On Laplace transform of derivative

[psie]

TL;DR Summary

I'm reading Ordinary Differential Equations by Adkins and Davidson. In it, they prove a theorem on the Laplace transform of the derivative of a function. I do not understand a part of the proof.

The following three results are used in the proof of the theorem I have a question about.

Lemma 2. Suppose ##f## is of exponential type of order ##a##. Let ##s>a##, then $$\lim_{t\to\infty}f(t)e^{-st}=0.$$

Proposition 3. Let ##f## be a continuous function of exponential type of order ##a##. Then the Laplace transform ##F(s)=\mathcal{L}\{f(t)\}(s)## exists for all ##s>a## and, moreover, ##\lim_{s\to\infty}F(s)=0##.

Lemma 4. Suppose ##f## is a continuous function defined on ##[0,\infty)## of exponential type of order ##a\geq 0##. Then any antiderivative of ##f## is also of exponential type and has order ##a## if ##a>0##.

Now follows the theorem and its proof I have a question about.

Theorem 6. Suppose ##f(t)## is a differentiable function on ##[0,\infty)## whose derivative ##f'(t)## is continuous and of exponential type of order ##a\geq 0##. Then $$\mathcal{L}\{f'(t)\}(s)=s\mathcal{L}\{f(t)\}(s)-f(0),\quad s>a.$$

Proof. By Lemma 4, ##f(t)## is of exponential type. By Proposition 3, both ##f(t)## and ##f'(t)## have Laplace transforms. Using integration by parts [...], we get \begin{align} \mathcal{L}\{f'(t)\}(s)&=\int_0^\infty e^{-st} f'(t) dt \nonumber \\ &=e^{-st}f(t)\rvert_0^\infty-\int_0^\infty -se^{-st}f(t)dt \nonumber \\ &=-f(0)+s\int_0^\infty e^{-st} f(t)dt=s\mathcal{L}\{f(t)\}(s)-f(0).\nonumber\end{align}

I do not understand why ##e^{-st}f(t)\rvert_0^\infty=-f(0)##. By Lemma 2, this is only possible if ##f## is of exponential type of order ##a## and by assumption we only know that ##f'(t)## is of exponential type of order ##a\geq 0##. If ##f'(t)## is of order ##a>0##, then yes, by Proposition 3, ##f(t)## will also be of order ##a##. However, what about the case ##a=0##? Which order does ##f(t)## then have? How is the theorem modified in regards to the condition ##s>a##? (if say ##f'(t)## is of order ##a## and ##f(t)## is of order ##b##, then we need ##s>b##, right?)

 

[fresh_42]

Let me sort this out. For ##a=0## we get that ##f' ## is constant and ##f## is linear. Then ##\left.e^{-st}f(t)\right|_0^\infty =0=-f(0)## as long as ##s>0=a.## In any other case, we get
\begin{align*}
\operatorname{ord} f' \geq a > 0\quad&\text{by assumption of theorem 6}\\
\operatorname{ord} f' = \bar{a} \geq a > 0 \quad&\text{my setting}\\
\operatorname{ord} f > \bar{a}\geq a > 0 \quad&\text{lemma 4}\\
\lim_{t \to \infty}e^{-st}f(t) =0 \;\forall \;s >\operatorname{ord} f > a\quad&\text{lemma 2}\\
\left.e^{-st}f(t)\right|_0^\infty =0-e^0f(0)=-f(0)
\end{align*}

 

[psie]

Thank you. I apologize for not having stated their definition of exponential type of order ##a##, but it reads $$|f(t)|\leq Ke^{at},$$ for the domain ##[0,\infty)## and some constant ##K##. Then,

For ##a=0## we get that ##f' ## is constant and ##f## is linear.

But ##\sin{t}## is of exponential type of order ##a=0##, since its absolute value is bounded by ##K=1##. I guess all we can conclude from ##a=0## is that ##f'(t)## is bounded.

Anyway, ##e^{-st}f(t)\rvert_0^\infty =-f(0)## still holds for ##s>0=a## I would say, although it is a bit strange not really knowing how the function ##f(t)## behaves in the limit as ##t\to\infty##.

 

[fresh_42]

Anyway, ##e^{-st}f(t)\rvert_0^\infty =-f(0)## still holds for ##s>0=a## I would say, although it is a bit strange not really knowing how the function ##f(t)## behaves in the limit as ##t\to\infty##.

The idea is that the behavior of ##e^{st}## dominates every factor that is less "explosive". And with negative powers, it eliminates them faster.

We do not actually need linearity. If ##|f(t)|<Ke^{0\cdot t}=K## then ##[e^{-st}f(t)]_0^\infty = 0- e^0\cdot f(0)=-f(0).##

 

[psie]

Hmm, ok. I'm probably not seeing something you are, but to summarize;

We know ##f'(t)## is of exponential type of order ##a=0##. We know that ##f(t)## is also of exponential type, but we don't know which order, it could be say ##1##. Then the derivative formula in theorem 6 does not hold for ##s>a=0##, but rather ##s>1## (because the limit ##\lim_{t\to\infty}f(t)e^{-st}## blows up for, e.g., ##s=1/2##).

If this observation is correct, then I feel like this is an important detail missing in the text, but maybe I'm misunderstanding something obvious.

 

[pasmith]

However, what about the case ##a=0##? Which order does ##f(t)## then have?

If [itex]f'[/itex] is of order 0, then [itex]|f'(t)| < K[/itex] on [itex][0,\infty)[/itex]. But then, for [itex]t \geq 0[/itex], [tex]\begin{split}
|f(t)| &= \left|f(0) + \int_0^t f'(x)\,dx\right| \\
&\leq |f(0)| + \int_0^t |f'(x)|\,dx \\
&\leq |f(0)| + Kt \\
&\leq (|f(0)| + K)e^{at} \end{split}[/tex] is of order [itex]a > 0[/itex].

 

[fresh_42]

The order does not change in case it is positive according to Lemma 4. That leaves us with ##\operatorname{ord}f' =0## which means that ##|f'(x)|<K.## We want to see why
$$
\left. e^{-st}f(t)\right|_0^\infty =-f(0) \Longleftrightarrow \lim_{t \to \infty}e^{-st}f(t)=0
$$
for ##s>0.## We get from the calculation in the previous post #6 by @pasmith that ##|f(t)|<|f(0)|+Kt.## Thus
$$
\left|e^{-st}f(t)\right| \leq \left|e^{-st}\right|\cdot |f(0)|+K \left|e^{-st}t\right| \stackrel{t\to \infty }{\longrightarrow }0 \text{ for any }s>0
$$

 

[psie]

Thank you both, means a lot. I have to ask, @pasmith, how did you obtain that last inequality, i.e. $$|f(0)| + Kt\leq (|f(0)| + K)e^{at},$$ for ##a>0##?

 

[fresh_42]

Thank you both, means a lot. I have to ask, @pasmith, how did you obtain that last inequality, i.e. $$|f(0)| + Kt\leq (|f(0)| + K)e^{at},$$ for ##a>0##?

This is only true for large values of ##t##, but we are only interested in large values of ##t##.

 

[WWGD]

Doesn't just the fact that ##K>0## and ##e^{at}>1## for ##a,t>0## do it?

 

[psie]

Doesn't just the fact that ##K>0## and ##e^{at}>1## for ##a,t>0## do it?

Maybe you were suggesting the following:

We have that ##t<t+1\leq e^t## for all ##t\in\mathbb R##, where the second inequality follows from Bernoulli's inequality, i.e. $$1+t\leq \left(1+\frac{t}{n}\right)^n\to e^t,$$ as ##n\to\infty## and for ##t\geq -1## (for ##t<-1## the inequality holds trivially since ##e^t>0##). Of course, ##1\leq e^{Kt}## for ##K,t## non-negative. So in that case, $$|f(0)| + Kt\leq (|f(0)| + 1)e^{Kt}.$$

 

[WWGD]

Maybe you were suggesting the following:

We have that ##t<t+1\leq e^t## for all ##t\in\mathbb R##, where the second inequality follows from Bernoulli's inequality, i.e. $$1+t\leq \left(1+\frac{t}{n}\right)^n\to e^t,$$ as ##n\to\infty## and for ##t\geq -1## (for ##t<-1## the inequality holds trivially since ##e^t>0##). Of course, ##1\leq e^{Kt}## for ##K,t## non-negative. So in that Herr.

Edit: Please ignore.

$$|f(0)| + Kt\leq (|f(0)| + 1)e^{Kt}.$$

Yes, pretty much this.
Edit: Please ignore.

 

[fresh_42]

Yes, pretty much this.

But the ##a## makes the difference. The values of ##t## have first to outrun the smallness of ##a=\varepsilon >0.##

 

[WWGD]

But the ##a## makes the difference. The values of ##t## have first to outrun the smallness of ##a=\varepsilon >0.##

I didn't see anywhere where a approached 0. Maybe I misread. Edit: Since ## t \rightarrow \infty##, why not just select ## t > 1/a##?

 

[fresh_42]

I didn't see anywhere where a approached 0. Maybe I misread.

It does not 'approach' zero. The second half of the discussion was basically all about whether ##a## close to zero (as in ##a>0##) is possible, or if ##a>1## has to be assumed. I thought your post was about whether ##t<e^{at}## may be assumed or not since this was the question in the posts before yours.

 

[WWGD]

It does not 'approach' zero. The second half of the discussion was basically all about whether ##a## close to zero (as in ##a>0##) is possible, or if ##a>1## has to be assumed. I thought your post was about whether ##t<e^{at}## may be assumed or not since this was the question in the posts before yours.

Ah, my bad, I'll just bow out of this one. I will just quickly edit my first post here.

 

[fresh_42]

Ah, my bad, I'll just bow out of this one. I will just quickly edit my first post here.

Don't. I guess that I am totally confused by now. Too many threads within this thread. At least I convinced myself that the book has no error, although case ##a=0## seems a bit academic to me.

I possibly made my life easier by just demanding ##a>0## if I were the author. Are there examples for
$$
\mathcal{L}\{f'(t)\}(s)=s\mathcal{L}\{f(t)\}(s)-f(0),\quad s>0
$$
where ##f'## is only continuous and bounded?

 

[pasmith]

It does not 'approach' zero. The second half of the discussion was basically all about whether ##a## close to zero (as in ##a>0##) is possible, or if ##a>1## has to be assumed. I thought your post was about whether ##t<e^{at}## may be assumed or not since this was the question in the posts before yours.

[itex]a > 1[/itex] is required, I think.

But to answer the OP's original question, we need [itex]\lim_{t \to \infty} e^{-st}f(t) = 0[/itex] for some [itex]s >0[/itex] in order for [itex]f[/itex] to have a Laplace transform.