I am currently reading a document on analytical solution of second-order PDE's which reads the following:

We could define new independent variables $\xi(x,y)$ and $\eta(x,y)$.... As before we compute the chain rule derivations:

$$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial x}+\frac{\partial u}{\partial \eta}\frac{\partial \eta}{\partial x}$$

$$\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 u}{\partial \xi^2}\left(\frac{\partial \xi}{\partial x}\right)^2+2 \frac {\partial^2 u}{\partial \xi \partial \eta}\frac{\partial \xi}{\partial x}\frac{\partial \eta}{\partial x}+\frac{\partial^2 u}{\partial \eta^2}\left(\frac{\partial \eta}{\partial x}\right)^2+\frac{\partial u}{\partial \xi}\frac{\partial^2 \xi}{\partial x^2}+\frac{\partial u}{\partial \eta}\frac{\partial ^2 \eta}{\partial x^2}$$

With similar computations for $\frac {\partial u}{\partial y}$. Regarding the second derivative, I am not happy about the last two terms. I am not sure why they appear. According to me, the second derivative can be taken as follows, since $u$ depends on $x$ and $y$ which in turn depend each on $\eta $ and $\xi$.

$$\frac{\partial^2 u}{\partial x^2} = \frac {\partial}{\partial x }\left(\frac {\partial u} {\partial x}\right)$$

We can reuse the very first chain rule as follows:

$$\frac {\partial}{\partial x }\left(\frac {\partial u} {\partial x}\right)=\left(\frac{\partial }{\partial \xi}\frac{\partial \xi}{\partial x}+\frac{\partial }{\partial \eta}\frac{\partial \eta}{\partial x}\right)\left(\frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial x}+\frac{\partial u }{\partial \eta}\frac{\partial \eta}{\partial x}\right)$$ which when multiplied through using multiplicative diistribution would yield:

$$\frac{\partial^2 u}{\partial \xi^2}\left(\frac{\partial \xi}{\partial x}\right)^2+2 \frac {\partial^2 u}{\partial \xi \partial \eta}\frac{\partial \xi}{\partial x}\frac{\partial \eta}{\partial x}+\frac{\partial^2 u}{\partial \eta^2}\left(\frac{\partial \eta}{\partial x}\right)^2$$

Comparing my answer with the books answer I am missing two terms. Why and how do these appear?


1 Answer 1


Picking up close to where you left, $$ \begin{alignedat}{4} \frac {\partial}{\partial x } \left(\frac {\partial u} {\partial x} \right) &= \frac {\partial}{\partial x } \left(\frac{\partial u}{\partial \xi} \frac{\partial \xi}{\partial x}+\frac{\partial u }{\partial \eta}\frac{\partial \eta}{\partial x}\right) \\&= \frac {\partial}{\partial x} \left(\xi_x u_\xi+\eta_x u_\eta\right) \\&= \xi_{xx}u_\xi + \xi_x\frac{\partial u_\xi}{\partial x} + \eta_{xx}u_\eta + \eta_x\frac{\partial u_\eta}{\partial x} \\ &= \xi_{xx}u_\xi + \xi_x \left(\frac{\partial u_\xi}{\partial \xi}\frac{\partial \xi}{\partial x} + \frac{\partial u_\xi}{\partial \eta}\frac{\partial \eta}{\partial x}\right) + \eta_{xx}u_\eta + \eta_x \left(\frac{\partial u_\eta}{\partial \xi}\frac{\partial \xi}{\partial x} + \frac{\partial u_\eta}{\partial \eta}\frac{\partial \eta}{\partial x}\right) \\&= \xi_{xx}u_\xi + \xi_x \left(\xi_x u_{\xi\xi} + \eta_x u_{\xi\eta}\right) + \eta_{xx}u_\eta + \eta_x \left(\xi_x u_{\xi\eta} + \eta_x u_{\eta\eta}\right) \\&= \xi^2_x u_{\xi\xi} + 2\xi_x\eta_x u_{\xi\xi} + \eta_x^2 u_{\eta\eta} + \xi_{xx} u_\xi + \eta_{xx} u_\eta. \end{alignedat} $$

Thus we got $$\bbox[1.5ex, border:solid 2pt #e10000]{ \frac{\partial^2 u}{\partial x^2} =\left(\frac{\partial\xi}{\partial x}\right)^2\frac{\partial^2u}{\partial\xi^2} + 2\left(\frac{\partial \xi}{\partial x}\frac{\partial \eta}{\partial x}\right) \frac {\partial^2 u}{\partial \xi \partial \eta} + \left(\frac{\partial \eta}{\partial x}\right)^2 \frac{\partial^2 u}{\partial \eta^2} + \frac{\partial u}{\partial \xi}\frac{\partial^2 \xi}{\partial x^2} + \frac{\partial u}{\partial \eta}\frac{\partial ^2 \eta}{\partial x^2}} $$

  • $\begingroup$ Key word: product rule (smack myself on the back of the head).... Thanks a bunch $\endgroup$
    – user32882
    Commented Apr 3, 2017 at 12:39

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