6.3: Del notation

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

\(\begin{array} {rcl} {(\text{i})} & & {\nabla = (\dfrac{\partial}{\partial x}, \dfrac{\partial}{\partial y})} \\ {(\text{ii})} & & {\text{grad} u = \nabla u = (u_x, u_y)} \\ {(\text{iii})} & & {\text{curl} F = \nabla \times F = (v_x - u_y)} \\ {(\text{iv})} & & {\text{div} F = \nabla \cdot F = u_x + v_y} \\ {(\text{v})} & & {\text{div grad } u = \nabla \cdot \nabla u = \nabla ^2 u = u_{xx} + u_{yy}} \\ {(\text{vi})} & & {\text{curl grad } u = \nabla \times \nabla u = 0} \\ {(\text{vii})} & & {\text{div curl } F = \nabla \cdot \nabla \times F = 0} \end{array}\)

Analytic Functions have Harmonic Pieces

The connection between analytic and harmonic functions is very strong. In many respects it mirrors the connection between \(e^z\) and sine and cosine.

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

Theorem \(\PageIndex{1}\)

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 \(u_x = v_y\) we have

\[u_{xx} = v_{yx}. \nonumber \]

Likewise, \(u_y = -v_x\) implies

\[u_{yy} = -v_{xy}. \nonumber \]

Since \(v_{xy} = v_{yx}\) we have

\[u_{xx} + u_{yy} = v_{yx} - v_{xy} = 0. \nonumber \]

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. \(v_{xy} = v_{yx}\).

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 \(\PageIndex{2}\)

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 \(f' = u_x - iu_y\). So, let's define
   \[g = u_x - iu_y. \nonumber \]
   This is our candidate for \(f'\).
2. Show that \(g(z)\) is analytic:
   Write \(g = \phi + i\psi\), where \(\phi = u_x\) and \(\psi = -u_y\). Checking the Cauchy-Riemann equations we have
   \[\begin{bmatrix} \phi_x & \phi_y \\ \psi_x & \psi_y \end{bmatrix} = \begin{bmatrix} u_{xx} & u_{xy} \\ -u_{yx} & -u_{yy} \end{bmatrix} \nonumber \]
   Since \(u\) is harmonic we know \(u_{xx} = -u_{yy}\), so \(\phi_x = \psi_y\). It is clear that \(\phi_y = -\psi_x\). 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 \(z_0\) in \(A\). Define the antiderivative of \(g(z)\) by
   \[f(z) = \int_{z_0}^{z} g(z)\ dz + u(x_0, y_0). \nonumber \]
   (Again, by Cauchy’s theorem this integral can be along any path in \(A\) from \(z_0\) to \(z\).)
4. Show that the real part of \(f\) is \(u\).
   Let's write \(f = U + iV\). So, \(f'(z) = U_x - i U_y\). By construction
   \[f'(z) = g(z) = u_x - iu_y. \nonumber \]
   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,
   \[f(z_0) = u(x_0, y_0) = U(x_0, y_0) + iV(x_0, y_0), \nonumber \]
   So, \(U(x_0, y_0) = u(x_0, y_0)\) (and \(V(x_0, y_0) = 0\)). 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 \(-v\).