2nd Order PDE problem

**bugatti79** · August 16th 2010, 08:44 AM

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

However I am stuck on calculating $u_x_y$ given that

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

I can calculate uxx which is

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

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

**Danny** · August 16th 2010, 09:22 AM

I get

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

**bugatti79** · August 16th 2010, 10:26 PM

Originally Posted by Danny

I get

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

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

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

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

**yeKciM** · August 17th 2010, 12:02 AM

you don't need to check "general form" for $u_{xy}$ if you can find $u_x$ to check in the book

and if your $u_x$ is correct where you have problem with $(u_x)'_y$ derivate that function on "y" ?

**Danny** · August 17th 2010, 06:16 AM

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
$\frac{\partial}{\partial x}[u_y]=\frac{\partial}{\partial x} [\omega_\xi \cdot \xi_y+ \omega_\eta \cdot \eta_y]$ then my next step goes to pieces..

Thanks

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

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

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

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

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

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

Now bring in the x transform

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

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

Substituting and expanding will give you the desired result.

**bugatti79** · August 17th 2010, 12:52 PM

Originally Posted by Danny

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

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

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

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

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

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

Now bring in the x transform

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

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

Substituting and expanding will give you the desired result.

So i get

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

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

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

Is it a setup for the chain rule?

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

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 $\partial y$ is not a function of x but only $\xi$ and $\eta$ .....and even the denominator $\partial\xi$ is not a function of x for the matter I dont think.....?
Im learning slowly but surely!
bugatti

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