Periodic Functions & Orthogonal Functions — Question 7

Question 7 – Solution

Strategy. Track the boundary expression before dividing by an eigenvalue difference.

Step 1: Derive the boundary identity. Multiply the equation for $u$ by $v$ and that for $v$ by $u$, then subtract. The $q$ terms cancel, and integration gives $$\boxed{(\lambda-\mu)\int_a^b wuv\,dx =[p(uv'-u'v)]_a^b.}$$ This identity is valid before any orthogonality conclusion is drawn.

Step 2: Use the actual endpoint conditions. At an endpoint with $B_c=0$, $A_c\ne 0$ and both function values are zero, so the boundary expression is zero. If $B_c\ne 0$, then $pu'=-A_cu/B_c$ and $pv'=-A_cv/B_c$; substitution again gives zero. Therefore $\lambda\ne\mu$ implies $\boxed{\int_a^b wuv\,dx=0}$. Equal eigenvalues give only the identity $0=0$.

Step 3: Normalize two distinct Dirichlet modes. Here $p=w=1,q=0$, with eigenvalues $\pi^2$ and $4\pi^2$. The theorem gives orthogonality, and direct integration of each square gives $1/2$. Thus $\boxed{\sqrt 2\sin\pi x,\ \sqrt 2\sin 2\pi x}$ are orthonormal. The normalization is separate from the boundary proof of orthogonality.

Step 4: Handle a repeated eigenvalue. Both $u=\sin x$ and $v=\sin x+\cos x$ have eigenvalue one and satisfy periodic matching on $[0,2\pi]$. Nevertheless, $\langle u,v\rangle=\pi\ne 0$. Since $\|u\|^2=\pi$, subtracting the projection gives $v-\langle v,u\rangle u/\|u\|^2=\cos x$. The resulting orthonormal pair is $$\boxed{\sin x/\sqrt\pi,\qquad\cos x/\sqrt\pi.}$$ Distinct eigenvalues ensure orthogonality under the hypotheses; a repeated eigenspace may instead require an orthogonal choice of basis.