numerical solution of differential equations

[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.

[Hitchhiker 42]

Originally Posted by Ether

In post14 you wrote:


I am assuming what you actually meant by that is

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

Is that correct?

Yes, sorry about that. That was what I meant.

[Ether]

Originally Posted by Hitchhiker 42

Yes, sorry about that. That was what I meant.

OK.

x(t) = 1/4*t^3 + 1 is the analytical solution to the Initial Value Problem

a(t) = sqrt(3*v(t))

with initial values*

x(0.01) = 1.00000025

and

v(0.01) = 7.5e-5

As you can see from the spreadsheet, the Midpoint method gives much better results for that IVP than Forward Euler and Forward/Backward Euler.


* selected to avoid the stationary point at t=0

[Ether]

Originally Posted by Hitchhiker 42

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

Originally Posted by Ether

I am assuming what you actually meant by that is

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

Is that correct?

Originally Posted by Hitchhiker 42

Yes, sorry about that. That was what I meant.

Originally Posted by Hitchhiker 42

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

Yes, but you got x and t confused.

given:

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

... take time derivative of x(t) to get:

x'(t) = (3/4)*t^2

...take time derivative of x'(t) to get:

x''(t) = (3/2)*t


Now solve for the differential equation:

solve x'(t) for t:

x'(t) = (3/4)*t^2 => t=sqrt(4x'(t)/3)

... and then substitute for t in x''(t):

x''(t) = (3/2)*t = (3/2)*sqrt(4x'(t)/3) = sqrt(3*x'(t))