Book at page 80 says ”it is usually true that $$S^{\prime \prime }\lll \left ( S^{\prime }\right ) ^{2}$$” as $$x\rightarrow x_{0}$$. What are the exceptions to this? Since it is easy to ﬁnd counter examples.

For $$y^{\prime \prime }=\sqrt{x}y$$, as $$x\rightarrow 0^{+}$$. Using $$y\left ( x\right ) =e^{S_{0}\left ( x\right ) }$$ and substituting, gives the ODE$$S_{0}^{\prime \prime }+\left ( S_{0}^{\prime }\right ) ^{2}=\sqrt{x}\tag{1}$$

If we follow the book, and drop $$S_{0}^{\prime \prime }$$ relative to $$\left ( S^{\prime }\right ) ^{2}$$ then

\begin{align*} \left ( S_{0}^{\prime }\right ) ^{2} & =x^{\frac{1}{2}}\\ S_{0}^{\prime } & =\omega x^{\frac{1}{4}} \end{align*}

So, lets check the assumption. Since $$S_{0}^{\prime \prime }=\omega \frac{1}{4}x^{-\frac{3}{4}}$$. Therefore (ignoring all multiplicative constants) \begin{align*} & S_{0}^{\prime \prime }\overset{?}{\lll }\left ( S_{0}^{\prime }\right ) ^{2}\\ & \frac{1}{x^{\frac{3}{4}}}\overset{?}{\lll }x^{\frac{1}{2}} \end{align*}

No. Did not check out. Since when $$x\rightarrow 0^{+}$$ the LHS blow up while the RHS goes to zero. So in this example, this ”rule of thumb” did not work out, and it is the other way around. Assuming I did not make a mistake, when the book says ”it is usually true”, it will good to know under what conditions is this true or why did the book say this?

This will help in solving these problem.