Proving that one solution always lies above the other

urbanist
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.

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

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

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

Show:

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

Homework Helper
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.

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

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

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

Show:

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

Starting from
$$H_0' + kH_0 = - y_0 \\ H_1' + kH_1 = -y_1$$
if we subtract the second from the first we obtain
$$(H_0 - H_1)' + k(H_0 - H_1) = y_1 - y_0$$
and if we multiply by $e^{kr}$ we find that
$$\frac{d}{dr} \left(e^{kr} (H_0 - H_1) \right) = e^{kr}(y_1 - y_0)$$
Now the right hand side is strictly positive for all $r \in (a,b]$, and so $e^{kr}(H_0 - H_1)$ is strictly increasing on $(a,b]$. Since it is initially strictly positive ($H_0(a) > H_1(a)$), it therefore remains strictly positive. It follows that $H_0(r) > H_1(r)$ for all $r \in (a,b]$ as required.

1 person
urbanist
Thanks a lot! That was a very elegant explanation :)

Now, can I prove the same for:

$H'(r)=-y(r)-H(r) k(-H(r))$ ,
where k is a strictly increasing positive function?

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Homework Helper
Thanks a lot! That was a very elegant explanation :)

Now, can I prove the same for:

$H'(r)=-y(r)-H(r) k(-H(r))$ ,
where k is a strictly increasing positive function?

Yes, assuming that continuous solutions of the now non-linear ODEs exist on $[a,b]$.

Starting from the same point as before, we obtain
$$(H_0 - H_1)' + k(-H_0)H_0 - k(-H_1)H_1 = y_1 - y_0$$
Now $k(-H_0)H_0 - k(-H_1)H_1$ is not obviously $\frac{f'(r)}{f(r)}(H_0 - H_1)$ for some strictly positive $f(r)$, so we can't use the previous method. However, we can use a different idea.

Suppose there exists $r_0 \in (a,b]$ such that $H_0(r_0) = H_1(r_0)$. It then follows, under the assumptions of the previous problem, that
$$(H_0 - H_1)'(r_0) = y_1(r_0) - y_0(r_0) > 0$$
which means that at that point $H_0 - H_1$ is strictly increasing. Thus locally $H_0(r) < H_1(r)$ if $r < r_0$ and $H_0(r) > H_1(r)$ if $r > r_0$.

It follows that there exists at most one such $r_0 \in (a,b]$, since once $H_0 > H_1$ the solutions can't intersect again in that interval; if they did then at that point $(H_0 - H_1)'$ would not be strictly positive, which is impossible.

Thus if $H_0(a) > H_1(a)$ then again it must follow that $H_0(r) > H_1(r)$ for all $r \in (a,b]$.

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urbanist
Beautiful solution, once again :)

I was wondering, whether the fact that y is not necessarily continuous matters.

Can I speak about $(H_0-H_1)'(r_0)$ and $y_0(r_0)-y_1(r_0)$?
Or do I need to deal with $(H_0-H_1)'(r\rightarrow r_0^-)$.
And then, can I claim $y_0(\rightarrow r_0^-)-y_1(r\rightarrow r_0^-)>0$? I think I can, since $y_0(r_0)-y_1(r_0)>0$ everywhere... But I'm not sure how to write that formally.

Homework Helper
Beautiful solution, once again :)

I was wondering, whether the fact that y is not necessarily continuous matters.

How badly behaved are the functions you are considering? If y is at least piecewise continuous then the above arguments will hold on each subinterval on which y is continuous.

I think there also constraints on y in order for H to exist in the first place.

Homework Helper
On closer inspection, it appears that if $y_1 - y_0$ and $k$ are not continuous then we can't necessarily show that $H_0 > H_1$.

We want to use the sign of $(H_0 - H_1)'$ to show that if $(H_0 - H_1)(r_0) = 0$ then $H_0 - H_1$ is strictly increasing in an open neighborhood of $r_0$. That is enough for us to conclude that if $H_0 - H_1$ is initially strictly positive then it remains strictly positive.

More formally, we want to show that there exists $\epsilon > 0$ such that if $|r - r_0| < \epsilon$ then $(H_0 - H_1)'(r) > 0$. Then, by the mean value theorem, $H_0 - H_1$ will be strictly increasing on that interval. It is not necessary for the application of the mean value theorem that $(H_0 - H_1)'$ be continuous. We do however need continuity, or at least suitable limiting behaviour, at $r_0$ in order to show that a suitable $\epsilon > 0$ exists.

Now if $(H_0 - H_1)'$ is continuous at $r_0$ then, since it is strictly positive at $r_0$, it follows that such an $\epsilon$ exists.

However if $(H_0 - H_1)'$ is not continuous at $r_0$ but
$$\lim_{r \to r_0^{+}} (H_0 - H_1)'(r) = \lim_{r \to r_0^{+}} (y_1 - y_0)(r) > 0$$ and
$$\lim_{r \to r_0^{-}} (H_0 - H_1)'(r) = \lim_{r \to r_0^{-}} (y_1 - y_0)(r) > 0$$
then that again guarantees the existence of such an $\epsilon$ (I am assuming here that
$$\lim_{r \to r_0} (k(-H_0)H_0 - k(-H_1)H_1)(r) = 0$$
and if that is not the case then it is not the case that the limits of $(H_0 - H_1)'$ and $y_1 - y_0$ are equal, and I don't see how to make further progress).

But the most that the condition $y_1(r) > y_0(r)$ gives us is that
$$\lim_{r \to r_0^{+}} (y_1 - y_0)(r) \geq 0$$
and
$$\lim_{r \to r_0^{-}} (y_1 - y_0)(r) \geq 0$$
if those limits actually exist. If one of those limits is zero then it might be that $(H_0 - H_1)'$ approaches zero only through negative values, and if one of those limits doesn't exist then we can't say anything at all.

urbanist
Well, we know that $y_1(r)>y_0(r)$ for all r.

If y is defined for all r, isn't it necessarily piecewise continuous?

k, at any rate, is continuous.

It seems to me that there has to be a neighborhood to the left of $r_0$ in which $y_1(r)-y_0(r)>k(-H_0)H_0-k(-H_1)H_1$...

Homework Helper
Well, we know that $y_1(r)>y_0(r)$ for all r.

If y is defined for all r, isn't it necessarily piecewise continuous?

In general, no. But since y is constrained to be strictly increasing https://www.physicsforums.com/newreply.php?do=newreply&p=4527279 [Broken] that the only discontinuities it can have are jump discontinuities, and there can only be a countable number of them. It follows that the only discontinuities $y_1 - y_0$ can have are jump discontinuities, and there can only be a countable number of them. It also follows that the necessary one-sided limits exist.

k, at any rate, is continuous.

Excellent; my approach should work.

It seems to me that there has to be a neighborhood to the left of $r_0$ in which $y_1(r)-y_0(r)>k(-H_0)H_0-k(-H_1)H_1$...

I'm not convinced that this is necessarily the case.

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