relation (6.30) can be transformed to the equality
i 1
i 1 0
j 1
T
T
t0
0
2  y2 [u]  z  2 dt    hp1dt.
Put it to the equality (6.26). We get
T
T
I (u   h)  I (u )    hp1dt    2  dt.
2
0
(6.33)
0
This is the analogue of the equality (6.12).
Prove the smoothness of the second term at the right side of this formula. Return to the
equality (6.7). Chose i  i , i  1, 2 here. We get
2
di

dt


 dt i i
 aij ji dt    h1dt.
i 1
, j 1
2 T
T
T
0
0
0
We have
1 d i 2
1
1
2
2 T
di

dt

dt


 i (T )


i
0 dt i 0 2 dt
0
2
2
because of the equality (6.28). So the previous formula can be transformed to the equality
T
2 T
1
2
(6.34)
i (T )    aij ji dt    h1dt.
2
i , j 1 0
0
Suppose the matrix
T
T
 a11 a12 


 a21 a22 
is positive, that is
2
 a 
ij i
i , j 1
2
j
   (i )2 1 ,2  ,
i 1
where  is a positive constant. Then the formula (6.34) can be transformed to the inequality
T
T
2
2
  1   2   dt    h1dt.
0


Using the inequality
 h1 

1 
2
2
0

1 2 2
h ,
2
we have

2
T
T
 1  dt    2  dt 
2
2
0
0
2 T 2
h dt.
2 0
So we get
T
 2  dt 
2
0
2 T 2
h dt.
2 2 0
Then
0  lim
 0
1

 T 2
h dt  0,
 0 2 2 
0
T
 2  dt  lim
2
t0
namely
lim
 0
1
T
 

2
2
dt  0.
t0
After devising the equality (6.33) by  and passing to the limit we have
T
I (u   h)  I (u )
lim
  I (u ), h    p1hdt.
 0

0
Hence we find the derivative of the functional I at the point u
I (u)  p1.
We can apply the gradient method know. Let k-iteration value vk of the control is known.
We determine the state function y1k, y2k from the system
 y1k  a11 y1k  a12 y2 k  v
,

 y2 k  a21 y1k  a22 y2 k  f
yik (0)  yi0 , i  1, 2.
Then we determine the solution of the adjoint system at the iteration k
 p1k  a11 p1k  a21 p2 k  0
,

 p2 k  a12 p1k  a22 p2 k  2  y2 k  z 
pik (T )  0, i  1, 2.
The control at the next iteration is determined by the formula
vk 1  vk  k p1k .
6.3. Differential equation with unknown coefficient
We consider the system be described by the equation
y  vy  f
(6.35)
with initial condition
(6.36)
y(0)  y 0 ,
0
where y  y (t ) is the state function, f  f (t ) is a known smooth function, y is a known
parameters, and v is the unknown positive parameter. For all parameter v the problem (6.35),
(6.36) has a unique solution y  y[ v ]. We have also the equality
(6.37)
y (t )  z (t ), t0  t  T ,
where z  z (t ) is a given function (result of measuring). Determine the following inverse
problem.
Problem 6.3. Find the function v such that the respective state function y[ v ] satisfies the
equality (6.37).
This problem can be transformed to the optimization control problem. Determine the
functional
T
I (v)    y(t )  z (t ) dt.
2
t0
Problem 6.3'. Minimize the functional I.
We will solve this problem with using the gradient method. Find the derivative of the
functional. Determine the value
T
I (u   h)    y[u   h]  z  dt ,
2
t0
where u, h and  are numbers. So we can find the difference
T
T
I (u   h)  I (u )  2  y[u ]  z  y[u   h]  y[u ] dt    y[u   h]  y[u ] dt ,
2
t0
(6.38)
t0
that is an analogue of the equality (6.4).
Consider this system for the functions v  u   h and v  u . We have
    u   h  y[u   h]  uy[u]  0,
 (0)  0,
where   y[u   h]  y[u ]. The equality (6.39) can be transformed to
   u   hy[u ]   h.
Multiply this equality by an arbitrary function  and integrate by t. We obtain
T
T
T
0
0
0
    u   dt   h y[u ] dt   h   dt.
(6.39)
(6.40)
(6.41)
After the integration by parts we get
T
T
T
0
0
   dt   0    dt   (T ) (T )    dt
T
0
because of the equality (6.40). So we obtain
T
T
T
0
0
0
 (T ) (T )       u   dt   h  y[u ] dt   h   dt.
Chose the function  equal to the solution p of the adjoint system
 p(t )  up(t )  g (t ), 0  t  T
(6.42)
(6.43)
p (T )  0,
(6.44)
where
2  y[u ]  z  , if t  t0 ,
g (t )  
0,
if t  t0 .

Then the formula (6.42) can be transformed to the equality
T
T
T
t0
0
0
2  y[u ]  z   dt   h  y[u ] pdt   h   pdt.
Put it to the equality (6.38). We get
T
T
T
0
0
t0
I (u   h)  I (u )   h  y[u ] pdt   h   pdt    2dt.
Return to the equality (6.41). Chose    here. We get
T
T
(6.45)
T
    u  dt   h y[u ] dt   h   dt.
2
2
0
0
0
Transform it to the equality
 (T )
2
T
T
T
0
0
 u   dt   h  y[u ] dt   h   2 dt.
2
0
So we have
T
 (T )
2
T
 (u   h)   dt   h  y[u ] dt.
2
0
0
The number u and h are positive. So the term u   h will be positive for small enough .
Then we obtain the inequality
T
h T
2

dt


y[u ] dt.
0
u   h 0
By Schwartz inequality we have
2
T
T
T
2
 y[u] dt    dt   y[u] dt.
2
0
0
0
Then we obtain
1/2
1/2

h T 2  T
2

dt

   dt     y[u ] dt  .
0
u  h  0
 0

T
2
So we get
1/2
T 2 
   dt 
0

1/2

h T
2

   y[u ] dt  .
u  h  0

Then
2
 h 
0  dt   u   h  0  y[u] dt.
Devise the equality (6.45) by . We obtain
T
T
T
I (u   h)  I (u )
1
 h  y[u ] pdt  h   pdt    2 dt.
2 T
T
2

We have the equality
0
0

t0
(6.46)
(6.47)
lim
 0
1

T
  dt  0
2
0
because of the estimate (6.46). Besides we get
T
T
T
0
0
0
2
2
  pdt    dt  p dt.
So
T
lim   pdt  0
 0
0
because of the estimate (6.46) too. Pass to the limit at the equality (6.47). We have
T
I (u   h)  I (u )
lim
  I (u ), h   h  y[u ] pdt.
 0

0
Then we find the derivative of the functional I at the point u
T
I (u )    y[u ] pdt.
0
Determine the gradient method. Let the value vk is known. The state function yk is determine
from Cauchy problem
yk  uk yk  f ,
yk (0)  y 0 .
Then we find the adjoint state from the system
 pk (t )  uk pk (t )  gk (t ), 0  t  T ,
pk (T )  0,
where
2  yk  z  , if t  t0 ,
g k (t )  
0,
if t  t0 .

The next iteration of the control is determined by the equality
T
vk 1  vk   k  yk pk dt.
0
6.4. Differential equation with two unknown parameters
We consider the system be described by the equation
y  ay  f
(6.48)
with initial condition
(6.49)
y(0)  y 0 ,
where y  y (t ) is the state function, f  f (t ) is a known smooth function. We do know the pair
of numbers v  (a, y 0 ), besides a is positive. For all value v the problem (6.48), (6.49) has a
unique solution y  y[ v ]. We have also the equality
y (t )  z (t ), t0  t  T ,
where z  z (t ) is a given function. Determine the following inverse problem.
(6.50)
Problem 6.4. Find the function v such that the respective state function y[ v ] satisfies the
equality (6.50).
This problem can be transformed to the optimization control problem. Determine the
functional
T
I (v)    y(t )  z (t ) dt.
2
t0
Problem 6.4'. Minimize the functional I.
Find the derivative of the functional. Determine the value
T
I (u   h)    y[u   h]  z  dt ,
2
t0
where u  (a, y ), h  (b, c), and  is a number. So we can find the difference
0
T
T
I (u   h)  I (u )  2  y[u ]  z  y[u   h]  y[u ] dt    y[u   h]  y[u ] dt ,
2
t0
(6.51)
t0
that is an analogue of the equalities (6.4) and (6.38).
Consider this system for the functions v  u   h and v  u . We have
    a   b  y[u   h]  ay[u]  0,
 (0)   c,
where   y[u   h]  y[u ]. The equality (6.52) can be transformed to
   a   by[u ]   b .
Multiply this equality by an arbitrary function  and integrate by t. We obtain
T
T
(6.53)
T
    a   dt   b y[u ] dt   b   dt.
0
(6.52)
0
(6.54)
0
After the integration by parts we get
T
T
T
0
0
   dt   0    dt   (T ) (T )   c (0)    dt
T
0
because of the equality (6.53). So we obtain
T
T
T
 (T ) (T )       a   dt   z (0)   b  y[u ] dt   b   dt.
0
0
(6.55)
0
Chose the function  equal to the solution p of the adjoint system
 p(t )  ap(t )  g (t ), 0  t  T
p (T )  0,
where
2  y[u ]  z  , if t  t0 ,
g (t )  
0,
if t  t0 .

It is equal to the problem (6.43), (6.44). Then the formula (6.55) can be transformed to the
equality
T
T
T
t0
0
0
2  y[u ]  z   dt   cp(0)   b  y[u ] pdt   b   pdt.
Put it to the equality (6.51). We get
T
T
T
I (u   h)  I (u )   cp(0)   b  y[u ] pdt   b   pdt    2dt.
0
This is the analogue of the formula (6.45).
Return to the equality (6.54). Chose    here. We get
0
t0
(6.56)
T
T
T
    a  dt   b  y[u ] dt   b   dt.
2
2
0
0
0
It can be transformed to
 (T )
2
T
T
T
0
0
 a   dt   c   b  y[u ] dt   b   2 dt.
2
2 2
0
Then we obtain the inequality
T
2
  dt 
0
 2c 2
b T

y[u ] dt ,
u   b u   b 0
that is the analogue of the formula (6.46). Therefore we can obtain again the conditions
T
T
1 2
lim   dt  0, lim   pdt  0.
 0

 0
0
0
Then we have
lim
 0
I (u   h)  I (u )

T
 cp(0)  b  y[u ] pdt.
0
This is the scalar product  h, I (u)  of the second order Euclid space between the vector
h  (b, z ) and the derivative I (u ) . So this derivative is second order vector with components
T
p (0) and   y[u ] pdt.
0
Determine now the gradient method. Let the parameters ak , yk0 are known. The state
function yk is determine from Cauchy problem
yk  ak yk  f ,
yk (0)  yk0 .
Then we find the adjoint state from the system
 pk (t )  ak pk (t )  gk (t ), 0  t  T ,
pk (T )  0,
where
2  yk  z  , if t  t0 ,
g k (t )  
0,
if t  t0 .

The next iteration of the unknown parameters is determined by the equalities
T
ak 1  ak   k  yk pk dt ,
0
y
0
k 1
 y   k pk (0).
0
k
6.5. Differential equation with pointwise measuring
We consider the system be described by the equation
y  ay  v
(6.57)
with initial condition
(6.58)
y(0)  y 0 ,
0
where y  y (t ) is the state function, a  0 and y are known parameters, and v  v(t ) is the
unknown function. For all smooth enough function v the problem (6.57), (6.58) has a unique
solution y  y[ v ]. We have also the equalities
(6.59)
y (ti )  zi , i  1,..., m,
where t1 ,..., tm are times of the experiments from the interval (0,T), and z1 ,..., zm are a given
values (results of measuring). Determine the following inverse problem.
Problem 6.5. Find the function v such that the respective state function y[ v ] satisfies the
equality (6.59).
This problem can be transformed to the optimization control problem. Determine the
functional
m
I (v)    y (ti )  zi  .
2
i 1
Problem 6.5'. Minimize the functional I.
Find the derivative of the functional. Determine the value
m
I (u   h)    y[u   h](ti )  zi  .
2
i 1
where u and h are functions,  is a number. So we can find the difference
m
m
i 1
i 1
I (u   h)  I (u )  2  y[u ](ti )  zi  (ti )    (ti )  ,
where   y[u   h]  y[u ] .
Determine -function by the equality
2
T
 F (t ) (t   )dt  F ( ).
0
Then we transform the previous equality to the formula
T m
T m
0 i 1
0 i 1
I (u   h)  I (u )  2    y[u ](t )  z (t ) (ti ) (t  ti )dt     (t )   (t  ti )dt ,
2
where z is an arbitrary function such that z (ti )  zi , i  1,..., m.
This  difference satisfies the problem
   a   h,  (0)  0.
This is the system (6.5), (6.6). So we determine the equality (6.7)
T
T
0
0
    a   dt    h dt.
It can be transform to the equality (6.8)
T
T
0
0
     a   dt  (T ) (T )    h dt .
Determine the adjoint system
m
 p(t )  ap(t )    y[u ](t )  z (t )  (t  ti ),
i 1
p (T )  0.
Put   p at the previous equality. We get
T
T m
I (u   h)  I (u )    hpdt     (t )   (t  ti )dt.
2
0 i 1
0
We can prove the equality
lim
 0
1
T m
  (t )

0 i 1
2
 (t  ti )dt  0.
(6.60)
After devising the equality (6.60) by  and passing to the limit we have
T
I (u   h)  I (u )
lim
  phdt.

 0
0
So we find
I (u )  p.
Then we can determine the gradient method. Let the function vk is known. Then we find the
function yk from Cauchy problem
yk  ayk  vk ,
yk (0)  y 0 .
Determine the adjoint state from the system
m
 pk (t )  apk (t )    yk (t )  z (t ) (t  ti ),
i 1
pk (T )  0.
The next iteration of the control can be determined by the formula
vk 1  vk  k pk , k  0,1,... .
6.6. Nonlinear differential equation with unknown absolute term
We consider the system be described by the equation
y  f ( y , v )
(6.61)
with initial condition
(6.62)
y(0)  y 0 ,
0
where y  y (t ) is the state function, f is known function of two arguments, y are known
parameters, and v  v(t ) is the unknown function. Let this problem has a unique solution
y  y[ v ]. We have also the equality
(6.63)
y (t )  z (t ), t0  t  T ,
where z  z (t ) is a given function (result of measuring). Determine the following inverse
problem.
Problem 6.6. Find the function v such that the respective state function y[ v ] satisfies the
equality (6.63).
This problem can be transformed to the optimization control problem. Determine the
functional
T
I (v)    y(t )  z (t ) dt.
2
t0
Problem 6.6'. Minimize the functional I.
Find its derivative. Determine the value
T
I (u   h)    y[u   h]  z  dt ,
2
t0
where u and h are functions,  is a number. So we can find the difference
T
T
t0
t0
I (u   h)  I (u )  2  y[u]  z   dt    2dt ,
where   y[u   h]  y[u ].
Consider the system with respect to the function 
(6.64)
   f  y[u h],u  h   f  y[u ],u  ,
 (0)  0.
(6.65)
(6.66)
Let the function f be smooth enough. Than we can transform the term
f  y[u  h],u  h   f  y[u ] ,u  h   f  y[u ],u  h   f y  y[u ],u  h    o( ) 
 f  y[u ],u  h   f y  y[u ],u    o( )   f y  y[u ],u  h   f y  y[u ],u  ,
where fy is the partial derivative of the function f with respect to the first argument, o( ) is the
high term with respect to . Put it to the equality (6.65); we get
    f  y[u ],u  h   f  y[u ],u   f y  y[u ],u    o( )   f y  y[u ],u  h   f y  y[u ],u .
After multiplication by an arbitrary function  and integration by t we obtain
T
T
    f  y[u ],u    dt    f  y[u ],u  h   f  y[u ],u  dt 
y
0
0
T

 f
y
 y[u ],u  h   f y  y[u ],u   o( )  dt.
0
After the integration by parts with using the initial condition (6.66) we get
T
T
     f y  y[u ],u   dt   (T ) (T )    f  y[u ],u  h   f  y[u ],u   dt 
0
0
T

 f
y
 y[u ],u  h   f y  y[u ],u   o( )  dt.
0
Chose the function  equal to the solution p of the system
p  f y  y[u ],u  p   g (t ), 0  t  T ,
p (T )  0,
where
2  y[u ]  z  , if t  t0 ,
g (t )  
0,
if t  t0 .

Therefore the previous formula can be transformed to the equality
T
T
t0
0
2   y[u ]  z   dt    f  y[u ],u  h   f  y[u ],u  pdt 
T

 f
y
 y[u ],u  h   f y  y[u ],u   o( ) pdt.
0
Put it to the equality (6.4). We get
T
I (u   h)  I (u )    f  y[u ],u  h   f  y[u ],u  pdt 
0
T



T
 f y  y[u],u  h   f y  y[u ],u   o( ) pdt    2dt.
0
t0
It is possible to obtain the inequality
T
  dt  c
2
2
,
0
where the positive constant c does not depend from . Then we devise the previous equality by 
and pass to the limit as   0. We get
lim
 0
I (u   h)  I (u )

T
  I (u ), h    f v  y[u ],u  phdt ,
0
where fv is the partial derivative of the function f with respect to the second argument. Hence we
find the derivative of the functional I at the point u
I (u)  f v  y[u],u  p.
We have the following iterative method. Let k-iteration value vk of the control is known.
Then we can find the appropriate value of the state function yk from Cauchy problem
yk  f ( yk , vk ),
yk (0)  y 0 .
Next step is the determination of the adjoint state pk from the system
pk  f y  yk ,uk  pk   g k (t ),
pk (T )  0,
where
2  yk  zk  , if t  t0 ,
g k (t )  
0,
if t  t0 .

Then we find the new iteration of the control
vk 1  vk  k fv  yk ,uk  pk .
Our next step is the analysis of inverse problems for mathematical physics problems.
Task
Find the derivatives of the following functionals:
1
1) I (v)   x 2 dt , where x   2v, x(0)  0.
0
1
2) I (v)  2  x 2 dt , where x   v, x(0)  1.
0
1
1 2
x dt , where x   v, x(0)  1.
2 0
Determine the gradient method for these problems.
3) I (v) 