An insulated Fourier series problem
Use separation of variables and Fourier series to solve the heat problem with insulated ends
$$u_t = 2u_{xx}, \: u_x(0,t)=u_x(1,t) = 0, \: u(x,0) = 1-x^2.$$
Feel free to use a computer algebra system to compute the Fourier series but do show your work. In particular, be sure to indicate how to use two integration by parts steps to compute the Fourier series.
Comments
Using separation of variables, our partial differential equation becomes
$$(XT)_t = 2(XT)_{xx},$$
or
$$XT' = 2X''T.$$
Rearranging,
$$\frac{T'}{2T} = \frac{X''}{X} = -\lambda.$$
Thus, $T' = -2\lambda T,$ so
$$T= e^{-2\lambda t},$$
while $X'' = -\lambda X,$ so
$$X = a\cos\left(\sqrt{\lambda}x\right) + b\sin\left(\sqrt{\lambda}x\right).$$
From our initial conditions, we know
$$X'(0) = 0 + b\sqrt{\lambda} = 0,$$
so $b = 0,$ and
$$X'(1) = -a\sqrt{\lambda}\sin\left(\sqrt{\lambda}\right) = 0,$$
so
$$\sqrt{\lambda} = n\pi.$$
Thus, our full solution is the Fourier cosine series
$$u(x,t) = \frac{a_0}{2} + \sum_{n=1}^{\infty}a_ne^{-2n^2\pi^2t}\cos(n\pi x),$$
where
$$a_n = 2\int_{0}^{1} (1 - x^2)\cos(n\pi x)dx.$$
First, we find $\frac{a_0}{2}$:
$$\frac{a_0}{2} = \int_{0}^{1}\left(1-x^2\right)dx = \left[x-\frac{x^3}{3}\right]_0^1 = \frac{2}{3}.$$
Next, we find $a_n$:
$$a_n = 2\int_{0}^{1}\left(1-x^2\right)\cos(n\pi x)dx = 2\int_{0}^{1}\cos(n\pi x)dx - 2\int_{0}^{1}x^2\cos(n\pi x)dx.$$
$$2\int_{0}^{1}\cos(n\pi x)dx = \frac{2}{n\pi}\left[\sin(n\pi x)\right]_0^1 = 0,$$
while $-2\int_{0}^{1}x^2\cos(n\pi x)dx$ can be evaluated by parts, setting $u=x^2$, $du=2xdx$, $v = \frac{1}{n\pi}\sin(n\pi x)$, and $dv = \cos(n\pi x)dx$:
$$ -2\int_{0}^{1}x^2\cos(n\pi x)dx = -\frac{2}{n\pi}\left(\left[x^2\sin(n\pi x)\right]_0^1 - 2\int_{0}^{1}x\sin(n\pi x)dx\right) = -\frac{2}{n\pi}\left(0 - 2\int_{0}^{1}x\sin(n\pi x)dx\right) = \frac{4}{n\pi}\int_{0}^{1}x\sin(n\pi x)dx.$$
We integrate by parts once again, setting $u=x$, $du=dx$, $v=-\frac{1}{n\pi}\cos(n\pi x)$, and $dv = sin(n\pi x)dx$:
$$\frac{4}{n\pi}\int_{0}^{1}x\sin(n\pi x)dx = -\frac{4}{n^2\pi^2}\left(\left[xcos(n\pi x)\right]_0^1 - \int_{0}^{1}\cos(n\pi x)dx\right) = -\frac{4}{n^2\pi^2}\left(cos(n\pi) - 0\right) = -\frac{4(-1)^n}{n^2\pi^2}.$$
Therefore,
$$a_n = 2\int_{0}^{1}\left(1-x^2\right)\cos(n\pi x)dx = -\frac{4(-1)^n}{n^2\pi^2}.$$
Thus,
$$u(x,t) = \frac{2}{3} - \frac{4}{\pi^2}\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}e^{-2n^2\pi^2t}\cos(n\pi x).$$