Solve Linear PDE via a change of variables

**bugatti79** · February 11th 2011, 08:06 AM

$ \omega(r,\theta)=u(r cos (\theta),r sin (\theta)) $ for some function u(x,y)

$ \displaystyle y \frac{\partial u}{\partial x}-x\frac{\partial u}{\partial y}=1 $

Rewrite the PDE and IC's in terms of $\omega,r, \theta$ by expressing

$ \displaystyle \frac{\partial \omega}{\partial r}$ and $ \displaystyle \frac{\partial \omega}{\partial \theta}$ in terms of $ \displaystyle \frac{\partial u}{\partial x}$ and $\displaystyle \frac{\partial u}{\partial y} $ and solve.

My attempt: I believe the first equation is of the form

$ \displaystyle \omega(r,\theta) = u(x(r,\theta),y(r,\theta))$ therefore $\begin{cases} \frac{\partial \omega}{\partial r} = \frac{\partial \omega}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial \omega}{\partial y} \frac{\partial y}{\partial r}\\\frac{\partial \omega}{\partial \theta} = \frac{\partial \omega}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial \omega}{\partial y} \frac{\partial y}{\partial \heta}\end{cases}$ but they are not in terms of $ \displaystyle \frac{\partial u}{\partial x}$ and $\displaystyle \frac{\partial u}{\partial y} $

Have I gone one step too far?

**Ackbeet** · February 11th 2011, 08:11 AM

Could be way off here, but couldn't you just use this:

$\displaystyle\begin{cases} \frac{\partial \omega}{\partial r} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial r}\\\frac{\partial \omega}{\partial \theta} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial \theta}\end{cases}?$

**bugatti79** · February 11th 2011, 08:21 AM

Originally Posted by Ackbeet

Could be way off here, but couldn't you just use this:

$\displaystyle\begin{cases} \frac{\partial \omega}{\partial r} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial r}\\\frac{\partial \omega}{\partial \theta} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial \theta}\end{cases}?$

Hi Ackbeet,

You could be right, I noticed my error...the dependant variable on the RHS should be u not omega!...I think

**Ackbeet** · February 11th 2011, 08:43 AM

Right. And, of course, the

$\displaystyle\frac{\partial x}{\partial r},\;\frac{\partial x}{\partial \theta},\;\frac{\partial y}{\partial r},\;\frac{\partial y}{\partial \theta}$

are all easy to compute.

**bugatti79** · February 11th 2011, 08:57 AM

Originally Posted by bugatti79

Hi Ackbeet,

You could be right, I noticed my error...the dependant variable on the RHS should be u not omega!...I think

$\displaystyle\begin{cases} \frac{\partial \omega}{\partial r} = \frac{\partial u}{\partial x} Cos(\theta) + \frac{\partial u}{\partial y} Sin(\theta)\\\frac{\partial \omega}{\partial \theta} = -\frac{\partial u}{\partial x} r Sin(\theta) + \frac{\partial u}{\partial y} r Cos (\theta)\end{cases}$

The task is to rewrite the PDE and IC's in terms of $\omega,r, \theta$ but $\dispalysytle \frac{\partial u}{\partial x}$ and $\dispalysytle \frac{\partial u}{\partial y}$ both appear twice in the above equations. My idea was to make $\dispalysytle \frac{\partial u}{\partial x}$ the subject of both equations and then equate...I dont think that would work...I have ran out of ideas on how to solve this!

**Ackbeet** · February 11th 2011, 09:03 AM

Originally Posted by bugatti79

$\displaystyle\begin{cases} \frac{\partial \omega}{\partial r} = \frac{\partial u}{\partial x} Cos(\theta) + \frac{\partial u}{\partial y} Sin(\theta)\\\frac{\partial \omega}{\partial \theta} = -\frac{\partial u}{\partial x} r Sin(\theta) + \frac{\partial u}{\partial y} r Cos (\theta)\end{cases}$

The task is to rewrite the PDE and IC's in terms of $\omega,r, \theta$ but $\dispalysytle \frac{\partial u}{\partial x}$ and $\dispalysytle \frac{\partial u}{\partial y}$ both appear twice in the above equations. My idea was to make $\dispalysytle \frac{\partial u}{\partial x}$ the subject of both equations and then equate...I dont think that would work...I have ran out of ideas on how to solve this!

Yeah, I think it's going to take some tricks. How about this (since I see a rotation matrix present):

$\begin{bmatrix}\omega_{r}\\ \omega_{\theta}/r\end{bmatrix}=\begin{bmatrix}\cos(\theta) &\sin(\theta)\\ -\sin(\theta) &\cos(\theta)\end{bmatrix}\begin{bmatrix}u_{x}\ \ u_{y}\end{bmatrix}.$

Inverting a rotation matrix is straight-forward: just transpose it! We get

$\begin{bmatrix}\cos(\theta) &-\sin(\theta)\\ \sin(\theta) &\cos(\theta)\end{bmatrix}\begin{bmatrix}\omega_{r }\\ \omega_{\theta}/r\end{bmatrix}=\begin{bmatrix}u_{x}\\ u_{y}\end{bmatrix}.$

How's that?

**bugatti79** · February 11th 2011, 10:34 AM

Originally Posted by Ackbeet

Yeah, I think it's going to take some tricks. How about this (since I see a rotation matrix present):

$\begin{bmatrix}\omega_{r}\\ \omega_{\theta}/r\end{bmatrix}=\begin{bmatrix}\cos(\theta) &\sin(\theta)\\ -\sin(\theta) &\cos(\theta)\end{bmatrix}\begin{bmatrix}u_{x}\ \ u_{y}\end{bmatrix}.$

Inverting a rotation matrix is straight-forward: just transpose it! We get

$\begin{bmatrix}\cos(\theta) &-\sin(\theta)\\ \sin(\theta) &\cos(\theta)\end{bmatrix}\begin{bmatrix}\omega_{r }\\ \omega_{\theta}/r\end{bmatrix}=\begin{bmatrix}u_{x}\\ u_{y}\end{bmatrix}.$

How's that?

Wow, that looks good I must say...never saw this implemented in pde's before. Transposing of the matrix involves making the first row the first column and the second row the scond column. so if im right then

$u_x=-Sin(\theta)\omega_r+Cos(\theta)\omega_\theta /r$

$u_y=Cos(\theta)\omega_r+Sin(\theta)\omega_\theta /r$ . Rewriting the pde becomes

$\displaystyle r Sin (\theta)[-Sin(\theta)\omega_r+Cos(\theta)\omega_\theta /r]-r Cos(\theta)[Cos(\theta)\omega_r+Sin(\theta)\omega_\theta /r]$ =1

The IC being $\omega(r Cos(\theta),0)=0$ for $0< r Cos(\theta)< \infty$

Is this correct...Can this be solved using the method of characteristics? ill try..

**Ackbeet** · February 11th 2011, 10:45 AM

From

$\begin{bmatrix}\cos(\theta) &-\sin(\theta)\\ \sin(\theta) &\cos(\theta)\end{bmatrix}\begin{bmatrix}\omega_{r }\\ \omega_{\theta}/r\end{bmatrix}=\begin{bmatrix}u_{x}\\ u_{y}\end{bmatrix},$

you should do regular matrix multiplication to arrive at

$u_{x}=\omega_{r}\cos(\theta)-\omega_{\theta}\sin(\theta)/r,$

$u_{y}=\sin(\theta)\omega_{r}+\omega_{\theta}\cos(\ theta)/r.$

Where does that take you?

**bugatti79** · February 11th 2011, 10:53 AM

Originally Posted by Ackbeet

From

$\begin{bmatrix}\cos(\theta) &-\sin(\theta)\\ \sin(\theta) &\cos(\theta)\end{bmatrix}\begin{bmatrix}\omega_{r }\\ \omega_{\theta}/r\end{bmatrix}=\begin{bmatrix}u_{x}\\ u_{y}\end{bmatrix},$

you should do regular matrix multiplication to arrive at

$u_{x}=\omega_{r}\cos(\theta)-\omega_{\theta}\sin(\theta)/r,$

$u_{y}=\sin(\theta)\omega_{r}+\omega_{\theta}\cos(\ theta)/r.$

Where does that take you?

Sorry I multiplied from the wrong side

$\displaystyle r Sin (\theta)[Cos (\theta)\omega_r-Sin(\theta)\omega_\theta /r]-r Cos(\theta)[Sin(\theta)\omega_r+Cos(\theta)\omega_\theta /r]$ =1

I am not sure if this can be tackled using method of characteristics..I will try later

**Ackbeet** · February 11th 2011, 10:56 AM

Assuming you've plugged everything in correctly, doesn't that simplify down quite a bit? You've got a cancellation, and if I'm not mistaken, the identity $\sin^{2}(\theta)+\cos^{2}(\theta)=1$ might be useful.

**bugatti79** · February 11th 2011, 12:48 PM

Originally Posted by Ackbeet

Assuming you've plugged everything in correctly, doesn't that simplify down quite a bit? You've got a cancellation, and if I'm not mistaken, the identity $\sin^{2}(\theta)+\cos^{2}(\theta)=1$ might be useful.

I think reduces quite considerably, I get

$\displaystyle \frac{\partial \omega}{\partial \theta}[Sin^2 \theta +Cos^2 \theta]+\frac{\partial \omega}{\partial r}[rSin \theta Cos \theta-rSin \theta Cos \theta]=-1$ therefore

$\omega (\theta)=-\theta+C$ would this translate to

$\omega (\theta)=u(x(\theta),y(\theta))$ ...?

**Ackbeet** · February 11th 2011, 12:50 PM

No, I think when you integrate $\omega_{\theta}=-1,$ you're going to have to have

$\omega=-\theta+f(r),$ right?

**bugatti79** · February 11th 2011, 01:01 PM

Originally Posted by Ackbeet

No, I think when you integrate $\omega_{\theta}=-1,$ you're going to have to have

$\omega=-\theta+f(r),$ right?

Ok is that because the initial statement says that omega is a function of 2 variables even though $\frac{\partial\omega}{\partial \theta}=-1$ is explicitly a function of theta only?

From your equation using the IC $\omega(r Cos(\theta),0)=0$ i get

$0=-rCos \theta+f(r)$ therefore $\omega(r, \theta)=- \theta+rCos \theta$

That theta doesnt look right on its own!...

**Ackbeet** · February 11th 2011, 01:05 PM

It's because the integration you did was sort of "partial integration w.r.t. $y$ ", not a total integration like you'd have in an ODE setting. You remember doing exact ODE's? When you integrated w.r.t. $y$ , you'd get an arbitrary function of $x$ . In your case, $\omega=\omega(r,\theta),$ period. It is a function of two variables. If you doubt the extra function $f(r)$ , then just do a partial differentiation and see if you get back to the original DE. You do, right?

What is the IC, exactly?

**bugatti79** · February 11th 2011, 01:15 PM

Originally Posted by Ackbeet

It's because the integration you did was sort of "partial integration w.r.t. $y$ ", not a total integration like you'd have in an ODE setting. You remember doing exact ODE's? When you integrated w.r.t. $y$ , you'd get an arbitrary function of $x$ . In your case, $\omega=\omega(r,\theta),$ period. It is a function of two variables. If you doubt the extra function $f(r)$ , then just do a partial differentiation and see if you get back to the original DE. You do, right?

What is the IC, exactly?

No I agree with you..I actually had your expression intially but I somehow changed my mind

Its $u(x,0)=0 for 0<x<\infty$ which I believe translates to $\omega(r Cos(\theta),0)=0$ for $0< r Cos(\theta)< \infty$

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