Notes on complex analysis

We'll recall some definitions. Let \(A\) be a domain in \(\R^2\) (which we will identify with \(\C\) when in the analytic setting). A scalar field on \(A\) is a function \(u(x, y): A \to \R\) - in this context, we'll think of such a function as assigning a value to each point in \(A\text{.}\)

A vector field on \(A\) is a function \(V:A \to \R^2\) that assigns to each point in \(A\) a vector \(V(x,y) = (v_1(x,y), v_2(x,y))\text{.}\) If \(u\) happens to be continuously differentiable, then a natural vector field associated with \(u\) is given by the gradient field \(\nabla u = \operatorname{grad} u = \ip{u_x}{u_y}\text{.}\) Continuous gradient fields have the property that line integrals in the field are path independent (which should remind you of the Cauchy-Goursat theorem). A function \(u\) with continuous gradient field \(\nabla u\) is called a potential function.

There is useful geometry connecting potentials and their gradient fields. Recall that a level curve for \(u\) is the set of points \((x,y) \in A\) with \(u(x,y) = C\) for constants \(C \in \R\text{.}\) Given that the gradient vectors measure the direction of greatest change in a scalar field, it shouldn't be surprising that the gradient vector at a point in \(A\) is orthogonal to the level curve through the point. If the vector field varies continuously, we might expect that we can trace flow lines through \(A\text{.}\) A flow line (say parametrized by \(\gamma\)) should have the property that at every point \((x,y) \in A\text{,}\) \(\gamma\dd(t) = \nabla u(x,y)\text{.}\) Typically, the problem of computing flows lines explicitly is hard. However, for certain functions \(u\text{,}\) this problem turns out to be much easier.

In particular, if \(u\) is harmonic, the associated gradient field has some very nice features. First, since \(u\) is \(C^2\text{,}\) it isn't hard to calculate that the curl of \(\operatorname{grad} u\) is zero. (One can do this by extending \(\gd u\) to a function in \(\R^3\) by putting 0 in the \(z\) coordinate, computing \(\nabla \times \nabla u\text{,}\) and applying Clairaut's theorem.)

Because the gradient field arising from a harmonic is function is so nice, we get a magical solution to the flow line problem.

It turns out that there is a very nice solution to the Dirichlet problem in the case that the domain \(A\) is a disk. A famous result called the Poisson integral theorem gives us the desried solution. Harmonic functions have a mean value property much like Gauss's mean value theorem for analytic functions. We'll restrict our attention to the case of the unit disk (which implies the result for any other disk).

Poisson's formula extends the mean value property to give the value of \(u\) at every point in the disk in terms of the values of \(u\) on the boundary.

The proof of the theorem above goes essentially by applying the conformal automorphism of \(\D\) that takes \(z \to 0\text{.}\) Then, we apply the harmonic mean value property and chase through the effect on the integral of the conformal map.

The solution to the Dirichlet problem is basically a converse to Poisson's formula - it asserts that given a function \(u_0\) continuous on the unit circle \(T\text{,}\) we can construct a unique harmonic function \(u\) on \(\D\) that agrees with \(u_0\) on the boundary.

Note that while we have a standard solution to the Dirichlet problem on the disk for any continuous function \(u\) on the circle bounding the disk, we have much stricter requirements for our choice of \(u\) in the Neumann case.

Thus we consider boundary functions \(\phi\) with \(\int_\gamma \phi = 0.\) In the context of these notes, we actually will restrict ourselves to the simplest case, \(\partial u/\partial n = 0\text{,}\) which corresponds to zero-flux across the boundary, a condition known as an insulated boundary. This means that flows will be parallel the the boundary in this case (which corresponds to the simple case of uniform fluid flow, e.g.).

Our essential approach will be to construct a standard solution that describes some physical situation and then conformally map it to more interesting regions, which will also transform the solution.

The vector field \(V\) is the gradient of the potential function \(\phi = \alpha x\text{,}\) which in turn is the real part of the complex function \(f(z) = \alpha z\text{.}\) Since \(\phi\) is the real part of an analytic function, it is harmonic. Further, as the flow is parallel to the boundary, the Neumann condition \(\partial \phi/\partial n = 0 \) is met.

We'll need the following highly useful interaction between conformal maps and harmonic functions.

An exercise will ask you to evaluate this more carefully.

In the Dirichlet setting, we already have the Poisson integral formula (which can be applied to any domain conformally equivalent to a disk). However, this is often way more than necessary, as frequently boundary conditions are specified as constant on a finite number of pieces of the boundary. In this sort of case, a standard solution is most easily defined on the upper half plane.

We want to find a harmonic function \(u\) on the upper half plane with the boundary condition that \(u(x,0)=1\) when \(x \lt 0\) and \(u(x,0) = 0\) when \(x > 0\text{.}\)

Finally, we'll look an example application. In the study of heat, the harmonic function \(T\) on a domain represents temperature, and its level curves \(Tx, y) = c\) are called isotherms. The harmonic conjugate of \(T\text{,}\) which we can denote \(T\ad\) has level curves that represent the flux lines induced by the gradient field \(\nabla T\text{.}\) Our task is to find a harmonic function on the unit disk with the left half circle on the boundary held at temperature \(T = 0\) and the right circle held at \(T = 1\text{.}\)

Now, we can look at the level curves of \(\RE f\) and \(\IM f\) to detect the isotherms and flux lines for our function on the disk. The flux lines in the plane and disk setting are visualized in the figure below.