Sentence examples for applying the divergence from inspiring English sources

Applying the divergence operator ∇⋅ to the first equation of (1.1) produces the expression of the pressure p = − 1 ∇ ⋅ ( v ⋅ ∇ v ).

Applying the divergence free theorem in its integral form leads to the equality int_{vartheta_{Omega_{alpha}}}frac{partial{V}}{partial{r}} ds=int_{vartheta_{Omega_{beta}}} frac{partial{V}}{partial{r}}ds.

Integrating (2.7) with (u=u_{k}) over G and then applying the divergence theorem, we observe that begin{aligned} M[u_{k}] geq& int_{G} A x) biggl[bigglvert nabla u_{k} -frac {u_{k}B x)}{A x)}biggrvert ^{alpha+1}+alphabigglvert frac{u_{k}}{v}{nabla v}biggrvert ^{alpha+1} &- (alpha+1) biggl nabla u_{k} -frac{u_{k}B x)}{A x)} biggr)Phibiggl(frac{u_{k}}{v}{nabla v}biggr) biggr],dx geq0.

Using the equations of motion (8 2 and (8 3 for the difference solutions and applying the divergence theorem, because of null boundary conditions, we are led to begin{aligned} int_{Omega } ( tau_{ij}dot{varepsilon }_{ij}+sigma_{ij} dot{gamma }_{ij}+ mu_{ijk}dot{chi }_{ijk} ),dV=- int_{ Omega } ( varrho ddot{u}_{i} dot{u}_{i}+I_{jk}ddot{varphi } _{js}dot{varphi }_{ks} ),dV.

end{aligned}We now integrating this identity over (N) and apply the divergence theorem.

end{aligned} (21) Integrate equality (21) over (D x_{3})), apply the divergence theorem, and use the lateral conditions (14); we get the equality (16).

end{aligned} (25) Integrate equality (25) over (D x_{3})), apply the divergence theorem, and if we use the lateral conditions (14) we get the equality (18).

end{aligned} (23) Integrate equality (23) over (D x_{3})), apply the divergence theorem; if we use the lateral conditions (14) we get the equality (17).

The proof follows the reasoning in Section 2.2: taking into account that when we apply the divergence theorem in (2.19), the Neumann boundary condition must be used, we get (2.21).

end{aligned} (27) Integrate equality (27) over (D x_{3})), apply the divergence theorem; if we use the lateral conditions (14) we get the equality (19) and the proof of Theorem 1 is completed.

We integrate (3.15) over G and then apply the divergence theorem to obtain 0geq V_{G}[u]+int_{G} A x) biggl nabla u - frac{u}{v} nabla v biggr)^{2},dx geq0, and therefore nabla u- frac{u}{v} nabla vequiv0 quadtextit{i.e.e