Undergrad Derivative and Parameterisation of a Contour Integral

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SUMMARY

The forum discussion centers on evaluating a contour integral defined as $$\Omega \equiv \oint_{\Omega} \mathbf{f}(\mathbf{s}) \cdot \mathrm{d}\mathbf{s}$$ along a surface parameterized by local curvilinear coordinates ##\mathbf{s}(u,v)##. The author explores the implications of parameterizing the integral, discretizing it, and taking the derivative with respect to a continuous parameter ##q##. The discussion highlights the importance of the order of operations in differentiation and parameterization, ultimately revealing that the results differ based on the sequence of these actions. The author seeks clarification on the correct approach to ensure a rigorous justification of their method.

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  • Familiarity with parameterization techniques in calculus
  • Knowledge of Leibniz's rule for differentiation under the integral sign
  • Basic concepts of vector calculus and curvilinear coordinates
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aphirst
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As part of the work I'm doing, I'm evaluating a contour integral:
$$\Omega \equiv \oint_{\Omega} \mathbf{f}(\mathbf{s}) \cdot \mathrm{d}\mathbf{s}$$
along the border of a region on a surface ##\mathbf{s}(u,v)##, where ##u,v## are local curvilinear coordinates, and where the surface itself is part of a family of surfaces ##\mathbf{s}(u,v,q)##, with ##q## being a continuous parameter.

The border ##\mathbf{s}## is actually the set of solutions to an equation (which I won't bother to specify, but you can take for granted has solutions of the required form) in ##\mathbf{s}(u,v)## (at each ##q## held constant). ##\mathbf{f}(\mathbf{s})## is shorthand for a more complex expression which depends entirely on ##\mathbf{s}## and other unstated constants (i.e. it doesn't depend on ##u,v,q## other than through ##\mathbf{s}##.

In my work I do three things to this expression:
  1. I explicitly parameterise the contour integral into a definite integral in ##\lambda \in [\lambda_1, \lambda_2]##
  2. I discretise the contour integral (I actually have a "trick" for my case which let's me apply it directly to the contour form)
  3. I take the derivative of this w.r.t. the aforementioned ##q##
It occurred to me that I should be able to get the same result no matter the order I do them. Step 2. is only relevant if you want to numerically evaluate this for a specific set of cases, so the question is "1. then 3." or "3. then 1.".

If I parameterise then differentiate, I get the following:

$$\begin{align}
\Omega &= \int_{\lambda_1}^{\lambda_2} \left( \mathbf{f}(\mathbf{s}) \cdot \frac{\mathrm{d}\mathbf{s}}{\mathrm{d}\lambda} \right) \mathrm{d}\lambda \\
\frac{\mathrm{d} \Omega}{\mathrm{d} q} &= \frac{\mathrm{d}}{\mathrm{d} q} \int_{\lambda_1}^{\lambda_2} \left( \mathbf{f}(\mathbf{s}) \cdot \frac{\mathrm{d}\mathbf{s}}{\mathrm{d}\lambda} \right) \mathrm{d} \lambda \\
\mathrm{(Leibniz'~rule)} &= \int_{\lambda_1}^{\lambda_2} \frac{\mathrm{d}}{\mathrm{d} q} \left( \mathbf{f}(\mathbf{s}) \cdot \frac{\mathrm{d}\mathbf{s}}{\mathrm{d}\lambda} \right) \mathrm{d} \lambda \\
\mathrm{(Product~rule)}&\Rightarrow \int_{\lambda_1}^{\lambda_2} \left( \frac{\mathrm{d} \mathbf{f}}{\mathrm{d} q}\cdot\frac{\mathrm{d}\mathbf{s}}{\mathrm{d}\lambda} + \mathbf{f}\frac{\mathrm{d}}{\mathrm{d} q} \cdot \frac{\mathrm{d} \mathbf{s}}{\mathrm{d} \lambda}\right) \mathrm{d} \lambda \\
\mathrm{(Integration~by~parts)} &\Rightarrow \int_{\lambda_1}^{\lambda_2}\left(\frac{\mathrm{d} \mathbf{f}}{\mathrm{d} q}\cdot\frac{\mathrm{d}\mathbf{s}}{\mathrm{d}\lambda}\right)\mathrm{d}\lambda + \left[ \mathbf{f} \frac{\mathrm{d}\mathbf{s}}{\mathrm{d}q}\right]_{\lambda_1}^{\lambda_2} - \int_{\lambda_1}^{\lambda_2} \left( \frac{\mathrm{d}\mathbf{s}}{\mathrm{d} q} \cdot \frac{\mathrm{d}\mathbf{f}}{\mathrm{d} \lambda}\right) \mathrm{d}\lambda \\
\left( \mathbf{s}(\lambda_1) \equiv \mathbf{s}(\lambda_2)\right) &\Rightarrow \int_{\lambda_1}^{\lambda_2} \left( \frac{\mathrm{d} \mathbf{f}}{\mathrm{d} q}\cdot\frac{\mathrm{d}\mathbf{s}}{\mathrm{d}\lambda} - \frac{\mathrm{d}\mathbf{s}}{\mathrm{d} q} \cdot \frac{\mathrm{d} \mathbf{f}}{\mathrm{d} \lambda}\right) \mathrm{d} \lambda
\end{align}$$
where I stop, since it seems clear that from here you need to use the actual form of ##\mathbf{f}(\mathbf{s})##.

However, if I try to differentiate then parameterise:
$$\begin{align}
\frac{\mathrm{d} \Omega}{\mathrm{d} q} &\equiv \frac{\mathrm{d}}{\mathrm{d} q} \oint_{\Omega} \mathbf{f}(\mathbf{s}) \cdot \mathrm{d} \mathbf{s} \\
\mathrm{(Leibniz'~rule)} &= \oint_{\Omega} \frac{\mathrm{d}}{\mathrm{d} q} \left( \mathbf{f}(\mathbf{s}) \right) \cdot \mathrm{d} \mathbf{s}
\end{align}$$
I'm straight away suspicious of whether I'm on the right track, since if I try to parameterise from here:

$$
\frac{\mathrm{d} \Omega}{\mathrm{d} q} = \int_{\lambda_1}^{\lambda_2} \left( \frac{\mathrm{d} \mathbf{f}}{\mathrm{d} q} \cdot \frac{\mathrm{d} \mathbf{s}}{\mathrm{d} \lambda} \right) \mathrm{d}\lambda
$$

I'm obviously missing a whole term from the alternate approach; assuming that that was right.

I'm pretty sure that in the second approach, the ##\mathrm{d}\mathbf{s}## is supposed to be affected by the ##\frac{\mathrm{d}}{\mathrm{d}q}## but I'm not sure how to write that as a valid contour integral (prior to explicit parameterisation).

To me this is only of academic importance - since I convert directly from the contour to a discretisation, and differentiate that - and I've been able to independently verify that what I'm doing is correct. However, if there's anyone who can point out the flaw in my reasoning, and/or recommend how best to express this process, I'd very much appreciate it, since it means I'd be able to make a clearer, more water-tight case for what I "actually" do, and can avoid "sloppy" justifications in the future.
 
Last edited:
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It's probably better to talk explicitly in terms of ##\mathbf{s}(u,v,q)## and ##\mathbf{f}(u,v,q)##, and restate that the path "##\mathrm{d}\mathbf{s}##" is actually a path through ##u,v## (where the path itself depends on ##q## due to ##q##'s impact on ##\mathbf{s}(u,v,q)## and on the unstated equations defining the location of the path). The expression "##\mathbf{f}(\mathbf{s})##" probably just leads to unnecessary confusion.
 
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