DE Change of Variables

• Jan 9th 2011, 02:14 PM
dwsmith
DE Change of Variables
$u_x-u_y+u=\exp{(x+2y)}$

$u(x,0)=0$

The solutions is $u(x,y)=-y\exp{(x+2y)}$

$\displaystyle\omega_{\xi}(\cos{\alpha}-\sin{\alpha})-\omega_{\eta}(\sin{\alpha}+\cos{\alpha})+\omega=ex p{[\xi(\cos{\alpha}+2\sin{\alpha})+\eta(2\cos{\alpha}-\sin{\alpha})]$

$\displaystyle\cos{\alpha}=-\frac{\sqrt{2}}{2} \ \ \mbox{and} \ \ \sin{\alpha}=\frac{\sqrt{2}}{2}$

$\displaystyle -\sqrt{2}\omega_{\xi}+\omega=\exp{\left(\frac{\sqrt {2}}{2}\xi-\frac{3\sqrt{2}}{2}\eta}\right)\Rightarrow \omega_{\xi}-\frac{1}{\sqrt{2}}\omega=\frac{\exp{\left(\frac{\s qrt{2}}{2}\xi-\frac{3\sqrt{2}}{2}\eta}\right)}{-\sqrt{2}}$

$x=A\xi+B\eta \ \ \mbox{and} \ \ y=C\xi+D\eta$

$\omega_{\xi}=Au_x+Cu_y=u_x-u_y\Rightarrow A=1 \ \ B=-1$

$x=\xi+B\eta \ \ \mbox{and} \ \ y=-\xi+D\eta$

$\omega_{\xi}+\omega=\mbox{?}$

Should this be set equal to $\displaystyle\exp{(x+2y)} \ \mbox{or} \ \exp{\left(\frac{\sqrt{2}}{2}\xi-\frac{3\sqrt{2}}{2}\eta}\right)$
• Jan 9th 2011, 08:00 PM
dwsmith
$\omega_{\xi}+\omega=\exp{(x+2y)}$

$\displaystyle\int\left(\frac{\partial}{\partial\xi }\left[\exp{(\xi)}\omega\right]\right)d\xi=\int\exp{(\xi+x+2y)}d\xi$

$\exp{(\xi)}\omega=\exp{(\xi+x+2y)}+g(\eta)\Rightar row\omega(\xi,\eta)=\exp{(x+2y)}+\exp{(-\xi)}g(\eta)$

$x=\xi+B\eta \ \ \mbox{and} \ \ 0=-\xi+D\eta$

Let D = 0 and B = 1 since y = 0, both D and B are arbitrary, and $AD - BC\neq 0$.

$x=\xi+\eta \ \ \mbox{and} \ \ y=0=-\xi$

$\eta=x-\xi=x+y$

$\omega(\xi,\eta)=u(x,y)=\exp{(x+2y)}+\exp{(y)}g(x+ y)$

$u(x,0)=\exp{(x)}+g(x)=0\Rightarrow g(x)=-\exp{(x)}$

$u(x,y)=\exp{(x+2y)}-\exp{(y)}\exp{(x+y)}=0$

This is obviously wrong but I don't know what to do differently.
• Jan 10th 2011, 07:05 AM
Jester
You got two things going on that seems to be confusing you - (1) a rotation of coordinates and (2) a general change of coordinates. Let me see if I can help.

$u_x - u_y + u = e^{x+2y}$

$u(x,0)=0$

You have introduced a linear change of variables (this can be more general but it's not necessary here)

$x = A \xi + B \eta,\;\;\;y = C \xi + D \eta$.

What you want is to transform your PDE to essentially an ODE by choosing the constant $A,B,C$ and $D$ with the requirement that $AD - BC \ne 0$ (why?).

As you've said

$u_{\xi} = Au_x + Cu_y$

and picked $A = 1$ and $C = -1$ (that's good!) so

$u_{\xi} = u_x - u_y$.

You still have the flexibility to choose $B$ and $D$ anything. You further chose $B =1$ and $D = 0$ so

$x = \xi + \eta,\;\;\;y = -\xi$.

This is also good b/c the boundary

$y = 0\;\;\;\Rightarrow\;\;\; \xi = 0$.

Now you must switch everthing over in terms of your new variables so since

$u_{\xi} = u_x - u_y$

$u_{\xi} + u = e^{\eta - \xi}$ with $u = 0$ when $\xi = 0$.

Now follow what you did in post #2.
• Jan 10th 2011, 12:06 PM
dwsmith
Quote:

Originally Posted by Danny

What you want is to transform your PDE to essentially an ODE by choosing the constant $A,B,C$ and $D$ with the requirement that $AD - BC \ne 0$ (why?).

$u_{\xi} + u = e^{\eta - \xi}$ with $u = 0$ when $\xi = 0$.

The book said that $\displaystyle\begin{vmatrix}A&B\\C&D\end{vmatrix}\ ne 0\Rightarrow AD-BC\ne 0$.

I don't understand how you obtain $e^{\eta-\xi}$
• Jan 10th 2011, 12:37 PM
Jester
The right hand side is

$e^{x+2y}$

$x = \xi + \eta, y = - \xi$
$x+2y = \xi + \eta - 2 \xi = \eta - \xi$.