275- Harmonic functions and harmonic conjugates

Recall that a function u(x,y) of two variables defined on an open domain D is harmonic iff  u is C^2 (i.e., all four second order derivatives u_{xx},u_{xy},u_{yx},u_{yy} exist and are continuous in D), and u satisfies Laplace equation


As mentioned in problem 6 of the Fall 2008 Calculus III final exam, a function v is a harmonic conjugate of u iff v is defined on D, v_x and v_y exist, and the Cauchy-Riemann equations hold:

u_x=v_y and u_y=-v_x.

It follows immediately from the Cauchy-Riemann equations that if v is a harmonic conjugate of a harmonic function u, then v is also C^2, with v_{xx}=-u_{yx}, v_{xy}=-u_{yy}, v_{yx}=u_{xx} and v_{yy}=u_{xy}. It is also immediate that v satisfies Laplace equation because v_{xx}+v_{yy}=-u_{yx}+u_{xy}=0, since continuity guarantees that the mixed partial derivatives commute. Thus v is also harmonic.

In fact, modulo continuity of the second order derivatives, the harmonic functions are precisely the functions that (locally) admit harmonic conjugates.

To see this, assume first that u is C^2 in D and that it admits a harmonic conjugate v. Then u_{xx}=v_{yx} and u_{yy}=-v_{xy} so u_{xx}+u_{yy}=0 and u was harmonic to begin with.

Conversely, assume that u is harmonic in D. Suppose first that D is (connected and) simply connected. I claim that then u admits a harmonic conjugate v in D. To see this, letting {\mathbf F}=(-u_y,u_x), notice that the existence of v is equivalent to the claim that {\mathbf F} is a gradient vector field, since {\mathbf F}=\nabla v iff v is a harmonic conjugate of u. But, since D is simply connected, then {\mathbf F} is a gradient iff it is conservative, i.e., \displaystyle \oint_\gamma {\mathbf F}\cdot d{\mathbf r}=0 for any simple piecewise smooth loop \gamma in D. Fix such a \gamma, and let R denote its interior. Then, by Green’s theorem,

\displaystyle \oint_\gamma {\mathbf F}\cdot d{\mathbf r}=\pm\iint_R u_{xx}+u_{yy}\,dA=0,

where the \pm sign is to be chosen depending on the orientation of \gamma. It follows that {\mathbf F} is indeed conservative and therefore a gradient, so u admits a harmonic conjugate.

Finally, if D is not simply connected, we cannot guarantee that such a v exists in all of D, but the argument above shows that it does in any open (connected) simply connected subset of D, for example, any open ball contained in D.  That we cannot extend this to all of D follows from considering, for example, u(x,y)=\log(x^2+y^2) in D={\mathbb R}^2\setminus\{(0,0)\}. This is a harmonic function but it does not admit a harmonic conjugate in D, since there is no continuous \arctan(y/x) in D. This example can be easily adapted (via a translation) to any non-simply connected D.

I close by remarking that, as mentioned in my previous post on average values of harmonic functions, one can use Green’s theorem to prove that harmonic functions u satisfy the average (or mean) value property, and this property characterizes harmonicity as well, implies that u is actually C^{\infty} (i.e., u admits partial derivatives of all orders, and they are all continuous) and has the additional advantage that it only requires that u is continuous, rather than C^2. Similarly, one can show that the Cauchy-Riemann equations on D suffice to guarantee that u and v are harmonic (and in particular, C^\infty). However, one needs to require that the equations hold everywhere on D. A pointwise requirement would not suffice. But I won’t address this issue here (I mention it in the notes in complex analysis that I hope to post some day).


One Response to 275- Harmonic functions and harmonic conjugates

  1. jabbar says:

    Iam abdul jabbar from Iraq.This site is very good to understand complex variables.thanks alot

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