Analytic functions · Definition

Harmonic function

complex analysisharmonic functions

Before you start you need to know

The real and imaginary parts of a holomorphic function are harmonic, which ties complex analysis to electric potential, steady-state heat and ideal fluid flow.

Definition

A real function \(u\) of two variables, with continuous second partial derivatives on an open set \(U\) of the plane, is harmonic if its Laplacian \(\Delta u\) is zero throughout \(U\).

\[\Delta u=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0\]

The affine functions \(ax+by+c\), \(x^2-y^2\), \(xy\), \(e^x\cos y\) and \(\ln\sqrt{x^2+y^2}\) (the last one on \(\mathbb{C}\setminus\{0\}\)) are harmonic. By contrast \(x^2+y^2\) has Laplacian \(4\). Sums and multiples of harmonic functions are harmonic, but products generally are not: \(u=v=x\) are harmonic while \(uv=x^2\) is not.

If \(f=u+iv\) is holomorphic, Cauchy-Riemann gives \(u_x=v_y\) and \(u_y=-v_x\). Differentiating once more, \(u_{xx}=v_{yx}=-u_{yy}\), that is \(\Delta u=0\), and likewise \(\Delta v=0\). This uses that \(f\) has derivatives of all orders, a fact proved later with Cauchy's integral formula.

Conversely, if \(u\) is harmonic on a simply connected open set, there is a harmonic \(v\), unique up to a constant, with \(u+iv\) holomorphic; it is called the harmonic conjugate of \(u\). It is found by integrating: \(v_y=u_x\) determines \(v\) up to a function of \(x\), and \(v_x=-u_y\) fixes that function. For \(u=x^2-y^2\) one gets \(v=2xy\) and \(f=z^2\).

The topological hypothesis matters: \(u=\ln|z|\) is harmonic on \(\mathbb{C}\setminus\{0\}\), but its conjugate would be \(\arg z\), which is not single-valued on that whole domain. Harmonic functions also satisfy the mean value property and have no strict interior local maxima, consequences seen later on.

Think about it

Is \(u=x^3-3xy\) harmonic?

No: \(u_{xx}=6x\) and \(u_{yy}=0\), so \(\Delta u=6x\), which is not zero. The harmonic one is \(x^3-3xy^2\).

Which holomorphic function has real part \(e^x\cos y\)?

\(f(z)=e^z\), since \(e^{x+iy}=e^x\cos y+ie^x\sin y\); the harmonic conjugate is \(e^x\sin y\).

Exercises 10

References

  • Elias M. Stein, Rami Shakarchi, Complex Analysis, 1st edition, Princeton University Press, 2003. Cap. 1, ejercicio 11
  • James Ward Brown, Ruel V. Churchill, Complex Variables and Applications, 9th edition, McGraw-Hill, 2014. §26–27
  • Lars V. Ahlfors, Complex Analysis, 3rd edition, McGraw-Hill, 1979. Cap. 4, §6