Question

Prove that the real and imaginary parts of an analytic function satisfy laplace equation

Answer 1

Answer:

show that real and imaginary parts of and anliytic satisty aplace fuction

Answer 2

Given :  z = u(x,y) + i v(x,y) is an analytic function

To Prove : The real and imaginary parts of an analytic function satisfy Laplace equation

Proof : It is given that the complex valued function

z = u(x,y) + i v(x,y) is an analytic function

Since f is analytic it will be differentiable throughout the domain

and it will also satisfy Cauchy Reiman equation

so By Cauchy Riemann equation

$\frac{du}{dx} = \frac{dv}{dy}$ and $\frac{du}{dy} = -\frac{dv}{dx}$     equation 1)

Now

$\frac{d^{2}u }{dx^{2} } + \frac{d^{2}u }{dy^{2} }$ =

$\frac{d}{dx}(\frac{du}{dx}) +\frac{d}{dy}(\frac{du}{dy})$

= $\frac{d}{dx}(\frac{dv}{dy}) +\frac{d}{dy}(\frac{-dv}{dx})$   (replacing $\frac{du}{dx}$ by $\frac{dv}{dy}$ and $\frac{du}{dy}$ by $-\frac{dv}{dx}$)

= $\frac{d^{2}v }{dxdy} - \frac{d^{2}v }{dx}dy$

= $\frac{d^{2}u }{dx^{2} } + \frac{d^{2}u }{dy^{2} }$ = 0

So u(x,y) satisfies Laplace's equation

Similarly

$\frac{d^{2}v }{dx^{2} } + \frac{d^{2}v }{dy^{2} }$ =

$\frac{d}{dx}(\frac{dv}{dx}) +\frac{d}{dy}(\frac{dv}{dy})$

= $\frac{d}{dx}(-\frac{du}{dy}) +\frac{d}{dy}(\frac{du}{dx})$   (replacing $\frac{du}{dx}$ by $\frac{dv}{dy}$ and $\frac{du}{dy}$ by $-\frac{dv}{dx}$)

= $-\frac{d^{2}u }{dxdy} - \frac{d^{2}u }{dx}dy$

= $\frac{d^{2}v }{dx^{2} } + \frac{d^{2}v }{dy^{2} }$ = 0

So v(x,y) satisfies Laplace's equation

Hence real and imaginary parts of an analytic function satisfy Laplace equation

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