# Difficulty Understanding Conjugate points of a differential equation.

I was struggling to understand this footnote from the book by Gelfand. I couldnot understand how he takes the solution of the differential equation outside the interval $$[a,b]$$ and also how will the solution of $$(26)$$ with initial condition $$h(a-\epsilon)=0$$ $$h'(a-\epsilon)=1$$ Cannot vanish on whole interval $$[a,b]$$

• A conjugate point is a second/the next greater root of $h$? Nov 6 '19 at 13:20

## 1 Answer

$$P,Q$$ are defined outside $$[a,b]$$ by the implicit assumptions of the cited footmark. Also, if $$P$$ is non-zero on $$[a,b]$$, then it is also non-zero on some open neighborhood of that interval. A linear ODE has solutions that can be extended to the full intervals over which the coefficients are continuous and the leading coefficient is non-zero.

This all makes it possible to take the initial condition at $$a−ϵ$$ for a sufficiently small positive $$ϵ$$.

For the overall claim, the conjugate point is a continuous function of the initial point. If it is greater than $$b$$ for initial point $$a$$, then that same is true for initial points in a neighborhood of $$a$$, simply by continuity.

• "the conjugate point is a continuous function of the initial point", i dont know about this can you shed some light about it. Nov 7 '19 at 8:57
• This is the meaning of the first sentence: "The solution ... depends continuously on the initial conditions" and by the implicit function theorem, the root depends continuously on the solution, as it has to be a simple one (as all roots of a non-trivial solution of a 2nd order linear DE have to be). Nov 7 '19 at 9:35