The elementary computations are left to the reader. Then this converges to 0 as R_{m}→∞. It
follows that
{∫ R }
0 nf (t)e−stdt
_{n=1}^{∞} is a Cauchy sequence and so it converges to
I ∈ ℝ. The above computation shows that if
Rˆ
_{n} also converges to ∞ as n →∞,
then
∫ Rn −st ∫ Rˆn −st
ln→im∞ f (t)e = nli→m∞ f (t)e
0 0
and so the limit does indeed exist and this is the definition of the improper integral
∫_{0}^{∞}f
(t)
e^{−ts}dt. ■
Certain properties are obvious. For example,
If a,b scalars and if g,f have exponential growth, then for all s large enough,
ℒ (af + bg)(s) = aℒ (f)(s)+ bℒ (g)(s)
If f^{′}
(t)
exists and has exponential growth, and so does f
(t)
then for s large
enough,
ℒ(f′)(s) = − f (0) +sℒ (f)(s)
One can also compute Laplace transforms of many standard functions without
much difficulty. That which is most certainly not obvious is the following major
theorem. This is the thing which is omitted from virtually all ordinary differential
equations books, and it is this very thing which justifies the use of Laplace transforms.
Without it or something like it, the whole method is nonsense. I am following [37].
This theorem says that if you know the Laplace transform, this will determine the
function it came from at every point of continuity of this function. The proof is fairly
technical but only involves the theory of the integral which was presented in this
chapter.
Theorem 9.7.3Let ϕ have exponential growth and have finitely many discontinuitieson every interval
[0,R]
and let f
(s)
≡ℒ
(ϕ)
(s)
. Then if t is a point of continuity of ϕ, itfollows that
is determined by its Laplace transform at every point of continuity.
Proof: First note that for k a positive integer, you can change the variable letting ku = t
and obtain
k+1∫ ∞ ( ) k+1 ∫ ∞ ( )k
k--- e−uu kdu = k--- e− t t- 1-dt
k! 0 k! 0 k k
The details involve doing this on finite intervals using the theory of the Riemann integral
developed earlier and then passing to a limit. Thus the above equals
Now this converges to 0 as k →∞. In fact, for a < 1,lim_{k→∞}
k+1
kk!-
(e− aa)
^{k} = 0 because of the
ratio test which shows that for a < 1,∑_{k}
kkk+!1
(e−aa)
^{k}< ∞ which implies the k^{th} term
converges to 0. Here a = 1 −δ. Next consider the last integral. This obviously converges to 0
because of the exponential growth of ϕ. In fact,
Now changing the variable letting uk = t, and doing everything on finite intervals
followed by passing to a limit, the absolute value of the above is dominated by
∫ ∞ k+1 ( )k ( )
k--- e− t t- 1- a +beλ(t∕k) dt
∫k(1+δ) k! k k
= ∞ -1e−ttk(a+ beλ(t∕k)) dt for some a,b ≥ 0
k(1+δ)k!
However, the limit as k →∞ of the integral on the right equals the improper integral on the
left. Thus this converges to 0 as k →∞. Thus all that is left to consider is the middle integral
in which δ was chosen such that
|ϕ(u)− ϕ(1)|
< ε. Thus
| |
||∫ 1+δ kk+1( −u )k || ∫ ∞ kk+1( −u )k
|| 1− δ k! e u (ϕ (u) − ϕ (1))du|| ≤ ε 0 k! e u du = ε
It follows that if ϕ is continuous at 1,
∫ ∞ kk+1 ( −u )k
kli→m∞ 0 k! e u (ϕ(u)− ϕ (1))du = 0
and so ∫_{0}^{∞}
kk+1-
k!
(e−uu)
^{k}ϕ
(u )
du = ϕ
(1)
. Now you simply replace ϕ
(u)
with ϕ
(tu)
where ϕ is
continuous at t. This function of u still has exponential growth and is continuous at u = 1.
Thus we obtain
I think the approach given above is really interesting because it gives an explicit
description of ϕ
(t)
at every point. However, it may be easier to use the Weierstrass
approximation theorem to prove this major result which shows that the function is
determined by its Laplace transform so here is another proof.
Lemma 9.7.4Suppose q is a continuous function defined on
[0,1]
. Also suppose that forall n = 0,1,2,
⋅⋅⋅
,
∫ 1 n
0 q(x)x dx = 0
Then it follows that q = 0.
Proof: By assumption, for p
(x)
any polynomial, ∫_{0}^{1}q
(x)
p
(x)
dx = 0. Now let
{pn(x)}
be a sequence of polynomials which converge uniformly to q
because s + s_{0} is large enough for this to happen. It follows from Lemma 9.7.5 that
ˆϕ
= 0.
But this implies that ϕ = 0 also. This proves the following fundamental theorem.
Theorem 9.7.6Suppose ϕ has exponential growth and is continuous on [0,∞).Suppose also that for all s large enough, ℒ
(ϕ)
(s)
= 0. Then ϕ = 0.
This proves the case where ϕ is continuous. Can one still recover ϕ at points of continuity?
Suppose ϕ is continuous at every point but finitely many on each interval
[0,t]
and has
exponential growth and ℒ
(ϕ)
(s)
= 0 for all s large enough. Does it follow that ϕ
(t)
= 0 for t
a point of continuity of ϕ? Approximating with finite intervals
[0,R ]
in place of [0,∞) and
then taking a limit, (details left to you.)