Harmonic Conjugates: Order Matters!

July 16, 2011

I was skimming over Brown’s Complex Analysis book, and I came across a neat little exercise I thought I would share.  The solution is not difficult — indeed, it is just a manipulation of equations — but the idea is interesting and especially telling about the strange kinds of not-so-symmetric things that go on in complex analysis. 


Terms and Things.

Let’s quickly define some concepts and terms; if you already know what analytic, Cauchy-Riemann, and Harmonic mean, then skip to the next section.  Complex functions can be written as f(x,y) = u(x,y) + v(x,y)i where $u,v$ are real functions.  Notice that little i in the corner there next to the v function; this makes v the imaginary part of the function f.  If a function is analytic at some point then it is expressible as a power series in a neighborhood of that point.  There are a few ways to check to see if a function is analytic, but the one we will be using is the following:


Theorem.  A function f:{\mathbb C}\rightarrow{\mathbb C} is analytic at some point (a,b) if when f(x,y) = u(x,y) + v(x,y)i we have that the partial derivative u_{x}, u_{y},v_{x},v_{y} exist in a neighborhood of (a,b) and are continuous at (a,b), and u_{x} = v_{y} and u_{y} = -v_{x} at (a,b) (i.e., that the function satisfies the Cauchy-Riemann Criteria).


So, this looks really complicated and all, but it turns out to be relatively easy in practice.  For example,


Example.  f(x,y) = x^{2} - y^{2} + 2xyi.  Then u(x,y) = x^{2} - y^{2} and v(x,y) = 2xy.  Therefore, we have that u_{x} = 2x, u_{y} = -2y, v_{x} = 2y and v_{y} = 2x.  This is seen to satisfy the CR Criteria as above, and we note that all of the partials exist everywhere and are continuous everywhere; this implies f is actually analytic everywhere on the complex plane!  Nice.


Additionally, we have the notion that a function u(x,y) is Harmonic if we have that u_{xx} + u_{yy} = 0, where these are the second partials of u (that is, take u and differentiate it with respect to x twice to get u_{xx}). 


One cool thing is that every analytic function f(x,y) = u(x,y) + iv(x,y) is that both u and $v$ are harmonic.  We can prove this pretty easily.  Note by the CR equations (and Fubini) that

u_{xx} = \frac{d}{dx}(u_{x}) = \frac{d}{dx}(v_{y}) = v_{yx}

= v_{xy} = \frac{d}{dy}v_{x} = -\frac{d}{dy}u_{y} = -u_{yy}

and hence u_{xx} + u_{yy} = u_{xx} - u_{xx} = 0


On the other hand, given some harmonic function u(x,y), is there some analytic function f(x,y) such that u(x,y) is the real part, and there is some v(x,y) which is the imaginary part of f?  If such a v(x,y) exists to make an analytic f(x,y) with u(x,y), then we say that v(x,y) is the harmonic conjugate of u(x,y).  Notice the order here!  Can we also say that u is the harmonic conjugate of v in this case?  This will be the main point of the next section.


Does Order Matter for Harmonic Conjugates?

In the last section, we noted that if u(x,y) was harmonic, then if we found some v that is the imaginary part of an analytic function f = u + iv then v is the harmonic conjugate of u.  But what does it mean to say this?  It means that, in particular,

u_{x} = v_{y}

u_{y} = -v_{x}

by the Cauchy-Riemann equations.  Now suppose that we said that u is the harmonic conjugate of v.  This means there is some function g = v + ui (Notice which is the real and which is the complex part!) which is analytic.  What does this mean?  In particular, it says that,

v_{x} = u_{y}

v_{y} = -u_{x}

by the Cauchy-Riemann equations.


Thus, what does it mean if u is a harmonic conjugate of v and v is a harmonic conjugate of u?

It says, by the above two equations, that u_{x} = v_{y} = -u_{x} and u_{y} = -v_{x} = -u_{y}, which implies u_{x} = u_{y} = 0.  Similarly, v_{x} = v_{y} = 0.  Hence, by integrating, we must have that u(x,y) and v(x,y) must be constant (and therefore f(x,y) must also be constant). 


Thus, we can throw away the order of harmonic conjugate pairs if and only if we only care about constant functions.  In other words, yeah, order matters.

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