Laplace transforms 671̃g(s) ̃h(s) = lim
a→∞∫a0dxe−sxg(x)∫a−x0dye−syh(y). (D.4)The use of the upper limita−xrather than simplyain theyintegral is permissible
because of the lima→∞and the fact that the integrals are assumed to decay rapidly.
The use of a triangular integration region rather than a square onesimplifies the
algebra. We then introduce the change of variablesx=t−z,y=zinto the integral,
which leads to
g ̃(s) ̃h(s) = lim
a→∞∫a0dte−st∫t0dz g(t−z)h(z). (D.5)Whena→∞, eqn. (D.5) is the Laplace transform of a convolution.
Laplace transforms are particularly useful for solving ordinary linear differential
and linear integro-differential equations. In particular, an ordinary linear differential
equation for a functionf(t) can be converted into a simple algebraic equation for
f ̃(s). This conversion is accomplished by expressing derivatives as simplealgebraic
expressions via Laplace transformation. Consider first the Laplace transform of the
functiong(t) =f′(t) = df/dt, which is given by
̃g(s) =∫∞
0dte−st
df
dt. (D.6)
An integration by parts shows that
g ̃(s) = e−stf(t)∣
∣
∣
∣
∞0+s∫∞
0dte−stf(t)=sf ̃(s)−f(0), (D.7)where it is assumed that lima→∞e−saf(a) = 0. Similarly, ifh(t) =f′′(t) = d^2 f/dt^2 ,
then by the same analysis
̃h(s) =s^2 f ̃(s)−sf(0)−f′(0). (D.8)In general, ifF(t) = dnf/dtn, then
F ̃(s) =snf ̃(s)−sn−^1 f(0)−sn−^2 f′(0)−···−f(n−1)(0). (D.9)In Section 15.2, we showed how these relations can be used to convert a linear
second-order differential equation forf(t) into an algebraic equation forf ̃(s), which
can then be easily solved. However, once an expression forf ̃(s) is found, obtaining
f(t) requires performing an inverse Laplace transform, and the inversion of Laplace
transforms is considerably less straightforward than the forward Laplace transform.
Of course, the simplest way to invert a Laplace transform is to look up the given
form forf ̃(s) in a table of Laplace transforms and find the corresponding form for
f(t). However, iff ̃(s) cannot be found in the table, then an explicit inversion must
be performed. Unfortunately, as we noted earlier,f(t) can diverge exponentially and
its Laplace transform will still exist, a fact that makes Laplace inversion rather tricky.