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Mathematics LibreTexts

6.3: Del notation

( \newcommand{\kernel}{\mathrm{null}\,}\)

Here’s a quick reminder on the use of the notation . For a function u(x,y) and a vector field F(x,y)=(u,v), we have

(i)=(x,y)(ii)gradu=u=(ux,uy)(iii)curlF=×F=(vxuy)(iv)divF=F=ux+vy(v)div grad u=u=2u=uxx+uyy(vi)curl grad u=×u=0(vii)div curl F=×F=0

Analytic Functions have Harmonic Pieces

The connection between analytic and harmonic functions is very strong. In many respects it mirrors the connection between ez and sine and cosine.

Let z=x+iy and write f(z)=u(x,y)+iv(x,y).

Theorem 6.3.1

If f(z)=u(x,y)+iv(x,y) is analytic on a region A then both u and v are harmonic functions on A.

Proof

This is a simple consequence of the Cauchy-Riemann equations. Since ux=vy we have

uxx=vyx.

Likewise, uy=vx implies

uyy=vxy.

Since vxy=vyx we have

uxx+uyy=vyxvxy=0.

Therefore u is harmonic. We can handle v similarly.

Note

Since we know an analytic function is infinitely differentiable we know u and v have the required two continuous partial derivatives. This also ensures that the mixed partials agree, i.e. vxy=vyx.

To complete the tight connection between analytic and harmonic functions we show that any har- monic function is the real part of an analytic function.

Theorem 6.3.2

If u(x,y) is harmonic on a simply connected region A, then u is the real part of an analytic function f(z)=u(x,y)+iv(x,y).

Proof

This is similar to our proof that an analytic function has an antiderivative. First we come up with a candidate for f(z) and then show it has the properties we need. Here are the details broken down into steps 1-4.

  1. Find a candidate, call it g(z), for f(z):
    If we had an analytic f with f=u+iv, then Cauchy-Riemann says that f=uxiuy. So, let's define
    g=uxiuy.
    This is our candidate for f.
  2. Show that g(z) is analytic:
    Write g=ϕ+iψ, where ϕ=ux and ψ=uy. Checking the Cauchy-Riemann equations we have
    [ϕxϕyψxψy]=[uxxuxyuyxuyy]
    Since u is harmonic we know uxx=uyy, so ϕx=ψy. It is clear that ϕy=ψx. Thus g satisfies the Cauchy-Riemann equations, so it is analytic.
  3. Let f be an antiderivative of g:
    Since A is simply connected our statement of Cauchy’s theorem guarantees that g(z) has an antiderivative in A. We’ll need to fuss a little to get the constant of integration exactly right. So, pick a base point z0 in A. Define the antiderivative of g(z) by
    f(z)=zz0g(z) dz+u(x0,y0).
    (Again, by Cauchy’s theorem this integral can be along any path in A from z0 to z.)
  4. Show that the real part of f is u.
    Let's write f=U+iV. So, f(z)=UxiUy. By construction
    f(z)=g(z)=uxiuy.
    This means the first partials of U and u are the same, so U and u differ by at most a constant. However, also by construction,
    f(z0)=u(x0,y0)=U(x0,y0)+iV(x0,y0),
    So, U(x0,y0)=u(x0,y0) (and V(x0,y0)=0). Since they agree at one point we must have U=u, i.e. the real part of f is u as we wanted to prove.
Important Corollary

u is infinitely differentiable.

Proof

By definition we only require a harmonic function u to have continuous second partials. Since the analytic f is infinitely differentiable, we have shown that so is u!

Harmonic conjugates

Definition: Harmonic Conjugates

If u and v are the real and imaginary parts of an analytic function, then we say u and v are harmonic conjugates.

Note

If f(z)=u+iv is analytic then so is if(z)=v+iu. So, if u and v are harmonic conjugates and so are u and v.


This page titled 6.3: Del notation is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Jeremy Orloff (MIT OpenCourseWare) via source content that was edited to the style and standards of the LibreTexts platform.

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