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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?

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

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

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

$a=\pm \pi$

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

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