# Sturm-Liouville Problem/Unit Weight Function/Lemma

## Theorem

Let $\alpha \left ( { x } \right ) : \R \to \R$ such that $\alpha \left ( { x } \right ) \in C^2 \left [ { a \,. \,. \, b } \right ]$.

Suppose:

$\displaystyle \forall h \left ( { x } \right ) \in C^2 \left [ { a \,. \,. \, b } \right ] : h \left ( { a } \right ) = h \left ( { b } \right ) = h' \left ( { a } \right ) = h' \left ( { b } \right ) = 0 : \int_a^b \alpha \left ( { x } \right ) h'' \left ( { x } \right ) \mathrm d x = 0$

Then:

$\forall x \in \left [ { a \,. \,. \, b } \right ] : \alpha \left ( { x } \right ) = c_0 + c_1 x$

where $c_0, c_1 \in \R$ are constants.

## Proof

Let $c_0, c_1$ be defined by the conditions:

$\displaystyle \int_a^b \left [ { \alpha \left ( { x } \right ) - c_0 - c_1 x } \right ] \mathrm d x = 0$
$\displaystyle \int_a^b \mathrm d x \int_a^x \left [ { \alpha \left ( { \xi } \right ) - c_0 - c_1 \xi } \right ] \mathrm \xi = 0$

Suppose:

$h \left ( { x } \right ) = \int_a^x \xi \int_a^\xi \left [ { \alpha \left ( { t } \right ) - c_0 - c_1 t } \right ] \mathrm d t$

This form satisfies conditions on $h$ in the theorem.

Then:

 $\displaystyle \int_a^b \left [ { \alpha \left ( { x } \right ) - c_0 - c_1 x } \right ] h'' \left ( { x } \right ) \mathrm d x$ $=$ $\displaystyle \int_a^b \alpha \left ( { x } \right ) h'' \left ( { x } \right ) \mathrm d x - c_0 \left [ { h' \left ( { b } \right ) - h' \left ( { a } \right ) } \right ] - c_1 \int_a^b x h'' \left ( { x } \right ) \mathrm d x$ $\displaystyle$ $=$ $\displaystyle - c_1 \left [ { b h' \left ( { b } \right ) - a h' \left ( { a } \right ) } \right ] - c_1 \left [ { h \left ( { b } \right ) - h \left ( { a } \right ) } \right ]$ $\displaystyle$ $=$ $\displaystyle 0$

On the other hand:

 $\displaystyle \int_a^b \left [ { \alpha \left ( { x } \right ) - c_0 - c_1 x } \right ] h'' \left ( { x } \right ) \mathrm d x$ $=$ $\displaystyle \int_a^b \left [ { \alpha \left ( { x } \right ) - c_0 - c_1 x } \right ]^2 \mathrm d x$ $\displaystyle$ $=$ $\displaystyle 0$

Therefore:

$\displaystyle \alpha \left ( { x } \right ) - c_0 - c_1 x = 0$

or:

$\displaystyle \alpha \left ( { x } \right ) = c_0 + c_1 x$

$\blacksquare$