6 Construction of the Solutions of Non-stationary …
93
Taking into account the orthogonality of the system of the eigenforms {V
(k)
(x) }
and assuming its completeness in the space of vector functions which are differentiated into twice and satisfying the conditions (6.16), we will be able to write the
solution of the problem (6.7)–(6.9) in the form
U(x, s) = U
(0)
(x, s) +
∞
k=1
V
(k)
(x)η k (s) I
(k)
, η k (s) = −
β
2
(s)
β 2 (s) + χ
2
k
(6.19)
I
(k)
=
U
(0)
(x, s)V
(k)
(x)d{
[V
(k)
(x)]
2 d}
−1
,
(6.20)
The work (Pshenichnov 2013) establishes sufficient conditions of the absence of
branch points on the complex plane at the components of vector U(x, s) without
assuming that ν = const. Here, the essence of this conditions is that the set of
eigenvalues of the problem (6.16), (6.17) is no more than countable one and there
are no branch points in all components of the vectors F(x, s), P
(1)
(x, s), P
(2)
(x, s),
as well as function (s). If these conditions are fulfilled, after application of the
methods of contour integration and the theory of residues the value for u(x, t) will
be simpler if the poles U
(0)
(x, s) and η k (s) have the first order.
If F(x, s), P
(1)
(x, s), P
(2)
(x, s), (s) lead to a situation when no eigenvalue of
s
(v)
km for a viscoelastic body, defined from the Eq. (6.18), is a singularity of U
(0)
(x, s),
the order of the pole of vector components U(x, s) in points s = s
(v)
km will be the same
as of the function η k (s).
Let us consider a case where the hereditary kernel is written in the form
T (t) = ae
−bt
, 0 < a < b, ,(s) = a/(s + b)
(6.21)
Then, the Eq. (6.18) for all χ k ∈ R is a cubic one with respect to s:
s
3
+ bs
2
+ χ
2
k s + χ
2
k (b − a) = 0, k = 1, 2, 3, . . .
(6.22)
It was proved that under the condition a/b < 8/9 this equation for any k =
1, 2, 3, . . . will have exactly one real root s = z k and two complex conjugate ones
s = α k ± iω k , ω k > 0, with z k < 0, α k < 0, z k = −b, z k = a − b. Moreover, it was
established that in points z k and α k ± iω k the functions η k (s) have simple poles.
6.4 Example
As an example, let us consider a problem of propagation of a non-stationary longitudinal wave in a cross section of a viscoelastic infinitely long cylinder being initially
in a non-perturbed state (Fig. 6.1). The interior surface of the cylinder is rigidly
93
Taking into account the orthogonality of the system of the eigenforms {V
(k)
(x) }
and assuming its completeness in the space of vector functions which are differentiated into twice and satisfying the conditions (6.16), we will be able to write the
solution of the problem (6.7)–(6.9) in the form
U(x, s) = U
(0)
(x, s) +
∞
k=1
V
(k)
(x)η k (s) I
(k)
, η k (s) = −
β
2
(s)
β 2 (s) + χ
2
k
(6.19)
I
(k)
=
U
(0)
(x, s)V
(k)
(x)d{
[V
(k)
(x)]
2 d}
−1
,
(6.20)
The work (Pshenichnov 2013) establishes sufficient conditions of the absence of
branch points on the complex plane at the components of vector U(x, s) without
assuming that ν = const. Here, the essence of this conditions is that the set of
eigenvalues of the problem (6.16), (6.17) is no more than countable one and there
are no branch points in all components of the vectors F(x, s), P
(1)
(x, s), P
(2)
(x, s),
as well as function (s). If these conditions are fulfilled, after application of the
methods of contour integration and the theory of residues the value for u(x, t) will
be simpler if the poles U
(0)
(x, s) and η k (s) have the first order.
If F(x, s), P
(1)
(x, s), P
(2)
(x, s), (s) lead to a situation when no eigenvalue of
s
(v)
km for a viscoelastic body, defined from the Eq. (6.18), is a singularity of U
(0)
(x, s),
the order of the pole of vector components U(x, s) in points s = s
(v)
km will be the same
as of the function η k (s).
Let us consider a case where the hereditary kernel is written in the form
T (t) = ae
−bt
, 0 < a < b, ,(s) = a/(s + b)
(6.21)
Then, the Eq. (6.18) for all χ k ∈ R is a cubic one with respect to s:
s
3
+ bs
2
+ χ
2
k s + χ
2
k (b − a) = 0, k = 1, 2, 3, . . .
(6.22)
It was proved that under the condition a/b < 8/9 this equation for any k =
1, 2, 3, . . . will have exactly one real root s = z k and two complex conjugate ones
s = α k ± iω k , ω k > 0, with z k < 0, α k < 0, z k = −b, z k = a − b. Moreover, it was
established that in points z k and α k ± iω k the functions η k (s) have simple poles.
6.4 Example
As an example, let us consider a problem of propagation of a non-stationary longitudinal wave in a cross section of a viscoelastic infinitely long cylinder being initially
in a non-perturbed state (Fig. 6.1). The interior surface of the cylinder is rigidly