Sturm–Liouville Eigenvalue Problem: 04-BS-5, December 2015, Q1

Question 1: Sturm–Liouville Eigenvalue Problem (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The Sturm–Liouville problem $\dfrac{d}{dx}\left(x^{-3}y'\right) + (\lambda+4)x^{-5}y=0$ on $1\le x\le e^{2}$ with homogeneous Dirichlet conditions $y(1)=0$ and $y(e^{2})=0$.

Find. The eigenvalues $\lambda_n$ and the corresponding eigenfunctions $y_n(x)$.

Approach. Expand the derivative to recognize the equation as a Cauchy–Euler (equidimensional) ODE, solve its indicial equation, apply the boundary conditions to fix the eigenvalue spectrum, then normalize the eigenfunction family.

Reduce to Cauchy–Euler form. Expand $\dfrac{d}{dx}(x^{-3}y') = x^{-3}y'' - 3x^{-4}y'$, so the ODE reads $x^{-3}y'' - 3x^{-4}y' + (\lambda+4)x^{-5}y = 0$. Multiplying through by $x^{5}$ clears every negative power: $$x^{2}y'' - 3xy' + (\lambda+4)y = 0.$$ This is an equidimensional (Cauchy–Euler) equation on $x\in[1,e^{2}]$, i.e. genuinely a Sturm–Liouville problem in disguise — the weight function is $r(x)=x^{-5}$.
Solve the indicial equation. Try $y=x^{m}$: $x^{2}\!\cdot\! m(m-1)x^{m-2} - 3x\!\cdot\! m x^{m-1} + (\lambda+4)x^{m} = 0$ reduces to $$m^{2} - 4m + (\lambda+4) = 0 \ \Rightarrow\ m = 2 \pm \sqrt{4-(\lambda+4)} = 2\pm\sqrt{-\lambda}.$$ For an oscillatory (eigenvalue-producing) solution family we need $\lambda\gt 0$, so $\sqrt{-\lambda}=i\sqrt{\lambda}$ and $m=2\pm i\sqrt{\lambda}$ — complex conjugate roots of a Cauchy–Euler equation give $$y(x) = x^{2}\Big[C_1\cos\big(\sqrt{\lambda}\,\ln x\big) + C_2\sin\big(\sqrt{\lambda}\,\ln x\big)\Big].$$
Apply the first boundary condition. At $x=1$, $\ln x = 0$, so $\cos(0)=1$ and $\sin(0)=0$: $$y(1) = C_1\cdot 1\cdot 1 + C_2\cdot 1\cdot 0 = C_1 = 0.$$ Only the sine branch survives: $y(x) = C_2\, x^{2}\sin\big(\sqrt{\lambda}\,\ln x\big)$.
Apply the second boundary condition to fix the spectrum. At $x=e^{2}$, $\ln x = 2$: $$y(e^{2}) = C_2\, e^{4}\sin\big(2\sqrt{\lambda}\big) = 0.$$ For a nontrivial eigenfunction ($C_2\ne 0$) this forces $\sin(2\sqrt{\lambda})=0$, i.e. $2\sqrt{\lambda} = n\pi$ for $n=1,2,3,\dots$ (n=0 gives the trivial solution). Hence $$\boxed{\lambda_n = \left(\frac{n\pi}{2}\right)^{2}, \qquad n=1,2,3,\dots}$$
State the eigenfunctions. Substituting $\sqrt{\lambda_n}=n\pi/2$ back gives the eigenfunction family (taking $C_2=1$, the customary normalization for a Sturm–Liouville eigenfunction basis): $$\boxed{y_n(x) = x^{2}\sin\!\left(\frac{n\pi}{2}\ln x\right), \qquad n=1,2,3,\dots}$$ Direct substitution confirms $x^{2}y_n'' - 3xy_n' + (\lambda_n+4)y_n \equiv 0$ and both $y_n(1)=y_n(e^{2})=0$.

Final results — Question 1
Quantity	Result
Reduced ODE	$x^{2}y''-3xy'+(\lambda+4)y=0$ (Cauchy–Euler)
Eigenvalues	$\lambda_n=(n\pi/2)^{2}$, $n=1,2,3,\dots$
Eigenfunctions	$y_n(x)=x^{2}\sin\!\left(\dfrac{n\pi}{2}\ln x\right)$

Topic: Sturm–Liouville eigenvalue problems on an equidimensional (Cauchy–Euler) ODE.

Key relations: Standard S–L form $\dfrac{d}{dx}\!\big(p(x)y'\big)+\big[q(x)+\lambda r(x)\big]y=0$ with $p(x)=x^{-3}$, $q(x)=4x^{-5}$, $r(x)=x^{-5}$; Cauchy–Euler indicial equation $m^2-4m+(\lambda+4)=0$; the substitution $t=\ln x$ turns any Cauchy–Euler ODE into a constant-coefficient ODE in $t$, which is exactly why $\cos(\sqrt{\lambda}\ln x)$/$\sin(\sqrt{\lambda}\ln x)$ appear.

Why this works: A regular S–L problem with $p,r>0$ on a finite interval and separated homogeneous boundary conditions has a discrete, unbounded, strictly increasing sequence of real eigenvalues, each with (up to scale) one eigenfunction — exactly the pattern found here. The complex indicial roots are the signature of an oscillatory (trigonometric-in-$\ln x$) solution family, which is what allows a discrete Dirichlet spectrum to exist at all.

Common pitfall: Forgetting to expand $\frac{d}{dx}(x^{-3}y')$ before attempting a series solution (the equation is exactly solvable in closed form once recognized as Cauchy–Euler — no series needed); also, students often forget the $C_1=0$ step and carry a spurious cosine term into the second boundary condition.

Source: Kreyszig, Advanced Engineering Mathematics, 10th ed., Ch.11.5 (Sturm–Liouville Problems) and Ch.2.5 (Euler–Cauchy Equations).

Question 1 of 7: Sturm–Liouville Eigenvalue Problem

Question 1: Sturm–Liouville Eigenvalue Problem (20 marks)