Determine all real numbers \(L>1\) for which \[ x^2y''(x)+y(x)=0, \qquad 1\le x\le L, \] with boundary conditions \[ y(1)=y(L)=0 \] has a nonzero solution.
Solution
1. The indicial equation for a solution of the form\[ y=x^m \]is\[ m(m-1)+1=0. \]
For \(y=x^m\), \[ y''=m(m-1)x^{m-2}. \] Thus \[ x^2y''+y = \bigl(m(m-1)+1\bigr)x^m. \] The differential equation holds for nonzero \(x^m\) exactly when the displayed coefficient vanishes.
2. The roots of the indicial equation are\[ m_\pm = \frac12 \pm \frac{i\sqrt3}{2}. \]
The equation in step 1 is \[ m^2-m+1=0. \] Its discriminant is \(-3\), so the quadratic formula gives the two displayed roots.
3. Every real solution on \((0,\infty)\) has the form\[ y(x) = x^{1/2} \left( A\cos\left(\frac{\sqrt3}{2}\log x\right) +B\sin\left(\frac{\sqrt3}{2}\log x\right) \right), \]with \(A,B\in\RR\).
The two complex solutions associated with step 2 are \[ x^{m_\pm} = x^{1/2} e^{\pm i(\sqrt3/2)\log x}. \] Taking their real and imaginary parts gives the displayed real basis.
4. The boundary condition\[ y(1)=0 \]forces\[ A=0. \]
Since \(\log1=0\), step 3 gives \[ y(1) = A. \] Thus \(y(1)=0\) is equivalent to \(A=0\).
5. After imposing \(y(1)=0\), a nonzero solution satisfies \(y(L)=0\) ifand only if\[ \frac{\sqrt3}{2}\log L = k\pi \]for some positive integer \(k\).
By step 4, every solution satisfying the first boundary condition is \[ y(x) = B x^{1/2} \sin\left(\frac{\sqrt3}{2}\log x\right). \] It is nonzero exactly when \(B\neq0\). Hence \[ y(L)=0 \] is equivalent to \[ \sin\left(\frac{\sqrt3}{2}\log L\right)=0. \] Therefore \[ \frac{\sqrt3}{2}\log L=k\pi \] for some integer \(k\). Since \(L>1\), the logarithm is positive, so \(k\geq1\).
6. The complete set of admissible values is\[ \boxed{ L = \exp\left(\frac{2\pi k}{\sqrt3}\right), \qquad k=1,2,3,\ldots }. \]
Solving the equation in step 5 for \(L\) gives the displayed values. Conversely, for any positive integer \(k\), the function \[ y(x) = x^{1/2} \sin\left(\frac{\sqrt3}{2}\log x\right) \] is nonzero, solves the differential equation by step 3, vanishes at \(x=1\), and vanishes at the displayed value of \(L\).
7. Q.E.D.
Step 6 gives all and only the required values of \(L\).