Let \(I=(a,b)\) be an interval containing \(1\), and let \(y=y(x)\) be a \(C^\infty\) function on \(I\) satisfying \[ y'=2^y-\frac1x. \] Prove rigorously that:
if \(y(1)>0\), then \(y\) is strictly increasing on \([1,b)\);
the same conclusion holds if \(y(1)=0\).
Solution
1. If \(c>1\) and \(y(c)=0\), then\[ y'(c)=1-\frac1c>0. \]
Substituting \(y(c)=0\) into the differential equation gives \[ y'(c) = 2^0-\frac1c = 1-\frac1c. \] Since \(c>1\), this quantity is positive.
2. Suppose \(y(1)>0\). Then\[ y(x)>0 \]for every \(x\in[1,b)\).
Assume otherwise. By continuity, there is some \(x_0>1\) with \(y(x_0)\leq0\), and the set \[ Z \coloneqq \{x\in[1,x_0]:y(x)=0\} \] is nonempty and closed. Let \[ c\coloneqq\min Z. \] Then \(c>1\) and \[ y(x)>0 \] for \(1\leq x<c\).
For every sufficiently small \(h<0\), \[ \frac{y(c+h)-y(c)}{h} <0, \] because the numerator is positive and the denominator is negative. Since \(y'(c)\) exists, passage to the limit \(h\to0^-\) gives \[ y'(c)\leq0. \] This contradicts step 1.
3. If \(y(1)>0\), then \(y\) is strictly increasing on \([1,b)\).
By step 2, \(y(x)>0\) for every \(x>1\). Hence \[ y'(x) = 2^{y(x)}-\frac1x > 1-\frac1x > 0 \] for every \(x>1\). For any \[ 1\leq x_1<x_2<b, \] the mean value theorem gives some \(c\in(x_1,x_2)\) with \[ y(x_2)-y(x_1) = y'(c)(x_2-x_1) > 0. \] This proves part (a).
4. Suppose \(y(1)=0\). Then\[ y'(1)=0 \]and\[ y''(1)=1. \]
The differential equation gives \[ y'(1) = 2^0-1 = 0. \] Differentiating the differential equation yields \[ y'' = (\log 2)2^y y' + \frac1{x^2}. \] Evaluating at \(x=1\) gives \[ y''(1) = (\log 2)2^0y'(1)+1 = 1. \]
5. If \(y(1)=0\), then there is \(\delta>0\) such that\[ y(x)>0 \]for every \(x\in(1,1+\delta]\).
Since \(y''(1)=1\) and \(y''\) is continuous, there is \(\delta>0\) such that \[ y''(x)>0 \] for \(1\leq x\leq1+\delta\), after shrinking \(\delta\) so that this interval lies in \(I\). Hence \(y'\) is strictly increasing there. By step 4, \[ y'(1)=0, \] so \[ y'(x)>0 \] for \(1<x\leq1+\delta\). Therefore \(y\) is strictly increasing on this interval, and since \(y(1)=0\), one has \(y(x)>0\) for every \(x\in(1,1+\delta]\).
6. If \(y(1)=0\), then\[ y(x)>0 \]for every \(x\in(1,b)\).
Step 5 gives positivity immediately to the right of \(1\). If \(y\) vanished again at some point, choose its first zero \(c>1\) after this initial positive interval. Then \(y(x)>0\) for \(x<c\) sufficiently close to \(c\). Exactly as in step 2, the left derivative at \(c\) would satisfy \[ y'(c)\leq0, \] contradicting step 1.
7. If \(y(1)=0\), then \(y\) is strictly increasing on \([1,b)\).
By step 6, \(y(x)>0\) for every \(x>1\). The same calculation as in step 3 gives \[ y'(x)>0 \] for every \(x>1\). The mean value theorem then shows that \[ y(x_2)>y(x_1) \] whenever \(1\leq x_1<x_2<b\). This proves part (b).
8. Q.E.D.
Step 3 proves part (a), and step 7 proves part (b).