Separation of Variables — Question 8

Question 8 – Solution

Strategy. A Robin boundary term can change the sign of the eigenvalue; analyze zero before treating the oscillatory and hyperbolic cases.

Step 1: Retain the boundary contribution. Integration by parts gives $\lambda\int X^2=\int(X')^2-[XX']_0^L=\int(X')^2+hX(L)^2$. For $h\ge 0$, equality at zero would force $X'=0$ and then $X=0$; nonzero factors therefore have $\lambda>0$. At $\lambda=0$, $X=Ax$ and the right condition is $A(1+hL)=0$. Thus a nonzero zero mode exists exactly when $\boxed{q=-1}$.

Step 2: Separate the positive and negative cases. For $\lambda>0$, $X=A\sin(\sqrt\lambda x)$ and $$\boxed{z\cos z+q\sin z=0,\quad z>0.}$$ For $\lambda=-\mu^2<0$, $X=A\sinh(\mu x)$ and the right condition becomes $\mu\cosh(\mu L)+h\sinh(\mu L)=0$. Since $\rho=\mu L>0$, $$\boxed{\rho\coth\rho=-q,\qquad \lambda=-\rho^2/L^2.}$$ Dividing by $\sinh\rho$ here is valid only in this strictly negative case.

Step 3: Prove the threshold and uniqueness. Let $F(\rho)=\rho\coth\rho$. It has limit $1$ as $\rho\downarrow 0$ and $F(\rho)\to\infty$. Its derivative is $F'=(\sinh\rho\cosh\rho-\rho)/\sinh^2\rho>0$: the numerator is zero at zero and has derivative $2\sinh^2\rho>0$ for $\rho>0$. Therefore its range is $(1,\infty)$, giving exactly one negative eigenvalue iff $q<-1$. That mode grows as $e^{\rho^2t/L^2}$. At $q=-1$, the zero mode $X=x$ is stationary; it is not a negative eigenmode.

Step 4: Quantify the concrete crossing. For $q=-2$, numerical substitution gives $F(1.915)<2<F(1.916)$. Strict monotonicity certifies one root in this interval, approximately $\rho=1.915008$. Thus $\lambda\approx-3.667256/L^2$. The open point at $(0,1)$ records the excluded endpoint limit of $F$.