PHYS20672 Summary 7

Laplace Transforms

  1. The Laplace transform of $f(t)$ is $$F(s)=\int_{0}^\infty \e^{-st} f(t) \d t.$$
  2. This definition is less subject to convention than the Fourier transform, though $x$ may be used in place of $t$ and $p$ in place of $s$. The Laplace transform is particularly suited to time-dependent problems with specified initial conditions though.
  3. The Laplace transform is insensitive to $f(t)$ for $t<0$. It will be unchanged if we replace $f(t)$ by $\theta(t)f(t)$, where $\theta(t)$ is the unit step function: 0 for $t<0$ and 1 for $t>0$.
  4. Convergence of the integral may require restrictions on $s$; for instance if $f(t)=e^{a s}$, $F(s)=1/(s-a)$ only for $s>a$.
  5. In practice the inverse is normally found via look-up tables such as given below. If $F(s)$ is the L.T. of $f(t)$, then $\theta(t)f(t)$ is the inverse Laplace transform of $F(s)$.
  6. The shift theorems are particularly useful: $$F(s-s_0)\to e^{s_0t}f(t)\theta(t) \quad\text{and}\quad e^{-st_0}F(s)\to f(t-t_0) \theta(t-t_0)$$
  7. The following expressions can be used to recast differential equations for $f(t)$ as algebraic ones for $F(s)$: $$f'(t)\to sF(s) - f(0) \quad\text{and}\quad f''(t)\to s^2F(s) -s f(0)- f'(0).$$ The initial conditions on $f(t)$ are incorporated directly.
  8. For Laplace transforms, the convolution $h(t)$ of two functions $f(t)$ and $g(t)$ is given by $h(t)= \int_{0}^t f(u)g(t-u)\d u$. This agrees with the previous definition (with an integral from $-\infty\to\infty$) if both functions vanish for negative values of their argument.
  9. The Laplace transform of a convolution $h(t)=f*g$ is a product: $H(s)=F(s)G(s)$.
  10. The inverse Laplace transform may be computed using the Bromwich integral, in which $s$ is treated as a complex variable lying on a line parallel to the imaginary axis: $$f(t)=\frac{1}{2\pi i}\int_{\lambda-i\infty}^{\lambda+i\infty} \e^{st} F(s) \d s.$$ The offset $\lambda$ must be positive, and large enough so that the line of integration lies to the right of all poles of $F(s)$.
  11. image of contour for Bromwich integral If $F(s)$ has only poles, and no more complicated analytic structure such as branch points, we can close the contour of integration as shown in the diagram (green arc). Provided $F(s)$ tends to zero as $|s|\to\infty$, if $t>0$ Jordan's Lemma ensures that contribution of the arc vanishes as the radius is taken to infinity. If $t<0$, we close the contour to the right instead. Hence $f(t)$ is just given by $\theta(t)$ times the sum of the residues of $\e^{st} F(s)$. If $F(s)$ has a branch cut the contour will be more complicated.









Arfken 15.8-12, Riley 11.2, 18.20, Boas 8.8-10, 14.7

Table of Laplace Transforms

$f(t)$ $F(s)=\int_0^\infty e^{-st}f(t)dt$ Restrictions
$1$ $\frac{1}{s}$ $s >0$
$e^{at}$ $\frac{1}{s-a}$ $ s >a$
$t^n$ $\frac{n!}{s^{n+1}}$ $ s>0$, $n$ a positive integer
$t^{-1/2} $ $\sqrt{\frac \pi s}$ $ s>0$
$\sin(at)$ $ \frac{a}{s^2+a^2}$ $ s>0$
$\cos(at)$ $ \frac{s}{s^2+a^2} $ $s >0$
$\sinh(at)$ $ \frac{a}{s^2-a^2}$ $ s>a$
$\cosh(at)$ $ \frac{s}{s^2-a^2} $ $s >a$
$t^n f(t)$ $ (-1)^n \frac{d^n}{ds^n}\left(F(s)\right)$ $ s>0$, $n$ a positive integer
$\frac {f(t)} t$ $ \int_s^\infty F(y) dy$ $ s>0$
$f(at)$ $ \frac 1 a\left(F(\frac s a)\right)$ $ s>0$
$f'(t)$ $sF(s) - f(0)$ $ s>0$
$f''(t)$ $s^2F(s) -s f(0)- f'(0)$ $ s>0$
$\theta(t-t_0)$ $\frac{e^{-s t_0}}{s}$ $t_0,s >0$
$\theta(t-t_0)f(t-t_0)$ $ e^{-st_0}F(s)$ $t_0,s>0$
$e^{s_0t} f(t)$ $F(s-s_0)$ $ s>s_0$
$\delta(t-t_0)$ $e^{-s t_0}$ $t_0,s >0$