Page 101 - IJOCTA-15-1
P. 101

On a robust stability criterion in the Cattaneo–Hristov diffusion equation

            This definition of robust stability constitutes a  must satisfy:
            generalization of the concept of stability under                   α
                                                                κ 2 (1 − α)β  2 cf  D y j + ˙y j
            constant-action perturbations used for systems                j  0  t
                                                                                  2
            of ordinary differential equations, and which was                + κ 1 β y j = v j (t), y j (0) = 0. (8)
                                                                                  j
            first proposed by Duboshin and Malkin; see e.g. 26
                                                              Here, v j ∈ V δ j  is an external perturbation.
            This concept of robust stability has been es-
                                                              The set of solutions y j obtained by choosing an
            tablished in another partial differential equation
            model; see. 34                                    external perturbation v j ∈ V δ j  on the right-hand
                                                                                           . This set is non-
                                                              side of (8) is denoted by Y δ j
            4. Application of the Fourier method              empty, since if we choose the null external pertur-
                                                              bation ¯v j ≡ 0, then we obtain the trivial solution
            If we use the method of separation of variables   ¯ y j ≡ 0. The solution ¯y j is called unperturbed. It
            to find the solution of (1)–(3), then we can de-  follows that Y δ j  ⊂ Y.
            termine a sequence of eigenvalues {λ j } j∈N such
                                                              We now determine other elements of the set Y δ j
                           2
                       π
            that λ j = ( j) and a sequence of eigenfunctions                               31
                       R                                      using the method described in.
            {X j } j∈N such that X j (x) = sin(β j x), x ∈ [0, R],
                          1/2                                 It is well known that, according to the concept
            where β j = λ j  . Since {X j } j∈N is a basis of  of the Caputo–Fabrizio fractional derivative, af-
              2
            L (0, R), it follows that the solutions of (1)–(3)  ter integrating by parts we obtain the following
            admit the representation given in (5).            equivalent representation
            It is well known that the method to justify the ex-                      Z  t  µ 2
                                                               cf  α        µ α           α −µ α(t−η)
            istence of solutions for the initial-boundary value  0 D y j (t) =  α  y j (t) −  α  e  y j (η)dη.
                                                                  t
            problem (1)–(3) that depends on a heat source                              0
                                                                                  R  t µ αη
                  δ
            v ∈ V , consists in verifying the uniform conver-  If we choose z j (t) =  0  e  y j (η)dη, then we ob-
                  Ω
            gence on Ω of the series on the right-hand side   tain ˙z j (t) = e µ αt y j (t), since y j is continuous be-
                                                                          1
            of (5), and the uniform convergence on Ω of four  cause y j ∈ H (0, T). This allows us to obtain the
            other series defining the functions               following expression
                                              2
                                  2
                       ∂y   ∂y   ∂ y       α  ∂ y                cf  α          1   −µ αt
                          ,    ,     ,  cf 0 D t  .     (7)       0 D y j (t) =    e     ( ˙z j (t) − µ α z j (t)) .
                                                                     t
                       ∂t   ∂x   ∂x 2       ∂x 2                              1 − α
            In the following result we justify the application  We observe that ˙y j (t) = e −µ αt  (¨z j (t) − µ α ˙z j (t)).
            of the method of separation of variables.         By substituting in equation (8) we get
                                         P  ∞
            Theorem 1. Let v(x, t) =        j=1  sin(β j x)v j (t)  ¨ z j (t) + 2a j ˙z j (t) − b j z j (t)
                                                        for                      µ αt
            a heat source for (1)–(3), where v j ∈ V δ j                     = e   v j (t), z j (0) = ˙z j (0) = 0,
            all j ∈ N.    Then the function y expressed in
                                                                                                           2
                                                                                     2
                                                                         1
                                                              where a j = ((κ 1 +κ 2 )β −µ α ) and b j = κ 2 µ α β .
            the form (5) is a unique solution to the initial-            2           j                     j
                                                              Since β j → ∞ as j → ∞, then we can assume
            boundary value problem (1)–(3) provided that the
                  P  ∞                                        that a j > 0 for all j ∈ N. Therefore, if we denote
            series   j=1 j is convergent, that is, y ∈ Y Ω .
                        δ
                                                                        2
                                                              by c j = (a + b j ) 1/2 , then the solution of (10) is
                                                                        j
            We present the proof of Theorem 1 in Appen-       expressed as
            dix A.                                                    Z  t  1
                                                              z j (t) =    e µ αη−a j (t−η)  sinh(c j (t − η))v j (η)dη.
            We now obtain the main properties of the Fourier
                                                                       0  c j
            coefficients. We choose an arbitrary external per-                                 −µ αt
                            δ
            turbation v ∈ V and assume that y is the solu-    In this case, since y j (t) = ˙z j (t)e  , then the
                            Ω
            tion admitting the representation given in (5). If  solution of the (8) can be expressed as
            we substitute the functions y and v in (1) and                     Z  t
                                                                        y j (t) =  φ j (η)v j (t − η)dη,  (9)
            equate the Fourier coefficients with respect to
                                                                                0
            each eigenfunction, then we obtain the following
                                                              where
            equality:
                                                                                   e
               ∞                                                φ j (η) = 1 − d 2 1/2 −(µ α+a j )η ·
                                                                              j

               X                           α
                  sin(β j x) κ 2 (1 − α)β 2 cf 0 D y j (t)+                       · cosh (c j η − arctanh(d j )) ,
                                           t
                                      j
               j=1                                                       a j
                                          ∞                   with d j =   . We observe that φ j (η) > 0 for
                                        X                               c j
                               2
                     ˙ y j (t) + κ 1 β y j (t) =  sin(β j x)v j (t).
                               j                              η ∈ (0, t) since d j ∈ (−1, 1). We conclude that
                                         j=1                                          on the right-hand side of
                                                              for every choice v j ∈ V δ j
            If in this equality we compare the terms of each  (8), the solution y j defined in (9) associated with
            series, we conclude that the functions y j and v j  this choice is an element of Y δ j .
                                                            95
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