91Ó°ÊÓ

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 $\mathbf{t}→\mathbf{+}∞$ 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,

ii. If ${\mathbf{r}}_{\mathbf{1}}\mathbf{,}{\mathbf{r}}_{\mathbf{2}}\mathbf{=}\mathbf{Î\pm }Â\pm \mathbf{β¾\pm }$,$\mathbf{β}≠0$ , then $\mathbf{Î\pm }\mathbf{=}\frac{{\mathbf{r}}_{\mathbf{1}}\mathbf{+}{\mathbf{r}}_{\mathbf{2}}}{\mathbf{2}}â\copyright ½0$.

Given that,

$\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{λ³\ae }\mathbf{-}\mathbf{y}$ … (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.

$\left(\mathbf{D}\mathbf{-}\mathbf{λ}\right)\left[\mathbf{x}\right]\mathbf{+}\left[\mathbf{y}\right]\mathbf{=}\mathbf{0}$ … (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 $\mathbf{D}\mathbf{-}\mathbf{λ}$on equation (4). Then, add them together.

$$\begin{gathered} \mathbf{3}\left(\mathbf{D}\mathbf{-}\mathbf{λ}\right)\left[\mathbf{x}\right]\mathbf{+}\mathbf{3}\left[\mathbf{y}\right]\mathbf{-}\mathbf{3}\left(\mathbf{D}\mathbf{-}\mathbf{λ}\right)\left[\mathbf{x}\right]\mathbf{+}\left(\mathbf{D}\mathbf{-}\mathbf{λ}\right)\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \\ \left(\mathbf{D}\mathbf{-}\mathbf{λ}\right)\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left[\mathbf{y}\right]\mathbf{+}\mathbf{3}\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \\ \left({\mathbf{D}}^{\mathbf{2}}\mathbf{-}\mathbf{D}\mathbf{-}\mathbf{λ\P Ù}\mathbf{+}\mathbf{λ}\mathbf{+}\mathbf{3}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \\ \left({\mathbf{D}}^{\mathbf{2}}\mathbf{-}\mathbf{D}\left(\mathbf{1}\mathbf{+}\mathbf{λ}\right)\mathbf{+}\mathbf{λ}\mathbf{+}\mathbf{3}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \\ \left({\mathbf{D}}^{\mathbf{2}}\mathbf{-}\mathbf{D}\left(\mathbf{1}\mathbf{+}\mathbf{λ}\right)\mathbf{+}\mathbf{λ}\mathbf{+}\mathbf{3}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0}            …\left(5\right) \end{gathered}$$

Since the auxiliary equation to the corresponding homogeneous equation is;

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

If ${\mathbf{r}}_{\mathbf{1}}\mathbf{,}{\mathbf{r}}_{\mathbf{2}}$are conjugate complex, then we know that ${\mathbf{r}}_{\mathbf{1}}·{\mathbf{r}}_{\mathbf{2}}â\copyright ¾0$, and thus we see that for x(t) to be bounded, we need ${\mathbf{r}}_{\mathbf{1}}·{\mathbf{r}}_{\mathbf{2}}â\copyright ¾0$and ${\mathbf{r}}_{\mathbf{1}}\mathbf{+}{\mathbf{r}}_{\mathbf{2}}â\copyright ½0$, 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,

${\mathbf{r}}_{\mathbf{1}}\mathbf{+}{\mathbf{r}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{λ}â\copyright ½0$and ${\mathbf{r}}_{\mathbf{1}}·{\mathbf{r}}_{\mathbf{2}}\mathbf{=}\mathbf{λ}\mathbf{+}\mathbf{3}â\copyright ¾0$

$\mathbf{λ}â\copyright ½\mathbf{-}\mathbf{1}$ and $\mathbf{λ}â\copyright ¾\mathbf{-}\mathbf{3}$

So, $\mathbf{λ}∈\left[\mathbf{-}\mathbf{3}\mathbf{,}\mathbf{-}\mathbf{1}\right]$.

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

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