Solving a Partial Differential Equations System

Abstract

The idea for solving a system of partial differential equations (PDEs) using numerical methods is to transform it into a system of equations that are easier to solve, such as algebraic equations or ordinary differential equations (ODEs), for which numerical solutions were presented in Chaps. 5 and 6 of this book.

The original version of this chapter was revised. An erratum to this chapter can be found at https://doi.org/10.1007/978-3-319-66047-9_8

Appendices

Appendix 7.1

Figures A.7.1, A.7.2, and A.7.3 show the VBA code developed to solve the ODE system generated in Sect. 7.4.3 (Eqs. 7.47, 7.48, 7.49, 7.50, 7.51, 7.52, 7.53, 7.54, 7.55 and 7.56). Observe that the code is almost identical to the one presented in Chap. 6 (Figs. 6.18b, 6.17, and 6.20). The function Derivative is changed to account for the new system of ODEs. The function RungeKutta4 is identical, except for the dimensions of variables k1, k2, k3, k4, and ytran, which have changed from 5 to 10, to account for the 10 ODEs. The main program RK4 is modified only in the dimension of the variables and values for the initial conditions and integration step, as highlighted in Fig. A.7.1.

Fig. A.7.1

The main program in Visual Basic for Applications (VBA) code to solve an ordinary differential equation (ODE) system (Eqs. 7.47, 7.48, 7.49, 7.50, 7.51, 7.52, 7.53, 7.54, 7.55 and 7.56) using the fourth-order Runge–Kutta (RK4) method

Fig. A.7.2

Function of the fourth-order Runge–Kutta (RK4) method called in the main program (Fig. A.7.1)

Fig. A.7.3

The function Derivative with ten ordinary differential equations (ODEs) (Eqs. 7.47, 7.48, 7.49, 7.50, 7.51, 7.52, 7.53, 7.54, 7.55 and 7.56) called in the function RungeKutta4

Appendix 7.2

Figure 7.11 shows a curve for the analytical solution of Eq. (4.15), rewritten below:

$$ \frac{\partial T}{\partial t}=\frac{k}{\rho {c}_p}\frac{\partial^2T}{\partial {x}^2} $$

(4.15)

At t = 0 h, T(x, 0) = 50 °C, for 0 ≤L≤ 1 m

At x = 0 m, T(0, t) = 70 °C, for t> 0 h

$$ \mathrm{At}\ x=1\ \mathrm{m},\frac{dT}{dx}=-\frac{h}{k}\left(T-{T}_{\mathrm{env}}\right),\mathrm{for}\ t>0\kern0.28em \mathrm{h} $$

(7.57)

The analytical solution of Eq. (4.15) can be obtained using the Fourier method, to yield:

$$ {\displaystyle \begin{array}{l}T=T\left(0,t\right)+\left(\frac{h}{h+k}\right)\left({T}_{\mathrm{env}}-T\left(0,t\right)x\right)\\ {}\kern1.6em +\sum \limits_{n=1}^{\infty}\frac{2}{\left(1+\frac{k}{h}{\left(\cos {\beta}_n\right)}^2\right)\ }\left(\frac{T\left(x,0\right)-T\left(0,t\right)}{\beta_n}\right.\left.-\left(T\left(x,0\right)-{T}_{\mathrm{env}}\right)\frac{\cos {\beta}_n}{\beta_n}\right)\\ {}\kern1.5em \exp \left(-\frac{k}{\rho\ \mathrm{cp}}\ {\beta}_n^2\ t\right)\;\sin \left({\beta}_nx\right)\end{array}} $$

in which β _n are the solutions of:

$$ \tan {\beta}_n+\frac{k}{h}{\beta}_n=0\kern1em \mathrm{For}\ \mathrm{n}=1,2,3,\dots $$