Feb 7, 2017 by Hugo Milan

Analytical solutions of the hyperbolic heat equation can be used to solve the hyperbolic diffusion equation because these equations are similar.

We define the hyperbolic diffusion equation in time-domain as:

\begin{equation} \frac{\partial C}{\partial t} = -\nabla q + S \label{eqD1} \end{equation}

with the following definition of flux

\begin{equation} \bar{q} + \tau\frac{\partial \bar{q}}{\partial t}= -D\nabla C \label{eqD2} \end{equation}

where \(C\) is concentration, \(\bar{q}\) is concentration flux, \(t\) is time, \(D\) is diffusivity, \(\tau\) is relaxation time, and \(S\) is source. Eqs. \ref{eqD1} and \ref{eqD2} can be combined to yield:

\begin{equation} \tau\frac{\partial^2 C}{\partial t^2} + \frac{\partial C}{\partial t} = D\nabla^2 C + S + \tau\frac{\partial S}{\partial t} \label{eqD3} \end{equation}

We define the heat equation in time-domain as:

\begin{equation} \rho c_{p}\frac{\partial T}{\partial t} = -\nabla q + S \label{eqH1} \end{equation}

with the following definition of flux

\begin{equation} \bar{q} + \tau\frac{\partial \bar{q}}{\partial t}= -k\nabla T \label{eqH2} \end{equation}

where \(T\) is temperature, \(\bar{q}\) is heat flux, \(t\) is time, \(\rho\) is density, \(c_{p}\) is specific heat, \(k\) is heat conductivity, \(\tau\) is thermal relaxation time, and \(S\) is source. Eqs. \ref{eqH1} and \ref{eqH2} can be combined to yield:

\begin{equation} \tau\rho c_{p}\frac{\partial^2 T}{\partial t^2} + \rho c_{p}\frac{\partial T}{\partial t} = k\nabla^2 T + S + \tau\frac{\partial S}{\partial t} \label{eqH3} \end{equation}

Obviously, when \(\tau \rightarrow 0 \), Eqs. \ref{eqD3} and \ref{eqH3} return to their classical form, which you can see here

Note that the diffusion equation and the heat equation have the same form when \(\rho c_{p} = 1\).

1) One-dimension

Click here to see how to solve the problem with the following initial condition and boundary conditions:

2) Two-dimensions

Click here to see how to solve the problem with the following initial condition and boundary conditions:

3) Three-dimensions

Click here to see how to solve the problem with the following initial condition and boundary conditions:

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