# Proving that one solution always lies above the other

• Oct 2nd 2013, 11:40 AM
urbanist
Proving that one solution always lies above the other
Hi all,

I'd be very happy if you could help me solve a problem in my research.

I need to prove the following:
$H'(r) = -y(r) - k H(r)$

k is a constant.

y is strictly increasing, but not continuous.

Let $(a,b]\subset R$.

$(H_x, y_x)$ denotes solution x.

$H_1(a).

$H_0(s)<0, H_1(s)<0$ for all $s\in(a,b]$.

$y_1(s)>y_0(s)$ for all $s \in (a,b]$.

Show:

$H_1(r) for all $r \in (a,b]$.
• Oct 2nd 2013, 11:51 AM
HallsofIvy
Re: Proving that one solution always lies above the other
What does "solution x" mean? What is "x" here?
• Oct 2nd 2013, 01:52 PM
urbanist
Re: Proving that one solution always lies above the other
x is just an indicator. A solution is a pair of functions y and H.

Ultimately, I need to prove it also for:

$H'(r)=-y(r)-H(r) k(-H(r))$,

where k is a strictly increasing positive function.

Thanks a lot!
• Oct 6th 2013, 12:53 PM
HallsofIvy
Re: Proving that one solution always lies above the other
The solutions of a differential equation obviously must be "differentiable" so must be continuous. So if $H_1(x_0)< H_2(x_0)$ for some $x_0$ and $H_1(x_1)> H_2(x_1)$ for some $x_1$, there must exist $x_2$ between $x_0$ and $x_1$ such that $H_1(x_2)= H_2(x_2)$.

But by the "existence and uniqueness" theorem, since there can be only one solution to this first order differential equation having a given value at $x_2$.