02. The Fredholm Alternative

2.1 Analogy with Linear Algebra

Motivation for Fredholm Alternative

Consider homogeneous and inhomogeneous problems

$$\begin{array}{rlrl}A\mathbf{x}& =0& & (\mathcal{H})\\ A\mathbf{x}& =\mathbf{b}& & (\mathcal{N})\end{array}$$

Linear Algebra Version $A\in {\mathbb{R}}^{n\times n}$ Let $\mathbf{x},\mathbf{b}\in {\mathbb{R}}^n$

Let

If $A$ is invertible (non-zero determinant) then
$(\mathcal{H})$ has only trivial solution $\mathbf{x}=0$
So $(\mathcal{N})$ has unique solution $\mathbf{x}=A^{-1}\mathbf{b}$

However if $\mathcal{H}$ has solution $\mathbf{x}={\mathbf{x}}_1\ne 0$ then
$A$ is singular and $\mathbf{b}$ doesn’t have a general solution for $(\mathcal{N})$

If for a specific $\mathbf{b}$ has solution $\mathbf{x}$ for $(\mathcal{N})$ then there are infinite solutions in form

$$\mathbf{x}+\alpha {\mathbf{x}}_1$$

Linear Transformation Version

Let $A$ be a linear transformation on ${\mathbb{R}}^n$ then
Let $A^{\ast }$ be the corresponding adjoint transformation such that

$$⟨A\mathbf{x},\mathbf{w}⟩\equiv ⟨\mathbf{x},A^{\ast }\mathbf{w}⟩$$

where $⟨\mathbf{x},\mathbf{w}⟩\equiv \mathbf{x}\cdot \mathbf{w}$ denotes the usual Cartesian Inner Product

If $(\mathcal{H})$ has non-trivial solutions for $\mathbf{x}$ then corresponding adjoint problem

$$A^{\ast }\mathbf{w}=0$$

also has non-trivial solutions for $\mathbf{w}$

Taking the inner product of $\mathcal{N}$ with $\mathbf{w}$ and using adjoint inner product relation then

Necessary Condition for $\mathcal{N}$ to be solvable is

$$⟨\mathbf{b},\mathbf{w}⟩=0\text{ for all }\mathbf{w}\text{ satisfying }({\mathcal{H}}^{\ast })$$

Link to original

Fredholm Alternative Theorem ( ${\mathbb{R}}^n$ version)

Consider homogeneous and inhomogeneous problems

$$\begin{array}{rlrl}A\mathbf{x}& =0& & (\mathcal{H})\\ A\mathbf{x}& =\mathbf{b}& & (\mathcal{N})\end{array}$$

Exactly one of the following occur

1. Homogeneous Equation $(\mathcal{H})$ $A\mathbf{x}=0$ only has zero solution
   Hence solution of $(\mathcal{N})$ $A\mathbf{x}=\mathbf{b}$ is unique

2. Homogeneous Equation $(\mathcal{H})$ $A\mathbf{x}=0$ has non-trivial solutions and so does $({\mathcal{H}}^{\ast })$ $A^{\ast }\mathbf{w}=0$
   Hence there are two sub-possibilities:

   2a) If $⟨\mathbf{b},\mathbf{w}⟩=0$ for all $\mathbf{w}$ satisfying $({\mathcal{H}}^{\ast })$ then $(\mathcal{N})$ has a non-unique solution

   2b) Otherwise $(\mathcal{N})$ has no solution

Link to original

---

2.2 Adjoint Operator and Boundary Conditions

Inner Product

Inner Product of two (suitably smooth) functions defined on an interval $[a,b]$ by

$$⟨u,v⟩:={\int }_a^{b}u(x)\overline{v(x)}dx$$

where $\overline{v(x)}$ denotes the complex conjugate of $v(x)$

Link to original

Adjoint Operator

Let $L$ be a linear operator then

Corresponding adjoint operator $L^{\ast }$ is defined by inner product relation

$$⟨Ly,w⟩=⟨y,L^{\ast }w⟩$$

for all $y,w$ in suitable inner product space

Solving for the Linear Operator

1. Using integration by parts move the derivatives of operator from $y$ to $w$

2. Using the boundary conditions of $y$ ensure that the boundary terms vanish to get boundary conditions for $w$

Link to original

Properties of Adjoint Operator

1. Calculate adjoint $L^{\ast }$ of operator $L$ without worrying about boundary conditions

2. If $L^{\ast }=L$ then operator $L$ is self-adjoint

3. Operator $L$ combined with homogenous boundary condition give problem in form

$$L+BC$$

So corresponding adjoint boundary conditions give adjoint problem

$$L^{\ast }+B{C}^{\ast }$$

4. If $L=L^{\ast }$ and $BC=B{C}^{\ast }$ then problem is fully self-adjoint

5. If $L=L^{\ast }$ but $BC\ne B{C}^{\ast }$ then problem is formally self-adjoint

Link to original

General Form for Adjoint Operator

Let $L$ be a linear operator then

$$Ly=P_2{y}^{\prime \prime }+P_1{y}^{\prime }+P_{0}y⟺L^{\ast }w=(P_{2}w{)}^{\prime \prime }-(P_{1}w{)}^{\prime }+P_{0}w$$

Property for the Boundary

As

$$\begin{array}{rl}wLy-y{L}^{\ast }w& =w[P_2{y}^{\prime \prime }+P_1{y}^{\prime }+P_{0}y]-y[(P_{2}w{)}^{\prime \prime }-(P_{1}w{)}^{\prime }+P_{0}w]\\ & =[P_{2}w{y}^{\prime }-(P_{2}w{)}^{\prime }y+P_{1}wy{]}^{\prime }\end{array}$$

Hence

$$⟨Ly,w⟩-⟨y,L^{\ast }w⟩=[P_{2}w{y}^{\prime }-(P_{2}w{)}^{\prime }y+P_{1}wy{]}^{\prime }$$

Link to original

02 - Fredholm Alternative (Linear ODE Version)

Fredholm Alternative Theorem (Linear ODE Version)

Consider homogeneous and inhomogeneous problems

$$\begin{array}{rlrl}Ly& =0& & (H)\\ Ly& =f\ne 0& & (N)\end{array}$$

for $0<x<a$ with linear homogenous boundary conditions of form

$$\begin{array}{rl}{B}_{1}y& ={\alpha }_{1}y(a)+{\alpha }_2{y}^{\prime }(a)+{\beta }_{1}y(b)+{\beta }_2{y}^{\prime }(b)=0\\ B_{2}y& ={\alpha }_{3}y(a)+{\alpha }_4{y}^{\prime }(a)+{\beta }_{3}y(b)+{\beta }_4{y}^{\prime }(b)=0\end{array}$$

where $({\alpha }_1,{\alpha }_2,{\beta }_1,{\beta }_2)$ and $({\alpha }_3,{\alpha }_4,{\beta }_3,{\beta }_4)$ are linearly independent

Define homogeneous adjoint equation

$$L^{\ast }w=0$$

and corresponding adjoint boundary conditions ($B{C}^{\ast }$) then

Exactly one of the following possibilities occur

1. Homogeneous Problem ($H+BC$) only has the zero solution
   Solution of $(N+BC)$ is unique

2. Homogeneous Problem ($H+BC$) admit non-trivial solution and so does $(H^{\ast }+B{C}^{\ast })$
   Hence there are two sub-possibilities

   2a) If $⟨f,w⟩=0$ for all $\mathbf{w}$ satisfying $(H^{\ast }+B{C}^{\ast })$ then $(N+BC)$ has a non-unique solution

   2b) Otherwise $(N+BC)$ has no solution

Link to original

---

2.3 Inhomogeneous Boundary Conditions and FAT

Dealing with Inhomogeneous Boundary Conditions for FAT

Consider homogeneous and inhomogeneous problems

$$\begin{array}{rlrl}Ly& =0& & (H)\\ Ly& =f\ne 0& & (N)\end{array}$$

for $0<x<a$

Suppose the linear homogenous boundary conditions $(NBC)$ are of form

$$\begin{array}{c}\begin{array}{c}{B}_{1}y={\alpha }_{1}y(a)+{\alpha }_2{y}^{\prime }(a)+{\beta }_{1}y(b)+{\beta }_2{y}^{\prime }(b)={\gamma }_1\\ B_{2}y={\alpha }_{3}y(a)+{\alpha }_4{y}^{\prime }(a)+{\beta }_{3}y(b)+{\beta }_4{y}^{\prime }(b)={\gamma }_2\end{array}\}\text{(NBC)}\end{array}$$

where $({\alpha }_1,{\alpha }_2,{\beta }_1,{\beta }_2)$ and $({\alpha }_3,{\alpha }_4,{\beta }_3,{\beta }_4)$ are linearly independent and constants ${\gamma }_1$ and ${\gamma }_2$

Let $v(x)$ be a twice differentiable function that satisfies conditions $(NBC)$
Let

$$\tilde{y}(x)=y(x)-v(x)$$

Hence $\tilde{y}$ satisfies

$$L\tilde{y}=f-Lv=\tilde{f}$$

with homogeneous boundary conditions

$$B_1\tilde{y}=0=B_2\tilde{y}$$

Hence can apply Fredholm Alternative (Linear ODE Version)

Link to original