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Second order partial differential of implicite function

Posted December 3rd, 2009 by Structure

Let $ G(x, y, z)=F(\frac{x}{z},\frac{y}{z})=0 $ an equation defined by a function $ F\in \mathcal{C}^2 $
Suppose we can apply the implicit function theorem in a neighborhood of a given point (a,b,c).
So we suppose

$$G(a,b,c)=F(\frac{a}{c},\frac{b}{c})=0$$

and

$$\frac{\partial G}{\partial z}(a,b,c)=\frac{\partial F}{\partial u}(\frac{a}{c},\frac{b}{c})(-\frac{a}{c^2})+\frac{\partial F}{\partial v}(\frac{a}{c},\frac{b}{c})(-\frac{b}{c^2})\ne 0$$

Then there is a function z=f(x,y) with f(a,b)=c; G(x,y,f(x,y))=0 or

$$ F(\frac{x}{f(x,y)},\frac{y}{f(x,y)})=0$$

We have

$$\frac{\partial f}{\partial x}(x,y)=<br /> -\frac{\frac{\partial G}{\partial x}(x,y,f(x,y))}{\frac{\partial G}{\partial z}(x,y,f(x,y))}=<br /> -\frac{\frac{1}{f(x,y)}\frac{\partial F}{\partial u}(\frac{x}{f(x,y)},\frac{y}{f(x,y)})}{-\frac{x}{f^2(x,y)}\frac{\partial F}{\partial u}(\frac{x}{f(x,y)},\frac{y}{f(x,y)})-\frac{y}{f^2(x,y)}\frac{\partial F}{\partial v}(\frac{x}{f(x,y)},\frac{y}{f(x,y)})}=$$
$$=\frac{f(x,y)\frac{\partial F}{\partial u}(\frac{x}{f(x,y)},\frac{y}{f(x,y)})}{x\frac{\partial F}{\partial u}(\frac{x}{f(x,y)},\frac{y}{f(x,y)})+y\frac{\partial F}{\partial v}(\frac{x}{f(x,y)},\frac{y}{f(x,y)})}$$
$$\frac{\partial f}{\partial y}(x,y)=-\frac{\frac{\partial G}{\partial y}(x,y,f(x,y))}{\frac{\partial G}{\partial z}(x,y,f(x,y))}=-\frac{\frac{1}{f(x,y)}\frac{\partial F}{\partial v}(\frac{x}{f(x,y)},\frac{y}{f(x,y)})}{-\frac{x}{f^2(x,y)}\frac{\partial F}{\partial u}(\frac{x}{f(x,y)},\frac{y}{f(x,y)})-\frac{y}{f^2(x,y)}\frac{\partial F}{\partial v}(\frac{x}{f(x,y)},\frac{y}{f(x,y)})}=$$
$$=\frac{f(x,y)\frac{\partial F}{\partial v}(\frac{x}{f(x,y)},\frac{y}{f(x,y)})}{x\frac{\partial F}{\partial u}(\frac{x}{f(x,y)},\frac{y}{f(x,y)})+y\frac{\partial F}{\partial v}(\frac{x}{f(x,y)},\frac{y}{f(x,y)})}$$

It is easy to see that

$$x\frac{\partial f}{\partial x}(x,y)+y\frac{\partial f}{\partial y}(x,y)=f(x,y)$$
$$\frac{\partial^2 f}{\partial x^2}(x,y)=\frac{\partial }{\partial x}(\frac{f(x,y)\frac{\partial F}{\partial u}(\frac{x}{f(x,y)},\frac{y}{f(x,y)})}{x\frac{\partial F}{\partial u}(\frac{x}{f(x,y)},\frac{y}{f(x,y)})+y\frac{\partial F}{\partial v}(\frac{x}{f(x,y)},\frac{y}{f(x,y)})})=$$
$$=\frac{\frac{\partial f}{\partial x}\frac{\partial F}{\partial u}(\frac{x}{f},\frac{y}{f})+f(\frac{\partial^2 F}{\partial u^2}(\frac{x}{f},\frac{y}{f})\frac{f-x\frac{\partial f}{\partial x}}{f^2}+\frac{\partial ^2F}{\partial v\partial u}(\frac{x}{f},\frac{y}{f})\frac{-y\frac{\partial f}{\partial x}}{f^2}) }<br /> {x\frac{\partial F}{\partial u}(\frac{x}{f(x,y)},\frac{y}{f(x,y)})+y\frac{\partial F}{\partial v}(\frac{x}{f(x,y)},\frac{y}{f(x,y)})}-$$
$$-\frac{ f\frac{\partial F}{\partial u}(\frac{x}{f},\frac{y}{f})[\frac{\partial F}{\partial u}(\frac{x}{f},\frac{y}{f})+x(\frac{\partial^2 F}{\partial u^2}(\frac{x}{f},\frac{y}{f})\frac{f-x\frac{\partial f}{\partial x}}{f^2}+\frac{\partial^2 F}{\partial v \partial u}(\frac{x}{f},\frac{y}{f})\frac{-y\frac{\partial f}{\partial x}}{f^2})+y(\frac{\partial^2 F}{\partial u\partial v}(\frac{x}{f},\frac{y}{f})\frac{f-x\frac{\partial f}{\partial x}}{f^2}+\frac{\partial^2 F}{\partial v ^2}(\frac{x}{f},\frac{y}{f})\frac{-y\frac{\partial f}{\partial x}}{f^2})] }{(x\frac{\partial F}{\partial u}(\frac{x}{f(x,y)},\frac{y}{f(x,y)})+y\frac{\partial F}{\partial v}(\frac{x}{f(x,y)},\frac{y}{f(x,y)}))^2}$$
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