Problem of the Week #84 - November 4th, 2013

[Chris L T521]

Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: Consider a differential equation of the form
\[A(x)y^{\prime\prime} + B(x)y^{\prime} + C(x)y + \lambda D(x)y = 0.\]
Show that you can express this equation in Sturm-Liouville form, given by
\[\frac{d}{dx}\left[p(x)\frac{dy}{dx}\right] - q(x)y -\lambda r(x)y = 0.\]

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Hint: [sp]First divide every term by $A(x)$ and then multiply the equation by $\exp\left(\displaystyle \int\frac{B(x)}{A(x)}\,dx\right)$.[/sp]

 

[Chris L T521]

This week's problem was correctly answered by Ackbach, MarkFL, and mathbalarka. You can find Mark's solution below.

[sp]We are given the following ODE:

\(\displaystyle A(x)y^{\prime\prime} + B(x)y^{\prime} + C(x)y + \lambda D(x)y = 0\)

Dividing through by $A(x)$ we obtain:

\(\displaystyle y^{\prime\prime} + \frac{B(x)}{A(x)}y^{\prime} + \frac{C(x)}{A(x)}y + \lambda \frac{D(x)}{A(x)}y = 0\)

Using the integrating factor:

\(\displaystyle \mu(x)=\exp\left(\displaystyle \int\frac{B(x)}{A(x)}\,dx\right)\)

we obtain:

\(\displaystyle \exp\left(\displaystyle \int\frac{B(x)}{A(x)}\,dx\right)y^{\prime\prime} + \frac{B(x)}{A(x)}\exp\left(\displaystyle \int\frac{B(x)}{A(x)}\,dx\right)y^{\prime} + \frac{C(x)}{A(x)}\exp\left(\displaystyle \int\frac{B(x)}{A(x)}\,dx\right)y + \lambda \frac{D(x)}{A(x)}\exp\left(\displaystyle \int\frac{B(x)}{A(x)}\,dx\right)y = 0\)

If we use the following definitions:

\(\displaystyle p(x)\equiv \exp\left(\displaystyle \int\frac{B(x)}{A(x)}\,dx\right)\implies p'(x)=\frac{B(x)}{A(x)}\exp\left(\displaystyle \int\frac{B(x)}{A(x)}\,dx\right)\)

\(\displaystyle q(x)\equiv -\frac{C(x)}{A(x)}\exp\left(\displaystyle \int\frac{B(x)}{A(x)}\,dx\right)\)

\(\displaystyle r(x)\equiv -\frac{D(x)}{A(x)}\exp\left(\displaystyle \int\frac{B(x)}{A(x)}\,dx\right)\)

we may write:

\(\displaystyle p(x)y^{\prime\prime} + p^{\prime}(x)y^{\prime} - q(x)y - \lambda r(x)y = 0\)

Observing that:

\(\displaystyle p(x)y^{\prime\prime} + p^{\prime}(x)y^{\prime}=\frac{d}{dx}\left[p(x)\frac{dy}{dx}\right]\)

we may then write:

\(\displaystyle \frac{d}{dx}\left[p(x)\frac{dy}{dx}\right] - q(x)y -\lambda r(x)y = 0 \)

Shown as desired.[/sp]