Heat equation

This is the heat equation:

$ \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} $

Now let's assume $\frac{\partial u}{\partial t}=-1$

Then the equation would become:

$ -\Delta u = 1 $

Solving PDE

The equation $-\Delta u = 1$ is solved here. It can be solved for a variety of input 2D or 3D meshes.

For example, consider input to be a 2D square disc with a hole in it:


Result screenshot

Question: how to justify the result

Static solution to an implicitly dynamic problem

Now, I cannot justify the final results. What do the results mean?

The assumption of $\frac{\partial u}{\partial t}=-1$ on heat equation would mean that the temperature is going down with a linear slope/rate of -1 then we are getting a static final heat distribution. The final distribution does not vary with time. It's static. What does it mean? Why the temperature is not reducing with a linear slope pf -1? I know the FEM tool we used doesn't consider the time variation. But what's the physical meaning of getting a static solution to a problem that is implicitly dynamic?

I know this might be a stupid question to ask. But I guess I'm missing something here.

  • $\begingroup$ @Jon I think for the $u=f(x,y,z,t)$ when I set $\frac{\partial u}{\partial t}=-1$ then I am assuming $u=f(x,y,z) - t$ , right? $\endgroup$
    – Megidd
    Commented May 4 at 9:51
  • 1
    $\begingroup$ Please don't use rainbow plots, they're difficult to read and often don't even convey what you want it to convey. $\endgroup$
    – Kyle Kanos
    Commented May 4 at 12:49
  • $\begingroup$ @KyleKanos Wow, I didn't know that. $\endgroup$
    – Megidd
    Commented May 4 at 12:53
  • 1
    $\begingroup$ Also, what are your boundary conditions? $\endgroup$
    – Kyle Kanos
    Commented May 4 at 12:56
  • $\begingroup$ @KyleKanos Good point. While solving the PDE, we are assuming homogeneous Dirichlet boundary conditions corresponding, for example, to zero temperature on the whole boundary. Taken from here: mfem.org/tutorial/fem $\endgroup$
    – Megidd
    Commented May 4 at 13:25

2 Answers 2


For the 3D space, generically, I have:

$ u=f(x,y,z,t) $

As pointed out by @Jon when I set:

$ \frac{\partial u}{\partial t}=-1 $

$ \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} = -1 $

Actually, I'm making an assumption here:

$ u=f(x,y,z) - t $

So, for the 2D example, the resulted graphs are displaying the $f(x,y)$ part of the $u$:

$ u=f(x,y) - t $

Therefore, the displayed graphs are initial temperatures at $t=0$

Am I right? Looks like I got the answer to my question.

  • 1
    $\begingroup$ This site is NOT a forum/thread. $\endgroup$ Commented May 4 at 9:56
  • $\begingroup$ @VincentThacker Right. I guess this is the answer to my question. $\endgroup$
    – Megidd
    Commented May 4 at 9:58
  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented May 4 at 10:02

Some interpretations that may be useful:

  • The transient conductive heat equation says that if the temperature is decreasing in some region $\left(\frac{dT}{dt}<0\right)$, then the net curvature of the temperature profile must be downward in that region $\left(\nabla^2 T<0\right)$. (Memorably, in 1D, a frown cools things down, whereas a smile warms things up. Equilibrium corresponds to a straight line connecting fixed-temperature boundaries or satisfying fixed-flux boundaries.)

  • A constant curvature could arise with a large temperature gradient that decreases somewhat locally, or it could feature a gentle temperature gradient near a maximum where the gradient is zero.

  • Reflecting this, near a constant-temperature boundary with higher temperatures in the interior, we'd expect a large temperature gradient to direct all the heat from the interior into the boundary. Conversely, in the interior, a gentler gradient may be suitable, as heat has more directions to travel. Near a maximum, all heat flow is outward, so the gradient is relatively small.

Given all this, what temperature profile must exist to produce a uniform rate of temperature decrease everywhere at that moment? One that exhibits a constant net downward curvature: A steep temperature increase near boundaries, leveling off to symmetric maxima in interior regions. Which is what your numerical simulation shows. It's a static image because you've removed any temporal dependence by replacing the time-containing term with a constant.

  • $\begingroup$ I enjoy studying your post. $\endgroup$
    – Megidd
    Commented May 4 at 20:31

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