The following Heat equation:
$u_t=4 u_{xx} $, $(t,x) \in (0, \infty)\times (0,1)$
$u(0,x)=x(1-x), x \in (0,1)$
$u(t,0)=0=u(t,1), t \in (0, \infty)$
should have the following solution:
$u(t,x)= \frac {8}{\pi^3} \sum_{n=1} ^{\infty} \frac {1}{(2n-1)^3} e^{-4(2n-1)^2 \pi^2 t} \sin((2n-1)\pi x) $
So to get this answer I used seperation of variables:
$u(t,x)=X(x)T(t)$
$\frac{X''}{X}=\frac{T'}{4T}=-\lambda$
Now I set ${\lambda = \alpha^2}$:
- $(I): X''+\alpha^2 X=0$
- $(II): T'+\alpha^2 T=0$
solving $(I)$:
$X(x)=c_1 \cos(\alpha x)+ c_2 \sin(\alpha x)$
Now using the boundary conditions $u(t,0)=0=u(t,1)$:
$c_1=0$ and $\sin(\alpha)=0$ so $\alpha= n \pi$
It follows that $X(x)=a_n \sin(n \pi x)$
Now I get the following solution, which is not identical to the given solution:
$u(t,x)=\sum_{n=0} ^{\infty} c_n e^{-4n^2 \pi^2 t} \sin(n\pi x)$
I know how to proceed from here on, but I don't understand how $\alpha = (2n-1) \pi $ in the given solution. We are asked to show that we get the given solution as a final solution for the equation.
We know that $u(x, 0) =x(1-x)$ and from the solution we Have that $u(x, 0) =\sum c_k\sin(k\pi x) $, so the unique posibility is that the coefficients $c_n$ are the same as the ones on the sine Fourier series of $x(1-x) $. If you compute that series you will get that $$x(1-x) = \frac{4}{\pi^3}\sum_{n=1}^{\infty}\frac{\left(1-\left(-1\right)^{n}\right)}{n^{3}}\sin\left(n\pi x\right),$$ since $1-(-1) ^n$ is zero for even $n$ and $2$ for odd $n$, we get that $c_k=0$ if $k=2n$ and $c_k=\frac{8}{\pi^3}\frac{1}{(2n-1) ^3}$ if $k=2n-1$.