Lecture 25

**Remark:** \(v\) is harmonic conjugate of \(u\) \(\nRightarrow\) \(u\) is a
harmonic conjugate of \(v\).

**Example:** \(u(x,y) = x^2-y^2\),
\(v(x,y) = 2xy\).
Since \(u+iv = z^2\) is entire \(\implies\) \(v\) is a harmonic conjugate of \(u\).

But if \(u\) were a harmonic conjugate of \(v\), then \(g = v+iu\) would be analytic: Check via C/R, nowhere analytic.

**Remark:** Suppose \(u\) is harmonic on a
simple connected domain \(\Omega\). Then \(u\) has a harmonic conjugate on \(\Omega\).

"Physical" configurations often modelled by solutions of partial differential equations (PDE).

Generally, interested in solving a PDE subject to associated initial/boundary conditions.

\( \text{(D) } \left\{ \begin{array}{cc} \Delta u = 0 & \text{in } \Omega\\ u|_{\partial \Omega} = \varphi & (*) \end{array} \right. \)

\((*)\) Says \(u(\mathbf x)= \varphi(\mathbf x)\) for all \(\mathbf x \in \partial \Omega\)

(\( \varphi: \partial \Omega \ra \R\)).

\(\Omega, \varphi\) are known/given, and \(u\) is the unknown.

\(\text{(D)}\) is called the *Dirichlet problem* for Laplace's equation, a. k. a.
*boundary problem of the first kind.*

One way to "solve" \(\text{(D)}\) is to find \(u\) that minimises \[\int_{\Omega} \left|\nabla u\right|^2 d\mathbf x\] subject to \(u|_{\partial \Omega} = \varphi\).

Also important: Boundary conditions of the second kind, called *Neumann boundary
conditions*

\( \text{(N) } \left\{ \begin{array}{cc} \Delta u = 0 & \\ \ds\frac{\partial u}{\partial \mathbf \nu} = \Psi & \text{on } \partial \Omega \end{array} \right. \)

In practice often have homogeneous Neumann boundary conditions, i. e., \(\Psi = 0\).

**Theorem:** If \(f\) is conformal and \(h\) is harmonic in
\(\Lambda\), then \(H\) is harmonic in \(\Omega\), where
\[H(x,y) = h\left(u(x,y), v(x,y)\right).\]

**Proof:** Mesy in general, but straightforward if
\(\Lambda\) is simply connected.

**Example:** \(h(x,y) = e^{-v}\sin u\) is harmonic in UHP (upper half plane).

Define \(w=x^2\) on \(\Omega =\) 1st quadrant.

**Example (cont):** So \(u = x^2 -y^2\) and \( v = 2xy\).

The previous Theorem implies \[H(x,y) = e^{-2xy}\sin \left(x^2-y^2\right). \] is harmonic on \(\Omega\).