Question

Show that v(x, y) = x2 + 2y - y2 is harmonic conjugate of u = = 2x – 2xy.​

Answer 1

Step-by-step explanation:

given :

• Show that v(x, y) = x2 + 2y - y2 is harmonic conjugate of u = = 2x – 2xy.

to find :

• conjugate of u = = 2x – 2xy.

solution :

• 2xy is the answer

Answer 2

$\large\underline{\sf{Solution-}}$

Given functions are

$\rm :\longmapsto\:v(x,y) = {x}^{2} + 2y - {y}^{2}$

and

$\rm :\longmapsto\:u(x,y) = 2x - 2xy$

We know

If u and v are two functions then v is said to be Harmonic Conjugate of u, iff their first order partial derivatives satisfied Cauchy Reimann Equations.

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \: }}}$

and

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \frac{\partial u}{\partial y} = - \: \frac{\partial v}{\partial x} \: }}}$

So, Consider

$\rm :\longmapsto\:v(x,y) = {x}^{2} + 2y - {y}^{2}$

Differentiating partially w. r. t. x and y, we get

$\rm :\longmapsto\:\dfrac{\partial v}{\partial x} = 2x$

and

$\rm :\longmapsto\:\dfrac{\partial v}{\partial y} = 2 - 2y$

Now, Consider

$\rm :\longmapsto\:u(x,y) = 2x - 2xy$

On differentiating partially w. r. t. x and y, we get

$\rm :\longmapsto\:\dfrac{\partial u}{\partial x} = 2 - 2y$

and

$\rm :\longmapsto\:\dfrac{\partial u}{\partial y} = - 2x$

Hence, we concluded that

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \: }}}$

and

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \frac{\partial u}{\partial y} = - \: \frac{\partial v}{\partial x} \: }}}$

Hence,

• v is Harmonic Conjugate of u.

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Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$