numerical solution of differential equations

[Ether]

Originally Posted by Richard Wallace

maybe just antsy for the 2017 FIRST Robotics Competition kickoff.

Yes I am very much looking forward to the Kickoff.

And I have some long-overdue hardware to return to you

[Hitchhiker 42]

Originally Posted by Richard Wallace

Hi Russ!
Should we expect, in general, that Euler integration will be well suited to approximate monotonic systems, while Midpoint integration gives better results for periodic systems? If so, why?

As I understand it, the main difference between the Euler method and the Midpoint method is that the midpoint method takes the slope by connecting a point behind and a point ahead of the given point. The Euler method just takes the slope at that point at extrapolates for that step. Please, do correct me if I'm wrong.

So, a monotonic function only increases (or decreases). It seems Euler integration would be better suited (read: more accurate) because the midpoint method depends on points behind the given point that's being calculated, which will tend to keep the slope smaller than it should be, whereas because the function doesn't tend to change direction (up or down) as much (it can only go one direction - monotonic), approximating ahead will tend to be better than approximating while taking into account behind the current point as well.

[GeeTwo]

What surprises me more than the gain in amplitude for Euler (which is pretty easy to guess if you consider what happens to energy at different points) is the excellent prediction of the period. I'll have to give this a look.

I was able to do a version without the two extra columns that tracked pretty closely, using the parabolic formula for constant acceleration to calculate the next position, and the average acceleration assuming constant jerk (x''') to calculate the next velocity.

x[n+1]   =  x[n]  + dt*(x'[n]  + x''[n]*dt/2)
x''[n+1] = -x[n+1] 
x'[n+1]  =  x'[n] + dt*(x''[n] + x''[n+1])/2

[Ether]

Originally Posted by Hitchhiker 42

As I understand it,...the midpoint method takes the slope by connecting a point behind and a point ahead of the given point...Please, do correct me if I'm wrong.

Where did you come by this understanding?

It seems Euler integration would be better suited [for monotonic function]...

You could easily modify the XLS I posted to test this hypothesis

[Hitchhiker 42]

Originally Posted by Ether

You could easily modify the XLS I posted to test this hypothesis

I've attached your spreadsheets modified for 1/4*x^3 + 1.
It seems that the Euler method does approximate it better.

[Ether]

@Mark: Where did you get the accel formula for columns D and G.

[Hitchhiker 42]

Originally Posted by Ether

@Mark: Where did you get the accel formula for columns D and G.

Not sure. Must have been really tired . I did that over, and also added an error column to show the difference between Xa and the predicted X (Xm or Xe).
I set dt = 0.01 s and compared the errors at t = 1, 2, 4 s

For 1s:
Midpt error: 0.598
Euler error: 0.585

For 2s:
Midpt error: 2.834
Euler error: 2.748

For 4s:
Midpt error: 50.070
Euler error: 48.113

Seems that, at least for this function, the Euler method holds up better. Tomorrow, I'll try it with a function where the second derivative isn't always positive on the interval I'm testing.

[Hitchhiker 42]

And the new sheets:

[Ether]

Originally Posted by Hitchhiker 42

And the new sheets:

where did you get the revised accel formula?

[Hitchhiker 42]

Originally Posted by Ether

where did you get the revised accel formula?

I took the derivative of the actual velocity function. Is this not correct?

[GeeTwo]

Originally Posted by Hitchhiker 42

I took the derivative of the actual velocity function. Is this not correct?

No, as I read it you're setting x''=3x/2. This would be solved by a function of the form x=Ae^3x/2

[Hitchhiker 42]

Originally Posted by GeeTwo

No, as I read it you're setting x''=3x/2. This would be solved by a function of the form x=Ae^3x/2

Could you explain what this means? I'm not sure I understand...

[Ether]

Originally Posted by Hitchhiker 42

Could you explain what this means? I'm not sure I understand...

In post14 you wrote:

Originally Posted by Hitchhiker 42

I've attached your spreadsheets modified for 1/4*x^3 + 1.

I am assuming what you actually meant by that is

x(t) = 1/4*t^3 + 1

Is that correct?

If so, then in order to investigate the performance of Euler vs Midpoint numerical solution (in the context of this thread) you need to convert that to an initial value problem (IVP) consisting of a differential equation (involving only x, x', and x'') plus initial values for x(t) and x'(t) at some point t=to.