# Thread: Is the Laplace transform an 1 to 1 function?

1. ## Is the Laplace transform an 1 to 1 function?

2. Originally Posted by Arturo_026
It depends on in what sense are functions the same. For example:

Let

$f(t)=t$ and

$g(t)=\begin{cases}t, \text{ if } 0 \le t < 10 \\ 50, \text{ if } t = 10 \\ t, \text{ if } t > 10 \end{cases}$

are "different" functions but they have the same Laplace transform.

For the 2nd question you really need to specify to co-domain.

Remember that s is a complex variable where as t is a real number

3. Originally Posted by TheEmptySet
It depends on in what sense are functions the same. For example:

Let

$f(t)=t$ and

$g(t)=\begin{cases}t, \text{ if } 0 \le t < 10 \\ 50, \text{ if } t = 10 \\ t, \text{ if } t > 10 \end{cases}$

are "different" functions but they have the same Laplace transform.

For the 2nd question you really need to specify to co-domain.

Remember that s is a complex variable where as t is a real number
Yes you are right. My question needs to be more especific. For example we are solving differential equations in class using the Laplace Transform, and we need to find f(t) by solving the inverse Laplace ... L^-1[F(t)] ... Then I wondered. Could it be possible that i could come up with two different solutions from the Laplace inverse.
i.e.

If L[f_1] = L[f_2] is it necessary for f_1 to be equal to f_2 ?

4. As my example shows if you change only a "few" points on a function it will not change its Laplace transform. Functions like this are considered the same if they only differ on a set of measure zero. Formally functions like this are considered the same function as they are in the same equivilence class. So in this sense they are equal.