# Showing inequalities for solutions to ODE

Let $f_1$ and $f_2$ be two continuous functions on $[a,b]\times\Bbb R$ such that $$f_1(x,y) < f_2(x,y).$$

Let $\varphi_1$ and $\varphi_2$ be two $C^1$ functions on $[a,b]$ which are respectively solutions of the following equations $$y'=f_1(x,y)\; and\; y'=f_2(x,y)$$ Assume there is $x_0 \in [a,b)$ such that $\varphi_1(x_0)=\varphi_2(x_0)$

1. Show that there is a constant $\delta>0$ such that $\varphi_1(x)<\varphi_2(x)$ for $x \in (x_0,x_0+\delta].$

2. Deduce that $\varphi_1(x) \le \varphi_2(x)$ for $x \in [x_0,b].$

3. Show that $\varphi_1(x) \ge \varphi_2(x)$ for $x \in [a,x_0].$

By hypotesis we have

$$\varphi_1'(x)<\varphi_2'(x)$$

Integrating on $[x_0, t]$ where $t<b$ $$\int_{x_0}^{t}\varphi_1'(x)dx\leq\int_{x_0}^{t}\varphi_2'(x)dx$$

then by the fundamental theorem of calculus

$$\varphi_1(t)-\varphi_1(x_0)\leq \varphi_2(t)-\varphi_2(x_0)$$ by hipotesis $$\varphi_1(x_0)<\varphi_2(x_0)$$ adding the last two inequalities we get $$\varphi_1(t)<\varphi_2(t)$$

since t was arbitrary we get 1 and 2.

The last property is not correct, take $\varphi_1(t)=t^2$ and $\varphi_2(t)=t$and $t \in[0,1/4]$ then we have $$\varphi_1(1/8)<\varphi_2(1/8)$$ $$\varphi_1'(t)<\varphi_2'(t)$$

but $$\varphi_1(t)<\varphi_2(t) \;\;\; t\in [0,1/4]$$

• sorry, I have changed the condition such that $\varphi_1(x_0)=\varphi_2(x_0)$ – Joash Sep 3 '14 at 6:14
• when you meant by the hypothesis, do you refer to this line: Let $f_1$ and $f_2$ be two continuous functions on $[a,b]\times\Bbb R$ such that $$f_1(x,y) < f_2(x,y).$$ – Joash Sep 3 '14 at 6:20