04. Eigenfunctions Expansions

4.1 Introduction

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4.2 Eigenfunctions of Linear BVP

Eigenvalues

Consider BVP

$$Ly=\lambda y$$

$\lambda$ is an eigenvalue if $Ly=\lambda y$ has non-trivial solutions

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Eigenvalue of the Adjoint ODE

Consider BVP

$$Ly=\lambda y$$

Then adjoint BVP

$$L^{\ast }w=\overline{\lambda }w$$

has the same eigenvalues

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Orthogonality of Eigenfunctions of Original vs Adjoint

Consider Original and Adjoint BVP

$$Ly=\lambda y\text{ and }{L}^{\ast }w=\overline{\lambda }w$$

If

$$\begin{array}{rl}L{y}_j& ={\lambda }_j{y}_j\text{ and so }{L}^{\ast }{w}_j=\overline{{\lambda }_j}{w}_j\\ L{y}_k& ={\lambda }_k{y}_k\text{ and so }{L}^{\ast }{w}_k=\overline{{\lambda }_k}{w}_k\end{array}$$

Then

$$⟨y_j,w_k⟩=0\text{ for }{\lambda }_j\ne {\lambda }_k$$

Proof

$$\begin{array}{rl}0& =⟨L{y}_j,w_k⟩-⟨y_j,L^{\ast }{w}_k⟩\\ & =⟨{\lambda }_j{y}_j,w_k⟩-⟨y_j,{\overline{\lambda }}_k{w}_k⟩\\ & ={\lambda }_j⟨y_{j,w_k}⟩-{\lambda }_k⟨y_{j,w_k}⟩\\ & =({\lambda }_j-{\lambda }_k)⟨y_j,w_k⟩\end{array}$$

Hence if ${\lambda }_j\ne {\lambda }_k$ so $⟨y_j,w_k⟩=0$

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4.3 Inhomogeneous Solution Process

Eigenvalue Expansion

Consider Inhomogeneous BVP

$$Ly=f(N)$$

with linear homogenous boundary conditions $(BC)$

Denote eigenvalue problem

$$Ly=\lambda y(E)$$

Let $\{({\lambda }_j,y_j){\}}_{j=1,2,\cdots }$ and $\{({\lambda }_j,w_j){\}}_{j=1,2,\cdots }$ be the eigenvalue-eigenfunction pairs

Then

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

Proof

1. Solve the eigenvalue problem $(E+BC)$ to get eigenvalue-eigenfunction pairs

$$\{({\lambda }_j,y_j){\}}_{j=1,2,\cdots }$$

2. Solve adjoint eigenvalue problem $(E^{\ast }+B{C}^{\ast })$ to obtain

$$\{({\lambda }_j,w_j){\}}_{j=1,2,\cdots }$$

(since ${\lambda }_j$ is already found)

3. Assume solution to inhomogeneous problem $(N+BC)$ of form

$$y(x)=\sum _i{c}_i{y}_i(x)$$

Find coefficients $c_i$, take inner product with $w_k$

$$\begin{array}{rl}⟨f,w_k⟩& =⟨Ly,w_k⟩\\ & =⟨y,L^{\ast }{w}_k⟩\\ & =⟨y,{\overline{\lambda }}_k{w}_k⟩\\ & ={\lambda }_k⟨\sum _i{c}_i{y}_i,w_k⟩\\ & ={\lambda }_k{c}_k⟨y_k,w_k⟩\end{array}$$

Hence can solve for $c_k$

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4.4 Eigenfunction Expansion and Green’s Function

Green's Function from Eigenvalue Expansion

Let $Ly=f$ have Eigenvalue Expansion

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

Let constant $n_k:=⟨y_k,w_k⟩$
Then

$$\begin{array}{rl}y(x)& =\sum _{k=1}^{\infty }\frac{1}{{\lambda }_k{n}_k}\left({\int }_a^{b}f(\xi )w_k(\xi )d\xi \right)y_k(x)\\ & ={\int }_a^b\left(\sum _{k=1}^{\infty }\frac{1}{{\lambda }_k{n}_k}{w}_k(\xi )y_k(x)\right)f(\xi )dx\\ & ={\int }_a^{b}g(x,\xi )f(\xi )d\xi \end{array}$$

where

$$g(x,\xi )=\sum _{k=1}^{\infty }\frac{1}{{\lambda }_k{n}_k}{w}_k(\xi )y_k(x)$$

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4.5 Eigenfunction Expansion and FAT

Types of Solutions with Zero-Eigenvalues

Let $\{({\lambda }_j,y_j){\}}_{j=1,2,\cdots }$ and $\{({\lambda }_j,w_j){\}}_{j=1,2,\cdots }$ be the eigenvalue-eigenfunction pairs

If ${\lambda }_k=0$ then

1. If $⟨f,w_k⟩=0$ then $c_k$ is arbitrary so solution is non-unique
2. If $⟨f,w_k⟩\ne 0$ then solution does not exist

Note that ${\lambda }_k=0$ corresponds to $Ly=0$ so inline with FAT

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4.6 Inhomogeneous Boundary Conditions

Eigenvalue Expansion of Inhomogeneous ODEs

Consider

$$Ly=f,B_{i}y={\gamma }_i(i=1,2,\cdots ,n)$$

Decomposition Approach $u$ such that

Find

$$Lu=0,B_{i}u={\gamma }_i(i=1,2,\cdots ,n)$$

So $\tilde{y}=y-u$ satisfies homogenous ODE
Hence can continue as above

Normal Approach

As eigenfunctions are always determined using homogeneous boundary conditions
Eigenfunctions won’t change under decomposition

However care must be taken in integration by parts when finding $⟨Ly,w_k⟩-⟨y,L^{\ast }{w}_k⟩$
As boundary terms will generally still be present

Thus extra boundary terms carry through to formula for coefficients $c_k$
When substituting $⟨Ly,w_k⟩$ in the formula for $c_k$

Refer to page $42/43$ of lecture notes for examples

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4.7 Sturm-Liouville Theory

4.7.1 Homogenous SL Problem

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

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

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4.7.2 Properties of SL Eigenfunctions and Eigenvalues

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 $j\ne k$

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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

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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

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

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4.7.3 Singular SL Problems

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

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4.7.4 Inhomogeneous SL Problems

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

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

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4.7.5 Transforming an Operator to SL Form

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 $\hat{L}y=rf$

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