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Thread: Solve Linear PDE via a change of variables

  1. #16
    Senior Member bugatti79's Avatar
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    I was hoping that by solving the pde the normal way which is part of the question that it might give a clue as to how the change of variables method should should turn out.

    Solving the original PDE $\displaystyle \displaystyle y \frac{\partial u}{\partial x}-x\frac{\partial u}{\partial y}=1$ I get

    $\displaystyle \displaystyle \frac{dy}{dx}=\frac{b(x,y)}{a(x,y)}=\frac{-x}{y} \implies \frac{y^2}{2}=\frac{-x^2}{2}+k$. Taking the positive solution of y

    $\displaystyle \displaystyle \frac{du}{dx}=\frac{h(x,y)}{a(x,y)}=\frac{1}{y} \implies u=\int \frac{1dx}{\sqrt{2k-x^2}}$

    which gives

    $\displaystyle \displaystyle u=Sin^{-1} [\frac{x}{\sqrt {2k}}]+f(\frac{y^2}{2} +\frac{x^2}{2})$

    The IC's are u(x,0)=0, therefore

    $\displaystyle \displaystyle 0=Sin^{-1} [\frac{x}{\sqrt {2(\frac{x^2}{2})}}]+f(\frac{x^2}{2})$. Let $\displaystyle \displaystyle t=\frac{x^2}{2}$ therefore

    $\displaystyle \displaystyle f(t)=\sqrt {2t}=-\frac {\pi}{2}$

    $\displaystyle \displaystyle u(x,y)=Sin^{-1} [\frac{x}{\sqrt {2(\frac{x^2}{2}+\frac{y^2}{2})}}}]-\frac {\pi}{2}$

    I am not sure of this is right either...
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  2. #17
    A Plied Mathematician
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    I'm afraid you take me beyond my province there. You see how little I know about pde's!
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  3. #18
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by bugatti79 View Post
    I was hoping that by solving the pde the normal way which is part of the question that it might give a clue as to how the change of variables method should should turn out.

    Solving the original PDE $\displaystyle \displaystyle y \frac{\partial u}{\partial x}-x\frac{\partial u}{\partial y}=1$ I get

    $\displaystyle \displaystyle \frac{dy}{dx}=\frac{b(x,y)}{a(x,y)}=\frac{-x}{y} \implies \frac{y^2}{2}=\frac{-x^2}{2}+k$. Taking the positive solution of y

    $\displaystyle \displaystyle \frac{du}{dx}=\frac{h(x,y)}{a(x,y)}=\frac{1}{y} \implies u=\int \frac{1dx}{\sqrt{2k-x^2}}$

    which gives

    $\displaystyle \displaystyle u=Sin^{-1} [\frac{x}{\sqrt {2k}}]+f(\frac{y^2}{2} +\frac{x^2}{2})$

    The IC's are u(x,0)=0, therefore

    $\displaystyle \displaystyle 0=Sin^{-1} [\frac{x}{\sqrt {2(\frac{x^2}{2})}}]+f(\frac{x^2}{2})$. Let $\displaystyle \displaystyle t=\frac{x^2}{2}$ therefore

    $\displaystyle \displaystyle f(t)=\sqrt {2t}=-\frac {\pi}{2}$

    $\displaystyle \displaystyle u(x,y)=Sin^{-1} [\frac{x}{\sqrt {2(\frac{x^2}{2}+\frac{y^2}{2})}}}]-\frac {\pi}{2}$

    I am not sure of this is right either...
    Referring to post #15

    $\displaystyle \displaystyle \omega_\theta=-\theta+f(r)$

    IC given is $\displaystyle \omega(rCos \theta,0)=0 $

    Using polar coordinates, when y is 0 (hence along the x line) theta is 0 therefore r Cos 0 =1 implies the f(r) =0. therefore

    $\displaystyle \omega(r,\theta)=-\theta$
    for $\displaystyle -\pi <\theta<\pi$

    Not sure how this domain of influence came about though.

    regarding post #16, Iv no idea how to solve it. I will try dx/dy instead of dy/dx and see where that gets me.
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  4. #19
    A Plied Mathematician
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    Using polar coordinates, when y is 0 (hence along the x line) theta is 0 therefore r Cos 0 =1 implies the f(r) =0.
    Why wouldn't it be $\displaystyle r\cos(\theta)=r?$ Why does $\displaystyle r=1?$ And why does $\displaystyle \theta=0?$ Couldn't it also equal $\displaystyle \pi?$ In that case, wouldn't

    $\displaystyle r\cos(\theta)=\pm r?$
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  5. #20
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Ackbeet View Post
    Why wouldn't it be $\displaystyle r\cos(\theta)=r?$ Why does $\displaystyle r=1?$ And why does $\displaystyle \theta=0?$ Couldn't it also equal $\displaystyle \pi?$ In that case, wouldn't

    $\displaystyle r\cos(\theta)=\pm r?$
    That is a typo, your right it should be r Cos 0 =r

    Good point...im not sure...this is where I get bogged down defining domain of influences.
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  6. #21
    A Plied Mathematician
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    Yeah, I think it's time for Danny, if he's willing.
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  7. #22
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    Quote Originally Posted by bugatti79 View Post
    I was hoping that by solving the pde the normal way which is part of the question that it might give a clue as to how the change of variables method should should turn out.

    Solving the original PDE $\displaystyle \displaystyle y \frac{\partial u}{\partial x}-x\frac{\partial u}{\partial y}=1$ I get

    $\displaystyle \displaystyle \frac{dy}{dx}=\frac{b(x,y)}{a(x,y)}=\frac{-x}{y} \implies \frac{y^2}{2}=\frac{-x^2}{2}+k$. Taking the positive solution of y

    $\displaystyle \displaystyle \frac{du}{dx}=\frac{h(x,y)}{a(x,y)}=\frac{1}{y} \implies u=\int \frac{1dx}{\sqrt{2k-x^2}}$

    which gives

    $\displaystyle \displaystyle u=Sin^{-1} [\frac{x}{\sqrt {2k}}]+f(\frac{y^2}{2} +\frac{x^2}{2})$

    The IC's are u(x,0)=0, therefore

    $\displaystyle \displaystyle 0=Sin^{-1} [\frac{x}{\sqrt {2(\frac{x^2}{2})}}]+f(\frac{x^2}{2})$. Let $\displaystyle \displaystyle t=\frac{x^2}{2}$ therefore

    $\displaystyle \displaystyle f(t)=\sqrt {2t}=-\frac {\pi}{2}$

    $\displaystyle \displaystyle u(x,y)=Sin^{-1} [\frac{x}{\sqrt {2(\frac{x^2}{2}+\frac{y^2}{2})}}}]-\frac {\pi}{2}$

    I am not sure of this is right either...
    What you have here is fine except replace the $\displaystyle \sqrt{2k}$ with $\displaystyle \sqrt{x^2+y^2}$ and simplify your final answer.

    You have a solution that satisifes the PDE and BC.

    Note: You can transform to polar coordinates if you like - the PDE becomes simple but you must interpret the BC, i.e.

    when $\displaystyle y = 0$, $\displaystyle u = 0$ for all $\displaystyle x $.

    Thus, $\displaystyle r \sin \theta = 0$ for all $\displaystyle r \cos \theta$ (This means that $\displaystyle r \ne 0$)
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