10 - Sturm-Liouville

Sturm-Liouville Problem

Sturm-Liouville Problem refers to self-adjoint linear ODEs in form

$$Ly(x)=\lambda r(x)y(x)$$

where $r(x)\ge 0$ is a weighting function and operator $L$ is in form

$$Ly(x)=-\frac{d}{dx}\left(p(x)\frac{dy(x)}{dx}\right)+q(x)y(x)\text{ for }a<x<b$$

Fully Self-Adjoint

Sturm-Liouville Problem is self-adjoint if the boundary conditions take separated form

$$\begin{array}{rl}{\alpha }_{1}y(a)+{\alpha }_2{y}^{\prime }(a)& =0\\ {\beta }_{1}y(b)+{\beta }_2{y}^{\prime }(b)& =0\end{array}$$

Orthogonality Property of Eigenfunctions for Sturm-Liouville

Orthogonality relation for SL eigenfunctions is

$${\int }_a^b{y}_j(x){\overline{y}}_k(x)r(x)dx\equiv ⟨y_j,r{y}_k⟩=0$$

for $j\ne k$

Realness of Eigenvalues of Sturm-Liouville

Consider Sturm-Liouville Problem
Assuming $p,q,r$ are real then $\overline{L}=L$

If $y_k$ is an eigenfunction of $\Lambda$ with eigenvalue ${\lambda }_k$ then

$${\overline{\lambda }}_k={\lambda }_k⟹\lambda \text{ is real}$$

Proof

Consider

$$\begin{array}{rl}0& =⟨L{y}_k,y_k⟩-⟨y_k,\lambda y_k⟩\\ & =⟨{\lambda }_{k}r{y}_k,y_k⟩-⟨y_k,{\lambda }_{k}r{y}_k⟩\\ & ={\lambda }_k⟨r{y}_k,y_k⟩-{\overline{\lambda }}_k⟨y_k,r{y}_k⟩\end{array}$$

As

$$⟨r{y}_k,y_k⟩=⟨y_k,r{y}_k⟩={\int }_a^b|y_k(x){|}^{2}r(x)dx>0$$

Then $({\lambda }_k-{\overline{\lambda }}_k)=0$ hence ${\overline{\lambda }}_k={\kappa }_k$ thus eigenvalues are real

Non-Negative Eigenvalues of Sturm-Liouville

If Sturm-Liouville system satisfies additional conditions

1. $p(x)>0$ and $r(x)>0$ on $a\le x\le b$

2. $q(x)\ge 0$ on $a\le x\le b$

3. Boundary Conditions have coefficients ${\alpha }_1{\alpha }_2\le 0$ and ${\beta }_1{\beta }_2\ge 0$

Then

$$\text{all eigenvalues }{\lambda }_k\ge 0$$

Proof

As $⟨y_k,L{y}_k-{\lambda }_{k}r{y}_k⟩=0$ then

$$\begin{array}{rl}-{\int }_a^b{y}_k(p(x)y_k^{\prime }(x){)}^{\prime }dx+{\int }_a^{b}q(x)y_k(x{)}^{2}dx-{\int }_a^b{\lambda }_{k}r(x)y_k(x{)}^{2}dx& =0\\ ⟹-[p{y}_k{y}_k^{\prime }{]}_a^b+{\int }_a^{b}p(x)y_k^{\prime }(x{)}^{2}dx+{\int }_a^{b}q(x)y_k(x{)}^{2}dx-{\lambda }_k{\int }_a^{b}r(x)y_k(x{)}^{2}dx& =0\end{array}$$

Hence

$${\lambda }_k=\frac{{\int }_a^{b}p(x)y_k^{\prime }(x{)}^{2}dx+{\int }_a^{b}q(x)y_k(x{)}^{2}dx-[p{y}_k{y}_k^{\prime }{]}_a^b}{{\int }_a^{b}r(x)y(x{)}^{2}dx}\ge 0$$

Singular Sturm-Liouville Problems

SL operator is singular at $x=a$ (or at the other end point) if

$$p(a)=0\text{ and }p>0\text{ on }(a,b]$$

Adjoint Boundary Conditions

By integration by parts

$$⟨Ly,w⟩-⟨y,Lw⟩=[p(y{w}^{\prime }-w{y}^{\prime }){]}_a^b=p(b)(y(b)w^{\prime }(b)-w(b)y^{\prime }(b))$$

Hence contribution from $x=a$ is zero regardless

Only need $y(x),y^{\prime }(x),w(x),w^{\prime }(x)$ to be bounded as $x\to a$

Zero at both end points

If

$$p(a)=0=p(b)\text{ and }p(x)>0\text{ for }x\in (a,b)$$

Then

$$⟨Ly,w⟩=⟨y,Lw⟩$$

with no boundary condition needed (except boundedness of course)

So $[a,b]$ is called the natural interval

Solving Inhomogeneous Sturm-Liouville Problems

Consider

$$Ly=f$$

Since Sturm-Liouville is self-adjoint then

$$L^{\ast }=L\text{ and }{w}_k\equiv y_k$$

With

$$y=\sum c_k{y}_k$$

Thus

$$\begin{array}{rl}⟨f,y_k⟩& =⟨Ly,y_k⟩\\ & =⟨y,L{y}_k⟩\\ & =⟨y,{\lambda }_{k}r{y}_k⟩\\ & ={\lambda }_k⟨\sum _i{c}_i{y}_i,r{y}_k⟩\\ & ={\lambda }_k{c}_k⟨y_k,r{y}_k⟩\end{array}$$

Hence

$$c_k=\frac{⟨f,y_k⟩}{{\lambda }_k,⟨y_k,r{y}_k⟩}$$

Transforming Second-Order Linear Operator to Sturm-Liouville Form

Consider Second-Order Linear Operator

$$Ly\equiv P_2{y}^{\prime \prime }+P_2{y}^{\prime }+P_{0}y$$

If $P_2\ne 0$ then

$$rLy=r{P}_{2}y+r{P}_1{y}^{\prime }+r{P}_{0}y$$

where

$$r(x)=-\frac{1}{P_2(x)}\exp \left(\int \frac{P_1(x)}{P_2(x)}dx\right)$$

so $rLy=-(p{y}^{\prime }{)}^{\prime }+qy$

Solving

Consider eigenvalue problem $Ly=\lambda y$ hence

$$\hat{L}y=r\lambda y=(-p{y}^{\prime }{)}^{\prime }+qy=\lambda ry$$

As $L$ is not self-adjoint then eigenfunctions $y_k$ are not orthogonal but

$$⟨y_j,w_k⟩=0\text{ for }j\ne k$$

However

$$⟨y_j,r{y}_k⟩=0\text{ for }j\ne k$$

For $Ly=f$ then there are two identical constructions

$$y(x)=\sum _{k=1}^{\infty }\frac{⟨f,w_k⟩}{{\lambda }_k⟨y_k,w_k⟩}{y}_k(x)\text{ or }y(x)=\sum _{k=1}^{\infty }\frac{⟨rf,y_k⟩}{{\lambda }_k⟨y_k,r{y}_k⟩}{y}_k(x)$$

where the latter uses equivalent Sturm-Liouville Problem $\hat{L}y=rf$