Fourier Series — Question 7

Question 7 – Solution

Strategy. Test the forcing against the periodic homogeneous modes before dividing by a spectral factor.

Step 1: Derive the compatibility conditions. For a periodic $C^2$ function, both $y$ and $y'$ match at the period endpoints. Two integrations by parts therefore give $$\int_0^{2\pi}(y''+4y)\cos 2x=0,\qquad \int_0^{2\pi}(y''+4y)\sin 2x=0.$$ Orthogonality makes the corresponding right sides $\pi\alpha$ and $\pi\beta$. Thus $$\boxed{\alpha=\beta=0}$$ is necessary. The constant and first harmonic do not contribute to these tests.

Step 2: Construct every compatible solution. With these parameters zero, a particular solution is $3/4+\cos x/3$. The homogeneous equation has basis $\cos 2x,\sin 2x$, both periodic. Thus $$\boxed{y(x)=\frac 34+\frac 13\cos x+A\cos 2x+B\sin 2x,\qquad A,B\in\mathbb R.}$$ This proves sufficiency as well as completeness: subtract the particular solution from any solution and solve the homogeneous equation.

Step 3: Enforce the additional normalization. The initial values give $3/4+1/3+A=0$ and $2B=0$. Hence $$\boxed{y(x)=\frac 34+\frac 13\cos x-\frac{13}{12}\cos 2x.}$$ It has $y(0)=0$, $y'(0)=0$, and is $2\pi$-periodic together with all derivatives. Substitution gives $y''+4y=3+\cos x$ because the frequency-two term is annihilated.

Step 4: Identify the resonance obstruction. For frequency $n$, the mode equation multiplies each coefficient of $y$ by $4-n^2$. At $n=2$ it reads $0=\alpha$ or $0=\beta$: it is a compatibility condition, not a formula for an unknown coefficient. When it holds, the two coefficients are free until extra conditions are supplied. When it fails, no periodic $C^2$ solution exists. Although an initial-value problem still has a unique solution, its resonant response cannot be periodic; altering homogeneous constants does not change either integral obstruction.