Contour Integrals

The basic differential relations for complex variables are the Cauchy - Riemann equations. For a function f(z) = u(x,y) + i v(x,y) of the complex variable z = x + i y,

$$\frac{\partial u}{\partial x} \frac{\partial v}{\partial y}$$ = (125)

$$\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y},$$ = (126)

whose extension to the requirement that u and v be harmonic has already been discussed in terms of finite differences. But the Cauchy - Riemann equations can be approximated directly in finite differences

$$\frac{u(x+\Delta x,y) - u(x,y)}{\Delta x} \frac{v(x,y+\Delta y) - v(x,y)}{\Delta y}$$ = (127)

$$\frac{v(x+\Delta x,y) - v(x,y)}{\Delta x} \frac {u(x,y) - v(x,y+\Delta y)}{\Delta y}$$ = (128)

Suppose that f has been defined along the real axis and that we want to know about its behavior in the upper half plane, y>0. Then

$$u(x,y+\Delta y) u(x,y) - \frac{\Delta y}{\Delta x} \left\{v(x+\Delta x,y)-v(x,y)\right\}$$ = (129)

which means that to get u on a line just above the real axis, it should be decremented everywhere by horizontal differences in v, weighted by the ratio of $\Delta y$ to $\Delta x$, which could be 1. The new u depends only on the old u and the v differences, both taken along the real axis. But similar results would apply to any contour.

Meanwhile v is changing according to

$$v(x,y+\Delta y) v(x,y) + \frac{\Delta y}{\Delta x} \left\{u(x+\Delta x,y)-u(x,y)\right\},$$ = (130)

to which similar comments apply.

Taken together, these two equations tell how to work up the positive y-axis strip by strip. Somehow that says that an analytic function is defined everywhere else by knowing its two halves on the real axis. Actually some kind of line drawn anywhere else ought to work, but this approach avoids worrying about corners and the like.

This is a kind of initial value problem, where both u and v are specified and are leapfrogged to get values somewhere else. But numerical values could be given for u and symbolic values for v. In the next strip, the v's would get numerical increments, the u's symbolic increments, but of first degree in v. Repeating, first degree expressions keep recurring until at last, in some final strip some exact values could be proposed. Then some linear equations woud have to be solved to find out what the values of the v's should have been to get the required values. The matrix just has +1's or -1's because of the sums and differences, and has to be raised to a power to get the number of parallel lines used in climbing.

Rather than flipping between u's and v's, the fact that each is harmonic can be used to work up the ladder of strips using averages. Then a row depends on two predecessors, so initial values and derivatives could be used (derivatives of u's gotten from values of v by using one of the Cauchy - Riemann equations). Also, instead of strips, think of any area, such as a rectangle. Points outside the rectangle aren't part of the differential equation but define averages, so one could try to accomodate the interior to a prescribed boundary.

Reasoning with finite differences means getting the matrices in detail and getting a good symbolic description. Usually ordinary calculus is applied to finding values at one place in terms of another.

First, line integrals.

In real analysis, an integral (at least a Riemann integral) is defined

$$\int_a^bf(x)dx \lim_{\Delta x \rightarrow 0} \sum_{i=0}^nf(x_i)\Delta x_i$$ = (131)

where an interval is subdivided and the function averaged over the points of subdivision. Much anguish can be expended over where to evaluate the function, how large to take the intervals, and so on. To do the same thing with an analytic function one could write

$$\int_a^bf(z)dz \lim_{\Delta z \rightarrow 0} \sum_{i=0}^nf(z_i)\Delta z_i$$ = (132)

with the proviso that $\sum \Delta z_i = b-a$ and subject to the ambiguity that many sequences of $\Delta z_i$'s lead from a to b. If that mattered, the integral would depend explicitly on the path. Thus the first order of business is to decide whether it does or it does not.

The dependence can be checked out beginning with a $\Delta x \Delta y$ rectangle. From the definition, we need

$$f(z)\Delta z (u+iv)(\Delta x + i \Delta y)$$ = (133)

$$(u\Delta x - v \Delta y) + i (u\Delta y + v\Delta x)$$ = (134)

With $\Delta z$ decomposed into its real and imaginary increments, consider the alternative orders in which they can be taken. In keeping with the spirit of Riemann integration, the function should be evaluated at the beginning of each interval of subdivision and not be allowed to retain the same value from one step to the next.

Using $\Delta x$, then $\Delta y$ we get

$$f(z)\Delta x + i f(z+\Delta x)\Delta y u\Delta x + i v\Delta x + i (u+\frac{\partial u}{\partial x}\Delta x) \Delta y - (v+\frac{\partial v}{\partial x}\Delta x) \Delta y.$$ = (135)

The other order, $\Delta y$, then $\Delta x$, leads to

$$i f(z)\Delta y + i f(z+\Delta y)\Delta x i u\Delta y - v \Delta y + (u+\frac{\partial u}{\partial y}\Delta y) \Delta x + i (v+\frac{\partial v}{\partial y}\Delta y) \Delta x.$$ = (136)

 

Figure 11: An integral can be shifted across a rectangle without changinging its value.

It is now a matter of comparing

$$\frac{\partial u}{\partial x} {\rm with} \frac{\partial v}{\partial y}$$ (137)

$$-\frac{\partial v}{\partial x} {\rm with} \frac{\partial u}{\partial y}$$ (138)

and the Cauchy - Riemann equations say that they are the same. Of course, this is based on one definition of a Riemann integral, in which the function at the left endpoint is multiplied by the length of the interval, and must be corrected at the midpoint if the interval is split in two. Otherwise terms proportional to the arc length will be neglected and the limiting sum will not be correct. In general, function values anywhere in the interval can be used in the definition, but first order effects will accrue if values from outside the interval creep in.

In general, the contour can be shifted across any rectangle for which a linear approximation to the function is valid, which would exclude any singularities or branch points - points where the function is not invertible and so making its definition suspect.

A consequence of path-shifting is that when the endpoint matches the initial point, the value of the integral is truly zero. Again the requirement is that the contour never has to cross over a singularity.

A useful side effect of the way Riemann integrals are defined is that the function can be defined in different ways in different places - just so long as the definitions overlap sufficiently that the Cauchy - Riemann equations hold for both definitions.

A common source of multiple definitions is to have two different Taylor's series with finite radii of convergence, centered on different points; the definitions can be compared where their disks of convergence overlap.

The vanishing of a closed contour integral is usually verified by using Green's formula, which is a useful technique in its own right. Note that the harmonicity condition makes the average over an area vanish, so that a double integral over a plane region should be zero except for boundary values which are not compensated in the sum.

  

Figure 12: A harmonic function can be extrapolated beyond a neighborhood as long as there is enough information to approximate a derivative.

So add a rim to any region and fill it with values to get the zero average. To make this work, the sum over the interior - an area integral - must be the negative of the sum of the boundary values - a contour integral. In the form needed for complex variable theory, suppose that u(x,y) and v(x,y) are two real valued functions such as the real and imaginary parts of an analytic function. Further suppose that it is desired to calculate

$${\int\int}_R \left(\frac{\partial u}{\partial y} - \frac{\partial v}{\partial x}\right) dx dy$$

and that is possible (although not explicitly shown in the formula) to parameterize x and y as functions of t. To avoid confusion, write functions $x(\alpha)$ and $y(\beta)$ using separate variables, and rewrite the integral

$${\int\int}_R \left(\frac{\partial u}{\partial y} - \frac{\p... ...}\right) \frac{dx}{d\alpha} \frac{dy}{d\beta} d\alpha d\beta$$

and then integrate. Note

$$\int_{\rm bottom}^{\rm top} \frac{\partial u}{\partial y}\frac{dy}{d\beta}d\beta u({\rm top}) - u({\rm bottom})$$ = (139)

$$\int_{\rm left}^{\rm right} \frac{\partial v}{\partial x}\frac{dx}{d\alpha}d\alpha u({\rm right}) - u({\rm left}).$$ = (140)

It remains to evaluate

$$\int\left[u({\rm top}) - u({\rm bottom})\right] dx$$

and to match it with

$$\int\left[v({\rm right}) - v({\rm left})\right] dy$$

to get

$${\int\int}_R \left(\frac{\partial u}{\partial y} - \frac{\partial v}{\partial x} \right) dx dy - \oint_{\partial R} \left( u dx + v dy\right)$$ = (141)

Then, by using the Cauchy - Riemann equations in the area integral to make it vanish, the vanishing of the contour integral follows.