# Thread: 2nd Order PDE problem

1. ## 2nd Order PDE problem

Hi all,
I am in the process of working out a solution to a 2nd order PDE.

However I am stuck on calculating $\displaystyle u_x_y$ given that

$\displaystyle \xi=y-sin(x)-x$ and $\displaystyle \eta=y-sin(x)+x$ and $\displaystyle \omega(\xi,\eta)=u(x,y)$

I can calculate uxx which is

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

and uy,uxx and uyy etc however I get stuck on uxy. I just cant get the answer in the book which leads me to belive my uxy is wrong. Can anyone give me a start on this uxy derivation like above for uxx?
Thanks

2. I get

$\displaystyle u_{xy} = \xi_{y}\xi_y u_{\xi \xi} + (\xi_x \eta_y + \xi_y \eta_x)u_{\xi \eta} + \eta_x \eta_y u_{\eta \eta} + \xi_{xy} u_\xi + \eta_{xy} u_\eta$.

3. Originally Posted by Danny
I get

$\displaystyle u_{xy} = \xi_{y}\xi_y u_{\xi \xi} + (\xi_x \eta_y + \xi_y \eta_x)u_{\xi \eta} + \eta_x \eta_y u_{\eta \eta} + \xi_{xy} u_\xi + \eta_{xy} u_\eta$.
hmmmm, can you explain how you got this? Is it a chain rule within a product rule? I started my calculation to be
$\displaystyle \frac{\partial}{\partial x}[u_y]=\frac{\partial}{\partial x} [\omega_\xi \cdot \xi_y+ \omega_\eta \cdot \eta_y]$ then my next step goes to pieces..

The final answer in the book is given as uxy =
$\displaystyle \omega_\xi_\xi(-cos(x)-1)+\omega_\xi_\eta(-cos(x)+1)+\omega_\eta_\xi(-cos(x)-1)+\omega_\eta_\eta(-cos(x)+1)$

it doesnt show the general form of uxy before the above line. The calculation of this seems to be different to that of uxx, uyy etc.
Any help is appreciated
Thanks

4. you don't need to check "general form" for $\displaystyle u_{xy}$ if you can find $\displaystyle u_x$ to check in the book and if your $\displaystyle u_x$ is correct where you have problem with $\displaystyle (u_x)'_y$ derivate that function on "y" ?

5. Originally Posted by bugatti79
hmmmm, can you explain how you got this? Is it a chain rule within a product rule? I started my calculation to be
$\displaystyle \frac{\partial}{\partial x}[u_y]=\frac{\partial}{\partial x} [\omega_\xi \cdot \xi_y+ \omega_\eta \cdot \eta_y]$ then my next step goes to pieces..

Thanks
I can. First you'll need the first derivative transforms

$\displaystyle \dfrac{\partial}{\partial x} = \xi_x \dfrac{\partial}{\partial \xi} + \eta_x \dfrac{\partial}{\partial \eta}$

$\displaystyle \dfrac{\partial}{\partial y} = \xi_y \dfrac{\partial}{\partial \xi} + \eta_y \dfrac{\partial}{\partial \eta}$.

Now expand what you have (although I prefer to the keep the u's)

$\displaystyle \frac{\partial}{\partial x}[u_y]=\frac{\partial}{\partial x} [u_\xi \cdot \xi_y+ u_\eta \cdot \eta_y]$

$\displaystyle =\frac{\partial}{\partial x}\left(u_\xi \right)\cdot \xi_y + u_\xi \cdot \xi_{xy} + \frac{\partial}{\partial x}\left(u_\eta \right) \cdot \eta_y+ u_\eta \cdot \eta_{xy}$.

Now bring in the x transform

$\displaystyle \frac{\partial}{\partial x}\left( u_\xi \right) = \xi_x \dfrac{\partial}{\partial \xi} \left(u_\xi\right) + \eta_x \dfrac{\partial}{\partial \eta} \left(u_\xi\right)$

$\displaystyle \frac{\partial}{\partial x}\left(u_\eta \right) = \xi_x \dfrac{\partial}{\partial \xi} \left(u_\eta\right) + \eta_x \dfrac{\partial}{\partial \eta} \left(u_\eta\right)$.

Substituting and expanding will give you the desired result.

6. Originally Posted by Danny
I can. First you'll need the first derivative transforms

$\displaystyle \dfrac{\partial}{\partial x} = \xi_x \dfrac{\partial}{\partial \xi} + \eta_x \dfrac{\partial}{\partial \eta}$

$\displaystyle \dfrac{\partial}{\partial y} = \xi_y \dfrac{\partial}{\partial \xi} + \eta_y \dfrac{\partial}{\partial \eta}$.

Now expand what you have (although I prefer to the keep the u's)

$\displaystyle \frac{\partial}{\partial x}[u_y]=\frac{\partial}{\partial x} [u_\xi \cdot \xi_y+ u_\eta \cdot \eta_y]$

$\displaystyle =\frac{\partial}{\partial x}\left(u_\xi \right)\cdot \xi_y + u_\xi \cdot \xi_{xy} + \frac{\partial}{\partial x}\left(u_\eta \right) \cdot \eta_y+ u_\eta \cdot \eta_{xy}$.

Now bring in the x transform

$\displaystyle \frac{\partial}{\partial x}\left( u_\xi \right) = \xi_x \dfrac{\partial}{\partial \xi} \left(u_\xi\right) + \eta_x \dfrac{\partial}{\partial \eta} \left(u_\xi\right)$

$\displaystyle \frac{\partial}{\partial x}\left(u_\eta \right) = \xi_x \dfrac{\partial}{\partial \xi} \left(u_\eta\right) + \eta_x \dfrac{\partial}{\partial \eta} \left(u_\eta\right)$.

Substituting and expanding will give you the desired result.
So i get

$\displaystyle u_x_y=\omega_\xi \cdot \xi_x_y+\xi_y \cdot [\xi_x \cdot \omega_\xi_\xi+\eta_x \cdot \omega_\eta_\xi] +\omega_\eta \cdot \eta_x_y+\eta_y \cdot[\xi_x \cdot \omega_\xi_\eta+\eta_x \cdot \omega_\eta_\eta]$

I can finally get the answer in the book. However, I just have 2 small queries:

1) How did you anticipate $\displaystyle \dfrac{\partial}{\partial x} = \xi_x \dfrac{\partial}{\partial \xi} + \eta_x \dfrac{\partial}{\partial \eta}$

Is it a setup for the chain rule?

2) As part of the above calculations we had $\displaystyle \frac{\partial}{\partial x}[\frac{\partial u}{\partial \xi} \cdot \frac{\partial \xi}{\partial y}]$ and $\displaystyle \omega(\xi,\eta)=u(x(\xi,\eta), y(\xi, \eta)))$

I am not 100% clear why this is differentiated as a product rule because I understand the product rule to be of the form y=uv where u and v are both functions of x.
Yet in the above equation I dont see how the product should be used because the denominator $\displaystyle \partial y$ is not a function of x but only $\displaystyle \xi$ and $\displaystyle \eta$.....and even the denominator $\displaystyle \partial\xi$ is not a function of x for the matter I dont think.....?
Im learning slowly but surely!
bugatti