Sentence examples for boundary functions can be from inspiring English sources

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and boundary functions can be given arbitrary.

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It is shown that in both the isothermal and non-isothermal cases, with either Dirichlet or third kind boundary conditions, two sequences of functions can be obtained, one of which converges to the maximal solution and the other to the minimal solution.

The present Green's functions can be used in boundary element method analysis of more complicated problems.

After some lengthy algebra, all of the above coefficients of the wave functions can be obtained from the boundary conditions.

Finally, the derived Green's functions can be applied for the boundary element method to solve the transport equation for an arbitrary geometry [ 20].

Then, the multiscale base functions can be constructed numerically under the boundary conditions through the micro scale computations.

It is found that after the boundary condition is enforced, the coefficient of every term in the Laurent series expansions for the potential functions can be explicitly determined by evaluating the Fourier transform of the boundary effect function and its conjugate on the exterior boundary.

Co-operation between functions can be limited.

With additional boundary conditions for the common boundaries of different materials from the continuous conditions of deformation and traction across the interior boundary, the torsion function can be solved numerically from the second boundary-value problem of potential theory.

To overcome this problem and convert it to a homogenous boundary condition equation, the pressure function can be assumed to be composed of two parts of one being time-independent and the other part being time dependent (Gonzalez-Velasco 1996): Pleft( {x, t} right) = vleft( x right) + w x, t) (43).

For example assuming there are three volumes ℋ 1, ℋ 2 and ℋ 3 and we choose to evolve the boundary of the first one, the energy function can be expressed as J 1 = ∫ ℋ 1 V d 1 ( x, Δ p 1 ) d x + ∫ ℋ 2 V d 2 ( x, Δ p 2 ) d x + ∫ ℋ 3 V d 3 ( x, Δ p 3 ) d x ⏟ ∫ H 1 V ¯ d 1 out ( x ) d x (17) = ∫ ℋ 1 V d 1 ( x, Δ p 1 ) d x + ∫ ℋ 1 V ¯ d 1 out ( x ) d x, (18).

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