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:

$\displaystyle H'(r) = -y(r) - k H(r) $

k is a constant.

y is strictly increasing, but not continuous.

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

$\displaystyle (H_x, y_x) $ denotes solution x.

$\displaystyle H_1(a)<H_0(a)<0 $.

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

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

Show:

$\displaystyle H_1(r)<H_0(r) $ for all $\displaystyle r \in (a,b] $.

Re: Proving that one solution always lies above the other

What does "solution x" mean? What is "x" here?

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:

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

where k is a strictly increasing positive function.

Thanks a lot!

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 $\displaystyle H_1(x_0)< H_2(x_0)$ for some $\displaystyle x_0$ and $\displaystyle H_1(x_1)> H_2(x_1)$ for some $\displaystyle x_1$, there must exist $\displaystyle x_2$ between $\displaystyle x_0$ and $\displaystyle x_1$ such that $\displaystyle H_1(x_2)= H_2(x_2)$.

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 $\displaystyle x_2$.