What is the Inverse Laplace Transform of e^(-sx^2/2)?

[Haku]

Homework Statement

Find the inverse Laplace transform of 1/s^2 * e^(-sx^2/2)

Relevant Equations

Laplace convolution theorem equation.

My attempt at finding this was via convolution theorem, where we take F(s) = 1/s^2 and G(s) = e^(-sx^2/2). Then to use convolution we need to find the inverses of those transforms. From a table of Laplace transforms we know that f(t) = t. But I am sort of struggling with e^(-sx^2/2). My 'guess' is that the inverse Laplace transform of e^(-sx^2/2) is δ(t - (x^2)/2). This is from the fact that the inverse Laplace transform of e^sc is δ(t+c). But then integrating this in the convolution integral proves difficult.
Am I on the right track or should I use another method of finding the inverse Laplace transform of 1/s^2 * e^(-sx^2/2)?
Note: Here I have used * to denote multiplication and not convolution.

 

[vela]

It seems a bit puzzling to me that you say the presence of the Dirac delta function makes evaluating the integral difficult. I find it makes evaluating integrals simple and straightforward. Can you elaborate more about why you find it difficult?

Also, are you wedded to using convolution? A straightforward application of the established properties of Laplace transforms will get you the answer to the problem.

 

[Haku]

It seems a bit puzzling to me that you say the presence of the Dirac delta function makes evaluating the integral difficult. I find it makes evaluating integrals simple and straightforward. Can you elaborate more about why you find it difficult?

Also, are you wedded to using convolution? A straightforward application of the established properties of Laplace transforms will get you the answer to the problem.

I am happy to use any method, preferably the easiest too.
The reason I said it made it difficult is because you would have to evaluate the Dirac delta function at ((t-T)-x^2/2) would you? (Where T is tau).

 

[vela]

Don't let the argument of the delta function confuse you. You have
$$\delta\left[(t-\tau) - \frac{x^2}{2}\right] = \delta\left[\left(t - \frac{x^2}{2}\right)-\tau\right].$$ You can treat the stuff in the parentheses as a constant since you're integrating with respect to $\tau$.

Also, you should be a little careful here. The inverse Laplace transform of $F(s)$ is $f(t) = t u(t)$ where $u(t)$ is the unit step function.

 

[Haku]

Don't let the argument of the delta function confuse you. You have
$$\delta\left[(t-\tau) - \frac{x^2}{2}\right] = \delta\left[\left(t - \frac{x^2}{2}\right)-\tau\right].$$ You can treat the stuff in the parentheses as a constant since you're integrating with respect to $\tau$.

Also, you should be a little careful here. The inverse Laplace transform of $F(s)$ is $f(t) = t u(t)$ where $u(t)$ is the unit step function.

Ah okay, I will try to integrate that now.
As for F(s), why would it be tu(t)? the Laplace transform of t is 1/s^2. So why would the inverse of that change the original function?
Also what was your easier way to calculate the inverse?
Thank you!

 

[vela]

As for F(s), why would it be tu(t)? the Laplace transform of t is 1/s^2. So why would the inverse of that change the original function?

If you recall, when you calculate $F(s)$, you calculate
$$F(s) = \int_0^\infty t\,e^{-st}dt.$$ All information about $f(t)$ when $t<0$ doesn't contribute to $F(s)$. Hence, when you say the inverse Laplace transform of $F(s) = 1/s^2$ is $f(t) = t$, it's really only true for $t>0$. Quite often, this detail doesn't matter as it's understood that $t>0$ (for example, when solving a differential equation). In problems like this, where the factor $e^{-as}$ represents a shift in time by $a$, it becomes important to remember that $f(t) = 0$ for $t<0$. When you shift the function to the right by $a$, the result is $0$ for $t<a$. (Note this is the simpler way: use the time-shift property of Laplace transforms.)

Another way to see this is to try calculating the Laplace transforms of
$$f(t) = \begin{cases} t-a, & t \ge a \\ 0, & t<a \end{cases}$$ and
$$g(t) = t-a$$ where $t>0$. You'll see $f$ yields the Laplace transform $e^{-as}/s^2$ whereas $g$ doesn't.

 

[Haku]

If you recall, when you calculate $F(s)$, you calculate
$$F(s) = \int_0^\infty t\,e^{-st}dt.$$ All information about $f(t)$ when $t<0$ doesn't contribute to $F(s)$. Hence, when you say the inverse Laplace transform of $F(s) = 1/s^2$ is $f(t) = t$, it's really only true for $t>0$. Quite often, this detail doesn't matter as it's understood that $t>0$ (for example, when solving a differential equation). In problems like this, where the factor $e^{-as}$ represents a shift in time by $a$, it becomes important to remember that $f(t) = 0$ for $t<0$. When you shift the function to the right by $a$, the result is $0$ for $t<a$. (Note this is the simpler way: use the time-shift property of Laplace transforms.)

Another way to see this is to try calculating the Laplace transforms of
$$f(t) = \begin{cases} t-a, & t \ge a \\ 0, & t<a \end{cases}$$ and
$$g(t) = t-a$$ where $t>0$. You'll see $f$ yields the Laplace transform $e^{-as}/s^2$ whereas $g$ doesn't.

Ahhhh yes, it is because the lower limit of the integral is 0 right?

So in this case you get; $$F(s) = \int_0^\infty t\,e^{-st}dt.$$ = 1/s^2
And we have the s-shifting variable e^-as, where as = sx^2/2 right?
But in my definition of the s-shifting prop we used e^at as the shifting 'variable', in this case we have t = 1 but also an x term so how does that all fit together?

Also, do we consider the case of t < 0 because once it is shifted, then it is no long t < 0, it becomes t < a, which could be > 0 right? Since the actual t < 0 would not matter since t denotes time right?

Thanks!

 

[vela]

There are two shifting properties: one in the time domain and one in the s-domain. You're mixing them up.

Even with the shift, your solution needs to be defined for $0 \le t \lt a$. That's why the unit step function is included.

 

[Haku]

There are two shifting properties: one in the time domain and one in the s-domain. You're mixing them up.

Even with the shift, your solution needs to be defined for $0 \le t \lt a$. That's why the unit step function is included.

Ah yes I see,

I think I understand now, so by the second shifting property we have that the inverse Laplace transform of e^(-sc)F(s) = f(t-c)H(t-c), and in this example c = x^2/2 and F(s) = 1/s^2.

So we get the inverse Laplace transform of e^(-sx^2/2)•1/s^2 = f(t-c)H(t-c), where c = x^2/2 and H is the unit step function.
That means the original function u(x, t) = the inverse Laplace transform of e^(-sx^2/2)•1/s^2 = f(t-x^2/2)H(t-x^2/2) = (t-x^2/2)u(t-x^2/2) right?
Thanks!

 

[vela]

That means the original function u(x, t) = the inverse Laplace transform of e^(-sx^2/2)•1/s^2 = f(t-x^2/2)H(t-x^2/2) = (t-x^2/2)u(t-x^2/2) right?

It's not correct to write "e^(-sx^2/2)•1/s^2 = f(t-x^2/2)H(t-x^2/2)" but you got the idea.

 

[Haku]

It's not correct to write "e^(-sx^2/2)•1/s^2 = f(t-x^2/2)H(t-x^2/2)" but you got the idea.

Ah I see.
Thank you! Helped a lot!