8.3.9 Time-Dependent Higher-Order Partial Differential Equations by Substitution and Separation of Variables: The Time-Dependent Diffusion Equation in Three Dimensions

8.3.9.1 Introduction

We already solved the time-dependent diffusion equation in section 8.3.8. We will reconsider this equation now deriving a general schematic for the solution of higher-order partial differential equations by the substitution and separation of variables approach we introduced in section 8.3.3, using the example of the heat equation. We will use this approach later on for solving the Navier-Stokes equation, which is a time-dependent inhomogeneous PDE.

8.3.9.2 General Approach

We have already derived the diffusion equation (see Eq. 6.90). In time-dependent form in three dimensions it is given by

$\frac{\partial f}{\partial t}=c^2\left(\frac{{\partial }^{2}f}{\partial x^2}+\frac{{\partial }^{2}f}{\partial y^2}+\frac{{\partial }^{2}f}{\partial z^2}\right)$

We now assume that the function f is a function of four functions that only depend on one of the four independent variables, respectively: T (t), X (x), Y (y), and Z (z)

$f\left(t,x,y,z\right)=T\left(t\right)X\left(x\right)Y\left(y\right)Z\left(z\right)$

Inserting this into Eq. 8.136 results in

$\begin{array}{lll}\frac{\partial T}{\partial t}XYZ& =& c^2\left(\frac{{\partial }^{2}X}{\partial x^2}TYZ+\frac{{\partial }^{2}Y}{\partial y^2}TXZ+\frac{{\partial }^{2}Z}{\partial z^2}TXY\right)\\ \frac{1}{T}\frac{\partial T}{\partial t}& =& c^2\left(\frac{1}{X}\frac{{\partial }^{2}X}{\partial x^2}+\frac{1}{Y}\frac{{\partial }^{2}Y}{\partial y^2}+\frac{Z}{T}\frac{{\partial }^{2}Z}{\partial z^2}\right)\end{array}$

Here again we have the case that a differential in time should be equal to a differential in space, which can only be true if the individual terms sum up to constants, which we term − λ_T, − λ_X², − λ_Y², and − λ_Z². The fact that we take the squared constants for the x, y, and z terms seems strange, but the reason will become obvious in a moment. We can now rewrite Eq. 8.138 as

$-{\lambda }_T=-c^2\left({\lambda }_X^2+{\lambda }_Y^2+{\lambda }_Z^2\right)$

and derive the four ODEs

$\frac{1}{T}\frac{\partial T}{\partial t}={\lambda }_T$

$\frac{1}{X}\frac{{\partial }^{2}X}{\partial x^2}=-{\lambda }_X^2$

$\frac{1}{Y}\frac{{\partial }^{2}Y}{\partial y^2}=-{\lambda }_Y^2$

$\frac{1}{Z}\frac{{\partial }^{2}Z}{\partial z^2}=-{\lambda }_Z^2$

Eq. 8.140 is solved by partial integration (section 8.2.2) to result in

$\begin{array}{lll}T\left(t\right)& =& {\mathrm{e}}^{{\lambda }_{T}t}\\ & =& c_0{\mathrm{e}}^{-\left({\lambda }_X^2+{\lambda }_Y^2+{\lambda }_Z^2\right)c^{2}t}\end{array}$

where we have used Eq. 8.139 to replace λ_T. For Eq. 8.141, Eq. 8.142, and Eq. 8.143 we find characteristic polynomials and solutions according to Tab. 8.2 depending on the boundary conditions. Assuming that the domain is bounded on all sides 0 ≤ x ≤ l, 0 ≤ y ≤ w, and 0 ≤ z ≤ h and that X (0) = X (l) = 0, Y (0) = Y (w) = 0, and Z (0) = Z (h) = 0 we obtain the solutions:

$X\left(x\right)={\displaystyle \sum _{m=1}^{\infty }{c}_1}sin\left(\frac{n\pi }{x}l\right){\lambda }_X=\frac{n\pi }{l}$

$Y\left(y\right)={\displaystyle \sum _{n=1}^{\infty }{c}_2}sin\left(\frac{m\pi }{y}w\right){\lambda }_Y=\frac{m\pi }{w}$

$Z\left(z\right)={\displaystyle \sum _{o=1}^{\infty }{c}_3}sin\left(\frac{o\pi }{z}h\right){\lambda }_Z=\frac{o\pi }{h}$

Using Eq. 8.144, Eq. 8.145, Eq. 8.146, and Eq. 8.147 we can now rewrite Eq. 8.137 as

$f(t,x,y,z)={\displaystyle \sum _{n=1}^{\infty }{\displaystyle \sum _{m=1}^{\infty }{\displaystyle \sum _{o=1}^{\infty }{c}_{mno}}}}{\text{e}}^{-\left({\left(\frac{n\pi }{l}\right)}^2+{\left(\frac{m\pi }{w}\right)}^2+{\left(\frac{o\pi }{h}\right)}^2\right)c^{2}t}\sin \left(\frac{n\pi }{x}l\right)\sin \left(\frac{m\pi }{y}w\right)\sin \left(\frac{o\pi }{z}h\right)$

where we have combined all constants into the constants c_mno. Obviously we are still missing the initial values that will account for the value of these constants.

8.3.9.3 Inhomogeneous Partial Differential Equations

Now that we have the solution of a very complex PDE that is time-dependent and three-dimensional, we can turn to the question of how to solve an inhomogeneous PDE, i.e., a differential equation that has a term on the right-hand side. An example of such an equation would be the Poisson equation, which we will need to solve later in form of the Navier-Stokes equation.

However, we will stick with the diffusion equation for the moment. Introducing an additional term transforms Eq. 8.136 to

$\begin{array}{lll}\frac{\partial f}{\partial t}& =& c^2\left(\frac{{\partial }^{2}f}{\partial x^2}+\frac{{\partial }^{2}f}{\partial y^2}+\frac{{\partial }^{2}f}{\partial z^2}\right)+\rho \left(t,x,y,z\right)\\ \frac{\partial f}{\partial t}-c^2\left(\frac{{\partial }^{2}f}{\partial x^2}+\frac{{\partial }^{2}f}{\partial y^2}+\frac{{\partial }^{2}f}{\partial z^2}\right)& =& \rho \left(t,x,y,z\right)\end{array}$

where we have added the term ρ to account for a source (if positive) or a sink (if negative) of concentration. These sources or sinks could be substance added or removed as well as chemical reactions which reduce or increase the concentration of a substance in the volume. Now obviously we are faced with an inhomogeneous solution. As discussed, in such a case we would begin by first solving the homogeneous case before proceeding to solve the inhomogeneous. If we insert Eq. 8.148 into Eq. 8.149 we obtain

$\begin{array}{l}\frac{\partial f}{\partial t}-c^2\left(\frac{{\partial }^{2}f}{\partial x^2}+\frac{{\partial }^{2}f}{\partial y^2}+\frac{{\partial }^{2}f}{\partial z^2}\right)=\rho \left(t,x,y,z\right)\\ {\displaystyle \sum _{n=1}^{\infty }{\displaystyle \sum _{m=1}^{\infty }{\displaystyle \sum _{o=1}^{\infty }{c}_{mno}}}}{{\mathrm{e}}^{-}}^{\left.\left({\left(\frac{n\pi }{l}\right)}^2+{\left(\frac{m\pi }{w}\right)}^2+{\left(\frac{o\pi }{h}\right)}^2\right)\frac{1}{C^{L}k}\right)}sin\left(\frac{n\pi }{x}l\right)sin\left(\frac{m\pi }{y}w\right)sin\left(\frac{o\pi }{z}h\right)=\rho \left(t,x,y,z\right)\end{array}$

In the next step, we would need to expand ρ (t, x, y, z) to a sine function in three dimensions and compare coefficients in order to determine c_mno. Having done so we have our solution to the three-dimensional diffusion equation.