My Math Forum Question about solving ODE

 Differential Equations Ordinary and Partial Differential Equations Math Forum

 February 19th, 2016, 10:54 PM #1 Newbie   Joined: Feb 2016 From: universe Posts: 7 Thanks: 0 Question about solving ODE hi guys, Hope it's okay, but I typed out my question to make the math equations more clear. I had a general question about solving ODEs. So as we know, if you have an equation of the form: thanks in advance for your help
 February 20th, 2016, 02:00 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,828 Thanks: 2160 I would suggest solving for each set of conditions separately. It may be possible to spot whether the solutions obtained are compatible.
February 20th, 2016, 09:14 PM   #3
Newbie

Joined: Feb 2016
From: universe

Posts: 7
Thanks: 0

Quote:
 Originally Posted by skipjack I would suggest solving for each set of conditions separately. It may be possible to spot whether the solutions obtained are compatible.

Yes, the solutions do intersect. However, the solution point is a singularity which is something I don't want. I would like to solve for both simultaneously, if possible, so that I get a smooth continuous function.

I was sure that there would be some kind of approximation method to do this, but I've searched and can't find one. Maybe I'm missing something? Any other ideas?

 February 21st, 2016, 03:18 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,828 Thanks: 2160 If there is a "solution" that "has a singularity", it's usually the case that every solution has a singularity somewhere. I don't understand your remark "the solution point is a singularity" as a function and its derivatives aren't defined at a singularity.
February 23rd, 2016, 09:10 PM   #5
Newbie

Joined: Feb 2016
From: universe

Posts: 7
Thanks: 0

Quote:
 Originally Posted by skipjack If there is a "solution" that "has a singularity", it's usually the case that every solution has a singularity somewhere. I don't understand your remark "the solution point is a singularity" as a function and its derivatives aren't defined at a singularity.
Maybe I worded it incorrectly, but I think the below graph will show I'm talking about. If I use the numerical integration method at point x0 right now I am getting the blue curve. If I then use the method again at point x1, I get the red curve. As you can see, they intersect at a singularity point.

What I'm looking for is a method for numerical integration that SIMULTANEOUSLY uses the information at BOTH points x0, v0, a0 and x1, v1, a1 to create a smoother curve, shown below.

There has to be some way to do multi-point numerical integration. The method would need to somehow use the information for both points and at the same time or in some intelligent way to create the overall curve. Are there advanced methods that do this?

 February 23rd, 2016, 11:13 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,828 Thanks: 2160 The two curves in your first illustration show the common situation that they can't be joined where they intersect because the original equation wouldn't be satisfied at that point. You can't change the curves to produce a smooth join without producing a function that doesn't satisfy the equation you solved. In particular cases, it sometimes happens that the two curves intersect at a point where they have the same slope (and the same second derivative), in which case, more than one set of conditions can be satisfied by a single "smooth" solution.
February 24th, 2016, 01:06 AM   #7
Newbie

Joined: Feb 2016
From: universe

Posts: 7
Thanks: 0

Quote:
 Originally Posted by skipjack The two curves in your first illustration show the common situation that they can't be joined where they intersect because the original equation wouldn't be satisfied at that point. You can't change the curves to produce a smooth join without producing a function that doesn't satisfy the equation you solved. In particular cases, it sometimes happens that the two curves intersect at a point where they have the same slope (and the same second derivative), in which case, more than one set of conditions can be satisfied by a single "smooth" solution.
Right. This is a consequence of performing numerical integration on x0, and then separately on x1. That's why I thought there must be some method of performing it simultaneously using both x0 and x1.

For example, if I start at x0 and iterate forward with +e. At the same time I start at x1 and iterate backward with -e. Then using some intelligent way, perhaps using weighted distance or some other approach, I could form the overall green curve shown above.

1) Are there any methods at all that attempt to tackle/investigate this issue?

2) What, in your opinion, is the best way to create a smooth curve using both x0(blue curve) and x1(red curve)?

thanks again for all your help. Really appreciate it.

 February 24th, 2016, 03:46 AM #8 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra Given the picture you give, there must be a stationary point somewhere near the intersection of the two approximations. Can you use that fact? Is there any benefit in writing $y=\dot x$ and studying the phase space (x,y) of the system? Last edited by skipjack; February 24th, 2016 at 04:39 AM.
 February 24th, 2016, 04:33 AM #9 Global Moderator   Joined: Dec 2006 Posts: 20,828 Thanks: 2160 In general, no "green curve" exists that satisfies the equation. If, for example, your differential equation has an arbitrary straight line as its general solution, there's little point in trying to join a "red" line that satisfies one set of conditions to a separate "blue" line that satisfies another set of conditions.
February 24th, 2016, 08:19 PM   #10
Newbie

Joined: Feb 2016
From: universe

Posts: 7
Thanks: 0

Quote:
 Originally Posted by v8archie Given the picture you give, there must be a stationary point somewhere near the intersection of the two approximations. Can you use that fact? Is there any benefit in writing $y=\dot x$ and studying the phase space (x,y) of the system?
What exactly do you mean by stationary point? The plots I show are x position versus velocity.

Quote:
 Originally Posted by skipjack In general, no "green curve" exists that satisfies the equation. If, for example, your differential equation has an arbitrary straight line as its general solution, there's little point in trying to join a "red" line that satisfies one set of conditions to a separate "blue" line that satisfies another set of conditions.
I see. The solutions to my problem all have a "parabolic" shape to them. So for the purposes of my work, I needn't worry about straight lines or other funky curve shapes.

I'm just surprised that there aren't other approximate methods besides the simple numerical/perturbation approach I stated in my first post. I have found things like the Newmark-beta method, but unless I'm mistaken, that is simply a variation. I would need some kind of multi-point technique.

 Tags ode, question, solving

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post ahsanalisony New Users 0 May 21st, 2014 07:27 PM dthomas86 Algebra 1 January 29th, 2013 02:21 PM xnarutox Algebra 5 September 10th, 2010 06:04 PM Striker2 Algebra 7 December 3rd, 2009 08:39 PM Houshmandz Algebra 2 June 11th, 2009 10:22 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top