Eigenvalues and Eigenfunctions — Question 9

Question 9 – Solution

Strategy. Recover the potential where the target is nonzero, then use endpoint regularity and a factored energy identity to certify the design.

Step 1: Reject an incompatible target. For $y=x(1-x)$, division inside the interval would require $q=\lambda+y''/y=\lambda-2/[x(1-x)]$, which is unbounded at both ends. Equivalently, continuity of the equation at zero would require $-y''(0)+q(0)y(0)=2=\lambda y(0)=0$. No continuous potential can work.

Step 2: Construct a regular potential. Write $c=\cos\pi x$. The target is $y=\sin\pi x(1+c/2)$ and $y''=-\pi^2\sin\pi x(1+2c)$. Consequently $$\boxed{q(x)=-\frac{3\pi^2c}{2+c},\qquad q(0)=-\pi^2,\quad q(1)=3\pi^2.}$$ Since $2+c\geq 1$, the formula is continuous on the closed interval. Substitution gives $qy=(\lambda y+y'')$, verifying the equation inside; continuity verifies it at the endpoints as well.

Step 3: Factor the energy. The factor $1+c/2$ lies between $1/2$ and $3/2$, so $y>0$ inside. The endpoint derivatives are $y'(0)=3\pi/2$ and $y'(1)=-\pi/2$, both nonzero. With $v=yz$ and $y''=(q-\lambda)y$, expansion gives $$v'^2+(q-\lambda)v^2=y^2z'^2+(yy'z^2)'.$$ For zero-endpoint $C^2$ functions $v$, the simple zeros of $y$ imply that $z=v/y$ and $z'$ stay bounded near either endpoint, by Taylor expansion. Thus $[yy'z^2]_0^1=0$ and integration proves the requested identity.

Step 4: Certify the lowest eigenvalue and its eigenspace. If $v$ is a nonzero eigenfunction with eigenvalue $\nu$, integration by parts and the identity yield $$(\nu-\pi^2)\int_0^1v^2dx=\int_0^1y^2z'^2dx\geq 0.$$ Thus $\nu\geq\pi^2$, and the constructed nonzero $y$ attains this bound. Equality forces $z'=0$ inside, hence $v=Cy$. Therefore $\pi^2$ is the lowest real eigenvalue and its eigenspace is precisely $\operatorname{span}\{y\}$.