How to compare the solutions of Jacobi's equations?

Question

If $f$ is the solution of Jacobi's equation $$f^{\prime\prime}(t) + K(t) f(t) =0,$$ $$ f(0)=0,\quad f^{\prime}(0)=1,$$ where $K$ is a real-valued continous function of $t$, and $f_1$ is the solution of $$f_1^{\prime\prime}(t) + Cf_1(t)=0 ,$$ $$ f_1(0)=0,\quad f_1^{\prime}(0)=1,$$ where $C$ is a constant, and $K\geq C$. Then how can I show that $f\leq f_1$?

Answer 1

Since $$ f(0)=f_1(0),f'(0)=f'_1(0)=1$$ there is $t^*>0$ such that $f(t)>0, f_1(t)>0$ for $t\in(0,t^*)$. Multiplying the first eqn by $f_1$ and then integrating from $0$ to $t$ give $$ \int_0^tf''(s)f_1(s)ds+\int_0^tK(s)f(s)f_1(s)ds=0 $$ or $$ f'(s)f_1(s)\bigg|_0^t-\int_0^tf'(s)f_1'(s)ds+\int_0^tK(s)f(s)f_1(s)ds=0 $$ or $$ f'(t)f_1(t)-\int_0^tf'(s)f_1'(s)ds+\int_0^tK(s)f(s)f_1(s)ds=0. \tag{1}$$ Similarly $$ f_1'(t)f(t)-\int_0^tf'(s)f_1'(s)ds+\int_0^tCf(s)f_1(s)ds=0. \tag{2}$$ (1)-(2) gives $$ f'(t)f_1(t)-f(t)f_1'(t)=\int_0^t(C-K(t))f(s)f_1(s)ds. $$ So for $t\in(0,t^*)$, $$ \bigg(\frac{f(t)}{f_1(t)}\bigg)'=\frac{1}{f_1^2(t)} \int_0^t(C-K(t))f(s)f_1(s)ds\le0. $$ Namely, $\frac{f(t)}{f_1(t)}$ is decreasing for $t\in(0,t^*)$. Thus for $0<s<t<t^*)$, $$ \frac{f(t)}{f_1(t)}\le\frac{f(s)}{f_1(s)} $$ Letting $s\to0^+$ gives $$ \frac{f(t)}{f_1(t)}\le\frac{f'(0)}{f_1'(0)}=1$$ which implies $$ f(t)\le f_1(t), \text{ for }t\in(0,t^*$). $$