Fourier Series — Question 6

Question 6 – Solution

Strategy. Solve the harmonic response and the piecewise differential equation independently.

Step 1: Expand the input. The periodic extension of $q$ is odd. Direct integration gives zero mean and zero cosine coefficients, with $$\boxed{b_n^{(q)}=\frac{2(1-(-1)^n)}{\pi n} =\begin{cases}4/(\pi n),&n\text{ odd},\\0,&n\text{ even}.\end{cases}}$$ The constant in $1+q$ is one.

Step 2: Solve the coupled mode equations. Write $y=m+\sum(A_n\cos nx+B_n\sin nx)$. Integration by parts is legitimate piecewise: continuity and periodicity cancel the internal and endpoint terms. Thus $2m=1$, $nB_n+2A_n=0$, and $-nA_n+2B_n=b_n^{(q)}$. Hence $$\boxed{m=\frac 12,\quad A_n=-\frac{4}{\pi(n^2+4)},\quad B_n=\frac{8}{\pi n(n^2+4)}\quad(n\text{ odd}),}$$ and both coefficients vanish for even $n$.

Step 3: Verify a closed-form periodic response. Put $r=e^{-2\pi}$. Solving the constant-forcing equations and matching gives $$\boxed{y(x)= \begin{cases} 1-\dfrac{e^{-2x}}{1+r},&0\leq x\leq\pi,\\[3pt] \dfrac{e^{-2(x-\pi)}}{1+r},&\pi\leq x\leq 2\pi. \end{cases}}$$ Both expressions give $y(\pi)=1/(1+r)$, and $y(0)=y(2\pi)=r/(1+r)$. A difference of two responses is $Ce^{-2x}$ across the whole period by continuity; periodicity forces $C=0$. The verified closed form therefore has the coefficients determined above.

Step 4: Interpret the response. The first branch increases and the second decreases, so the extrema are $r/(1+r)$ and $1/(1+r)$. Integrating the equation over a period gives mean $1/2$. The response is continuous, but $y'$ jumps by $-2$ at $\pi$ and by $2$ at $0$ modulo $2\pi$; the assigned input values do not require a derivative there. Differentiation couples sine and cosine coefficients, creating a phase shift. Absolute coefficient summability and the Fourier theorem identify the plotted series with the continuous response.