Let us consider the functions K and U defined on the interval by the relations
(9)
In the next theorem we prove a relationship between the functions K and U.
Theorem 1 Assume that the symmetry relations (5) are satisfied. Then we have
(10)
where we have denoted by the following expression:
(11)
for all .
Proof Using the constitutive equations (4) and the symmetry relations (5), we obtain
(12)
Here we have suppressed, for convenience, the dependence of the functions on argument x, because there is no likely confusion.
By using the equations of motion (1), the balance of equilibrated forces (2), the energy equation (3) and the geometric equations, we deduce
(13)
Similarly,
(14)
Now, we integrate, over B, both sides of the equality (12) and then use (5), (13), (14), and the divergence theorem, and it follows that
(15)
Finally, we integrate the equality (15) from 0 to t and arrive at the desired result (10) so that the theorem is demonstrated. □
In the next lemma we prove another relation between the functions U and K defined by (9).
Lemma 1 Assume that the symmetry relations (5) are satisfied. Then we have the following relations:
(16)
(17)
for all .
The function
P
is defined by
(18)
Proof Taking into account the equations of motion (1), the balance of equilibrated forces (2), the energy equation (3), and the constitutive equations (4), we obtain
From this equality it is easy to deduce that
By integrating this relation over the interval we obtain
(19)
If we add (10) and (19), term by term, (16) follows.
If we subtract (10) from (19), term by term, (17) follows and this concludes Lemma 1. □
The uniqueness result from the next theorem is based on the results from Theorem 1 and Lemma 1.
Theorem 2 Assume that:
-
(i)
the symmetry relations (5) are satisfied;
-
(ii)
ϱ, and J are strictly positive;
-
(iii)
d is strictly positive or strictly negative;
-
(iv)
the conductivity tensor is positive semi-definite.
Then the mixed initial boundary value problem of thermoelasticity of microstretch materials consisting of (1)-(3), the initial condition (7), and the boundary condition (8) has at most one solution.
Proof Suppose, to the contrary, that our mixed problem has two solutions,
Let us denote by
the difference of two solutions, where
Because of linearity, this difference is also a solution of our problem, but it corresponds to null data.
Thus, from (17) we obtain
(20)
By using the hypotheses (ii) and (iv) of the theorem, (20) implies that
(21)
and
(22)
But , , and ω vanish initially, so that from (21) we deduce
(23)
Taking into account (22) and (23), (16) reduces to
Since or , this relation yields
From (23) and (24) we deduce that the difference of the two solutions is null, i.e., we have the uniqueness of solution and Theorem 2 is demonstrated. □
Consider two scalar functions u and v which are defined on and continuous with respect to time.
As is well known, the convolution product of the function u and v is defined by the integral
Also, let us consider the functions and defined by
(25)
For a continuous function h defined on , we denote by the convolution product , i.e.
(26)
Using these considerations, we can write the energy equation (3) and the initial condition
in the equivalent form
(27)
where
Consider two external data systems , which act on the thermoelastic-microstretch material, defined by
and denote by a solution of the mixed problem which corresponds to
Also, we use the following notations:
(29)
Lemma 2 Assume that the symmetry relations (5) are satisfied. Consider the functions defined by
(30)
Then we have
(31)
for all .
Proof We introduce the notation
(32)
From (32) and the constitutive equations (4) we deduce
(33)
Using the symmetry relations (5), from (33) we obtain
(34)
On the other hand, using the equations of motion (1), the balance of equilibrated forces (2), and the energy equation (3) in (27) and (32), we are lead to the following expression of :
(35)
Now, we integrate (35) over B, then we use the symmetry relations (34) and the divergence theorem so that we obtain the desired result (31) and Lemma 2 is proved. □
Based on the result of Lemma 2 we can prove the reciprocal result from the next theorem.
Theorem 3 Assume that the symmetry relations (5) are satisfied. Let be a solution corresponding to the external data , . Then the following reciprocal relation is valid:
(36)
where we have used the notations
(37)
Proof We use the substitution and in (31) and integrate the resulting relation on the interval such that, by using the symmetry relations (5), we obtain
(38)
It is easy to prove that
(39)
Taking the convolution of (38) with g and using (39), we obtain the reciprocal relation (36) and the proof of Theorem 3 is complete. □
Remark In the case of null boundary data, from (36) we deduce that the operator of the thermoelastodynamics of microstretch bodies is symmetric with regard to the convolution.
Based on this symmetry, we can obtain some variational theorems of Gurtin type in classical thermoelasticity.
Also, based on the symmetry relations (31) we can obtain a minimum principle similar to those obtained by Reiss [19] in the classical isothermal case.
Theorem 4 Assume that the symmetry relations (5) are satisfied. Let us consider the function
(40)
for all .
Then we have
(41)
Proof Using the result on from Lemma 2, we obtain
(42)
Let us apply (42) to the process
From (30) and (40), we obtain the equality
(43)
Similarly,
(44)
It is easy to prove the relations
(45)
Using the symmetry relations (5) and (45), from (42), (43), and (44) we obtain the desired result (41) and the proof of Theorem 4 is complete. □
A similar reciprocal result has been obtained in [11], but using some strong hypotheses on the thermoelastic coefficients.
If the conductivity tensor is assumed be positive definite, then the result of Theorem 4 can be used to obtain the uniqueness result established in [18].
Now, our intention is to give another proof of Theorem 2 by using the result of Theorem 4. Let us consider that the mixed problem formulated above has two solutions,
Let us denote by
the difference of two solutions, where
If we apply (41) for the difference, we are lead to
From this equality, by using the hypotheses (ii) and (iv) of Theorem 2, we obtain
Then, from (16), (17), and hypothesis (iii) of Theorem 2, we obtain
which proves Theorem 4.
Remark Using a similar procedure as in [18] and [9], we can use (16), (17), and (41) to obtain some continuous dependence results.