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Math Help - Solve Linear PDE via a change of variables

  1. #1
    Senior Member bugatti79's Avatar
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    Solve Linear PDE via a change of variables

     <br />
\omega(r,\theta)=u(r cos (\theta),r sin (\theta))<br />
for some function u(x,y)

     <br />
\displaystyle y \frac{\partial u}{\partial x}-x\frac{\partial u}{\partial y}=1<br />


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

     <br />
\displaystyle \frac{\partial \omega}{\partial r} and  <br />
\displaystyle \frac{\partial \omega}{\partial \theta}in terms of  <br />
\displaystyle \frac{\partial u}{\partial x} and  \displaystyle \frac{\partial u}{\partial y}<br />
and solve.

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

     <br />
\displaystyle \omega(r,\theta) = u(x(r,\theta),y(r,\theta)) therefore \begin{cases}<br />
\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  <br />
\displaystyle \frac{\partial u}{\partial x} and  \displaystyle \frac{\partial u}{\partial y}<br />

    Have I gone one step too far?
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  2. #2
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    Could be way off here, but couldn't you just use this:

    \displaystyle\begin{cases}<br />
\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}?
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  3. #3
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Ackbeet View Post
    Could be way off here, but couldn't you just use this:

    \displaystyle\begin{cases}<br />
\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
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  4. #4
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    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.
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  5. #5
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by bugatti79 View Post
    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}<br />
\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!
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    Quote Originally Posted by bugatti79 View Post
    \displaystyle\begin{cases}<br />
\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?
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  7. #7
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Ackbeet View Post
    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..
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    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?
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  9. #9
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Ackbeet View Post
    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
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  10. #10
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    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.
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  11. #11
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Ackbeet View Post
    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))...?
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  12. #12
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    No, I think when you integrate \omega_{\theta}=-1, you're going to have to have

    \omega=-\theta+f(r), right?
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  13. #13
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    Quote Originally Posted by Ackbeet View Post
    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!...
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  14. #14
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    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?
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  15. #15
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Ackbeet View Post
    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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