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Thread: u_{xx}+u_{yy}=0

  1. #1
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    u_{xx}+u_{yy}=0

    Separation of Variables

    $\displaystyle u_{xx}+u_{yy}=0$

    $\displaystyle \displaystyle e^{rx+sy}=e^{rx}e^{sy}$

    $\displaystyle u(x,y)=\varphi(x)\psi(y)$

    $\displaystyle \varphi''(x)\psi(y)+\varphi(x)\psi''(y)=0$


    $\displaystyle \displaystyle\frac{\varphi''(x)\psi(y)+\varphi(x)\ psi''(y)}{\varphi(x)\psi(y)}=0$

    $\displaystyle \displaystyle \frac{\varphi''(x)}{\varphi(x)}+\frac{\psi''(y)}{\ psi(y)}=0$

    $\displaystyle \displaystyle \frac{\varphi''(x)}{\varphi(x)}=-\frac{\psi''(y)}{\psi(y)}$

    $\displaystyle \displaystyle \frac{\varphi''(x)}{\varphi(x)}=\lambda \ \mbox{and} \ \frac{\psi''(y)}{\psi(y)}=-\lambda$

    $\displaystyle \Rightarrow \varphi''(x)-\lambda\varphi(x)=0\Rightarrow \mbox{sol.s} \ \cos{(x\sqrt{\lambda})} \ \mbox{and} \ \sin{(x\sqrt{\lambda})}$

    $\displaystyle \Rightarrow \psi''(y)+\lambda\psi(y)=0\Rightarrow \mbox{sol.s} \ e^{\pm y\sqrt{\lambda}}$

    How are those solutions obtained?

    Also, I can't tell if the books has $\displaystyle e^{\pm y\sqrt{\lambda}}$ or $\displaystyle e^{\pm y^{\sqrt{\lambda}}}$.
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  2. #2
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    Quote Originally Posted by dwsmith View Post
    Separation of Variables

    $\displaystyle u_{xx}+u_{yy}=0$

    $\displaystyle \displaystyle e^{rx+sy}=e^{rx}e^{sy}$

    $\displaystyle u(x,y)=\varphi(x)\psi(y)$

    $\displaystyle \varphi''(x)\psi(y)+\varphi(x)\psi''(y)=0$


    $\displaystyle \displaystyle\frac{\varphi''(x)\psi(y)+\varphi(x)\ psi''(y)}{\varphi(x)\psi(y)}=0$

    $\displaystyle \displaystyle \frac{\varphi''(x)}{\varphi(x)}+\frac{\psi''(y)}{\ psi(y)}=0$

    $\displaystyle \displaystyle \frac{\varphi''(x)}{\varphi(x)}=-\frac{\psi''(y)}{\psi(y)}$

    $\displaystyle \displaystyle \frac{\varphi''(x)}{\varphi(x)}=\lambda \ \mbox{and} \ \frac{\psi''(y)}{\psi(y)}=-\lambda$

    $\displaystyle \Rightarrow \varphi''(x)-\lambda\varphi(x)=0\Rightarrow \mbox{sol.s} \ \cos{(x\sqrt{\lambda})} \ \mbox{and} \ \sin{(x\sqrt{\lambda})}$

    $\displaystyle \Rightarrow \psi''(y)+\lambda\psi(y)=0\Rightarrow \mbox{sol.s} \ e^{\pm y\sqrt{\lambda}}$

    How are those solutions obtained?

    Also, I can't tell if the books has $\displaystyle e^{\pm y\sqrt{\lambda}}$ or $\displaystyle e^{\pm y^{\sqrt{\lambda}}}$.
    The two final DEs are second-order linear constant coefficient homogeneous. I suggest you read Differential Equations Tutorial to refresh your memory how to solve DEs of this type.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    The two final DEs are second-order linear constant coefficient homogeneous.
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