0
$\begingroup$

Page 11, Nonlinear Dynamics and Chaos, by Strogatz, says the following:

This is the domain of classical applied mathematics and mathematical physics where the linear partial differential equations live. Here we find Maxwell's equations of electricity and magnetism, the heat equation, Schrödinger's wave equation in quantum mechanics, and so on. These partial differential equations involve an infinite "continuum" of variables because each point in space contributes additional degrees of freedom. Even though these systems are large, they are tractable, thanks to such linear techniques as Fourier analysis and transform methods.

I'm wondering what is meant by the following:

These partial differential equations involve an infinite "continuum" of variables because each point in space contributes additional degrees of freedom.

"Each point in space contributes additional degrees of freedom"? We have three spacial dimensions, but since we're talking about an "infinite" continuum, I'm assuming he's referring to the infinite possible coordinates for these dimensions? I still don't understand why this infinite number of coordinates means that there are infinite variables? In mathematics, the number of variables has nothing to do with the number of possible coordinate combinations (which would be uncountably infinite, since there are an uncountably infinite number of real numbers).

Am I misinterpreting something here?

I would greatly appreciate it if people could please take the time to clarify this.

$\endgroup$
1
$\begingroup$

Partial differential equations solve for fields, that is, typically functions $\mathbb{R}^3\rightarrow \mathbb{R}$. How many scalar "variables" do you need to represent such a function?

What is the value of the field at point $(0,0,0)$ ? This is one variable.

What is the value of the field at point $(0,1,0)$ ? This is another variable.

What is the value of the field at point $(e,\sqrt{2},\pi)$ ? A third variable.

How many variables do we have in total, then? That would be the cardinality of the continuum, $\aleph_1$, hence a continuum of variables.

$\endgroup$
  • $\begingroup$ Thanks for the answer. What do you mean by "scalar variables to represent a function"? I don't understand the point you're trying to make by then giving examples of 3 arbitrary variables. Specifically, I don't see why we require an infinite continuum of variables to represent such a function. $\endgroup$ – The Pointer Jan 22 '19 at 10:33
  • 1
    $\begingroup$ The point is that you need a new variable for every point in space, of which I gave just three examples. $(0,0,0)$ is not a variable, it is a point at which a variable exists. The variable is the value of $f(0,0,0)$. $\endgroup$ – Wouter Jan 22 '19 at 11:32
  • $\begingroup$ Hmm, why is it that we need a new variable for every point in space? $\endgroup$ – The Pointer Jan 22 '19 at 11:37
  • 1
    $\begingroup$ Because the solution of a PDE is not a scalar variable (a number), but a function. Specifying a function is equivalent to specifying its values at every point - those values-at-every-point are the "variables" that your source material is talking about. $\endgroup$ – Wouter Jan 22 '19 at 12:00
1
$\begingroup$

See the answer to this question. For completeness, the relevant bit:

For PDEs, the situation is much more complex. This is because, whereas an ODE defines a flow in an $n$-dimensional phase space, a PDE defines a flow in a function space. I'm skipping a lot of very important and interesting details here, but the important thing is that function spaces are infinite dimensional. To make the connection with the ODE case, take the example of the advection equation, given by $$ c \frac{\partial u}{\partial t} + \frac{\partial u}{\partial x} = 0. $$ You can use the method of characteristics to write down the general solution to this PDE. The solution is given by $$ u(x,t) = f(x - c t), $$ where $f(x)$ is a function that plays the role of the initial condition, since $u(x,0) = f(x)$. So, you can pick any function $f(x)$ you like (again, skipping a couple of important and interesting details here), and you get a solution to the PDE. As there are, roughly speaking, infinitely many functions to choose from, you have infinitely many 'degrees of freedom' in picking an initial condition (initial function) for your PDE.

The order of the PDE mainly influences the particular function space it acts on, so the function space in which you can pick a function as initial condition. However, typically, all these function spaces are infinite dimensional -- in the sense that they are not finite-dimensional. The field of functional analysis addresses questions that arise in this context.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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