The Laplace Transform is another form of integral transform which is used in applied mathematics to solve differential equations. Although it lacks some of the mathematical elegance of the Fourier Transform, it is still extremely useful. It is very popular in engineering, and unlike the Fourier Transform can be used without knowledge of complex numbers.
Definition
Let f(t) be a function which does not grow exponentially as t increases. Then the Laplace Transform is defined as
F(p) = integral(f(t)exp(-pt)dt, t=0...inf).
The Laplace Transform F(p) is often denoted as f with a circumflex ("^"). Since the Laplace Transform does not depend on the values of f for negative t, we usually assume f(t)=0 in this region.
Basic properties
Using elementary properties of integration we can prove the following useful results.
- Linearity. Let h(t) = f(t) + g(t). Then H(p) = F(p) + G(p).
- Translation. Let g(t) = f(t-a) for a>0. Then G(p) = exp(-ap) F(p).
- Frequency shift. Let g(t) = exp(lt) f(t). Then G(p) = F(p-l) for p>l.
- Scaling. Let g(t) = f(at) for a>0. Then G(p) = F(p/a)/a.
- Derivative. Let g(t) = f'(t). Then G(p) = pF(p) - f(0).
- Second derivative. Let g(t) = f''(t). Then G(p) = p2F(p)-pf(0)-f'(0).
- Factors of t. Let g(t) = tf(t). Then G(p) = - dF(p)/dp.
Many of these properties are similar to the Fourier Transform. Results 5 and 6 are critical in solving differential equations: see the Laplace Transformation node for more details.
Inversion
Inversion is one place where the Laplace Transform lacks mathematical elegance, since it requires an ugly contour integral in the complex plane. Consequently most people who use Laplace Transforms for practical purposes look up Laplace inversions in tables.
If f(t) has Laplace Transform F(p), then we have the relation
F(t) = 1/(2πi) integral(F(p)exp(pt)dp, p=γ-i∞...γ+i∞)
where γ is a large real number, chosen so that the path of integration lies to the right of all the singularities of F(p). This is known as the Bromwich inversion integral; see this node for a full mathematical treatment of where this result comes from.
Laplace versus Fourier
Both Laplace Transforms and Fourier Transforms can be used to solve differential equations, so a natural question to ask is "which one is better?".
Because of the negative exponential term in the Laplace Transform integration, convergence is a lot stronger, and polynomial expressions which could not be handled with the Fourier Transform can be happily considered. However Fourier Transforms make up for this with a greater mathematical elegance, and tie in beautifully with many areas of applied mathematics, particularly quantum mechanics. The Fourier Transform also has many applications beyond the solution of differential equations.
The Laplace Transform picks out precisely the solution to a differential equation that obeys certain initial conditions at t=0. Hence it is ideally suited to initial value problems, like the one given above.
The Fourier Transform picks out precisely the solution to a differential equation which decays to zero at large distances. In many applied problems this is exactly the solution we want, since a function which grows at large distances would be considered unphysical.
Often a mixture of the two methods proves the most effective. We are often interested in functions that depend on time and space. The most physical solution to an equation is one which obeys initial conditions at t=0, and which tends to zero at large distances. This directly corresponds to Fourier transforming with respect space and Laplace transforming with respect to time.