A proof of Theorem 1, page 266, is outlined below. The goal is to show that\({\bf{f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{) = g(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{) = 0}}\). Justify each step

1. From the given hypotheses, deduce that\(\mathop {\lim }\limits_{t \to \infty } {\bf{x'(t) = f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}\) and\(\mathop {\lim }\limits_{t \to \infty } {\bf{y'(t) = g(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}\).
2. Suppose\({\bf{f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{) > 0}}\). Then, by continuity,\({\bf{x'(t) = }}\frac{{{\bf{f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}}}{{\bf{2}}}\)for all large t(say, for\({\bf{t}} \ge {\bf{T}}\)). Deduce from this that\({\bf{x(t) > }}\frac{{{\bf{tf(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}}}{{\bf{2}}}{\bf{ + C}}\)fort>I, where Cis some constant.
3. Conclude from part (b) that\(\mathop {\lim }\limits_{t \to \infty } {\bf{x(t)}} = + \infty \), contradicting the fact that this limit is the finite number x*. Thus, f(x*, y*) cannot be positive.
4. Argue similarly that the supposition thatf(x*, y*) < 0 also leads to a contradiction; hence,f(x*, y*) must be zero.

In the same manner, argue that g(x*, y*) must bezero. Therefore, f(x*, y*) = g(x*, y*) = 0, and (x*, y*) isa critical point.

Suppose here (x(t),y(t)) is a solution on \(\left[ {0,\infty } \right.)\)of the system;;

\(\begin{array}{c}\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ = f(x,y)}}\\\frac{{{\bf{dy}}}}{{{\bf{dt}}}}{\bf{ = g(x,y)}}\end{array}\)

Where f and g are continuous functions and the limits are:

\(\begin{array}{c}{{\bf{x}}^{\bf{*}}}{\bf{ = }}\mathop {{\bf{lim}}}\limits_{{\bf{t}} \to \infty } {\bf{x(t)}}\\{{\bf{y}}^{\bf{*}}}{\bf{ = }}\mathop {{\bf{lim}}}\limits_{{\bf{t}} \to \infty } {\bf{y(t)}}\end{array}\)

Exist and are finite.

Here f and g are continuous functions so,

\(\begin{array}{c}\mathop {{\bf{lim}}}\limits_{{\bf{t}} \to \infty } {\bf{x'(t) = }}\mathop {{\bf{lim}}}\limits_{{\bf{t}} \to \infty } {\bf{f(x(t),y(t))}}\\{\bf{ = f(}}\mathop {{\bf{lim}}}\limits_{{\bf{t}} \to \infty } {\bf{(x(t),}}\mathop {{\bf{lim}}}\limits_{{\bf{t}} \to \infty } {\bf{y(t))}}\\{\bf{ = f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}\\\mathop {{\bf{lim}}}\limits_{{\bf{t}} \to \infty } {\bf{y'(t) = }}\mathop {{\bf{lim}}}\limits_{{\bf{t}} \to \infty } {\bf{g(x(t),y(t))}}\\{\bf{ = g(}}\mathop {{\bf{lim}}}\limits_{{\bf{t}} \to \infty } {\bf{(x(t),}}\mathop {{\bf{lim}}}\limits_{{\bf{t}} \to \infty } {\bf{y(t))}}\\{\bf{ = g(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}\end{array}\)

Suppose that \({\bf{f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{) > 0}}\), since

\({\bf{f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{) = }}\mathop {{\bf{lim}}}\limits_{{\bf{t}} \to \infty } {\bf{x'(t)}}\)

Apply the definition of continuity then;

\(\left| {{\bf{x'(t) - f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}} \right|{\bf{ < \varepsilon }}\)

For \({\bf{\varepsilon = }}\frac{{{\bf{f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}}}{{\bf{2}}}\)then;

\(\begin{array}{c}\left| {{\bf{x'(t) - f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}} \right|{\bf{ < }}\frac{{{\bf{f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}}}{{\bf{2}}}\\{\bf{ - }}\frac{{{\bf{f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}}}{{\bf{2}}}{\bf{ < x'(t) - f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{) < }}\frac{{{\bf{f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}}}{{\bf{2}}}\\\frac{{{\bf{f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}}}{{\bf{2}}}{\bf{ < x'(t) < }}\frac{{{\bf{3f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}}}{{\bf{2}}}\end{array}\)

Now, one gets\(\frac{{{\bf{f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}}}{{\bf{2}}}{\bf{ < x'(t)}}\) for\({\bf{t}} \ge {\bf{T}}\).

Apply the fundamental theorem of calculus.

\(\begin{array}{c}{\bf{x(t) - x(T) = }}\int\limits_{\bf{T}}^{\bf{t}} {{\bf{x'(s)ds}}} \\{\bf{x(t) = }}\int\limits_{\bf{T}}^{\bf{t}} {{\bf{x'(s)ds}}} {\bf{ + x(T)}}\\{\bf{ > }}\int\limits_{\bf{T}}^{\bf{t}} {\frac{{{\bf{f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}}}{{\bf{2}}}{\bf{ds}}} {\bf{ + x(T)}}\\{\bf{ = }}\frac{{{\bf{f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}}}{{\bf{2}}}{\bf{(t - T) + x(T)}}\\{\bf{ = }}\frac{{{\bf{tf(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}}}{{\bf{2}}}{\bf{ - }}\frac{{{\bf{Tf(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}}}{{\bf{2}}}{\bf{ + x(T)}}\\{\bf{x(t) > }}\frac{{{\bf{tf(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}}}{{\bf{2}}}{\bf{ + C}}\end{array}\)

Continuing the part (b) then;

\(\begin{array}{c}\mathop {{\bf{lim}}}\limits_{{\bf{t}} \to \infty } {\bf{x(t) > }}\mathop {{\bf{lim}}}\limits_{{\bf{t}} \to \infty } \frac{{{\bf{tf(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}}}{{\bf{2}}}{\bf{ + C = }}\infty \\\mathop {{\bf{lim}}}\limits_{{\bf{t}} \to \infty } {\bf{x(t) = }}\infty \\{\bf{f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}} \le 0\end{array}\)

Suppose that \({\bf{f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}} < 0\) then;

\(\begin{array}{l}{\bf{x(t) < }}\frac{{{\bf{f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}}}{{\bf{2}}}\\{\bf{x(t) < }}\frac{{{\bf{tf(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}}}{{\bf{2}}}{\bf{ + C}}\end{array}\)

So, \(\mathop {{\bf{lim}}}\limits_{{\bf{t}} \to \infty } {\bf{x(t)}} \le \mathop {{\bf{lim}}}\limits_{{\bf{t}} \to \infty } \le \frac{{{\bf{tf(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}}}{{\bf{2}}}{\bf{ + C = - }}\infty \)

Therefore \(\mathop {{\bf{lim}}}\limits_{{\bf{t}} \to \infty } {\bf{x(t) = - }}\infty \)

Hence,

\(\begin{array}{l}{\bf{f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}} \ge 0\\{\bf{f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}} \le 0\\{\bf{f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{) = 0}}\end{array}\)

Now repeating the same procedure for y instead of x and g instead of f gives the result. Thus\({\bf{(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}\)is a critical point.