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=(vx−uy)(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).
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=vyx−vxy=0.
Therefore u is harmonic. We can handle v similarly.
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.
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.
- 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′=ux−iuy. So, let's define
g=ux−iuy.
This is our candidate for f′. - Show that g(z) is analytic:
Write g=ϕ+iψ, where ϕ=ux and ψ=−uy. Checking the Cauchy-Riemann equations we have
[ϕxϕyψxψy]=[uxxuxy−uyx−uyy]
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. - 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.) - Show that the real part of f is u.
Let's write f=U+iV. So, f′(z)=Ux−iUy. By construction
f′(z)=g(z)=ux−iuy.
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.
- Find a candidate, call it g(z), for f′(z):
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
If u and v are the real and imaginary parts of an analytic function, then we say u and v are harmonic conjugates.
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.