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Solve the heat equation

$\displaystyle k\frac{\partial^2 u}{\partial x^2} = \frac{\partial u}{\partial t}, 0< x < L, t > 0$

$\displaystyle u(0,t) = 0, u(L,t) = 0, t > 0$

$\displaystyle u(x,0) = \begin{cases} x, &0 < x < \frac{L}{2}\\ L-x, &\frac{L}{2}<x<L\\ \end{cases}$

using

$\displaystyle u(x,t) = \frac{2}{L}\displaystyle\sum\limits_{n=1}^\infty \left( \int_0^L u(x,0)\sin\left( \frac{n\pi}{L}x\right) dx\right) e^{\left( -k\frac{n^2\pi^2}{L^2}t\right)} \sin\left( \frac{n\pi}{L}x\right)$

$\displaystyle =\frac{2}{L}\displaystyle\sum\limits_{n=1}^\infty b_n e^{\left( -k\frac{n^2\pi^2}{L^2}t\right)} \sin\left( \frac{n\pi}{L}x\right)$

$b_n =$

$\displaystyle\int$

$\frac{L}{2}$

$0$

$\hskip 1pt dx\hskip 35pt + \hskip 35pt$

$\displaystyle\int$

$L$

$\frac{L}{2}$

$\hskip 1pt dx$

$\hskip 30pt$ u = dv = $\hskip 30pt$ u = dv =

$\hskip 30pt$ du = v = $\hskip 30pt$ du = v =

$\hskip 12pt =$

$\Bigg\vert$

$\frac{L}{2}$

$0$

$+\displaystyle\frac{L}{n\pi}$

$\displaystyle\int$

$\frac{L}{2}$

$0$

$dx+$

$\Bigg\vert$

$L$

$\frac{L}{2}$

$-\displaystyle\frac{L}{n\pi}$

$\displaystyle\int$

$L$

$\frac{L}{2}$

$dx$

$\hskip 12pt =$

$+$

$\Bigg\vert$

$\frac{L}{2}$

$0$

$+$

$-$

$\Bigg\vert$

$L$

$\frac{L}{2}$

$\hskip 12pt =$

You can earn partial credit on this problem.