Cauchy Riemann Equations

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wolly

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I have to find ∂t∂φ\frac{\partial t}{\partial \varphi}t=x2+y2t=x^2+y^2and φ(t(x,y))=x2+y2\varphi(t(x,y))=x^2+y^2How did 4x^2 appear?
 
wolly said:
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I have to find ∂t∂φ\frac{\partial t}{\partial \varphi}t=x2+y2t=x^2+y^2and φ(t(x,y))=x2+y2\varphi(t(x,y))=x^2+y^2How did 4x^2 appear?
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You have two functions:
φ(t)\displaystyle \varphi(t)
t(x,y)\displaystyle t(x,y)

You want to find ∂2φ∂x2\displaystyle \frac{\partial^2 \varphi}{\partial x^2}

Apply the chain rule.

∂φ∂x=∂φ∂t∂t∂x=2x∂φ∂t\displaystyle \frac{\partial \varphi}{\partial x} = \frac{\partial \varphi}{\partial t}\frac{\partial t}{\partial x} = 2x\frac{\partial \varphi}{\partial t}

You can change ∂φ∂t\displaystyle \frac{\partial \varphi}{\partial t} to dφdt\displaystyle \frac{d \varphi}{d t} if you want because it has only one variable.

∂2φ∂x2=∂∂x(2x∂φ∂t)\displaystyle \frac{\partial^2 \varphi}{\partial x^2} = \frac{\partial}{\partial x}\left(2x\frac{\partial \varphi}{\partial t}\right)

Apply the product rule.

∂2φ∂x2=2∂φ∂t+2x∂∂x(∂φ∂t)\displaystyle \frac{\partial^2 \varphi}{\partial x^2} = 2\frac{\partial \varphi}{\partial t} + 2x\frac{\partial}{\partial x}\left(\frac{\partial \varphi}{\partial t}\right)

Apply the chain rule.

∂2φ∂x2=2∂φ∂t+2x∂∂t(∂φ∂t)∂t∂x=2∂φ∂t+2x∂2φ∂t22x=2∂φ∂t+4x2∂2φ∂t2\displaystyle \frac{\partial^2 \varphi}{\partial x^2} = 2\frac{\partial \varphi}{\partial t} + 2x\frac{\partial}{\partial t}\left(\frac{\partial \varphi}{\partial t}\right)\frac{\partial t}{\partial x} = 2\frac{\partial \varphi}{\partial t} + 2x\frac{\partial^2 \varphi}{\partial t^2}2x = 2\frac{\partial \varphi}{\partial t} + 4x^2\frac{\partial^2 \varphi}{\partial t^2}

That's how 4x2\displaystyle 4x^2 appeared.
 
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