Question

If e^ax cosy is harmonic then a is

Answer 1

$\textbf{Given:}$

$f(x,y)=e^{ax}\,cosy\,\text{is harmonic}$

$\textbf{To find:}$

$\text{The value of a}$

$\textbf{Solution:}$

$\boxed{\textbf{A function is called harmonic when}\,\bf\dfrac{{\partial}^2f}{{\partial}x^2}+\dfrac{{\partial}^2f}{{\partial}y^2}=0}$

$f=e^{ax}\,cosy$

$\dfrac{{\partial}f}{{\partial}x}=a\,e^{ax}\,cosy$

$\dfrac{{\partial}^2f}{{\partial}x^2}=a^2\,e^{ax}\,cosy$

$\dfrac{{\partial}f}{{\partial}y}=e^{ax}\,(-siny)$

$\dfrac{{\partial}^2f}{{\partial}y^2}=-e^{ax}\,cosy$

$\text{Since f is harmonic, we have}$

$\dfrac{{\partial}^2f}{{\partial}x^2}+\dfrac{{\partial}^2f}{{\partial}y^2}=0$

$\implies\,a^2\,e^{ax}\,cosy+(-e^{ax}\,cosy)=0$

$\implies\,e^{ax}\,cosy(a^2-1)=0$

$\implies\,a^2-1=0$

$\implies\,a^2=1$

$\implies\,a=\pm\,1$

$\textbf{Answer:}$

$\textbf{The value of 'a' are \pm\,1 }$