# Using convolution for Laplace transform

1. Apr 3, 2007

### sensitive

1. The problem statement, all variables and given/known data
Using convolution theorem for Laplace theorem,, show that

2. Relevant equations
inverse Laplace transform (1/(S^3/2*(s-1)) = (2*e^t)/Pi^1/2 intregral (from 0 to t) e^-x*x^1/2dx.

3. The attempt at a solution

The inverse Laplace above is a product of 1/s^3/2 and 1/(s-1)

and both terms are the Laplace transform of 2/Pi^1/2*t^1/2 and e^t respectively.

changing variable t = u gives:2/Pi^1/2*u^1/2 and e^u

using convolution: integral (from 0 to infinity) e^-st integral (from 0 to t)
(2/Pi^1/2*u^1/2 )(e^u)du dt.

I dont know how to go about with the integral from here onwards..Help!

2. Apr 4, 2007

### HallsofIvy

Staff Emeritus
The "convolution theorem" you are told use says that the Laplace transform of a convolution of functions is the product of the Laplace transforms of the functions. Isn't that exactly what you need here?

3. Apr 5, 2007

### sensitive

I am sorry but I dont really get you. Laplace transform of a convolution of functions is the product of the Laplace transforms of the functions.

Isnt that what I am trying to do. finding the product of the inverse Laplace (which as given) and converting the products to its functions.

Is there a rule or formula for convolution theorem of Laplace theorem?

4. Apr 5, 2007

### HallsofIvy

Staff Emeritus
Good. You are saying that $\frac{1}{s^{3/2}}\frac{1}{s-1}$ is the product of the Laplace transforms of $\frac{2}{\sqrt{\pi t}}$ and $e^t$. You also know that the Laplace transform of a convolution is the product of the individual functions. So $\frac{1}{s^{3/2}}\frac{1}{s-1}$ is the Laplace transform of the convolution of $\frac{2}{\sqrt{\pi t}}$ and $e^t$. If you can show that the convolution of those function is equal to $\frac{2e^t}{\sqrt{\pi}}\int_0^t e^{-x\sqrt{x}}dx$, you are done!

By the way, what "Precalculus" class deals with the Laplace transform?

5. Apr 5, 2007

### sensitive

My problem now is to prove it because now the function has integral form 0 to t where does that come from?

Is there a convolution rule that i need to know to be able to prove it?

And im so sorry if I post it in the wrong section. Under which section should post this Laplace problems? thx.

6. Apr 5, 2007

### HallsofIvy

Staff Emeritus
First what is the DEFINITION of the convolution of two functions? I think that is the formula you need.

Second, this should be in "Calculus and Beyond". I will go ahead and move it.

7. Apr 5, 2007

### sensitive

Oh I just found an appropriate section for this problem. Sorry for posting the problem at the wrong area. I'll make sure Ill post any problem/s in the appropraite sectionnext time.

8. Apr 5, 2007

### sensitive

There was some typing error, in the original equation.

inverse Laplace transform (1/(S^3/2*(s-1)) = (2*e^t)/Pi^1/2 intregral (from 0 to t) (e^-x)*(x^1/2)dx.

The inverse Laplace above is a product of 1/s^3/2 and 1/(s-1)

and both terms are the Laplace transform of (2/Pi^1/2)*t^1/2 and e^t respectively.

Using convolution theorem,
integral (from 0 to infinity) e^-st (integral (from 0 to t) f (t - x)*g(x)dx)dt

putting the Laplace transform it becomes
integral (from 0 to infinity) e^-st (integral (from 0 to t) (e^ (t - x))*(2/Pi^1/2)*x^1/2dx)dt

But can I do this,
(integral (from 0 to infinity) e^-st dt)*(integral (from 0 to t) (e^ (t - x))*(2/Pi^1/2)*x^1/2dx)dt

If i can do that then it is already proven
(2*e^t)/Pi^1/2 intregral (from 0 to t) (e^-x)*(x^1/2)dx.

9. Apr 5, 2007

### sensitive

Well, actually i don't think I can do the following:
(integral (from 0 to infinity) e^-st dt)*(integral (from 0 to t) (e^ (t - x))*(2/Pi^1/2)*x^1/2dx)dt

I think the integration
integral(from 0 to infinity) of e^-st*(integral (from 0 to t) (e^ (t - x))*(2/Pi^1/2)*x^1/2dx)dt using integration by parts

integration of e^-st (from 0 to infinity), will give 1.

hence and leaving (integral (from 0 to t) (e^ (t - x))*(2/Pi^1/2)*x^1/2dx)

am I right? I hope I am making sense.