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Q29E

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### 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

# In Problems 29 and 30, determine the range of values (if any) of the parameter that will ensure all solutions x(t), and y(t) of the given system remain bounded as $\mathbf{t}\mathbf{\to }\mathbf{+}\mathbf{\infty }$. $\frac{\mathbf{dx}}{\mathbf{dt}}{\mathbf{=}}{\mathbf{\lambda x}}{\mathbf{-}}{\mathbf{y}}{\mathbf{,}}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{dy}}{\mathbf{dt}}{\mathbf{=}}{\mathbf{3}}{\mathbf{x}}{\mathbf{+}}{\mathbf{y}}$

The parameter $\mathbf{\lambda }$ of the given system remains bounded as $\mathbf{t}\to \mathbf{+}\infty$ is $\mathbf{\lambda }\in \left[\mathbf{-}\mathbf{3}\mathbf{,}\mathbf{-}\mathbf{1}\right]$.

See the step by step solution

## Step 1: General form

Elimination Procedure for 2 × 2 Systems:

To find a general solution for the system

${\mathbf{L}}_{\mathbf{1}}\left[\mathbf{x}\right]\mathbf{+}{\mathbf{L}}_{\mathbf{2}}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{f}}_{\mathbf{1}}\mathbf{,}\phantom{\rule{0ex}{0ex}}{\mathbf{L}}_{\mathbf{3}}\left[\mathbf{x}\right]\mathbf{+}{\mathbf{L}}_{\mathbf{4}}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{f}}_{\mathbf{2}}\mathbf{,}$

Where ${\mathbf{L}}_{\mathbf{1}}\mathbf{,}{\mathbf{L}}_{\mathbf{2}}\mathbf{,}{\mathbf{L}}_{\mathbf{3}}\mathbf{,}$ and ${\mathbf{L}}_{\mathbf{4}}$ are polynomials in $\mathbf{D}\mathbf{=}\frac{\mathbf{d}}{\mathbf{dt}}$

1. Make sure that the system is written in operator form.

1. Eliminate one of the variables, say, y, and solve the resulting equation for x(t). If the system is degenerating, stop! A separate analysis is required to determine whether or not there are solutions.

1. (Shortcut) If possible, use the system to derive an equation that involves y(t) but not its derivatives. [Otherwise, go to step (d).] Substitute the found expression for x(t) into this equation to get a formula for y(t). The expressions for x(t), and y(t) give the desired general solution.

1. Eliminate x from the system and solve for y(t). [Solving for y(t) gives more constants- twice as many as needed.]

1. Remove the extra constants by substituting the expressions for x(t) and y(t) into one or both of the equations in the system. Write the expressions for x(t) and y(t) in terms of the remaining constants.

Vieta’s formulas for finding roots:

For function y(t) to be bounded when $\mathbf{t}\to \mathbf{+}\infty$ we need for both roots of the auxiliary equation to be non-positive if they are reals and, if they are complex, then the real part has to be non-positive. In other words,

1. If ${\mathbf{r}}_{\mathbf{1}}\mathbf{,}{\mathbf{r}}_{\mathbf{2}}\in \mathbf{R}$ , then ${\mathbf{r}}_{\mathbf{1}}·{\mathbf{r}}_{\mathbf{2}}⩾\mathbf{0}\mathbf{,}{\mathbf{r}}_{\mathbf{1}}\mathbf{+}{\mathbf{r}}_{\mathbf{2}}⩽\mathbf{0}$,

ii. If ${\mathbf{r}}_{\mathbf{1}}\mathbf{,}{\mathbf{r}}_{\mathbf{2}}\mathbf{=}\mathbf{\alpha }±\mathbf{\beta i}$,$\mathbf{\beta }\ne 0$ , then $\mathbf{\alpha }\mathbf{=}\frac{{\mathbf{r}}_{\mathbf{1}}\mathbf{+}{\mathbf{r}}_{\mathbf{2}}}{\mathbf{2}}⩽0$ .

## Step 2: Evaluate the given equation

Given that,

$\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{\lambda x}\mathbf{-}\mathbf{y}$ … (1)

$\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{y}$ … (2)

Let us rewrite the given system of equations into operator form.

$\left(\mathbf{D}\mathbf{-}\mathbf{\lambda }\right)\left[\mathbf{x}\right]\mathbf{+}\left[\mathbf{y}\right]\mathbf{=}\mathbf{0}$ … (3)

$\mathbf{-}\mathbf{3}\left[\mathbf{x}\right]\mathbf{+}\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0}$ … (4)

Multiply 3 on equation (3) and multiply $\mathbf{D}\mathbf{-}\mathbf{\lambda }$ on equation (4). Then, add them together.

$\mathbf{3}\left(\mathbf{D}\mathbf{-}\mathbf{\lambda }\right)\left[\mathbf{x}\right]\mathbf{+}\mathbf{3}\left[\mathbf{y}\right]\mathbf{-}\mathbf{3}\left(\mathbf{D}\mathbf{-}\mathbf{\lambda }\right)\left[\mathbf{x}\right]\mathbf{+}\left(\mathbf{D}\mathbf{-}\mathbf{\lambda }\right)\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0}\phantom{\rule{0ex}{0ex}}\left(\mathbf{D}\mathbf{-}\mathbf{\lambda }\right)\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left[\mathbf{y}\right]\mathbf{+}\mathbf{3}\left[\mathbf{y}\right]\mathbf{=}\mathbf{0}\phantom{\rule{0ex}{0ex}}\left({\mathbf{D}}^{\mathbf{2}}\mathbf{-}\mathbf{D}\mathbf{-}\mathbf{\lambda D}\mathbf{+}\mathbf{\lambda }\mathbf{+}\mathbf{3}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0}\phantom{\rule{0ex}{0ex}}\left({\mathbf{D}}^{\mathbf{2}}\mathbf{-}\mathbf{D}\left(\mathbf{1}\mathbf{+}\mathbf{\lambda }\right)\mathbf{+}\mathbf{\lambda }\mathbf{+}\mathbf{3}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0}\phantom{\rule{0ex}{0ex}}\left({\mathbf{D}}^{\mathbf{2}}\mathbf{-}\mathbf{D}\left(\mathbf{1}\mathbf{+}\mathbf{\lambda }\right)\mathbf{+}\mathbf{\lambda }\mathbf{+}\mathbf{3}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \dots \left(5\right)$

## Step 3: Substitution method

Since the auxiliary equation to the corresponding homogeneous equation is;

${\mathbf{r}}^{\mathbf{2}}\mathbf{-}\left(\mathbf{1}\mathbf{+}\mathbf{\lambda }\right)\mathbf{r}\mathbf{+}\mathbf{\lambda }\mathbf{+}\mathbf{3}\mathbf{=}\mathbf{0}$

If ${\mathbf{r}}_{\mathbf{1}}\mathbf{,}{\mathbf{r}}_{\mathbf{2}}$ are conjugate complex, then we know that ${\mathbf{r}}_{\mathbf{1}}·{\mathbf{r}}_{\mathbf{2}}⩾0$, and thus we see that for x(t) to be bounded, we need ${\mathbf{r}}_{\mathbf{1}}·{\mathbf{r}}_{\mathbf{2}}⩾0$ and ${\mathbf{r}}_{\mathbf{1}}\mathbf{+}{\mathbf{r}}_{\mathbf{2}}⩽0$, independent of whether ${\mathbf{r}}_{\mathbf{1}}\mathbf{,}{\mathbf{r}}_{\mathbf{2}}$ are reals or they are complex numbers.

Using Vieta’s formulas, we now have,

${\mathbf{r}}_{\mathbf{1}}\mathbf{+}{\mathbf{r}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{\lambda }⩽0$ and ${\mathbf{r}}_{\mathbf{1}}·{\mathbf{r}}_{\mathbf{2}}\mathbf{=}\mathbf{\lambda }\mathbf{+}\mathbf{3}⩾0$

$\mathbf{\lambda }⩽\mathbf{-}\mathbf{1}$ and $\mathbf{\lambda }⩾\mathbf{-}\mathbf{3}$

So, $\mathbf{\lambda }\in \left[\mathbf{-}\mathbf{3}\mathbf{,}\mathbf{-}\mathbf{1}\right]$ .

Similarly, if we eliminate y(t) from the system we get, $\left(\mathbf{-}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\left(\mathbf{1}\mathbf{+}\mathbf{\lambda }\right)\mathbf{D}\mathbf{-}\left(\mathbf{\lambda }\mathbf{+}\mathbf{3}\right)\right)\left[\mathbf{x}\right]\mathbf{=}\mathbf{0}$ .

And for both x(t) and y(t) to be bounded when $\mathbf{t}\to \mathbf{+}\infty$ one needs $\mathbf{\lambda }\in \left[\mathbf{-}\mathbf{3}\mathbf{,}\mathbf{-}\mathbf{1}\right]$.

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