Difference Equation

1. Feb 10, 2016

Cyrus

I have a difference equation which is given as:

ΔP = e^P [1]

where we can re-write ΔP as: Δ P = P_2 - P_1, where the subscripts indicate two distinct discrete time indices.

What I would like to do: is to convert this into a continuous time expression and solve it, if possible.

In order to help give some insight, I will solve a similar type of problem where I know the solution.

ΔP = c_1 [2]

Note here, that in all cases we are running the recursive algorithm at a fixed data rate. Therefore, I can rewrite equation [1] as:

Δ P = P_2 - P_1 = c_2 ⋅ Δ t

where c_1 = c_2 ⋅ Δ t

This allows me to divide both sides by [equation] \Delta t [/equation]:

ΔP /Δt = c_2

And in the limit:

dP/dt = c_2

which then becomes:

P(t) - P(0) = c_2⋅(t - t_0)

And so the result is that this recursive equation [2] gives us a linear ramp if we were to implement it. What I am trying to do for equation [1] is figure out what this expression will look like.

2. Feb 10, 2016

deskswirl

To which discrete time index is e^P refering?

3. Feb 10, 2016

HallsofIvy

You can approximate it by a "continuous derivative" with different results depending upon what "continuity" assumptions you make.

4. Feb 10, 2016

BvU

looks explosive if I try a few numbers:
Code (Text):

n   P1           exp(P1)         P2
0   0.0001       1.0001      1.0002
1   1.0002       2.718826    3.719026
2   3.719026    41.2242     44.94323
3  44.94323      3.3E+19     3.3E+19

I find it hard to believe this is what's intended ... ?

5. Feb 10, 2016

Cyrus

Sorry all I forgot it's e^p1

6. Feb 10, 2016

Cyrus

Good point, it's actually e^(p1 - c_3)

Which is a negative exponential, but we can rewrite that as e^p1 / e^c_3 so I left out the denominator since it was a constant value. c_3 is just a constant

7. Feb 10, 2016

Cyrus

I'm open to any solutions that are simple with some reasonable assumptions. :-)

8. Feb 10, 2016

Greg Bernhardt

Hey look who's back

9. Feb 11, 2016

BvU

Still explodes, at some point P1 > c3.
Let C3 = 100 and P1 start at 1:

10. Feb 11, 2016

RUber

It looks like you have, for a fixed data rate
$\Delta P = e^P,$ so $P_2 - P_1 = e^{P_1 } \Delta t$?
In the same way you simplified the first problem,

$\frac{\delta P}{\delta t} = \frac{ P(t+\delta t) - P(t) }{ \delta t} = e^{ P(t) }.$
In the limit, this will give you $P' = e^{P(t)}$

As BvU has pointed out, no matter what your starting P_1 is, your function will eventually blow up. Is there some other context for this problem? Maybe some initial values?