# Laplace self-transform

1. Aug 12, 2007

### bobbyk

1. The problem statement, all variables and given/known data

Find a function of time f(t) whose Laplace transform is f(s). Other than zero!

2. Relevant equations

3. The attempt at a solution

I've searched the internet and haven't fouind any.

2. Aug 12, 2007

### Kummer

Neither have I.

I believe this is a consequence of the Fredholm Integral Equation, in which, the solutions take trivial form, i.e. only zero.

Have you ever though about a similar question? Which functions (non-trivial) are its own Fourier Transforms?

Last edited: Aug 12, 2007
3. Aug 12, 2007

### bobbyk

Oh, yeah! There are many Fourier self-transforms. Gaussian, for example.

4. Aug 12, 2007

### Kummer

The Hankel functions are the eigenfunctions of this Integral equation if you are interested.

5. Aug 12, 2007

### bobbyk

Yes, I'm aware of the Hankel functions.
But let's look for the Laplace self-transforms. Do you claim there aren't any?

Way to go, Tiger!

Bob.

6. Aug 12, 2007

### Kummer

I really do not know, I wish to know myself. I do not think there are, but I cannot prove it, nor have I tried to prove it. I asked some of the professors in the University and I do not have an answer. Next time I go, which is in two weeks, I try again for you.

Here is your problem stated formally.

Problem: Does there exist a function $$f:[0,\infty)\to \mathbb{R}$$ such that $$F(s)=\int_0^{\infty} e^{-st}f(t) dt$$ is integrable and converges for each $$s\geq 0$$ with the property that $$f(t) = F(s)$$.

7. Aug 13, 2007

### HallsofIvy

Staff Emeritus
No, f(t) is never equal to F(s).
You want a function f(t) such that
$$f(s)=\int_0^{\infty} e^{-st}f(t) dt$$
or, using your F, f(s)= F(s).

8. Aug 13, 2007

### bobbyk

Yes, I want that function. Other than f(t) identically equal to zero, of course.
Do you know of such a function? Or how to go about finding one?
Or how to prove there isn't one?

Bob

9. Aug 13, 2007

### Dick

How about 1/sqrt(t)? The laplace transform is sqrt(pi)/sqrt(s). It's an eigenfunction, though not with unit eigenvalue...

10. Aug 14, 2007

### bobbyk

Yes, 1/sqrt(t) is close, and suggests that multiplying it by some other function might do the trick.
Bob

11. Feb 8, 2008

### bobbyk

Kummer, did you ever get an answer to my question about a Laplace Self-Transform?
You said you were going to ask the profs.

How about you, Dick, do you know of any?

Bob

12. Feb 8, 2008

### Dick

No. But then I didn't drive myself all that crazy thinking about the eigenvalue spectrum of the laplace transform either. Is this important?

13. Feb 8, 2008

### bobbyk

Dick, wouldn't it be important to find a Laplace Self-Transform - something that most
people say doesn't exist? I think it's very exciting!

Bob

14. Feb 8, 2008

### Dick

I think the world's gotten along pretty well since August without a Laplace Self-Transform. Can't say I'm all that fired up about it. Why do you say 'most people say doesn't exist'? Has anyone given you a reason?

15. Feb 8, 2008

### bobbyk

No, they don't give a reason - they just think it doesn't exist - just like you apparently
do. Look, Dick, as you know, it's not a matter of opinion - either there is such a function or there isn't. I belive there is.

Thanks for responding in a friendly manner!

Bob

16. Feb 8, 2008

### Dick

I'm just guessing it doesn't. It's a pure hunch. Now I have to ask, why do you believe it does? I'll look into it again.

17. Feb 8, 2008

### Dick

Ok, I've got you something. http://mathforum.org/kb/thread.jspa?forumID=13&threadID=1535681&messageID=5522747 It's a reference to a thread on the sci.math mailing list. It doesn't answer your question but it gives you a lot more eigenfunctions. But none with eigenvalue 1 (near as I can tell). I would suggest you post your question on the sci.math forum. Those people are about as smart as they come, aside from the cranks, and Robert Israel is REALLY SMART. He'll probably pick up on the question and if there is a reason why the laplace transform doesn't have an eigenvalue of 1, he'll know. Try it.

Last edited: Feb 8, 2008
18. Feb 9, 2008

### bobbyk

Thanks, Dick, I will!

19. Feb 9, 2008

### Ben Niehoff

hmm...two clues here:

1. Hankel functions are eigenfunctions of the Fourier transform.

2. $1/\sqrt(t)$ is an eigenfunction of the Laplace transform, but with non-unit eigenvalue.

This makes me wonder, could the Bessel functions be Laplace self-transforms?

20. Feb 9, 2008

### coomast

In the book "Laplace transforms" by Murray R. Spiegel, is in chapter 1 an exercise on this subject. There are 135 exercises of which I was unable to find 6 of them. One of these 6 is off course the one were it is asked wheter a given function could be it's own transform.

Anyway, the exercise asked to prove that

$$L\left[ F(t) \right] = L\left[ at^{-\alpha}+bt^{-\beta \right] =\Lambda \cdot \left(as^{-\alpha}+bs^{-\beta}\right)$$

In which:

$$\alpha+\beta =1$$

and

$$\Lambda = \pm\sqrt{\pi csc(\alpha \pi)}$$

And as a second part it was necessary to answer the question if F(t) was it's own transform and to explain why.

Maybe this will shed some light on the post.