Q30E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

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

Step-by-Step Solution

Answer

The parameter of the given system remains bounded as $\mathbf{t}\to \mathbf{+}\infty$  is $\mathbf{\lambda }⩽\mathbf{1}$.

Step 1: General form

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 $\mathbf{D}\mathbf{=}\frac{\mathbf{d}}{\mathbf{dt}}$

 

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

Step 2: Evaluate the given equation

Given that,

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

  $\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{y} \dots \left(2\right)$       

 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).

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

Step 3: Substitution method

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 $\mathbf{\lambda }⩽\mathbf{1}$.