Q8E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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 ${\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{and}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}$?

Step-by-Step Solution

Answer

$$\begin{gathered} {\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{a} \\ {\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{p}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{b} \end{gathered}$$

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

$$\begin{gathered} {\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{y}\mathbf{,} \\ {\mathbf{x}}_{\mathbf{2}}\mathbf{=}\mathbf{py}\mathbf{\text{'}} \end{gathered}$$ 

The equation transforms as:

 $$\begin{gathered} \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{2}}\mathbf{+}{\mathbf{qx}}_{\mathbf{1}}\mathbf{=}\mathbf{0} \\ \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{2}}\mathbf{=}\mathbf{-}{\mathbf{qx}}_{\mathbf{1}} \\ \mathbf{y}\mathbf{\text{'}}\mathbf{=}\frac{{\mathbf{x}}_{\mathbf{2}}}{\mathbf{p}} \\ \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{1}}\mathbf{=}\frac{{\mathbf{x}}_{\mathbf{2}}}{\mathbf{p}} \end{gathered}$$

 

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;

 $$\begin{gathered} {\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{a} \\ {\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{p}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{p}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{b} \end{gathered}$$

This is the required result.