If the functions $x(t)$ and its derivatives $x'(t), x''(t), \ldots, x^n(t)$ are continuous* and $x(0^+) = x'(0^+) = x''(0^+) \ldots = x^{n-1}(0^+)=0$ ($0^+$ denotes the right side limit when the independent variable approaches 0) and they don't have continuity problems at $0$, then the Laplace Transform of nth derivative of the function $x$ is

$$ \mathscr{L}\left [ x^{n}(t) \right ]= s^n X(s) $$

*Note: the nth derivative $x^n$ can be piecewise continuous.

The terms $-s^{n-1}x(0^+)-s^{n-2}x'(0^+) \ldots -s x^{n-2}(0^+) - x^{n-1}(0^+)$ are 0.

$\lim_{s \to \infty} s^n X(s) = 0$

This implies that $X(s) \to 0$ faster than



$$X(s) = o \left (\frac {1}{s^n} \right )$$

a) Why does the function $x(t)$ and its derivatives $x'(t)$, $x"(t)$, $\ldots x^n(t)$ need to be continuous and $x^n(t)$ could also be piecewise continuous?

b) Why can't the function $x(t)$ and its derivatives $x'(t)$, $x"(t)$, $\ldots x^n(t)$ have discontinuity problems at 0? Here, one is taking the right-side limit when the variable approaches 0. So I think that it doesn't matter if these functions have discontinuities at $t=0$.


If the functions $x(t)$ and its derivatives $x'(t)$, $x"(t)$, $\ldots , x^n(t)$ are continuous.

And $x(0)$ = $x'(0)$= $x"(0) \ldots = x^{n-1}(0)=0$ (Here I replaced $(0^+)$ by $(0)$)

and they don't have discontinuity problems at $t=0$

then the $\mathscr{L}\{x^{(n)}\}(s) = s^{n}X(s) - s^{n-1}x(0) - \ldots - x^{(n-1)}(0)$ exists.

In this case. It is necessary that the functions don't have discontinuity problems at 0. Because if they do one can't evaluate the function or its derivatives up to $x^{(n-1)}$ at t=0.

But if one says that if $x(t)$ and its derivatives $x'(t)$, $x"(t)$, $\ldots , x^n(t)$ are continuous.

And $x(0^+)$ = $x'(0^+)$= $x"(0^+) \ldots = x^{n-1}(0^+)=0$.

Do you need the third assumption "and they don't have discontinuity problems at $t=0$", or you can completely remove it?

Because one is taking the right-side limits of the function $x(t)$ and its derivatives when $t$ approaches zero.

Note: The function is not continuous when it says that the function has discontinuity problems.

  • $\begingroup$ They're probably using the Riemann integral. $\endgroup$ – DisintegratingByParts May 10 at 0:38

Simple case. If $x$ is $n-1$ times differentiable at $0$, then, according to Taylor's theorem

$$ x(t) = x(0) + \dot{x}(0)t + \frac{\ddot{x}(0)}{2!}t^2 + \ldots + \frac{x^{(n-1)}(0)}{(n-1)!}t^{n-1} + r(t), $$

for $t\geq 0$ and $r(t) = o(t^{n-1})$ as $t\to 0$.

Under the assumption that $x(0) = \dot{x}(0) = \ldots = x^{(n-1)}(0) = 0$, we have $x(t) = o(t^{n-1})$ as $t\to 0$. This means that for all $c>0$, there is a $T>0$ so that $|x(t)| \leq c t^{n-1}$ for $t\in [0, T)$. If we multiply by $e^{-s\tau}$ and integrate from $0$ to $\infty$, we obtain

$$X(s) = o\left(\tfrac{1}{s^{n}}\right) \text{ as } s\to\infty,$$

where $s\in\mathbb{R}$.

Just to be clear, this is because

\begin{align} X(s) {}={}& \int_0^\infty e^{-s\tau}x(\tau) \mathrm{d}\tau \\ {}={}& \int_0^\infty e^{-\xi}x(\xi/s) \mathrm{d}\xi \\ {}\approx{}& \int_0^\infty e^{-\xi} \cdot o\left(|{\xi}/{s}|^{n-1}\right)\mathrm{d}\xi \text{ as } s\to \infty \\ {}={}& o\left(\tfrac{1}{s^{n}}\right) \text{ as } s\to \infty \end{align}

Proposition. Suppose that $x$ is $n-1$ times differentiable for all $t\in (0, t_0)$ for some $t_0 > 0$. Suppose that the following limits exist and are equal to zero: $x(0^+) = \ldots = x^{(n-1)}(0^+) = 0$. Suppose also that $x$ has a Laplace transform, which is denoted by $X$. Then, we have that

\begin{align} X(s) = o \left(\tfrac{1}{s^{n}}\right), \text{ as } s\to\infty, \end{align}

with $s\in\mathbb{R}$.

Proof. Since $x(0^+) = \ldots = x^{(n-1)}(0^+) = 0$, for all $\epsilon > 0$ there is a $T>0$ such that $|x(t)| < \epsilon$, $|\dot{x}(t)| < \epsilon$, $\ldots$, $|x^{(n-1)}(t)| < \epsilon$ for all $t \in (0, T)$. Then, choose a $\zeta \in (0, T)$ and apply Taylor's theorem on $x$ at $\zeta$:

\begin{align} x(t) = x(\zeta) + \dot{x}(t)(\zeta)(t-\zeta) + \frac{\ddot{x}(\zeta)}{2}(t-\zeta)^2 + \ldots + \frac{x^{(n-1)}}{(n-1)!}(t-\zeta)^{n-1} + r(t), \end{align}

where $r(t) = o(|t-\zeta|^{n-1})$. But then,

\begin{align} x(t) = \underbrace{\epsilon \left(1 + T + \tfrac{T^2}{2} + \ldots + \tfrac{T^{n-1}}{(n-1)!}\right)}_{\beta} + r(t), \end{align}

and proceeding as above we conclude that $X(s) = o \left(\tfrac{1}{s^{n}}\right)$. $\Box$

About the assumptions. In the above statement we did not use the assumptions of piecewise continuity, although these are often imposed (along side exponential order) to guarantee that $x$ has a Laplace transform. By the way, since $x$ is $n-1$ times differentiable on $(0, t_0)$, its derivatives up to order $n-2$ must be continuous, so only $x^{(n-1)}$ can be assumed to be discontinuous.

If this is an exercise in a textbook, I guess that the assumption that all derivatives are continuous is given to allow to use the fact that $\mathscr{L}\{x^{(n)}\}(s) = s^{n}X(s) - s^{n-1}x(0^+) - \ldots - x^{(n-1)}(0^+)$, which can also be used to prove the above result.

Update. Here is an example of a function $x(t)$ that satisfies the assumptions of the above proposition, but is not continuous at $0$.

$$ x(t) = \begin{cases} 1, &\text{ if } t = 0 \\ t^2\sin(t), & \text{ if } 0 < t \leq 1 \\ e^{-t}, & \text{ if } t > 1 \end{cases} $$

This has a discontinuity at $t=0$ (it doesn't matter) and another discontinuity at $t=1$ (again, it doesn't matter). It is continuous and infinitely many times differentiable over $(0,1)$ and

$$ x(0^+) = \dot{x}(0^+) = \ddot{x}(0^+) = 0, $$

but $x^{(3)}(0^+) = 6 \neq 0$. Note that $1 = x(0) \neq x(0^+) = 0$.


$$ X(s) = o(1/s^3), \text{ as } s\to \infty. $$

  • $\begingroup$ If $x(t)$ is supported on $t \ge 0$ and it is $n$-times differentiable and $x^{(n)}(t)e^{-ct} \in L^1$ then $\mathcal{L}[x^{(n)}(t)](s)=s^n \mathcal{L}[x(t)](s)\to 0$ uniformly as $\Re(s) \to \infty$. Conversely if $s^3 \mathcal{L}[x(t)](s)\to 0$ uniformly as $\Re(s) \to \infty$ then $s \mathcal{L}[x(t)](s)$ is $L^1$ on vertical lines so that $x'(t)= \frac{1}{2i\pi} \int_{a-i\infty}^{a+i\infty} s \mathcal{L}[x(t)](s) e^{st}ds$ is bounded and continuous. Thus OP's statement fails with $x(t) = \sin(e^t)e^{-1/t}1_{t > 0}$ which is $C^\infty$. $\endgroup$ – reuns May 12 at 2:18
  • $\begingroup$ @reuns Thanks for the comment. Function $x(t)=\sin(e^t)e^{-1/t}1_{t>0}$ has the property $0=x(0^+) = \dot{x}(0^+) = \ldots$ for all derivatives. Its Laplace transform goes to 0 as $s \to \infty$ as $o(1/s^k)$ for any $k$. Isn't it? $\endgroup$ – Pantelis Sopasakis May 12 at 2:39
  • $\begingroup$ Oh, I see (the penny dropped). In my answer I assumed $s\to\infty$ over the reals. It would be nice if you could also provide an answer. $\endgroup$ – Pantelis Sopasakis May 12 at 2:48
  • 1
    $\begingroup$ For the asymptotic on the real line : if $x(t)$ is smooth and bounded then $x(t) = y(z)+z(t)$ where $z(t)$ is bounded supported on $t > 1$ and $y(t)$ is smooth and supported on $[0,2]$, thus for every $n, y^{(n}(t)\in L^1$ so the Laplace transform of $y(t)$ decays faster than $s^{-n}$ uniformly as $\Re(s) \to \infty$, and the Laplace transform of $z(t)$ is bounded by $C |e^{-s}|$. $\endgroup$ – reuns May 12 at 2:52
  • $\begingroup$ Thanks a lot for your detailed answer. About the assumptions in the OP I have doubts if they are right. Can you check my edit in the OP? $\endgroup$ – roy212 May 12 at 20:18

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