Differential equations, solution with Laplace transforms

Differential equations and their solutions will be stated for the elementary models with the main lands of inputs. Since the ODEs are linear, solutions by Laplace transforms are feasible. 

Do not worry if you have forgotten the significance of the characteristic equation. We will come back to this issue again and again. We are just using this example as a prologue. Typically in a class on differential equations, we learn to transform a linear ordinary equation into an algebraic equation in the Laplace-domain, solve for the transformed dependent variable, and finally get back the time-domain solution with an inverse transformation. 

The solution of Fick s second law gives the variation of flux, and thence diffusion-limited current, with time, it being important to specify the conditions necessary to define the behaviour of the system (boundary conditions). Since the second law is a partial differential equation it has to be transformed into a total differential equation, solved, and the transform inverted1. The Laplace transform permits this (Appendix 1). 

Linear first order hyperbolic partial differential equations are solved using Laplace transform techniques in this section. Hyperbolic partial differential equations are first order in the time variable and first order in the spatial variable. The method involves applying Laplace transform in the time variable to convert the partial differential equation to an ordinary differential equation in the Laplace domain. This becomes an initial value problem (IVP) in the spatial direction with s, the Laplace variable, as a parameter. The boundary conditions in x are converted to the Laplace domain and the differential equation in the Laplace domain is solved by using the techniques illustrated in chapter 2.1 for solving linear initial value problems. Once an analytical solution is obtained in the Laplace domain, the solution is inverted to the time domain to obtain the final analytical solution (in time and spatial coordinates). This is best illustrated with the following example. 

Solution of Differential Equations with Laplace Transforms... 

Linear differential equations with constant coefficients can be solved by a mathematical technique called the Laplace transformation . Systems of zero-order or first-order reactions give rise to differential rate equations of this type, and the Laplaee transformation often provides a simple solution. 

The preceding two equations are examples of linear differential equations with constant coefficients and their solutions are often found most simply by the use of Laplace transforms [1]. 

This partial differential equation is most conveniently solved by the use of the Laplace transform of temperature with respect to time. As an illustration of the method of solution, the problem of the unidirectional flow of heat in a continuous medium will be considered. The basic differential equation for the X-direction is ... 

In this section, we will outline only those properties of the Laplace transform that are directly relevant to the solution of systems of linear differential equations with constant coefficients. A more extensive coverage can be found, for example, in the text book by Franklin [6]. 

Let us first state a few important points about the application of Laplace transform in solving differential equations (Fig. 2.1). After we have formulated a model in terms of a linear or linearized differential equation, dy/dt = f(y), we can solve for y(t). Alternatively, we can transform the equation into an algebraic problem as represented by the function G(s) in the Laplace domain and solve for Y(s). The time domain solution y(t) can be obtained with an inverse transform, but we rarely do so in control analysis. 

We now derive the time-domain solutions of first and second order differential equations. It is not that we want to do the inverse transform, but comparing the time-domain solution with its Laplace transform helps our learning process. What we hope to establish is a better feel between pole positions and dynamic characteristics. We also want to see how different parameters affect the time-domain solution. The results are useful in control analysis and in measuring model parameters. At the end of the chapter, dead time, reduced order model, and the effect of zeros will be discussed. 

Other integral transforms are obtained with the use of the kernels e" or xk among the infinite number of possibilities. The former yields the Laplace transform, which is of particular importance in the analysis of electrical circuits and the solution of certain differential equations. The latter was already introduced in the discussion of the gamma function (Section 5.5.4). 

Burns and Curtiss (1972) and Burns et al. (1984) have used the Facsimile program developed at AERE, Harwell to obtain a numerical solution of simultaneous partial differential equations of diffusion kinetics (see Eq. 7.1). In this procedure, the changes in the number of reactant species in concentric shells (spherical or cylindrical) by diffusion and reaction are calculated by a march of steps method. A very similar procedure has been adopted by Pimblott and La Verne (1990 La Verne and Pimblott, 1991). Later, Pimblott et al. (1996) analyzed carefully the relationship between the electron scavenging yield and the time dependence of eh yield through the Laplace transform, an idea first suggested by Balkas et al. (1970). These authors corrected for the artifactual effects of the experiments on eh decay and took into account the more recent data of Chernovitz and Jonah (1988). Their analysis raises the yield of eh at 100 ps to 4.8, in conformity with the value of Sumiyoshi et al. (1985). They also conclude that the time dependence of the eh yield and the yield of electron scavenging conform to each other through Laplace transform, but that neither is predicted correctly by the diffusion-kinetic model of water radiolysis. 

The statement cA = c0/ (1 + K) in Eqs. (157a and b) above is tantamount to saying that cA + cB = Co, where c0 is the total concentration of both species of the dissolved solute. If the diffusivities SDA8 and DBs are assumed to be equal, then cB can be eliminated from Eqs. (155) and (156) and a fourth-order, linear partial-differential equation is obtained. The solution of this equation consistent with the conditions in Eq. (157) is obtainable by Laplace transform techniques (S9). Sherwood and Pigford discuss the results in terms of the behavior of the liquid-film mass transfer coefficient. 

They may be obtained by means of the matrix IET but only together with the kernel E(f) = F(t) specified by its Laplace transformation (3.244), which is concentration-independent. However, from the more general point of view, Eqs. (3.707) are an implementation of the memory function formalism in chemical kinetics. The form of these equations shows the essentially non-Markovian character of the reversible reactions in solution the kernel holds the memory effect, and the convolution integrals entail the prehistoric evolution of the process. In the framework of ordinary chemical kinetics S(/j = d(t), so that the system (3.707) acquires the purely differential form. In fact, this is possible only in the limit when the reaction is entirely under kinetic control. 

The occurrence of partial differential equations in electrochemistry is due to the variation of concentration with distance and with time, which are two independent variables, and are expressed in Fick s second law or in the convective-diffusion equation, possibly with the addition of kinetic terms. As in the resolution of any differential equation, it is necessary to specify the conditions for its solution, otherwise there are many possible solutions. Examples of these boundary conditions and the utilization of the Laplace transform in resolving mass transport problems may be found in Chapter 5. 

INTRODUCTION TO LINEAR ALGEBRA AND DIFFERENTIAL EQUATIONS, John W. Dettman. Excellent text covers complex numbers, determinants, orthonormal bases, Laplace transforms, much more. Exercises with solutions. Undergraduate level. 416pp. 5k x 8k. 65191-6 Pa. 8.95... 

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