The Laplace Transformation (after the French mathematician Pierre Simon Marquis de Laplace) is a method for solving differential equations, and the corresponding initial and boundary value problems.

Finding a solution of a differential equation using the Laplace Transformation consists of three steps:

  1. Transformation of the equation into a subsidiary equation (moving the equation from the so called t-space to s-space)
  2. Solving the subsidiary equation by simple algebraic manipulations
  3. Transforming the subsidiary equation back into t-space to obtain the solution of the given problem
The Laplace Transformation is widely used in engineering, especially in problems where the (mechanical/electrical) driving force has discontinuities. Another common use of the Laplace Transformation is in Process Control.

I will only describe the basic definition of the Laplace Transform, some basic properties, and a basic example on the use of this method.

Consider a function f(t), defined for all t≥0. Multiply this function by e-st, and integrate with respect to t from zero to infinity. The resulting integral F(s) is given by:

          F(s) = ∫ e-st f(t) dt
The function F(s) of the variable s is called the Laplace Transform, denoted by L(f). Furthermore, the original function f(t) is called the inverse transform, denoted by L-1(F):
          L(f) = ∫ e-st f(t) dt

          f(t) = L-1(F)
Some simple general transforms are given in the following table. More complex transforms can be found in books of mathematical tables.
  f(t)        L(f)
  ----        ----
  1           1/s
  t           1/s2
  tn          n!/(sn+1)    n=1, 2, ...
  eat         1/(s-a)

An important property of the Laplace Transform is its linearity:

          L{af(t) + bg(t)} = aL{f(t)} + bL{g(t)}
In order to solve differential equations using Laplace Transforms, the transform of the derivative of a function f(t) is needed. This is given by:
          L(f') = sL(f) - f(0)
This definition can be extended to transforms of derivatives of arbitrary higher orders:
          L(fn) = snL(f) - sn-1 f(0) - sn-2 f'(0) - ... - f(n-1)(0)

Example: solve the initial value problem:

          y" + 4y' +3y = 0    y(0) = 3, y'(0) = 1
  1. First set up the subsidiary equations, with Y(s) = L(y):
              L(y') = sY - y(0) = sY - 3
              L(y") = s2Y -sy(0) - y'(0) = s2Y - 3s - 1
              s2Y + 4sY +3Y = 3s + 1 + 4*3
  2. Solve the subsidiary equation for Y:
              (s + 3)(s + 1)Y = 3s + 13
                      3s + 13      -2     5
              Y = -------------- = --- + ---
                  (s + 3)(s + 1)   s+3   s+1
  3. Solve the given problem by using the inverse transform:
              L-1{1/(s + 3)} = e-3t
              L-1{1/(s + 1)} = e-t
              y(t) = -2e-3t + 5e-t
Of course, this differential equation would have been quite easy to solve using a substitution method. However, the Laplace Transformation allows for a systematic method to solve more complex differential equations, and also systems of differential equations.