# Laplace transform Problem

• yy205001
Can you think of a way to use the definition of the Laplace transform to show that it doesn't exist?In summary, the first question asks if a given function has a Laplace transform for large values of s, and the answer is no due to the function not being of exponential order. The second question asks to find a function that has a Laplace transform equal to es, or explain why no such function exists. Using Mellin's formula, it can be shown that no such function exists, as the integral does not converge for γ = 0. The convolution integral equation can also be used to solve this problem.

## Homework Statement

Can people help me on these two questions, please.

Q1)
Does f(t) have a Laplace transform F(s) for sufficiently large real value of s, where

f(t) = et/(t4-1).

Q2)
Either find a function f(t) for which F(s) = L{f(t);t→s} = es, or explain why no such function f(t) exists.

## The Attempt at a Solution

For Q1:
Since there is no such a real constants K s.t f(t)≤ K*f(t), therefore f(t) is not of exponential order. So, Laplace transform does not exist.

For Q2:
My guess for the answer is No but I can't explain it mathematically.

Any help is appreciated!

Start out by writing out the definition of the laplace transform explicitly in terms of the unknown function ##f(t)##. Knowing that you want the integral to evaluate to ##e^s##, you should be able to see what kind of function would be necessary to satisfy the equation. (*Hint: it's the kind of "function" you would see in a time shift*)

So, i start at ∫e-stf(t)dt from 0 to ∞ and set it to es?

yy205001 said:

## Homework Statement

Can people help me on these two questions, please.

Q1)
Does f(t) have a Laplace transform F(s) for sufficiently large real value of s, where

f(t) = et/(t4-1).

Q2)
Either find a function f(t) for which F(s) = L{f(t);t→s} = es, or explain why no such function f(t) exists.

## The Attempt at a Solution

For Q1:
Since there is no such a real constants K s.t f(t)≤ K*f(t), therefore f(t) is not of exponential order. So, Laplace transform does not exist.
This can't be correct. f(t) ≤ K f(t) is satisfied by K=1. What did you really mean to say?

For Q2:
My guess for the answer is No but I can't explain it mathematically.

Any help is appreciated!

yy205001 said:
So, i start at ∫e-stf(t)dt from 0 to ∞ and set it to es?

Regarding Q2: Yup! It's almost the definition of the function I'm thinking of itself: have you seen it before?

Except there's a crutial part to the limits of your integral that you would have to pay close attention to in order to understand whether the function is actually a viable solution...

That is to say, you can think about what the function would need to be, and then perhaps why such a function can or can't exist with respect to the Laplace transform

vela:

But when happen if t4<1? Then K=1 does not satisfy the inequality.

PhysicsandSuch:

So i need to make assumption for Question 2?

yy205001 said:
vela:

But when happen if t4<1? Then K=1 does not satisfy the inequality.

PhysicsandSuch:

So i need to make assumption for Question 2?

I don't think you have to make any assumptions, just examine closely for the necessary functional form and whether it's possible given the domain of the integral...

What did you have in mind? Maybe write out your line of thinking so I can get a better idea of where you are getting stuck...

yy205001 said:
But when happen if t4<1? Then K=1 does not satisfy the inequality.
I don't think you're reading what you wrote. Surely you agree that f(t) = f(t) for all t. If K=1, then f(t) = K f(t). So f(t) ≤ K f(t) for all t.

yy205001 said:

## Homework Statement

Can people help me on these two questions, please.

Q1)
Does f(t) have a Laplace transform F(s) for sufficiently large real value of s, where

f(t) = et/(t4-1).

Q2)
Either find a function f(t) for which F(s) = L{f(t);t→s} = es, or explain why no such function f(t) exists.

## The Attempt at a Solution

For Q1:
Since there is no such a real constants K s.t f(t)≤ K*f(t), therefore f(t) is not of exponential order. So, Laplace transform does not exist.

For Q2:
My guess for the answer is No but I can't explain it mathematically.

Any help is appreciated!

For Q2: have you tried using Mellin's formula for the inverse Laplace transform of F(s)? That formula states that
$$f(t) = {\cal{L}}^{-1}(F)(t) = \frac{1}{2 \pi i} \lim_{T \to \infty} \int_{\gamma-iT}^{\gamma+iT} e^{st} F(s) \, ds,$$
where the integration is done along the vertical line ##\text{Re}(s) = \gamma## in the complex s-plane, with ##\gamma## lying to the right of all singularities of F(s). For the case of F(s) = e^s you can take γ = 0 and just do the integral. Do you get a finite result?

vela:
I think I wrote something wrong above, it should be abs(f(t)) ≤ K*eσt. So, if there exist real constants K and σ such that for all sufficiently large values of t, the inequality holds then Laplace transform exists.

PhysicsandSuch:
I think the convolution integral equation might help! I will try this out first.

Ray Vickson:
Um..I don't think I have not gone this far yet in my course. So, this might not be the way to get the answer.

yy205001 said:
vela:
I think I wrote something wrong above, it should be abs(f(t)) ≤ K*eσt. So, if there exist real constants K and σ such that for all sufficiently large values of t, the inequality holds then Laplace transform exists.
I thought so, but I wasn't sure what criterion you were referring to.

This particular function satisfies the above condition because it only cares about what happens at large t. In other words, the function goes to 0 fast enough so as not to cause problems with convergence as you integrate out to ∞.

You need to find a different reason for why F(s) doesn't exist.