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Math Help - 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 y \frac{\partial u}{\partial x}-x\frac{\partial u}{\partial y}=1 I get

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

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

    \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 y \frac{\partial u}{\partial x}-x\frac{\partial u}{\partial y}=1 I get

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

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

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

    IC given is \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

    \omega(r,\theta)=-\theta
    for -\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 r\cos(\theta)=r? Why does r=1? And why does \theta=0? Couldn't it also equal \pi? In that case, wouldn't

    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 r\cos(\theta)=r? Why does r=1? And why does \theta=0? Couldn't it also equal \pi? In that case, wouldn't

    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 y \frac{\partial u}{\partial x}-x\frac{\partial u}{\partial y}=1 I get

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

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

    \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 \sqrt{2k} with \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 y = 0, u = 0 for all x .

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