# Asymptotic behaviour of Laplace transform

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

$$\frac{1}{s^n}$$

Therefore

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

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 $$t=0$$.

Edit:

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.

• They're probably using the Riemann integral. – 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 $$1 = x(0) \neq x(0^+) = 0$$.

Then,

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

• 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 $C^\infty$. – reuns May 12 at 2:18
• @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? – Pantelis Sopasakis May 12 at 2:39
• 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. – Pantelis Sopasakis May 12 at 2:48
• 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 $C |e^{-s}|$. – reuns May 12 at 2:52
• 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? – roy212 May 12 at 20:18