Laplace Transforms

For any given function f(t) (assumed to equal zero for t less than
zero) we can define another function F(s) by the definite integral

                     +inf
                      /
             F(s)  =  | f(t) exp(-st) dt                      (1)
                      /
                     t=0

The function F(s) is called the Laplace transform of the function
f(t).  Note that F(0) is simply the total area under the curve f(t)
for t=0 to infinity, whereas F(s) for s greater than 0 is a "weighted" 
integral of f(t), since the multiplier exp(-st) is a decaying exponential
function equal to 1 at t=0.  Thus as s increases, F(s) represents the 
area under f(t) weighted more and more toward the initial region near 
t=0.  Knowing the value of F(s) for all s is sufficient to fully 
specify f(t), and conversely knowing f(t) for all t is sufficient to 
determine F(s).  Thus there is a unique "mapping" between functions 
in the "t domain" and the corresponding functions in the "s domain".  

The benefit of considering the Laplace transform of a function is 
that it sometimes enables us to solve problems easily in the "s 
domain" that would be difficult to solve in the "t domain".  To
see why, it's helpful to review some fundamental properties of
these transforms.  First, the Laplace transform is a *linear*
operator, which means that for any functions f,g, and constant c 
we have

        L[f+g] = L[f] + L[g]          L[cf] = c L[f]

Now, the key to the usefulness of Laplace transforms arises from the 
following indefinite integral


  /                  exp(ax) /       f'(x)   f"(x)   f"'(x)       \
  |f(x) exp(ax) dx = -------( f(x) - ----- + ----- - ------ + ...  )
  /                     a    \         a      a^2      a^3        /


This shows that the Laplace transform of a function f(t) can be
expressed as

                 exp(-st)  /       f'(t)   f"(t)       \   |inf
    L[f(t)]  = - -------- ( f(t) + ----- + ----- + ...  )  |
                     s     \         s      s^2        /   |t=0

With t=+inf the right hand quantity vanishes, so the Laplace transform
is simply given by the negative of the indefinite integral evaluated
at t=0, i.e.,

                   f(0)     f'(0)     f"(0)
      L[f(t)]  =   ----  +  -----  +  -----  +  ...            (2)
                     s       s^2       s^3        


This makes it easy to see that if f(t) is a constant A, then 
L[f] = A/s. Also, if f(t) = At+B, then f'(t)=A and the Laplace 
transform is L[f] = B/s + A/s^2.  For a less trivial example, 
suppose f(t) = A sin(wt + u) for some constants A, W, and u.  
This function has the derivatives

        f'(t) =  Aw  cos(wt + u)

        f"(t) = -Aw^2 sin(wt + u)

        f"'(t) = -Aw^3 cos(wt + u)

                  etc.

Substituting these derivatives, at t=0, into equation (2) gives

           A sin(u)   Aw cos(u)   Aw^2 sin(u)
   L[f] =  -------- + --------- - -----------  -  ...
              s           s^2         s^3

Collecting terms by sines and cosines, this can be written as

          _                _  _                                      _
         | sin(u)   wcos(u) ||      / w \2    / w \4    / w \6        |
 L[f] = A| ------ + ------- || 1 - ( --- ) + ( --- ) - ( --- ) + ...  |
         |_   s       s^2  _||_     \ s /     \ s /     \ s /        _|


The right-hand factor is just a geometric series in -(w/s)^2, so we
have
                _                _  _             _
               | sin(u)   wcos(u) ||       1       |
    L[f]  =  A | ------ + ------- || ------------- |
               |_   s       s^2  _||_ 1 + (w/s)^2 _|


                  s sin(u) + w cos(u)
             =  A -------------------
                      s^2 + w^2

In the special case when the phase angle u is zero, this reduces
to the familiar L[Asin(wt)] = Aw/(s^2 + w^2).

Another example, and one that is often useful, is f(t) = Aexp(rt) for
constants A and r, which has the derivatives

          f'(t) = Ar exp(rt)

          f"(t) = Ar^2 exp(rt)

               etc.

Thus by equation (2) we have the Laplace transform

                A     Ar     Ar^2               A
       L[f] =  --- + ---- + ----- + ....  =  -------
                s    s^2     s^3              s - r


Likewise if we have a function 

              f(t) = A exp(r1 t) + B exp(r2 t)

the Laplace transform is

                  A         B        (A + B)s - (A r2 + B r1)
      L[f]  =  ------- + -------  =  ------------------------
                s - r1    s - r2         (s - r1)(s - r2)


So, we've seen how equation (2) enables us to determine the Laplace
transforms of typical functions, but the most important point to
notice about (2) is that it can be written in the form

             f(0)      1   /f'(0)     f"(0)         \
   L[f]  =   ----  +  --- ( -----  +  -----  +  ...  )
               s       s   \  s        s^2          /

and the quantity in parentheses is L[f'].  Therefore, rearranging
terms, we can write this as

                L[f']  =  s L[f] - f(0)

By the same token we can write (2) as

             f(0)     f'(0)      1   /f"(0)     f"'(0)         \
   L[f]  =   ----  +  -----  +  --- ( -----  +  ------  +  ...  )
               s       s^2      s^2  \  s        s^2           /

and, noting that the quantity in parentheses is L[f"], we can multiply
through by s^2 and rearrange terms to give

     L[f"]  =  s^2 L[f]  -  s f(0)  -  f'(0)

Likewise we can deduce

     L[f"']  =  s^3 L[f]  -  s^2 f(0)  -  s f'(0)  -  f"(0)

and so on.  In this way we can express the nth derivative of f(t) as
a polynomial in s with coefficients given by the derivatives of f(t)
at t=0 and a leading coefficient of L[f].

To see how this can be used to solve differential equations, consider
the homogeneous equation

           a f"(x) + b f'(t) + c f(t)  =  0

In order to determine a particular solution for this 2nd-order equation
we must specify two conditions.  Often we are able to specify the
"initial conditions", i.e., the values of f(0) and f'(0).  So, taking 
the Laplace transform of both sides of our differential equation, (and 
recalling that L is a linear operator), we have

            a L[f"]  +  b L[f']  +  c L[f]  =  0

Substituting the previous expressions for the transforms of these
derivatives, we have

  a(s^2 L[f] - s f(0) - f'(0)) + b(s L[f] - f(0)) + c L[f]  =  0

Collecting terms, this gives

  (a s^2 + b s + c) L  =  (a s + b) f(0)  + a f'(0)

Solving for L = L[f] gives

                 a f(0) s + [b f(0) + a f'(0)]
        L[f]  =  -----------------------------
                       a s^2 + b s + c

We recognize this as the Laplace transform of Aexp(r1 t)+Bexp(r2 t)
where r1,r2 are the roots of the characteristic equation in the
denominator, and A,B are constants satisfying the conditions

                A + B = a f(0)         

              A r2 + B r1 =  b f(0) + a f'(0)

Thus for any coefficients a,b,c we can determine the characteristic 
roots r1,r2, and then for any given initial conditions f(0) and f'(0) 
we can solve for A and B to give the complete solution.

The above example was based on a homogeneous equation, i.e., an 
equation with no forcing function, because the right hand side was
zero.  Laplace transforms can also handle inhomogeneous equations,
simply by taking the transform of the forcing function along with
the other terms of the equations, and then solving for L[f] just
as above.

We can also deduce several other interesting facts about Laplace
transforms from equation (2).  For example, it's easy to show that
if L[f(t)] = F(s), then  L[ t f(t) ] = -dF(s)/ds.

Although Laplace transforms are certainly interesting, and can be
useful in some circumstances, they also have some drawbacks.  Most
obvious is that fact that they require "initial" conditions, i.e.,
we must stipulate the value and derivatives of f(t) at t=0, whereas
we often wish to impose conditions at different points of the 
function, such as specifying f(1), f'(3), and f"(17).  These three 
conditions are just as suitable for fixing the solution of a 3rd-
order differential equation as are f(0), f'(0), and f"(0), but since 
they are imposed at different values of the independent variable we 
cannot use the conventional Laplace transform method.

Another drawback of Laplace transform methods is that they tend to
be taught as a "canned" technique, enabling one to solve equations
by a simple recipe, without really understanding how it works.  The 
problem with this approach is that it can easily give erroneous
results if applied inappropriately.  It's also worth noting that any
equation that can be solved by Laplace transforms can also be solved
(often more efficiently and perspicuously) by other methods.