Laplace Tranform

Laplace transform is useful in equation solving. By definition, Laplace transform transforms a function $f(x)$ defined on $x\geq 0$ into another function by calculating the convolution

$$ F(s) = \int_0^\infty e^{-s x} f(x) dx, $$

or symbolically,

$$ F(s) = \mathcal L_s (f(x)). $$

A table of important Laplace transforms can be found on mathworld.wolfram.com. Here we steal some of the commonly used.

$$ \begin{align} \mathcal L_s (1) =& \frac{1}{s}\\ \mathcal L_s (x^n) =& \frac{n!}{s^{n+1}}\\ \mathcal \sin(kx) =& \frac{k}{k^2+s^2} \\ \mathcal \cos(kx) =& \frac{s}{k^2+s^2}. \end{align} $$

It can also be applied to differentials.

$$ \begin{align} \mathcal L_s \left(\frac{d f(x)}{dx}\right) =& s \mathcal L_s (f(x)) - f(0) \\ \mathcal L_s \left(\frac{d^2f(x)}{dx^2}\right) =& s^2 \mathcal L_s (f(x)) - s f(0) - \left.\frac{df(x)}{dx}\right\vert_{x=0}. \end{align} $$

Laplace Transform of Differentials

The general form is

$$ \mathcal L_s (f^{(n)}(x)) = s^n \mathcal L (f(x)) - \sum_{m=1}^n s^{n-m} f^{(m-1)}(0). $$

Integrals with upper limit as argument also transform nicely.

$$ \mathcal L_s \left( \int_0^x f(x') dx' \right) = \frac{\mathcal L_s(f)}{s}. $$

Laplace transform is useful in solving differential equations because it transforms many equations into fractions and polynomials. A simple example is the harmonic osicllators. The equation of motion is

$$ \ddot x(t) = - \omega^2 x(t), $$

which is transformed into

$$ s^2 X(s) - s x(0) - \dot x(0) = - \omega^2 X(s). $$

The solution to it is

$$ X(s) = \frac{ s x(0) + \dot x(0)}{s^2 + \omega^2} = x(0)\frac{ s }{s^2 + \omega^2} + \frac{\dot x(0)}{\omega} \frac{\omega}{s^2 + \omega^2}. $$

We can spot sin and cos from the solution, or more generally perform an inverse Laplace transform,

$$ x(t) = x(0) \cos(\omega t) + \frac{\dot x(0)}{\omega} \sin(\omega t). $$

This example is too simple sometimes naive. However, it shows the spirit.

Caveats

Laplace transform has a lot of counter intuitive expressions.

1. Laplace transform of product of two functions $f(x)g(x)$ is NOT the product of the Laplace transforms $\mathcal L_s(f)\mathcal L_s(g)$. However, if one of the functions is a constant, say $f(x)=5$, we can prove that the Laplace transform of $5g(x)$ is $5\mathcal L_s(g)$.

2. The product of two Laplace transforms $\mathcal L_s(f)\mathcal L_s(g)$ is the Laplace transform of a convolution

   $$ \int_0^x f(x’) g(x-x’) dx’. $$

3. Small s corresponds to large x, due to the nature of the exponential suppression in Laplace transform. For example, for small argument x, the function $e^{k t}$ becomes almost 1, meanwhile, the Laplace transform of the function $1/(s - k)$ becomes $1/s$ under large s. We can see that the two limits are consistant since the Laplace transform of 1 is $1/s$.

4. In so many circumstance the Laplace transform doesn’t exist simple because the integral doesn’t converge. Please beware of this and use the transform only when it exists.

Inverse Laplace Transform

We do not usually use the general form of inverse Laplace transform since we can find it in the table. Nevertheless we write it down here.

$$ f(x) = \frac{1}{2\pi i} \int_{-i\infty}^{i \infty} F(s) e^{sx}ds. $$

By defining $s=i s^\dagger$, we can rewrite the formula

$$ f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(is^\dagger) e^{is^\dagger x} ds^\dagger . $$

The more interesting application is to solve matrix differential equations. For any equations

$$ \partial_x \mathbf f(x) = \mathbf A \mathbf f(x), $$

Laplace transform takes it to the form

$$ s \mathbf F(x) - f(0) = \mathbf A \mathbf F(s). $$

The solution is

$$ \mathbf F(x) = \frac{1}{s \mathbf I - \mathbf A} \mathbf f(0) . $$

So the final solution for $f(x)$ is

$$ \mathbf f(x) = \mathcal L^{-1} \left(\frac{1}{s \mathbf I - \mathbf A} \right)\mathbf f(0). $$

We could work out the Taylor expansion of solution,

$$ \frac{1}{s \mathbf I - \mathbf A} = \frac{\mathbf I}{s}+ \frac{\mathbf A}{s^2} + \frac{\mathbf A^2}{s^3} \cdots. $$

The inverse Laplace transform can be done simply term by term,

$$ \mathcal L^{-1} \left(\frac{1}{s \mathbf I - \mathbf A} \right) = \mathbf I + x \mathbf A + \frac{1}{2!} (x \mathbf A)^2 + \cdots = e^{x \mathbf A}. $$

Finally we obtain the formal solution of the system, which is

$$ \mathbf f(x) = \exp\left( x \mathbf A \right)\mathbf f(0). $$

Only Works for Constant Coefficients

This result only works for constant coefficients. In general, if the matrix $A$ depends on the argument $x$, the solution can be systematically calculated using the so called Magnus Expansion. However, it is as tedious as a numerical solution.