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\noindent
{\bf  Ch.4. INTEGRAL EQUATIONS AND GREEN'S FUNCTIONS}

\noindent
Ronald B Guenther and John W Lee, Partial Differential Equations of
Mathematical Physics and Integral Equations.\\
Hildebrand, Methods of Applied Mathematics, second edition

\vspace{0.6cm}

\noindent
In the study of the partial differential equations of hyperbolic and
parabolic types we solved several initial and boundary value problems. While
solving these equations we used the method separation of variables which 
reduces the problem to one of the following types of {\it Sturm-Liouville
problems}

\vspace{0.3cm}

\noindent
{\bf Sturm-Liouville Eigen value problem}:\, Let $p(x) > 0, q(x) \ge 0,
r(x) \ge 0$ in $I=(a,b)$. Here we assume $b>a$. Let
$X \in C^{2}(I) \cap C^{1}(\bar{I})$, then the set of equations given below

\begin{eqnarray}
-{d \over dx}\, [p(x)\, {dX \over dx}]+q(x) X=\lambda \, r(x) X, \label{eq01}\\
X(a)-h_{0}X^{\prime}(0)=0,\\
X(b)+h_{1}X^{\prime}(b)=0.
\end{eqnarray}

\noindent
is called the {\it The Sturm-Liouville Eigenvalue Problem}
(SLEVP). Here $h_{0} \ge 0$ and  $h_{1} \ge 0$ are
constants. Let the Sturm-Liouville operator $L$ be defined by

\begin{equation}
L=-{d \over dx}\, p(x) \, {d \over dx}+q(x), \label{eq02}
\end{equation}

\noindent
then the {\it Sturm-Liouville Eigenvalue  equation becomes}

\begin{equation}
L\,X=\lambda \, r(x) \, X,~~X(a)-h_{0}X^{\prime}(a)=0,
~~X(b)+h_{1}X^{\prime}(b)=0.
\end{equation}

\noindent
Note that $L$ is self-adjoint, i.e., $L=L^{*}$

\vspace{0.3cm}

\noindent
{\bf The Sturm-Liouville Boundary Value Problem}: With the same
$p,q,r$ defined above we have the Sturm-Liouville Boundary Value
Problem (SLBVP)

\begin{eqnarray}
L\, X=f(x),~~ in~~ I, \label{eq03}\\
X(a)-h_{0}X^{\prime}(a)=0,\\
X(b)+h_{1}\,X^{\prime}(b)=0.
\end{eqnarray}

\noindent
Here $f(x)$ is a continuous function in $I$.

\vspace{0.3cm}

\noindent
{\bf Sturm-Liouville Problem}: Combination of SLEVP and SLBVP may
be given as


\begin{eqnarray}
L\,X=\lambda r(x) X+ f(x), ~~ in ~~ I, \label{eq04}\\
X(a)-h_{0} x^{\prime}(a)=0,\\
X(b)+h_{1} x^{\prime}(b)=0.
\end{eqnarray}


\vspace{0.5cm}

\noindent
{\bf 4.1. The Sturm Liouville Boundary Value Problem and Green's Functions}

\vspace{0.3cm}

\noindent
Let us formulate the SLBVP as

\begin{eqnarray}
L\,u(x)=f(x),~~~ x \in I, \label{eq06}\\
u(a)-h_{0} u^{\prime}(a)=0,\\
u(b)+h_{1} u^{\prime}(b)=0.
\end{eqnarray}

\noindent
We shall solve this problem by the use of Green's function technique.
Let $u,v \in C^{2}(I) \cap C^{1}(\bar{I})$ then it is easy to prove the
{\it Lagrange identity}

\begin{equation}
v L u-u L v=-{d \over dx } [p(x) (v u^{\prime}-u v^{\prime})], \label{eq07}
\end{equation}

\noindent
Now letting $v=G(x,y)$ and

\begin{eqnarray}
L_{x}\, G(x,y)= \delta (x-y), ~~~ x,y \in I, \label{eq08}\\
G(a,y)-h_{0}\, {dG(x,y) \over dx}|_{x=a}=0, \label{eq09}\\
G(b,y)+h_{1}\, {dG(x,y) \over dx}|_{x=b}=0. \label{eq10}
\end{eqnarray}

\noindent
then we have

\begin{equation}
u(x)=\int_{a}^{b}\, G(x,y)\, f(y) dy,  \label{eq11}
\end{equation}

\noindent
Hence the SLBVP reduces to the construction of the Green's function
$G(x,y)$ satisfying (\ref{eq08})-(\ref{eq10}). In addition
to these properties $G(x,y)$ satisfy also the following properties

\vspace{0.3cm}

\noindent
{\bf 1}. $G(x,y)$ is continuous in $x,y \in I$  ({\it prove it})
in particular we have as $\varepsilon \rightarrow 0_{+}$

\begin{equation}
G(x,y)|_{x=y-\varepsilon}=G(x,y)|_{x=y+\varepsilon}. \label{eq12}
\end{equation}

\noindent
{\bf 2}. ${dG(x,y) \over dx}$ is discontinuous at $x=y$. As
$\varepsilon \rightarrow 0_{+}$ we have the jump condition
{\it prove it}

\begin{equation}
{dG(x,y) \over dx}|_{x=y-\varepsilon}-{dG(x,y) \over dx}|_{x=y+\varepsilon}=
p(y) \label{eq13}
\end{equation}

\noindent
Since $G(x,y)$ satisfies the homogeneous equation for $x<y$ and $x>y$
we then have

\begin{eqnarray}
G(x,y)=\left\{ \begin{array}{ll}
a_{1}(y)\,v_{0}(x)+a_{2}(y)\,v_{1}(x), ~~ a \le x < y \le b\\
b_{1}(y)\,v_{0}(x)+b_{2}(y)\,v_{1}(x), ~~ a \le y < x \le b
\end{array}
\right.
\end{eqnarray}

\noindent
where $v_{0}(x)$ and $v_{1}(x)$ are the solutions of the homogeneous
SL-equation and $a_{1}, a_{2}, b_{1}$ and $b_{2}$ are functions of $y$
to be determined through the conditions (\ref{eq09}), (\ref{eq10}),
(\ref{eq12}), and (\ref{eq13}). If we choose $v_{0}$ as a solution
of the homogeneous SL equation satisfying the {\it left boundary condition}

\begin{equation}
v_{0}(a)-h_{0}\,v_{0}^{\prime}(a)=0, \label{lbc}
\end{equation}

\noindent
and $v_{1}(x)$ as the solution of the SL equation satisfying the {\it
right boundary condition}

\begin{equation}
v_{1}(b)+h_{1} v_{1}^{\prime}(b)=0 , \label{rbc}
\end{equation}

\noindent
then $a_{2}=b_{1}=0$ and due to the symmetry $G(x,y)=G(y,x)$ and
remaining conditions we show that

\begin{eqnarray}
G(x,y)=\left\{ \begin{array}{ll}
v_{0}(x)\,v_{1}(y), ~~ a \le x < y \le b\\
v_{0}(y)\,v_{1}(x), ~~ a \le y < x \le b
\end{array}
\right. \label{eq14}
\end{eqnarray}

\noindent
The jump condition (\ref{eq13}) becomes

\begin{equation}
p(x)\, W(v_{0},v_{1})=-1
\end{equation}

\noindent
where $W(u,v)=uv^{\prime}-vu^{\prime}$ is the Wronskian of $u$ and $v$.
We then  have the following theorem:

\vspace{0.3cm}

\noindent
{\bf Theorem 1}.\, {\it The SLBVP (\ref{eq06}) has a Green's function
if and only if the corresponding homogeneous SLEVP [with $f(x)=0$] has only
the trivial solution, in which case the Green's function is given in
(\ref{eq14}) with the boundary conditions in (\ref{lbc}) and (\ref{rbc}).}

\vspace{0.3cm}

\noindent
{\bf proof}:\, {\bf (a)} If we assume the existence of the Green's function
we have the solution (\ref{eq11}) for each function $f(x)$ in $I$. The
corresponding homogeneous solution (with $f=0$ in (\ref{eq11}) goes to the
trivial solution $u=0$. {\bf (b)} Let us assume that the only solution of
homogeneous SLBVP is only the trivial one.  Hence we should be able to find
nontrivial solutions so that $pW=C$. Let $C=0$. This means $v_{0}$ and
$v_{1}$ are proportional hence , for example $v_{0}$ satisfies the
homogeneous SLBVP which is nontrivial, hence we obtain a contradiction. This
means $C=\ne 0$. By a simple scaling $C=-1$ (by redefining $v_{0}$ for
instance). Hence $pW=-1$. hence we have $v_{0}$ and $v_{0}$ solving the
homogeneous SL equation, $v_{0}$ satisfying the left boundary condition
(\ref{lbc}) and $v_{1}$ the right boundary condition (\ref{rbc}). Then
there exist a unique Green's function given in (\ref{eq14}).

\vspace{0.3cm}

\noindent
To summarize all properties of the Green's function we formulate
the following theorem

\vspace{0.3cm}

\noindent
{\bf Theorem 2}.\, {\it Assume that the homogeneous SLBVP ($f=0$
 problem has only the trivial solution. Let $G(x,y)$ be the function
 given in (\ref{eq14}) and with the conditions (\ref{lbc}) and (\ref{rbc}).
 Then

 \vspace{0.3cm}

 \noindent
 {\bf {i}}\, G(x,y) is continuous on the square $ a \le x,y \le b$ and
 has continuous second derivatives on each of the triangles $ a \le x < y \le b$
 and $a \le y < x \le b$. On each triangle $L_{x}\,G(x,y)=0$\\
 {\bf (ii)}\, G(x,y) satisfies the boundary conditions at $x=a$ and $x=b$,\\
 {\bf (iii)}\, For each $y \in (a,b)$ , G(x,y) satisfies the jump condition
 (\ref{eq13})
}
 \vspace{0.2cm}

 \noindent
 Conversely , properties (i)-(iii) uniquely determine $G$ such that
 (\ref{eq11}) solves the SLBVP in (\ref{eq11}).

\vspace{0.3cm}

\noindent
We have remarked that , since $L$ is self adjoint, the Green's function
is symmetric. If $p(x) > 0, q \ge 0 $ and $h_{0}, h_{1} \ge 0$ the we have
the following result

\vspace{0.3cm}

\noindent
{\bf Theorem 3}.\, {\it The Green's function of the SLBVP exists if $ p(x) > 0,
q(x) \ge 0$, and $h_{0}, h_{1} \ge 0$}.

\vspace{0.3cm}

\noindent
{\bf proof}.\, Th.1 says that the Green's function exists if and only if
the homogeneous SLBVP has only the trivial solution. The homogeneous SLBVP
is given as follows. Let $z \in C^{2}(I) \cap C^{1}(\bar{I})$ satisfy

\begin{eqnarray}
-[p(x) z^{\prime}]^{\prime}+q(x) z=0, ~~x \in I,\\
z(a)-h_{a} z^{\prime}(a)=0,~~~z(b)+h_{1} z^{\prime}(b)=0.
\end{eqnarray}

\noindent
Multiplying the differential equation by $Z$ and integrating over $I$  
and using integration parts we obtain

\begin{equation}
p(b) h_{1} {z^{\prime}(b)}^2+p(a) h_{0} {z^{\prime}(a)}^2+
\int_{a}^{b}\,[p(x) {z^{\prime}}^2+q(x) z^2] dx=0
\end{equation}

\noindent
Here, since $p > 0$ in $I$ then $z^{\prime}(x)=0$ in $I$. Hence 
$z=$ a constant everywhere in $I$,
but this constant should vanish at the boundary points $a$ and $b$, then
$z(x)$ must vanish at all pints in $I$. This proves that under
the given circumstances the only solution of the SLBVP is the trivial one
,$z=0$.

\vspace{0.3cm}

\noindent
{\bf Example 1}.\,Solve $u^{\prime \prime}=f(x), ~~x \in (a,b)$
and $u(a)=u(b)=0$.

\vspace{0.3cm}

\noindent
{\bf solution}.\, Let
\begin{eqnarray}
G(x,y)=\left\{ \begin{array}{ll}
a_{1} x+ a_{2}, ~~ a \le x < y \le b \\
b_{1} x+ b_{2}, ~~ a \le y < x \le b
\end{array}
\right.
\end{eqnarray}

\noindent
where the coefficient are to be determined through the conditions
(\ref{eq09}), (\ref{eq10}), (\ref{eq12}) and (\ref{eq13}). We find

\begin{eqnarray}
G(x,y)=\left\{ \begin{array}{ll}
{(x-a)(y-b) \over b-a}, ~~ a \le x < y \le b\\
{(y-a)(x-b) \over b-a}, ~~ a \le y < x \le b
\end{array}
\right.
\end{eqnarray}

\noindent
Hence the solution is given by

\begin{eqnarray}
u(x)=\int_{a}^{b}\, G(x,y) f(y) dy=\\
\int_{a}^{x}{(y-a)(x-b) \over b-a}\, f(y) dy+\int_{x}^{b}\, {(y-b)(x-a)
\over b-a}\, f(y) dy,\\
={x-b \over b-a}\, \int_{a}^{x}\, (y-a)\, f(y) dy+ {x-a \over
b-a}\, \int_{x}^{b}\, (y-b)\, f(y) dy
\end{eqnarray}

\vspace{0.3cm}

\noindent
{\bf Example 2}. Solve the same problem by changing the boundary
conditions to $u(a)=0,~~ u^{\prime}(b)=0$.

\vspace{0.3cm}

\noindent
{\bf Example 3}. Solve the same problem by changing the boundary conditions
to $u(a)=0,~~u^{\prime}(a)=0$.

\vspace{0.3cm}

\noindent
{\bf Example 4}. Solve the same problem by changing the boundary conditions
to $u(a)-\alpha^2\, u^{\prime}(a)=0$ and $u(b)+\beta^2\, u^{\prime}(b)=0$.
Here $\alpha$ and $\beta $ are some constants.

\vspace{0.3cm}

\noindent
{\bf 4.2 The Neumann Series}

\vspace{0.2cm}

\noindent
Let us now consider the general SL problem

\begin{eqnarray}
L\,u=\lambda\,r(x)\,u+f(x), ~~x \in (a,b), \label{gsl0}\\
u(a)-h_{0}u^{\prime}(a)=0,~~~u(b)+h_{1} u^{\prime}(b)=0. \label{gsl1}
\end{eqnarray}


\noindent
Since the Green's function $G(x,y)$ outlined in the first section
belongs to the operator $L$ we then have

\begin{equation}
u(x)=\int_{a}^{b}\, G(x,y)\,[f(y)+\lambda\, r(y) u(y)] dy
\end{equation}

\noindent
or, equivalently we have

\begin{equation}
u(x)=g(x)+\lambda\, \int_{a}^{b}\, G(x,y)\, r(y)\, u(y) dy,
\label{fr0}
\end{equation}

\noindent
where
$$g(x)=\int_{a}^{b} G(x,y) f(y) dy$$

\noindent
Eq(\ref{fr0}) is called the {\it The Fredholm integral equation
of the second kind} which can be put into a more symmetrical one.
Let

\begin{equation}
y(x)=\sqrt{r(x)}\, u(x),~~ g(x,s)=\sqrt{r(x)} \sqrt{r(s)} \, G(x,s),~~
f_{1}(x)=\sqrt{r(x)}\, g(x).
\end{equation}

\noindent
then

\begin{equation}
y(x)=f_{1}(x)+\lambda \int_{a}^{b}\,g(x,s)\,y(s) ds, \label{fr1}
\end{equation}

\noindent
In our future analysis we let $f_{1} \rightarrow f$. Here $g(x,s)$ is called
the {\it kernel} of the integral equation, $f(x)$ is given and $\lambda$
is in general a complex parameter. In most cases it is real. And also
we can assume that  $g(x,s)$
is continuous in $a \le x,s \le b$ and $f(x)$ is continuous in
$ a \le x \le b $. The integral equation given in (\ref{fr1})  may be solved
by using several methods. Let us assume that $y$ is a power series in
$\lambda$

\begin{equation}
y(x)=\sum_{n=0}^{\infty}\, \lambda^{n}\, y_{n}(x), \label{pw0}
\end{equation}


\noindent
where the coefficients $y_{n}(x)$ will be determined by the use of the
integral equation (\ref{fr1}) (we let $f_{1}=f$). We obtain

\begin{eqnarray}
y_{0}(x)=f(x), \label{ser0}\\
y_{1}(x)=\int_{a}^{b}\,g(x,s)\, y_{0}(s) ds, \label{ser1}\\
y_{2}(x)=\int_{a}^{b}\,g(x,s)\, y_{1}(s) ds,\label{ser2}\\
y_{n+1}=\int_{a}^{b}\, g(x,s)\,y_{n}(s) ds, \label{ser3}
\end{eqnarray}

\noindent
for $n=0,1,\cdots$
Hence all $y_{n}(x)$ can be calculated recursively. We have shown that
the series (\ref{pw0}) must be uniformly convergent to justify the
term by term integration in (\ref{fr1}).

\vspace{0.3cm}

\noindent
{\bf Proposition 4}.\, {\it Assume that $g(x,s)$ is continuous in
$ a \le x,s \le b$ and $f(x)$ is continuous in $ a \le x \le b$.
The series in (\ref{fr1}) is uniformly convergent if
$|\lambda| M (b-a) \le 1$}

\vspace{0.3cm}

\noindent
{\bf proof}. Since both $g$ and $f$ are continuous we
have
$$M=max|g(x,s)|, ~~~ N=max |f(x)|$$
in  $ a \le x,s \le b$.
Using (\ref{ser0}), (\ref{ser1}, (\ref{ser2}),
and (\ref{ser3}) we obtain

\begin{eqnarray}
|y_{0}(x)| \le N,\\
|y_{1}(x)| \le MN (b-a),\\
|y_{2}(x)| \le N M^2 (b-a)^2,
|y_{n}(x)| \le N M^{n} (b-a)^{n}|
\end{eqnarray}

\noindent
which leads to
$$|\sum_{n=0}^{\infty}\, \lambda^{n} y^{n}(x)| \le N\,\sum_{n=0}^{\infty}
[|\lambda| M (b-a)|^{n} < \infty$$


\noindent
This result provides also a solution of the integral equation (\ref{fr1})

\vspace{0.3cm}

\noindent
{\bf Proposition 5}.\, {\it Let 

\begin{equation}
y(x)=f(x)+\lambda \int_{a}^{b}\, g(x,s)\,y(s) ds \label{fr2}
\end{equation}

\noindent
be the integral equation. Let $g(x,s)$ and $f(x)$ be continuous in 
$ a \le x,s \le b$. If $|\lambda| < {1 \over M(b-a)}$ then

\begin{equation}
y(x)=f(x)+\lambda \int_{a}^{b}\, \gamma(x,s)\,f(s) ds
\end{equation}

\noindent
where $\gamma(x,s)$ is called the {\it resolvent} kernel and given by 

\begin{equation}
\gamma(x,s)=\sum_{n=1}^{\infty}\, \lambda^{n-1}\, k_{n}(x,s)
\end{equation} 
}
\vspace{0.3cm}

\noindent
{\bf proof}: Using (\ref{ser0})-(\ref{ser3}) we obtain

\begin{eqnarray}
y_{0}(x)&=&f(x),\\
y_{1}(x)&=&\int_{a}^{b}\, g(x,s)\, f(s) ds,\\
y_{2}(x)&=&\int_{a}^{b}\, g(x,s)\,y_{s} ds,\\
& =& \int_{a}^{b}\, g_{2}(x,s)\, f(s) ds,\\
\end{eqnarray}

\noindent
where $k_{2}(x,s)=\int_{a}^{b}\, g(x,t)\,g(t,s) dt$.
Defining, in general

\begin{eqnarray}
g_{1}(x,s)&=&g(x,s),\\
g_{n}(x,s)&=&\int_{a}^{b}\, g(x,t) g_{n-1}(t,s) dt, ~~n=2,3, \cdots
\end{eqnarray}

\noindent
then it is straightforward to establish the relation
$$y_{n}(x)=\int_{a}^{b}\, g_{n}(x,s)\, f(s) ds, ~~n=1,2, \cdots$$
and hence

\begin{eqnarray}
y(x)&=&\sum_{n=0}^{\infty}\, \lambda^{n}\, y_{n}(x)\\
&=&f(x)+\sum_{n=1}^{\infty}\, \lambda^{n}\, y_{n}(x)\\
&=&f(x)+\sum_{n=1}^{\infty} \lambda^{n}\, \int_{a}^{b}\, y_{n}(s) f(s) ds\\
&=&f(s)+\lambda \int_{a}^{b}\, \gamma(x,s) f(s) ds , \label{gam0}
\end{eqnarray}

\noindent
where

\begin{equation}
\gamma(x,s)=\sum_{n=1}^{\infty}\, \lambda^{n} g_{n}(x,s)
\end{equation}

\noindent
Convergence of the series in (\ref{gam0}) is guaranteed by Prop.4, and
the solution given in Prop.5 is unique. 

\vspace{0.3cm}

\noindent
{\bf Theorem 6}.\, {\it Let g(x,s) and f(x) be continuous on 
$ a \le x,s \le b$. Let $M= max|g(x,s)|$ for all $(x,s) \in [a,b]$.
If $|\lambda| < {1 \over M(b-a)}$ the the solution
$$y(x)=f(x)+\lambda \int_{a}^{b}\, \gamma(x,s)\,f(s) ds$$
is unique and continuous.}

\vspace{0.3cm}

\noindent{\bf proof}:\, Let us assume that there are two different solutions
$z_{1}(x)$ and $ z_{2}(x)$ of the integral equation (\ref{fr2}). Then their
difference $w(x)=z_{2}(x)-z_{1}(x)$ satisfies
$$w(x)=\lambda \int_{a}^{b}\, g(x,s) w(s) ds$$
Hence we have for all $x \in [a,b]$ we have
$$|w(x)| \le |\lambda| M W (b-a)$$
where $W=max |w(x)|$. Then 
$$[1-|\lambda|M(b-a)] W \le 0$$
This implies that $W=0$. Hence $z_{1}=z_{2}$ everywhere in $[a,b]$.

\vspace{0.3cm}

\noindent
{\bf Example 5}.\, Find the solution of
$$y(x)=f(x)+\lambda \int_{a}^{b}\, e^{x-s}\, y(s) ds$$

\vspace{0.3cm}

\noindent
{\bf Solution}: Try to find the following solution. Given a continuous
 $f(x)$ for $x \in [a,b]$ we have
$$y(x)=f(x)+{\lambda\, e^{x} \over 1-\lambda\,(b-a)}\, 
\int_{a}^{b} e^{-s}\, f(s) ds$$
where $ \lambda \ne {1 \over b-a}$. (i)\, Discuss the case 
$\lambda ={1 \over b-a}$ (ii)\, and as an example let $f(x)=x^2$.
Discuss also the application of (iii)\, Prop.4, (iv)\, Prop.5 and 
(v)\, Thm.6 to this example.
(vi)\, find also the Neumann series corresponding to this example.

\vspace{0.3cm}

\noindent
{\bf 4.3. Fredholm Equation with Separable Kernels}.

\vspace{0.3cm}

\noindent
A separable kernel is given  as

\begin{equation}
g(x,s)=\sum_{n=1}^{N}\, p_{n}(x)\, q_{n}(s) \label{sep0}
\end{equation}

\noindent
Here we assume that the $N$ functions $p_{n}$ are linearly independent.
Then the integral equation (\ref{fr2}) takes the form

\begin{eqnarray}
y(x)&=&f(x)+\lambda \int_{a}^{n}\, g(x,s)\,y(s) ds,\\
&=& f(x)+\lambda \int_{a}^{b}\, \left[ \sum_{n=1}^{N}\, p_{n}(x) 
\, q_{n}(s) \right] \,y(s) ds ,\\
&=& f(x)+\lambda \sum_{n=1}^{N} p_{n}(x)\, \left[ \int_{a}^{b}\,
q_{n}(s) y(s) ds \right],\\
&=& f(x)+\lambda \sum_{n=1}^{N}\, c_{n}\, p_{n}(x), \label{denk0}
\end{eqnarray}

\noindent
where

\begin{equation}
c_{n}=\int_{a}^{b}\, q_{n}(s) y(s) ds , ~~~~n=1,2, \cdots \label{denk1}
\end{equation}

\noindent
Using (\ref{denk0}) in (\ref{denk1}) we obtain the following algebraic
linear equations for the constants $c_{n}$'s

\begin{equation}
c_{n}=\beta_{n}+\lambda \sum_{m=1}^{N}\, \alpha_{nm}\,c_{m}, \label{denk2}
\end{equation}

\noindent
where for all $n,m=1,2, \cdots$

\begin{eqnarray}
\beta_{n}&=&\int_{a}^{b}\, q_{n}(s)\, f(s) ds,\\
\alpha_{mn}&=&\int_{a}^{b}\, p_{n}(s)\, q_{m}(s) ds
\end{eqnarray}

\noindent
let $A$ denote the $N \times N$ matrix corresponding to $\alpha_{mn}$
 $B$ be the column $N$-vector corresponding to $\beta_{n}$, and
 $C$ be the column unknown $N$-vector to be determined  then
(\ref{denk2}) simply becomes

\begin{equation}
[I-\lambda A] C= B , \label{denk3})
\end{equation}

\noindent
where $I$ is the $ N \times N$ unit matrix. The above linear equation
for $C$ is easily solved , but we have to consider all possible cases.
In the above equation (\ref{denk2}) we need to determine the unknown
coefficients $c_{n}$'s in terms of the known coefficients $\beta_{n}$ and
$\alpha_{mn}$ for all $m,n=1,2,\cdots$

\vspace{0.2cm}

\noindent
{\bf case (a)}. $f(x)=0$ or $B=0$ the equation (\ref{denk3}) becomes
homogeneous. For nontrivial solutions $det [I-\lambda A]$ must vanish.
Otherwise there is only the trivial solution $c_{n}=0$, for all
$n=1,2,\cdots$. If $ det [I-\lambda A]=0$ at least
(depending upon the rank of matrix $A$ one of the $c_{n}$'s is left
arbitrary. In such a case there are infinitely many solutions.
To remind you the terminology: Those values of $\lambda$ where
 $det [I-\lambda A]=0$ are called characteristic or eigenvalues and
 any nontrivial solution of the homogeneous integral equation is called
  the corresponding characteristic or eigenfunction. If there are $k$ number
  of constants $c_{n}$'s $n=1,2, \cdots , k$ for a given eigenvalue
  $\lambda$, then $k$ number of linearly independent eigenfunctions 
are obtained.


\vspace{0.3cm}

\noindent
{\bf case (b)}.  $f(x) \ne 0$ but $\beta_{n}=0, ~~n=1,2, \cdots$ this means
that $f(x)$ is orthogonal to all functions $q_{n}(x)$ , $n=1,2, \cdots$.
Hence $B=0$.  The case {\bf (a)} applies also here except for the
fact that here the solution (\ref{denk0}) contains the function $f(x)$. Hence the
trivial solution $C=0$ correspond to the solution $y(x)=f(x)$.  Solutions
corresponding to eigenvalues of $\lambda$  should be expresses  as the sum of
$f(x)$ and linear sum of the corresponding eigenfunctions.

\vspace{0.3cm}

\noindent
{\bf case (c)}. $B \ne 0$. We assume that at least for some $n$ ,
$\beta_{n} \ne 0$.{\bf if $det [I-\lambda A] \ne 0$} a unique nontrivial
solution of (\ref{denk3}) exists,
leading to a unique nontrivial solution $y(x)$ of the integral equation
 (\ref{denk0}). {\bf if$ det [I-\lambda A]=0$} either there is no solution
 or the solution is not unique meaning that there are infinitely
 many solutions.

 \vspace{0.3cm}

 \noindent
 {\bf Example 6}.\, Let $g(x,s)=1-3xs$. Solve the corresponding
 integral equation by considering all three cases above


\vspace{0.3cm}

\noindent
{\bf 4.4. Hilbert-Schmidt Theory}

\vspace{0.2cm}

\noindent
When the kernel $g(x,s)$ is not of type (\ref{sep0}) there are , in
general, infinitely many eigenvalues and eigenfunctions of the homogeneous
Fredholm equation. In addition, there may also be more than one
eigenfunctions corresponding to one eigenvalue. This is called
the degenaracy and the number of eigenfunctions corresponding to a
single eigenvalue is called the multiplicity. In this section we assume
that the kernel $g(x,s)$ is symmetric with respect to the variables
$x$ and $s$. We also assume that  multiplicity is one.

\vspace{0.3cm}

\noindent
{\bf Remark}: \, The homogeneous Fredholm equation

\begin{equation}
y(x)=\lambda\, \int_{a}^{b}\, g(x,s)\, y(s)\, ds
\end{equation}

\noindent
can not have zero eigenvalue , because the corresponding eigenfunction
is also zero. We the have the following result.

\vspace{0.3cm}

\noindent
{\bf Proposition 7}.\, {\it Eigenfunctions of the homogeneous Fredholm
integral equation with a symmetric kernel corresponding to different
eigenvales are orthogonal.  }

\vspace{0.3cm}

\noindent
{\bf proof}:\, By its definition we have that

\begin{equation}
y_{m}=\lambda_{m}\, \int_{a}^{b}\, g(x,s)\, y_{m}\,ds , ~~~m=1,2, \cdots
\end{equation}

\noindent
Multiplying by $y_{n}(x)$  and integrating over $(a,b)$ we get

\begin{eqnarray}
\int_{a}^{n}\, y_{m}(a)\,y_{n}(x)\, dx &=
& \lambda_{m}\, \int_{a}^{b}\,y_{n}(x) dx \, \int_{a}^{b}\, g(x,s)\, 
y_{m}(s) ds,\\
&=& \lambda_{m} \int_{a}^{b}\, y_{m}(s) ds \,
 \int_{a}^{b}\, y_{n}(x)\, g(x,s) dx,\\
&=&{ \lambda_{m} \over \lambda_{n}}\, \int_{a}^{b}\, y_{m}(s)\, y_{n}(s) ds
\end{eqnarray}

\noindent
which leads to

\begin{equation}
\int_{a}^{b}\, y_{m}(x)\, y_{n}(x) \, dx= 0
\end{equation}

\noindent
when $\lambda_{m} \ne \lambda_{n}$

\vspace{0.3cm}

\noindent
{\bf Remark 1}: If there are more than one eigenfunctions corresponding to an
eigenvalue $\lambda_{n}$ then orthogonalization of such a set is performed
by the standard Gram-Schmidt procedure. In the sequel we assume
that such sets are orthogonalized.

\vspace{0.3cm}

\noindent
{\bf Remark 2}:\, In the case of complex functions we generalize the Prop. 7
in the following way

\vspace{0.3cm}

\noindent
{\bf Proposition 7$^{\prime}$}. {\it Eigenfunctions of the homogeneous Fredholm
integral equation with a hermitian kernel , $g(x,s)=\bar{g}(s,x)$
corresponding to different eigemvales are orthogonal.}

\vspace{0.3cm}

\noindent
Note that the inner product in this case is defined by
$$ < f, g> =\int_{a}^{b}\, f(x) \bar{g}(x) dx$$
where bar over a letter denotes complex conjugation. This proposition
implies that
$$\int_{a}^{b}\, y_{m}(x)\, \bar{y}_{n}(x) dx =0$$
for $m \ne n$. A Corollary of this proposition is that  
$\lambda_{n}$ for such homogeneous Fredholm integral equation with
hermition kernel and hence also for real symmetric kernels the
eigenvales are real {\it (Prove these statements)}.

\vspace{0.3cm}

\noindent
We have now a theorem which will be used very often in our future analaysis.

\vspace{0.3cm}

\noindent
{\bf Theorem 8}. \, {\it Let $g(x,s)$ be a real and symmetric continuous
kernel over $(a,b)$. Let $H(x)$ be any continuous function over $(a,b)$
the any function defined as

\begin{equation}
h(x)=\int_{a}^{b}\, g(x,s) H(s) ds
\end{equation}

\noindent
can be represented as a linear superposition of the eigenfunctions
$y_{n}(x)$, $n=1,2, \cdots$,
of the homogeneous Fredholm equations with same kernel over $(a,b)$.}

\vspace{0.3cm}

\noindent
Hence we have
$$h(x)=\sum_{n=1}\, a_{n}\, y_{n}(x)$$
and
$$a_{n}={1 \over \rho_{n}}\, \int_{a}^{b}\, h(x) y_{n}(x) ds$$
with $\rho_{n}=\int_{a}^{b}\, {y_{n}(x)}^2\,dx$.

\vspace{0.3cm}

\noindent{Remark 3}:\, If there arefinite number of eigenfunctions then
the functions generated by the operation
$$\int_{a}^{b}\,g(x,s)\, H(s) ds$$
form a restiricted class of functions , irrespctive the form of the function
$H(x)$.

\vspace{0.3cm}

\noindent
{\bf Example 7}. Let $g(x,s)=sin(x+s)$ with $(a,b)=(0, 2\pi)$ then it is
easy to show that

\begin{eqnarray}
\lambda_{1}&=&{1 \over \pi}, ~~~y_{1} =\sin x+ \cos x, \\
\lambda_{2}&=&-{1 \over \pi},~~~y_{2}=\sin x- \cos x.
\end{eqnarray}

\noindent
Hence we have finite number of eigenfunctions. Then any function $h(x)$
given as
$$h(x)  =\int_{0}^{2\pi}\, g(x,s)\,H(s) ds$$
takes the form
$$g(x)=C_{1} \sin x + C_{2}\, \cos x$$
whatever the function $H(x)$ is. It is obvious that $h(x)$ can also
be written as
$$h(x)= a_{1} y_{1}(x)+a_{2} y_{2}(x)$$


\vspace{0.3cm}

\noindent
{\bf Remark 4}. \, In some cases the eigenfunctions
of the homogeneous Fredholm integral equation  may not form a complete set
(see DK for the definition). This means that any continuous function
$f(x)$  defined in $(a,b)$ may not be represented over the same interval
by a series of $y_{n}$'s. In the sequel we assume that $y_{n}$'s form a
complete set over the interval $(a,b)$.

\vspace{0.3cm}

\noindent
The essence of the eigenfunctions of the homogeneous Fredholm integral
eequation with a real and symmetric kernel
shows up when we wish to solve the inhomogeneous Fredholm integral
equation

\begin{equation}
y(x)=f(x)+\lambda\, \int_{a}^{b}\, g(x,s) y(s) ds \label{frd4})
\end{equation}


\noindent
where $f(x)$ is a given continuous function over the interval $(a,b)$.
First we shall use the orthonormalized set of eigenfunctions $\phi_{n}(x)$
which are defined by
$$y_{n}=C_{n} y_{n}(x)$$
where $C_{n}={1 \over \rho_{n}}$ and
$$\int_{a}^{b}\, \phi_{m}(x)\, \phi_{n}(x) dx=0, ~~~m \ne n$$
Then from (\ref{frd4}) by letting
$$\int_{a}^{b}\, \lambda\, g(x,s)\, y(s) ds=
\sum_{n=1}\, a_{n}\, \phi_{n}(x)$$
Hence we get
$$y(x)=f(x)+\sum_{n=1}\, a_{n}\, \phi_{n}(x), ~~~ a \le x \le b$$
where
$$a_{n}=\int_{a}^{b}\, [y(s)-f(s)] \phi_{n}(s) ds$$
By defining
$$c_{n}=\int_{a}^{b}\, y(s)\, \phi_{n} ds,~~~ \beta_{n}=
\int_{a}^{b}\, f(s)\, \phi_{n}(s) ds$$
then
$$a_{n}=c_{n}-\beta_{n}$$
Multiplying (\ref{frd4}) by $\phi_{n}(x)$ over $(a,b)$ and
using the symmetry of the kernel we obtain

\begin{equation}
[1-{\lambda \over \lambda_{n}}]\, c_{n}=\beta_{n}, ~~~n=1,2, \cdots
\label{eqc0}
\end{equation}

\vspace{0.3cm}

\noindent
{\bf case a}. If $\lambda=\lambda_{k}$ where $\lambda_{k}$ is of the
eigenvalues then $c_{k}$ becomes arbitrary and the solution
and $\beta_{k}=0$ or $\int_{a}^{b}\,f(s)\, \phi_{k}(s) ds=0$
and the solution becomes

\begin{equation}
y(x)=f(x)+c_{k}\, \phi_{k}(x)+\lambda_{k}\, \sum_{n \ne k}\, {\beta_{n}
\over \lambda_{n}-\lambda}\, \phi_{n}(x)
\end{equation}

\vspace{0.3cm}

\noindent
{\bf case b}. \, If $\lambda \ne $ any one of the eigenvalues. Then the
solution is unique.

$$y(x)=f(x)+\lambda\, \sum_{n=1}\, {\beta_{n}\, \over \lambda_{n}-\lambda}
\phi_{n}(x)$$

\vspace{0.3cm}

\noindent
{\bf 4.4 Singular Integral Equations}

\vspace{0.2cm}

\noindent
So far we have assumed that {\bf (i)}\, the interval $(a,b)$ of
the integral equations are finite was the kernel $g(x,s)$ was
continiuous. If an integral equation has either an infinite interval
or has a discontiniuous kernel then such an integral equation
is called {\it singular integral equation}. The first two of the
following are singular Fredholm equations of the first kind and the last one
is the singular Volterra integral equation of the
second kind (known also as the Abel's equation).
Here $y(x)$ are  unkown functions to be determined in each cases.

\begin{eqnarray}
a)~~~~F(x)&=&\int_{0}^{\infty}\, e^{-xs}\, y(s) ds,\\
b)~~~~F(x)&=&\int_{0}^{\infty}\, \sin(xs)\, y(s) ds,\\
c)~~~~F(x)&=&\int_{0}^{x}\, { y(s) \over \sqrt{x-s}} ds
\end{eqnarray}

\noindent
Here in each case $F(x)$ is a given function. The solutions of
the homogeneous integral equations

\begin{equation}
y(s)=\lambda \int_{a}^{b}\, g(x,s) y(s) ds, \label{frd5}
\end{equation}

\noindent
with

\vspace{0.2cm}

\noindent
{\bf (i)}: $(a,b)=(0, \infty)$ or\\
{\bf (ii)}: $g(x,s) $ is not continuous in $(a,b)$

\vspace{0.2cm}

\noindent
do not share the same properties as solutions of the homogeneous equations
in the previous sections. For example the first singular equation
above has continuous eigenvalues, the second one has two eigenvalues with
infinite multiplicity and the last one can be solved exactly. Here we shall
only present the first case. Recall the definition of the Gamma function

\begin{equation}
\int_{0}^{\infty}\, e^{-xs}\, s^{\alpha-1}\, ds=\Gamma(\alpha)\, x^{-\alpha},
~~\alpha > 0
\end{equation}

\noindent
Changing $\alpha$ to $1-\alpha$ and rewriting the above equation once more
we get

\begin{equation}
\int_{0}^{\infty}\, e^{-xs}\, s^{-\alpha}\, ds=\Gamma(1-\alpha)\, x^{\alpha-1},
~~\alpha < 1
\end{equation}

\noindent
Dividing first one by $\sqrt{\Gamma(\alpha)}$ the second one by
$\sqrt{\Gamma(1-\alpha)}$ and adding them we obtain

\begin{eqnarray}
\int_{0}^{\infty}\, [{ 1\over \sqrt{\Gamma(\alpha)}}\, s^{\alpha-1}+
{1 \over \sqrt{\Gamma(1-\alpha)}}\, s^{-\alpha}] ds\\
=\sqrt{\Gamma(\alpha)}\, x^{-\alpha}+ \sqrt{\Gamma(1-\alpha)}\, x^{\alpha-1},
~~~ 0 < \alpha < 1
\end{eqnarray}

\noindent
This last equation gives us the eigenvalues and the eigenfunctions
of the homogeneous singular Fredholm integral equatuion (\ref{frd5})

\begin{eqnarray}
\lambda_{\alpha}&=&{1 \over \sqrt{\Gamma(\alpha)\, \Gamma(1-\alpha)}},\\
y_{\alpha}&=&\sqrt{\Gamma(1-\alpha)}\, x^{\alpha-1}+\sqrt{\Gamma(\alpha)}\,
x^{\alpha}
\end{eqnarray}

\noindent
By using the identity
$$\Gamma(\alpha)\, \Gamma(1-\alpha)={\pi \over \sin \alpha\, \pi},~~~
0 < \alpha < 1$$

\noindent
then the eigenvalues become more simpler
$$\lambda_{\alpha}=\sqrt{{\sin \alpha \pi \over \pi}}$$
As we observe that $\alpha$ takes any values in the interval $(0,1)$, hence
eigenvalues take also continuous values.

\end{document}






















