# Differential Operator Notation

1. Jun 14, 2013

### muzialis

I am finding it unintuitive to follow a calculation in a certain notation I am not too familair with.
To write down the equation
$$z(t) = - y(t) + \tau \frac{\partial y}{\partial t}$$ the following notation is employed
$$z(t) = -(1-\tau \frac{\partial }{\partial t}) y$$
So far so gud. But then, the following happens
$$\frac{ z(t)}{1-\tau \frac{\partial }{\partial t}} = - y$$
I am not making much sense of this.
The author continues by noting that
$$\frac{ 1}{1-\tau \frac{\partial }{\partial t}} \approx 1 + \tau \frac{\partial }{\partial t}$$
which moves the time derivative to the function z.
I am unsure on the procedure, considering operators as objects on which even division works, can anybody shed any light?

Many thanks

2. Jun 14, 2013

### HallsofIvy

Staff Emeritus
You are right, that is very unintuitive and what some professors I have had would call "abuse of terminology" (perhaps too abusive!)

What is happening here, and the reason why the "fraction like" notation $df/dx$ is so evocative is that while df/dx is NOT a fraction, it is the limit of a fraction and so can be treated like one (to prove that, in a particular calculation, you can treat a derivative as a fraction, go back before the limit, apply the fraction operation to the "difference quotient", then take the limit again). The introduction of "differentials", df=(df/dx)dx, is another attempt to make use of the fact that the derivative can be "treated like a fraction".

Some texts use "operator notation" and would write your example as $$z(t)= -y(t)+ \tau Dy(t)$$ where "Dy(t))" is the derivative of y. Then write that as $$z(t)= (\tau D- 1)y(t)$$ and $$y(t)= \frac{z(t)}{\tau D- 1}$$, treating the "differential operator", D, as if it were a number.

Frankly, this sort of thing is done more often in Engineering texts or "Differential Equations for Engineers" where they are not terribly concerned with mathematical rigor.

3. Jun 14, 2013

### muzialis

HallsofIvy,

I understand the notation, what is puzzling me is ots very audacious use.
In the paper I am reading, by a physicist, the starting point is the ODE
$$q =- G + \frac{\partial G }{\partial t}$$ (nothing else than Fourier's Law, for the curious), presented in the particular notation we discussing as
$$q = - (1 - \frac{\partial }{\partial t}) G$$, or even
$$\frac{q}{(1 - \frac{\partial }{\partial t}) } = -G$$
Then, by noting that
$$\frac{1}{1-\frac{\partial}{\partial t}} \approx 1 + \frac{\partial}{\partial t}$$
the following ODE is obtained
$$\dot{q} + q = -G$$, well, I do not see how this and the initial ODE could be related in more rigorous way, the two seem very different to me. If for example the given q is identically zero, the first one is solved by an exponent, while the second one is solved by the zero constant function.
Apparently the physicist Cattaneo used this procedure in the late '40 to derive the hyperbolic heat equation.

Many thanks

4. Jun 14, 2013

### Mandelbroth

Actually, if we treat the curve satisfying the function's relation as a subset of the Cartesian product of its domain and codomain (as a submanifold of $\mathbb{R}^2$) and consider its differentials as covectors in the cotangent space, the derivative is the quotient of differentials. That's an idea from differential geometry, but it is still trivially and rigorously a fraction for a 1-manifold because the cotangent space is one dimensional, making all covectors scalar multiples of each other. I thought you knew that. :uhh:

Because the differential operator is a linear operator, it is acceptable to treat it as a matrix.

5. Jun 14, 2013

### micromass

Staff Emeritus
Just because $\mathbf{v} = a\mathbf{w}$ makes sense in a vector space, doesn't mean that we define $\frac{\mathbf{v}}{\mathbf{w}}$ in some way. Although you can do it, I have never seen a book define division of covectors or differential forms. It's just not a useful idea.

The problem is that the space is infinite dimensional. And thus matrices can no longer be used as such.

6. Jun 14, 2013

### Mandelbroth

I got this from here. Read mathwonk's comment.

I'm not saying we evaluate it using matrices. I'm saying it follows properties of matrices. Is that wrong?

7. Jun 14, 2013

### micromass

Staff Emeritus
I know you got it from there, but I don't think it's correct. No single book does it that way.

I'm just saying you need to be careful when extrapolating properties of matrices to properties of linear operators. Things will start to behave very strangely.

8. Jun 14, 2013

### muzialis

Gentlemen,

it is a truly interesting discussion, but I may I in parallel stress that the question posed, albeit probably far less intriguing, is still unanswered?

Thank you very much for all

9. Jun 14, 2013

### Mandelbroth

Maybe they just didn't have enough room in the margin. :tongue:

In all seriousness, I think it seems useful. We might think of it as applying a function of two vectors, defined as

$$f(\vec{v},\vec{w}) = \left\{\begin{matrix} \frac{\|\vec{v}\|}{\|\vec{w}\|}, & \exists \alpha\in\mathbb{R}: \vec{v}=\alpha\vec{w} \\ \operatorname{Undefined}, & \not\exists \alpha\in\mathbb{R}: \vec{v}=\alpha\vec{w} \end{matrix}\right.$$

Strangely? How? I'd like to see an example to understand this better, if you please. I'm assuming you know more about this than I do, Micro, so I'd like to see what you mean.

We may also consider treating $\partial/\partial t$ as the basis vector for the tangent space such that it, when multiplied by f, gives the directional derivative of f?

Last edited: Jun 14, 2013
10. Jun 14, 2013

### Mute

What's going on, basically, is an abuse of notation in which $\frac{1}{1-\frac{\partial}{\partial t}}$ is a shorthand for the inverse operator $L^{-1}= \left(1-\frac{\partial}{\partial t}\right)^{-1}$, where $L = 1 - \frac{\partial}{\partial t}$.

Under certain conditions, you can write the inverse operator as a power series, so what your reference is doing is writing

$$L^{-1} = 1 + \sum_{k=1}^\infty \left(\frac{\partial}{\partial t}\right)^k = 1 + \sum_{k=1}^\infty \frac{\partial^k}{\partial t^k},$$
and then keeping only the lowest order derivative term, presumably because the function q or G vary slowly and the higher derivatives can be neglected to first order.

I believe this approximation procedure is called (or at least related to) the Resolvent formalism, which comes up in the study of Green's functions (Here, the inverse operator $L^{-1}$ should be a Green's function, such that when you apply it to $q = -LG \Rightarrow L^{-1}q = -L^{-1}LG$, there is actually an integration involved $\int dt~ L^{-1} q = -\int dt~L^{-1}L G = -\int dt~\delta(t-t') G(t') = -G(t)$, so it's really $\int dt~ L^{-1} q \approx q + \dot{q}$.)

Edit: To make this clearer, hopefully, let's do this in Fourier space. The Fourier transformed equation

$$q(t) = -G(t) - \frac{dG}{dt}(t)$$
becomes
$$\hat{q}(\omega) = -\hat{G}(\omega) + i\omega \hat{G}(\omega).$$
Note that the time derivative effectively got replace by a factor of $i\omega$. We can solve this equation for $\hat{G}(\omega)$,

$$-\hat{G}(\omega) = \frac{\hat{q}(\omega)}{1-i\omega}.$$
Inverse Fourier transforming this will give $G(t)$. However, suppose the integration over the left hand side is hard to do, but, we expect that it is the low frequency components of $q(t)$ that dominate its behavior. Then, we can expand the denominator and write

$$-\hat{G}(\omega) \approx (1 + i\omega) \hat{q}(\omega).$$
This equation is much easier to inverse transform, and note that just as $d/dt$ became $i\omega$ in Fourier space, $i\omega$ will become $d/dt$ when back in the time domain, and so we find

$$-G(t) \approx q(t) + \frac{dq}{dt}.$$

So what the author did in the reference linked to in the first post was an abuse of notation which provided a shortcut to this result.

Last edited: Jun 14, 2013
11. Jun 14, 2013

### muzialis

That is really helpful, many heartily thanks!

12. Jun 14, 2013

### Staff: Mentor

The crux of the question, I believe, is that operator notation appears to indicate multiplication, but that isn't so.

I have revised what muzialis wrote to use ordinary derivative notation, rather than partial derivation notation, since I don't see any indication that the functions involved have more than one variable.
I believe that your confusion with the line above stems from a misconception of what the right side means. It does NOT mean $(1-\tau \frac{d}{dt})$ times y.

It's more akin to function notation where f(x) is read "f of x", not f times x. If we expand this expression we get 1(y) and $-\tau~d/dt(y)$. For the sake of simplicity, I am ignoring the initial minus sign.

The first should be interpreted to mean "1 of y", or better yet, "1 operating on y," with the idea being that "1" represents the identity operator.
Nor am I. The thing in the denominator is an operator (the identity operator minus $\tau$ times the derivative operator. You can't divide a number (z(t)) by an operator.

13. Jun 14, 2013

### Mute

I edited the post to also present the discussion in terms of Fourier transform solution methods, which may make the issue a bit clearer. I hope that also helps!

14. Jun 14, 2013

### lurflurf

If v=aw it makes complete sense to define a=v/w
The trouble is such a definition depends upon v and w being linearly dependent and as such cannot be extended to the whole space.
Division of covectors or differential forms is a useful idea. It is just not very wide reaching. The contents of the small amount of books you have seen is not relevant, nor is content of the presumably larger set of all books on earth.

Infinite matrices are and can be used. That is not important as a linear operator is still useful without a matrix representation. Often we can restrict to finite dimension and use a matrix if we like.

(1/A) is not an abuse of notation it is a notation. It means the same thing as A^-1. That is like the laughable argument that 3/5 does not exist it is just an abuse of (5^-1)3. I know it can seem like it, but rigor and writing things a dumb way are not the same.

These types of manipulations are useful and common in subjects like umbral calculus, differential equations, algebra, operator theory, discrete math, and applied math. They can be made rigorous, but the more common and more useful approach is to preform the manipulation and take the result to be tentative. Later obtain rigor more directly and easily by checking if the result is true. The value of the manipulations is that correct results are easily derived that might not be obvious using other methods.

15. Jun 14, 2013

### lurflurf

So the extent to which
$$\frac{ 1}{1-\tau \frac{\partial }{\partial t}} \approx 1 + \tau \frac{\partial }{\partial t}$$
In which as usual multiplicative notation denotes operator composition.
Depends closely upon the extent to which
$$1 \approx \left( 1-\tau \frac{\partial }{\partial t} \right) \left( 1 + \tau \frac{\partial }{\partial t} \right)=\left( 1-\tau^2 \frac{\partial^2 }{{\partial t}^2}\right)$$
As mentioned previously including more terms from
$$\frac{ 1}{1-\tau \frac{\partial }{\partial t}} \approx \sum_{k=0}^n \tau^k \frac{\partial^k }{{\partial t}^k}$$
can often provide more accuracy.