Question

Easy Question :)

Derive Laplace Equation from Cauchy Riemann Equations for a complex analytical function given by ;

f ( z ) = u + iv

where , u ( x , y ) and v ( x , y )

Laplace Equation you had to derive is :-

${\quad \leadsto \quad \bf \dfrac{\partial² u}{\partial x²} + \dfrac{\partial² u}{\partial y²} = 0 }$ ​

Answer 1

Step-by-step explanation:

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Answer 2

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

Given complex analytic function is

$\purple{\rm :\longmapsto\:f ( z ) = u + iv \: \: where , \: u ( x , y ) \: and \: v ( x , y )}$

We have to prove that,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \frac{ {\partial }^{2} u}{ {\partial x}^{2} } + \frac{ {\partial }^{2} u}{ {\partial y}^{2} } = 0}}}$

and

$\purple{\rm :\longmapsto\:\boxed{\tt{ \frac{ {\partial }^{2} v}{ {\partial x}^{2} } + \frac{ {\partial }^{2} v}{ {\partial y}^{2} } = 0}}}$

Now, From Cauchy Reimann Equations, we have

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

and

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

Now, From first equation of Cauchy Reimann, we have

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

Differentiate partially w. r. t. x, we get

$\rm :\longmapsto\: \: \dfrac{\partial ^{2} u}{\partial {x}^{2} } \: = \dfrac{\partial^{2} v}{\partial x\partial y} - - - - (1)$

Also, from 2nd equation of Cauchy Reimann, we have

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

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

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

can be rewritten as

$\rm :\longmapsto\: \: \dfrac{\partial ^{2} u}{\partial {y}^{2} } \: = \: - \: \dfrac{\partial^{2} v}{\partial x\partial y} - - - - (2)$

So, on equating equation (1) and (2), we get

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

$\rm :\longmapsto\: \boxed{\tt{ \: \dfrac{\partial ^{2} u}{\partial {x}^{2} } \: + \: \dfrac{\partial^{2} u}{\partial \: {y}^{2} } = 0}}$

Also, Again From first equation of Cauchy Reimann, we have

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

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

$\rm :\longmapsto\: \: \dfrac{\partial ^{2} v}{\partial {y}^{2} } \: = \dfrac{\partial^{2} u}{\partial x\partial y} - - - - (3)$

From Second equation of Cauchy Reimann, we have

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

On differentiating partially both sides w. r. t. x, we get

$\rm :\longmapsto\: \: - \: \dfrac{\partial ^{2}v}{\partial {x}^{2} } \: = \dfrac{\partial^{2} u}{\partial x\partial y} - - - - (4)$

On equating equation (3) and (4), we get

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

$\rm :\longmapsto\: \: \boxed{\tt{ \dfrac{\partial ^{2} v}{\partial {x}^{2} } \: + \: \dfrac{\partial^{2} v}{\partial \: {y}^{2} } = 0}}$

Hence, Proved