# 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:

# 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.

• @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? Commented May 4 at 9:51
• Please don't use rainbow plots, they're difficult to read and often don't even convey what you want it to convey. Commented May 4 at 12:49
• @KyleKanos Wow, I didn't know that. Commented May 4 at 12:53
• Also, what are your boundary conditions? Commented May 4 at 12:56
• @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 Commented May 4 at 13:25

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.

• This site is NOT a forum/thread. Commented May 4 at 9:56
• @VincentThacker Right. I guess this is the answer to my question. Commented May 4 at 9:58
– 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.

• I enjoy studying your post. Commented May 4 at 20:31