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given that $\zeta _{(x,y)}= \ln y+\frac{1}{x}$ and $\phi_{(x,y)}= 4\ln y-\frac{1}{x}$ ,

I've been asked to find $\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial^2 u}{\partial x^2}, \frac{\partial^2 u}{\partial x \partial y}, \frac{\partial^2 u}{\partial y^2}$ in terms of $x, y, \frac{\partial u}{\partial \zeta}, \frac{\partial y }{\partial \phi}, \frac{\partial^2 u}{\partial \zeta^2}, \frac{\partial^2 u}{\partial \zeta \partial \phi}, \frac{\partial^2 u}{\partial \phi^2}$.

This is what I have so far:

$\frac{\partial \zeta}{\partial x}=-\frac{1}{x^2}$

$\frac{\partial \zeta}{\partial y}=\frac{1}{y}$

$\frac{\partial \phi}{\partial x}=\frac{1}{x^2}$

$\frac{\partial \phi}{\partial y}=\frac{4}{y}$

$\frac{\partial u}{\partial x}=-\frac{1}{x^2}\frac{\partial u}{\partial \zeta}+\frac{1}{x^2}\frac{\partial u}{\partial \phi}$

$\frac{\partial u}{\partial y}=\frac{1}{y}\frac{\partial u}{\partial \zeta}+\frac{4}{y}\frac{\partial u}{\partial \phi}$

$\frac{\partial^2 u}{\partial x^2}=\frac{1}{x^4}\frac{\partial^2 u}{\partial \zeta^2}-\frac{2}{x^4}\frac{\partial^2 u}{\partial \zeta \partial \phi}+\frac{1}{x^4}\frac{\partial^2 u}{\partial \phi^2}+\frac{2}{x^3}\frac{\partial u}{\partial \zeta}-\frac{2}{x^3}\frac{\partial u}{\partial \phi}$

$\frac{\partial^2 u}{\partial y^2}=\frac{1}{y^2}\frac{\partial^2 u}{\partial \zeta^2}+\frac{8}{y^2}\frac{\partial^2 u}{\partial \zeta \partial \phi}+\frac{16}{y^2}\frac{\partial^2 u}{\partial \phi^2}-\frac{1}{y^2}\frac{\partial u}{\partial \zeta}+\frac{4}{y^2}\frac{\partial u}{\partial \phi}$

$\frac{\partial^2 u}{\partial x \partial y}=\frac{1}{x^4}\frac{\partial^2 u}{\partial \zeta^2}+\frac{2}{x^4}\frac{\partial^2 u}{\partial \zeta \partial \phi}+\frac{1}{x^4}\frac{\partial^2 u}{\partial \phi^2}$

I'm 100% convinced my second derivatives are flat-out wrong, but I'm not sure how to fix this. Can anyone point me in the right direction here?

Edit:

$u_{(x,y)}$ satisfies the PDE $4x^4\frac{\partial^2 u}{\partial x^2}+3x^2y\frac{\partial^2 u}{\partial x \partial y}-y^2\frac{\partial^2 u}{\partial y^2}+4x^2(1+2x)\frac{\partial u}{\partial x}-2y\frac{\partial u}{\partial y}=0$

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  • $\begingroup$ What is $u$ supposed to be a function of? $\endgroup$ – Mattos Mar 3 at 12:43
  • $\begingroup$ $u_{x,y}$ satisfies this PDE. $4x^4\frac{\partial^2 u}{\partial x^2}+3x^2y\frac{\partial^2 u}{\partial x \partial y}-y^2\frac{\partial^2 u}{\partial y^2}+4x^2(1+2x)\frac{\partial u}{\partial x}-2y\frac{\partial u}{\partial y}=0$ $\endgroup$ – C. Wolfe Mar 3 at 12:48
  • $\begingroup$ I'm aware there were some incorrect signs in my LaTeX, I think I've fixed them now $\endgroup$ – C. Wolfe Mar 3 at 13:02
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I guess your main source of confusion here is how to apply the multivariable chain rule.

First round is simple enough

\begin{align} u_x &= \zeta_xu_\zeta + \phi_xu_\phi \\ u_y &= \zeta_yu_\zeta + \phi_yu_\phi \end{align}

Second round, take the first round and repeat the operation. For example

\begin{align} u_{xx} &= (\zeta_xu_\zeta + \phi_xu_\phi)_x \\ &= \zeta_{xx}u_\zeta + \zeta_x(u_\zeta)_x + \phi_{xx}u_\phi + \phi_x(u_\phi)_x \\ &= \zeta_{xx}u_\zeta + \zeta_x(\zeta_xu_{\zeta\zeta} + \phi_xu_{\zeta\phi}) + \phi_{xx}u_\phi + \phi_x(\zeta_xu_{\phi\zeta}+\phi_xu_{\phi\phi}) \\ &= (\zeta_x)^2u_{\zeta\zeta} + 2\zeta_x\phi_xu_{\zeta\phi} + (\phi_x)^2u_{\phi\phi} + \zeta_{xx}u_{\zeta} + \phi_{xx}u_{\phi} \end{align}

Repeat for the other derivatives, you'll obtain

\begin{align} u_{yy} &= (\zeta_y)^2u_{\zeta\zeta} + 2\zeta_y\phi_yu_{\zeta\phi} + (\phi_y)^2u_{\phi\phi} + \zeta_{yy}u_{\zeta} + \phi_{yy}u_{\phi} \\ u_{xy} &= \zeta_x\zeta_yu_{\zeta\zeta} + (\zeta_x\phi_y+\zeta_y\phi_x)u_{\zeta\phi} + \phi_x\phi_yu_{\phi\phi} + \zeta_{xy}u_\zeta + \phi_{xy}u_\phi \end{align}

Your results aren't completely wrong. $u_{yy}$ has a sign error and the mixed derivative needs some $y$'s in there.

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  • $\begingroup$ Ok, if I followed correctly...$\frac{\partial^2 u}{\partial x^2}=\frac{1}{x^4}\frac{\partial^2 u}{\partial \zeta^2}-\frac{2}{x^4}\frac{\partial^2 u}{\partial \zeta \partial \phi}+\frac{1}{x^4}\frac{\partial^2 u}{\partial \phi^2}+\frac{2}{x^3}\frac{\partial u}{\partial \zeta}-\frac{2}{x^3}\frac{\partial u}{\partial \phi}$ $\endgroup$ – C. Wolfe Mar 3 at 14:48
  • $\begingroup$ $\frac{\partial^2 u}{\partial y^2}=\frac{1}{y^2}\frac{\partial^2 u}{\partial \zeta^2}+\frac{8}{y^2}\frac{\partial^2 u}{\partial \zeta \partial \phi}+\frac{16}{y^2}\frac{\partial^2 u}{\partial \phi^2}-\frac{1}{y^2}\frac{\partial u}{\partial \zeta}-\frac{4}{y^2}\frac{\partial u}{\partial \phi}$ $\endgroup$ – C. Wolfe Mar 3 at 14:51
  • $\begingroup$ $\frac{\partial^2 u}{\partial x \partial y}=\frac{-1}{x^2y}\frac{\partial^2 u}{\partial \zeta^2}-\frac{3}{x^2y}\frac{\partial^2 u}{\partial \zeta \partial \phi}+\frac{4}{x^2y}\frac{\partial^2 u}{\partial \phi^2}$ $\endgroup$ – C. Wolfe Mar 3 at 14:55
  • $\begingroup$ I think I may have some signs wrong, trying to figure out where $\endgroup$ – C. Wolfe Mar 3 at 15:05
  • $\begingroup$ This looks right to me. If you can't simplify the PDE, maybe the problem is your characteristics? $\endgroup$ – Dylan Mar 3 at 15:08

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