Integral Inequalities

Consider a reactant undergoing an nth-order reaction with rate kC in dispersed-phase droplets. The concentration c is assumed to vary between 0 and Cq. If the distribution of solute concentration is denoted/(c), which is nonnegative and such that jo°f(c)dc = l, the average concentration in the drop phase is then 

On the other hand, the reaction rate in terms of the average concentration is given by 

It is easy to show that the rate of reaction as described by the lumped dispersed phase analysis underestimates the actual rate by using the Holder s inequality. Let g c) = and h(c) Thus, 

The second integral on the right-hand side of the inequality is unity by definition. Hence, with p = n and q = (n - 1), 

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A.A. Martynjuk and R. Gutovskii, Integral Inequalities and Stability of Motion, Naukova Dumka, Kiev, 1979 (in Russian). 

Wolfgang, W., Differential and Integral Inequalities, Berlin Spring-Verlag, 1970. 

Thus, M has the penalty function e W(I —k), which is positive and enlarges M whenever any integral inequality is violated. 

Consider the batch reactor problem in Example 7.1 (p. 192) subject to the following integral inequality constraints ... 

Figure 7.19 The optimal states when the integral inequality constraints are satisfied upon convergence in Example 7.8...

This generalized cosine index is often called the Carbo index. Naturally the Carbo index is limited to the range (0,1), where Cab = 1 means perfect similarity. Still more (dis)similarity measures have been introduced and will be discussed further in this chapter. The range of the Carbo index naturally agrees with the Schwartz integral inequality " ... 

Integral Inequalities and the Theory of Nonlinear Oscillations. Nauka, Moscow, 1975. 

Let us assume that kij = 0 on L. This enables us to integrate by parts in the second and the third terms of (2.231) and to obtain the inequality... 

Hence, the integration of (3.12) with respect to t from 0 to T — ft gives the inequality... 

In this case the boundary conditions (5.81) are included in (5.84). At the first step we get a priori estimates. Assume that the solutions of (5.79)-(5.82) are smooth enough. Multiply (5.79), (5.80) by Vi, Oij — ij, respectively, and integrate over fl. Taking into account that the penalty term is nonnegative this provides the inequality... 

We do not show the dependence of v,a,, f on t in (5.87). The integration by parts in the third term of the left-hand side of (5.87) can be done. Recall that satisfies the equation (5.68). As a result the following inequality is obtained ... 

Boundary conditions (5.81) can be taken into account here in order to integrate by parts in the left-hand side. Next we can integrate the inequality obtained in t from 0 to t. This implies... 

In so doing we have omitted the nonnegative term containing the penalty operator. Using the formula (5.181), the integration by parts can be done in the third and the fifth terms of the left-hand side of (5.189). Also, note that Mij satisfy equation (5.175). Integration of (5.189) in t from 0 to t results in the inequality... 

Taking into account the conditions (5.187), we first integrate by parts in the left-hand side of the inequality obtained and next we integrate in t. Simultaneously, the integration by parts in t is fulfilled. This gives the inequality... 

Assuming a sufficient regularity of the solution to (5.247)-(5.252), we can deduce relations considered as a corollary from the exact formulation of the problem. In what follows the theorem of existence of these relations is established. Substituting the values, 7] from (5.248), (5.249) in (5.251) and summing the resulting inequality with (5.247), we obtain, after integration over J,... 

This transformation has removed the elastic stress work, whose integral around the cycle is zero, leaving only the inelastic stress work done during the cycle. Using this result, the work inequality (5.37) can be written in the form... 

For, suppose that this inequality is not true, and that becomes negative at t = t. The smoothness assumptions that have been made imply that the integrand must continue to be negative for at least a short finite-time interval Ar. Now a second cycle may be chosen which is identical with tj) over the time interval (tf, tf + At), but in which elastic unloading commences at time t f + At. Since is zero during elastic portions of the cycle, the integral of is negative for this second cycle, which contradicts (5.50). 

Finally we note that the value of the integral appearing in (3-75) can be further decreased only by replacing ( — m)2 by the smallest value it can take on in the — m e namely e2. Making this replacement, we obtain the inequality... 

It is commonly believed that K (t ) may be carried outside the integral without lack of accuracy if inequality (2.23) is satisfied. This is the same way that was used in Chapter 1 to obtain the non-Markovian differential equation... 

When (Xfpi )/ ((Z-Xf)pL) > 1 the integral (10.40) is much smaller than the integral (10.41) and it can be neglected. This inequality is fulfilled for the majority of physically realistic conditions when Xf > L 10. ... 

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