Results 1 to 11 of 11

Math Help - x^{2}u_{x}-u_{y}+u=0

  1. #1
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5

    x^{2}u_{x}-u_{y}+u=0

    x^{2}u_{x}-u_{y}+u=0

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

    x^2\varphi'(x)\psi(y)-\varphi(x)\psi'(y)+\varphi(x)\psi(y)=0

    x^2\varphi'(x)\psi(y)+\varphi(x)[\psi(y)-\psi'(y)]=0

    \displaystyle\frac{x^2\varphi'(x)}{\varphi(x)}=-\left[\frac{\psi(y)-\psi'(y)}{\psi(y)}\right]

    \displaystyle x^2\varphi'(x)-\lambda\varphi(x)=0\Rightarrow x^2m-\lambda=0\Rightarrow m=\frac{\lambda}{x^2}

    \displaystyle\lambda=\frac{\psi'(y)}{\psi(y)}-1

    \psi'(y)-\lambda\psi(y)-1=0\Rightarrow n-\lambda-1=0\Rightarrow n=\lambda+1

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

    Can this method be used with variable coefficients?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,324
    Thanks
    8
    You need to learn about the method of charactistics. Not everything admits a separation of variables.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    3
    Quote Originally Posted by dwsmith View Post
    x^{2}u_{x}-u_{y}+u=0

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

    x^2\varphi'(x)\psi(y)-\varphi(x)\psi'(y)+\varphi(x)\psi(y)=0

    x^2\varphi'(x)\psi(y)+\varphi(x)[\psi(y)-\psi'(y)]=0

    \displaystyle\frac{x^2\varphi'(x)}{\varphi(x)}=-\left[\frac{\psi(y)-\psi'(y)}{\psi(y)}\right]

    \displaystyle x^2\varphi'(x)-\lambda\varphi(x)=0\Rightarrow x^2m-\lambda=0\Rightarrow m=\frac{\lambda}{x^2}

    \displaystyle\lambda=\frac{\psi'(y)}{\psi(y)}-1

    \psi'(y)-\lambda\psi(y)-1=0\Rightarrow n-\lambda-1=0\Rightarrow n=\lambda+1

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

    Can this method be used with variable coefficients?

    Thanks.
    Dear dwsmith,

    No. Suppose we have a differential equation such as, \displaystyle\varphi'(x)-\lambda\varphi(x)=0 We take the trail solution as, \varphi(x)=Ae^{mx}
    Why do we take this exponential function as our trail solution? This is based on the realization that the exponential function keeps its original shape even after differentiation. That is even after taking the derivative you get a the same exponential function with a different coefficient. This is necessary to substract the two functions, \varphi(x)~and~\varphi'(x). The same arguement is valid for any constant coefficient linear differential equation(either partial or ordinary). But when the equation is, \displaystyle x^2\varphi'(x)-\lambda\varphi(x)=0 taking the solution as \varphi(x)=Ae^{mx} is not justifiable since the equation is not a constant coefficient one. For more information refer, Linear differential equation - Wikipedia, the free encyclopedia

    Hope this helps to clarify matters.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by Danny View Post
    You need to learn about the method of charactistics. Not everything admits a separation of variables.
    Only on page 14 of the book. I am learning as it is presented.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by Sudharaka View Post
    Dear dwsmith,

    No. Suppose we have a differential equation such as, \displaystyle\varphi'(x)-\lambda\varphi(x)=0 We take the trail solution as, \varphi(x)=Ae^{mx}
    Why do we take this exponential function as our trail solution? This is based on the realization that the exponential function keeps its original shape even after differentiation. That is even after taking the derivative you get a the same exponential function with a different coefficient. This is necessary to substract the two functions, \varphi(x)~and~\varphi'(x). The same arguement is valid for any constant coefficient linear differential equation(either partial or ordinary). But when the equation is, \displaystyle x^2\varphi'(x)-\lambda\varphi(x)=0 taking the solution as \varphi(x)=Ae^{mx} is not justifiable since the equation is not a constant coefficient one. For more information refer, Linear differential equation - Wikipedia, the free encyclopedia

    Hope this helps to clarify matters.
    How am I suppose to approach this problem then? My book hasn't covered it yet.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    \displaystyle\frac{x^2\varphi'(x)}{\varphi(x)}=-\left[\frac{\psi(y)-\psi'(y)}{\psi(y)}\right]

    x^2\varphi'(x)-\lambda\varphi(x)=0

    \varphi(x)=x^m

    x^2mx^{m-1}-x^m\lambda=0\Rightarrow x^m[x^2mx^{-1}-\lambda]=0\Rightarrow x^m[xm-\lambda]=0

    Can I have an x in my m solution?

    \displaystyle m=\frac{\lambda}{x}\mbox{?}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,324
    Thanks
    8
    Nope. To solve this separate variables

    \dfrac{d \varphi}{\varphi} = \dfrac{\lambda}{x^2} and integrate.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    \displaystyle\int\frac{\varphi'(x)}{\varphi(x)}=\i  nt\frac{\lambda}{x^2}\varphi'(x)\Rightarrow \ln{(\varphi(x))}=\frac{-\lambda}{x}+c\Rightarrow \varphi(x)=C_2\exp{\left(\frac{-\lambda}{x}\right)}

    \displaystyle u(x,y)=C_2\exp{\left(\frac{-\lambda}{x}\right)}C_3\exp{(y(\lambda+1))}=C_4\exp  {\left(\frac{-\lambda}{x}\right)}\exp{(y(\lambda+1))}

    Correct?
    Last edited by dwsmith; January 4th 2011 at 11:43 AM. Reason: removed +
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    3
    Quote Originally Posted by dwsmith View Post
    x^{2}u_{x}-u_{y}+u=0

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

    x^2\varphi'(x)\psi(y)-\varphi(x)\psi'(y)+\varphi(x)\psi(y)=0

    x^2\varphi'(x)\psi(y)+\varphi(x)[\psi(y)-\psi'(y)]=0

    \displaystyle\frac{x^2\varphi'(x)}{\varphi(x)}=-\left[\frac{\psi(y)-\psi'(y)}{\psi(y)}\right]

    \displaystyle x^2\varphi'(x)-\lambda\varphi(x)=0\Rightarrow x^2m-\lambda=0\Rightarrow m=\frac{\lambda}{x^2}

    \displaystyle\lambda=\frac{\psi'(y)}{\psi(y)}-1

    \psi'(y)-\lambda\psi(y)-1=0\Rightarrow n-\lambda-1=0\Rightarrow n=\lambda+1

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

    Can this method be used with variable coefficients?

    Thanks.
    Dear dwsmith,

    There is a mistake, but somehow you have obtained the correct answer. The correct method is,

    \displaystyle\lambda=\frac{\psi'(y)}{\psi(y)}-1\Rightarrow \psi'(y)-\lambda\psi(y)-\psi(y)=0\Rightarrow n-\lambda-1=0\Rightarrow n=\lambda+1
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    I have it written down on paper correctly though
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    3
    Quote Originally Posted by dwsmith View Post
    \displaystyle\int\frac{\varphi'(x)}{\varphi(x)}=\i  nt\frac{\lambda}{x^2}\varphi'(x)\Rightarrow \ln{(\varphi(x))}=\frac{-\lambda}{x}+c\Rightarrow \varphi(x)=C_2\exp{\left(\frac{-\lambda}{x}\right)}

    \displaystyle u(x,y)=C_2\exp{\left(\frac{-\lambda}{x}\right)}C_3\exp{(y(\lambda+1))}=C_4\exp  {\left(\frac{-\lambda}{x}\right)}\exp{(y(\lambda+1))}

    Correct?
    Dear dwsmith,

    Well there are some problems. The integration should be, \displaystyle\int\frac{d(\varphi(x))}{\varphi(x)}=  \int\frac{\lambda}{x^2}dx and you have forgotten the absolute value of the natural logarithm.

     \ln{\mid\varphi(x)\mid}=\dfrac{-\lambda}{x}+c

    This wont change the answer but still it matters.
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum