• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

Suggested languages for you:

Americas

Europe

Q8E

Expert-verified
Found in: Page 259

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# Sturm–Liouville Form. A second-order equation is said to be in Sturm–Liouville form if it is expressed as $\left[p\left(t\right)y\text{'}\left(t\right)\right]\mathbf{\text{'}}\mathbf{+}\mathbf{q}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}$. Show that the substitutions ${{\mathbf{x}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{y}}{\mathbf{,}}{{\mathbf{x}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{py}}{\mathbf{\text{'}}}$ result in the normal form ${\mathbf{x}}{{\mathbf{\text{'}}}}_{{\mathbf{1}}}{\mathbf{=}}\frac{{\mathbf{x}}_{\mathbf{2}}}{\mathbf{p}}{\mathbf{,}}{{\mathbf{x}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{-}}{{\mathbf{qx}}}_{{\mathbf{1}}}$. If $\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{a}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{b}$ are the initial values for the Sturm–Liouville problem, what are ${{\mathbf{x}}}_{{\mathbf{1}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{and}{{\mathbf{x}}}_{{\mathbf{2}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}$?

${\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{a}\phantom{\rule{0ex}{0ex}}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{p}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{b}$

See the step by step solution

## Step 1: Express the equation in form of x

Here given. $\left[\mathbf{p}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\right]\mathbf{\text{'}}\mathbf{+}\mathbf{q}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}$

And

${\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{y}\mathbf{,}\phantom{\rule{0ex}{0ex}}{\mathbf{x}}_{\mathbf{2}}\mathbf{=}\mathbf{py}\mathbf{\text{'}}$

The equation transforms as:

$\mathbf{x}{\mathbf{\text{'}}}_{\mathbf{2}}\mathbf{+}{\mathbf{qx}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\phantom{\rule{0ex}{0ex}}\mathbf{x}{\mathbf{\text{'}}}_{\mathbf{2}}\mathbf{=}\mathbf{-}{\mathbf{qx}}_{\mathbf{1}}\phantom{\rule{0ex}{0ex}}\mathbf{y}\mathbf{\text{'}}\mathbf{=}\frac{{\mathbf{x}}_{\mathbf{2}}}{\mathbf{p}}\phantom{\rule{0ex}{0ex}}\mathbf{x}{\mathbf{\text{'}}}_{\mathbf{1}}\mathbf{=}\frac{{\mathbf{x}}_{\mathbf{2}}}{\mathbf{p}}$

## Step 2: The initial conditions.

The given initial conditions are $\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{a}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{b}{\mathbf{.}}$

Initial conditions after transformations;

${\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{a}\phantom{\rule{0ex}{0ex}}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{p}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{p}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{b}$

This is the required result.