I have been exploring finite differences and heat transfer using the 2D heat equation to further expand my knowledge. So far I think it is going well.

I am running into some confusion around grid spacing for the finite difference method.

Basically if I have a metal plate measuring 100mm x 100mm it would seem natural to establish a grid spacing of 1mm which would yield 100 x 100 nodes or so (I haven't fully explored the different grid spaces yet to see what works best). If I run the simulation I get results I would expect after a certain time given the initial conditions. At time, t, the system has evolved to a certain state.

Now, I want to try a larger metal plate made of the exact same material with the exact same properties. The only difference is that it is larger, 1m x 1m. For this plate I would establish the grid spacing at 1cm yielding the same number of notes, 100 x 100. I might be naive here but I can't see how the scale would affect the finite differences. I would expect the plate to take longer to reach the same state (or pretty close, provided the initial conditions and boundary conditions were the same) as the smaller plate. But from what I see in my calculations is that it doesn't. I might be missing something and would appreciate any pointers to the relevant literature.


For the purposes of this question there are constant heat sources defined at the boundaries, nothing fancy. It really is a very basic problem definition. I had set the $\alpha$ value to 1 do that I could investigate the algorithm. I didn't pay attention to the units that it was defined with.

I think my problem is in regards to the unit that are used to define the spacing. In the initial problem I had the units in my mind as millimeter. I really should have paid closer attention to the units used in defining the heat capacity, material density and thermal conductivity. They were defined with meters. To me this would imply that I need to define the grid spacing in terms of meters. This would give me the difference that I am looking for. For example the first plate of 100mm x 100mm would have a spacing of 1mm = 0.001m. The second plate of 1m x 1m would have a spacing of 1 cm = 0.01m.

Is this logic correct?

  • 1
    $\begingroup$ Hi Bluebill and welcome to scicomp! It would help if you completely define the problem that your are studying. What exactly are your boundary conditions? Do you have any heat sources? $\endgroup$
    – Paul
    Commented Dec 10, 2016 at 21:33
  • $\begingroup$ @Paul, there are constant heat sources at the boundaries. But my question was about the grid spacing and how it would scale to different size plates of exactly the same material and boundary conditions, so I didn't include them. Conceptually I was confused by the units that I was using. I should have looked more closely at the units with regards to the heat capacity, density and thermal conductivity of the material. They are all defined using the meter. To me that means that my grid spacing should be defined in meters. I think that helps me understand how to set the spacing better. $\endgroup$
    – Bluebill
    Commented Dec 11, 2016 at 14:00
  • $\begingroup$ @Paul, I updated my question with a better description of my issue. $\endgroup$
    – Bluebill
    Commented Dec 11, 2016 at 14:08

2 Answers 2


Let's assume the following heat transient heat transfer equation in 1D : $$ \frac{\partial T} {\partial t} = \alpha \frac{\partial^2 T}{\partial x^2} $$

If we take its finite difference approximation using an explicit time scheme we obtain : $$ \frac{T_i^{t+\Delta t} - T_i^{t}}{\Delta t} = \frac{\alpha}{(\Delta x)^2} (T_{i-1}^{t}-2T_i^{t}+T_{i+1}^{t}) $$

So as you see, the diffusion you will obtain in a time step will be proportional to : $ \frac{\Delta t\alpha}{(\Delta x)^2}$. As such, if you increase the size of your domain, then $\Delta x$ will be larger (if you use the same number of points). Therefore, in a single time step, the diffusion of the temperature will be less pronounced because the diffusion term of your finite difference scheme will be smaller.

  • $\begingroup$ I had thought much the same thing. I realized what my problem was. I was confusing units. Taking a look at the $\alpha$ which contains the heat capacity, thermal conductivity and density. All of which are defined by meters. To make the grid spacing correct I would need to specify it in meters so all of the units line up? That would solve my conceptual problem, I think. I'll have to play around with it a little more. $\endgroup$
    – Bluebill
    Commented Dec 11, 2016 at 13:54
  • 2
    $\begingroup$ You do not need to define it in meters, you just need to use coherent units. $\alpha$ is $Length^2/Time$. Therefore, there is a length scale in the definition of $\alpha$. So if you take $\alpha$ in m^2/s, then yes, your entire grid must be defined in meter. However, you can define alpha is whatever unit you want, and then the same units should be applied to your grid. For instance, $\alpha$ could be in $mm^2/s$, then your grid would have to be defined in $mm$, etc. Usually we try to either render the equation dimensionless or use SI units. $\endgroup$
    – BlaB
    Commented Dec 11, 2016 at 14:39

A very good understanding about functioning of $dt$, $dx$ and $\alpha$ in discretized PDE can be developed if you look into the stability analysis for the heat equation. Here is the link for one such tutorial about that. I hope it will help.


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