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Math Help - Separation of Variables

  1. #1
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    Separation of Variables

    u_{xyz}-xyzu=0

    \varphi'(x)\psi'(y)\omega'(z)-xyz\varphi(x)\psi(y)\omega(z)=0

    \displaystyle\frac{\varphi'(x)}{\varphi(x)}=\frac{  xyz\psi(y)\omega(z)}{\psi'(y)\omega'(z)}

    \varphi'(x)-\lambda\varphi(x)=0\Rightarrow m=\lambda

    \varphi(x)=\exp{(x\lambda)}

    \lambda\psi'(y)\omega'(z)=xyz\psi(y)\omega(z)

    \displaystyle\frac{\psi'(y)}{\psi(y)}=\frac{xyz\om  ega(z)}{\lambda\omega'(z)}

    \psi'(y)-\mu\psi(y)=0\Rightarrow n=\mu

    \psi(y)=\exp{(y\mu)}

    \displaystyle\int\frac{\omega'(z)}{\omega(z)}=\int  \frac{xyz}{\lambda\mu}dz\Rightarrow \ln{|\omega(z)|}=\frac{xyz^2}{2\lambda\mu}+c

    \displaystyle\omega(z)=C_2\exp{\left(\frac{xyz^2}{  2\lambda\mu}\right)}

    \displaystyle u(x,y,z)=\varphi(x)\psi(y)\omega(z)=C\exp{\left(x\  lambda+y\mu+\frac{xyz^2}{2\lambda\mu}\right)}

    However, this solution doesn't check out.
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  2. #2
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    Quote Originally Posted by dwsmith View Post
    u_{xyz}-xyzu=0

    \varphi'(x)\psi'(y)\omega'(z)-xyz\varphi(x)\psi(y)\omega(z)=0

    \displaystyle\frac{\varphi'(x)}{\varphi(x)}=\frac{  xyz\psi(y)\omega(z)}{\psi'(y)\omega'(z)}

    \varphi'(x)-\lambda\varphi(x)=0\Rightarrow m=\lambda

    \varphi(x)=\exp{(x\lambda)}

    \lambda\psi'(y)\omega'(z)=xyz\psi(y)\omega(z)

    \displaystyle\frac{\psi'(y)}{\psi(y)}=\frac{xyz\om  ega(z)}{\lambda\omega'(z)}

    \psi'(y)-\mu\psi(y)=0\Rightarrow n=\mu

    \psi(y)=\exp{(y\mu)}

    \displaystyle\int\frac{\omega'(z)}{\omega(z)}=\int  \frac{xyz}{\lambda\mu}dz\Rightarrow \ln{|\omega(z)|}=\frac{xyz^2}{2\lambda\mu}+c

    \displaystyle\omega(z)=C_2\exp{\left(\frac{xyz^2}{  2\lambda\mu}\right)}

    \displaystyle u(x,y,z)=\varphi(x)\psi(y)\omega(z)=C\exp{\left(x\  lambda+y\mu+\frac{xyz^2}{2\lambda\mu}\right)}

    However, this solution doesn't check out.
    There's a mistake in this line

    \displaystyle\frac{\varphi'(x)}{\varphi(x)}=\frac{  xyz\psi(y)\omega(z)}{\psi'(y)\omega'(z)}

    It should be

    \displaystyle\frac{\varphi'(x)}{x\varphi(x)}=\frac  {yz\psi(y)\omega(z)}{\psi'(y)\omega'(z)}

    Notice the location of the x.
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  3. #3
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    So for psi, the y should be with it as well then?
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  4. #4
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    Yep you got it! BTW - what book are you going through and are you doing it independently?
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  5. #5
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    Quote Originally Posted by Danny View Post
    Yep you got it! BTW - what book are you going through and are you doing it independently?
    Elementary Partial Differential Equations by Berg and McGregor 1966. Yes, independently.
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  6. #6
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    Quote Originally Posted by dwsmith View Post
    Elementary Partial Differential Equations by Berg and McGregor 1966. Yes, independently.
    Wow - that's an old book. Couldn't you find anything more recent?
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  7. #7
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    Quote Originally Posted by Danny View Post
    Wow - that's an old book. Couldn't you find anything more recent?
    That is the one my prof likes.
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  8. #8
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    I don't mind the book but it lacks in examples.
    Last edited by dwsmith; January 5th 2011 at 04:21 PM.
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