**Example 1.**

*1. Let as y*1 ≥ *τ*1 ≥ 0, *the domain Q lies inside the rotation body* |*y*-| ≤ *M*2 (*y*1 + <sup>1</sup>), *i.e., <sup>λ</sup>*(*y*1) ≤ *<sup>M</sup>*(*y*1 + <sup>1</sup>), *M* > 0. *We have from (15)*

$$
\mu(y\_1) = \frac{\pi c(y\_1)}{M(y\_1 + 1)}, \quad c(y\_1) = \frac{d\_0}{\beta(y\_1)}.
$$

*Suppose that <sup>c</sup>*(*<sup>x</sup>*1) = *c* = *const* > 0.

*In this case, from the inequality (6) we have*

$$\int\_{Q\_1} E(u) \, dx \, dy \, dt \le \Phi^{-1}(\tau\_1, \tau\_2) \int\_{Q\_2} E(u) \, dx \, dy \, dt \le \left(\frac{\tau\_1 + 1}{\tau\_2 + 1}\right)^{\pi c} \int\_{Q\_2} E(u) \, dx \, dy \, dt.$$

*2. Consider an example of Q for which*

$$
\lambda(y\_1) \le \pi c \left[ (y\_1 + 1)^{k-1} \right]^{-1}, \\
k = const > 0.
$$

*It is clear that if k* > 1, *the domain Q is narrowing at x*1 → +<sup>∞</sup>. *If k* = 1, *then <sup>λ</sup>*(*<sup>x</sup>*1) ≤ *πc and this case includes domains lying in the band with the width πc*. *If* 0 < *k* < 1, *then Q can be extended respectively at x*1 → +<sup>∞</sup>. *For this kind of domains, we can assume*

$$
\mu(y\_1) \le (y\_1 + 1)^{k-1}.
$$

Then the estimate (6) is valid for considered domains if

$$\Phi^{-1}(\tau\_1, \tau\_2) = 2 \exp \left[ - (\tau\_2 + 1)^k + (\tau\_1 + 1)^k \right].$$

As a corollary of the Saint-Venant principle, we have the uniqueness theorem for the problem (1), (3) in unlimited domain *Q* for classes of functions increasing in infinity depending from *<sup>λ</sup>*(*τ*).

**Theorem 2.** *Let f*(*<sup>x</sup>*, *y*, *t*) = 0 *in Q and conditions of theorem 1 hold. If <sup>u</sup>*(*<sup>x</sup>*, *y*, *t*) *is a generalized solution of the problem (1), (3) in Q and for a sequence τm* → +∞ *at m* → +∞ *and some r*∗ = *const* > 0,

$$\int\_{Q\_{\text{tw}}} \mathbb{E}(\boldsymbol{u})d\boldsymbol{x} \, d\boldsymbol{y} \, dt \le \varepsilon(\boldsymbol{\tau}\_{\text{w}})\Phi(\boldsymbol{r}\_{\*}, \boldsymbol{\tau}\_{\text{w}}) \tag{13}$$

*where <sup>ε</sup>*(*<sup>τ</sup>m*) → 0 *at τm* → +<sup>∞</sup>, *then u* = 0 *in Qr*∗ .

**Proof.** We have from (6) considering (13)

$$\int\_{Q\_{\mathbb{T}}} E(u)dx\,dy\,dt \le \Phi^{-1}(r\_\*, \tau\_{\mathfrak{m}}) \int\_{Q\_{\mathbb{T}}} E(u)dx\,dy\,dt \le \varepsilon(\tau\_{\mathfrak{m}}) \to 0$$

at *τm* → +<sup>∞</sup>. Hence *u* = 0 in Ω*d*∗ .

> Further for any fixed *r*1 > *r*<sup>∗</sup>, we have

$$\Phi(r\_{\ast \prime}, \tau\_{\text{tt}}) = e^{\int\_{r\_{\ast}}^{r\_{\text{tt}}} \mu(s) ds} = e^{\int\_{r\_{1}}^{r\_{\text{tt}}} \mu(s) ds} e^{\int\_{r\_{\ast}}^{r\_{1}} \mu(s) ds} = c \Phi(r\_{1}, \tau\_{\text{tt}})$$

Therefore

$$\int\_{Q\_1} E(u)dx \, dy \, dt \le \Phi^{-1}(r\_1, \tau\_m) \int\_{Q\_{\tau\_m}} E(u)dx \, dy \, dt \le \Phi^{-1}(r\_1, \tau\_m) \varepsilon(\tau\_m) \Phi(r\_\*, \tau\_m) = $$
 
$$c^{-1}\varepsilon(\tau\_m) \to 0 \text{ as } \tau\_m \to +\infty.$$

Hence, *u* = 0 in *Qr*1. Since *r*1 was chosen arbitrary, *u* = 0 in *Q*.
