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\begin{center}

{\bf Application to Vibrating String}

\end{center}

${}$

We have a uniform string of length $l$, mass per unit length $m$,
and equilibrium tension $T$.  Let there be a disturbing
(transverse) force $f(x,t)$ per unit length.

With the assumptions that

\begin{itemize}
\item[1)]
There are no longitudinal body forces (or negligible forces).

\item[2)]
Displacement $y(x,t)$ is purely transverse.

\item[3)]
$\ds\frac{\p y}{\p x}$ is small compared with 1.
\end{itemize}

it can be shown that to the first order $T$ is uniform along the
string, and the equation of motion is

$\ds\frac{\p^2y}{\p x^2}=\frac{1}{c^2}\frac{\p^y}{\p
t^2}-\frac{1}{T}f(x,t) \hfill (1)$

when $\ds c=\left(\frac{T}{m}\right)^\frac{1}{2}$

${}$

For free motion $f=0$ and we have

$\ds\frac{\p^2y}{\p x^2}=\frac{1}{c^2}\frac{\p^2y}{\p t^2} \hfill
(2)$

${}$

Seeking solutions of the form

$\hspace{1.5in}y=X(x)Y(t)$

$\ds\frac{1}{X}\frac{\p^X}{d
x^2}=\frac{1}{c^2}\frac{1}{Y}\frac{d^2Y}{dt^2}$

${}$

Hence both sides are constant and we write the constant as
$\ds-\frac{w^2}{c^2}$.

${}$

This gives $\ds\frac{d^2Y}{dt^2}=-w^2Y$

${}$

$Y=A\cos wt+b\sin wt \hfill (3)$

${}$

and $\ds\frac{d^2X}{dt^2}=-\frac{w^2}{c^2}X \hfill (4)$

${}$

$\ds X=C\cos\frac{w}{c}x+D\sin\frac{w}{c}x \hfill (4a)$

${}$

For fixed ends we require $y(0,t)=y(l,t)=0$

Therefore $\left.\begin{array}{l} X(0)=0\\
X(l)=0\end{array}\right\} \hfill (5)$

${}$

The differential equation (4), and the conditions (5) constitute a
two-point boundary problem.  The differential equation includes a
parameter $\ds\lambda=\frac{w^2}{c^2}$ not yet determined.

In fact we show below that $\lambda$ must have one of a set of
values

$\ds\lambda_1=\frac{\pi^2}{l^2}, \,\,
\lambda_2=\frac{(2\pi)^2}{l^2}\cdots$.

These are the eigenvalues, the corresponding solutions
$X(x,\lambda)$ are the eigenfunctions.

i.e. (4) and (5) constitute an eigenvalue problem.

$\ds X=C\cos\frac{w}{c}x+D\sin\frac{w}{c}x$

The conditions (5) give $C=0$ and $\ds D\sin\frac{wl}{c}=0$

Hence for a non-trivial solution $\ds\sin\frac{wl}{c}=0$,
therefore $\ds w=\frac{n\pi c}{l} \hspace{0.2in} n=1,2\cdots$

Hence solutions satisfying (5) are

$\ds y=\sin\frac{n\pi x}{l}\left[A_n\cos\frac{n\pi
ct}{l}+B_n\sin\frac{n\pi ct}{l}\right] \hfill (6)$

${}$

These are the normal modes of vibration and the values of
$\ds\frac{w}{2\pi}$, i.e. $\ds\frac{c}{2l}, \frac{2c}{2l}\cdots$
are the normal frequency.

Formally a general solution satisfying (2) and the end conditions
(5) is

$\ds y=\sum_{n=1}^\infty\sin\frac{n\pi
x}{l}\left[A_n\cos\frac{n\pi ct}{l}+B_n\sin\frac{n\pi
ct}{l}\right]$

If this series converges and is twice differentiable term-by-term
then the function $y$ is a solution.

${}$

{\bf Given initial conditions}

Suppose that $\dot{y}(x,0)=0$ and $y(x,0)=F(x) \hspace{0.3in}
0\leq x\leq l$

Now a solution satisfying $\dot{y}(x,0)=0$ is

$\ds y=\sum_1^\infty A_n\sin\frac{n\pi x}{l}\cos\frac{n\pi ct}{l}
\hfill (1)$

when $t=0$ this gives $\ds y(x,0)=\sum_1^\infty A_n\sin\frac{n\pi
x}{l}$.

Hence $\ds\sum_1^\infty A_n\sin\frac{n\pi x}{l}$ is the sine
series expansion of $F(x)$ in $0\leq x\leq l$.

i.e. $\ds A_n=\frac{2}{l}\int_0^lF(x)\sin\frac{n\pi x}{l}dx \hfill
(2)$

${}$

{\bf Interpretation of solutions in terms of progressive waves}

We have $\ds y=\frac{1}{2}\sum_1^\infty
A_n\left[\sin\frac{n\pi}{l}(x+ct)+\sin\frac{n\pi}{l}(x-ct)\right]$

Write $\ds F_s(x)=\sum_1^\infty A_n\sin\frac{n\pi x}{l}$

$F_s(x)=\left\{\begin{array}{cr} F(x) &amp; 0\leq x\leq l\\ -F(-x) &amp;
-l\leq x\leq0\end{array}\right.$

${}$

$F_s(x+2l)=F_s(x)$

$y=\frac{1}{2}\{F_s(x+ct)+F_s(x-ct)\}$

$\ds\frac{\p^2y}{\p t^2}=\frac{1}{2}\{F_s''(x+ct)+F_s''(x-ct)\}$

$\ds\frac{1}{c^2}\frac{\p^2y}{\p t^2}=
\frac{1}{2}\{F_s''(x+ct)+F_s''(x-ct)\}$

${}$

Whenever $\zeta=x+ct$ and $\eta=x-ct$ are such that $F(x)$ is
twice differentiable at the values concerned.

Also $y(x,0)=\frac{1}{2}\{F_s(x)+F_s(x)\}=F(x)$

${}$

$y(0,t)=0 \hspace{0.3in} (F_s(x)$ odd)

$\ds y(l,y)=\frac{1}{2}\{F_s(l+ct)+F_s(l-ct)\}$

$\ds\hspace{0.5in}=\frac{1}{2}\{F_s(l+ct)+F_s(-[l+ct])\}=0
\hspace{0.3in} (F$ periodic $2l)$.

${}$

{\bf Plucked String}

$\ds F(x)=\frac{h}{a}x \hspace{0.95in} 0\leq x\leq a$

$\ds\hspace{0.4in}=\frac{h}{l-a}(l-x)\hspace{0.3in} a\leq x\leq l$

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\begin{eqnarray*} A_n &amp;=&amp; \frac{2}{l}\int_0^lF(x)\sin\frac{n\pi
x}{l}dx\\ &amp;=&amp; \frac{2}{l}
\left[-\frac{F(x)}{\frac{n\pi}{l}}\cos\frac{n\pi x}{l}\right]_0^l+
\frac{2}{l}\int_0^lF'(x)\frac{\cos\frac{n\pi x}{l}}
{\frac{n\pi}{l}}dx\\ &amp;=&amp;
0+\frac{2}{l}\left[F'(x)\frac{\sin\frac{n\pi x}{l}}
{\left(\frac{n\pi}{l}\right)^2}\right]_0^a+
\frac{2}{l}\left[F'(x)\frac{\sin\frac{n\pi x}{l}}
{\left(\frac{n\pi}{l}\right)^2}\right]_a^l
-\int_0^lF''(x)\frac{\sin\frac{n\pi x}{l}}
{\left(\frac{n\pi}{l}\right)^2}dx\\ &amp;=&amp;
\frac{2}{l}\frac{l^2}{n^2\pi^2}\sin\frac{n\pi a}{l}
[f'(a-0)-f'(a+0)] \hspace{0.5in} {\rm as\ } f''\equiv0\\ &amp;=&amp;
\frac{2l}{n^2\pi^2}\sin\frac{n\pi a}{l}
\left[\frac{h}{a}+\frac{h}{l-a}\right]\\ &amp;=&amp;
\frac{2hl^2}{n^2\pi^2a(l-a)}\sin\frac{n\pi a}{l}\end{eqnarray*}

Therefore $\ds y(x,t)=\frac{2l^2h}{\pi^2a^2(l-a)}\sum_1^\infty
\frac{\sin\frac{n\pi x}{l}\sin\frac{n\pi a}{l}}{n^2}
\cos\frac{n\pi ct}{l}$

Note that any normal mode which has a node at $x=a$ is absent from
the series since in that case $\ds\sin\frac{n\pi a}{l}=0$ for all
$n$.

This occurs when $\ds\frac{a}{l}=$ rational $\ds\frac{r}{s}
\hspace{0.3in} r&lt;s$.

Then $\ds\sin\frac{n\pi a}{l}$ vanishes for $n=s,2s,3s\cdots$.

The corresponding modes are absent.

${}$

{\bf Physical Illustration of Parseval's Theorem}

$$K.E=\frac{1}{2}\int_0^lm\left(\frac{\p y}{\p t}\right)^2dx$$

The P.E of the element $\triangle x$ is the work done by the
tension at the ends in extending the element from $\triangle x$ to
$\ds\left(\triangle x^2+\triangle y^2\right)^\frac{1}{2}$

i.e. to $\triangle x\left(1+\left(\frac{\triangle y}{\triangle
x}\right)^2\right)^\frac{1}{2}=\triangle x\left(1+\frac{1}{2}
\left(\frac{\triangle y}{\triangle x}\right)^2+\cdots\right)$

Therefore the extension is $\ds\frac{1}{2}\left(\frac{\p y}{\p
x}\right)^2dx$

The work done is $\ds\frac{1}{2}T\left(\frac{\p y}{\p
x}\right)^2dx$

Therefore the total work done is
$\ds\frac{1}{2}\int_0^lT\left(\frac{\p y}{\p x}\right)^2dx$

$\ds y=\sum_1^\infty C_n\sin\frac{n\pi x}{l}\cos\left(\frac{n\pi
ct}{l}+\alpha_n\right)$

$\ds\frac{\p y}{\p t}=\sum_1^\infty-\frac{n\pi c}{l}
\sin\frac{n\pi x}{l}C_n\sin\left(\frac{n\pi
ct}{l}+\alpha_n\right)$

$\ds\frac{\p y}{\p x}=\sum_1^\infty\frac{n\pi}{l}\cos\frac{n\pi
x}{l}C_n\cos\left(\frac{n\pi ct}{l}+\alpha_n\right)$

Applying Parseval's formula

\begin{eqnarray*} \int_0^l\left(\frac{\p y}{\p t}\right)^2dx &amp;=&amp;
\frac{\pi^2c^2}{l^2}\sum_1^\infty n^2\left(C_n\sin
\left[\frac{n\pi ct}{l}+\alpha_n\right]\right)^2\frac{l}{2}\\
\int_0^l\left(\frac{\p y}{\p t}\right)^2dx &amp;=&amp;
\frac{\pi^2}{l^2}\sum_1^\infty n^2\left(C_n\cos\left[\frac{n\pi
ct}{l}+\alpha_n\right]\right)^2\frac{l}{2}\end{eqnarray*}

Hence $\ds\hspace{0.35in}
K.E=\frac{1}{2}m\frac{\pi^2c^2}{l^2}\sum_1^\infty
n^2C_n^2\sin^2\left(\frac{n\pi ct}{l}+\alpha_n\right)\frac{l}{2}$

$T=mc^2$ so $\ds P.E=
\frac{1}{2}m\frac{\pi^2c^2}{l^2}\sum_1^\infty
n^2C_n^2\cos^2\left(\frac{n\pi ct}{l}+\alpha_n\right)\frac{l}{2}$

Sum=$\ds\frac{1}{4}\frac{m\pi^2c^2}{l}\sum_1^\infty n^2C_n^2$=
constant=$\bar{K.E}+\bar{P.E}$

${}$

{\bf Forced Motion}

$\ds\frac{\p^2y}{\p x^2}=\frac{1}{c^2}\frac{\p y}{\p
t^2}-f(x,t){T} \hfill (1)$

$f(x,t)$= force / unit length at distance $x$, and time $t$.

Assume a simple harmonic forcing term $f(x,t)=F(x)\cos wt$.

We seek a solution of (1), simple harmonic with the same
frequency.

i.e. $y=Y(x)\cos wt \hfill (2)$

By substitution we find that with $\ds G(x)=-\frac{F(x)}{T}$

$\left.\begin{array}{l}
\ds\frac{d^2Y}{dx^2}+\frac{w^2}{c^2}Y=G(x)\\ {\rm We\ must\ also\
have\ } Y(0)=0=Y(l)\end{array}\right\} \hfill (3)$

We seek a solution in which $Y$ and $Y'$ are continuous, and we
shall assume that $G(x)$ is continuous in $0\leq x\leq l$.

[Note that the general solution of (1) is of the form $Y(x)\cos
wt+z$, where $z$ satisfies the homogeneous equation
$\ds\frac{\p^2z}{\p x^2}=\frac{1}{c^2}\frac{\p^z}{\p t^2}$ this
added term would be needed in general, in order that the initial
conditions should be satisfied, since the particular solution
$Y(x)\cos wt$, would not, in general, satisfy these.]

Write $\ds\lambda=\frac{w^2}{c^2}$

$\ds\lambda_n=\left(\frac{n\pi}{l}\right)^2 \hfill
\left[w_n=\frac{n\pi c}{l}, {\rm \ i.e\ }
\lambda_n=\frac{wn^2}{c^2}\right]$

Then (3) is

$\left.\begin{array}{ccc} Y''+\lambda Y &amp;=&amp; G(x)\\ Y(0) &amp;=&amp; 0\\
Y(l) &amp;=&amp; 0\end{array}\right\} \hfill (3a)$

Write $\ds u_n=\sin\frac{n\pi x}{l}$, so that

$u_n''+\lambda_nu_n=0 \hfill (4)$

From $(3a)u_n-(4)Y$ we get

$u_nY''-Yu_n''+(\lambda-\lambda_n)Yu_n=Gu_n$

i.e. $\ds\frac{d}{dx}(u_nY'-u_n'Y)+(\lambda-\lambda_n)u_nY=Gu_n$

integrating from 0 to $l$, we have that

$\ds[u_nY'-u_n'Y]_0^l+(\lambda-\lambda_n)\int_0^lu_nYdx=\int_0^lu_nGdx$

[The evaluation of the first term is between the end limits only
as $Y,Y',u_n,u_n'$ are continuous in $[0,l]$.]

$u_n(0)=Y(0)=u_n(l)=Y(l)=0$ therefore $[u_nY'-u_n'Y]_0^l=0$

Therefore $\ds(\lambda-\lambda_n)\int_0^lu_nYdx=\int_0^lu_nGdx
\hfill (5)$

${}$

{\bf Case I} $\lambda\not=\lambda_n \hspace{0.3in} (n=1,2,\cdots)$

i.e. $\lambda_n$ is not an eigenvalue of the system

$Y''+\lambda Y=0 \hspace{0.3in} Y(0)=0 \hspace{0.3in} Y(l)=0$

$\ds\int_0^lu_nYdx=\frac{1}{\lambda-\lambda_n}\int_0^lu_nGdx
\hfill (6)$

i.e. $\ds Y_n=\frac{1}{\lambda-\lambda_n}G_n$ where $Y_n$ and
$G_n$ are the Fourier sine coefficients for $Y$ and $G$ in
$[0,l]$.

Hence if $\ds G(x)=\sum_1^\infty G_n\sin\frac{n\pi x}{l}$

then $\ds
Y(x)=\sum_1^\infty\frac{G_n}{\lambda-\lambda_n}\sin\frac{n\pi
x}{l}=\sum_1^\infty\frac{c^2G_n}{w^2-w_n^2}\sin\frac{n\pi x}{l}$
is (formally) a solution of the differential equation and the end
conditions.

[Note that if $G(x)$ is continuous in $0\leq x\leq l$ and $G(0)=0,
\,\,\, G(l)=0$, then the coefficients are at most
$O(\frac{1}{n^2})$ and the coefficients of $Y(x)$ are at most
$O(\frac{1}{n^4})$.  Hence the series for $Y(x)$ is certainly
twice differentiable term-by-term since the derived series has
coefficients of order $\frac{1}{n^2}$ and hence converges
absolutely and uniformly.]

${}$

Note that when $w$ is near to $w_m$ the dominant term in $Y$ is
then

$\ds\frac{c^2}{w^2-w_m^2}G_m\sin\frac{n\pi x}{l}$ if $G_m\not=0$.

When $w=w_m$ the solution fails unless $G_m=0$.

${}$

{\bf Case II} $\lambda=\lambda_m \hspace{0.3in}$ i.e. $w=w_m$

From $\lambda-\lambda_n\int_0^lY_{u_n}ax=\int_0^lGu_ndx$

We have $\ds0=\int_0^lGu_mdx=\frac{l}{2}G_m$

Therefore $G_m=0$ is a necessary condition for the existence of a
solution of the type $y=Y(x)\cos wt$.

If $\ds G_m=0 \hspace{0.3in} Y_n=\frac{1}{\lambda_m-\lambda_n}G_n
\hspace{0.3in} n\not=m$

Consider $\ds Y(x)=\sum_{n=1,\,n\not=m}^\infty
\frac{1}{\lambda_m-\lambda_n}G_n\sin\frac{n\pi x}{l}$

By formal differentiation term by term

$\ds\left(\frac{d^2}{dx^2}+\lambda_m\right)Y=
\sum_{n=1,\,n\not=m}^\infty G_n\sin\frac{n\pi x}{l}= G(x) \hfill
\left(\frac{n^2\pi^2}{l^2}=\lambda_n\right)$

Since the series is the sine series for $G(x)$ where there is no
term in $\ds\sin\frac{m\pi x}{l}$.

Hence the above expression for $Y(x)$ is a solution, but is not
unique since

$\ds Y(x)=\sum_{n=1,\,n\not=m}^\infty
\frac{G_n}{\lambda_m-\lambda_n}\sin\frac{n\pi x}{l}+
A\sin\frac{m\pi x}{l}$

is also a solution.

[In case I the solution for $Y$ is unique, for if $Y_1$ and $Y_2$
satisfy

$Y''+\lambda Y=G, \hspace{0.1in} Y(0)=Y(l)=0$, then putting
$Y_3=Y_1-Y_2,$

$Y_3''+\lambda Y=0, \hspace{0.1in} Y_3(0)=Y_3(l)=0$. This has a
non trivial solution only if $\lambda=\lambda_n$ for some $n$,
which is not so, i.e $Y_3=0$.]

${}$

{\bf Summary}

\begin{itemize}
\item[i)]
$\lambda\not=\lambda_n$ for every $n$, the solution for $Y$ exists
and is unique.

\item[ii)]
$\lambda=\lambda_m$, no solution $y=Y\cos wt$ when $G_m\not=0$. If
$G_m=0$ a solution exists but is not unique.
\end{itemize}

${}$

{\bf Case III} $\lambda=\lambda_m \hspace{0.3in} G_m\not=0$

The solution $\ds Y(x)=\sum_1,^\infty
\frac{G_n}{\lambda_n-\lambda_m}\sin\frac{n\pi x}{l}$ fails.

If $\lambda\not=\lambda_m$ for the moment,

$\ds y(x,t)=\left[\sum_{n=1,\,n\not=m}^\infty
\frac{G_n}{\lambda-\lambda_m}\sin\frac{n\pi x}{l}\right]\cos wt+
G_m\sin\frac{m\pi x}{l}\frac{\cos wt}{\lambda-\lambda_m}$

Consider $\ds y_1(x,t)=y(x,t)-G_m\sin\frac{m\pi x}{l}\frac{\cos
w_mt}{\lambda-\lambda_m}$

where the added term satisfies

$\ds\left(\frac{\p^2}{\p x^2}-\frac{1}{c^2}\frac{\p^2}{\p
t^2}\right)y=0 \hspace{0.3in} y(0,t)=y(l,t)=0$

and hence does not alter the forcing term $G(x)\cos wt$.  Then

$\ds y_1(x,t)=\left[\sum_{n=1,\,n\not=m}^\infty
\frac{G_n}{\lambda-\lambda_m}\sin\frac{n\pi x}{l}\right]\cos wt+
G_m\sin\frac{m\pi x}{l}\frac{\cos wt-\cos w_mt}
{\lambda-\lambda_m}$

Now $\ds\lim_{\lambda\to\lambda_m}\frac{\cos wt-\cos
w_mt}{\lambda-\lambda_m}=\frac{\p}{\p\lambda}\frac{(\cos wt-\cos
w_mt)_{\lambda=\lambda_m}}{l}$

$\ds\hspace{0.4in}=-t\sin w_mt
\left(\frac{dw}{d\lambda}\right)_{\lambda=\lambda_m}$

but $\ds\lambda=\frac{w^2}{c^2} \hspace{0.2in}$ therefore
$\ds\hspace{0.1in}\frac{2}{w_m}\left(\frac{dw}{d\lambda}\right)
_{\lambda=\lambda_m}= \frac{1}{\lambda_m}$

Therefore the above limit is $\ds-t\sin
w_mt\frac{w_m}{2\lambda_m}=-\frac{(w_mt)\sin(w_mt)}{2\lambda_m}$

Hence the limiting form of $y_1(x,t)$ is

$$\left[\sum_{n=1,\,n\not=m}^\infty \frac{G_n}{\lambda-\lambda_m}
\sin\frac{n\pi x}{l}\right]\cos w_mt-\frac{G_m}{2}\sin\frac{m\pi
x}{l}\frac{w_mt\sin w_mt}{\lambda_m}$$

This shows the phenomenon of resonance since the second term has
amplitude increasing with time (linearly).

${}$

{\bf Alternative method for solution of the non-homogeneous
equation (in case $\lambda\not=\lambda_n$)}

$Y''+\lambda Y=G(x) \hspace{0.3in} 0\leq x\leq l \hspace{0.3in}
Y(0)=y(l)=0 \hfill (1)$

Let $u,v$ be solutions of the homogeneous equation

$u''+\lambda u=0 \hfill (2a)$

$v''+\lambda v=0 \hfill (2b)$

Choose $u$ and $v$ so that

$u(0)=v(l)=0 \hfill (3)$

In this case $u=\sin\lambda^\frac{1}{2}x \hspace{0.2in}
(\sin\frac{wx}{l}) \hspace{0.2in} {\rm and} \hspace{0.2in}
v=\sin\lambda^\frac{1}{2}(l-x) \hspace{0.2in}
(\sin\frac{w}{c}(l-x))$

From $(1)u-(2a)Y$ we get $uY''-u''Y=uG$

i.e. $\ds\frac{d}{dx}(uY'-u'Y)=u(x)G(x) \hfill (4)$

Similarly $(1)v-(2b)Y$ gives $\ds\frac{d}{dx}(vY'-v'Y)=v(x)G(x)
\hfill (5)$

Finally $(2a)v-(2b)u$ gives $\ds\frac{d}{dx}(vu'-uv')=0 \hfill
(6)$

From (6) $v(x)u'(x)-v'(x)u(x)={\rm const}=v(0)u'(0)$

$\ds=\lambda^\frac{1}{2}\sin\lambda^\frac{1}{2}l=
\frac{w}{c}\sin\frac{wl}{c}=\triangle(\lambda) \hfill (7)$

Integrate (4) from 0 to $x$:

$\ds u(x)Y'(x)-u'(x)Y(x)=\int_0^xu(\zeta)G(\zeta)d\zeta \hfill
(8)$

since $u(0)=Y(0)=0$

Integrate (5) from $l$ to $x$:

$\ds v(x)Y'(x)-v'(x)Y(x)=\int_l^xv(\zeta)G(\zeta)d\zeta \hfill
(9)$

since $v(l)=Y(l)=0$.

Equations (8) and (9) are linear equations in $Y$ and $Y'$.

The determinant of the coefficients is

$\ds\left|\begin{array}{cc} u(x) &amp; -u'(x)\\ v(x) &amp;
-v'(x)\end{array}\right|=u'(x)v(x)-v'(x)u(x)=\triangle(\lambda)
\hfill$ from (7)

Hence (8) and (9) can be solved for $Y$ and $Y'$ (for any $G(x)$)
if

$\triangle(\lambda)\not=0, \hspace{0.2in}
\triangle=\lambda^\frac{1}{2}\sin\lambda^\frac{1}{2}l\not=0$ as
$\lambda$ is not an eigenvalue.

Hence solving for $Y(x)$

$(9)u-(8)v$ gives

$\ds\triangle(\lambda)Y(x)=u(x)\int_l^xv(\zeta)G(\zeta)d\zeta-
v(x)\int_0^xu(\zeta)G(\zeta)d\zeta \hfill (10)$

Similarly we have

$\ds\triangle(\lambda)Y'(x)=u'(x)\int_l^xv(\zeta)G(\zeta)d\zeta-
v'(x)\int_0^xu(\zeta)G(\zeta)d\zeta \hfill (10')$

[Note that $(10')$ follows from differentiation of 10]

Differentiating $(10')$ we find

$\ds\triangle(\lambda)Y''(x)=u''(x)\int_l^xv(\zeta)G(\zeta)d\zeta-
v''(x)\int_0^xu(\zeta)G(\zeta)d\zeta$

$\ds\hspace{0.5in}+[u'(x)v(x)-v'(x)u(x)]G(x) \hfill (10'')$

Therefore $\triangle\lambda(Y''+\lambda Y)= (u''+\lambda u)
\int_l^x-(v''+\lambda v)\int_0^x+\triangle(\lambda)G(x)$

$\ds\hspace{1in}=\triangle(\lambda)G(x)$ as $u''+\lambda u=
v''+\lambda v=0$

Since $\triangle(\lambda)\not=0 \hspace{0.5in} Y''+\lambda Y=0$.

We can write 10 as

$\ds\triangle(\lambda)Y(x)=-\int_0^lg(x,\zeta)G(\zeta)d\zeta
\hfill (10a)$

${}$

$\ds g(x,\zeta)=\left\{\begin{array}{cc}
\frac{1}{\triangle(\lambda)}v(x)u(\zeta) &amp; 0\leq\zeta\leq x\\
\frac{1}{\triangle(\lambda)}v(\zeta)u(x) &amp; x\leq\zeta\leq
l\end{array}\right.$

$g(x,\zeta)$ is continuous in $\zeta$ at $x$.

$\ds\frac{\p}{\p\zeta}g(x,\zeta)$ is discontinuous at $x$.

$\ds\left[\frac{\p}{\p\zeta}(g(x,\zeta))\right]_{x-0}^{x+0}=
\frac{1}{\triangle(\lambda)}[v'(\zeta)u(x)-v(x)u'(\zeta)]_{x=\zeta}=
-1$

$\ds\left(\frac{\p2}{\p\zeta^2}+\lambda\right)g=0$ also
$g(\zeta,x)=g(x,\zeta)$ since

$\ds g(x,\zeta)=\frac{1}{\triangle\lambda}[v({\rm max}(\zeta,x))
\cup({\rm min}(x,\zeta))]$

for $0\leq x\leq l, \hspace{0.1in} 0\leq\zeta\leq l$, and
max$(x,\zeta)={\rm max}(\zeta,x)$.


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