Eigenvalues and Eigenfunctions — Question 8

Question 8 – Solution

Strategy. Separate the existence of a particular solution from the nontrivial homogeneous space at a spectral parameter.

Step 1: Solve away from the spectrum. Since $-\bigl(\sin\pi x\bigr)''=\pi^2\sin\pi x$, $$\boxed{y=\frac{\sin\pi x}{\pi^2-\lambda}}$$ solves the equation and both endpoints whenever $\lambda\ne\pi^2$. If $\lambda$ is outside the entire spectrum, any difference of two solutions is a zero-endpoint homogeneous solution and is therefore zero. This proves uniqueness, including all zero and negative parameter values.

Step 2: Prove failure of compatibility at the first eigenvalue. Let $s=\sin\pi x$. For a proposed solution at $\lambda=\pi^2$, twice integrating by parts gives $$\int_0^1s(-y''-\pi^2y)dx =[-sy'+s'y]_0^1+\int_0^1(-s''-\pi^2s)y\,dx=0.$$ Both $s$ and $y$ vanish at the endpoints. The forcing instead requires this integral to equal $\int_0^1s^2dx=1/2$, a contradiction. No solution exists.

Step 3: Find every compatible resonant solution. For $n\geq 2$, the particular solution above is valid and the homogeneous space is spanned by $\sin(n\pi x)$. Therefore all solutions are $$\boxed{y=\frac{\sin\pi x}{(1-n^2)\pi^2}+C\sin(n\pi x),\quad C\in\mathbb R.}$$ The supplied product identity and $\int_0^1\sin^2(n\pi x)dx=1/2$ show that the extra integral condition is $C/2=0$, selecting exactly $C=0$.

Step 4: Distinguish two limiting behaviors. Outside the spectrum, $\max|y|=1/|\pi^2-\lambda|$, which diverges at the first eigenvalue. As $\lambda\to 4\pi^2$, the same solution converges uniformly to $-\sin(\pi x)/(3\pi^2)$. This is only the $C=0$ member of the infinite family at $4\pi^2$. A finite chosen limit does not remove the homogeneous freedom.