What are the differences between Heat equations and Poisson Equations?

Question

Am fairly new into heat equations and wanted to have some clarifications. What are the distinguishing features between the heat equation and the Poisson equation?

Answer 1

The heat equation describes a process of diffusion of heat in time. $u(x,t)$ represents there the temperature of your body at time $t$ and location $x$ (given some initial/boundary conditions and some source of heat). Poisson equation is about a stationary or equilibrium distribution of temperature, that is, you are looking for a function of $x$ alone that describes the temperature assuming it is not changing anymore. Poisson's equation describes the limit situation, when the heat is not flowing anymore (given some boundary conditions and sources).

If you start heating a body, its temperature is described by the heat equation but for very long times, the initial distribution of temperature is forgotten and your solution resembles more and more the solution of the corresponding Poisson equation (with the same sources and boundary conditions)

Answer 2

Quickly first, Laplace's equation has the form

$$\Delta u(\vec{x}) = 0.$$

Now the heat equation is of the form

$$\frac{\partial u}{\partial t} - c^2 \Delta u = 0$$

for some $c \in \mathbb{R}$. It's a little different from Laplace's equation in that it contains the time derivative.

Poisson's equation is, again, a little different from Laplace's equation in that it is nonhomogeneous. Poisson's equation is

$$-\Delta u(\vec{x}) = f(\vec{x}).$$

Some main distinctions between the heat equation and Poisson's equation are that the heat equation is a parabolic equation while Poisson's equation is elliptic. The heat solution is time-dependent whereas the Poisson solution is not. That is, the heat equation seeks solutions which are self-smoothing in time where solutions to Poisson's equation are statically fixed in space. A solution to the heat equation eventually reaches an equilibrium where $u_t$ is essentially zero. At this point, the heat solution also satisfies Laplace's equation. In Poisson's equation, $f(\vec{x})$ represents a heat distribution, and if $f \equiv 0,$ then Poisson's equation reduces to Laplace's equation. Of course, the solutions for all of them depend on the domain and initial/boundary conditions.

Answer 3

A Possion Eqn. has the form:

$$f(x)=\nabla^2\phi$$

which says: "There is something, $f(x)$, which depends on the rate of change of the local neighborhood of something else."

The Heat Eqn. has the form:

$$\frac{\partial\phi}{\partial t}=\nabla^2\phi \text{ (where }\phi\text{ is heat)}$$

which says: "The something is heat, and the rate of change of heat $\frac{\partial \phi}{\partial t}$ depends on the rate of change of its own local neighborhood.". This is what it means when we say heat moves (changes) from areas of hot to areas of cold (neighborhood).

If we understand that the $f(x)$ in Poisson Eqn. represents the "thing to solve for", and we substitute $f(x)=\frac{\partial\phi}{\partial t}$, then:

$$f(x)=\frac{\partial\phi}{\partial t}=\nabla^2\phi$$

In other words, the Heat Eqn. is a special case of the Poisson Eqn. in which the "thing to solve for" is the rate of change of heat. The Poisson Eqn. is a generalization that could solve for many kinds of things, including gravity (f=gravitational force=4 pi G), electrostatics (f=charge=-p/e), fluid pressure (f=pressure), or heat (f=rate of change of heat=∂Φ/∂t)

We can see how the Poisson Eqn. is general because there are two somethings, φ and f, one usually known and the other unknown. These are two different somethings. In the Heat Eqn., however, both these somethings are heat: φ. What does this mean? That the change of heat depends on the current heat itself. Remember again: "Heat flows from areas of hot to cold". So, heat moves based on itself. That is a pretty specific kind of motion not true of other physical things. Heat Eqn is a specific form of the Poisson Eqn.

Answer 4

Ok, there are several things I'll address here. Firstly, Poisson's equation is generally stated as $$f(\mathbf{r})=\nabla^2\Phi$$ Where $\mathbf{r}$ is a vector. You incorrectly stated in your post that $f(\mathbf{r})$ is the thing we are trying to solve for. This is not true. This would make the problem trivial - knowing $\Phi$, all we would have to do is take its Laplacian and we're done. When dealing with Poisson's equation we know what $f(\mathbf{r})$ is and we are solving a second order PDE to determine $\Phi.$ Notice that both $f$ and $\Phi$ are of the same "type" - that is, they both deal with a spacial quantity, the position vector, $\mathbf{r}$.

To contrast, the heat equation is $$\frac{\partial u}{\partial t}=\alpha\nabla^2u$$ The inclusion of the dimensional constant $\alpha$ is important, since the RHS deals with a spacial quantity whereas the LHS deals with a temporal quantity. So no, the heat equation is not a special case of Poisson's equation. In Poisson's equation, the LHS is a function of space and in the Heat equation the LHS is a function of space and time. That is, in Poisson's equation, we are looking for a time-independent solution in the form $\Phi(\mathbf{r})$ whereas in the heat equation we are looking for a time dependent solution of the form $u(\mathbf{r},t)$.