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Thread: PDE

  1. #1
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    PDE

    Need help with this problem. Problem is attached.

    Thanks
    Attached Thumbnails Attached Thumbnails PDE-problem-10.jpg  
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by meks08999 View Post
    Need help with this problem. Problem is attached.

    Thanks
    Something is wrong......


    $\displaystyle u(x,y)=\psi(s)$ where $\displaystyle s=x^2+y^2$

    taking some derivatives we get

    $\displaystyle u_x=\frac{d \psi}{ds}(2x)$ and

    $\displaystyle u_y=\frac{d \psi}{ds}(2y)$

    Now

    $\displaystyle yu_x+xu_y=2xy\frac{d \psi}{ds}+2xy\frac{d \psi}{ds}=4xy\frac{d \psi}{dx} \ne 0$
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  3. #3
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    What do you mean by something is wrong?
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  4. #4
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    Quote Originally Posted by meks08999 View Post
    What do you mean by something is wrong?
    The preposed solution is not a solution to the PDE.

    If you think about the problem geometrically

    $\displaystyle yu_x+xu_y=0$ the left hand side is the directional derviative of the function.

    This states the function is constant in the direction of the vector $\displaystyle y \vec{i} +x \vec{j}$

    This tells us that $\displaystyle \frac{dy}{dx}=\frac{x}{y} \iff y^2=x^2+c \iff y^2-x^2=c$

    This is a family of hyperbola's not circles as the problem asks......

    Your PDE needs to have one of the terms negative for that to be a solution.

    $\displaystyle u(x,y)=\psi(y^2-x^2)$ is a solution to the pde.

    I.e if your problem read

    $\displaystyle -yu_x+xu_y=0$ or $\displaystyle yu_x-xu_y=0$

    $\displaystyle u(x,y)=\psi(y^2+x^2)$ would be a solution.
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  5. #5
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    There is an error in the problem. Thanks for pointing it out. I just received an email from my teacher just a little bit ago saying there was an incorrect sign.

    It was supposed to be yu_x-xu_y=0.

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  6. #6
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    Quote Originally Posted by meks08999 View Post
    There is an error in the problem. Thanks for pointing it out. I just received an email from my teacher just a little bit ago saying there was an incorrect sign.

    It was supposed to be yu_x-xu_y=0.
    To change variables you need to use the chain rule

    $\displaystyle u(x,y)=u(r\cos(\theta),r\sin(\theta))$

    Hint: Find $\displaystyle \frac{\partial u }{\partial \theta}$
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  7. #7
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    Use these chain rules

    $\displaystyle
    u_x = u_r r_x + u_{\theta} \theta_x
    $

    $\displaystyle
    u_y = u_r r_y + u_{\theta} \theta_y.
    $
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