Analytical solutions of Diffusion and Heat equations

Feb 7, 2017 by Hugo Milan

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

We define the diffusion equation in time-domain as:

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

with the following definition of flux

\begin{equation} \bar{q} = -D\nabla C \end{equation}

where \(C\) is concentration, \(\bar{q}\) is concentration flux, \(t\) is time, \(D\) is diffusivity, and \(S\) is source.

And we define the heat equation in time-domain as:

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

with the following definition of flux

\begin{equation} \bar{q} = -k\nabla T \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, and \(S\) is source.

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

Cartesian coordinates

1) One-dimension

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

$$\begin{matrix} C(y, t = 0) = C_{c} & \text{ or } & T(y, t = 0) = T_{c}\\ C(y = 0, t) = C_{c} & \text{ or } & T(y = 0, t) = T_{c}\\ C(y = H, t) = C_{s} & \text{ or } & T(y = H, t) = T_{s} \end{matrix}$$

2) Two-dimensions

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

$$\begin{matrix} C(x,y, t = 0) = C_{c} & \text{ or } & T(x,y, t = 0) = T_{c}\\ C(x,y = 0, t) = C_{c} & \text{ or } & T(x,y = 0, t) = T_{c}\\ C(x,y = H, t) = C_{s} & \text{ or } & T(x,y = H, t) = T_{s}\\ \dfrac{\partial C}{\partial x} (x = 0, y, t) = 0 & \text{ or } & \dfrac{\partial T}{\partial x}(x = 0, y, t) = 0\\ \dfrac{\partial C}{\partial x}(x = L, y, t) = \dfrac{q_x}{D} & \text{ or } & \dfrac{\partial T}{\partial x}(x = L, y, t) = \dfrac{q_x}{k} \end{matrix}$$

3) Three-dimensions

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

$$\begin{matrix} C(x,y, z,t = 0) = C_{c} & \text{ or } & T(x,y,z, t = 0) = T_{c}\\ C(x,y = 0,z, t) = C_{c} & \text{ or } & T(x,y = 0,z, t) = T_{c}\\ C(x,y = H,z, t) = C_{s} & \text{ or } & T(x,y = H, z,t) = T_{s}\\ \dfrac{\partial C}{\partial x} (x = 0, y, z,t) = 0 & \text{ or } & \dfrac{\partial T}{\partial x}(x = 0, y, z,t) = 0\\ \dfrac{\partial C}{\partial x}(x = L, y, z,t) = \dfrac{q_x}{D} & \text{ or } & \dfrac{\partial T}{\partial x}(x = L, y, z,t) = \dfrac{q_x}{k}\\ \dfrac{\partial C}{\partial z} (x, y, z = 0,t) = 0 & \text{ or } & \dfrac{\partial T}{\partial z}(x, y, z=0,t) = 0\\ \dfrac{\partial C}{\partial z}(x, y, z = T_z,t) = \dfrac{q_z}{D} & \text{ or } & \dfrac{\partial T}{\partial z}(x, y, z = T_z,t) = \dfrac{q_z}{k} \end{matrix}$$

Cylindrical coordinates

1) One-dimension

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

$$\begin{matrix} C(r, t = 0) = C_{c} & \text{ or } & T(r, t = 0) = T_{c}\\ C(r = 0, t) = C_{c} & \text{ or } & T(r = 0, t) = T_{c}\\ C(r = R, t) = C_{s} & \text{ or } & T(r = R, t) = T_{s} \end{matrix}$$

Spherical coordinates

1) One-dimension

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