1. Once we obtain $\exp(-\lambda kt)$, where does the n come from in the exponential for the summation?

2. This is what I have.

$\displaystyle u(x,t)=\sum_{n=0}^{\infty}a_n\exp(-\lambda_nkt)\cos\left(\frac{(2n+1)\pi x}{2L}\right), \ \ \ \ \ \lambda_n=\left[\frac{(2n+1)\pi }{2L}\right]^2$

3. Originally Posted by dwsmith
This is what I have.

$\displaystyle u(x,t)=\sum_{n=0}^{\infty}a_n\exp(-\lambda_nkt)\cos\left(\frac{(2n+1)\pi x}{2L}\right), \ \ \ \ \ \lambda_n=\left[\frac{(2n+1)\pi }{2L}\right]^2$
Right Now you just need to solve for the $a_n$'s.

4. I understand the use of orthogonality because everything goes to zero accept for when m = n, but why is it used here?

5. Originally Posted by dwsmith
I understand the use of orthogonality because everything goes to zero accept for when m = n, but why is it used here?
$\displaystyle \varphi_m(x)u(x,0)=\varphi_{m}(x)(L-x)=\sum_{n=0}^{\infty}a_n\varphi_m(x)\varphi_n(x)$

So as you said when $m=n$

$\displaystyle \int_{0}^{L}\varphi_{n}(x)(L-x)dx=a_n\int_{0}^{L}\varphi_n(x)\varphi_n(x)dx$

Solving for what we want gives

$\displaystyle a_n=\frac{\int_{0}^{L}\varphi_{n}(x)(L-x)dx}{\int_{0}^{L}[\varphi_n(x)]^2dx}$

6. Originally Posted by TheEmptySet
$\displaystyle \varphi_m(x)u(x,0)=\varphi_{m}(x)(L-x)=\sum_{n=0}^{\infty}a_n\varphi_m(x)\varphi_n(x)$

So as you said when $m=n$

$\displaystyle \int_{0}^{L}\varphi_{n}(x)(L-x)dx=a_n\int_{0}^{L}\varphi_n(x)\varphi_n(x)dx$

Solving for what we want gives

$\displaystyle a_n=\frac{\int_{0}^{L}\varphi_{n}(x)(L-x)dx}{\int_{0}^{L}[\varphi_n(x)]^2dx}$
$a_n$ looks pretty brutal but here is what I obtained:

$\displaystyle a_n=(-1)^n\frac{4L}{(2m+1)\pi}+\left[(-1)^n-1\right]\left(\frac{8L(x-1)}{\left[(2m+1)\pi\right]^2}\right)$

Correct?

7. $\displaystyle \int_{0}^{L} \cos\left( \frac{(2n+1)\pi x}{2L}\right)(L-x)dx=\frac{4L^2}{\pi^2(2n+1)^2}$

8. Originally Posted by TheEmptySet
$\displaystyle \int_{0}^{L} \cos\left( \frac{(2n+1)\pi x}{2L}\right)(L-x)dx=\frac{4L^2}{\pi^2(2n+1)^2}$
$\displaystyle \int_0^L L\cos\left(\frac{(2m+1)\pi x}{2L}\right)dx-\int_0^Lx\cos\left(\frac{(2m+1)\pi x}{2L}\right)dx$

$\displaystyle L\left[\frac{2L\sin\left(\frac{(2m+1)\pi x}{2L}\right)}{(2m+1)\pi}\right]_0^L-\left[\frac{2Lx}{(2m+1)\pi}-\frac{2L}{(2m+1)\pi}\int_0^L\sin\left(\frac{(2m+1) \pi x}{2L}\right)dx\right]$

Is this much correct?

9. $\displaystyle u(x,t)=\frac{8L}{(2n+1)^2\pi^2}\sum_{n=0}^{\infty} \exp(-\lambda_n kt)\cos\left(\frac{(2n+1)\pi x}{2L}\right), \ \ \ \lambda_n=\left[\frac{(2n+1)\pi}{2L}\right]^2$

Correct?

10. Originally Posted by dwsmith
$\displaystyle u(x,t)=\frac{8L}{(2n+1)^2\pi^2}\sum_{n=0}^{\infty} \exp(-\lambda_n kt)\cos\left(\frac{(2n+1)\pi x}{2L}\right), \ \ \ \lambda_n=\left[\frac{(2n+1)\pi}{2L}\right]^2$

Correct?
I like to use tabular integration when I am doing Fourier series. This works if you of the functions will differentiate to 0 eventally.

$\begin{array}{|c|c|}\hline f(x) & g(x) \\ \hline (-1)\frac{d}{dx}f(x) & \int g(x) \\ \hline \frac{d^2}{dx^2}f(x) & \int\int g(x) \\ \hline \vdots & \vdots \\ \hline (-1)^n\frac{d^n}{dx^n}f(x) & \int_{\text{n times}} g(x) \\ \hline 0 & \int_{\text{n+1 times}} g(x)\\ \hline \end{array}$

So in your case we have

$\begin{array}{|c|c|}\hline L-x & \cos(\lambda_n x) \\ \hline(-1)[-1] & \frac{1}{\lambda_n}\sin(\lambda_n x) \\ \hline 0 & - \frac{1}{\lambda_n^2}\cos(\lambda_n x) \\ \hline \end{array}$

Now multiply diagonally down to get the antiderivative

$\displaystyle \frac{(L-x)}{\lambda_n }\sin(\lambda_n x)-\frac{1}{\lambda_n^2} \cos(\lambda_n x) \bigg|_{0}^{L}=-\frac{1}{\lambda_n^2}\left[ \cos(\lambda_n L)-1\right]=\frac{1}{\lambda_n^2}$

This is the same as yours

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