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Let $x = u^2-v^2$ and $y = 2uv$, and suppose that $z = f(x,y)$ is differentiable. Show that: $$ \left(\frac{\partial z}{\partial u}\right) ^2+\left(\frac{\partial z}{\partial v} \right)^2 = 4(u^2+v^2)\left(\left(\frac{\partial z}{\partial x} \right)^2+\left(\frac{\partial z}{\partial y} \right)^2\right)$$

By taking the derivative in respect to $u$ and $v$ then taking their composite values:

$\frac{d}{dx}=2u-2v; \frac{d}{dy} = 2v+2u$

Which are equivalent to: $2(u-v)2(u+v)=4(u^2-v^2)$ However the signs are different as I'm expected to get a positive between $u$ and $v$.

MangoPizza
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Stackcans
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    the expressions $\frac{d}{dx}=2u-2v$ and $\frac{d}{dy} = 2v+2u$ doesn't make sense – Masacroso Sep 04 '21 at 10:50
  • @Masacroso Perhaps you're right. What's the best way of expressing it as I'm still trying to get familiar with the logic? – Stackcans Sep 04 '21 at 10:53
  • Do you know chain rule? – Arctic Char Sep 04 '21 at 11:20
  • @ArcticChar $\frac{\partial z}{\partial u}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial u}$? – Stackcans Sep 04 '21 at 11:27
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    No.... where's the $\frac{\partial z}{\partial y}$? – Arctic Char Sep 04 '21 at 11:30
  • @ArcticChar Just as in illustration for $x$ is this going down the right direction? - $\frac{\partial z}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial z}{\partial x}\frac{\partial x}{\partial v} = \frac{\partial z}{\partial x}\left(\frac{\partial x}{\partial u} + \frac{\partial x}{\partial v} \right) = \frac{\partial z}{\partial x}\left(2u-2v \right)$? – Stackcans Sep 04 '21 at 12:08
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    Um... no. The correct formula is $\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \frac{\partial v}{\partial x}.$ I suggest you go back to the chapter on chain rule first... – Arctic Char Sep 04 '21 at 12:14
  • @ArcticChar This cleared it up for me, thanks – Stackcans Sep 04 '21 at 12:26

1 Answers1

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Sketch for the solution: let $\varphi (u,v):=(u^2-v^2,2uv)=(x,y)$, then you want to show that

$$ \Delta (f\circ \varphi )(u,v)=4(u^2+v^2)\Delta f(x,y) $$

where $\Delta $ is the Laplacian, defined for a differentiable function $g:\mathbb{R}^n\to \mathbb{R}$ by

$$ \Delta g(p)=\sum_{k=1}^n (\partial_k)^2g(p)=\sum_{k=1}^n\partial ^2g(p)[e_k,e_k] $$ where $\partial g(p)$ is the Fréchet derivative of $g$ at $p$, the $e_1,\ldots ,e_n$ is the standard orthonormal basis of $\mathbb{R}^n$ and $\partial _k$ is the $k$-th partial derivative. Now, the chain rule and the product rule gives

$$ \begin{align*} (\partial _k)^2 (f\circ \varphi )(p)&=\partial ^2(f\circ \varphi )(p)[e_k,e_k]\\&=\partial (\partial (f\circ \varphi)(p)e_k)e_k\\&=\partial ((\partial f\circ \varphi )(p)\partial \varphi (p)e_k)e_k\\ &=\partial ((\partial f\circ \varphi )(p)\partial _k\varphi (p))e_k\\&=\partial ((\nabla f\circ \varphi) (p)\cdot \partial _k\varphi (p))e_k\\ &=\partial(\nabla f\circ \varphi )(p)e_k\cdot \partial _k\varphi (p)+\nabla f(\varphi (p))\cdot (\partial _k)^2\varphi (p)\\ &=(\partial\nabla f\circ \varphi )(p)\partial _k \varphi (p)\cdot \partial _k\varphi (p)+\nabla f(\varphi (p))\cdot (\partial _k)^2\varphi (p)\\ &=H_f (\varphi(p))\partial _k \varphi (p)\cdot \partial _k\varphi (p)+\nabla f(\varphi (p))\cdot (\partial _k)^2\varphi (p)\\ &=\partial _k \varphi (p)^\top H_f (\varphi(p))\partial _k\varphi (p)+\nabla f(\varphi (p))\cdot (\partial _k)^2\varphi (p) \end{align*} $$ where $\nabla f(q)$ and $H_f(q)$ are the gradient and the Hessian of $f$ at $q$ respectively. Adding up you get $$ \Delta (f\circ \varphi )(p)=\nabla f(\varphi (p))\cdot \Delta \varphi (p)+\partial _1 \varphi (p)^\top H_f (\varphi(p))\partial _1\varphi (p)+\partial _2 \varphi (p)^\top H_f (\varphi(p))\partial _2\varphi (p) $$ Now just simplify the above knowing that $[\partial \varphi (u,v)]=\left[\begin{smallmatrix}\partial _1 \varphi (u,v)&\partial _2\varphi (u,v)\end{smallmatrix}\right]=2\left[\begin{smallmatrix}u&-v\\v&u\end{smallmatrix}\right]$ and $\Delta \varphi (u,v)=0$.

Masacroso
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