Say we have a bar centered at $x=0$, that is heated. We have the 1D Heat equation which we can solve to find a parabolic temperature profile :

$$\kappa \frac{d^2 T}{d x^2}=-\frac{Q}{L S}$$



$T_{\infty}$ is the ambient temperature at the end of the bar. This boundary condition alone is enough to solve the equation. However, now say we want to add a very tiny rod at the center of the bar that is connected to the environment to act as a path for heat flow. $R$ is the thermal resistance of the rod. $S$ is the cross sectional area of the bar. $\kappa$ is thermal conductivity of the system. $Q$ is the heat dissipated by joule heating. The corresponding boundary condition is:

$$\frac{T(0)-T_{\infty}}{R_{\rm thermal \ rod}}=\kappa S \left(\frac{d T}{dx}_{x=0^+}-\frac{d T}{d x}_{x=0^-} \right)$$

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I have fiddled around with this math for a while but cannot figure out how to implement it.

In particular I am following this paper on the T-Bridge Method for thermal conductivity.

  • $\begingroup$ cannot figure out how to implement it what is "it" and how have you been trying/thinking so far? $\endgroup$
    – Kyle Kanos
    Commented Jun 13, 2014 at 1:38
  • $\begingroup$ The first two equations I can solve - it yields a parabolic temperature profile. However coupling in the third, I can't solve. "it" is the third equation. $\endgroup$
    – Fire
    Commented Jun 13, 2014 at 2:12
  • $\begingroup$ A sketch would help. I do not understand your second case, especially not how you came to the boundary condition. Also, what is $S$, $R$, etc. I can guess, but I should not have to guess, because it is a source of misinterpretation $\endgroup$
    – Bernhard
    Commented Jun 13, 2014 at 7:16
  • $\begingroup$ Edited to include all this information. (its all in the referenced paper) $\endgroup$
    – Fire
    Commented Jun 13, 2014 at 11:05
  • $\begingroup$ If you include the bar, then I think you should the heat equation of the bar. $\endgroup$
    – jinawee
    Commented Jun 13, 2014 at 13:17

1 Answer 1


I think the problem is that while in the first case your differential equation applies to all your domain of interest and you can just use it, in the second situation the DE doesn't apply at x=0. This means you need to solve the DE at the domains $0<x<L/2$ and $-L/2<x<0$ separately, where it does still apply.

When applying it to those domains you should remember that the temperature $T$ is still continuous throughout all the domain because you don't want infinite $\frac{d T}{d x}$; that is, infinite heat fluxes. If you do that, after you solve the DE with $T(0)$ as an undeterminate boundary value, you should find that the right hand side of your third equation is a function of $T(0)$ and thus you can find it in terms of the problem constants $T_{\infty}$, $S$, $R_{\text{thermal rod}}$, etc.

  • $\begingroup$ This is what I was trying, but I can't seem to get anything other than $\kappa S \left(\frac{d T}{dx}_{x=0^+}-\frac{d T}{d x}_{x=0^-} \right)=0$. I am obviously making an incorrect assumption or handling the equations incorrectly. $\endgroup$
    – Fire
    Commented Aug 14, 2014 at 15:20
  • $\begingroup$ Why zero? The heat flows through the bar must match the heat flow through the rod as in your original post. The temperature profile in the rod is obviously linear, so the heat flow though the rod is $\propto T_0-T_\inf$. Then use the symmetry of the problem around $x=0$, the result is a Robin type of boundary condition for the bar: $2\partial_x T + T_0 = const$ With this it should be easy to find a solution of the form $T(x) = ax^2+bx+c$. $\endgroup$ Commented Aug 15, 2014 at 11:47

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