Laplace Transform of Composition of Functions: Floor(f(t))

In summary, you can take the Laplace transform of a composition of functions, but you don't know how to do it.
  • #1
Nocturne
4
0
Hello,

I am trying to take the Laplace transform of floor(f(t)) in order to solve the differential equation f'(t)=floor(f(t)).

I know that http://functions.wolfram.com/IntegerFunctions/Floor/22/04/" gives L(floor(t)) = (e^(-s))/(s(1-e^(-s))) instead -- are these equivalent?) and L(f(t)) = F(t) (of course), but I realized that I have no idea how to take the Laplace transform of a composition of functions, and no table I have been able to find contains L(floor(f(t))) or rules about compositions of functions. There is plenty of information on convolutions, but that isn't what I'm looking for unless this problem can be reformulated as one.

My question, at its essence, is this: given functions f and g, how do I determine L(f(g(t))? More specifically I want to know L(floor(f(t))), but any insight on the general case (if the problem does generalize) would be much appreciated.

In terms of background for the problem, I'm trying to model some cells for my cancer research, and my system of equations based upon the simple case f'=a*f(t) overestimates growth rates by effectively allowing hypothetical "fractions of cells" to divide. The discrete case based on f'=a*floor(f(t)) should solve this problem, but I do not know how to obtain an analytical solution for it.

I apologize if I am missing something obvious here, as well as for not knowing LaTeX.

Thank you!

Edit: I had originally posted a version of this thread in the Calculus and Analysis section https://www.physicsforums.com/showthread.php?p=2524246" since it was more concerned with the Laplace transform itself than solving the differential equation once I had it, but I realized that my query was still more germane to the Differential Equations forum.

Edit2: [tex]\mathcal{L}(\text{floor}(t))=\frac{e^{-s}}{s(1-e^{-s})}[/tex] or perhaps [tex]\mathcal{L}(\text{floor}(t))=\frac{1}{s(e^{s}-1)}[/tex] -- included for clarity now that I have learned some LaTeX.
 
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  • #2
You try to solve the nonlinear (!) differential equation f'(t)=floor(f(t)). So it is not concerned with the Laplace transform. The Laplace transform fits exceptionally for linear problems.

I doubt whether exist analytical solution for f'(t)=floor(f(t)). Such discontinuous right-hand side requires interminable energy. So, I think, it is not a good model for some cells for your cancer research.
 
  • #3
Thank you for your reply. I am aware that it is nonlinear; I just did not know that the Laplace transform could not be applied to nonlinear ordinary differential equations (as I mentioned, I thought I may have been missing something obvious, but I wasn't sure).

I am fairly confident that the equation is a good model for my work, though. In an idealized system, at least, cells can indeed divide indefinitely at an exponential rate (I believe this is the "interminable energy" to which you referred), and the only difference between my y'=a*floor(y) model and the standard (linear) y'=a*y model for cell growth is that mine does not overestimate the growth rate by allowing hypothetical "fractions of cells" to replicate. This may not matter much for large numbers of cells, but it becomes important for accurately modeling small clusters of them.

Regardless of method, is there any way to obtain an analytical solution to this equation? I don't even care if it's closed-form, honestly. Even an infinite series (assuming decently fast convergence) would be acceptable, since I am implementing this in a computer program anyway.

If not, though, I imagine I could use a numerical method and compare the results with the linear version to see what sort of differences might be expected. How can I know which numerical method would give the best results, though? I am concerned that some might not take the discrete nature of the right-hand side into account very well. Perhaps a carefully-selected step size and starting point could ensure that this is done properly, but I want to make sure that I don't end up ignoring the very same discrete behavior that I tried to incorporate into the model in the first place.

Edit: But it seems the step size would be dependent on the value of the dependent variable, unless I'm not understanding this correctly. Again, any assistance in developing any form of analytical solution or, failing that, an accurate numerical approximation, would be greatly appreciated. Thanks!
 
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  • #4
Hi, I'm new here but I'd like to help, sory for grammer, english isn't my first language.

First of all, your function isn't strictly speaking function becouse it's derivate acording to diff. equation should be only oneside discontinius in some points which isn't posible acording to standard definition of derivate. So it must be distribution rather then a function.

But, math aside if you want to compare behaviour of this so-called function f(t) to some continius functin g(t), in your case exponential or whatever, hear what you should do:

Sey you are interested in difrence in some point t0 you calculate g(t0). Now that is to my knowlige dificult but you could make comparison in neighbourhood of t0 and hire is how. If g and f are suposedly close aproximatino of ich other, you culd compare it in t1 when y1=f(t1)=g(t0) if you have t=f^-1(y) and fortunaly that "funciton" has cloused form.

so path is this t0->y1->t1 and then you cuold comapare f(t1)=y1 and g(t1)

closed form of t=f^-1(y) is t=H[floor(y)]+(y-floor(y))/(1+floor(y)) + const where H[n] is harmonic number defined here: http://en.wikipedia.org/wiki/Harmonic_number and const is arbitrary time shift dipendent of initial conditions of diff. equation.

I Houp, I could be of eny help.
 
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Related to Laplace Transform of Composition of Functions: Floor(f(t))

1. What is the Laplace Transform of Composition of Functions: Floor(f(t))?

The Laplace Transform of Composition of Functions: Floor(f(t)) is a mathematical operation that takes a function, in this case Floor(f(t)), and converts it into a different representation. It is often used in engineering and physics to solve differential equations and analyze systems.

2. How is the Laplace Transform of Composition of Functions: Floor(f(t)) calculated?

The Laplace Transform of Composition of Functions: Floor(f(t)) is calculated using the standard Laplace Transform formula, where the function Floor(f(t)) is substituted for the variable t. This results in a new function, known as the Laplace Transform of Floor(f(t)).

3. What is the significance of the Floor function in the Laplace Transform of Composition of Functions?

The Floor function, denoted as Floor(f(t)), is a mathematical function that rounds down the value of f(t) to the nearest integer. This function is important in the Laplace Transform of Composition of Functions as it allows for the transformation of non-integer functions into a form that is easier to manipulate and analyze.

4. Can the Laplace Transform of Composition of Functions: Floor(f(t)) be used to solve real-world problems?

Yes, the Laplace Transform of Composition of Functions: Floor(f(t)) can be used to solve a variety of real-world problems. It is commonly used in engineering and physics to analyze systems and solve differential equations. It can also be used in other fields, such as economics and finance, to model and solve complex problems.

5. Are there any limitations to using the Laplace Transform of Composition of Functions: Floor(f(t))?

One limitation of using the Laplace Transform of Composition of Functions: Floor(f(t)) is that it can only be applied to functions that are defined for all values of t. This means that functions with discontinuities or infinite values may not have a valid Laplace Transform. Additionally, the Laplace Transform can only be applied to linear systems, meaning that non-linear functions may not have a valid transform.

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