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Nonlinear equations in one variable

Here is the mathematical problem you must solve Given a set of chemicals, temperature and pressure, find the specific volume of the mixture. To do this, you must find the critical temperature and pressure of each chemical. Once you have the parameters, you must solve the cubic equation, Eq. (2.8), which is a nonlinear equation in one variable. Because it is a cubic equation, it is possible to find the solution in a series of analytical steps (Perry and Green, 1997, p. 3-114), but this is not usually done because it is quicker to find the solution numerically, albeit iteratively. [Pg.8]

Each of the partial derivatives when equated to zero may well yield a nonlinear equation. Hence, the minimization of /(x) is converted into a problem of solving a set of nonlinear equations in n variables, a problem that can be just as difficult to solve as the original problem. Thus, most engineers prefer to attack the minimization problem directly by one of the numerical methods described in Chapter 6, rather than to use an indirect method. Even when minimizing a function of one variable by an indirect method, using the necessary conditions can lead to having to find the real roots of a nonlinear equation. [Pg.155]

Can we reduce the two nonlinear coupled equations to one equation in one variable and the other is a simple explicit linear equation Yes, we can do that through the two alternative equivalent route shown in Table 3.1. From either of the two routes, we get one equation in terms of Y (one variable) we can solve it and get the value of Y. Then, the value of can be obtained easily from Eq. (3.101) or (3.102). [Pg.255]

Fig. 4.1. Newton s method for solving a nonlinear equation with one unknown variable. The solution, or root, is the value of x at which the residual function R(x) crosses zero. In (a), given an initial guess. vl0,), projecting the tangent to the residual curve to zero gives an improved guess v( l ). By repeating this operation (b), the iteration approaches the root. Fig. 4.1. Newton s method for solving a nonlinear equation with one unknown variable. The solution, or root, is the value of x at which the residual function R(x) crosses zero. In (a), given an initial guess. vl0,), projecting the tangent to the residual curve to zero gives an improved guess v( l ). By repeating this operation (b), the iteration approaches the root.
Other nonlinear equations involving one independent variable may also be converted to linear form but, in most cases, the required transformation will be different. Sometimes, of course, no simple transforma-... [Pg.238]

Equation (3.13.1-4) can be used to eliminate either Cs or Ts from one of the differential equations, with the result that, in general, only one (nonlinear) equation with one dependent variable must be solved. [Pg.224]

The term nonlinear in nonlinear programming does not refer to a material or geometric nonlinearity but instead refers to the nonlinearity in the mathematical optimization problem itself. The first step in the optimization process involves answering questions such as what is the buckling response, what is the vibration response, what is the deflection response, and what is the stress response Requirements usually exist for every one of those response variables. Putting those response characteristics and constraints together leads to an equation set that is inherently nonlinear, irrespective of whether the material properties themselves are linear or nonlinear, and that nonlinear equation set is where the term nonlinear programming comes from. [Pg.429]

These problems refer to models that have more than one (w>l) response variables, (mx ) independent variables and p (= +l) unknown parameters. These problems cannot be solved with the readily available software that was used in the previous three examples. These problems can be solved by using Equation 3.18. We often use our nonlinear parameter estimation computer program. Obviously, since it is a linear estimation problem, convergence occurs in one iteration. [Pg.46]

In general for a set of nonlinear equations the necessary condition for determinancy is that there exists at least one admissible set of output variables for the equations (C7, S4). We can think of an output variable as that variable for which a given equation is solved either by an iteration process or by an elimination process. The set of all such assigned pairs of variables and equations is called the output set. Clearly an admissible output set must satisfy the conditions (i) each equation has exactly one output variable, (ii) each variable appears as the output variable of exactly one equation, (iii) each output variable must occur in the assigned equations in such a manner that it can be solved for uniquely. Such an output set (circled entities) is illustrated in terms of an occurrence matrix in Fig. 4. Algorithms for finding output sets have been published by Steward (S4) and Gupta et al. (G8). [Pg.145]

An alternative method of solving the equations is to solve them as simultaneous equations. In that case, one can specify the design variables and the desired specifications and let the computer figure out the process parameters that will achieve those objectives. It is possible to overspecify the system or to give impossible conditions. However, the biggest drawback to this method of simulation is that large sets (tens of thousands) of nonlinear algebraic equations must be solved simultaneously. As computers become faster, this is less of an impediment, provided efficient software is available. [Pg.90]

Thus, it would be natural to attempt to extend the QMOM approach to handle a bivariate NDF. Unfortunately, the PD algorithm needed to solve the weights and abscissas given the moments cannot be extended to more than one variable. Other methods for inverting Eq. (125) such as nonlinear equation solvers can be used (Wright et al., 2001 Rosner and Pykkonen, 2002), but in practice are computationally expensive and can suffer from problems due to ill-conditioning. [Pg.282]

The model involves four variables and three independent nonlinear algebraic equations, hence one degree of freedom exists. The equality constraints can be manipulated using direct substitution to eliminate all variables except one, say the diameter, which would then represent the independent variables. The other three variables would be dependent. Of course, we could select the velocity as the single independent variable of any of the four variables. See Example 13.1 for use of this model in an optimization problem. [Pg.69]

One method of handling just one or two linear or nonlinear equality constraints is to solve explicitly for one variable and eliminate that variable from the problem formulation. This is done by direct substitution in the objective function and constraint equations in the problem. In many problems elimination of a single equality constraint is often superior to an approach in which the constraint is retained and some constrained optimization procedure is executed. For example, suppose you want to minimize the following objective function that is subject to a single equality constraint... [Pg.265]

Relations (fl)-(g) define the MINLP problem. It is important to note that the relations between the binary and continuous variables in Equation (/) are linear. It is possible to impose the desired relations nonlinearly. For example, one could replace Cl by Cl 71 everywhere Cl appears. Then if 71 = 0, Cl does not appear, and if 71 = 1, Cl does appear. Alternatively, one could replace Cl by the conditional expression (if 71 = 1 then Cl else 0). Both these alternatives create nonlinear models that are very difficult to solve and should be avoided if possible. [Pg.365]

The system thus obtained involves N — n + 1 variables, including t//, related by the same number of equations. Since N — n of these are nonlinear equations because of the aq term, an iteration procedure is needed. One starts from a set of q values obtained for a = 0. The equations then become linear and the Gauss elimination method may thus be used to obtain these starting q values. In a second round, these values are used in the aq term and a new set of q values are obtained by... [Pg.467]


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