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Thread: 2nd order PDE - help needed

  1. #1
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    2nd order PDE - help needed

    H''(n) + (1/2)n*H'(n) = 0.

    Boundary conditions: H(0) = 1, H(inf) = 0.

    I really don't know where to start ... any help will be greatly appreciated.
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  2. #2
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    Quote Originally Posted by Baluba View Post
    H''(n) + (1/2)n*H'(n) = 0.

    Boundary conditions: H(0) = 1, H(inf) = 0.

    I really don't know where to start ... any help will be greatly appreciated.
    First separate and integrate

    $\displaystyle \frac{H''}{H'} = - \frac{1}{2} n\, dn$

    so

    $\displaystyle \ln H' = \ln c_1 - \frac{1}{4} n^2\;\; \Rightarrow \;\;H' = c_1 e^{-n^2/4}$

    then integrate again (on the second integration we'll keep it in integral form)

    $\displaystyle
    H = c_1 \int_0^n e^{-t^2/4}dt + c_2
    $

    $\displaystyle
    H(0) = 1\; \text{ gives}\; c_2 = 1
    $ while $\displaystyle
    H(\infty) = 0\; \text{ gives}\; c_1 \pi + 1 = 0
    $

    giving the solution as

    $\displaystyle
    H(n) = 1 - \frac{1}{\pi} \int_0^n e^{t^2/4}dt
    $

    or in terms of error functions

    $\displaystyle
    H(n) = 1 - \text{erf} \left(\frac{n}{2}\right)$.
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  3. #3
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    Thanks a lot!
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