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:
- Transformation of the equation into a subsidiary equation (moving
the equation from the so called t-space to
s-space)
- Solving the subsidiary equation by simple algebraic manipulations
- 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
0
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
0
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
- 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
-
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
-
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.