Green s theorem

After the application of Green s theorem to the second order term in Equation (2.81) we get the weak form of the residual statement as... 

The described application of Green s theorem which results in the derivation of the weak statements is an essential step in the formulation of robu.st U-V-P and penalty schemes for non-Newtonian flow problems. 

Papanastasiou et al. (1992) suggested that in order to generate realistic solutions for Navier-Stokes equations the exit conditions should be kept free (i.e. no outflow conditions should be imposed). In this approach application of Green s theorem to the equations corresponding to the exit boundary nodes is avoided. This is eqvrivalent to imposing no exit conditions if elements with... 

In Equation (4.12) the discretization of velocity and pressure is based on different shape functions (i.e. NjJ = l,n and Mil= l,m where, in general, mweight function used in the continuity equation is selected as -Mi to retain the symmetry of the discretized equations. After application of Green s theorem to the second-order velocity derivatives (to reduce inter-element continuity requirement) and the pressure terms (to maintain the consistency of the formulation) and algebraic manipulations the working equations of the U-V-P scheme are obtained as... 

It is evident that application of Green s theorem cannot eliminate second-order derivatives of the shape functions in the set of working equations of the least-sc[uares scheme. Therefore, direct application of these equations should, in general, be in conjunction with C continuous Hermite elements (Petera and Nassehi, 1993 Petera and Pittman, 1994). However, various techniques are available that make the use of elements in these schemes possible. For example, Bell and Surana (1994) developed a method in which the flow model equations are cast into a set of auxiliary first-order differentia] equations. They used this approach to construct a least-sciuares scheme for non-Newtonian flow equations based on equal-order C° continuous, p-version hierarchical elements. 

The last teim in the right-hand side of Equation (4.143) represents boundary line integrals. These result from the application of Green s theorem to... 

Note that Green s theorem in the plane expressed as... 

By writing down the kinetic energy term T = — (ftV2m)V explicitly, and using Green s theorem, the transition matrix element is finally converted into a surface integral similar to Bardeen s, in terms of modified wavefunctions ... 

Using Green s theorem, it can be converted into a volume integral over fir, the tip side from the separation surface. Noticing that the sample wavefunction ip satisfies Schrodinger s equation, Eq. (3.2), in fir, and that the Green s function satisfies Eq. (3.8), we obtain immediately... 

To proceed with the proof, we first state a property of the Bardeen integral If both functions involved, i i and x> satisfy the same Schrodinger equation in a region fi, then the Bardeen integral J on the surface enclosing a closed volume w within fl vanishes. Actually, using Green s theorem, Eq. (3.2) becomes... 

Since r, — r2 f1 is the Green s function of the Laplacian operator, using the standard Green s theorem techniques (see Appendix A)... 

Finally, the homogeneous density (termed a here) for reaction of species with a vast excess of the other follows from Green s theorem (Appendix A, Sect 3) as... 

Applying Green s theorems to the right-hand side of Eq. (6), we obtain... 

Before particularizing, we can consider a number of simplifications. In the steady-state condition, all time derivatives are zero. At equilibrium, in addition to zero time derivatives, all potential differences and potential gradients are zero. If the system is sourceless and sinkless, S = 0. By Green s theorem... 

If equation (9) is averaged over the cross-section, the use of Green s theorem and the condition (11) reduces it to... 

In this last equation we have used Green s theorem and the first part of condition (c) which together make the integration of the last term in (39) vanish. Again we set m(0) = 1 and find that (39) and its associated conditions vanish if... 

Now the first part of the condition (44), together with the conditions on r0 and r2, determine the functions x and xi- The constants ej and e2 could be evaluated from the latter part of (44) and the condition m = 0. This, however, is unnecessary as it will appear that only the combination (ju.2e2 - fije,) is required in m(2>. To evaluate this we integrate (44) around the contour Ti using Green s theorem and the differential equations to obtain... 

The variational K-matrix is symmetric, as is required by unitarity, but the trial K-matrix is not K 2 = K21 but K 2 / K v This identity may be readily proved by applying Green s theorem to... 

A version of Green s theorem follows from partial integration of the symmetric integral... 

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