Derivative and Parameterisation of a Contour Integral

In summary, the conversation discussed evaluating a contour integral along the border of a region on a surface parameterized by local curvilinear coordinates. The surface is part of a family of surfaces, and the border is the set of solutions to an equation. The expression to be evaluated depends on the surface and other constants, and the process for evaluating it involves parameterizing the contour integral, discretizing it, and taking the derivative with respect to a continuous parameter. The question was whether the order of these steps matters, and it was concluded that it does not.
  • #1
aphirst
Gold Member
25
5
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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  • #2
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.
 

1. What is a contour integral?

A contour integral is a type of line integral that is evaluated along a path in the complex plane. It is used to calculate the area under a curve in the complex plane, and is often used in complex analysis and physics.

2. How is a contour integral parameterized?

A contour integral is parameterized by finding a function that maps the path of the integral onto a parameter such as time or distance. This function is then used to calculate the value of the integral at each point along the path.

3. What is the derivative of a contour integral?

The derivative of a contour integral is the rate of change of the integral with respect to the parameter used in the parameterization. It is calculated using the chain rule and is useful in solving problems involving contour integrals.

4. How is parameterization used in contour integration?

Parameterization is used in contour integration to simplify the calculation of the integral. By parameterizing the path of the integral, it can be broken down into smaller, more manageable sections that are easier to evaluate.

5. What are some real-world applications of contour integrals?

Contour integrals have many real-world applications, including calculating electric fields in physics, calculating fluid flow in engineering, and solving problems in complex analysis and number theory. They are also used in computer graphics to create smooth curves and surfaces.

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