Eigenvalues and Eigenfunctions — Question 4

Question 4

Let $h\in\mathbb R$ control the endpoint condition $$-y''=\lambda y,\qquad 0\leq x\leq 1,\qquad y(0)=0,\quad y'(1)=h\,y(1).$$ Unlike fixed zero endpoint values, this boundary condition can permit a negative eigenvalue.

Tasks

1. Determine precisely when zero is an eigenvalue and identify its eigenspace.

2. For $\lambda=-\mu^2<0$, derive the equation for $\mu$. Prove that there is exactly one negative eigenvalue when $h>1$ and none when $h\leq 1$.

3. For $\lambda=k^2>0$, derive the frequency equation without losing or adding modes by division. Locate all its roots by intervals between multiples of $\pi$, including the special first interval.

4. Derive the energy identity and explain why it does not always imply nonnegative eigenvalues. For $h=2$, bound the negative eigenvalue using $1<\mu<2$, and sketch the equation locating $\mu$.

Question 4 – Solution

Strategy. Separate all three signs and prove root counts by monotonicity on the correct domains.

Step 1: Check the zero parameter directly. At $\lambda=0$, $y=Ax$ and the right condition is $A=hA$. Thus zero is an eigenvalue exactly when $h=1$, with eigenspace $\operatorname{span}\{x\}$.

Step 2: Count the negative modes. For $\lambda=-\mu^2$, the nonzero solution is a multiple of $\sinh(\mu x)$, and its endpoint condition is $\boxed{\mu\coth\mu=h}$. The function $g(\mu)=\mu\coth\mu$ tends to one as $\mu\downarrow 0$ and to infinity as $\mu\to\infty$. Its derivative has numerator $\sinh\mu\cosh\mu-\mu>0$, since this numerator starts at zero and has derivative $2\sinh^2\mu>0$. Thus precisely $h>1$ gives one negative mode.

Step 3: Locate every positive mode. Here $y=A\sin(kx)$ and $k\cos k=h\sin k$. A positive multiple of $\pi$ cannot solve this equation, so division by $\sin k$ is safe at a root: $$\boxed{k\cot k=h.}$$ Its derivative is $(\sin k\cos k-k)/\sin^2k<0$ for $k>0$ away from poles. On each $(n\pi,(n+1)\pi)$, $n\geq 1$, it decreases from $+\infty$ to $-\infty$, giving exactly one root for every $h$. On $(0,\pi)$ it decreases from the unattained limit one to $-\infty$: one root if $h<1$, none if $h\geq 1$.

Step 4: Retain the endpoint energy term. Integration by parts gives $\lambda\int_0^1y^2dx=\int_0^1(y')^2dx-hy(1)^2$, which can be negative. For $h=2$, $g(1)<2<g(2)$: the first inequality follows from $e^2>3$ and $g(2)>2$ follows from $\coth 2>1$. Thus $1<\mu<2$ and $\boxed{-4<\lambda_-<-1}$. The graph locates the unique crossing.