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\begin{center}
{\bf Sturm - Liouville}

\end{center}

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{\bf Sturm - Liouville systems}

${}$


\begin{eqnarray} \frac{d}{dx} \left( k(x) \frac{dy}{dx} \right) +
(\lambda g() - l(x))y & = & 0 \hspace{.2in} a < x < b \\
\alpha_1y(a) + \alpha _2y(b) + \alpha_3 y'(a) \alpha_4y'(b) & = &
0 \\ \beta_1y(a) + \beta_2y(b) + \beta_3y'(a) + \beta_4y'(b) & = &
0 \end{eqnarray}

The above relations comprise a Sturm-Liouville system: $\lambda$
is a parameter to be determined.  The relation (2) and (3) are
linearly independent i.e. the vectors $(\alpha_1, \alpha_2,
\alpha_3,\alpha_4)$; $(\beta_1,\beta_2,\beta_3,\beta_4)$ are
linearly independent.

${}$

\underline{Examples}

\begin{enumerate}
\item String with tension $T(x)$ and density $m(x)$ variable along
the string, and subject to a transverse restoring force of
magnitude $s(x)$ per unit length per unit transverse displacement.
For displacement to varying with time as $\cos \omega t$ we find:
$$\frac{d}{dx} \left(t(x) \frac{dy}{dx} \right) + \left(
\frac{\omega^2}{c^2}m(x) - s(x) \right) y = 0$$
$\begin{array}{ccrcl} {\rm end\ conditions} & {\rm e.g.} & y(0) &
= & 0 \\ & & y(l) & = & 0 \end{array}$

\item Thermally conducting bar with slowly varying cross section
$A(x)$, heat loss along the surface $h(x)$ per unit length, no
internal generation of heat.  Variable conductivity  $K(x)$.
Variable heat capacity $c(x)$/unit vol. $$\frac{d}{dx} \left[ k(x)
A(x) \frac{dy}{dx}\right] + \left[ p c(x) - h(x) \right]y =0$$ for
solutions with a time variation $\alpha e^{-pt}$ and appropriate
end conditions.
\end{enumerate}

${}$
\newpage
\underline{Existence and Uniqueness of a Solution of a Linear
Second Order Equation}

$$y''(x) + q(x) y'(x) + r(x) y(x) = 0$$ $q(x), \, r(x)$ continuous
in $ a \leq x \leq b$

$y(a) , \, y'(a)$ assigned arbitrary.

Define the vector $\ds w = \left( \begin{array}{c} w_1 \\ w_2
\end{array} \right)$

Let norm $w \equiv ||w|| = |w_1 | + |w_2| = o \Leftrightarrow w =
0$

$||w_1|| + ||w_2|| \geq ||w_1 + w_2||$

If $\ds M = \left( \begin{array}{cc} m_11 & m_12 \\ m_21 & m_22
\end{array} \right)$ and $c = 2 \max |m_{ij}|$

$||Mw|| \leq c||w||$

Define $\ds w'(x) = \left( \begin{array}{c} w'_1(x) \\ w'_2(x)
\end{array} \right) \int_a^x w(t) dt = \left( \begin{array}{c}
\int_a^xw_1(t) dt \\ \int_a^x w_2(t) dt \end{array} \right)$

\begin{eqnarray*} \left| \left| \int_a^x  w(t)\,dt  \right|
\right| & = & \left| \int_a^x w_1(t) dt \right| + \left| \int_a^x
w_2(t) dt \right|\\ & \leq & \left| \int_a^x w_1(t) dt \right| +
\left| \int_a^x w_2(t) dt \right| \\ & = & \int_a^x ||w(t) dt ||
\end{eqnarray*}

${}$

Now define $v(x) = y'(x)$ then $v'(x) = -(x)v(x) - r(x)y(x)$

define $\ds w(x) = \left( \begin{array}{c} y(x) \\ v(x)
\end{array} \right)\hfill (1)$

$\ds \frac{d}{dx} w(x) = A(x) w(x) \hfill(2)$

where $\ds A(x) = \left( \begin{array}{cc} 0 & 1 \\ -r(x) & \-q(x)
\end{array} \right) \hfill (3)$

Let $c(x) = 2 \max \{ 1, \, |r(x) | , \, |q(x)|\} \, \, a \leq x
\leq b$

Let $w(a) = \alpha = \left( \begin{array}{c} y(a) \\ y'(a)
\end{array} \right)$, $\ds c = \sup_{a \leq x \leq b}c(x)$

From (2), integrating from $a$ to $x$

${}$

$\ds w(x) = \alpha + \int_a^x A(t) w(t) dt \hfill (4)$

${}$

[This is a vector intrgral equation]

Define the iterant $w^k(x)\hspace{.2in} k = 0,1, \ldots $ by $\ds
w^0(x) = \alpha$

$w^{k+1}(x) = \alpha + \int_a^xA(t) + w^k(t) dt \hspace{.2in} k =
0, 1, \ldots \hfill (5)$

\begin{enumerate}
\item By induction the $w^k$ all exist and are continuous in
$[a,b]$

\item By induction on the equation $$w^{k+1} (x) - w^k(x) =
\int_a^x A(t)(w^k(t) - w^{k-1}(t))dt$$ we can show that
$$||w^{k+1}(x) - w^k(x)|| \leq ||\alpha ||
\frac{[c(x-a)]^{k+1}}{(k+1)!}$$

\item We then have

$\ds \sum_{k=0}^\infty ||w^{k+1}(x) - w^k(x) ||$ converges
uniformly in $[a,b]$

therefore $\ds\sum_{k=0}^\infty (w^{k+1}(x) - w^k(x))$ converges
uniformly in $[a,b]$

$\ds w^k(x) = \alpha + \sum_{r=0}^{k-1}(w^{r+1}(x) - w^r(x))$

Hence $\ds \lim_{k \to \infty} w^k(x)$ exists $=w(x)$ uniformly in
$[a,b]$ and $w(x)$ is continuous in $[a,b]$ letting $k \to \infty$
in equation (5) we get $$w(x) = \alpha + \int_a^x A(t) w(t) dt$$
RHS is differentiable , therefore $$w'(x) = A(t) w(x) {\rm \ \
amd\ \ } w(\alpha) = \alpha$$
\end{enumerate}
${}$

\underline{Uniqueness}

Assume that $z(x)$ is a solution, continuous and bounded in
$[a,b]$, of the integral equation \begin{eqnarray*} z(x) & = &
\alpha + \int_a^x A(t) z(t) dt \\ w^{k+1}(x) & = & \alpha +
\int_a^x A(t) w^k(t) dt \\ w^{k+1}(x) - z(x) & = & \int_a^x A(t)
(w^k(t) - z(t)) dt \end{eqnarray*}

By induction we can show that $$||w^k(x) - z(x) || \leq \frac {
m[c(x-a)]}{k!}$$ where $m = \bar{bd} ||z(x) - a||$ in $[a,b]$

Therefore $\ds \lim_{k \to \infty} ||w^k(x) - z(x) || = 0$
uniformly in $[a,b]$

Therefore $\ds ||w(x) - z(x) || = 0 $ i.e. $|w(x) = z(x) $

($p_0 >0, \, p_0 p_1p_2$ continuous in $[a,b]$)

A solution of $L(y) =0$ exists in $[a,b]$ such that $y(v), \,
y'(c)$ have arbitrary values, $c \in [a,b]$ and this solution is
unique.

${}$

{\bf Wronskian of two solutions}

${}$

If $u(x), \, v(x), \, u'(x), \, v'(x)$ are continuous then $$W
=_{af} \left( \begin{array}{cc} u(x) & v(x) \\ u'(x) & v'(x)
\end{array} \right)$$ is the Wronskian determinant.

\begin{description}
\item[(i)] $W =  0$ in $[a,b]$ is the necessary and sufficient
condition that $u$ and $v$ are linearly dependent.
\item[(ii)] $W \not=  0$ in $[a,b]$ is the necessary and sufficient
condition that $u$ and $v$ are linearly independent.
\end{description}

If now $L(u) = 0 \, \, L(v)=0$ \begin{eqnarray*} 0 & = & vL(u) -
uL(v) \\ & = & p_0(vu'' - v''u) + p_1(vu' - v'u) \\ & = & p_0
\frac{d}{dx} (vu'-v'u) + p_1(vu' - v'u) \end{eqnarray*} Write
$$p(x) = \exp \int_a^x \frac{p_1(t)}{p_0(t)} dt$$ therefore $\ds
\frac{p'(x)}{px} = \frac{p_1(x)}{p_0(x)}$

Therefore the above equation is $$p\frac{d}{dx}(vu'-v'u) + p'(vu'
- v'u)=0$$ i.e. $\ds  p(vu' - v'u)=$constant.

i.e. $\ds \left| \begin{array}{cc} u & v \\ u' & v' \end{array}
\right|=$constant.

$\ds \int_a^x \frac{p_1(t)}{p_0(t)} \, dt$ is bounded in $[a.b]$.

Therefore $\ds p(x) = \exp \int_a^x \frac{p_1(t)}{p_0(t)} dt >0$
in $[a,b]$

Hence
\begin{description}
\item[(i)] $W=0$ at $x=c \in [a,b] \Rightarrow W=0 \, \, a \leq x
\leq b$
\item[(ii)] $W\not=0$ at $x=c \in [a,b] \Rightarrow W\not=0 \, \, a \leq x
\leq b$
\end{description}


${}$

{\bf Example of choice of linearly independent solutions.}

$\begin{array}{lll} u(x): & u(a) = 1 & u'(a) = 0 \\ v(x): & v(a)
=0 & v'(a) = 1 \end{array}$

$\ds W(u,v) = \left| \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}
\right| = 1 \not=0$ therefore $u$, $v$, are linearly independent
in $a \leq  x \leq b$

${}$

{\bf Fundamental System of solutions}

${}$

\underline{Definition} Any pair $u(x), \, v(x)$ of linearly
independent solutions constitute a fundamental system.

${}$

\underline{Theorem} Any solution of $L(y)=0$ is of the form $y =
Au+Bv$

${}$

\underline{Proof} We can choose $A, B$ such that $y(c) , \ y'(c)$
have any assigned values, $c$ in $[a,b]$ \begin{eqnarray*} y(c) &
= & A u(c) + Bv(c) \\y'(c) & = & A u'(c) + Bv'(c) \end{eqnarray*}
$\ds \left| \begin{array}{cc} u(c) & v(c) \\ u'(c) & v'(c)
\end{array}  \right| \not=0$ therefore $A$ and $B$ are uniquely
determined.

Consider $\ds z(x) = y(x) - A u(x) - Bv(x)$

$\,\,L(z) = 0$ \hspace{.35in} -(i) as $L$ is linear.

$\left. \begin{array}{c} z(c) = 0 \\ z'(c) = 0 \end{array}
\right\}$ \hspace{.1in} -(ii)

(i) and (ii) are satisfied by $z \equiv 0$,  Therefore by the
uniqueness theorem this is the only solution.

Therefore $z \equiv 0 \hspace{.2in} a \leq x \leq b$

Therefore $y(x) = A \cdot u(x) + B \cdot v(x) \hspace{.2in} a \leq
x \leq b$

\newpage
{\bf Adjoint (2nd order) linear differential operators}

${}$

$\ds L(u) = (p_0D^2 + p_1 D + p_2) u \hfill (1)$

${}$

Let $v = v(x), \, v'(x) \, v''(x)$ exist.

\begin{eqnarray*} vL(u) & = &  vp_0D^2u + vp_1Du + vp_2u  \\  vp_0D^2(u)
& = & uD^2(vp_0) + D[vp_0Du - uD(vp_0)] \\ vp_1 Du & = & -uD(vp_1)
+ D(vp_1u) \end{eqnarray*}

Hence $\ds vL(U) = u[D^2(vp_0) - D(vp_1) + vp_2] + D[vp_0 Du -
uD(vp_0) + vp_1u]$

Write $\ds M(v) = D^2(vp_0) - D(vp_1) + vp_2 \hfill (2)$

$\ds M(v) = p_0D^2v + (2p'_0 - p_1) Dv + (p''_0 - p'_1 + p_2)v
\hfill (3)$

$\begin{array}{rcl} vL(u) - uM(v) & = & \ds\frac{d}{dx}
(vp_0u'-u(p_0v'+p'_0v)+p_1uv) \\&=&
\ds\frac{d}{dx}(p_0(vu'-u'v)+(p_1-p'_0)uv)  \end{array}$ \hfill(4)

$M$ is said to be the adjoint of $L$ in view of the form of R.H.S.
Also $L = adj M$

${}$

{\bf Self adjoint Operator}

${}$

\underline{Definition} $L$ is self  adjoint if $M \equiv L$

${}$

From (1) and 93) the necessary and sufficient condition is $p_1 =
p'_0$

then $\ds L(u) = \frac{d}{dx} \left( p_0 \frac{du}{dx} \right) +
p_2u$

If $L$ is self adjoint the from (4) $\ds vL(u) - uL(v) =
\frac{d}{dx} p_0(vu' - u'v)$

${}$

{\bf Reduction to self-adjoint form}

${}$

if $L = p_0D^2 + p_1D + p_2$

Define $\ds p(x) = \exp \int_a^x \frac{p_1(t)}{p_0(t)} dt$ ($p_0
>0$ in $[a,b]$)

Then $\ds \frac{p'(x)}{p(x)} = \frac{p_1(x)}{p_0(x)}$

Hence \begin{eqnarray*} \frac{p}{p_0} L(y) & = & \left( pd^2 + p'D
+ \frac{pp_2}{p_0} \right) y\\ & = & \frac{d}{dx} \left( p
\frac{dy}{dx} \right) + \frac{pp_2}{p_0}y \end{eqnarray*}

${}$

{\bf Self adjoint System}

${}$

$$L(y) = \frac{d}{dx} \left( p \frac{dy}{dx} \right) + qy$$

Suppose we have a self adjoint operator $L$.  Consider the
homogeneous system:

$\ds L(y) = 0 \hspace{.3in} (a \leq x \leq b)$ \hfill (1)

$\left.\begin{array}{c}
0=U_1(y)=a_1y(a)+b_1y'(a)+c_1y(b)+d_1y(b)\\
0=U_2(y)=a_2y(a)+b_2y'(a)+c_2y(b)+d_2y(b)\end{array}\right\}$
\hfill(2)

[condition (2) constitutes 2-point boundary conditions]

Let $u,v$ be any two functions such that $u'v'$ are continuous in
$[a,b]$ (not necessarily satisfying $L(y)=0$)

$\ds vL(u) - uL(v) = \frac{d}{dx} p(vu' - v'u)$

Therefore

$\ds \int_a^x (vLu - uLv) dx = -p(b) \left|
\begin{array}{cc} u(b) & v(b) \\ u'(b) & v'(b) \end{array} \right|
+ p(a) \left| \begin{array}{cc} u(a) & v(a) \\ u'(a) & v'(a)
\end{array} \right|$ \hfill (3)

${}$

\underline{Definition} The boundary conditions (2) are said to be
self-adjoint if R.H.S of (3) vanishes whenever $u$ and $v$ satisfy
(2) i.e. $U_i(v) = 0$, $U_i(v) =0\, i = 1, 2$ and $u$ and $v$ are
linearly independent.

${}$

\underline{Theorem} The necessary and sufficient condition for
this to occur is

$\ds p(a) \left| \begin{array}{cc} c_1 & d_1 \\ c_2 & d_2
\end{array} \right| = p(b) \left| \begin{array}{cc} a_1 & b_1 \\
a_2 & b_2\end{array} \right|$ \hfill (4)

${}$

${}$

\underline{Proof} The equations $U_i = 0 \, \, V_i = 0$ may be
written:

$\ds \left( \begin{array}{cc} a_1 & b_1 \\ a_ 2 & b_2 \end{array}
\right) \left( \begin{array}{c} u(a) \\ u'(a) \end{array} \right)
+  \left( \begin{array}{cc} c_1 & d_1 \\ c_2 & d_2 \end{array}
\right)\left( \begin{array}{c} u(b) \\ u'(b) \end{array} \right) =
0$

$\ds \left( \begin{array}{cc} a_1 & b_1 \\ a_ 2 & b_2 \end{array}
\right) \left( \begin{array}{c} v(a) \\ v'(a) \end{array} \right)
+ \left( \begin{array}{cc} c_1 & d_1 \\ c_2 & d_2 \end{array}
\right)\left( \begin{array}{c} v(b) \\ v'(b) \end{array} \right) =
0$

$\ds \Leftrightarrow  \left( \begin{array}{cc} a_1 & b_1 \\ a_ 2 &
b_2 \end{array} \right) \left( \begin{array}{cc} u(a) & v(a) \\
u'(a) & v'(a) \end{array} \right) +  \left( \begin{array}{cc} c_1
& d_1 \\ c_2 & d_2 \end{array} \right)\left( \begin{array}{cc}
u(b) & v(b) \\ u'(b) & v'(b) \end{array} \right) = 0$

$AW_a + BW_b = 0$

Therefore $AW_a = -BW_b$

Taking determinants:

$|A| |W_a| = |B||W_b|$\hfill(5)

\hspace{.5in} ($|-B| = |B|$ as $B$ is of even order)

Therefore $\ds p(a) |W_a| = p(b)|W-b|$

$\Leftrightarrow p(a) |B| = p(B)|A|$

${}$

\underline{Examples}

\begin{description}
\item[(i)] $\ds \begin{array}{l} y(a) = 0 \\ y(b) =0 \end{array}
\left( \begin{array}{cc} a_1 & b_1 \\ a_2 & b_2 \end{array}
\right) = \left( \begin{array}{cc} 1 &  0 \\ 0 & 0
\end{array}\right) \left( \begin{array}{cc} c_1 & d_1 \\ c_2 & d_2
\end{array} \right) = \left( \begin{array}{cc} 0 & 0 \\ 1 & 0
\end{array}\right)$

(string with fixed ends.)
\item[(ii)]$\ds \begin{array}{l} y'(a) = 0 \\ y'(b) =0 \end{array}
\left( \begin{array}{cc} a_1 & b_1 \\ a_2 & b_2 \end{array}
\right) = \left( \begin{array}{cc} 0 &  1 \\ 0 & 0
\end{array}\right) \left( \begin{array}{cc} c_1 & d_1 \\ c_2 & d_2
\end{array} \right) = \left( \begin{array}{cc} 0 & 0 \\ 0 & 1
\end{array}\right)$

(string with free ends.)
\item[(iii)]$\ds \begin{array}{l} y(a) = 0 \\ y'(b) =0 \end{array}
\left( \begin{array}{cc} a_1 & b_1 \\ a_2 & b_2 \end{array}
\right) = \left( \begin{array}{cc} 1 &  0 \\ 0 & 0
\end{array}\right) \left( \begin{array}{cc} c_1 & d_1 \\ c_2 & d_2
\end{array} \right) = \left( \begin{array}{cc} 0 & 0 \\ 0 & 1
\end{array}\right)$

(string with 1 fixed end, and 1 free end.)
\item[(iv)]$\ds \begin{array}{l} y'(a)  + \alpha y(a)= 0 \\ y'(b) + \beta y(b) =0 \end{array}
\left( \begin{array}{cc} a_1 & b_1 \\ a_2 & b_2 \end{array}
\right) = \left( \begin{array}{cc} \alpha &  1 \\ 0 & 0
\end{array}\right) \left( \begin{array}{cc} c_1 & d_1 \\ c_2 & d_2
\end{array} \right) = \left( \begin{array}{cc} 0 & 0 \\ \beta & 1
\end{array}\right)$

(elasticity constrained ends.)
\item[(v)]$\ds \begin{array}{l} y(a) - y(b) = 0 \\ y'(a) - y'(b) =0 \end{array}
\left( \begin{array}{cc} a_1 & b_1 \\ a_2 & b_2 \end{array}
\right) = \left( \begin{array}{cc} 1 &  0 \\ 0 & 1
\end{array}\right) \left( \begin{array}{cc} c_1 & d_1 \\ c_2 & d_2
\end{array} \right) = \left( \begin{array}{cc} -1 & 0 \\ 0 & -1
\end{array}\right)$

(Periodic boundary conditions.)
\end{description}

In (i) - (iv) $|A| = |B| =0$

In (v) $|A| = |B| =1$ and we require $p(a) = p(b)$

${}$

{\bf Sturm Loiuville Systems}

${}$

$\ds \frac{d}{dx} \left( p \frac{dy}{dx} \right) + (\lambda q(x) -
r(x) ) y =0$ \hfill (1)

where $\lambda$ is a parameter.  i.e. $L(y) - \lambda q(y) =0$
\hfill (2)

$\ds L(y) = \frac{d}{dx} \left( p \frac{dy}{dx} \right) - r(x) y$

We assume $p>0$ in $[a,b]$ (later also $r>0, \, q>0$)

The boundary conditions are:

$\ds U_i(y) = a_1y(a) + b_iy'(a) + c_iy(b) + d_iy(b) = 0 ,\, \, \,
\, i = 1,2$ \hfill (3)

The system is assumed to be self adjoint.


\newpage
{\bf Eigenvalues and Eigenfunctions}

${}$

Let $,v$ be any linearly independent pair of solutions of

$L(y) + \lambda q \cdot y = 0$ \hfill (1)

Then any other solution $y$ is a linear combination of $u$ and
$v$.

$y(x) = \alpha u(x) + \beta v(x)$ where $\alpha$ and $\beta$ are
constants.  Now $U_1$ and $U_2$ are linear and homogeneous.
\begin{eqnarray*} U_1 & = & \alpha U_1 (U) + \beta U_1(v) \\U_2
& = & \alpha U_2 (U) + \beta U_2(v) \end{eqnarray*} $U_1(y) = 0
\hspace{.2in} U_2(y) =0$  gives

$\ds \left( \begin{array}{cc} U_1(u) & U_1(v) \\ U_2(u) & U_2(v)
\end{array} \right) \left( \begin{array}{c} \alpha \\ \beta
\end{array} \right) = 0$ \hfill (2)

For a non trivial solution for $\alpha $ and $beta$ the determint
must vanish.

$\Delta(\lambda) =\left| \begin{array}{cc} U_1(u) & U_1(v) \\
U_2(u) & U_2(v) \end{array} \right|= 0$ \hfill (3)

The determinant is a function of $\lambda$ since both $y$ and $v$
satisfy $L(y) + \lambda q y =0$

i.e. $u = u(x,\lambda) \hspace{.2in} v = v(x,\lambda)$

$\ds U_i(u) = a_iu(a, \lambda)+b_iu_x(a, \lambda)+ c_iu(b,
\lambda) + d_iu_x(b, \lambda)$

and

$\ds U_i(v) = a_iv(a, \lambda)+b_iv_x(a, \lambda)+ c_iv(b,
\lambda) + d_iv_x(b, \lambda)$

The equation $\Delta(\lambda) =0$ is the characteristic equation
for $\lambda$

${}$

\underline{Definition} the value of $\lambda$ satisfying
$\Delta(\lambda)=0$ are the eigenvalue of the system.

${}$

When $\lambda = \lambda_n$ (an eigenvalue) there is a non-trivial
solution for $\alpha, \beta$ from (2) and the solution $y = \phi_n
= \alpha_nu(x, \lambda_n) + \beta_n(x, \lambda_n)$ is said to be
the eigenfunction belonging to $\lambda_n$.  $\phi_n$ is uniquely
determined apart from any non-zero constant.

We assume

(1) There exists an infinite set $\lambda_1, \lambda_2, \ldots $
of eigenvalues.

(2) there exist only one linearly independent eigenfunction
belonging to $\lambda_n$. This may not be true in special cases)


\underline{Example}
\begin{enumerate}
\item Suppose the boundary condition is $y(a) = 0$, then $y'(a)
\not=0$.  Suppose $\phi'\, \phi^2$ were 2 linearly independent
eigenfunctions belonging to $\lambda$.


Let $\ds \phi^3 = \phi' - \phi^2 \frac{\phi'(a) }{\phi^2(a)}$

$\phi^3(a) = 0$ as $\phi'(a) = \phi^2(a) =0$

Also $\phi'^3(a)=0.$  Therefore as $L(\phi^3) + \lambda q(\phi^3)
=0$

$\phi^3(x) =0$ in the whole interval.  Therefore $\phi' = \phi^2$
therefore (2) holds.

\item \begin{eqnarray*} y'' + \lambda y & = & 0 \hspace{.2in} p =
1\, \, q = 1 \\ y(0) & = & y(l) \\ y'(0) & = & y'(l)
\end{eqnarray*} The solution of the differential equation is

$\ds y = A \cos x \lambda ^\frac{1}{2} + B \sin x \lambda
^\frac{1}{2}$

Then \begin{eqnarray*} A &=& A \cos l \lambda ^\frac{1}{2} + B
\sin l \lambda ^\frac{1}{2} \\ \lambda^\frac{1}{2}B & = &
\lambda^\frac{1}{2}(-A \sin l \lambda^\frac{1}{2} + B \sin
l\lambda ^\frac{1}{2})\end{eqnarray*}

Hence (rejecting $\lambda =0$ as trivial) we have

\begin{eqnarray*} \left| \begin{array}{cc} \cos \theta -1 & \sin \theta \\
-\sin \theta & \cos \theta -1 \end{array} \right| & = & 0
\hspace{.2in} \theta = l \lambda ^\frac{1}{2} \\ (1 - cos
\theta)^2 + \sin^2 \theta & = & 0\\ 2(1 - \cos \theta & = & 1
\hspace{.2in} \theta = \pm 2n\pi \, \, \, n = 1, 2, \ldots \\
\lambda_n & = & \frac{4\pi^2}{l^2} n^2 \, \, \, \, \, n =
1,2,\ldots \end{eqnarray*}

For $\lambda = \lambda_n$ the equations a for A and B are
\begin{eqnarray*} 0A + 0B = 0 \\ 0 A + 0B = 0 \end{eqnarray*} i.e.
$A$ and $B$ are arbitrary.

Therefore $\ds y = A \cos \frac{2n\pi x}{l} + B \frac{2n\pi x}{l}$
is an eigenfunction belonging to $\lambda n$

i.e. $\ds\cos \frac{2n \pi x}{l} , \, \sin\frac{2n\pi x}{l} $ are
eigenfunctions belonging to $\lambda _n$ and are linearly
independent: (2) doesn't hold.

\end{enumerate}

\newpage

{\bf Independence of eigenvalues with respects to the choice of
$u$ and $v$}

${}$

Let $\bar u, \bar v$ be another linearly independent pair of
solutions of $\ds L(y) + \lambda q \cdot y =0$

Then $\begin{array} {rcl} u & = & c_{11} \bar u + c_{12} \bar v \\
v & = & c_{21} \bar u + \_{22} \bar v \end{array} \, \, \, \,
c_{11}, \ldots$ constant and $\left| \begin{array}{cc} c_{11} &
c_{12} \\ c_{21} & c_{22} \end{array} \right| \not=0$

$\begin{array}{rcl} U_i(u) & = & c_{11} U_i(\bar u) +
c_{12}U_i(\bar u) \\ U_i(v) & = & c_{21} U_i(\bar v) +
c_{22}U_i(\bar v) \hspace{.2in} i = 1,2 \end{array}$

i.e. $\left( \begin{array}{cc} U_1(u) & U_2(u) \\ U_1(v) & U_2(v)
\end{array} \right) = \left( \begin{array}{cc} c_{11} & c_{12} \\
c_{21} & c_{22} \end{array} \right) \left( \begin{array}{cc}
U_1(\bar u) & U_2(\bar u) \\ U_1(\bar v) & U_2(\bar v)\end{array}
\right)$

${}$

${}$

Taking determinants $\Delta (\lambda) = \left| \begin{array}{cc}
c_{11} & c_{12} \\ c_{21} & c_{22} \end{array} \right| \left|
\begin{array}{cc} U_1(\bar u) & U_2(\bar u) \\ U_1(\bar v) &
U_2(\bar v)\end{array} \right| = |C| \bar \Delta(\lambda)$

and $|C| \not=0$ therefore $\Delta(\lambda) = 0 \Leftrightarrow
\bar \Delta(\lambda) =0$

${}$

\underline{Example} Uniform string under constant tension and with
fixed ends, and no restraining force.

The differential equation is $$y'' + \lambda y = 0 \hspace{.3in}
\lambda = \frac{w^2}{c^2}$$ $a = 0, \, b =l \Rightarrow
\begin{array}{l} U_1(y) = y(0) =0 \\ U_2(y) = y(l) =0 \end{array}$

Two linearly independent solutions of the equation are $$ u = \cos
x \lambda^\frac{1}{2} \hspace{.5in} v = \sin x
\lambda^\frac{1}{2}$$ $\ds u(0) =1 \hspace{.2in} u'(0) =0
\hspace{.2in} v(0) =0 \hspace{.2in} v'(0) = \lambda^\frac{1}{2}$

$\ds \Delta(\lambda) = \left| \begin{array}{cc} 1 & 0 \\ \cos
\lambda^\frac{1}{2} & \sin l \lambda^\frac{1}{2} \end{array}
\right| = \sin l \lambda^\frac{1}{2}$

Therefore the equation for $\ds\lambda$ is $\sin l
\lambda^\frac{1}{2}=0 \Rightarrow \lambda = \frac{n^2\pi^2}{l^2}
\cdot \frac{w}{c} = \frac{n\pi}{l}$

${}$

{\bf Properties of eigenvalues and Eigenfunctions}

${}$

We assume now that $q(x) >0, \,r(x)>0$ in $[a,b]$ in addition to
$p(x) >0$ in $[a,b]$

\begin{enumerate}
\item The eigenvalues are real

\item If $\phi_m \, \phi_n$ belong to £$\lambda_m\, \lambda_n$
$$\int_a^bq(x) \phi_n(x) \phi_m(x) dx =0 \hspace{,2in} (m
\not=n)$$ i.e. $\phi_1,\, \phi_2 , \ldots$ are orthogonal over
$[a,b]$ with weighting function $q(z)$

\item if the boundary conditions are suitably restricted (and $p,q
>0 \, \, r \geq 0$ in $[a,b]$) then the eigenvalues are positive
\end{enumerate}

${}$

\underline{Proofs}
\begin{enumerate}
\item
\item Let $\lambda_n, \, \lambda_n$ be any two eigenvalues and let
$\phi_m, \, \phi_n$ belong to them.  Then \begin{eqnarray*}
L(\phi_m) + \lambda_m q \phi_m & = & 0 \\ L(\phi_n) + \lambda_n q
\phi_n & = & 0 \\ U_i (\phi _ m) =0 \hspace{.2in} U_i(\phi_n) = 0
& & \hspace{.2in} i = 1, 2\end{eqnarray*}
$$\int_a^b\{\phi_nL(\phi_m) - \phi_mL(\phi_n)\}dx = p(x)
[\phi_n\phi'_m - \phi_m\phi'_n]_a^b$$ and the R.H.S =0 if the
boundary conditions are self adjoint.

Therefore $\ds \int_a^b b\{\phi_nL(\phi_m) - \phi_mL(\phi_n)\}dx =
0$

Therefore $\ds \int_a^b \{ \phi_m \lambda_n q \phi_n - \phi_n
\lambda_m q \phi_m \} dx =0$

Therefore $\ds (\lambda_n - \lambda_m) \int_a^b q \phi_m \phi_n
\,dx =0$ \hfill(i)

Therefore $\ds \int_a^b q \phi_m \phi_n \, dx =0 \hspace{.3in} m
\not=n $ \hfill(ii)

(i) If $\lambda = \rho + i \sigma$ is an eigenvalue and $\phi = X
+ iY$ is an eigenfunction belonging to it then $\bar \lambda$ is
also an eigenvalue and $\bar \phi$ will belong to it.

For $L(\phi) + \lambda q\phi =0$

$\Rightarrow \overline{L(phi)} + \overline{\lambda q \cdot q} =0$

$\Rightarrow L(\bar\phi) + \bar \lambda q \cdot \bar \phi =0$

Since $L$ is a real linear operator and $q$ is real.

Also $U_i(\phi) =0 \Rightarrow U_i(\bar \phi) =0$ as $U_i$ is real
and linear.

Hence in (i) above take $\lambda = \lambda_n$ and $\bar \lambda =
\lambda_m$

Then \begin{eqnarray*} (\bar \lambda - \lambda \int_a^b q \phi
\bar \phi dx & = & 0 \\ (\bar \lambda - \lambda) \int_a^b
q|\phi|^2 dx & = & 0 \hspace{.2in} q>0 \, \, \, \, |\phi|
\not\equiv 0
\end{eqnarray*} therefore $\bar \lambda - \lambda =0$ i.e. $\lambda$ is
real.

\item $$L(\phi_n) + \lambda  q \cdot \phi_n=0$$ Therefore $\ds
\lambda_n\int_a^bq \phi^2_n dx = -\int_a^b \phi_nL(\phi_n) dx$

\begin{eqnarray*} \phi_nL(\phi_n) & = & \phi \left( \frac{d}{dx} p
\frac{d\phi_n}{dx} - r \phi_n\right) \\ & = & \frac{d}{dx}
\phi_np\frac{d\phi_n}{dx} - p \left(\frac{d\phi_n}{dx} \right)^2 -
r\phi_n^2\end{eqnarray*}

therefore $\ds \lambda_n \int_a^b a \phi^2_n dx = \int_a^b(p
\phi_n^{2'} + r\phi_n^2) dx - [p\phi_n\phi'_n]_a^b$

Therefore if the boundary conditions are such that

$[p\phi_n \phi'_n]_a^b \leq 0 \hspace{.2in} \lambda_n>0$
\end{enumerate}


${}$

\underline{Examples}

${}$

\begin{description}
\item[(i)] $\ds y(a) =0 \hspace{.2in} y(b) =0 \Rightarrow \phi_n(a)=
\phi_n(b) =0 {\rm \ \ \ and \ \ } [p\phi_n \phi'_n]_a^b=0$

\item[(ii)] $\ds y'(a) =0 \hspace{.2in} y'(b) =0 \Rightarrow \phi'_n(a)=
\phi'_n(b) =0 {\rm \ \ \ and \ \ } [p\phi_n \phi'_n]_a^b=0$

\item[(iii)] \begin{eqnarray*} y'(a) - h_1y(a) & = & 0
\hspace{.3in} h_1>0 \\ y'(b) + h_2y(b) & = & 0 \hspace{.3in} h_2>0
\\ -[p \phi_n \phi'_n]_a^b & = & -p(b) (-h_2\phi^2_n(b)) +
p(a)(h_1\phi^2_n(a)) >0 \end{eqnarray*}

\item[(iv)] $\ds y(a) = y(b) \hspace{.3in} y'(a) = y'(b)$

Here, with the condition for self adjointness $\ds p(a) = p(b),
[p\phi_n\phi'_n]_a^b=0$

\end{description}

\newpage

{\bf Formal Explanations in Eigenfunctions}

${}$

Consider the homogeneous system $$L(y) + \lambda q \cdot y = 0
\hspace{.3in}\left(L = \frac{d}{dx} p\frac{d}{dx} +r\right)$$
$$U_i(y) =0 \hspace{.5in} i = 1,2.$$

with eigenvalues $\lambda_1, \lambda_2, \ldots$

and eigenfunctions $\phi_1, \phi_2 \ldots$

We assume that the $\phi_n$ have been normalised $${\rm i.e.\ \ }
\int_a^b q \phi_n^2 \, dx = 1$$ If a function $F(x)$ defined in
$[a,b]$ has a uniformly convergent expansion $$F(x) =
\sum_1^\infty A_n \phi_n(x)$$ then $$\int_a^b q(x) \phi_m(x) F(x)
dx = A_m$$ Consider the non-homogeneous system \begin{eqnarray*}
L(y) + \lambda q \cdot y & = & f(x) \\ U_i(y) &  = & 0
\hspace{.3in} i = 1,2 \end{eqnarray*}
\begin{enumerate}
\item If $\lambda$ is not an eigenvalue of the homogeneous system
the solution is unique.
\item If $\lambda = \lambda_m$ a necessary condition for existence
of a solution is that $$\int_a^b\phi_m(x) f(x) dx=0$$ i.e. $f$
must be orthogonal to $\phi_m$.
\item if $\lambda = \lambda_m$ and $f$ is orthogonal to $\phi_m$,
then the solution is not unique.
\end{enumerate}

\newpage

\underline{Proofs}

${}$

\begin{itemize}
\item[1.] Suppose $y$ and $z$ are two solutions
$$\begin{array}{rclrcll} L(y) + \lambda q \cdot y & = & f(x) &
U_i(y) & = & 0 \hspace{.2in}& i = 1,2 \\ L(z) + \lambda q \cdot z
& = & f(x) & U_i(z) & = & 0 & i = 1,2 \\ L(y-z) + \lambda q \cdot
(y-z) & = & 0 & \hspace{.2in}U_i(y-z) & = & 0 & i =
1,2\end{array}$$i.e. $L(y-z)$ is a solution of the homogeneous
system.  If $\lambda$ is not an eigenvalue this must be zero.
Therefore $y=z$ in $[a,b]$
\item[3.] If $\lambda = \lambda_m$, $y-z = A\phi_m(x)$ where $A$
is an arbitrary constant i.e. $y = z+A\phi_m(x)$ and the solution
is not unique.
\item[2.] $\ds \int_a^b[\phi_mL(y) - yL(\phi_m)]dx =
[p(x)[\phi_my'-\phi'_my]]_a^b = 0$

since $y_v\phi_m$ satisfies the boundary conditions which are self
adjoint.  i.e. $\ds \int_a^b [\phi_m\{f(x) - \lambda q \cdot y\} -
y\{-\lambda_m q \phi_m\}]dx =0$

$\ds  \lambda - \lambda_n \int_a^b q \phi_m y dx = \int_a^b \phi_m
f \, dx$ \hfill (1)

Therefore $\lambda$ is an eigenvalue  i.e. $\lambda = \lambda_m$,
$$\int_a^b \phi_m f \, dx =0$$
\end{itemize}

${}$

{\bf Formal Series Solution}

${}$

If we assume $y$ has an expansion in eigenfunction we have from
(1) above. $$(\lambda - \lambda_n)a_n = \int_a^b \phi_nf\, dx =
b_n$$ hence if $\lambda$ is \underline{not} an eigenvalue $\ds a_n
= \frac{b_n}{\lambda - \lambda_n} \hspace{.2in} n = 1, 2, \ldots$

i.e.$\ds y = \sum_{n=1}^\infty \frac{\phi_n(x)}{\lambda-\lambda_n}
\int_a^b \phi_n(\xi)f(\xi)\, d \xi$

If $\lambda = \lambda_m$ then $\ds a_n = \frac{b_n}{\lambda_m -
\lambda_m} \hspace{.2in} m \not=n$

and $0 \cdot a_m = 0$ Therefore
$$\sum_{\begin{array}{c}_{n=1}\\^{n\not=m}\end{array}}^\infty
\frac{\phi_n(x)}{\lambda_m - \lambda_n} \int_a^b \phi_n(\xi)
f(\xi) \, d\xi + A\phi_m(x), {\rm \ \ A\ arbitrary}$$



{\bf Stationary Property of Eigenvalues}

${}$

When the boundary conditions are such that $\ds [p(x)yx)y'(x)]_a^b
=0$ we have $$\lambda_n = \frac{\int_a^b (p\phi_n^{'2} +
r\phi_n^2)dx}{\int_a^bq \phi_n^2 dx}$$ Write


$\left.\begin{array}{rcl} I(\phi,\psi) & = & \ds
\int_a^b(p\phi'\psi' + r\phi\psi)dx \\ \\  J(\phi,\psi) & = & \ds
\int_a^b q \phi\psi \end{array}\right\}$ \hfill(1)

$\ds \lambda_n  = \frac{I(\phi_n,\phi_n)}{J(\phi_n,\phi_n)}$

We suppose that $\phi_n$ are normalised i.e. $J(\phi_n, \phi_n)
=1$

then $\lambda_n=I(\phi_n.\phi_n)$ \hfill(2)

Consider $\lambda_n = I(\phi,\phi)$ where $J(\phi,phi)=1$
\hfill(3)

and
\begin{itemize}
\item[(i)] $\phi'$ continuous in $[a,b]$
\item[(ii)] $\phi$ satisfied the boundary conditions
\end{itemize}

(it does not necessarily satisfies $L(y) + \lambda q (y) =0$)

We show that $\lambda = I(\phi, \phi)$ is stationary for small
variations of $\phi$ from $\phi_n$.  This is the extremum property
of the integral $I(\phi,\phi)$ subject to the normalising
condition $j(\phi,\phi=0$ and $ to (i), (ii)$.

Write $\phi(x) = \phi_n(x) + \epsilon\psi(x)$, where $\epsilon$ is
a constant and $\psi$ satisfies $(i)$ and also the boundary
conditions since $U_i$ is linear.  We show that $$I(\phi\phi) -
I(\phi_n\phi_n)=O(\epsilon^)$$ The normalising condition on $\phi$
is

$\ds 1 = J(\phi,phi) = J(\phi_n\phi_n) + 2\epsilon J(\phi_n\psi)
+\epsilon^2J(\psi\psi)$

$1 = 1+ 2\epsilon J(\phi_n\psi)+\epsilon^2J(\psi\psi)$

Therefore $2\epsilon J(\phi_n\psi) + \epsilon^2J(\psi\psi)=0$

\begin{eqnarray*} I(\phi\phi) &  = & I(\phi_n\phi_n) 2\epsilon
I(\phi_n\psi) + \epsilon^2I(\psi\psi) \\ I(\phi_n\psi) & = &
\int_a^b(p\phi'_n\psi'_n + r\phi_n \psi)dx \\ & = &
[p\phi'_n\psi'_n]_a^b + \int_a^b \left\{-\psi \frac{d}{dx}
p\phi'_n + r \phi_n\psi\right\} \\ 0 = [p\phi\phi']_a^b & = &
[p\phi_n\phi'_n]_a^b + [\epsilon p(\phi_n\psi' + \phi'_n\psi)]_a^b
+ [\epsilon^2 p \phi\phi']_a^b \end{eqnarray*}

From the self adjoint condition $$[p(\phi_n\psi -
\psi'_n\phi)]_a^b =0$$ Therefore
$$[p\phi'_n\psi]_a^b=[p\phi_n\psi']_a^b=0$$ Therefore
\begin{eqnarray*}I(\phi_n\psi) & = & \int_a^b \psi [-L(\phi_n)]_a^b=0 \\
& = & \lambda_n \int_a^b q \psi \phi_n \,dx \\ & = & \lambda_n
J(\phi_n \psi) \end{eqnarray*} Therefore \begin{eqnarray*}
I(\phi\phi) - I(\phi_n\phi_n) & = & 2\epsilon\lambda_n
J(\phi_n\psi) + \epsilon^2I(\psi\psi) \\ & = &
\epsilon^2[I(\psi\psi) \lambda_nJ(\psi\psi)] \end{eqnarray*} This
establishes the stationary property.

${}$

{\bf Illustration}

${}$

$$y'' + \lambda y = 0\hspace{.2in} y(0) = 0 \hspace{.2in} y(1)
=0$$

The exact solution for $\lambda_1$ and $\phi_1$ is $\ds \phi_1 x
\sin \pi x \hspace{.2in} \lambda_1 = \pi^2 \approx 9.87$

The normalised $\phi_1 is 2^\frac{1}{2} \sin \pi x$

Take $\phi = Cx(1-x)$

$\ds \int_0^1 \phi^2 \, dx = C^2 \int_0^1 x^2(1-x)^2 = C^2
\frac{\Gamma(3) \Gamma(3)}{\Gamma(6)} = \frac{C^2}{30}$

Therefore $\ds \phi = \sqrt{30}x(1-x)$

$\ds I(\phi\phi) = \int_0^1\phi'^2 \, dx = 30 \int_0^1(1-2x)^2\,
dx = 30 \frac{1}{6} 2 = 10$

Compare with 9.87 thus the error is $\approx 1.4\%$

\newpage

{\bf Formulation of the eigenvalue }

{\bf problem as an \lq \lq isoperimetric problem\rq \rq}

${}$

The eigenvalues of the system $L(y) + \lambda q \cdot y =0$ with
$y(a) = y(b) =0$ are the extrema of $\ds I = \int_a^b(p\phi'^2 +
r\phi^2) dx$

subject to the normalising condition $\ds J = \int_a^b q\phi^2 dx
= 1$ and
\begin{itemize}
\item[(i)] $\phi'$ continuous in $[a,b]$
\item[(ii)] $\phi(a) = \phi(b) =0$
\end{itemize}


\end{document}