Transform the following equation to new independent variables $u = x$, $v = x^2+y^2$

$$ y\frac{\partial z}{\partial x} - x \frac{\partial z}{\partial y} = 0 $$

Notce you don't have to actually solve the pde, the problem is to transform it. The official answer is $ \frac{\partial z}{\partial u} = 0 $

My work:

$$ y\frac{\partial z}{\partial x} = x\frac{\partial z}{\partial y} \longrightarrow \frac{1}{x}\frac{\partial z}{\partial x} = \frac{1}{y}\frac{\partial z}{\partial y} $$

$$ \frac{\partial z}{\partial y} = \frac{\partial z}{\partial v}\frac{\partial v}{\partial y} \longrightarrow \frac{\partial z}{\partial y} = \frac{\partial z}{\partial v}2y $$

Substituting, you get

$$ \frac{1}{x} \frac{\partial z}{\partial x} = 2\frac{\partial z}{\partial v} $$

By the same procedure you can get

$$ \frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} $$

Finally, I got

$$ \frac{1}{u}\frac{\partial z}{\partial u} = 2\frac{\partial z}{\partial v} $$

Which is obviously not the official answer. Am I doing something wrong and if not, how do I finish the problem?



In your procedure the step $\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u}$ does not true.

By chain rule:

$$\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial x}$$

$$\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial y}$$

$\implies$ $$\frac{\partial z}{\partial x} =\frac{\partial z}{\partial u}(1)+\frac{\partial z}{\partial v}(2x)=\frac{\partial z}{\partial u}+2x\frac{\partial z}{\partial v}$$ and

$$ \frac{\partial z}{\partial y} =\frac{\partial z}{\partial u}(0)+\frac{\partial z}{\partial v}(2y)=2y\frac{\partial z}{\partial v}$$

So $y\frac{\partial z}{\partial x} - x \frac{\partial z}{\partial y} = 0\implies y\frac{\partial z}{\partial u}+2yx\frac{\partial z}{\partial v}-2xy\frac{\partial z}{\partial v}=0\implies \frac{\partial z}{\partial u}=0$

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  • $\begingroup$ This is a nice illustration of the fact that a partial derivative depends not only on what variable you're taking the derivative with respect to, but also on what else you're holding constant. Even though $u = x$, it is not the case that$$\left(\frac{\partial z}{\partial x}\right)_y = \left(\frac{\partial z}{\partial u} \right)_v.$$ $\endgroup$ – Michael Seifert May 28 '19 at 3:01

$\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u}\frac{d u}{d x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial x}$

$$\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u}\cdot1 + \frac{\partial z}{\partial v}\cdot2x $$


$$\frac{\partial z}{\partial y} = 0 + \frac{\partial z}{\partial v}\cdot2y$$

Now, $y\frac{\partial z}{\partial x} - x\frac{\partial z}{\partial y} = 0$

$y\frac{\partial z}{\partial u} + 2xy\cdot\frac{\partial z}{\partial v} -2xy\cdot \frac{\partial z}{\partial v}=0\implies y\frac{\partial z}{\partial u}=0$

As, $y$ need not be necessarily $0$,

$$\frac{\partial z}{\partial u}=0$$

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$${\partial z\over\partial x}={du\over dx}{\partial z\over\partial u}+{dv\over dx}{\partial z\over\partial v}=(1){\partial z\over\partial u}+(2x){\partial z\over\partial v}$$ $${\partial z\over\partial y}={du\over dy}{\partial z\over\partial u}+{dv\over dy}{\partial z\over\partial v}=(0){\partial z\over\partial u}+(2y){\partial z\over\partial v}$$ $$0=y{\partial z\over\partial x}-x{\partial z\over\partial y}=y\left({\partial z\over\partial u}+2x{\partial z\over\partial v}\right)-x\left(2y{\partial z\over\partial v}\right)=y{\partial z\over\partial u}$$ Hence ${\partial z\over\partial u}=0$

The more important point is that I don't think you understand how to use the chain rule in multivariable calculus - perhaps you should check how it works

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  • $\begingroup$ Indeed I have almost zero practice with the multivariable chain rule despite reading about it.. I'll practice it. Thanks. $\endgroup$ – Victor May 27 '19 at 17:14
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    $\begingroup$ @VictorS. I think the hurdle is understanding that although $u=x$, the partial derivatives with respect to them are not the same. The reason is by the implied fixing of $v$ and $y$ respectively. When we write $\partial z/\partial x$ we mean $\partial z(x,y)/\partial x$ for fixed $y$, but $\partial z/\partial u$ means $\partial z(u,v)/\partial u$ for fixed $v$. Indeed, if we also had $v=y$, then this would be trivial. But here, fixing $y$ is different to fixing $v$. $\endgroup$ – stanley dodds May 27 '19 at 17:23

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