How to solve this question on harmonic functions?

How can I determine $a$ so that the given function is harmonic, and find its harmonic conjugate?

$$u = e^{\pi x}\cos(av)$$

Where $v$ is itself a real valued function of x,y.

Is there any other method than using Laplace Equation and taking double derivative and solving the equation as it tends to become too complicated?

• $v$ is a real valued function of what?. – Nosrati Aug 16 '17 at 18:44
• Is $v$ harmonic? – Robert Lewis Aug 16 '17 at 18:58
• There must be some hypothesis on $v$; if $v = y$, $a = \pi$ makes $u$ harmonic, but if $v = x$ there is no solution for $a$. – Robert Lewis Aug 16 '17 at 19:01
• And in light of my last comment, $v$ harmonic is not the right assumption. – Robert Lewis Aug 16 '17 at 19:03
• @RobertLewis v is a function of x and y. – Sherry Aug 17 '17 at 16:05

Here is my attempt.

Use definition of Laplace equation. Solutions of Laplace's equation are harmonic functions.

$f(x,v)=e^{\pi x}cos(av)$

$\Delta f=\frac{\partial^2f }{\partial x^2}+\frac{\partial^2f }{\partial v^2}=0$

After taking second partials and plugging them into Laplace we get:

$\pi ^{2}cos(av)e^{\pi x}-a^{2}e^{\pi x}cos(av)=0$

$e^{\pi x}cos(av)(\pi ^{2}-a^{2})=0$

$a=\pm \pi$

• Your answer is correct but are you considering v to be independent of x? – Sherry Aug 17 '17 at 16:31
• Yes, its the only way that I can make sense of the problem. – Erock Brox Aug 18 '17 at 18:07