Question

Suppose that y(x) is a solution of the autonomous equationdy/dx = f(y) and is bounded above and below by two consecutive critical points as in sub region R₂ of Figure 2.1.6(b). If f(y) > 0 in the region, then$\underset{\mathbf{x}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}\mathit{y}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{c}}_{\mathbf{2}}$ Discuss why there cannot exist a number $\mathit{L}\mathbf{<}{\mathbf{c}}_{\mathbf{2}}$such that.$\underset{\mathbf{x}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}\mathit{y}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{L}$ As part of your discussion, consider what happens to $\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$as.$\mathit{x}\mathbf{\to }\mathit{\infty }$

Short answer

L is a critical point. It is new critical point is between $c_1$and $c_2$we reach a contradiction since $c_1$and$c_2$ are consecutive critical points. So it must be the case$L=c_2$ .

Step-by-step solution

Step 1:Critical points

The zeros of the function f in $dy/dx=f\left(y\right)$are of special importance. Wesay that a real number c is a critical point of the autonomous differential equation$dy/dx=f\left(y\right)$ if it is a zero of f—that is, $f\left(c\right)=0$. A critical point is also called an equilibrium point or stationary point.

Step 2:Solution for the function dy/dx = f(y)

For$\frac{dy}{dx}=f\left(y\right)$ , we are assuming f and $f\text{'}$are continuously functions of y. If $f\left(y\right)$changes signs within R, then$f\left(y\right)=0$ for some $y_0$within R. However this implies $y_0$is a critical point, which cannot happen since $c_1$and $c_2$are consecutive critical point. Therefore, the function $f\left(y\right)$is either always positive or always negative.

In this case we are given,$f\left(y\right)>0$for all y in R. This implies$c_2$is a horizontal asymptote for.$y\left(x\right)$

$\begin{array}{l}\underset{x\to \infty }{\lim }y\left(x\right)=c_2\\ \underset{x\to \infty }{\lim }y\text{'}\left(x\right)=0\end{array}$

Critical points

Let,

$\underset{x\to \infty }{\lim }y\left(x\right)=L\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

Where.$L\le c_2$ We want to show.$L=c_2$

Since L is a horizontal asymptote for.$y\left(x\right)$

$$\begin{gathered} \underset{x\to \infty }{\lim }y\text{'}\left(x\right)=0\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right) \\ \begin{array}{c}f\left(L\right)=f\left(\underset{x\to \infty }{\lim }y\right(x\left)\right)\\ =\underset{x\to \infty }{\lim }f\left(y\right(x\left)\right)\\ =\underset{x\to \infty }{\lim }y\text{'}\left(x\right)=0\end{array} \end{gathered}$$

So, we have that L is a critical point. It is new critical point is between $c_1$and $c_2$we reach a contradiction since $c_1$and $c_2$are consecutive critical points. So it must be the case that.

Hence, L is a critical point. It is new critical point is between $c_1$and $c_2$we reach a contradiction since $c_1$and $c_2$are consecutive critical points. So it must be the case.$L=c_2$