Fourier Cosine Series — Question 9

Question 9 – Solution

Strategy. Track the constant mode separately: it becomes singular as the regularization vanishes.

Step 1: Solve the forced modes. The constant is divided by $\alpha^2$, while the first cosine is divided by $1+\alpha^2$. Therefore $$\boxed{u_\alpha=\frac 1{\alpha^2}+\frac{\cos x}{1+\alpha^2}.}$$ Its derivative is $-\sin x/(1+\alpha^2)$, zero at both endpoints. Substitution recovers both the constant forcing one and the cosine term.

Step 2: Establish uniqueness and inspect the mean. For a homogeneous difference, integration by parts gives $\int_0^\pi[(w')^2+\alpha^2w^2]dx=0$, so $w=0$. The mean is $\boxed{1/\alpha^2}$ and diverges as $\alpha\downarrow 0$; the full family cannot have a finite uniform limit.

Step 3: Control the normalized family. Subtracting the mean gives $v_\alpha=\cos x/(1+\alpha^2)$. The first two derivatives of cosine, as well as cosine itself, have supremum norm one. Thus for $j=0,1,2$, $$\boxed{\|v_\alpha^{(j)}-(\cos x)^{(j)}\|_\infty =\frac{\alpha^2}{1+\alpha^2}\longrightarrow 0.}$$ This proves function and derivative convergence on the whole interval.

Step 4: Identify the correct limiting problem. The original forcing has integral $\pi\ne 0$, violating the Neumann condition $\int f=0$ when $\alpha=0$. Hence the unregularized original problem is impossible. Subtracting the mean also removes the constant forcing: $-v_\alpha''+\alpha^2v_\alpha=\cos x$. The limit solves $$\boxed{-v''=\cos x,\quad v'(0)=v'(\pi)=0,\quad \int_0^\pi v=0, \qquad v=\cos x.}$$ It is the unique zero-mean solution of this compatible problem.