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Math Help - DE Power Series

  1. #1
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    DE Power Series

    The coefficients of a certain power series

    \displaystyle\mbox{P(s,t)}=\sum_{m=0}^{\infty}\sum  _{n=0}^{\infty}a_{m,n}s^mt^n

    satisfy

    3(m+1)a_{m+1,n}-(n+1)a_{m,n+1}+a_{m,n}=0

    and it is known that \mbox{P(t,t)}=\exp{(2t)}. Find \mbox{P(s,t)}.

    \displaystyle\mbox{P}_s=\sum_{m=1}^{\infty}\sum_{n  =0}^{\infty}a_{m,n}ms^{m-1}t^n

    \displaystyle\mbox{P}_t=\sum_{m=0}^{\infty}\sum_{n  =1}^{\infty}a_{m,n}ns^{m}t^{n-1}

    \displaystyle c_1\sum_{m=1}^{\infty}\sum_{n=0}^{\infty}a_{m,n}ms  ^{m-1}t^n+c_2\sum_{m=0}^{\infty}\sum_{n=1}^{\infty}a_{  m,n}ns^{m}t^{n-1}+c_3\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a_{m,n  }s^mt^n=0

    Is this how it should be approached?
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  2. #2
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    Try c_1=3, c_2 = -1, c_3 = 1
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  3. #3
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    Quote Originally Posted by snowtea View Post
    Try c_1=3, c_2 = -1, c_3 = 1
    Ok, so that is how it is done with those types of questions?
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  4. #4
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    Same as your previous question:

    <br />
\displaystyle \sum_{m=1}^{\infty}\sum_{n=0}^{\infty}a_{m,n}ms^{m-1}t^n<br />
= \sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a_{m+1,n}(m+  1)s^mt^n<br />
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  5. #5
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    Quote Originally Posted by snowtea View Post
    Same as your previous question:

    <br />
\displaystyle \sum_{m=1}^{\infty}\sum_{n=0}^{\infty}a_{m,n}ms^{m-1}t^n<br />
= \sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a_{m+1,n}(m+  1)s^mt^n<br />
    I understand that. I wasn't sure if doing this (see below) was the correct procedure.

    \displaystyle c_1\sum_{m=1}^{\infty}\sum_{n=0}^{\infty}a_{m,n}ms  ^{m-1}t^n+c_2\sum_{m=0}^{\infty}\sum_{n=1}^{\infty}a_{  m,n}ns^{m}t^{n-1}+c_3\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a_{m,n  }s^mt^n=0
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  6. #6
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    Quote Originally Posted by dwsmith View Post
    I understand that. I wasn't sure if doing this (see below) was the correct procedure.

    \displaystyle c_1\sum_{m=1}^{\infty}\sum_{n=0}^{\infty}a_{m,n}ms  ^{m-1}t^n+c_2\sum_{m=0}^{\infty}\sum_{n=1}^{\infty}a_{  m,n}ns^{m}t^{n-1}+c_3\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a_{m,n  }s^mt^n=0
    Well, it is the correct approach for this problem.

    If you do not know the form it is better to take your recurrence:
    3(m+1)a_{m+1,n}-(n+1)a_{m,n+1}+a_{m,n}=0

    Multiply every term by s^mt^n, and do the double summation of the entire expression, then simplify.
    It becomes the form of the summation you are asking about.
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  7. #7
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    Now, my trouble is how do I use the initial condition?
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  8. #8
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    You do realize that the summation you wrote gives you the DE:

    3P_s - P_t + P = 0

    Solve it like any other DE.
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  9. #9
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    Yes.
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  10. #10
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    I can just solve that like any other PDE than?
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  11. #11
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    So do you already have the general form of the solutions for the DE?

    If so, just plug-in P(t,t) to solve with initial conditions right? Perhaps I'm missing something?
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  12. #12
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    Quote Originally Posted by snowtea View Post
    So do you already have the general form of the solutions for the DE?

    If so, just plug-in P(t,t) to solve with initial conditions right? Perhaps I'm missing something?
    You aren't. I was just confused for no reason.
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  13. #13
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    I am struggling to obtain the book solution.

    \displaystyle\exp{\left(\frac{s+7t}{4}\right)}

    P(s,t)=P(\xi\cos{\alpha}-\eta\sin{\alpha},\xi\sin{\alpha}+\eta\cos{\alpha})  =u(\xi,\eta)

    u_{\xi}(3\cos{\alpha}-\sin{\alpha})-u_{\eta}(3\sin{\alpha}+\cos{\alpha})+u=0

    \displaystyle\sqrt{10}u_{\xi}+u=0\Rightarrow u_{\xi}+\frac{u}{\sqrt{10}}=0

    s=3\xi+B\eta \ \ \ t=-\xi+D\eta

    \exp{(\xi)u=g(\eta)\Rightarrow u(\xi,\eta)=\exp{(-\xi)}g(\eta)

    I tried B = 0 and D = 1/3; and D = 0 and B = 3 but nothing is happening.

    Never mind I forgot to divide out a 2.

    s=3\xi+3\eta \ \ \ t=-\xi

    P(s,t)=\exp{(t)}g\left(\frac{s}{3}+t\right)

    P(t,t)=\exp{(t)}g\left(\frac{t}{3}+t\right)=\exp{(  2t)}

    P(t,t)=g\left(\frac{t}{3}+t\right)=\exp{(t)}

    \displaystyle g(t)=\exp{\left(\frac{3}{4}t\right)}

    \displaystyle P(s,t)=\exp{(t)}\exp{\left[\frac{3}{4}*\left(\frac{s}{3}+t\right)\right]}=\exp{\left(\frac{s+7t}{4}\right)}
    Last edited by dwsmith; January 11th 2011 at 07:30 PM.
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