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Math Help - u_{xy}+u_{yz}+u_{zx}-u=0

  1. #1
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    u_{xy}+u_{yz}+u_{zx}-u=0

    Using separation of variables

    u_{xy}+u_{yz}+u_{zx}-u=0

    u(x,y,z)=\varphi(x)\psi(y)\omega(z)

    \varphi'(x)\psi'(y)\omega(z)+\varphi(x)\psi'(y)\om  ega'(z)+\varphi'(x)\psi(y)\omega'(z)-\varphi(x)\psi(y)\omega(z)=0

    \varphi'(x)[\psi'(y)\omega(z)+\psi(y)\omega'(z)]+\varphi(x)[\psi'(y)\omega'(z)-\psi(y)\omega(z)]=0

    \displaystyle\frac{\varphi'(x)}{\varphi(x)}=-\left(\frac{\psi'(y)\omega(z)+\psi(y)\omega'(z)}{\  psi'(y)\omega'(z)-\psi(y)\omega(z)}\right)

    Now what?
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  2. #2
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    Well, pick a constant of separation:

    \displaystyle\frac{\varphi'(x)}{\varphi(x)}=\lambd  a=-\left(\frac{\psi'(y)\omega(z)+\psi(y)\omega'(z)}{\  psi'(y)\omega'(z)-\psi(y)\omega(z)}\right).

    The first equality gives you a DE for \varphi. The second equality gives you another equation that, I believe separates out thus:

    \psi'(y)\omega(z)+\psi(y)\omega'(z)=-\lambda(\psi'(y)\omega'(z)-\psi(y)\omega(z)).

    You can do the same sort of trick you just did. That is, solve for \psi'(y)/\psi(y), and choose another separation constant.

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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by dwsmith View Post
    u(x,y,z)=\varphi(x)\psi(y)\omega(z)
    How do you know this?
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  4. #4
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    \varphi'(x)-\lambda\varphi(x)=0\Rightarrow m-\lambda=0\Rightarrow m=\lambda

    \displaystyle\lambda=-\left(\frac{\psi'(y)\omega(z)+\psi(y)\omega'(z)}{\  psi'(y)\omega'(z)-\psi(y)\omega(z)}\right)}

    \displaystyle\Rightarrow\lambda\psi'(y)\omega'(z)-\lamba\psi(y)\omega(z)}\right)+\psi'(y)\omega(z)+\  psi(y)\omega'(z)=0

    \Rightarrow\psi'(y)[\lambda\omega'(z)+\omega(z)}]+\psi(y)[\omega'(z)-\lambda\omega(z)]=0

    \displaystyle\Rightarrow\mu=\frac{\psi'(y)}{\psi(y  )}=-\left(\frac{\omega'(z)-\lambda\omega(z)}{\lambda\omega'(z)+\omega(z)}\rig  ht)

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

    \displaystyle\Rightarrow\mu=-\left(\frac{\omega'(z)-\lambda\omega(z)}{\lambda\omega'(z)+\omega(z)}\rig  ht)

    \omega'(z)-\lambda\omega(z)+\mu\lambda\omega'(z)+\mu\omega(z)  =0

    \displaystyle\omega'(z)[1+\mu\lambda]+\omega(z)[\mu-\lambda]=0\Rightarrow a(1+\mu\lambda)+\mu-\lambda=0\Rightarrow a=\frac{\lambda-\mu}{1+\mu\lambda}

    \displaystyle u(x,y,z;\lambda,\mu)=\exp{(x\lambda)}*\exp{(y\mu)}  *\exp{\left(z\frac{\lambda-\mu}{1+\mu\lambda}\right)}

    I tried the substitution of \displaystyle\mu=s \ \mbox{and} \ \lambda=\frac{1-\mu t}{\mu+t} in order to manipulate the exponential method to the separations method but it fell short.

    Exponential solution:

    \displaystyle u(x,y,z;s,t)=\exp{\left(x\frac{1-st}{s+t}\right)}*\exp{(ys)}*\exp{(zt)}

    What substitution do I need to make?

    Thanks.
    Last edited by dwsmith; January 2nd 2011 at 12:30 PM.
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  5. #5
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    Quote Originally Posted by chiph588@ View Post
    How do you know this?
    This is from Elementary PDE by Berg and McGregor 1966.

    ...different technique for finding a particular solutions of homogeneous linear PDE is called the method of separation of variables. The exponential solution \exp{(rx+sy)}=\exp{(rx)}*\exp{(sy)}, are products of functions of the separable variables.... We seek a solution of the form u(x,y)=\varphi(x)\psi(y) (pg. 14).
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  6. #6
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    Quote Originally Posted by dwsmith View Post
    \varphi'(x)-\lambda\varphi(x)=0\Rightarrow m-\lambda=0\Rightarrow m=\lambda

    \displaystyle\lambda=-\left(\frac{\psi'(y)\omega(z)+\psi(y)\omega'(z)}{\  psi'(y)\omega'(z)-\psi(y)\omega(z)}\right)}

    \displaystyle\Rightarrow\lambda\psi'(y)\omega'(z)-\lamba\psi(y)\omega(z)}\right)+\psi'(y)\omega(z)+\  psi(y)\omega'(z)=0

    \Rightarrow\psi'(y)[\lambda\omega'(z)+\omega(z)}]+\psi(y)[\omega'(z)-\lambda\omega(z)]=0

    \displaystyle\Rightarrow\mu=\frac{\psi'(y)}{\psi(y  )}=-\left(\frac{\omega'(z)-\lambda\omega(z)}{\lambda\omega'(z)+\omega(z)}\rig  ht)

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

    \displaystyle\Rightarrow\mu=-\left(\frac{\omega'(z)-\lambda\omega(z)}{\lambda\omega'(z)+\omega(z)}\rig  ht)

    \omega'(z)-\lambda\omega(z)+\mu\lambda\omega'(z)+\mu\omega(z)  =0

    \displaystyle\omega'(z)[1+\mu\lambda]+\omega(z)[\mu-\lambda]=0\Rightarrow a(1+\mu\lambda)+\mu-\lambda=0\Rightarrow a=\frac{\lambda-\mu}{1+\mu\lambda}

    \displaystyle u(x,y,z;\lambda,\mu)=\exp{(x\lambda)}*\exp{(y\mu)}  *\exp{\left(z\frac{\lambda-\mu}{1+\mu\lambda}\right)}

    I tried the substitution of \displaystyle\mu=s \ \mbox{and} \ \lambda=\frac{1-\mu t}{\mu+t} in order to manipulate the exponential method to the separations method but it fell short.

    Exponential solution:

    \displaystyle u(x,y,z;s,t)=\exp{x\left(\frac{1-st}{s+t}\right)}*\exp{(ys)}*\exp{(zt)}

    What substitution do I need to make?

    Thanks.
    Dear dwsmith,

    Think about it like this. You have the two solutions,

    \displaystyle u(x,y,z;\lambda,\mu)=\exp{(x\lambda)}*\exp{(y\mu)}  *\exp{\left(z\frac{\lambda-\mu}{1+\mu\lambda}\right)}--------(Seperation method)

    \displaystyle u(x,y,z;s,t)=\exp{x\left(\frac{1-st}{s+t}\right)}*\exp{(ys)}*\exp{(zt)}---------(Exponential method)

    Consider the exponential part with y as the variable. Since the coefficient of y must be equal in both of these equations,

    \mu=s---------(1)
    Now move on to the exponential term with z as the variable. Using the same reasoning,

    t=\dfrac{\lambda-\mu}{1+\mu\lambda}\Rightarrow t=\dfrac{\lambda-s}{1+s\lambda}\Rightarrow \lambda=\dfrac{t+s}{1-ts}--------(2)

    (1) and (2) gives the substitutions necessary to convert the solution obtained from the seperation method to the solution obtained from the exponential method. Hope you understood.
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  7. #7
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    By making that substitution, we don't obtain the same solutions.

    \displaystyle u(x,y,z;\lambda,\mu)=\exp{(x\lambda)}*\exp{(y\mu)}  *\exp{\left(z\frac{\lambda-\mu}{1+\mu\lambda}\right)}

    \displaystyle s=\mu, \ t=\frac{\lambda-\mu}{1+\mu\lambda}, \ \mbox{and} \ \lambda=\frac{t+s}{1-st}

    \displaystyle u(x,y,z;s,t)\rightarrow u(x,y,z;\lambda,\mu)=\exp{\left(x\frac{1}{\lambda}  \right)}*\exp{(\mu s)}*\exp{\left(z\frac{\lambda-\mu}{1+\mu\lambda}\right)}

    Any thoughts?

    I could say \displaystyle \frac{1}{\lambda}=\lambda_2 but would this be ok?
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  8. #8
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    Quote Originally Posted by dwsmith View Post
    By making that substitution, we don't obtain the same solutions.

    \displaystyle u(x,y,z;\lambda,\mu)=\exp{(x\lambda)}*\exp{(y\mu)}  *\exp{\left(z\frac{\lambda-\mu}{1+\mu\lambda}\right)}

    \displaystyle s=\mu, \ t=\frac{\lambda-\mu}{1+\mu\lambda}, \ \mbox{and} \ \lambda=\frac{t+s}{1-st}

    \displaystyle u(x,y,z;s,t)\rightarrow u(x,y,z;\lambda,\mu)=\exp{\left(x\frac{1}{\lambda}  \right)}*\exp{(\mu s)}*\exp{\left(z\frac{\lambda-\mu}{1+\mu\lambda}\right)}

    Any thoughts?

    I could say \displaystyle \frac{1}{\lambda}=\lambda_2 but would this be ok?
    Dear dwsmith,

    You should get the same results. The problem is that there is a mistake in your calculations. Its in your first post. The fifth line should be,

    \displaystyle\frac{\varphi'(x)}{\varphi(x)}=-\left(\frac{\psi'(y)\omega'(z)-\psi(y)\omega(z)}{\psi'(y)\omega(z)+\psi(y)\omega'  (z)}\right)

    You will have to do it all over again.....
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  9. #9
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    Quote Originally Posted by Sudharaka View Post
    Dear dwsmith,

    You should get the same results. The problem is that there is a mistake in your calculations. Its in your first post. The fifth line should be,

    \displaystyle\frac{\varphi'(x)}{\varphi(x)}=-\left(\frac{\psi'(y)\omega'(z)-\psi(y)\omega(z)}{\psi'(y)\omega(z)+\psi(y)\omega'  (z)}\right)

    You will have to do it all over again.....
    Thanks. I am not redoing the problem though haha.
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