Properties of Laplace transform

The properties of the Laplace transform are the algebraic identities that make it useful for solving ODEs. Together they give: how derivatives transform, how shifts and exponential modulations transform, and what conditions $f$ must satisfy for the transform to exist.

Existence conditions

The Laplace transform $\mathcal{L}\{f\}(s)={\int }_0^{\infty }{e}^{-st}f(t)dt$ exists (converges) for $s>a$ if:

1. $f$ is piecewise continuous on $[0,\infty )$.
2. $f$ has exponential order: there exist constants $T,M>0$ and $a$ such that

$|f(t)|\le M{e}^{at}\text{for all }t\ge T$

The two conditions together ensure $f$ doesn’t grow faster than some exponential.

Piecewise continuity

A function $f(t)$ is piecewise continuous on $[0,\infty )$ if for any interval $[0,N]$, you can split it into a finite number of subintervals where $f$ is continuous, except possibly at jump discontinuities at the boundaries.

A jump discontinuity is when $f$ has a discontinuity at $t_0$ but the one-sided limits ${\lim }_{t\to t_0^{-}}f(t)$ and ${\lim }_{t\to t_0^{+}}f(t)$ both exist and are finite.

Functions with worse discontinuities (like infinite jumps) can’t be Laplace-transformed in the standard sense.

Exponential order

This rules out functions that grow too fast. $f(t)=t^2$ is exponential order ($a=1$ works). $f(t)=e^{t^2}$ is not — it grows faster than any $e^{at}$. So $\mathcal{L}\{e^{t^2}\}$ doesn’t exist.

Derivative property

The most important property, and what makes Laplace useful for ODEs:

$\mathcal{L}\{f^{\prime }(t)\}(s)=sF(s)-f(0)$

By induction:

$\mathcal{L}\{f^{(n)}(t)\}(s)=s^{n}F(s)-s^{n-1}f(0)-s^{n-2}{f}^{\prime }(0)-\cdots -f^{(n-1)}(0)$

In particular:

$\mathcal{L}\{y^{\prime }(t)\}=sY(s)-y(0)$

$\mathcal{L}\{y^{\prime \prime }(t)\}=s^{2}Y(s)-sy(0)-y^{\prime }(0)$

This is what turns differentiation into multiplication by $s$. ODEs with constant coefficients become polynomial-coefficient algebraic equations in $Y(s)$.

The minus signs (with initial values) mean initial conditions get baked in automatically — you don’t have to apply them separately at the end.

Linearity

For constants $c_1,c_2$:

$\mathcal{L}\{c_{1}f(t)+c_{2}g(t)\}=c_{1}F(s)+c_{2}G(s)$

Trivial from the linearity of integration.

First shifting theorem (s-shift)

Multiplying by $e^{at}$ in the time domain shifts the s-variable:

$\mathcal{L}\{e^{at}f(t)\}(s)=F(s-a)\text{for }\mathrm{Re}(s-a)>a_0$

where $a_0$ is the abscissa of convergence of $f$ (the smallest real $\alpha$ such that $f$ is exponential-order with rate $\alpha$, in the sense of the existence condition above). In words: the original transform $F$ converges for $\mathrm{Re}(s)>a_0$; multiplying by $e^{at}$ in time shifts that region to $\mathrm{Re}(s)>a_0+a$. The "$T$" appearing in some texts is the time after which the exponential-order bound holds and is unrelated to the convergence variable — don’t confuse the two.

Inverse:

${\mathcal{L}}^{-1}\{F(s-a)\}(t)=e^{at}{\mathcal{L}}^{-1}\{F(s)\}(t)$

Useful corollary: ${\mathcal{L}}^{-1}\left\{\frac{1}{(s-a{)}^k}\right\}(t)=\frac{t^{k-1}{e}^{at}}{(k-1)!}$.

Second shifting theorem (t-shift)

Multiplying by $u(t-a)$ (Heaviside step function) and shifting time:

$\mathcal{L}\{u(t-a)f(t-a)\}(s)=e^{-as}F(s)$

Inverse:

${\mathcal{L}}^{-1}\{e^{-as}F(s)\}(t)=u(t-a)f(t-a)$

This handles time-shifted forcing in IVPs — for example, when a force is applied at $t=a$ rather than $t=0$.

Convolution

Products in the s-domain correspond to convolutions in time:

$\mathcal{L}\{f\ast g\}(s)=F(s)G(s)$

where $(f\ast g)(t)={\int }_0^{t}f(t-\tau )g(\tau )d\tau$. This is huge for solving ODEs with arbitrary forcing.

Multiplication by $t^n$

$\mathcal{L}\{t^{n}f(t)\}(s)=(-1{)}^n{F}^{(n)}(s)$

Differentiating $F$ in $s$ corresponds to multiplying $f$ by $-t$. Useful occasionally.

Worked example

Compute $\mathcal{L}\{e^{at}\}$ via the integral and verify against the table:

$F(s)={\int }_0^{\infty }{e}^{-st}{e}^{at}dt={\int }_0^{\infty }{e}^{(a-s)t}dt={\left[\frac{e^{(a-s)t}}{a-s}\right]}_0^{\infty }$

For $s>a$: the integrand $e^{(a-s)t}$ decays since $a-s<0$, so the integral converges to $\frac{0-1}{a-s}=\frac{1}{s-a}$. ✓

For $s\le a$: the integrand grows or stays bounded but doesn’t decay, integral diverges.

So $\mathcal{L}\{e^{at}\}=\frac{1}{s-a}$ for $s>a$.

Verifying derivative property

Compute $\mathcal{L}\{\sin (3t)\}$ directly: $\frac{3}{s^2+9}$.

Now use the derivative property. $f(t)=\sin (3t)$, $f^{\prime }(t)=3\cos (3t)$, $f(0)=0$.

$\mathcal{L}\{f^{\prime }(t)\}=sF(s)-f(0)=s\cdot \frac{3}{s^2+9}-0=\frac{3s}{s^2+9}$

Direct check: $\mathcal{L}\{3\cos (3t)\}=3\cdot \frac{s}{s^2+9}=\frac{3s}{s^2+9}$. ✓

In context

These properties together make the Laplace transform a complete tool for solving linear ODEs with constant coefficients and any reasonable forcing. The procedure is in Method of Laplace transform.

For inverting transforms (the reverse direction, often the harder step), see Inverse Laplace transform and the partial-fractions techniques used there.