Laplace transform of derivatives

[LagrangeEuler]

I have a question regarding Laplace transforms of derivatives
\mathcal{L}[f'(t)]=p\mathcal{L}[f(t)]−f(0^−)
Can anyone explain me why $0^-$?

 

[wrobel]

because it is a mistake

 

[jasonRF]

I have a question regarding Laplace transforms of derivatives
\mathcal{L}[f'(t)]=p\mathcal{L}[f(t)]−f(0^−)
Can anyone explain me why $0^-$?

This convention is often used in some engineering disciplines - it is the approach I learned in my signals and systems class as an electrical engineering major and continue to use in my job in industry. In this approach, the Laplace transform is defined as
$$ \mathcal{L}[f(t)] = \int_{0^-}^\infty \, f(t) \, e^{-pt} \, dt $$
Where the lower limit of integration is simply the left-sided limit. This does two things: first, it makes it clear that the Laplace transform of the delta function is unity (of course the integral is symbolic in this case). Second, it allows us to use pre-initial conditions when using Laplace transforms to solve circuit problems that include a switch that is thrown at t=0. A good summary of the benefits of this approach is

https://math.mit.edu/~hrm/papers/lmt.pdf

jason

 

[wrobel]

Consider a function $$f(t)=e^{-1/t}.$$
There is no problem with the standard formula here:
$$\mathcal L(f')=p\mathcal L(f)-f(0+)$$
And what about the formula
$$\mathcal L(f')=p\mathcal L(f)-f(0-)$$
?

 

[jasonRF]

Consider a function $$f(t)=e^{-1/t}.$$
There is no problem with the standard formula here:
$$\mathcal L(f')=p\mathcal L(f)-f(0+)$$
And what about the formula
$$\mathcal L(f')=p\mathcal L(f)-f(0-)$$
?

Good question.

If I were considering that kind of problem I would use the transform defined by $$ \mathcal{L}[f(t)] = \int_{0^+}^{\infty} f(t) \, e^{-pt} \, dt $$
which of course has the derivative rule you state. I’m a firm believer in selecting the right tool for the job.

The convention the OP mentioned fails for this case of course. It is more convenient for a different class of problems - especially those described by systems of constant coefficient ODEs. First, we are often interested in solutions for delta function inputs. Second, for linear circuit problems with a switch that is thrown at $t=0$, the voltage across an inductor and the current through a capacitor can be discontinuous. We usually know the values at $t=0^-$ prior to throwing the switch, and it takes real work to figure out what the values will be at $t=0^+$ in order to apply the derivative rule you stated.

Of course, your convention can work for the circuit problems as well, but it just involves a lot more work.

Jason

 

[wrobel]

I’m a firm believer in selecting the right tool for the job.

Then one must provide precise definitions, theorems and proofs. In this respect the text you cited above is very bad.
For example look at the bottom of page 5:

" A function f (t) defined on the
interval [0, ∞) is piecewise smooth if there is a sparse set
S ⊂ [0, ∞) such that all of the derivatives of f (t) exist on
the complement of S and have left and right limits at 0"

Once again: a function is defined on $[0,\infty)$ and it has left limit at 0.

And the text is full of such things!

 

[jasonRF]

Then one must provide precise definitions, theorems and proofs. In this respect the text you cited above is very bad.
For example look at the bottom of page 5:

" A function f (t) defined on the
interval [0, ∞) is piecewise smooth if there is a sparse set
S ⊂ [0, ∞) such that all of the derivatives of f (t) exist on
the complement of S and have left and right limits at 0"

Once again: a function is defined on $[0,\infty)$ and it has left limit at 0.

And the text is full of such things!

Fair enough. I primarily provided the reference in case folks were really interested in the motivation for the specific Laplace transform definition that the OP was asking about. For motivation I thought it was very good, even if it fails in the rigor department.

jason