Question

Prove that u(x) := e* (x COS Y – ysiny) is harmonic and find a conjugate function of u.

Answer 1

Answer:

$u(x)=\ =\ e^x(xcosy\ -\ ysiny)\ \textrm{is harmonic.}$

$\textrm{conjugate function},\ v\ =\ 2e^x(xsiny\ +\ ycosy)\ +\ c$

Step-by-step explanation:

As given in question,

• $u(x)\ =\ e^x(xcosy\ -\ ysiny)$

       $=\ e^x.xcosy\ -\ e^x.ysiny$

$=>\ u'_x(x)\ =\ e^x.cosy\ +\ xe^xcosy\ -\ ye^xsiny\\\\=>\ u''_x(x)\ =\ e^xcosy+e^xcosy+xe^xcosy-e^xysiny\\\\=>\ u'_y(x)\ =\ -e^x.xsiny\ -\ e^xsiny\ -\ e^x.ycosy\\\\=>\ u''_y(x)\ =\ -xe^xcosy\ -\ 2e^xcosy\ +\ e^x.ysiny$

For a system to be harmonic,

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

From the above calculated value we can see that

$\dfrac{\partial^2 u}{\partial x^2}\ +\ \dfrac{\partial^2 u}{\partial y^2}$

$=\ e^xcosy+e^xcosy+xe^xcosy-e^xysiny\ -\ xe^xcosy\ -\ 2e^xcosy\ +\ e^x.ysiny$

= 0

So, given function u(x) is harmonic.

Now to find conjugate function of u,

$dv\ =\ \dfrac{\partial v}{\partial x}.dx\ +\ \dfrac{\partial v}{\partial y}.dy$

According to Cauchy-Riemen Equation,

$\dfrac{\partial u}{\partial x}\ =\ \dfrac{\partial v}{\partial y}\\\\\dfrac{\partial u}{\partial y}\ =\ -\dfrac{\partial v}{\partial x}$

So, we4 can write the above equation as,

$dv\ =\ -\dfrac{\partial u}{\partial y}dx\ +\ \dfrac{\partial u}{\partial x}dy$

on  putting the values, we have

$dv\ =\ -(-e^x.xsiny\ -\ e^xsiny\ -\ e^x.ycosy)dx\ +\ (e^x.cosy\ +\ xe^xcosy\ -\ ye^xsiny)dy$

on integrating both side, we get

$\int{dv}\ =\ \int{e^x.xsiny\ +\ e^xsiny\ +\ e^x.ycosy)dx}\ +\ \int{(e^x.cosy\ +\ xe^xcosy\ -\ ye^xsiny)dy}$

$=>\ v\ =\ 2xe^xsiny\ +\ 2ye^xcosy\ +\ c$

        $=\ 2e^x(xsiny\ +\ ycosy)\ +\ c$

hence, the conjugate equation can be given by

$v\ =\ 2e^x(xsiny\ +\ ycosy)\ +\ c$