Let \(f,g:\mathbb R\to(0,\infty)\) be continuous with \(g\le f\). Suppose the initial-value problem \[ x'=f(x), \qquad x(0)=0, \] has a solution defined on all of \(\mathbb R\). Prove that \[ x'=g(x), \qquad x(0)=0, \] also has a solution defined on all of \(\mathbb R\).
Solution
Define \[ F(u)\coloneqq\int_0^u\frac{ds}{f(s)}. \]
1. The function \(F\) is strictly increasing and satisfies\[ F'(u)=\frac1{f(u)}>0. \]
The function \(f\) is positive and continuous, so \(1/f\) is positive and continuous. The fundamental theorem of calculus gives the derivative formula, and positivity of the derivative gives strict increase.
2. If \(x:\RR\to\RR\) is the assumed global solution of\[ x'=f(x), \qquad x(0)=0, \]then\[ F(x(t))=t \]for every \(t\in\RR\).
By the chain rule and step 1, \[ \frac{d}{dt}F(x(t)) = F'(x(t))x'(t) = \frac{1}{f(x(t))}f(x(t)) =1. \] At \(t=0\), \[ F(x(0))=F(0)=0. \] Therefore \(F(x(t))=t\).
3. The function \(F\) maps \(\RR\) onto \(\RR\), and hence\[ \lim_{u\to\infty}F(u)=\infty, \qquad \lim_{u\to-\infty}F(u)=-\infty. \]
Step 2 shows that every \(t\in\RR\) is a value of \(F\), namely \(F(x(t))\). Thus \(F\) is onto. Since \(F\) is strictly increasing by step 1, surjectivity forces the stated endpoint limits.
4. Define\[ G(u)\coloneqq\int_0^u\frac{ds}{g(s)}. \]Then\[ \lim_{u\to\infty}G(u)=\infty, \qquad \lim_{u\to-\infty}G(u)=-\infty. \]
Since \(0<g\le f\), \[ \frac1g\ge\frac1f. \] For \(u>0\), \[ G(u) = \int_0^u\frac{ds}{g(s)} \ge \int_0^u\frac{ds}{f(s)} =F(u), \] so step 3 gives \(G(u)\to\infty\).
For \(u<0\), \[ G(u) = -\int_u^0\frac{ds}{g(s)} \le -\int_u^0\frac{ds}{f(s)} =F(u), \] so step 3 gives \(G(u)\to-\infty\).
5. The function \(G:\RR\to\RR\) is a \(C^1\) bijection with \(G'(u)>0\) everywhere.
As in step 1, \[ G'(u)=\frac1{g(u)}>0, \] so \(G\) is strictly increasing. Step 4 gives both endpoint limits, hence \(G\) is onto. Strict increase gives injectivity.
6. Define\[ y(t)\coloneqq G^{-1}(t). \]Then \(y\) is defined for every \(t\in\RR\) and satisfies\[ y'=g(y), \qquad y(0)=0. \]
By step 5, the inverse is defined on all of \(\RR\). Since \(G'\) is everywhere nonzero, the inverse function theorem gives \[ y'(t) = \frac{1}{G'(y(t))} = g(y(t)). \] Also \(G(0)=0\), so \(y(0)=G^{-1}(0)=0\).
7. Hence the initial-value problem\[ x'=g(x), \qquad x(0)=0 \]has a solution defined on all of \(\RR\).
The function \(y\) constructed in step 6 is such a solution.
8. Q.E.D.
Step 7 is the required conclusion.