I'm confused on the notation with regards to partial derivatives and points of evaluation. In a Taylor series, the operations are applied to a function, then the point of evaluation is applied. For example:
$ u(x,t) = u|_{x=a} + \Delta x \frac{\partial u}{\partial x}|_{x=a} + \frac{(\Delta x)^2}{2} \frac{\partial^2 u}{\partial x^2}|_{x=a} + \ldots$
If I now want to apply an operator, $\frac{\partial }{\partial x}$, to the series I believe I have to treat those points of evaluation as constants. The rational is, for example, $u|_{x=a} = u(a,t)$ is no longer a function of $x$ and thus the derivative should view that term as a constant. If I carry out this logic on the above series I get
$ \frac{\partial u}{\partial x} = \frac{\partial u}{\partial x}|_{x=a} + \Delta x \frac{\partial^2 u}{\partial x^2}|_{x=a} + \ldots$
which is in agreement with the series expansion for the derivative had I started there.
If instead of the operator, $\frac{\partial }{\partial x}$, I want to apply an operator $\frac{\partial }{\partial t}$, things change (or so I believe). In this case $u|_{x=a} = u(a,t)$ is still a function of $t$ and therefore those terms should not be treated as constants.
$ \frac{\partial u}{\partial t} = \frac{\partial u|_{x=a}}{\partial t} + \Delta x \frac{\partial }{\partial t} \left(\frac{\partial u}{\partial x}|_{x=a}\right) + \Delta x \frac{\partial }{\partial t} \left(\frac{\partial^2 u}{\partial x^2}|_{x=a}\right) + \ldots$
Assuming the above is correct, this highlights my questions.
- Focusing on the first term ... is it wrong to say $\frac{\partial u|_{x=a}}{\partial t} = \frac{\partial u}{\partial t}|_{x=a}$ ? I'm confused on when the "evaluate at" must stay within the operator and when it is free to be applied after (assuming it can be).
- The second term emphasizes this confusion. Here I believe I'm not allowed to say $\frac{\partial u}{\partial x}|_{x=a} = \frac{\partial u|_{x=a}}{\partial x}$ because this is how the Taylor series is defined...as discussed prior.
- In the second term, am I allowed to pull in the time derivative such that $\frac{\partial }{\partial t} \left(\frac{\partial u}{\partial x}|_{x=a}\right) = \frac{\partial^2 u}{\partial t \partial x}|_{x=a}$, where the point of evaluation is applied after the second derivative is taken? I don't see how the two would differ.
I could carry on with the confusion, but I think I've made my point. Is there any freedom with the point of evaluation (i.e. apply before/after operators) or ways to manipulate it? Is there an order of operations rule in play here? If so, can you point me to a reference for further info? I've been reading some papers and it seems like they are rather loose with when "the point of evaluation" is being applied. Since they don't show all their steps, it is hard to tell whose is the one making the mistake.
EDIT:
I've been working more with these types of problems and I believe I have shed some light on an answer, albeit I'm new to this so I'm still hesitant.
Anyway, what I've found is that it depends on how $a$ is defined. For example, if $a$ is a constant, then you can push/pull the point of evaluation in/out as you please. However, if for example $a(x)$, then it does matter. This is shown via the chain rule. For example
$\displaystyle u|_{x=a(x)} \equiv u(a(x),t)$
If I define a vector function $\textbf{f}(x,t) = <a(x),t>$, then the expression
$\displaystyle u(a(x),t) \equiv u(\textbf{f}(x,t)) \equiv (u\circ \textbf{f})(x,t)$
The total derivative (by the chain rule) becomes
$\displaystyle \left[\frac{\partial u|_f}{\partial x} \quad \frac{\partial u|_f}{\partial t}\right] = \textbf{D}u|_{f}\textbf{D}f$
in which $\textbf{D}$ is the derivative operator. The expressions are
$\displaystyle \textbf{D}u|_f = \left[ \frac{\partial u}{\partial x}|_{f}\quad \frac{\partial u}{\partial t}|_{f}\right]$
$ \textbf{D}f = \begin{bmatrix} &\dfrac{\partial a}{\partial x} &\dfrac{\partial a}{\partial t} \\ &\dfrac{\partial t}{\partial x} &\dfrac{\partial t}{\partial t} \end{bmatrix} = \begin{bmatrix} &\dfrac{\partial a}{\partial x} & 0 \\ &0 &1 \end{bmatrix} $
Multiplying the two matrices gives
$\displaystyle \frac{\partial u|_f}{\partial x} = \frac{\partial u}{\partial x}|_{f}\dfrac{\partial a}{\partial x}$
and
$\displaystyle \frac{\partial u|_f}{\partial t} = \frac{\partial u}{\partial t}|_{f}$
Since the notation $|_f \equiv |_{x=a}$ applies, we can see that the interchange can only occur if the point of evaluation is a constant with respect to the variable of differentiation.
So, to address my points
- I think $\frac{\partial u|_{x=a}}{\partial t} = \frac{\partial u}{\partial t}|_{x=a}$ if $a$ is not a function of $t$.
- Again, it depends on how $a$ is defined.
- It's still a bit unclear, but I do know the function argument $\frac{\partial u}{\partial x}|_{x=a}$ is a function of $t$. The fact that $a$ is not makes me believe I can interchange the two, but I'