Proving that one solution always lies above the other

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Hi all,

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

I need to prove the following:
[itex] H'(r) = -y(r) - k H(r) [/itex]

k is a constant.

y is strictly increasing, but not continuous.

Let [itex] (a,b]\subset R [/itex].

[itex] (H_x, y_x) [/itex] denotes solution x.

[itex] H_1(a)<H_0(a)<0 [/itex].

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

[itex] y_1(s)>y_0(s) [/itex] for all [itex]s \in (a,b] [/itex].

Show:

[itex] H_1(r)<H_0(r) [/itex] for all [itex] r \in (a,b] [/itex].
 

Answers and Replies

  • #2
pasmith
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Hi all,

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

I need to prove the following:
[itex] H'(r) = -y(r) - k H(r) [/itex]

k is a constant.

y is strictly increasing, but not continuous.

Let [itex] (a,b]\subset R [/itex].

[itex] (H_x, y_x) [/itex] denotes solution x.

[itex] H_1(a)<H_0(a)<0 [/itex].

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

[itex] y_1(s)>y_0(s) [/itex] for all [itex]s \in (a,b] [/itex].

Show:

[itex] H_1(r)<H_0(r) [/itex] for all [itex] r \in (a,b] [/itex].
Starting from
[tex]
H_0' + kH_0 = - y_0 \\
H_1' + kH_1 = -y_1
[/tex]
if we subtract the second from the first we obtain
[tex]
(H_0 - H_1)' + k(H_0 - H_1) = y_1 - y_0
[/tex]
and if we multiply by [itex]e^{kr}[/itex] we find that
[tex]
\frac{d}{dr} \left(e^{kr} (H_0 - H_1) \right) = e^{kr}(y_1 - y_0)
[/tex]
Now the right hand side is strictly positive for all [itex]r \in (a,b][/itex], and so [itex]e^{kr}(H_0 - H_1)[/itex] is strictly increasing on [itex](a,b][/itex]. Since it is initially strictly positive ([itex]H_0(a) > H_1(a)[/itex]), it therefore remains strictly positive. It follows that [itex]H_0(r) > H_1(r)[/itex] for all [itex]r \in (a,b][/itex] as required.
 
  • #3
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Thanks a lot! That was a very elegant explanation :)

Now, can I prove the same for:

[itex]H'(r)=-y(r)-H(r) k(-H(r))[/itex] ,
where k is a strictly increasing positive function?
 
Last edited:
  • #4
pasmith
Homework Helper
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Thanks a lot! That was a very elegant explanation :)

Now, can I prove the same for:

[itex]H'(r)=-y(r)-H(r) k(-H(r))[/itex] ,
where k is a strictly increasing positive function?
Yes, assuming that continuous solutions of the now non-linear ODEs exist on [itex][a,b][/itex].

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

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

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

Thus if [itex]H_0(a) > H_1(a)[/itex] then again it must follow that [itex]H_0(r) > H_1(r)[/itex] for all [itex]r \in (a,b][/itex].
 
Last edited:
  • #5
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Beautiful solution, once again :)

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

Can I speak about [itex](H_0-H_1)'(r_0)[/itex] and [itex]y_0(r_0)-y_1(r_0)[/itex]?
Or do I need to deal with [itex](H_0-H_1)'(r\rightarrow r_0^-)[/itex].
And then, can I claim [itex]y_0(\rightarrow r_0^-)-y_1(r\rightarrow r_0^-)>0[/itex]? I think I can, since [itex]y_0(r_0)-y_1(r_0)>0[/itex] everywhere... But I'm not sure how to write that formally.
 
  • #6
pasmith
Homework Helper
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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.
 
  • #7
pasmith
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On closer inspection, it appears that if [itex]y_1 - y_0[/itex] and [itex]k[/itex] are not continuous then we can't necessarily show that [itex]H_0 > H_1[/itex].

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

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

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

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

But the most that the condition [itex]y_1(r) > y_0(r)[/itex] gives us is that
[tex]\lim_{r \to r_0^{+}} (y_1 - y_0)(r) \geq 0[/tex]
and
[tex]\lim_{r \to r_0^{-}} (y_1 - y_0)(r) \geq 0[/tex]
if those limits actually exist. If one of those limits is zero then it might be that [itex](H_0 - H_1)'[/itex] approaches zero only through negative values, and if one of those limits doesn't exist then we can't say anything at all.
 
  • #8
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Well, we know that [itex]y_1(r)>y_0(r)[/itex] 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 [itex]r_0[/itex] in which [itex]y_1(r)-y_0(r)>k(-H_0)H_0-k(-H_1)H_1[/itex]...
 
  • #9
pasmith
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Well, we know that [itex]y_1(r)>y_0(r)[/itex] 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 [itex]y_1 - y_0[/itex] 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 [itex]r_0[/itex] in which [itex]y_1(r)-y_0(r)>k(-H_0)H_0-k(-H_1)H_1[/itex]...
I'm not convinced that this is necessarily the case.
 
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