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 $\mathbf{t}\mathbf{\to }\mathbf{+}\mathbf{\infty }$.

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\mathbf{x}\mathbf{+}\mathbf{\lambda y}\mathbf{,} \\ \frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{y} \end{gathered}$$

Elimination Procedure for 2 × 2 Systems:

To find a general solution for the system

$$\begin{gathered} {\mathbf{L}}_{\mathbf{1}}\left[\mathbf{x}\right]\mathbf{+}{\mathbf{L}}_{\mathbf{2}}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{f}}_{\mathbf{1}}\mathbf{,} \\ {\mathbf{L}}_{\mathbf{3}}\left[\mathbf{x}\right]\mathbf{+}{\mathbf{L}}_{\mathbf{4}}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{f}}_{\mathbf{2}}\mathbf{,} \end{gathered}$$

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

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

Given that,

$\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\mathbf{x}\mathbf{+}\mathbf{\lambda y} \dots \left(1\right)$

role="math" localid="1664101250446" $\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{y} \dots \left(2\right)$

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

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

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

Multiply D+1 on equation (4). Then, subtract with equation (3).

$$\begin{gathered} \left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{x}\right]\mathbf{-}\mathbf{\lambda }\left[\mathbf{y}\right]\mathbf{-}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{x}\right]\mathbf{+}{\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)}^{\mathbf{2}}\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \\ {\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)}^{\mathbf{2}}\left[\mathbf{y}\right]\mathbf{-}\mathbf{\lambda }\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \\ \left({\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)}^{\mathbf{2}}\mathbf{-}\mathbf{\lambda }\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \\ \left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{D}\mathbf{+}\mathbf{1}\mathbf{-}\mathbf{\lambda }\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \\ \left({\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)}^{\mathbf{2}}\mathbf{-}\mathbf{\lambda }\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \cdots \left(5\right) \end{gathered}$$

Since the auxiliary equation to the corresponding homogeneous equation is .

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

Then,

$$\begin{gathered} {\left(\mathbf{r}\mathbf{+}\mathbf{1}\right)}^{\mathbf{2}}\mathbf{-}\mathbf{\lambda }\mathbf{=}\mathbf{0} \\ {\left(\mathbf{r}\mathbf{+}\mathbf{1}\right)}^{\mathbf{2}}\mathbf{=}\mathbf{\lambda } \\ \mathbf{r}\mathbf{+}\mathbf{1}\mathbf{=}\pm \sqrt{\mathbf{\lambda }} \\ \mathbf{r}\mathbf{=}\mathbf{-}\mathbf{1}\pm \sqrt{\mathbf{\lambda }} \end{gathered}$$

So, the roots are$\mathbf{r}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{+}\sqrt{\mathbf{\lambda }}$ and $\mathbf{r}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{-}\sqrt{\mathbf{\lambda }}$.

Then, the general solution of y is $\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{Ae}}^{\left(\mathbf{-}\mathbf{1}\mathbf{+}\sqrt{\mathbf{\lambda }}\right)\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\left(\mathbf{-}\mathbf{1}\mathbf{-}\sqrt{\mathbf{\lambda }}\right)\mathbf{t}} \dots \left(2\right)$

Now substitute the equation (6) in equation (4).

$$\begin{gathered} \left[\mathbf{x}\right]\mathbf{-}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \\ \mathbf{x}\mathbf{=}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{y}\right] \\ \mathbf{x}\mathbf{=}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[{\mathbf{Ae}}^{\left(\mathbf{-}\mathbf{1}\mathbf{+}\sqrt{\mathbf{\lambda }}\right)\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\left(\mathbf{-}\mathbf{1}\mathbf{-}\sqrt{\mathbf{\lambda }}\right)\mathbf{t}}\right] \\ \mathbf{=}\left(\mathbf{-}\mathbf{1}\mathbf{+}\sqrt{\mathbf{\lambda }}\right){\mathbf{Ae}}^{\left(\mathbf{-}\mathbf{1}\mathbf{+}\sqrt{\mathbf{\lambda }}\right)\mathbf{t}}\mathbf{+}\left(\mathbf{-}\mathbf{1}\mathbf{-}\sqrt{\mathbf{\lambda }}\right){\mathbf{Be}}^{\left(\mathbf{-}\mathbf{1}\mathbf{-}\sqrt{\mathbf{\lambda }}\right)\mathbf{t}}\mathbf{+}{\mathbf{Ae}}^{\left(\mathbf{-}\mathbf{1}\mathbf{+}\sqrt{\mathbf{\lambda }}\right)\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\left(\mathbf{-}\mathbf{1}\mathbf{-}\sqrt{\mathbf{\lambda }}\right)\mathbf{t}} \\ \mathbf{=}\mathbf{A}\sqrt{\mathbf{\lambda }}{\mathbf{e}}^{\left(\mathbf{-}\mathbf{1}\mathbf{+}\sqrt{\mathbf{\lambda }}\right)\mathbf{t}}\mathbf{-}\mathbf{B}\sqrt{\mathbf{\lambda }}{\mathbf{e}}^{\left(\mathbf{-}\mathbf{1}\mathbf{-}\sqrt{\mathbf{\lambda }}\right)\mathbf{t}} \\ \mathbf{x}\left(\mathbf{t}\right)\mathbf{=}\mathbf{A}\sqrt{\mathbf{\lambda }}{\mathbf{e}}^{\left(\mathbf{-}\mathbf{1}\mathbf{+}\sqrt{\mathbf{\lambda }}\right)\mathbf{t}}\mathbf{-}\mathbf{B}\sqrt{\mathbf{\lambda }}{\mathbf{e}}^{\left(\mathbf{-}\mathbf{1}\mathbf{-}\sqrt{\mathbf{\lambda }}\right)\mathbf{t}} \dots \left(7\right) \end{gathered}$$

Since x and y do not approach the $\infty$. When, $\mathbf{t}\to \mathbf{+}\infty$. So, the relation between them is taken as below.

$\left(\mathbf{-}\mathbf{1}\mathbf{+}\sqrt{\mathbf{\lambda }}\right)\left(\mathbf{-}\mathbf{1}\mathbf{-}\sqrt{\mathbf{\lambda }}\right)⩾\mathbf{0}$

Let us rearrange the terms. To get,

$$\begin{gathered} \left(\sqrt{\mathbf{\lambda }}\mathbf{-}\mathbf{1}\right)\left(\sqrt{\mathbf{\lambda }}\mathbf{+}\mathbf{1}\right)⩽\mathbf{0} \\ \left(\mathbf{\lambda }\mathbf{-}\mathbf{1}\right)⩽\mathbf{0} \\ \mathbf{\lambda }⩽\mathbf{1} \end{gathered}$$

Therefore, the values of$\mathbf{\lambda }$ for which the solution of x(t) and y(t) remains bounded are $\mathbf{\lambda }⩽\mathbf{1}$.