← 2017 Paper 2

UPSC 2017 Maths Optional Paper 2 Q3b — Step-by-Step Solution

15 marks · Section A

Harmonic functions and harmonic conjugate · Complex Analysis · asked 7× in 14 yrs · Read the full method →

Question

Let $f=u+iv$ be an analytic function on the unit disc $D=\{z\in\mathbb C:|z|<1\}$. Show that

at all points of $D$.

Technique

Differentiate the Cauchy–Riemann equations and add/subtract; invoke $C^\infty$ smoothness of analytic functions to equate mixed partials (Schwarz), giving $\nabla^2u=\nabla^2v=0$.

Solution

The claim is that $u$ and $v$ are harmonic. The two ingredients are: (1) an analytic function satisfies the Cauchy–Riemann equations, and (2) an analytic function is infinitely differentiable, so its real/imaginary parts have continuous second-order partials, justifying equality of mixed partials.

Step 1 — Cauchy–Riemann equations

Since $f=u+iv$ is analytic on $D$, at every point of $D$ the Cauchy–Riemann (CR) equations hold:

Step 2 — Smoothness justifies mixed-partial equality

A function analytic on an open set is infinitely complex-differentiable there; in particular $u$ and $v$ have continuous partial derivatives of all orders on $D$. By Clairaut/Schwarz’s theorem, continuity of the second partials gives equality of mixed partials:

Step 3 — $u$ is harmonic

Differentiate (CR-1) with respect to $x$ and (CR-2) with respect to $y$:

Add and use mixed-partial equality of $v$:

Step 4 — $v$ is harmonic

Differentiate (CR-1) with respect to $y$ and (CR-2) with respect to $x$:

Subtract the second from the first and use mixed-partial equality of $u$:

Conclusion

Answer