On Laplace transform of derivative

[psie]

TL;DR

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 f&#039; is of order 0, then |f&#039;(t)| &lt; K on [0,\infty). But then, for t \geq 0, \begin{split}<br /> |f(t)| &amp;= \left|f(0) + \int_0^t f&#039;(x)\,dx\right| \\<br /> &amp;\leq |f(0)| + \int_0^t |f&#039;(x)|\,dx \\<br /> &amp;\leq |f(0)| + Kt \\<br /> &amp;\leq (|f(0)| + K)e^{at} \end{split} is of order a &gt; 0.

 

[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.

a &gt; 1 is required, I think.

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