Proof of the convolution theorem for laplace transform

[O.J.]

My textbook provides a proof but there's one thing about the proof i do not understand

it starts assuming L{f(t)} = the laplace integral with the f(t) changed to f(a)

same goes with L{g(t)} as it changes it to g(b)

i understand the big picture>>starting from a product of 2 L transforms and working ur way back to an expression in terms of the two functions transformed, but how can u change the variable for f and g when we started off stating f of T and g of T.?

 

[O.J.]

Am I not presenting my questions clearly?

 

[Rainbow Child]

Let

$$F(s)=\mathcal{L}\{f(t)\}=\int_0^{+\infty}e^{-s\,t}\,d\,t\,f(t) \quad \text{and} \quad G(s)=\mathcal{L}\{g(t)\}=\int_0^{+\infty}e^{-s\,t}\,g(t)\,d\,t$$

be the two Laplace transormations and let's denote

$$(f\star g)(t)=\int_0^{+\infty}f(\tau)\,g(t-\tau)\,d\,\tau$$

the convolution of $$f,\,g$$.

Thus the Laplace transform for the convolution would be

$$\mathcal{L}\{(f\star g)(t)\}=\int_0^{+\infty}e^{-s\,t}\left( \int_0^{+\infty}f(\tau)\,g(t-\tau)\,d\,\tau \right)\,d\,t =\int_0^{+\infty}\int_0^{+\infty}e^{-s\,t}f(\tau)\,g(t-\tau)\,d\,\tau \,d\,t \quad (1)$$

The above double integral is to be evaluated in the domain $$\mathcal{D}=\{(t,\tau): t \in (0,+\infty),\tau \in (0,+\infty)\}$$.

Performing the change of variables
$$(t,\tau)\rightarrow (u,v): t=u+v, \, \tau=v$$ with $$u \in (0,+\infty),v \in (0,+\infty)$$ we have $$d\,\tau\,d\,t=d\,u\,d\,v$$ since the Jacobian equals to 1. Thus (1) becomes

$$\mathcal{L}\{(f\star g)(t)\}=\int_0^{+\infty}\int_0^{+\infty}e^{-s\,(u+v)}f(u)\,g(v)\,d\,u\,d\,v= \int_0^{+\infty}e^{-s\,u}f(u)\,d\,u \cdot \int_0^{+\infty}e^{-s\,v}\,g(v)\,d\,v$$

yielding to

$$\mathcal{L}\{(f\star g)(t)\}=\mathcal{L}\{f(t)\} \cdot \mathcal{L}\{g(t)\}$$

Nothing but double integrals!

 

[xenosicotte]

Isn't the laplace convolution between 0 and t rather than 0 and infinity?

 

[blue_raver22]

Thank you for your question. The convolution theorem for Laplace transforms states that the Laplace transform of the convolution of two functions f(t) and g(t) is equal to the product of their individual Laplace transforms. This theorem is a powerful tool in solving differential equations and other mathematical problems.

In the proof provided in your textbook, the authors have used a change of variables for the functions f(t) and g(t) to f(a) and g(b), respectively. This is a common practice in mathematical proofs and allows for easier manipulation of the functions.

To understand why this change of variables is valid, let's look at the definition of the Laplace transform:

L{f(t)} = ∫f(t)e^(-st)dt

Here, the variable of integration is t, but this does not mean that the function f(t) can only be evaluated at t. In fact, we can substitute any variable we want for t, such as a or b, and the result will still be the same. This is because the Laplace transform is an integral over the entire domain of the function, not just a specific value of t.

Therefore, in the proof, the authors have chosen to use a and b as the variables for f(t) and g(t), respectively, to make the manipulation of the functions easier. This does not change the fundamental definition of the Laplace transform.

I hope this helps clarify the change of variables in the proof. If you have any further questions, please do not hesitate to ask. Keep exploring and questioning, as that is the essence of science.