I am looking to solve the heat conduction equation in a semi-infinite solid with in-depth radiation on the domain $-\infty < x < 0$. The governing equation of this problem is:

$$\rho c \frac{\partial T}{\partial t} = k \frac{\partial^2 T}{\partial x^2} + q_r \kappa \mathrm{e}^{\kappa x}$$ where $$T(x,0) = T_0\\ T(-\infty,t) = T_0\\ \frac{\partial T}{\partial x}(0,t) = 0$$

Making the substitution of $\theta = T - T_0$, $\alpha = k/(\rho c)$, and $\beta = q_r \kappa /(\rho c)$, the problem is simplified to:

$$\theta_t = \alpha \theta_{xx} + \beta \mathrm{e}^{\kappa x}$$ where $$\theta(x,0) = 0\\ \theta(-\infty,t) = 0\\ \theta_x(0,t) = 0$$

Applying the Laplace Transform in time ($\mathscr{L}\{\theta(x,t)\} = \Theta(x,s)$) to the equation yields:

$$ s \Theta = \alpha \Theta_{xx} + \frac{\beta \mathrm{e}^{\kappa x}}{s} $$ where $$\Theta(-\infty,s) = 0\\ \Theta_x(0,s) = 0$$

Splitting the solution into a homogeneous and particular solution and substituting $\tau = \sqrt{s/\alpha}$ yields:

$$ \Theta_H = c_1 \mathrm{e}^{\tau x} + c_2 \mathrm{e}^{-\tau x}\\ \Theta_P = c_3 \mathrm{e}^{\kappa x}\\ $$

Plugging the particular solution back into the ODE for $\Theta$ yields: $$ c_3 = \frac{\beta}{s (\alpha \kappa^2 - s)} $$

Solving the constant temperature boundary condition at $x=-\infty$ yields: $$ c_2 = 0 $$

Solving the zero gradient boundary condition at $x=0$ yields: $$ c_1 = \frac{\beta}{\tau s (s - \alpha \kappa^2)} $$

Combining all of the above into a single equation: $$ \Theta(x,s) = \Theta_H + \Theta_P\\ \Theta_H = \frac{\beta}{\tau s (s - \alpha \kappa^2)}\mathrm{e}^{\tau x} \\ \Theta_P = \frac{\beta}{s (\alpha \kappa^2 - s)} \mathrm{e}^{\kappa x} $$

At this point the solution is found by simply taking the inverse Laplace Transform of $\Theta_H$ and $\Theta_P$ separately.

$$ \mathscr{L}^{-1}\{\Theta_P\} = \beta \mathrm{e}^{\kappa x} \mathscr{L}^{-1}\{(s(\alpha \kappa^2 - s))^{-1}\} = \frac{(\mathrm{e}^{\alpha \kappa^2 t} - 1)}{\alpha \kappa}\beta \mathrm{e}^{\kappa x} $$

Now, what is the Inverse Laplace Transform of $\Theta_H$?

  • $\begingroup$ DOH! I'll fix that and update the post in the morning. Not at my computer at the moment and editing LaTeX/Stack Exchange on my phone is painful. $\endgroup$
    – LWhitson2
    Nov 7, 2014 at 2:15
  • 1
    $\begingroup$ The derivation should be correct now through to determining $\Theta_H$. $\endgroup$
    – LWhitson2
    Nov 7, 2014 at 14:08

1 Answer 1


Using Maple I am obtaining the following result for the inverse Laplace transform that you need:

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Please do right click on the image to enlarge it.

Then, the complete solution is:

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Please do right click on the image to enlarge it.