Sturm-Liouville Problem/Unit Weight Function/Lemma

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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 $


$\blacksquare$