1. ## 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:

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

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

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

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

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.

2. Originally Posted by sataq
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:

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

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

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

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

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 $\displaystyle B_n$ as we increment $\displaystyle n$

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

so writing out a few terms gives

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

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

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