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Math Help - u_{xx}-u_{yy}+u_y=0

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

    Using exponential substitution:

    \exp{(rx+sy)}

    u_{xx}-u_{yy}+u_y=0

    \exp{(rx+sy)}(r^2-s^2+s)=0

    r=\pm\sqrt{s^2-s}

    u=\exp{(\pm x\sqrt{s^2-s}+ys)}
    -\infty,0]\cup [1,\infty)" alt=" \ \ s-\infty,0]\cup [1,\infty)" />




    Can this be simplified more or is there a more tactful way to represent this?
    Last edited by dwsmith; December 29th 2010 at 11:52 PM. Reason: error in exp term
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  2. #2
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    The answer is this

    \displaystyle U=\exp{\left(\frac{1\pm\sqrt{1-4\lambda}}{2}y\right)}*\sin{(x\sqrt{\lambda})}

    \displaystyle V=\exp{\left(\frac{1\pm\sqrt{1-4\lambda}}{2}y\right)}*\cos{(x\sqrt{\lambda})}

    Lambda is any constant.

    I am not sure how this is the solution.
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  3. #3
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    Quote Originally Posted by dwsmith View Post
    The answer is this

    \displaystyle U=\exp{\left(\frac{1\pm\sqrt{1-4\lambda}}{2}y\right)}*\sin{(x\sqrt{\lambda})}

    \displaystyle V=\exp{\left(\frac{1\pm\sqrt{1-4\lambda}}{2}y\right)}*\cos{(x\sqrt{\lambda})}

    Lambda is any constant.

    I am not sure how this is the solution.
    Dear dwsmith,

    Therefore the general solution would be,

    u=Ae^{(x\sqrt{s^2-s}+ys)}+Be^{(-x\sqrt{s^2-s}+ys)}

    Substitute, \sqrt{s^2-s}=\sqrt{-\lambda}

    s=\frac{1\pm\sqrt{1-4\lambda}}{2}

    u=\exp{\left(\frac{1\pm\sqrt{1-4\lambda}}{2}y\right)}\left(Ae^{ix\sqrt{\lambda}}+  Be^{-ix\sqrt{\lambda}}\right)

    u=\exp{\left(\frac{1\pm\sqrt{1-4\lambda}}{2}y\right)}\left[(A+B)\cos x\sqrt{\lambda}+i(A-B)\sin x\sqrt{\lambda}\right)

    Take A+B=F and i(A-B)=G,

    u=F\exp{\left(\frac{1\pm\sqrt{1-4\lambda}}{2}y\right)}\cos x\sqrt{\lambda}+G\exp{\left(\frac{1\pm\sqrt{1-4\lambda}}{2}y\right)}\sin x\sqrt{\lambda}\right)

    Hope you understood.
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  4. #4
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    Quote Originally Posted by Sudharaka View Post

    s=\frac{1\pm\sqrt{1-4\lambda}}{2}
    How would you know to let s = that expression without knowing the answer?
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  5. #5
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    Quote Originally Posted by Sudharaka View Post

    Substitute, \sqrt{s^2-s}=\sqrt{-\lambda}
    Why is this negative lambda?
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  6. #6
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    Quote Originally Posted by dwsmith View Post
    Why is this negative lambda?
    Dear dwsmith,

    I have taken \sqrt{s^2-s}=\sqrt{-\lambda} so that, I get required value for s;

    s^2-s=-\lambda\Rightarrow s^2-s+\lambda=0\Rightarrow{s=\frac{1\pm\sqrt{1-4\lambda}}{2}}

    If I didnt know the answer I would not present it as I had done in post #3. I would have given the answer as,

    u=Ae^{(x\sqrt{s^2-s}+ys)}+Be^{(-x\sqrt{s^2-s}+ys)}

    But since you had given thes answers in the book and told that you are not sure how they have been obtained I had shown that both anwers are equivelent. Does this clarify your doubts?
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  7. #7
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    Now I am solving the same equation using separation of variables. However, I have hit a snag.

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

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

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

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

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

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

    \displaystyle \psi''(y)-\psi'(y)-\lambda\psi(y)=0\Rightarrow n^2-n-\lambda=0

    The first equation is suppose to yield s^2+\lambda, and the second equation needs to be of the form s^2-s+\lambda in order to obtain the same solutions from the exponential method.

    I haven't come across this in the book but can I just set both equations equal to -\lambda\mbox{?}
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  8. #8
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    Quote Originally Posted by dwsmith View Post
    Now I am solving the same equation using separation of variables. However, I have hit a snag.

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

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

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

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

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

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

    \displaystyle \psi''(y)-\psi'(y)-\lambda\psi(y)=0\Rightarrow n^2-n-\lambda=0

    The first equation is suppose to yield s^2+\lambda, and the second equation needs to be of the form s^2-s+\lambda in order to obtain the same solutions from the exponential method.

    I haven't come across this in the book but can I just set both equations equal to -\lambda\mbox{?}
    Dear dwsmith,

    Yes you can. What is the reasoning behind setting \displaystyle \frac{\varphi''(x)}{\varphi(x)}=\lambda ~and~ \frac{\psi''(y)-\psi'(y)}{\psi(y)}=\lambda ?? It is that the two parts are independent of y and x respectively. So that if they are equal to each other there must be a common constant to which both are equal. In this case you had taken it to be \lambda. But you can take the constant as you like it to be. For example instead of -\lambda you can take \alpha if you dont like taking a the negative sign. Then there exists \lambda such that \alpha=-\lambda. Substitute for \alpha and you will get the same answer.

    Hope you got the idea. I am a non native speaker of english. So if you find anything that confuses you please do not hesitate to ask me.
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