Laplace Transforms

[A Tripolation]

I'm having difficulty understanding what this problem is asking of me.

 

"Does the Laplace transform have a fixed point? Is there a function [math]f(t)[/math] defined on [math] 0 < t < \infty [/math], such that

 

[math]

\int_{0}^{\infty} f(t)e^{-st} dt = f(s)

[/math]

 

There is an extra condition that [math]f[/math] is not identically 0."

 

I can use Laplace transforms as a general rule, but I do not quite understand what they mean mathematically. This question seems to be addressing that mathematical meaning, if I'm reading it correctly. How should I go about starting this problem? What do I need to understand to solve this?

 

Thanks.

[Schrödinger's hat]

I'm having difficulty understanding what this problem is asking of me.

 

"Does the Laplace transform have a fixed point? Is there a function [math]f(t)[/math] defined on [math] 0 < t < \infty [/math], such that

 

[math] \int_{0}^{\infty} f(t)e^{-st} dt = f(s) [/math]

 

There is an extra condition that [math]f[/math] is not identically 0."

 

I can use Laplace transforms as a general rule, but I do not quite understand what they mean mathematically. This question seems to be addressing that mathematical meaning, if I'm reading it correctly. How should I go about starting this problem? What do I need to understand to solve this?

 

Thanks.

 

Humm, bit of a tricky one.

Basically they are asking if you can find some function f(t) which, when you put it through a laplace transform, comes out as the same function.

 

I can get [math] \mathcal{L}(f(t)) = a F(s)[/math] (our old friend [math]\pi[/math] shows up, too).

 

Getting this far is as simple as looking at a table of standard Laplace transforms and looking for something that is of the same form in both columns (which I'll leave as an exercise for you to complete as this is the homework help section).

 

 

As a concrete example for the Fourier transform, you'd use the gaussian.

[math]\mathcal{F}(e^{\frac{-x^2}{2}}) = e^{\frac{-s^2}{2}}[/math] (give or take a pi depending on your definition of fourier transforms)

 

We've got a bit of wiggle room here because we can use the coefficient of [math]x^2[/math] to modify the coefficient in order to force it to be 1.

 

Unfortunately the function that works for the laplace transform isn't an exponential so I don't know if it's possible to get exactly the same function out (including the coefficient being one)

You may be able to get the answer by fiddling with shifting/scaling theorems and such, I'm not sure.

 

Making further progress without trial and error (and understanding what's going on further) would be aided by understanding of eigenfunctions of operators and hilbert spaces

Edited November 18, 2011 by Schrödinger's hat

[A Tripolation]

Basically they are asking if you can find some function f(t) which, when you put it through a laplace transform, comes out as the same function.

 

Alright. That makes sense to me. And I can read the math in that connotation now.

 

I can get [math] \mathcal{L}(f(t)) = a F(s)[/math] (our old friend [math]\pi[/math] shows up, too).

 

Getting this far is as simple as looking at a table of standard Laplace transforms and looking for something that is of the same form in both columns (which I'll leave as an exercise for you to complete as this is the homework help section).

 

So the only way to discern this to look at the tables? There are no manipulations of any sort that I can implement that will show me a general function yields it's own general function as a solution to the Laplace transform?

 

As a concrete example for the Fourier transform, you'd use the gaussian.

[math]\mathcal{F}(e^{\frac{-x^2}{2}}) = e^{\frac{-s^2}{2}}[/math] (give or take a pi depending on your definition of fourier transforms)

 

I haven't learned much about Fourier transforms. I don't quite understand what you're saying here.

 

 

We've got a bit of wiggle room here because we can use the coefficient of [math]x^2[/math] to modify the coefficient in order to force it to be 1.

 

Unfortunately the function that works for the laplace transform isn't an exponential so I don't know if it's possible to get exactly the same function out (including the coefficient being one)

You may be able to get the answer by fiddling with shifting/scaling theorems and such, I'm not sure.

 

 

We want to modify the coefficient to be 1 so that it has no impact on our [math]f(t) = f(s)[/math], correct?

 

Making further progress without trial and error (and understanding what's going on further) would be aided by understanding of eigenfunctions of operators and hilbert spaces.

 

We are only now covering eigenfunctions. What is an eigenfunction of an operator? I have never even used Hilbert spaces before, so I don't think I can use those to solve this problem.

 

Thanks for the help, by the way.

 

Also, could a mod delete the post above? My browser seems to have oddly submitted the edit.

Edited November 18, 2011 by A Tripolation

[mooeypoo]

Deleted the double-post.

[Schrödinger's hat]

Alright. That makes sense to me. And I can read the math in that connotation now.

 

So the only way to discern this to look at the tables? There are no manipulations of any sort that I can implement that will show me a general function yields it's own general function as a solution to the Laplace transform?

Well you could try some broad classes of functions to see if they work.

My thought process went something along these lines, edited slightly due to the benefit of hindsight.

Just put in the most general forms of exponentials/polynomials/etc you can integrate and see what comes out.

You'll find that basic exponentials and trig functions give you functions involving polynomials, so you can rule out those.

Things like [math]e^{f(x)}[/math] get quite difficult to integrate so maybe leave them out.

 

Polynomials in are going to give you polynomials out, so this looks promising.

High integer powers of t give you highly negative integer powers of s.

So this train of logic leads us to consider negative powers of t, fractional powers, or complex powers

[math]t^q[/math] where q is not a positive integer. Looks like our most promising candidate. Try some fractional powers around 1

You can try putting in set numbers by trial and error, or try it for a general case (will probably involve special functions and all sorts of nasty stuff).

Failing that, go back to [math]e^{f(x)}[/math] (already knowing the answer I can tell you it's the first case).

 

 

There are some more advanced ways of approaching this problem, but I am barely aware of them (let alone knowing them well enough to teach someone else).

 

I haven't learned much about Fourier transforms. I don't quite understand what you're saying here.

A fourier transform is another transform a lot like a laplace, but double sided and involving imaginaries.

 

 

 

 

We want to modify the coefficient to be 1 so that it has no impact on our [math]f(t) = f(s)[/math], correct?

Yes, unfortunately I don't know how to do that off hand, and I weakly suspect it may not be possible.

 

We are only now covering eigenfunctions. What is an eigenfunction of an operator? I have never even used Hilbert spaces before, so I don't think I can use those to solve this problem.

An eigenfunction is a way of generalising the concept of an eigenvector.

In linear algebra you get an operator (usually represented as a matrix). You find vectors which, when multiplied by this matrix will give you the same vector back, but scaled by the eigenvalue.

 

You can extend this concept to apply to functions (rather than vectors) and other operators (not just matrices)

You can think of your function (let's say the function [math]sin(t)[/math]) as a bit like an infinite dimensional vector which takes on a value for every index (value of t)

In this context you can think of a derivative or a laplace transform as an operator much like the matrix.

 

For example the class of functions [math]e^{ax}[/math] are the eigenfunctions of the derivative operator with eigenvalue [math]a[/math] [math]\frac{\partial}{\partial x} e^{ax} = a e^{ax}[/math] and [math]e^x[/math] has an eigenvalue [math]a=1[/math] so it is the fixed point.

 

The function I found (which is lying waiting for you in most tables of laplace transforms) is an eigenfunction, but its eigenvalue is not 1, so it is not a fixed point.

It's also unclear to me whether others exist or not (namely an eigenfunction with eigenvalue 1 which we're after). I suspect that they either don't, or are quite hideous.

Edited November 19, 2011 by Schrödinger's hat

[Schrödinger's hat]

I'll just leave these here:

[math] \Gamma(z)=\int_0^\infty t^{z-1}e^{-t} dt [/math]

 

[math] \mathcal{L}[t^{z-1}](s) = \int_0^\infty t^{z-1}e^{-st} dt [/math]

 

Where z is a complex number.

 

 

 

(It may also help to find that eigenfunction I was talking about first. Then see how this applies to finding more eigenfunctions, Then you might be able to find a fixed point).

Edited November 24, 2011 by Schrödinger's hat