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Thread: 2nd order PDE, using transformations...?

  1. #1
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    2nd order PDE, using transformations...?

    Please help me understand where i have gone wrong !

    The problem statement: A function u(r,t) satisfies the equation $\displaystyle \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial u}{\partial r}\right) = \frac{1}{c^2}\frac{\partial^2 u}{\partial t^2}$ ---(1)
    where c is constant.

    a) By introducing the new dependant variable v=ru(r,t) and writing
    $\displaystyle \alpha = r + ct, \beta = r-ct$,
    show that $\displaystyle \frac{\partial^2v}{\partial \alpha\partial \beta} =\,0$

    My work : I simplify (1); $\displaystyle \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial u}{\partial r}\right) = $$\displaystyle \,\frac{1}{r^2}\left[\frac{\partial}{\partial r}(r^2)\frac{\partial u}{\partial r}+\frac{\partial}{\partial r}(\frac{\partial u}{\partial r})r^2\right]$ = $\displaystyle \frac{1}{r^2}\left[2r\frac{\partial u}{\partial r} + r^2\frac{\partial^2 u}{\partial r^2}\right]$
    so that (1) = $\displaystyle \frac{2}{r}\frac{\partial u}{\partial r}+\frac{\partial^2 u}{\partial r^2}-\frac{1}{c^2}\frac{\partial^2 u}{\partial t^2} = 0$

    And now with the substitutions $\displaystyle \alpha = r + ct, \beta = r-ct$ i find the values( and switching notation,gonna be here all night!)
    $\displaystyle u_{r}=u_{\alpha} + u_{\beta}$
    $\displaystyle u_{t}=cu_{\alpha}-cu_{\beta}$
    $\displaystyle u_{rr}=u_{\alpha\alpha}+2u_{\alpha\beta}+u_{\beta\ beta}$
    $\displaystyle u_{tt}=c^2\left(u_{\alpha\alpha}-2u_{\alpha\beta}+u_{\beta\beta}\right)$
    So that (1) now becomes $\displaystyle \frac{2}{r}\left(u_{\alpha} + u_{\beta}\right) + \left(u_{\alpha\alpha}+2u_{\alpha\beta}+u_{\beta\b eta}\right)-\frac{1}{c^2}\left(c^2\left(u_{\alpha\alpha}-2u_{\alpha\beta}+u_{\beta\beta}\right)\right) = 0$
    which becomes $\displaystyle \frac{2}{r}\left(u_{\alpha} + u_{\beta}\right) + \left(u_{\alpha\alpha}+2u_{\alpha\beta}+u_{\beta\b eta}\right)-\left(u_{\alpha\alpha}-2u_{\alpha\beta}+u_{\beta\beta}\right) = 0$
    And finally $\displaystyle \frac{2}{r}\left(u_{\alpha} + u_{\beta}\right) + 4u_{\alpha\beta} = 0$ -----(2)

    Now there is precisely 1 example in my textbook,and hardly any theory behind this method! At this point they use the new dependant variable v in the transformation $\displaystyle u=ve^{\alpha\mu+\eta\beta}$ This is where i start getting stuck! I don't think i know how to use the new dependant variable properly!

    Do they always use this transformation $\displaystyle u=ve^{\alpha\mu+\eta\beta}$ and since i only need to prove$\displaystyle \frac{\partial^2v}{\partial \alpha\partial \beta} =\,0$ i only need $\displaystyle v(\alpha,\beta)$ right ? Let me go on.

    with the transformation $\displaystyle u=ve^{\alpha\mu+\eta\beta}$ i get the following values:
    $\displaystyle u_{\alpha}=v_{\alpha}e^{\alpha\mu+\eta\beta}+v\mu e^{\alpha\mu+\eta\beta}=e^{\alpha\mu+\eta\beta}\le ft(v_{\alpha}+v\mu\right)$
    $\displaystyle u_{\beta}=v_{\beta}e^{\alpha\mu+\eta\beta}+\eta ve^{\alpha\mu+\eta\beta}=e^{\alpha\mu+\eta\beta}\l eft(v_{\beta}+\eta v\right)$
    $\displaystyle u_{\alpha\beta}=\frac{\partial}{\partial \alpha}\left[e^{\alpha\mu+\eta\beta}\left(v_{\beta}+\eta v\right)\right]=\mu e^{\alpha\mu+\eta\beta}(v_{\beta}+\eta v) + v_{\alpha\beta}e^{\alpha\mu+\eta\beta}+\eta v_{\alpha}e^{\alpha\mu+\eta\beta}$
    = $\displaystyle e^{\alpha\mu+\eta\beta}\left(\mu v_{\beta}+\mu \eta v +v_{\alpha\beta}+\eta v_{\alpha}\right)$

    So making these substitutions,
    (2) becomes $\displaystyle \frac{2}{r}\left[e^{\alpha\mu+\eta\beta}\left(v_{\alpha}+v_{\beta}+ v\eta+v\mu \right)\right] + 4e^{\alpha\mu+\eta\beta}\left(\mu v_{\beta}+\mu \eta v +v_{\alpha\beta}+\eta v_{\alpha}\right) = 0$
    which becomes $\displaystyle e^{\alpha\mu+\eta\beta}\left[\frac{2}{r}\left(v_{\alpha}+v_{\beta}+v\eta+v\mu \right) + 4(\mu v_{\beta} +\mu \eta v +v_{\alpha\beta} +\eta v_{\alpha})\right] = 0$
    Which means $\displaystyle \frac{2}{r}\left(v_{\alpha}+v_{\beta}+v\eta+v\mu \right) + 4(\mu v_{\beta} +\mu \eta v +v_{\alpha\beta} +\eta v_{\alpha}) = 0$. Regrouping the order of these
    equations i get $\displaystyle 4v_{\alpha\beta}+v_{\alpha}\left(\frac{2}{r} + 4 \eta \right) + v_{\beta}\left(\frac{2}{r}+4 \mu \right) + v \left(\frac{2 \mu}{r} +\frac{2 \eta}{r} +4 \mu \eta \right)=0$
    Now to eliminate the first order equations, i can set
    $\displaystyle \frac{2}{r} + 4 \eta = 0,\, so\, \mu = -\frac{1}{2r} $
    $\displaystyle \frac{2}{r}+4 \mu =0, \,so\,\, \eta = -\frac{1}{2r}$
    I am left with the equation $\displaystyle 4v_{\alpha\beta} -\frac{1}{r^2}v=0$ -----(3)

    Now,i try to solve for v by seperation of variables:
    Let $\displaystyle v(\alpha,\beta) = X(\alpha)Y(\beta)$
    Then (3) becomes $\displaystyle 4X^{\prime}Y^{\prime}-\frac{1}{r^2}XY= 0$
    Which leads to $\displaystyle X^{\prime}-\frac{\lambda}{4}X=0$ and $\displaystyle Y^{\prime}-\frac{1}{\lambda r^2}Y$ from which i get
    $\displaystyle X=e^{\frac{\lambda \alpha}{4}}$ and Y = $\displaystyle e^{\frac{\beta}{\lambda r^2}}$
    And so $\displaystyle v(\alpha,\beta)=\left(e^{\frac{\lambda \alpha}{4}}\right)\left(e^{\frac{\beta}{\lambda r^2}}\right)\,=\,e^{\left(\frac{\lambda \alpha}{4}+\frac{\beta}{\lambda r^2}\right)}$

    But here's the problem ! $\displaystyle \frac{\partial^2 v}{\partial \alpha \beta} = \frac{1}{4r^2}e^{\left(\frac{\lambda \alpha}{4}+\frac{\beta}{\lambda r^2}\right)}$ which is never zero, so after all this i can not prove what is asked of me! can someone please tell me what i am doing wrong here?
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  2. #2
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    Quote Originally Posted by punkstart View Post
    Please help me understand where i have gone wrong !

    The problem statement: A function u(r,t) satisfies the equation $\displaystyle \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial u}{\partial r}\right) = \frac{1}{c^2}\frac{\partial^2 u}{\partial t^2}$ ---(1)
    where c is constant.

    a) By introducing the new dependant variable v=ru(r,t) and writing
    $\displaystyle \alpha = r + ct, \beta = r-ct$,
    show that $\displaystyle \frac{\partial^2v}{\partial \alpha\partial \beta} =\,0$

    My work : I simplify (1); $\displaystyle \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial u}{\partial r}\right) = $$\displaystyle \,\frac{1}{r^2}\left[\frac{\partial}{\partial r}(r^2)\frac{\partial u}{\partial r}+\frac{\partial}{\partial r}(\frac{\partial u}{\partial r})r^2\right]$ = $\displaystyle \frac{1}{r^2}\left[2r\frac{\partial u}{\partial r} + r^2\frac{\partial^2 u}{\partial r^2}\right]$
    so that (1) = $\displaystyle \frac{2}{r}\frac{\partial u}{\partial r}+\frac{\partial^2 u}{\partial r^2}-\frac{1}{c^2}\frac{\partial^2 u}{\partial t^2} = 0$

    And now with the substitutions $\displaystyle \alpha = r + ct, \beta = r-ct$ i find the values( and switching notation,gonna be here all night!)
    $\displaystyle u_{r}=u_{\alpha} + u_{\beta}$
    $\displaystyle u_{t}=cu_{\alpha}-cu_{\beta}$
    $\displaystyle u_{rr}=u_{\alpha\alpha}+2u_{\alpha\beta}+u_{\beta\ beta}$
    $\displaystyle u_{tt}=c^2\left(u_{\alpha\alpha}-2u_{\alpha\beta}+u_{\beta\beta}\right)$
    So that (1) now becomes $\displaystyle \frac{2}{r}\left(u_{\alpha} + u_{\beta}\right) + \left(u_{\alpha\alpha}+2u_{\alpha\beta}+u_{\beta\b eta}\right)-\frac{1}{c^2}\left(c^2\left(u_{\alpha\alpha}-2u_{\alpha\beta}+u_{\beta\beta}\right)\right) = 0$
    which becomes $\displaystyle \frac{2}{r}\left(u_{\alpha} + u_{\beta}\right) + \left(u_{\alpha\alpha}+2u_{\alpha\beta}+u_{\beta\b eta}\right)-\left(u_{\alpha\alpha}-2u_{\alpha\beta}+u_{\beta\beta}\right) = 0$
    And finally $\displaystyle \frac{2}{r}\left(u_{\alpha} + u_{\beta}\right) + 4u_{\alpha\beta} = 0$ -----(2)

    Now there is precisely 1 example in my textbook,and hardly any theory behind this method! At this point they use the new dependant variable v in the transformation $\displaystyle u=ve^{\alpha\mu+\eta\beta}$ This is where i start getting stuck! I don't think i know how to use the new dependant variable properly!

    Do they always use this transformation $\displaystyle u=ve^{\alpha\mu+\eta\beta}$ and since i only need to prove$\displaystyle \frac{\partial^2v}{\partial \alpha\partial \beta} =\,0$ i only need $\displaystyle v(\alpha,\beta)$ right ? Let me go on.

    with the transformation $\displaystyle u=ve^{\alpha\mu+\eta\beta}$ i get the following values:
    $\displaystyle u_{\alpha}=v_{\alpha}e^{\alpha\mu+\eta\beta}+v\mu e^{\alpha\mu+\eta\beta}=e^{\alpha\mu+\eta\beta}\le ft(v_{\alpha}+v\mu\right)$
    $\displaystyle u_{\beta}=v_{\beta}e^{\alpha\mu+\eta\beta}+\eta ve^{\alpha\mu+\eta\beta}=e^{\alpha\mu+\eta\beta}\l eft(v_{\beta}+\eta v\right)$
    $\displaystyle u_{\alpha\beta}=\frac{\partial}{\partial \alpha}\left[e^{\alpha\mu+\eta\beta}\left(v_{\beta}+\eta v\right)\right]=\mu e^{\alpha\mu+\eta\beta}(v_{\beta}+\eta v) + v_{\alpha\beta}e^{\alpha\mu+\eta\beta}+\eta v_{\alpha}e^{\alpha\mu+\eta\beta}$
    = $\displaystyle e^{\alpha\mu+\eta\beta}\left(\mu v_{\beta}+\mu \eta v +v_{\alpha\beta}+\eta v_{\alpha}\right)$

    So making these substitutions,
    (2) becomes $\displaystyle \frac{2}{r}\left[e^{\alpha\mu+\eta\beta}\left(v_{\alpha}+v_{\beta}+ v\eta+v\mu \right)\right] + 4e^{\alpha\mu+\eta\beta}\left(\mu v_{\beta}+\mu \eta v +v_{\alpha\beta}+\eta v_{\alpha}\right) = 0$
    which becomes $\displaystyle e^{\alpha\mu+\eta\beta}\left[\frac{2}{r}\left(v_{\alpha}+v_{\beta}+v\eta+v\mu \right) + 4(\mu v_{\beta} +\mu \eta v +v_{\alpha\beta} +\eta v_{\alpha})\right] = 0$
    Which means $\displaystyle \frac{2}{r}\left(v_{\alpha}+v_{\beta}+v\eta+v\mu \right) + 4(\mu v_{\beta} +\mu \eta v +v_{\alpha\beta} +\eta v_{\alpha}) = 0$. Regrouping the order of these
    equations i get $\displaystyle 4v_{\alpha\beta}+v_{\alpha}\left(\frac{2}{r} + 4 \eta \right) + v_{\beta}\left(\frac{2}{r}+4 \mu \right) + v \left(\frac{2 \mu}{r} +\frac{2 \eta}{r} +4 \mu \eta \right)=0$
    Now to eliminate the first order equations, i can set
    $\displaystyle \frac{2}{r} + 4 \eta = 0,\, so\, \mu = -\frac{1}{2r} $
    $\displaystyle \frac{2}{r}+4 \mu =0, \,so\,\, \eta = -\frac{1}{2r}$
    I am left with the equation $\displaystyle 4v_{\alpha\beta} -\frac{1}{r^2}v=0$ -----(3)

    Now,i try to solve for v by seperation of variables:
    Let $\displaystyle v(\alpha,\beta) = X(\alpha)Y(\beta)$
    Then (3) becomes $\displaystyle 4X^{\prime}Y^{\prime}-\frac{1}{r^2}XY= 0$
    Which leads to $\displaystyle X^{\prime}-\frac{\lambda}{4}X=0$ and $\displaystyle Y^{\prime}-\frac{1}{\lambda r^2}Y$ from which i get
    $\displaystyle X=e^{\frac{\lambda \alpha}{4}}$ and Y = $\displaystyle e^{\frac{\beta}{\lambda r^2}}$
    And so $\displaystyle v(\alpha,\beta)=\left(e^{\frac{\lambda \alpha}{4}}\right)\left(e^{\frac{\beta}{\lambda r^2}}\right)\,=\,e^{\left(\frac{\lambda \alpha}{4}+\frac{\beta}{\lambda r^2}\right)}$

    But here's the problem ! $\displaystyle \frac{\partial^2 v}{\partial \alpha \beta} = \frac{1}{4r^2}e^{\left(\frac{\lambda \alpha}{4}+\frac{\beta}{\lambda r^2}\right)}$ which is never zero, so after all this i can not prove what is asked of me! can someone please tell me what i am doing wrong here?
    A couple of things. First, (2) is right but you must remember that when you switch variables, everthing switches. Thus, $\displaystyle r = \alpha + \beta$ and (2) becomes

    $\displaystyle \frac{2}{\alpha + \beta}\left(u_{\alpha} + u_{\beta}\right) + 4u_{\alpha\beta} = 0$

    Next, a substitution in the form

    $\displaystyle
    u = v e^{p \alpha + q \beta},\;\;p, q\; \text{constant}
    $

    won't work in trying to achieve your goal.

    My suggestion. Start with the original PDE and substitute

    $\displaystyle
    u(r,t) = r^m v(r,t)
    $

    and see if you can choose a $\displaystyle m$ such all lower order terms vanish. If you can do that and obtain

    $\displaystyle
    v_{rr} = \frac{1}{c^2} v_{tt}
    $

    then the change of variable will work!
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  3. #3
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    Just use a and b to cut on the latex.

    First solve: $\displaystyle r=1/2(a+b)$ and $\displaystyle t=\frac{1}{2c}(a-b)$

    You fill in the steps:

    $\displaystyle \frac{\partial v}{\partial b}=1/2 r\frac{\partial u}{\partial r}-\frac{r}{2c}\frac{\partial u}{\partial t}+1/2 u$

    $\displaystyle \frac{\partial^2 v}{\partial a \partial b}=1/2\left[\frac{\partial}{\partial a}\left(r\frac{\partial u}{\partial r}\right)\right]-\frac{1}{2c}\frac{\partial}{\partial a}\left(r\frac{\partial u}{\partial t}\right)+1/2 \frac{\partial u}{\partial a}$

    You know how to do:

    $\displaystyle \frac{\partial}{\partial a}\left(r\frac{\partial u}{\partial r}\right)=r\left(\frac{\partial^2 u}{\partial r^2}\frac{\partial r}{\partial a}+\frac{\partial^2 u}{\partial t \partial r}\frac{\partial t}{\partial a}\right)+\frac{\partial u}{\partial r}\frac{\partial r}{\partial a}$

    Do that with the other one, combine everything and therefore (from original equation in u):

    $\displaystyle \frac{\partial^2 v}{\partial a \partial b}=1/4 r\frac{\partial^2 u}{\partial r^2}+1/4 \frac{\partial u}{\partial r}-\frac{r}{4c^2}\frac{\partial^2 u}{\partial t^2}+1/4\frac{\partial u}{\partial r}=0$
    Last edited by shawsend; Apr 27th 2010 at 05:48 AM. Reason: corrected notation
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  4. #4
    Super Member
    Joined
    Aug 2008
    Posts
    903
    Here's how I got $\displaystyle \frac{\partial v}{\partial b}$: If v is a function of r and t and each of r and t are functions of a and b, then we just use the chain rule to find:

    $\displaystyle \frac{\partial v}{\partial b}=\frac{\partial v}{\partial r}\frac{\partial r}{\partial b}+\frac{\partial v}{\partial t}\frac{\partial t}{\partial b}$

    but $\displaystyle v=r u(r,t)$ and therefore $\displaystyle \frac{\partial v}{\partial b}=\frac{\partial}{\partial b}\left(r u(r,t)\right)$

    $\displaystyle \begin{aligned}
    \frac{\partial v}{\partial b}&=r\frac{\partial u}{\partial b}+u\frac{\partial r}{\partial b} \\
    &=r\left[\frac{\partial u}{\partial r}\frac{\partial r}{\partial b}+\frac{\partial u}{\partial t}\frac{\partial t}{\partial b}\right]+u\frac{\partial r}{\partial b}
    \end{aligned}
    $

    $\displaystyle
    =r\left[\frac{\partial u}{\partial r} (1/2)-\frac{\partial u}{\partial t}(\frac{1}{2c})\right]+1/2 u $

    $\displaystyle
    =1/2 r\frac{\partial u}{\partial r}-\frac{r}{2c}\frac{\partial u}{\partial t}+1/2 u
    $
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