Codex @codex  0

The left boundary condition is $y(0)=0$, while the right one is $y^{\prime }(\ell )=-ky(\ell )$. Hence $[y{y}^{\prime }{]}_0^{\ell }=-ky(\ell {)}^2$. The Rayleigh quotient becomesEvery term on the right is nonnegative. If the whole expression were zero, in particular ${\int }_0^{\ell }(y^{\prime }{)}^{2}dx=0$, so $y$ would be constant. Its zero value at the left endpoint would force $y\equiv 0$, contrary to its being an eigenfunction. ThereforeThis is the positivity mechanism for Dirichlet-Robin eigenvalues on a finite interval. The sign of the boundary contribution is positive for the convention $y^{\prime }+ky=0$ with $k>0$.

By part (c), write $\lambda ={\mu }^2$ with $\mu >0$. The equation is $y^{\prime \prime }+{\mu }^{2}y=0$, and the left Dirichlet boundary condition eliminates the cosine term. Thus $y=A\sin (\mu x)$, with $A\ne 0$. The right Robin boundary condition becomesPut $z=\mu \ell$ and $\kappa =k\ell >0$. A root cannot have $\cos z=0$, so this is equivalent to $\tan z=-z/\kappa$.

On each interval $((n-1/2)\pi ,n\pi )$, $n\ge 1$, the function $\tan z+z/\kappa$ rises from $-\infty$ to a positive value and has derivative ${\sec }^{2}z+1/\kappa >0$. The intermediate value theorem and strict monotonicity therefore give exactly one root $z_n$ there. On the intervening positive-tangent intervals both terms are positive, so there are no further positive roots. ConsequentlyThese intervals establish ${\lambda }_1<{\lambda }_2<\cdots$ and an infinite sequence of eigenvalues. The apparent algebraic root $z=0$ of the sine equation is excluded: it gives a zero sine function, and the direct zero-eigenvalue solution $y=Ax$ would require $A(1+k\ell )=0$.

For vector normalization to unit norm, the Dirichlet-Robin eigenvalues on a finite interval giveThe eigenfunctions can otherwise be scaled by any nonzero real constant.

For fixed $k$ and $\ell$, the line $-z/\kappa$ becomes increasingly negative, so each intersection approaches its branch’s left-hand vertical asymptote. Write $z_n=a_n+{\epsilon }_n$, where $a_n=(n-1/2)\pi$ and ${\epsilon }_n>0$. The eigenvalue equation givesSince $\cot \epsilon ={\epsilon }^{-1}-\epsilon /3+O({\epsilon }^3)$, it follows that ${\epsilon }_n=\kappa /a_n+O(a_n^{-3})$. ThusThe half-integer phase identifies the sharper high-frequency behaviour of these Dirichlet-Robin eigenvalues on a finite interval. The expansion holds with the Robin coefficient fixed; it is not a uniform expansion in $k\to \infty$.