# Riccati Comparison Principle

This is a proposition in a book I'm reading whose proof I am unsure of how to fill in the details

Proposition

Suppose we have two smooth functions $$\rho_{1,2} : (0,b) \to \mathbb{R}$$ such that $$\rho_1'+ \rho_1^2 \leq \rho_2'+ \rho_2^2$$ then $$\rho_2 - \rho_1 \geq \limsup_{t \to 0}\,(\, \rho_2(t) - \rho_1(t)\,)$$

Question

The author says that this follows from the easily verified fact that the function $$(\rho_2 - \rho_1) e^F$$ is increasing where $$F$$ is the antiderivative of $$\rho_2 + \rho_1$$ on $$(0, b)$$. This would be true if $$\rho_2 - \rho_1$$ is increasing but I don't think that this is the case. How does the lim sup inequality follow from the fact that $$(\rho_2 - \rho_1) e^F$$ is increasing?

• Which book? 
– Dap
Jan 25, 2019 at 4:53
• Riemannian Geometry by Peter Petersen. In the third edition, the Riccati Comparison Principle is on page 254. Jan 29, 2019 at 3:08

The book is wrong and it's easy to find counterexamples. For example if $$ρ_1(x)=1/(x+2)$$ and $$ρ_2(x)=1/(x+1)$$ then $$ρ_i'+ρ_i^2=0$$ and $$ρ_2(0)-ρ_1(0)=1/2$$ but $$ρ_2(1)-ρ_1(1)=1/6.$$
The author may have been thinking of something like the following argument which I will paraphrase from "New comparison theorems in Riemannian geometry" by Y. Han and coauthors, Theorem 5.1. Suppose $$ρ_1$$ and $$ρ_2$$ are both $$1/t+O(1).$$ Define $$ϕ_i(t)=t\exp(\int_0^t (ρ_i(s)-1/s)ds).$$ Then each $$ϕ_i$$ is a $$C^1$$ function on $$[0,b)$$ (i.e. the right-sided derivative exists at zero - in fact it's $$1$$), smooth and strictly positive on $$(0,b),$$ with $$ϕ_i(0)=0.$$ Differentiating gives $$ϕ'_i(t)=ρ_i(t) ϕ_i(t)$$ and hence $$ϕ''_i(t)=(ρ'_i(t)+ρ_i(t)^2)ϕ_i(t).$$ This leads to $$(ϕ_2'ϕ_1-ϕ_1'ϕ_2)'=ϕ_2''ϕ_1-ϕ_1''ϕ_2=((ρ'_2+ρ_2^2)-(ρ'_1+ρ_1^2))ϕ_1ϕ_2\geq 0.$$ Using the fact that $$ϕ_2'(0)ϕ_1(0)-ϕ_1'(0)ϕ_2(0)=0$$ we get $$ϕ_2'ϕ_1\geq ϕ_1'ϕ_2.$$ Using our previous equation for $$ϕ_i',$$ this means $$ρ_2(t)\geq ρ_1(t).$$
Note that \begin{align}\frac d{dt}\left[(\rho_2-\rho_1)e^F\right]&=(\rho'_2-\rho'_1)e^F+(\rho_2-\rho_1)(\rho_2+\rho_1)e^F\\ &=e^F\left(\rho'_2-\rho'_1+\rho_2^2-\rho_1^2\right) \end{align} and since $e^F>0$, your inequality implies that the derivative of $(\rho_2-\rho_1)e^F$ is positive. So the function $(\rho_2-\rho_1)e^F$ is increasing.
On the other hand, a careless approach with some possible errors could be this: \begin{align} \rho_1'+ \rho_1^2 \leq \rho_2'+ \rho_2^2 &\implies\rho_1'-\rho_2' \leq \rho_2^2- \rho_1^2\\ &\implies-\dfrac{\rho_1'-\rho_2'}{\rho_1-\rho_2}\ge\rho_1+\rho_2\\ &\implies-\ln|\rho_1-\rho_2|\ge F\\ &\implies\frac 1{|\rho_1-\rho_2|}\le e^F\implies(\rho_2-\rho_1)e^F\ge 1 \end{align} or something like that....
• Right, but how does the fact that $\rho_2 - \rho_1 e^F$ is increasing imply that $\rho_2 - \rho_1 \geq limsup _{t \to 0}(\rho_2(t) - \rho_1(t))$ Jan 9, 2017 at 13:29