09 - Eigenfunctions of Linear BVP

Eigenvalues

Consider BVP

$$Ly=\lambda y$$

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

Eigenvalue of the Adjoint ODE

Consider BVP

$$Ly=\lambda y$$

Then adjoint BVP

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

has the same eigenvalues

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 $⟨y_j,w_k⟩=0$

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 $c_k$

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

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

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

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

Examples of Families of Orthogonal Polynomials as Solutions

Second Order Linear ODEs with Families of Orthogonal Polynomials as Solutions with

$${\int }_a^b{p}_m(x)p_n(x)r(x)d=0\text{ for }m\ne n$$

where fixed weighting function $r(x)$ can be inferred by appropriate Sturm-Liouville Eigenvalue Problem

Important Examples are

1. “Jacobi-like” polynomials in form (including Legendre, Chebyshev Polynomials)

$$(1-x^2)y^{\prime \prime }(x)+(a+bx)y^{\prime }(x)+\lambda y(x)=0$$

posed on interval $[-1,1]$ with constants $a,b$ and appropriate discrete set of values of $\lambda$

2. Associated Laguerre Polynomials satisfying Laguerre’s Equation

$$x{y}^{\prime \prime }(x)+(a+1-x)y^{\prime }(x)+\lambda y(x)=0$$

with polynomial solution $y(x)=L_n^a(x)$ where $\lambda \in {\mathbb{Z}}^{\ge 0}$
Satisfying Orthogonality Relation

$${\int }_0^{\infty }{L}_m^a(x)L_n^a(x)(x^a{e}^{-x})dx=0\text{ for }m\ne n$$

Laguerre Polynomials correspond to $a=0$ denoted by $L_n(x)\equiv L_n^0(x)$

3. Hermite Polynomials are solutions of Hermite Equation

$$y^{\prime \prime }(x)-2x{y}^{\prime }(x)+\lambda y(x)=0$$

which has polynomial solution $H_n(x)$ when $\lambda =2n$ for integer $n\ge 0$
Satisfying Orthogonality Relation

$${\int }_{-\infty }^{\infty }{H}_m(x)H_n(x)e^{-x^2}dx=0\text{ for }m\ne n$$