Fourier Sine Series — Question 8

Question 8 – Solution

Strategy. Divide each forcing coefficient by its eigenvalue, then use an independent closed-form solution to verify the spectral construction.

Step 1: Obtain two descriptions of the solution. Since $-\bigl(\sin nx\bigr)''=n^2\sin nx$, the proposed coefficients are $\boxed{a_n=b_n/n^2=2(-1)^{n+1}/n^3}$. Direct integration of $u''=-x$ and the two endpoint conditions gives $$\boxed{u(x)=\frac{\pi^2x-x^3}{6}.}$$ Its second derivative is $-x$ and both endpoint values vanish.

Step 2: Verify the actual coefficients. Let $\widehat u_n=(2/\pi)\int_0^\pi u\sin nx\,dx$. Twice integrating by parts, using zero endpoint values of $u$ and of the sine, gives $(2/\pi)\int_0^\pi(-u'')\sin nx\,dx=n^2\widehat u_n$. The left side is $b_n$, proving $\widehat u_n=a_n$. The continuous odd extension and the convergence theorem identify the series with this polynomial, including at the endpoints.

Step 3: Justify the first derivative and limit the second. The derivative coefficients have magnitude $2/n^2$, so their cosine series converges uniformly. The original series converges at zero, where it is zero; the stated differentiation theorem therefore applies once. The formal second derivative is $-\sum b_n\sin nx$. At $x=\pi$ every term is zero, whereas $u''(\pi)=-\pi$. Thus the second derivative series cannot equal $u''$ pointwise everywhere on the closed interval.

Step 4: State the valid convergence and smoothing. For $U_N=\sum_{n=1}^Na_n\sin nx$, $U_N''=-S_N$ exactly. The supplied forcing convergence gives $\boxed{\|U_N''+x\|_2\to 0}$; isolated endpoint disagreement does not change this norm. Moreover $$\boxed{\|u-U_N\|_\infty\leq 2\sum_{n>N}n^{-3}\leq 1/N^2.}$$ Dividing by $n^2$ suppresses high-frequency forcing coefficients and gives a uniformly convergent displacement with a uniformly convergent first derivative.