Sturm–Liouville Eigenvalue Problem: 04-BS-5, May 2014, 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 regular Sturm–Liouville problem $y''+2y'+(1+\lambda)y=0$ on $[0,1]$ with homogeneous Dirichlet conditions $y(0)=0$, $y(1)=0$.

Find. Every eigenvalue $\lambda_n$ and a corresponding eigenfunction $y_n(x)$.

Approach. Remove the first-derivative term with the substitution $y=e^{-x}u(x)$ (the standard reduction for a constant-coefficient operator), reducing the problem to the canonical form $u''+\lambda u=0$ with the same boundary conditions, then solve that classical problem case by case in $\lambda$.

Eliminate the $y'$ term. Let $y=e^{-x}u(x)$, so $y'=e^{-x}(u'-u)$ and $y''=e^{-x}(u''-2u'+u)$. Substituting into the ODE, $$e^{-x}\big[(u''-2u'+u)+2(u'-u)+(1+\lambda)u\big]=e^{-x}\big[u''+\lambda u\big]=0,$$ so the reduced equation is $u''+\lambda u=0$ (the $u'$ and non-$\lambda$ $u$ terms cancel exactly — a hallmark of this reduction whenever the coefficient of $y'$ is twice the exponent used).
Transform the boundary conditions. Since $e^{-x}\ne0$ everywhere, $y(0)=0\Rightarrow u(0)=0$ and $y(1)=0\Rightarrow u(1)=0$.
Rule out $\lambda\le0$. For $\lambda=0$, $u=A+Bx$; both boundary conditions force $A=B=0$ (trivial). For $\lambda<0$, write $\lambda=-\mu^2$ ($\mu>0$); $u=Ae^{\mu x}+Be^{-\mu x}$, and $u(0)=u(1)=0$ again forces $A=B=0$. Only $\lambda>0$ can give a nontrivial eigenfunction.
Solve for $\lambda>0$. Write $\lambda=\mu^2$; the general solution is $u=A\cos\mu x+B\sin\mu x$. $u(0)=0\Rightarrow A=0$. $u(1)=0\Rightarrow B\sin\mu=0$; since $B\ne0$ (else trivial), $\sin\mu=0\Rightarrow\mu=n\pi$, $n=1,2,3,\dots$ $$\boxed{\lambda_n=n^2\pi^2,\qquad n=1,2,3,\dots}$$
Back-substitute for the eigenfunctions. $u_n(x)=\sin(n\pi x)$, and $y_n=e^{-x}u_n$: $$\boxed{y_n(x)=e^{-x}\sin(n\pi x)}$$

Eigenvalues and eigenfunctions
n	λ_n	y_n(x)
1	π² ≈ 9.8696	e^−x sin(πx)
2	4π² ≈ 39.478	e^−x sin(2πx)
3	9π² ≈ 88.826	e^−x sin(3πx)
general n	n²π²	e^−x sin(nπx)

Topic: Regular Sturm–Liouville eigenproblems.

Key relations: self-adjoint (SL) form $(py')'+(q+\lambda w)y=0$; here multiplying the ODE by the integrating factor $e^{2x}$ gives $(e^{2x}y')'+(1+\lambda)e^{2x}y=0$, i.e. weight $w(x)=e^{2x}$ — SL theory then guarantees real, ordered eigenvalues $\lambda_n\to\infty$ and eigenfunctions orthogonal with respect to $w$.

Why this works: a linear 2nd-order ODE with constant coefficients and a $y'$ term reduces to normal form via $y=u\exp\!\big(-\tfrac12\int(\text{coeff. of }y')\,dx\big)$; here that exponent is $-x$, which converts the operator into the same $u''+\lambda u=0$ that governs a vibrating string. The eigenvalue count and spacing ($\lambda_n=n^2\pi^2$) is therefore identical to the elementary Dirichlet problem — only the eigenfunction shape picks up the $e^{-x}$ envelope.

Common pitfall: stopping after finding $\lambda_n=n^2\pi^2$ and forgetting to re-attach the $e^{-x}$ factor to $u_n$ when stating $y_n$; also failing to rule out $\lambda\le0$ explicitly (some solutions assume $\lambda>0$ without justification, which loses marks even though the conclusion is correct).

Source: Kreyszig, Advanced Engineering Mathematics, Sturm–Liouville problems and eigenfunction expansions (Ch. 11); reduction to normal form (Ch. 2).

Question 1 of 7: Sturm–Liouville Eigenvalue Problem

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