ON THE SOLUTION OF SOME HIGHER-ORDER INTEGRO-DIFFERENTIAL EQUATIONS OF SPECIAL FORM

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Introduction

Integro-differential equations are used to model physical phenomena and processes in engineering, physics, biology and economics [1–4]. Higher-order Integro-differential equations are also play an important role in mathematical modeling [5–7]. In most cases, they cannot be solved analytically and therefore approximate solutions are persuaded [8; 9]. However, exact solutions are always attractive and in some instances essential [10–12]. Based on the theory of extensions of operators [13–15], the authors have obtained closed form solutions of several types of boundary value problems for integro-differential equations with classical and nonlocal boundary conditions [16–21]. The present paper is a sequel to work of [19] and deals with the exact solution of one more class of boundary value problems of special form.

Let X be a Banach space of complex valued functions of x defined on Ω,

A� : X → X a bijective

linear differential operator incorporating some initial or boundary conditions. First, we consider the integro-

differential equation

 

Bu(x) =

m ∫

A�u(x) − ∑

j=1 Ω

Kj (x, t)A�u(t)dt = f (x), x ∈ Ω,

D(B) = D(A�) ⊂ X, (1)

where B : X → X is a linear operator, Kj (x, t) ∈ X(Ω × Ω) are known kernel functions, f (x) ∈ X is an input function and u(x) ∈ D(B) is the sought function describing the response of the system modeled by (1).

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Next, we contemplate the more involved integro-differential equation

4  m 

4 ∑ ∑ ∫ i

B4u(x) =

A� u(x) −

 

4

 

i=1



j=1

Kij (x, t)A� u(x)dt = f (x), x ∈ Ω,

Ω

D(B4) = D(A� ), (2)

i

where B4 : X → X is a linear operator,

A� , i = 1, 2, 3, 4, are powers, self-compositions, of the operator A�

and the kernels Kij (x, t) ∈ X(Ω × Ω) are known functions. We assume that the inverse operator I = A�−1

is known explicitly, and that the kernels Kj (x, t), Kij (x, t) are separable functions. We establish existence and uniqueness criteria and provide the solutions of the problems (1) and (2) in closed form. Lastly, we investigate under which conditions B4 = B4 and develop a decomposition technique for obtaining the exact solution to the problem B4u(x) = f (x).

The outline of the paper is as follows. In Section 1., the two problems are put in a convenient matrix form and exact solution formulae are obtained by a direct method. In Section 2., the conditions for factorizing problem (2) and a decomposition solution method are presented. Finally, two examples are solved in Section 3.

 

1. The Direct Method

   First, we deal with problem (1). We assume that the kernels Kj (x, t) are separable functions, i.e. Kj (x, t) =

   = gj (x)hj (t), j = 1, . . . , m, where gj (x), hj (x) ∈ X, and introduce the vector of functions

   g = ( g1 . . . gm ) ∈ Xm, gj = gj (x) ∈ X, j = 1, . . . , m, (3) and the vector of linear and bounded functionals

    Φ1

   Φ =  ..

   

    ∈ X∗ , Φ

   ∫

   = Φ (·) =

    

   h (t) · dt ∈ X∗, j = 1, . . . , m. (4)

   Ω

    

    .  m j j j

   Φm

   Also, by Φ(f ) and Φ(g) we denote the vector and the m × m matrix

    Φ1(f ) 

    Φ1(g1) · · · Φ1(gm) 

   Φ(f ) =  ..

    , Φ(g) =  ..

   . . . ..

    , (5)

    .   . . 

    

   respectively, and note that

   Φm(f )

   Φm(g1) · · · Φm(gm)

    

   Φ(gN) = Φ(g)N, (6)

   where N is an m × k, k ∈ N, constant matrix. Finally, 1m symbolizes the m × m identity matrix and 0 the zero column vector.

   Thus, the problem (1) can be put in the convenient form

   Bu = A�u − gΦ(A�u) = f, D(B) = D(A�). (7)

   The conditions for existence and uniqueness, the solution to problem and the correctness of the operator B

   are provided by the next theorem.

   It is recalled here that an operator P : X → X is said to be correct if P is bijective and its inverse P−1

   is bounded on X.

   Theorem 1. Let X be a complex Banach space,

   A� : X → X a bijective linear operator and I = A�−1 its

   m

    

   inverse, g ∈ Xm and Φ ∈ X∗

   as defined in (3) and (4), respectively, and B : X → X the linear operator

   Bu = A�u − gΦ(A�u), D(B) = D(A�). (8)

   Then the following statements are true:

   1. The operator B is bijective on X if and only if

      det W = det[1m − Φ(g)] ̸= 0, (9)

      and the unique solution to boundary value problem (7), for any f ∈ X, is given by the formula

      u = B−1f = If + IgW−1Φ(f ). (10)

   2. If in addition the inverse operator A�−1 is bounded on X, then the operator B is correct.

      Providas E., Parasidis I.N. On the solution of some higher-order integro-differential equations of special form

      16 Провидас E., Парасидис И.Н. О решении некоторых интегро-дифференциальных уравнений высшего порядка...

       

      Proof. (i) Let det W ̸= 0 and u ∈ ker B. Then,

      Bu = A�u − gΦ(A�u) = 0, (11)

      and after applying the vector Φ on both sides of (11) and utilizing (6),

      [1m − Φ(g)] Φ(A�u) = WΦ(A�u) = 0, (12)

      which implies that Φ(A�u) = 0. Substitution into (11) yields Bu = A�u = 0 and hence u = 0. This means that ker B = {0} and therefore the operator B is injective. Conversely, we prove that if B is an injective operator then det W ̸= 0, or equivalently, if det W = 0, then B is not injective. Let det W = 0. Then there exists a nonzero vector c = col(c1, . . . , cm) such that Wc = 0. Consider the element u0 = A�−1gc and note that u0 ̸= 0; otherwise u0 = 0 implies gc = 0 and then Wc = [1m − Φ(g)]c = c − Φ(gc) = c = 0 which

      contradicts the hypothesis that c is a nonzero vector. From equation (8), we get

      Bu0 = gc − gΦ(g)c = g[1m − Φ(g)]c = gWc = g0 = 0, (13) which means that ker B ̸= 0 and so B is not injective.

      Further, by multiplying from the left both sides of (7) by A�−1, we get

      u − A�−1gΦ(A�u) = A�−1f, (14)

      while acting by the vector Φ on (7), we have

      [1m − Φ(g)] Φ(A�u) = WΦ(A�u) = Φ(f ). (15)

      By hypothesis det W ̸= 0 and therefore equation (15) can be solved with respect to Φ(A�u). Substituting

      Φ(A�u) = W−1Φ(f ) into (14), we obtain the solution formula (10).

      Finally, since the input function f in (7) and (10) is any arbitrary f ∈ X, we have R(B) = X which

      means that B is bijective.

      1. Suppose that (9) is true and that the operator

         A�−1 is bounded. Then by (i) the operator B is

         bijective and the unique solution to problem (7) is given by (10). In (10) the operator A�−1 and the functionals

         Φ1, . . . , Φm are bounded. This means that the operator B−1 is bounded and hence the operator B is correct. D

         To find the solution of the boundary value problem (2), we now introduce the vectors of functions

         q = ( q1 . . . qm ) , r = ( r1 . . . rm ) , s = ( s1 . . . sm ) ,

         z = ( z1 . . . zm ) , qj , rj , sj , zj ∈ X, j = 1, . . . , m, (16)

         and write the problem (2) in the form

         4 2 3 4

         B4u =

         A� u − qΦ(A�u) − rΦ(A� u) − sΦ(A� u) − zΦ(A� u) = f,

         4

         D(B4) = D(A� ), (17)

         and prove the following theorem.

         Theorem 2. Let X be a complex Banach space,

         A� : X → X a bijective operator and I = A�−1 its inverse,

         m

          

         q, r, s, z ∈ Xm and Φ ∈ X∗

         as in (16) and (4), respectively, and the operator B4 : X → X defined by

         4 2 3 4

         B4u =

         A� u − qΦ(A�u) − rΦ(A� u) − sΦ(A� u) − zΦ(A� u),

         4

         D(B4) = D(A� ). (18)

         Then the following statements are true:

         1. If

             

            where

            det V ̸= 0, (19)

             Φ(I3q) − 1m Φ(I3r) Φ(I3s) Φ(I3z) 

            V = 

            Φ(I

            2q) Φ(I

            2r) − 1m Φ(I

            2s) Φ(I

            2z)

             , (20)

            

             

            

             

             Φ(Iq) Φ(Ir) Φ(Is) − 1m Φ(Iz) 

            Φ(q) Φ(r) Φ(s) Φ(z) − 1m

            then the operator B4 is bijective on X and the unique solution to problem (17) is given by

            4

             

            u = B−1f

             Φ(I3f ) 

            2

            = I4f − ( I4q I4r I4s I4z ) V−1  Φ(I

            f )  . (21)

            

             

            

             

             Φ(If ) 

            Φ(f )

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         2. Conversely, if the operator B4 is injective and also the vectors q, r, s, z are linearly independent, then

            det V ̸= 0.

         3. If in addition the inverse I = A�−1 is bounded on X, then the operator B4 is correct. Proof. (i) Let det V ̸= 0 and u ∈ ker B4, i.e.

            4 2 3 4

            A� u − qΦ(A�u) − rΦ(A� u) − sΦ(A� u) − zΦ(A� u) = 0, (22)

            4

            where u ∈ D(A� ). We introduce the vectors

             Φ(A�u) 

            2

             Φ(A� u) 

            q¯ = ( q r s z ) ,

            Φ¯ (u) = 

            

             , (23)

            3 

             

            and write (22) conveniently in the matrix form

             Φ(A� u) 

            �

             

            Φ(A4u)

            4

             

            A� u − q¯Φ¯ (u) = 0. (24)

            By applying the inverse operator I = A�−1 four times on both sides of (24), we get

            A

             

            4−n

            �

            u − I

            nq¯Φ¯ (u) = 0, n = 1, 2, 3, (25)

            u − I

            4q¯Φ¯ (u) = 0. (26)

            Acting by the vector Φ on both sides of (24) and the three equations (25), we obtain

            4−n

             

            Φ(A�

            u) − Φ(I

            nq¯)Φ¯ (u) = 0, n = 0, 1, 2, 3, (27)

            where I0 = 1. Equation (27) can be rearranged in the following configuration

             Φ(I3q) − 1m Φ(I3r) Φ(I3s) Φ(I3z) 

             Φ(I2q) Φ(I2r) − 1m Φ(I2s) Φ(I2z)

            

             

             Φ(Iq) Φ(Ir) Φ(Is) − 1m Φ(Iz)

            Φ(q) Φ(r) Φ(s) Φ(z) − 1m

            or

            

             

             Φ¯ (u) = 0, (28)

            

            VΦ¯ (u) = 0, (29)

            where V is the 4m × 4m matrix in (20). Since det V ̸= 0, it is implied from (29) that

            Φ¯ (u) = 0 and then

            from (26) that u = 0. This means that ker B = {0} and therefore the operator B is injective.

            To obtain the solution of (17), we write it in the form

            4

             

            A� u − q¯Φ¯ (u) = f, f ∈ X, (30)

            By applying the inverse operator I = A�−1 four times on both sides of (30), we get

            A

             

            4−n

            �

            u − I

            nq¯Φ¯ (u) = I

            nf, n = 1, 2, 3, (31)

            u − I4q¯Φ¯ (u) = I4f, (32)

            and then acting by the vector Φ on both sides of (30) and (31), we have

            4−n

             

            Φ(A�

            u) − Φ(I

            nq¯)Φ¯ (u) = Φ(I

            nf ), n = 0, 1, 2, 3, (33)

            or in matrix form

             Φ(I3f ) 

            2

            VΦ¯ (u) = −  Φ(I

            f )  . (34)

            

             

            

             

             Φ(If ) 

            Φ(f )

            By inverting (34) and substituting into (32), we obtain (21). In (17) and (30), f is an arbitrary element of

            X and therefore R(B4) = X and so the operator B4 is bijective and the problem (17) everywhere solvable.

            1. Let the vectors q, r, s, z be linearly independent. We will prove that if B4 is an injective operator then det V ̸= 0, or equivalently, if det V = 0 then B4 is not injective. Let det V = 0. Then there exists a

               4

                

               nonzero vector c = col(c1, c2, c3, c4) such that Vc = 0 where ci = col (ci1, ..., cim), i = 1, 2, 3, 4. Consider the element u0 = I4(qc1 + rc2 + sc3 + zc4) ∈ D(A� ).

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               18 Провидас E., Парасидис И.Н. О решении некоторых интегро-дифференциальных уравнений высшего порядка...

                

               Substituting this element into (18), we obain

               Bu0 = qc1 + rc2 + sc3 + zc4

               −q [Φ(I3q)c1 + Φ(I3r)c2 + Φ(I3s)c3 + Φ(I3z)c4]

               −r [Φ(I2q)c1 + Φ(I2r)c2 + Φ(I2s)c3 + Φ(I2z)c4]

               −s [Φ(Iq)c1 + Φ(Ir)c2 + Φ(Is)c3 + Φ(Iz)c4]

               −z [Φ(q)c1 + Φ(r)c2 + Φ(s)c3 + Φ(z)c4]

               = − ( q r s z ) Vc = 0. (35)

               This means that u0 ∈ ker B4. Note that u0 ̸= 0, because by hypothesis q, r, s, z are linearly independent and

               c ̸= 0. Hence, B4 is not injective.

            2. Let additionally I = A�−1 is bounded, i.e. A� is correct. Then the operators Ii, i = 2, 3, 4, are bounded

         4

          

         and since the functionals Φj , j = 1, . . . , m, are bounded on X, it is concluded that the operator B−1 is

         bounded too by means of (21). Hence, if the operator A� is correct and the condition (19) is satisfied then the operator B4 is correct. D

          

2. Decomposition Method

   Let the operator B : X → X be defined by

   Bu = A�u − zΦ(A�u), D(B) = D(A�), (36)

   �

    

   where z ∈ D(A3)m, and let the vectors

   q = A�r − zΦ(A�r), r = A�s − zΦ(A�s), s = A�z − zΦ(A�z). (37) In this case, problem (2) can be formulated as

   �

    

   B4u = B4u = f, D(B4) = D(A4). (38)

   It is recognized that conditions (37) are seldom met in real life problems, but when it happens the solution process is simplified further. Moreover, the results derived here are of theoretical importance and valuable for extra developments.

   Theorem 3. Let X be a complex Banach space,

   A� : X → X a bijective operator and I = A�−1 its inverse

   m

    

   and Φ ∈ X∗

   as in (4). Let q, r, s, z ∈ Xm satisfy (37) and the operator B : X → X be defined by (36).

   Then the following statements are true:

   1. The operator

      4

       

      2 3 4

      B4u =

      A� u − qΦ(A�u) − rΦ(A� u) − sΦ(A� u) − zΦ(A� u),

      4

      D(B4) = D(A� ), (39)

      is biquadratic, meaning B4u = B4u.

   2. If

      det W = det[1m − Φ(z)] ̸= 0, (40)

      then the operator B4 is bijective and the unique solution to boundary value problem (38) is obtained by

      4

       

      u = B−1f = u4,

      uk+1 = Iuk + IzW−1Φ(uk ), k = 0, 1, 2, 3, u0 = f. (41)

       

   3. Conversely, if the operator B4 is injective and the components of the vector z are linearly independent, then det W ̸= 0.

   4. Finally, if (40) holds true and in addition the inverse operator I = A�−1 is bounded on X, then B4 is correct.

      Proof. (i) From (37) and (36), we have

      s = Bz,

      r = Bs = B2z,

      q = Br = B3z, (42)

      2 3 3

      where the operators B2, B3 : X → X with D(B2) = D(A� ) and D(B ) = D(A� ). Substituting (42) into (39),

      we get

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      4 3 2 2 3 4

      B4u =

      A� u − B zΦ(A�u) − B zΦ(A� u) − BzΦ(A� u) − zΦ(A� u)

      3 [ 2 3

      = BA� u − B

      B2zΦ(A�u) + BzΦ(A� u) + zΦ(A� u)]

      = B [ � u − B zΦ(A�u) − BzΦ(A� u) − zΦ(A� u)]

      A3 2 2 3

      = B [

      A2u − B2zΦ(Au) − BzΦ(A2u)]

      B � � �

      = B2 [ � u − BzΦ(A�u) − zΦ(A� u)]

      A2 2

      = B2 [

      ]

      Au − BzΦ(Au)

      B � �

      Au

       

      = B3 [ �

      ]

      — zΦ(A�u)

      = B4u. (43)

      B4u	=	u0, u0 = f,
      B3u	=	B−1u0 = u1,	u1 = Iu0 + IzW−1Φ(u0),
      B2u	=	B−1u1 = u2,	u2 = Iu1 + IzW−1Φ(u1),
      Bu	=	B−1u2 = u3,	u3 = Iu2 + IzW−1Φ(u2),
      u	=	B−1u3 = u4,	u4 = Iu3 + IzW−1Φ(u3),

       

      1. Let det W ̸= 0. Then by Theorem 1 the operator B in (36) is bijective and consequently the operator B4 = B4 is again bijective as composition of bijective linear operators. Thus, the boundary value problem (38) may be solved as follows

          

         which yields (41).

          

         (44)

      2. Suppose that the components of the vector z are linearly independent and we will prove that if B4 is an injective operator then det W ̸= 0, or equivalently, if det W = 0 then B4 is not injective. Let det W = 0.

         Then there exists a nonzero vector c = col(c1, . . . , cm) such that Wc = 0. Consider the element u0 = Izc

         and notice that u0 ̸= 0 because the components of the vector z are linearly independent and c is nonzero.

         From (39) and (42), we obtain

          

         4 3 2 2 3 4

         B4u0 =

         A� u0 − B zΦ(A�u0) − B zΦ(A� u0) − BzΦ(A� u0) − zΦ(A� u0)

         3 3 2 2 3

         = A� zc − B zΦ(z)c − B zΦ(A�z)c − BzΦ(A� z)c − zΦ(A� z)c

         2 3 2 2

          

         = B [A� z − B zΦ(z) − BzΦ(A�z) − zΦ(A� z)] c

         2 2 2

          

         = B [BA�z − B2zΦ(z) − BzΦ(A�z)] c

         = B2 [A�z − BzΦ(z) − zΦ(A�z)] c

         = B2 [Bz − BzΦ(z)] c

         = B3z [1m − Φ(z)] c

         = B3zWc = 0. (45)

         From (45) it is implied that u0 ∈ ker B4 and hence B4 is not injective.

      3. Assume (40) holds true. Then the operator B4 is bijective and its inverse is given by (41). If in

         addition

         A�−1 is bounded and since the functionals Φ1, . . . , Φm in (41) are bounded, it is implied that the

         4

          

         operator B−1 is bounded and hence B4 is correct. D

          

3. Examples

In this section, we select and solve two representative example problems to explain the implementation as well as to show the efficiency of the techniques presented in the previous sections.

Example 1. Let Ω = {x ∈ R3 : |x| < 1} and ∂Ω = {x ∈ R3 : |x| = 1}. Consider the boundary value problem

j=1

 

△u(x) − ∑2

gj (x) ∫Ω

vj (y)△u(y)dy = f (x), x ∈ Ω,

u|∂Ω = 0, (46)

Providas E., Parasidis I.N. On the solution of some higher-order integro-differential equations of special form

20 Провидас E., Парасидис И.Н. О решении некоторых интегро-дифференциальных уравнений высшего порядка...

 

where △ is the Laplace operator in R3. We take X = L2(Ω) and

2

 

A�u(x) = △u(x), D(A�) = {u(x) ∈ W2(Ω) : u

 

|∂Ω

 

= 0},

Φi(·) =

∫

vi(x) · dx, i = 1, 2,

Ω

g = ( g1(x) g2(x) ) , (47)

and thus the problem (46) is put in the form

Bu(x) = A�u(x) − gΦ(A�u(x)) = f (x), D(B) = D(A�). (48)

It is known that the problem

|∂Ω 2

 

A�u(x) = △u(x) = f (x), u(x) = 0, u ∈ W2(Ω), f (x) ∈ L2(Ω). (49) is uniquely and everywhere solvable on L2(Ω) and

∫

u(x) = A�−1f (x) =

G(x, y)f (y)dy, ∀f ∈ L2(Ω), (50)

Ω

G G1 G1

 

where (x, y) = (x, y)+g(x, y) is the Green’s function, (x, y) = 1

4π|x−y|

and g(x, y) is a harmonic function

in Ω and belongs to C∞(Ω¯ ). Also, the operator A�−1 is bounded and hence

A� is correct. We compute,

det W = det [12 − Φ(g)] = det

[ 1 − Φ1(g1(x)) −Φ1(g2(x))

−Φ2(g1(x)) 1 − Φ2(g2(x))

]

. (51)

If det W ̸= 0, then the problem (46) admits a unique solution which is obtained by substituting into (10).

Finally, the operator B by Theorem 1 is correct.

Example 2. Consider the fourth order integro-differential equation

− −

 

1

 

u(4)(t) + 3 (5101t3 32978t2 + 102042t 148458) ∫ 64 0

1

xu′(x)dx

1 3

− 16 (593t

1

− 3834t2

∫

+ 11874t − 17266)

0

∫ 1

xu′′

(x)dx

4

 

+ (23t3 − 150t2 + 462t − 670)

∫ 1

xu′′′(x)dx

0

−(t3 − 6t2 + 18t − 26)

xu(4)(x)dx

0

279 3 2

= − 640 (5101t

subject to boundary conditions

− 32978t

+ 102042t − 148458), (52)

We take X = C[0, 1], consequently,

u(i)(0) = 2u(i)(1), i = 0, 1, 2, 3. (53)

A�u = u′(t), D(A�) = {u(t) ∈ C1[0, 1] : u(0) = 2u(1)}, (54)

2

 

A� u = u′′

2

 

(t), D(A� ) = {u ∈ C

2[0, 1] : u

(i)

(0) = 2u

(i)

(1), i = 0, 1} ,

3

 

A� u = u

′′′

3

 

(t), D(A� ) = {u ∈ C

3[0, 1] : u

(i)

(0) = 2u

(i)

}

(1), i = 0, 2 ,

 

and

�

 

A4u = u

(4)

�

 

(t), D(A4) = {u ∈ C

4[0, 1] : u

(i)

(0) = 2u

(i)

3

 

(1), i = 0, }

, (55)

�

 

Φ(Aiu) =

(∫ 1

 

0

xu(i)

)

(x)dx

 

, i = 1, 2, 3, 4,

−

 

q = ( 3 (5101t3

64

− 32978t2

)

+ 102042t − 148458) ,

( 1 )

− −

 

r = (593t3 3834t2 + 11874t 17266) ,

16

1

 

( )

(

 

3 2

s = − 4 23t

− 150t

+ 462t − 670) ,

z = (t3 − 6t2 + 18t − 26) ,

279 3 2

f = − 640 (5101t

− 32978

+ 102042t − 148458). (56)

Вестник Самарского университета. Естественнонаучная серия. 2020. Том 26, № 1. С. 14–22

Vestnik of Samara University. Natural Science Series. 2020, vol. 26, no. 1, pp. 14–22 21

 

It is easy to verify that the operator A� is correct and its inverse is given by

A�−1f (t) = ∫ t f (x)dx − 2 ∫ 1 f (x)dx, f ∈ C[0, 1]. (57)

0 0

and that the functional Φ is bounded, i.e. Φ ∈ C[0, 1]∗ = X∗. Thus, the boundary value problem (52), (53)

can be expressed now in the operator form (17).

To find its solution we can apply Theorem 2. However, by inspecting the operator

A�, the vector of

functionals Φ and the vectors q, r, s, z, we can verify that the conditions (37) are satisfied, and therefore the decomposition Theorem 3 is more appropriate, which is much easier to implement. It is straightforward to show that

det W = det [1 − Φ(z)]

[

= 1 −

∫ 1 ]

x(x3 − 6x2 + 18x − 26)dx

0

93

= 10 ̸= 0. (58)

4

 

Thus, the problem (52), (53) is correct and it can be formulated as B4u = f , where B is the first order operator in (36). By substituting into (41), we obtain the unique solution of the problem (52), (53), which is

u(t) =

2

 

4

 

t − 2t3 + 9t2 − 26t + 75 . (59)

About the authors

E. Providas

University of Thessaly

Author for correspondence.
Email: providas@uth.gr
ORCID iD: 0000-0002-0675-4351

Candidate of Technical Sciences, associate
professor

Greece

I. N. Parasidis

University of Thessaly

Email: paras@teilar.gr
ORCID iD: 0000-0002-7900-9256

Сandidate of Technical Sciences, associate professor

Greece

References

Supplementary files