Proving that one solution always lies above the other

  • Thread starter urbanist
  • Start date
In summary, we discussed the problem of proving H_1(r) < H_0(r) for all r \in (a,b] given the initial conditions H_1(a) < H_0(a) < 0 and H_0(s) < 0, H_1(s) < 0 for all s \in (a,b] and y_1(s) > y_0(s) for all s \in (a,b], where y is a strictly increasing but not necessarily continuous function. We showed that this can be proven using the non-linear ODE H'(r) = -y(r) - kH(r), where k is a constant, by subtracting the two equations for H_0
  • #1
urbanist
9
0
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].
 
Physics news on Phys.org
  • #2
urbanist said:
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.
 
  • Like
Likes 1 person
  • #3
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
urbanist said:
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
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
urbanist said:
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
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
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
urbanist said:
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.
 
Last edited by a moderator:

1. How do you prove that one solution always lies above the other?

There are several mathematical techniques that can be used to prove that one solution always lies above the other. One common approach is to use the method of induction, where you first show that the statement is true for a base case, and then show that if it is true for one case, it must also be true for the next case. Another approach is to use the properties of inequalities, such as the transitive property or the addition/subtraction property, to manipulate the expressions and show that one solution will always be greater than the other.

2. Can you provide an example of using induction to prove that one solution always lies above the other?

Yes, for example, let's say we want to prove that for all natural numbers n, the sum of the first n natural numbers is greater than n squared. We can first show that this statement is true for n=1, since 1 is a natural number and 1 is greater than 1 squared. Then, we assume that the statement is true for some arbitrary value k, and use this assumption to prove that it must also be true for k+1. By doing this, we have shown that if the statement is true for one case, it must also be true for the next case, and therefore it must be true for all natural numbers n.

3. How can the properties of inequalities be used to prove that one solution always lies above the other?

The transitive property of inequalities states that if a is greater than b, and b is greater than c, then a is greater than c. This can be used to show that if one solution is greater than another, and that other solution is greater than a third, then the first solution must also be greater than the third. The addition/subtraction property states that if a is greater than b, then adding or subtracting the same value from both sides will still result in a being greater than b. This can be used to manipulate expressions and show that one solution will always be greater than another.

4. Are there any other techniques that can be used to prove that one solution always lies above the other?

Yes, there are other techniques that can be used, such as using the properties of limits or derivatives in calculus. These techniques involve using mathematical concepts to show that as the values of the solutions approach infinity, one will always be greater than the other. However, these techniques may not always be applicable depending on the specific problem being solved.

5. Can a computer be used to prove that one solution always lies above the other?

Yes, a computer can be used to prove that one solution always lies above the other by using mathematical software or programming languages to perform calculations and manipulate expressions. However, it is important to note that the results of such calculations are only as accurate as the inputs and algorithms used, and may not necessarily constitute a rigorous mathematical proof.

Similar threads

  • Differential Equations
Replies
1
Views
588
Replies
3
Views
445
Replies
1
Views
701
  • Differential Equations
Replies
1
Views
623
  • Differential Equations
Replies
1
Views
716
  • Advanced Physics Homework Help
Replies
1
Views
814
  • Differential Equations
Replies
4
Views
1K
Replies
1
Views
524
  • Differential Equations
Replies
2
Views
3K
  • Calculus
Replies
2
Views
1K
Back
Top