Uniform bar

An infinitely long, uniform bar is perfectly insulated and one end is kept at a constant temperature ($T_s$). How does its temperature evolve in time and space?

Because the bar's cross-section is negligible W.R.T. its length, we can treat this as a quasi-$\text{1D}$ problem and Fourier's equation applies. We're looking for a function $T(x,t)$ that satisfies:

$$\partial_t T=\alpha \partial_{xx} T$$

and any boundary/initial conditions.

In addition we assume the bar's initial ($t=0$) temperature is $T_i$.

My question is inspired by this PSE question. It was initially framed for a constant heat flux $\mathbf{q}$ at $x=0$ but somewhat later the OP seems to have settled for the simpler case I presented above.

For a bar of finite length $L$ this becomes a fairly banal BVP (a second BC was used where the heat conduction at $x=L$ is $0$) with initial condition, and with solution of the form:

$$\boxed{T(x,t)=T_s+\displaystyle\sum_{n=1}^{\infty} B_n\exp\Big({-\alpha \Big(\frac{\pi n }{2L}\Big)^2 t}\Big)\cos\Big(\frac{\pi n x}{2L}\Big)}$$ For $n=1,2,3,...$ Determine the coefficients $B_n$ with the Fourier series: $$B_n=\frac{2}{L}\int_0^LT_i\cos\Big(\frac{\pi n x}{2L}\Big)\mathrm{d}x$$

For a very long bar, i.e. $L\to \infty$, this tends to the question the OP asked.

At one point he then declared in the comments that he'd found the answer to his question on a page titled '2.2. Transient Conduction in Semi-Infinite Slab' (find the link provided by G.Smith in the comments). It's a very poor quality resource that posits the solution without any derivation whatsoever as being:

$$\boxed{\frac{T(x,t)-T_s}{T_i-T_s}=\mathrm{erf}\Big(\frac{x}{2\sqrt{\alpha t}}\Big)}$$

This looks like an extremely neat solution to a complex problem (at least to me) and I haven't the slightest idea how it was arrived at or whether it is even correct. The equation is dimensionally consistent, always a good sign of course.

On way forward could be to compare the solution for the the finite bar and the web page's solution for ever increasing values of $L$ and see if they converge. But that seems like an awful amount of work.

So my question really is: 'does anyone have a(n) idea(s) on how to tackle the $L=\infty$ problem?'

This page does confirm the validity of the $\mathrm{erf}$ solution.

  • 3
    $\begingroup$ To check the answer, just substitute it into the differential equation and BCs, and see if it satisfies them. This is a so-called similarity solution or boundary layer solution applicable to a case in which there is no identifiable length scale. The way of solving a problem like this is to start with de-dimensionalizing the differential equation and BCs, to there point that there is a length scale that can be cancelled by combining two of the dimensionless independent variables into a single similarity variable. See Bird et al, Transport Phenomena. $\endgroup$ Mar 26, 2021 at 17:26
  • $\begingroup$ Ah. The LHS and the group $\frac{x}{\sqrt{\alpha t}}$ are indeed dimensionless groups. $\endgroup$
    – Gert
    Mar 26, 2021 at 17:33
  • $\begingroup$ You’ve linked to a file on a C drive. $\endgroup$
    – G. Smith
    Mar 26, 2021 at 17:46
  • $\begingroup$ @G.Smith Thanks! There seems to be no way to link this as a web page. I've provided the full title, so it will be easy to Google. $\endgroup$
    – Gert
    Mar 26, 2021 at 17:58
  • $\begingroup$ Are you talking about this PDF? $\endgroup$
    – G. Smith
    Mar 26, 2021 at 19:19

1 Answer 1


With thanks to 'Chet' in the comments it appears this problem (at least without convection) isn't really hard.

$$\partial_t T=\alpha \partial_{xx}T$$ Boundary conditions: $$T(x,0)=T_i$$


Introduce a Similarity Variable $\eta$:

$$\eta=\frac{x}{2\sqrt{\alpha t}}$$ This transforms the original PDE into an ODE: $$\frac{\mathrm{d}^2T(\eta)}{\mathrm{d}\eta^2}=-2\eta \frac{\mathrm{d}T(\eta)}{\mathrm{d}\eta}\tag{1}$$ Transform also the boundary conditions: $$T(x,0)=T_i\to T(\eta\to \infty)=T_i$$ $$T(0,t)=T_s\to T(\eta=0)=T_s$$ This then solves to: $$\frac{T(x,t)-T_s}{T_i-T_s}=\mathrm{erf}(\eta)$$ $$\boxed{\frac{T(x,t)-T_s}{T_i-T_s}=\mathrm{erf}\Big(\frac{x}{2\sqrt{\alpha t}}\Big)}$$

Solving $(1)$ -

Substitute $z=T'(\eta)$ and $z'=T''(\eta)$, so that:

$$z'=-2\eta z$$ $$\frac{\mathrm{d}z}{z}=-2\eta$$ $$\ln z=-\eta^2+c$$ $$z=c_1\exp{(-\eta^2)}$$ $$T'(\eta)=c_1\exp{(-\eta^2)}$$ Integrate again: $$T(\eta)=c_1\int \exp{(-\eta^2)}+c_2$$

Apply the BC and IV to determine $c_1$ and $c_2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.