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Math Help - Heat Equation

  1. #1
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    Heat Equation

    Hi guys. I am going through an example in my book, and I am stuck on the last step concerning the initial condition.

    The coefficient derived from the initial condition is:

    B_n = 4 \frac{1 - (-1)^n}{\pi n^3}

    and the linear sum of the particular solutions is:
    <br />
u(x, t) =  \displaystyle\sum_{n=1}^\infty B_n e^{(-a^2 k_n^2 t)} sin (\frac{n \pi x}{L})

    I think then B_n is substituted into the equation  u(x, t) above to get:

    u(x, t) = \frac{8}{\pi} \displaystyle\sum_{n=1}^\infty \frac{e^{- (2n-1)^2 a^2 t} }{(2n-1)^3} sin((2n-1)x)

    I am having difficulty working out the algebra on this one. Can anyone show it to me step by step. I can see that there is a factor (2n-1) but I don't know where they got that from. Can anyone explain? I am finding this really difficult to work out.
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  2. #2
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    Quote Originally Posted by sataq View Post
    Hi guys. I am going through an example in my book, and I am stuck on the last step concerning the initial condition.

    The coefficient derived from the initial condition is:

    B_n = 4 \frac{1 - (-1)^n}{\pi n^3}

    and the linear sum of the particular solutions is:
    <br />
u(x, t) = \displaystyle\sum_{n=1}^\infty B_n e^{(-a^2 k_n^2 t)} sin (\frac{n \pi x}{L})

    I think then B_n is substituted into the equation  u(x, t) above to get:

    u(x, t) = \frac{8}{\pi} \displaystyle\sum_{n=1}^\infty \frac{e^{- (2n-1)^2 a^2 t} }{(2n-1)^3} sin((2n-1)x)

    I am having difficulty working out the algebra on this one. Can anyone show it to me step by step. I can see that there is a factor (2n-1) but I don't know where they got that from. Can anyone explain? I am finding this really difficult to work out.
    Look at B_n as we increment n

    B_1 = 4 \frac{1 - (-1)^1}{\pi 1^3} = \frac{8}{\pi 1^3}
    B_2 = 4 \frac{1 - (-1)^2}{\pi 2^3} = 0
    B_3 = 4 \frac{1 - (-1)^3}{\pi 3^3} = \frac{8}{\pi 3^3}
    B_4 = 4 \frac{1 - (-1)^4}{\pi 4^3} = 0
    B_5 = 4 \frac{1 - (-1)^5}{\pi 5^3} = \frac{8}{\pi 5^3}
    B_6 = 4 \frac{1 - (-1)^6}{\pi 6^3} = 0

    so writing out a few terms gives

    <br />
u(x, t) = \frac{8}{\pi} \left( \frac{e^{-a^2 k_1^2 t}}{1^3} \sin \frac{1 \pi x}{L} + \frac{e^{-a^2 k_3^2 t}}{3^3} \sin \frac{3 \pi x}{L} + \frac{ e^{-a^2 k_5^2 t}}{5^3} \sin \frac{5 \pi x}{L} + \cdots \right)

    and b/c of the jump by 2's, we can write this as

    <br />
= \frac{8}{\pi}\sum_{n=1}^{\infty} \frac{e^{-a^2 k_{2 n -1}^2 t}}{(2n-1)^3} \sin \frac{{2 n -1} \pi x}{L}
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